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rJ! gives you more solved problems than any other guide rJ! demonstrates the best problem-solving strategies rJ! onimproves perfonnance exams rJ! helps cut study time mDA.NASAR rJ! cross-reference includes easy-to-read index and diagrams rJ! soprovides subiect coverage thorough, graduate students and professionals can use this guide, too SCHAUM'S SOLVED PROBLEMS SERIES 3000 SOLVED PROBLEMS IN ELECTRIC CIRCUITS by Syed A. Nasar University of Kentucky McGRA W-HILL New York St. Louis San Francisco Auckland Bogota Caracas Lisbon London Madrid Mexico City Milan Montreal New Delhi San Juan Singapore Sydney Tokyo Toronto I Syed A. Nasar, Ph.D., Professor of Electrical Engineering at the University of Kentucky. Dr. Nasar has written many books, including two Schaum's Outlines, ELECTRIC MACHINES AND ELECTROMECHANICS and BASIC ELECTRICAL ENGINEERING; a power-systems text for Macmillan; and a textbook for McGraw-Hill's College Division, INTRODUCTION TO ELECTRICAL ENGINEERING. Project supervision by The Total Book. Index by Hugh C. Maddocks, Ph. D. Library of Congress Cataloging-in-Publication Data Nasar, S. A. Schaum's 3000 solved problems in electric circuits. 1. Electric circuits-Problems, exercises, etc. I. Title. 11. Title: Schaum's three thousand solved problems in electric circuits. TK454.N36 1988 621.319'2076 87-25974 ISBN 0-07-045936-3 (Formerly published under ISBN 0-07-045921-5.) 12 13 14 15 16 17 18 19 20 YFM YFM 5 4 Copyright © 1988 by The McGraw-Hill Companies, Inc. All rights r.!served. Printed in the United States of America. Except as permitted under the United States Copyright Act of 1976, no part of this publication may be reproduced or distributed in any form or by any means, or stored in a data base or retrieval system, without the prior written permission of the publisher. McGraw-Hill A Division of The McGraw-HiU Companies (~r'7 6<-. CONTENTS Chapter 1 UNITS AND BASIC CONCEPTS 1 Chapter 2 RESISTANCE AND OHM'S LAW 6 Chapter 3 SERIES AND PARALLEL RESISTIVE CIRCUITS 16 Chapter 4 KIRCHHOFF'S LAWS 43 Chapter 5 NETWORK THEOREMS 72 Chapter 6 CAPACITORS 116 Chapter 7 INDUCTORS 129 Chapter 8 AC SOURCES, WAVEFORMS, AND CIRCUIT RELATIONSHIPS 137 Chapter 9 COMPLEX NUMBERS AND PHASORS 145 Chapter 10 AC CIRCUITS UNDER STEADY STATE 152 Chapter 11 MAGNETICALLY COUPLED CIRCUITS 231 Chapter 12 RESONANCE 268 Chapter 13 FREQUENCY RESPONSE AND FILTERS 291 Chapter 14 THREE-PHASE CIRCUITS 304 I I I I 351 Chapter 15 TRANSIENTS IN DC CIRCUITS Chapter 16 STEP, RAMP, AND IMPULSE FUNCTI NS 423 Chapter 17 DUALS AND ANALOGS 432 Chapter 18 TRANSIENTS IN AC CIRCUITS 441 Chapter 19 CIRCUITS WITH MULTIFREQUENCY I PUTS 450 Chapter 20 CIRCUITS WITH NONSINUSOIDAL SO 462 Chapter 21 LAPLACE TRANSFORM METHOD 491 Chapter 22 ST ATE VARIABLES METHOD 579 Chapter 23 TWO-PORT NETWORKS 594 Chapter 24 REVIEW PROBLEMS 620 INDEX 747 iii To the Student Think of it!-an expected score of 75% on any exam in Electric Circuits, with no other preparation! The reasoning is simple: There are only 4000 possible problems in the field (as you must know), and this book solves 3000 of them for you! Speaking seriously, you have here the most careful and complete anthology of examination-type problems on the market today. In using the book, you should, of course, concentrate on the area of your maximum weakness-the Laplace transform or whatever. But do not neglect to work problems involving familiar material, too; you might well learn more efficient methods of handling them. The heuristic value of a clear circuit diagram need not be stressed: if a problem in this book should carry a diagram but doesn't, be sure to sketch out one before undertaking the solution. May your success be electric. v /7 CHAPTER 1 L/Units and Basic Concepts 1.1 Powers of 10 appear frequently with units of measurements. These powers of 10 are written in abbreviated forms. If electric resistance is measured in ohms (fl), express the following values in powers of 10 and write them in their abbreviated forms: 2000 fl and 3,000,000 fl. I 1.2 2000 fl = 2 X 10 3 fl = 2 kiloohm = 2 kfl = 5 X 10- 6 F = 5 microfarad = 5 J-tF 0.0005 F = 0.5 X 10- 3 F = 0.5 millifarad = 0.5 mF = 500 J-tF 0.000000001 F = 1 X 10- 9 F = 1.0 picofarad = 1 pF 0.000005 F The unit of electric inductance is henry (H). Express the following values in powers of 10 and write them in their abbreviated forms: 0.01 Hand 0.003 H. I 1.4 0.01 H = 10 X 10- 3 H = 10 millihenry = 10 mH 1000 Hz 2 min = 0.1 gigahertz = 0.1 GHz 120 X 10- 3 10 3 = 2 x 60 s = 120 s = 5 km = 1.2 x 10 ms = 5 x 10 3 m = 5 x 10 3 X 10 2 cm = 5 x 10 5 cm 15 cm 15 3 10 2 x 10 = = 150 mm Electric current is measured in amperes (A). If an ampere is expressed as a flow of charge in coulombs per second (C/ s), how many electrons pass a given point in 30 s in a conductor carrying 8-A current. The charge on an electron is approximately 1.6 x 10- 19 C. I Charge 240 C correspond to =A x s = 8 x 30 = 240 C 1.6 x 10- 19 C correspond to 1 electron (1 x 240)/(1.6 x 10- 19 ) = 15 x 10 20 Find the current in a conductor through which approximately 1.6 x 10- 19 C. I electrons 2.5 x 10 20 electrons pass in 8 s if the charge on an electron is 1= (no. of electrons)\charge on electron, C) = 2.5 time, s 1.10 5 S Convert 15 centimeters to millimeters: I 1.9 10 Hz X = 5 MHz Convert 5 kilo meters to centimeters: I 1.8 9 Convert 2 minutes to milliseconds: I 1.7 = 3 X 10- 3 H = 3 millihenry = 3 mH 5,000,000 Hz = 5 X 10 6 Hz = 5 megahertz = 1 X 10 3 Hz = 1 kilohertz = 1 kHz 100,000,000 Hz = 0.1 1.6 0.003 H Electric frequency is measured in hertz (Hz). Express the following frequencies in powers of 10 and in their respective abbreviated forms: 1000 Hz, 5,000,000 Hz, and 100,000,000 Hz. I I.S = 3 X 10 6 fl = 3 megohm = 2 Mfl Electric capacitance is measured in farads (F). However, this is rather a large unit. Express the following values in powers of 10 and write them in their abbreviated forms: 0.000005 F, 0.0005 F, and 0.000000001 F. I 1.3 3,000,000 fl 10 2 () X 1.6 X 10- 19 =5A 8 A charge of 360 C passes through a conductor in 20 s. I X I What is the corresponding current in amperes? = ~ = 360 = 18 A s 20 2 1.11 0 CHAPTER 1 The current in an electric circuit rises exponentiall) as given by i = 10(1 - e~2t) A. Calculate the charge flowing through the circuit in 250 ms. °250 ( - 2t ) 0.250 I q = i dt = 10(1- e~2t) dt = 10 t + ~-;:,= 10(0.250 + 1e~2x0250 - 0 - D = 0.5326 C J 1.12 1 o L A 75-W bulb draws a 680-mA current. bulb? I 1.13 t= 1.14 = 50 x 10~" x 50 X (charge, C) (electric field, Vim) Work done = force x distance = 2.5 ~< SO x 1O~2 = 1.25 N· m = 10 3 = 2.5 N 1.251 Power is defined as the rate of work done or the rate of energy conversion. Thus, the unit of power is the joule per second (lis) which is equal to one watt (W). If the time taken to move the 50-J-LC charge of Prob. 1.14 through 50 cm is 10 ms, calculate the corresponding power. work clone Power = - - .---~ time 1.25 = 125 W 10 >< 1O~3 We observed in Prob. 1.14 that an electric charge experiences a force in an electric field. Electric potential difference (between two pOints) is measured in volt:; (V), and is defined as the work done in moving a unit positive charge (from one point to the other). What is the potential difference between two points if it requires 220 J-Ll to move a lO-J-LC charge from one point to the other? 1O~" 220 >< 1 V = 1 1I C or V == 10 x 10 " = 22 V From Prob. 1.16, V=lIC==(J/s)/(C/s)=W/A. Calculate the potential difference across a resistor dissipating 30 W of power while taking 2.5 A of current. Also, calculate the ohmic value of the resistance. W 30 V= - = =12V 1 2.5 30 W R = fi = (2.5)2 = 4.8 fl An energy of 121 is expended in moving a 2-C charge from infinity to a point A. Assuming infinity to be at zero potential, determine the potential difference bttween point A and infinity (i.e., the potential at A). I 1.19 = 6 x 2 x 60 = 720 C = I 1.18 s) Force I 1.17 A)(tiJno~. The unit of force is the newton (N) and work is measured in netwon-meters (N' m), which is also the unit of energy. Alternatively, energy is expressed in joule5 11), where 11 = 1 N· m. Determine the work done in moving a 50- J-L C electric charge (Q) through a distance of 50 cm in the direction of a uniform electric field (E) of 50 kV/m, if the force F is given by f= QE. I 1.16 How many coulombs of charge pass through the resistor in 2 min? q = (current, I 1.15 How much time will be required to pass a 30-C charge through the charge, C 31) . = ---~--- = 441.17 s = 7.35 mm 68 ~< 10--' current, A A current of 6 A flows in a resistor. I () V Ax work or energy, 1 12 = - =6V char:s(:, C 2 = ----~---- If an additional energy of 31 is required to move the 2-C charge of Prob. 1.18 from point A to another point B, calculate the potential difference between points A and B. Also determine the potential difference between point B and infinity. I 1.20 = work or energy, 1 ch arge, C :; = " 12 + 3 1.5 V Vnx = - 2 - =7.5V " The potential difference between two conductors is 110 V. from one conductor to the other? I 1.21 VAn How much work is done in moving a 5-C charge Work = energy =, 110 x 5 = 5501 Determine the charge that requires 1-kl energy to be moved from infinity to a point having a 12-V potential. 3 I cne rgy, 1 10 Charge C = ---~-- = , pote ltial, V 12 = 83 33 C . UNITS AND BASIC CONCEPTS D 3 1.22 A Car battery supplies 4S J of energy at 12 V over a certain period of time. this period. 4SJ I q == 12 V ==4C 1.23 Electric utilities employ as the unit of energy the kilowatt-hour (kWh). The power consumed in a household over a 24-h period is as follows: S A.M. to 2 p.M.-1.5 kW; 2 P.M. to 6 p.M.-O.5 kW; 6 P.M. to 11 p.M.-2.6 kW; and 11 P.M. to S A.M.-1.0 kW. What is the energy consumption in megajoules? I Total kWh = (power, kW)(time, h) = 1.5 x 6 + 0.5 x 4 + 2.6 x 5 + 1.0 x 9 = 33 kWh =33 x 10 3 x 60x 60W·s= IIS.S 1.24 An electric heater takes 1.2 kWh in 30 min at 120 V. ]_ I 1.25 - 10 6 J = llS.SMJ What is the current input to the heater? 3 Vlt _ 1.2 x 10 /0.5 _ 120 V - . output 99 X 10' Efficiency = -.- - = . = 0.99 Input Input - 20A t= . 99 X 10' Input V = 0 99 = 100 kJ . or V P= 100 X 10 3 120 x 20 = 41.67 s What is the ohmic value of the resistance of the heating element of the heater of Probs. 1.24 and 1.25? V 120 R=-=-=6fl I 1.27 X The heater of Prob. 1.24 has an efficiency of 99 percent. The heat energy required to boil a certain amount of water is 99 kJ. If the current taken by the heater is 20 A at 120 V, find the time required to boil the water. I 1.26 Determine the charge moved during ] 20 A 1l0-V light bulb takes 0.9-A current and operates 12 hi day. to operate the bulb for 30 days. I V = Pt = 110 x 0.9 X 1O~3 At the rate of 7 cents/kWh, determine the cost x 12 x 30 = 35.64 kWh Cost of operation = 35.64 x $0.07 = $2.50. 1.28 The voltage and current in a circuit element are respectively given by v = 100\1'2 sin t V A. Calculate the instantaneous power and the average power delivered to the circuit. I Instantaneous power p = vi = (100\1'2 sin t)(5\1'2 sin t) = 1000 sin 2 t W = 1000 x and i = 5\1'2 sin t ~ (1- cos 2t) P = 500 - 500 cos 2t W The cosine function averages to zero, so the average value of p, 1.29 Pay = 500 W. A resistor draws a current i = S sin wt A at a voltage v = 200 sin wt V. Calculate the energy consumed by the resistor per cycle (or over one period of the current wave). Hence, determine the average power dissipated in the resistor. . d 2rr O ne peno == - s I w 2"./W Energy W = 1o 12"./w (200 sin wt)(S sin wt) dt = _ 1600 vi dt = _ rr J w 0 W 1600rr Average power Pay = 2rrlw = w(2rrlw) 1.30 SOO W The energy capacity or rating of a battery is generally expressed in ampere-hour (Ah). supply 0.5 A continuously for three days. What must be the rating of the battery? I 1.31 = A battery is rated at 30 Ah. I Ah = ] X hr = 0.5 X A battery is required to 3 x 24 == 36 Ah For how many hours can it continuously supply a current of 2.5 A? . Ah 30 Time = = = 12 h ] 2.5 4 1.32 D CHAPTER 1 The capacity of a car battery depends on the ambient temperature as shown in Fig. 1-1. rated at 72 Ah at 25°e. For how long can the battelY supply a 16-A current at O°C? I 0.8 x 72 = 57.6 Ah. From Fig. 1-1, at O°C the rating of the battery reduces to . Time t Ah 57.6 6 A certain battery is Therefore, . = 1 =, -R;- = 3.6 h = 3 h 3 mm liD (DV - - 80 - - - - - - - - -- _. - - - - - - - - ... .~ f..:::. ~ - - - - - -- t 70 I \ bo I 50~______________ --+ -20 -10 c ----~I--__--~----~--~ :1." ,.'!i 30 10 --.. T, 1.33 °C Fig. 1-1 The capacity of a car battery depends on the current drawn (or discharge) from the battery, as shown in Fig. 1-2. The battery is rated at 70 Ah at a discharge rate of 5 A as shown. How long will the battery supply 20 A of current? I From Fig. 1-2, at 20 A the rating of the battery becomes 58 Ah. Hence, -r = :w58 =.2 9 h = 2 h 54 mm. · T Ime t == Ah 1.34 If the rating of the battery with a discharge charactt:ristic shown in Fig. 1-2 is not allowed to go below 64 Ah, for how long can the battery supply the rated current? I From Fig. 1-2 at 64 Ah, discharge rate = 12 A. Hence, Ah . · t =, - = -64 = 5 h 20 mm T Ime " 12 1.35 Combine the characteristics of Figs. 1-1 and 1-2 to obtain the rating of the battery at 17 A and at 10°C if the battery is rated at 100 percent at 5 A and at 25°C I From Fig. 1-2, rating at 10°C = 0.9 X 60 = 54 Ah. 1.36 17 A = 60 Ah which is considered as 100 percent at 25°C. The decay of charge in an electric circuit is given by I i = dq dt q = 50e -30or}Le. From Fig. 1-1, rating Determine the resulting current. = -50 x 300 x 10- 6 e- 300r = _15e- 3oor mA UNITS AND BASIC CONCEPTS o 5 80 t aO~ 1.37 ______~__- r____~__~__~~____~~______~__ Evaluate the current in Prob. 1.36 at the following instants: I At t = 0: t = 10 ms, and t=oo: p 00. 2 i = -15e ~300x 10- = -0.7468 mA i=-15e~~=0 The voltage v and current i in an ac circuit are respectively given by v = 34 sin 377t V 60°) A. Determine the instantaneous and average powers delivered to the circuit. I t= i = -15eo = -15 mA t = 1O~2 s: 1.38 t = 0, = vi = (34 sin 377t)[2 sin (377t - 60°)] = 68 sin 377t sin (377t - 60°) = 68 x ! [cos (377t - 377t + 60°) - cos (377t + 377t - 60°)] = 34 [cos 60° - and i = 2 sin (377t- cos (754t - 60°)] W Pav = 34 cos 60° = 17 W 1.39 The voltage v and current i at the pair of terminals of an electric circuit are given by v = 100 sin t V and i = -5 sin t A. Evaluate the average power and state if the circuit absorbs or delivers power. 2 I p = vi = (100 sin t)( -5 sin t) = -500 sin t W Pav = -500 x ! 2 (since the average value of sin t = !) = -250 W The negative sign indicates that negative power is absorbed by the circuit; i.e., the circuit delivers power. 1.40 The voltage v and current i in a circuit are given by v == 10 sin t V instantaneous and average powers, and explain your result. I and i = 2 cos t A. Determine the p = vi = (10 sin t)(2 cos t) = 20 sin t cos t = 10 sin 2t W The instanta~eous power pulsates with twice the frequency of the voltage or current. Pav = 0 W, since the average value of sin 2t = O. Zero average power indicates that the circuit is nondissipative or conservative. CHAPTER 2 '\\ Resistance and Ohm's I~aw~ 2.1 A copper conductor of circular cross section 5 mm in di:lIneter is 5 m long. Calculate its resistance at 20°C if the resistivity of copper at 20 DC is 1.72 X 10--" n· m. 2.2 R= - = 4 38 mfl . A 40-m metallic conductor of cross-sectional area 1 mnf' has a resistance of 12 fl. the metal. I a = - I 40 RA [Note that 2.3 (1.72:<10")5 = -------17"(5 :< 10- 3 )2/4 pi A I = - - - - -3- = (12)(10- )2 Calculate the conductivity of 3.33 MSlm 1 siemens (S) = 1 fl-I.j A cube of an alloy of resistivity 1.12 J.tfl . m is 2 cm on c. side. of the cube. Determine the resistance between any two faces I 2.4 We have two cubes-one measuring I m on one side and the other 21 m. Find the ratio of conductivities of the materials of the cubes so that the resistance between an y two faces of one cube is the same as that for the other cube. I and 1 2.5 Calculate the length of copper wire having a diameter of 5.8 x 10 7 S/m. i6 I 1= aRA 2.6 in = 10 ~ in and resistance of 2 fl. x 2.54 >< 10 -2 = 1.5875 = 5.8 x 10 7 x 2 x ~ (1.5875 X X Conductivity of copper is 10- 3 m 10- 3)2 = 229.6 m A rectangular bus bar made of aluminum is 0.9 m long. n. 15 m wide, and 1.3 cm thick. If current in the bus bar flows along its length. and the conductivity of aluminu111 is 3.57 x 10" Slm, calculate the bus bar resistance. I 2.7 or R = - I aA = O.q -----= 2 (3.57 x 10")(0.1:; x 1.3 x 10 1.293 J.tfl ) A transmission line cable consists of 19 strands of identical copper conductors, each 1.5 mm in diameter. The physical length of the cable is 2 km. But, because of the twist of each strand, the actual lengths of the conductors are increased by 5 percent. What is the resistance of the cable? Resistivity of copper is 1.72 x 10-" fl· m. I Allowing for twist, 1= (1.05)(2000)=21OOm. 10- 3 )2 = 33.576 X 10- 6 m 2• Area of cross section of 19 strands = 19(17"/4)(1.5 x = pi = I.72 x _~~~~ 2100 = 1.076 fl R 2.8 33.576 >: 10- 6 Variation of resistance with temperature is expressed in terms of temperature coefficient a. Explicitly, the resistance RT at a temperature T QC is related to the le,istance at 0 QC by RI = Ro(1 + aoT) as graphically depicted in Fig. 2-1, where aa is the temperature coefficient at 0 QC. The figure also shows the inferred absolute zero for copper. Using Fig. 2-1, find the resistance of a copper wire at -20 QC if its resistance at 0 QC is 20 n. I 6 A From Fig. 2-1 we have 234.5 + T, RI 234.5 + T2 R2 RESISTANCE AND OHM'S LAW R / D 7 / /' /' ___ -.!3._ R, 7 "/ ,/ -----_~27~~7.·C~~IC-2~'~~~.~~c---bo~~----~---------L~T > - - - - - T , - - - - - l.. ~1 ·C i ". "1 . Fig. 2-1 1-1- - - - - - - - - - - T, -------------1 From the data R2 2.9 = (234.5 - 20)20 234.5 + 0 = 18.290 Values of the temperature coefficient a for copper for different temperatures are plotted in Fig. 2-2, from which a200C = 0.00393°C-'. If the resistance of a given wire is 20 0 at 20°C, what is its resistance at 60°C? -![ xlO 4,,<0 42.J> 410 tiJV Ylo 7 !-l , ~ 0 - ?~o ?s t ,,1~6o ?5"0 ~r<> 0 I R2 = R,[ I + /0 ~o ~o - T, ·c fo so Fig. 2-2 a, (T2 - 11)] = 20[1 + 0.00393(60 - 20)J = 23.1440 8 2.10 D CHAPTER 2 A sample of copper wire has a resistance of 50 n at 10°C. What is the maximum operating temperature if the resistance of the wire is to increase by at most 10 pl~Jcent? I RI =50n, R z =50+0.1x50=55n. From Fg. 2-2, a at 1O°C=0.00409°C~1 =a l RI [1 + a l (T2 - T I )], we obtain 55 = 50[1 + 0.00409(Tz - 10)] or T z = 34.45 qc. 2.11 A metallic conductor has a resistance of 7 n at 0 °C. temperature coefficient of the metal at 20°C. At 20°C the resistance becomes 7.8 n. I or Hence, 2.12 Since R2 = Calculate the 7 = 7.8[1 + a l (-20)] a I = temperature coefficient at 20°C = 0.00513 °C ~ I. For the metal of the conductor of Prob. 2.11, determine the temperature coefficient at O°C. I 2.13 • or Obtain a general relationship between or aT th,~ and ao I R T = R, (1 + aoT) (1) R 1 (1- aTT) (2) Ro== Solving for aT respective temperature coefficients at O°C and at T QC. from Eq. (2) yields (3) Substituting RT from Eq. (1) into (3) gives (4) 2.14 Derive a general relationship between a l and a 2 the I rl~spective temperature coefficients at TI °C and at T z QC. From Eq. (4) of Prob. 2.13 we obtain 1 - = aT 1 + auT ---~-- 1 = - +T ao 0'0 and Thus, By subtraction, 1 1 ---=T-T al 2.15 a I 2 z or The temperature coefficient of carbon at 0 QC is -0.000515 QC~ 1 and that of platinum is 0.00357 QC- 1 at 40 QC. A carbon coil has a resistance of 15 n and a platinum coil has a resistance of 12 n each at 20°C. At what temperature will the two coils have the same resistance? Notice that the temperature coefficient for carbon is negative. I From Eq. (4) of Prob. 2.13: For platinum: 0.00357 _ ° ~I a o = 1 _ 40 x 0.()0357 - 0.00416 C For the two resistances to be equal at a temperature T QC, 12(1 + 0.00416T) == 15(1- 0.000515T) or 2.16 1 + 0.00416T= 1.25 - 0.OO(l64375T or The two coils of Prob. 2.15 are connected in series and operate at 20°C. coefficient a e of the combination at 40°C. I Calculate the "effective" temperature From Eq. (4) of Prob. 2.13, at 20 CO: For carbon: -0.00051S _ ° ~I a = 1 _ 0.000515 >aO - -0.000520 C RESISTANCE AND OHM'S LAW D 9 From the data of Prob. 2.15: Rcarbon = 15[1- 0.000520(40 - 20)] = 14.8440 At 20°C, we have (from Prob. 2.15): 0.00416 ° ~I a = 1 + 0.00416 x 20 = 0.00384 C For platinum: Rplatinum = 12[1 + 0.00384(40- 20)] = 12.92160 At 40°C: Re = 14.844 + 12.9216 = 27.7656 0 At 20°C: Re = 12.0 + 15.0 = 27.0 0 27.7656=27[1 + a e (40-20)] 2.17 a e =0.001418°C 1 or The minimum current required for the operation of a relay coil is 500 mA at 120 V. If the current taken by the coil at 20°C is 530 mA (at 120 V) and the temperature coefficient of the resistor material is 0.00427 °C~ I at 0° C, calculate the maximum temperature above which the relay will fail to operate. I R At 20°C: 20 = 120 530 x 10 3 264 0 = 2 . 1 3 2 0 = 40.0 At T °C (the maximum allowable temperature): R = T Since RT = Ro(1 + aoT) RT R 20 2.18 120 500 x 10 we have: 1 + aoT 1 + 20a o 240 1 + 0.00427T -22-6-.4-1 = -:-1-+-0-=-.-=0-=0--:42-:C7=-x-=2=-=-0 or or The resistance of a 25-0 resistor increases by 10 percent when its operating temperature increases from 15 to 50°C. Calculate the mean temperature rise of the resistor from an ambient temperature of 20 °C when its resistance is 300 and the temperature coefficient remains constant. I Solving for aD and Ro yield~: aD = 0.002985 °C~l and Ro = 23.9286 0 At a temperature T °C we have: 30 = 23.92861(1 + 0.002985T) or Temperature rise = 85 - 20 = 65°C. 2.19 "It has been experimentally found that the resistivity of conducting materials, such as copper and aluminum, varies linearly with temperature." Depict this statement graphically and mathematically. f,n·"", , :r. o , 1"1;. T, ·C Fig. 2-3 10 D CHAPTER 2 I The statement is shown graphically in Fig. 2-3, from which we have: tan () :: m pz - P1 = (1) T z - T1 pz = P1 + m(Tz .- 1'1) = PI [ 1 + ; (T z - T I ) ] and 2.20 The resistance of a silver wire is 0.10 at 20°C. At what temperature will its resistance decrease by 25 percent 1 if its temperature coefficient of resistance at 20°C IS 0.0038°C- • I 2.21 (2) 0.75 x O. L == 0.1[1 + 0.0038(T2 - 20)) or The resistivity of iron at 0 and 20°C is resistivity at 10 QC. I 8.68 X 10- 8 n .m and 9.75 x 10- 8 0. m or respectively. Calculate its From Eg. (1) of Prob. 2.19, 8 (9.75 - 8.68)10- = 0.0535 m= 20-1) X 10- 8 From Eg. (2) of Prob. 2.19, PlO 2.22 = PZO + melO - 20) = [9.75 + O.0535( -10)] x 10- 8 = 9.215 X 10- 8 0. m A piece of wire of uniform cross section has a resistance of 0.8 O. If the length of the wire is doubled and its area of cross section is increased four times, what is it~ resistance? The temperature variation of resistance may be neglected. ptl I Original wire: R, = -;;- =0.80 n l Wire with modified dimensions: 2.23 An electromagnet is wound with a copper coil having 150 turns and a mean length of 20 cm per turn. The coil wire has a rectangular cross section 10 x 2 mm. Cllculate the resistance of the coil at 55°C and determine the power dissipated in the coil at 55 °C if the coil current is 6 A. The resistance of a I-m long wire of 1 mm 2 cross section at 20°C is 0.00172 0 and a o = (1/234.5) QC I. I - ao Pzo= P55 = I 1/234.5 _ 1 0-1 20/234.5 - 254.5 C -I- eAR = 1 x In-" Ix 0.0172 = 1.72 1 P20[1 X 10- 8 0. m ' + a zo (55 - 20)] = 1.72 xJ(l·8 [ 1 + 254.5 (55 - 20) J= 1.96 X 10- 8 0. m R = 55 2.24 _ a zo - 1 + 20a o - iA = 1.96 x 10" x 150 x 0.20 = 2.94 x lO- z 0 10 x 2 x 10 6 Power = ]zR = 6 z(2.94 x lO- z) = 1.0584 W P55 The power taken by a resistive coil made of copper wire is 220 W at 110 V and 20°C. Calculate the power consumed by the coil at 110 V and 120°C. The tenperature coefficient at 20°C is 0.00393 °C- 1• Z Z z V V 110 I Pzo = R or R 20 = P = 220 = 55 0 zo R l20 = Rzo[l zo + a zo (120 - 20)1 "" 55[1 + 0.00393(100)] = 76.615 0 1 L(l2 P l20 = 76615 = 157.93 W 2.25 A flat aluminum ring 5 mm thick has a negligible air gap. If the inner and outer radii of the ring are 0.2 and 0.25 m respectively, determine the resistance of tlw ring at 20°C. At this temperature the resistivity of aluminum is 2.78 x 10- 8 0. m. RESISTANCE AND OHM'S LAW I 0 11 rmean = ~ (ro + rJ = HO.25 + 0.20) = 0.225 m Mean length 1= 27T'r mean 1= 271'0.225 = 1.4137 m Area of cross section R = -pi · R eSlstance A 2.26 =5 x 4 10- 3 (0.25 - 0.20) = 2.5 x 10- m 2 8 = 2.78xlO- x1.4137 2.5 x 10 4 1572·X 10- 4Hn = • A resistor made of aluminum wire dissipates 25 W of power at 50 V at 20°C. Calculate the current in a second resistor made of copper and having the same resistance as the first resistor and consuming four times the power of the first resistor. I 2.27 or Hot temperature R75 = 110 2 = 550 = R 20 [1 + l¥zo(75 or = 20 + 55 = 75°C 20)] = R 20 [1 + 0.0043(75 - 20)] or R 20 = 44.480 Rx = R75 - R 20 = 55 - 44.48 = 10.520 2 Conductor sizes (cross sections) in electric motors are chosen on the basis of current loadings expressed in A/m • 2 In a particular machine, the allowable current rating is 3 x 10 6 A/m in 0.5-m-long copper conductors. Calculate the conductor cross section if the loss in each conductor is not to exceed 1 W at 20°C. The resistivity of copper at 20 °C is 1.72 X 10- 8 0. m. I where Power P = /2R = (fA)2 J = I/A = current density or current loading, ~ = f2Apf or or 2.29 !WO = \' Wo = 1.0 A A resistive coil draws 2.0 A at 1tO V after operating for a long time. If the temperature rise is 55°C above the ambient temperature of 20 QC, calculate the external resistance which must be initially connected in series with the coil to limit the current to 2.0 A. The temperature coefficient of the material of the coil is 0.0043 °C- I at 20°C. I 2.28 /2 A = 12.92mm 2 Wire of a certain material x and a given cross section has a resistance of 100 O/km and a temperature coefficient of 0.0025°C- I • Wire of another material y of a given cross section has a resistance of 500/km and a temperature coefficient of 0.00075°C- I . It is desired to make a coil having a 1000-0 resistance and a temperature coefficient of 0.001 by using suitable lengths of the two wires in series. Calculate their respective lengths. I Let Rx and Ry be the respective resistances at the given temperatures. the total series resistance becomes R, = Rx(1 Then at a temperature change I:!..T, + 0.00251:!..T) + R/1 + 0.000751:!..T) (1) Since 0.001 is the temperature coefficient of the combination, we also have (2) Combining Eqs. (1) and (2) yields: Rx(1 + 0.00251:!..T) + R/1 + 0.000751:!..T) = (Rx + Ry)(l + 0.0011:!..T) R x(0.0015 I:!.. T) = Ry(0.00025 I:!.. T) or Thus, Rx=ijRy=~Ry, but The respective lengths are: fx 2.30 Rx+Ry=10000. 1km = 1000 6250 = 6.25 km Consequently, fy = R x =6250 and Ry=3750. lkm 500 3750=7.5km It is desired to maintain a 5-A constant current in a resistor made of copper wire through a temperature rise of 55°C from 20 °C ambient temperature. The value of resistance at 20 QC is 400 and the temperature coefficient is 0.00428 °C- I at O°C. Determine the minimum and maximum voltage that must be available from the power supply to maintain the desired current. 12 D CHAPTER 2 I Vrnin (at 20°C) ,= li'l,J = 40 x 5 = 200 V R75 R 20 1 + 75 x 0.00428 I + 20 x 0.00428 v'nax 2.31 =, 48.67 x 5 = 243.35 V Since the current is 5 A at both temperatures, Pzo = 12 R lo = 52 X 40 =I Ps = 12R75 = 52 x 48.67 = 1.21675 kW kW Determine the current through and the voltage aCIOs~;:he resistor of Prob. 2.30 if it is required that the power dissipated at 75°C is the same as that at 20 °C and 200 V. I P75 V75 or 2.33 = R7;1 = 1.217 x 40 = 48.67 n Calculate the power dissipated in the resistor of Proh. 2.30 at 20 and at 75°C. I 2.32 (at 75°C) R75 or = V 1000 x V~o 200 40 = PlO = ]f2 ) 48.67 = 220.61 V~5 2 48.67 and V 175 = 220.61 48.67 = 4.533 A Determine the ratio of powers dissipated in two resiston, each having the same length and each made of copper wire of circular cross section, but one having a diametn twice that of the other, and each being connected across the same voltage. V2 V' V 2A 'TT' V2D~ P = - = - - - , = -I - = - - I I RI pt/AI pt 4 pt V2 V2D~ 'TT' P = ,-- =, - - 2 R2 4 pt Similarly, PI = ,e~" 4D~ = 4 P2 D~ D~ 2.34 Find the ratio of powers in the two resistors of Prob. 2.33 when the resistors carry the same current. I = j2R P I = ,;' pt = i Al l' P = f2R = Similarly, 2 'TT' ~ D~ PI l,'D~ P, = \'1); 2.35 D~ 12pt 'TT' 2 Ilpe I 4 A 100-W IIO-V light bulb has a filament made of an alby having a temperature coefficient of 0.0055 °C - I at O°C. The normal operating temperature of the bulb is 2000 'c. How much current will the bulb draw at the instant it is turned on when the room temperature is 20 DC? From your result verify that burnout of bulbs is more frequent at the instant they are turned on. I The ratio of resistances at the two temperatures is given by 1 + 20a" 1 + 2000a\) I -' ,20 x 0.0055 _2 T~~2i)00 + 0.0055 = 9.25 x 10 2 110 100 = 121 At 20°C: Compare with 2.36 R 20 = 121 x 9.25 x 10- 2 = 11.20 f 2000 = I¥l and = n 121l = 110 11.2 = 9.82A 0.91 A The current loading of the heating element of a 11O-\' 750-W electric heater is not to exceed 2600 A/in 2 (cf. Prob. 2.28). The resistivity of the wire material is 12 x 10- 8 n· m. Calculate the length and the area of cross section of the heating element. I RESISTANCE AND OHM'S LAW iA Since R = ~= 2 110 750 P e= or 2.37 = 13 2600 Afinz = 2 2 62 = 6.818. 2600 III =. x Area A D 16.13 0~3.III 2 1 n = pe = A = 1.69 mm 2 12 x 1O~8e l.69 x 10 6 __ 16_._13_x_I._6_9--;x;-I_0_~6__ = 227.16 m 12 x 10 8 Heat energy is often measured in calories and 1 calorie (cal) = 4.184 joule (l). It is desired to design a heating element to boil a certain amount of water in 2 min requiring 40 kcal heat energy. If the heating element is to operate at 110 V, calculate its current and power ratings. I 1 kcal = 4.184 kl = 4.184 kW· s = 4184 W· s The heat energy required is Q = 40kcal = 40 x 4184 Let P be the power required. Then, P = 167,360 W . s = 1395 W I 120 s 2.38 167,360 W· s = P = V= 1395 110 = 12.7 A For the data of Prob. 2.37, determine the resistance of the heating element if the same amount of water is required to boil in 30 s. I The same amount of energy must be delivered in one-fourth the time; so the power is now P = 4(1395) = 5580 W = VI 2.39 5580 1= 110 or = 50.7 A R= V 7= 110 50.7 = 2.17 n As the temperature of a heating element changes, its resistance also changes, and so does the temperature coefficient. In a certain case, the temperature varies linearly with time and is given by T QC = (20 + lOt), where t is time in seconds. The temperature coefficient of the material is 0.0065 °C~l at O°C. If the initial resistance of the heating element is 2 n, find its resistance after 10 s. I At t = 0: and T = 20 + 10 x 10 = 120 QC At t = 10 s: R 120 = Rzo[1 + a 2o (120 - 20)] From Eq. (4) of Prob. 2.13, ~ a ZO = I + ao20 0.0065 0005 QC l 1 + 0.0065 x 20 =. 75 R 120 = 2[1 + 0.00575(120 - 20)] = 3.15 n 2.40 For the heating element of Prob. 2.39, express the resistance as a function of time. I Resistance at a temperature T is given by RT = Ro(1 + aoT) = R(t) From the data: Ro Hence 2.41 =RI [I + a20(0 - 20°)] R(t) = l.77[1 = 2(1 - 20 x 0.00575) = 1.77 n + 0.0065(20 + lOt)] = (2 T = 20 + lOt + 0.115t) n If the heating element of Prob. 2.39 or 2.40 is connected across a 11O-V source, calculate the initial and final powers. V2 110 2 R=2n P. = - = = 6050 W I At t = 0: R 2 I At t = 10 s: R=3.15n 110 2 Pf = 3.15 = 3841 W 14 2.42 D CHAPTER 2 For Probs. 2.39 through 2.41 determine the energy dissipated in the heating element over the lO-s period. dU == R(iY il = 2 + 0.115t dt 1102 U == 110 2 (HI dt == 0 5 [In (2 ).. 2 + 0.115t .11 or 2.43 110 2 V2 I + 0.115t)go == 47.795 kJ = 0.0133 kWh A block of iron is heated directly by dissipating power in the internal resistance of the block. Because of the temperature rise, the resistance increases exponentially with time and is given by R(t) == 0.5e 21 n, where t is in seconds. The block is connected across a 11O-V source and dissipates 1827 cal heat energy over a certain period of time. Calculate this period of time. I Let t be the required time. Then energy dissipcted is U= 2 2 V2 f.1 ----dt==-110 110 JI I--dt== eJo R(t) 05e 0.5 21 _ ~ ~0~.5 (e -'I ';, dt 2 110 (1- e- 21 J ) 1827 cal == 1827 x 4.184 == 7644 J == U Now 1 - e- 21 =, 76442 = 0.632 110 -2t In e == In 0.368 or Thus e -21 = 0.368 or Hence, 2.44 =, 2l 0 (j t = -2t == -1 or 0.5 s A light bulb, having a tungsten filament, draws 0.5 A at 110 V. The cold resistance of the filament is 20 n at 20 QC. At this temperature the temperature coefficient of resistance (for tungsten) is 0.005 QC-I. Determine the operating temperature of the bulb. I Resistance at the operating temperature T QC i~ RT == V 110 I = D.5 == 220 n == R2o\1 -+ ,:l2o(T - 20)] = 20[1 + 0.005(T - 20)] Solving for T yields T == 2020 QC. 2.45 The operating temperature of a tungsten-filament 1l0-V 40-W bulb is 2020 QC (cf. Prob. 2.44). The filament is made of a O.Ol-mm-diameter wire having a resistivity of 5.55 x 10- 8 n· m at 20 QC and a temperature coefficient of 0.005 QC-I. Calculate the length of the filament wire. I At 2020 QC: or R 2020 == V2 P R 20 == 110 2 40 = == 302.5 11 275 . + 0.005(2020 - 20)] == 11R 20 n ,= !!?cQ~ == A R 5.55 x lO- e 1T14[(0.01)2 X 10 6] 1T x 27.5 X 10- 2 e= - - - - - - == 3 89 cm 4 x 555 . Hence 2.46 == 302.5 n = 8 20 [1 A 60-mm-thick electrode is cut from a solid 70-mm-radius hemisphere made of copper, as shown in Fig. 2-4. Calculate the current through the electrode if 6 V is applied across it. Resistivity of copper is l.72 x 10- 8 n·m. I Let R be the resistance of the electrode. Then, for the infinitesimal disk shown in Fig. 2-4, or p ( tan h- 1 b a - tan h-10) ==;. I == ~ == R == . f r~a 1T )FO (b 2 dx - x2 ) ::)a == f (tanh- 1 1T b 0 8 P == ;. tan h-1 ab 0.702 ~ 10 8 :: l.72: 10- tanh- 1 60 == 0.702 x 10- 8 70 8.547 x 108 A = 854.7 MA n H RESISTANCE AND OHM'S LAW 0 15 I 6Y 2.47 Fig. 2-4 A carbon resistor dissipates 60 W of power while drawing 0.5 A of current at 20°C. How much power will be dissipated in the resistor at 100 °C if connected across a 120-V source? Temperature coefficient of carbon at 20°C is -0.0005°C- 1• 60 or R zo = --z = 240 n I At 20°C: 0.5 At 100°C: R100 == Rzo[l - 0.0005(100 - 20)] = 240(1 - 0.04) = 230.4 n Vl 120 2 Pzoo == R = 230.4 = 62.5 W zoo 2.48 We have two resistors wound with round copper wire. The length and the diameter of the first wire are t and A respectively and those of the second wire are 0.25t and 0.5A. Determine the ratios of currents and powers for the two resistors if they are connected across the same voltage source. I 2.49 R 1 = p(0.25t) 0.5A = 05 pt = 0 5R . A . 1 If the same current flows through the two resistors of Prob. 2.48, determine the ratios of voltages and powers. I 2.50 and From Prob. 2.48, RzfR 1 = 0.5, Obtain the exact and approximate ratios of the resistances of a coil at two temperatures Tl and T1 assuming that the only other given quantity is the temperature coefficient a o at O°c. I Let Ro be the resistance at O°C. Then at the temperatures T1 and Tl we have and which is the exact ratio. R1 R z = (1 or R1 Approximately, 1 : + a oT1)(1 + aOT1) = (1 + a oT 1)[1 - aoTz + (aoT z ) _ ... ] = 1 + ao(T, - T z ) CHAPTER 3 Series and Parallel Resistive Circuits 3.1 \\ "--~ How much current will flow through a 2-0 resistor eonnected in series with a 4-0 resistor, and the combination connected across a 12-V source? What is the voltage across each resistor? V 12 I =-=-=2A I Rs 6 I and 3.2 A 2-0 resistor is connected in parallel with a 4-0 n:sistor and the combination across a 12-V source. current through each resistor and the total current supplied by the source. I 3.3 12n =¥=6A =i- = 3 A Itotal =6t 3=9A What is the total resistance of the combination of a 2·0 and a 4-0 resistance in parallel? supplied by a 12-V source connected across the combination. 1 1 1 3 -=-t-=Rp 2 4 4 I 3.4 I' ll Find the Calculate the current V 12 I=R=4/3=9A or p Two resistors of ohmic values RI and R2 are connected in series, and the combination across a source of voltage V. How is this voltage divided across the resistors? I 3.5 Two resistors of ohmic values RI and R2 are connel;ted in parallel, and the combination across a source of current I. How is this current divided through the Iesistors? I Let or V:= IRp = IIRI = 12R 2. R2 IRp R IRp I = = I ---I ==I I I RI RI + RI 2 R2 RI t R2 V= voltage across the combination. Therefore, 3.6 Use the results of Prob. 3.4 to solve Prob. 3.1. I V =V I 3.7 RI 2 = 12 2 t 4 = 4 V RI t R2 I 2 R2 4 II = I R t R = 9 2 t 4 = 6 A I 2 Calculate the power in each resistor of Prob. 3.1 and verify that the total power supplied by the source is the sum of the powers in the resistors. I PI = I~RI = 22 X 2 =8 W P2 = I~R2 PI t P2 = 8 t 16 = 24 W 16 R2 4 V2 = V R t R = 12 2 t 4 = 8 V Use the results of Prob. 3.5 to solve Prob. 3.2. I 3.8 Then = 22 X 4 = 16 W (since Ps = VI = 12 x 2 = 24 W II = 12 = 2 A) SERIES AND PARALLEL RESISTIVE CIRCUITS 3.9 Determine the power in each resistor of Prob. 3.2. sum of the powers in the resistors. V2 I 3.10 122 V2 + P2 = 72 + 36 = 108 W 122 P =-=-=36W 2 R2 4 Ps = VI = 12 x 9 = 108 W A 3-0 and a 6-0 resistor are connected in parallel and the combination in series with an 8-0 resistor. the total resistance. I 17 Verify that the total power supplied by the source is the P =-=-=72W ] R] 2 p] D R tota ] = Rp Calculate + Rs = 2 + 8 = 10 0 3.11 A 20-V source is connected across the resistor combination of Prob. 3.10. What is the voltage across the 8-0 resistor? V 20 I=--=-=2A I VS!l = 8 x 2 = 16 V R tota ] 10 3.12 Determine the power absorbed by each resistor of Prob. 3.11. V2 2 I Pg!l = ~ n = 1~ = 32 W V3 n = V6 n = V-Vg n = 20 - 16 = 4 V Check: 3.13 Total power = 32 + ~ + ¥ = 40 W. Power from source = VI = 20 x 2 = 40 W. For the circuit shown in Fig. 3-1, find the value of k so that the resistance of the combination is a minimum. I For R min , aRlak = 0, R = ka + f!: 2 k which implies that 2k(2ak) - 2(ea + 2a) =0 R = ea + 2a 2k 2e - e - 2= 0 or or k= V2 = 1.414 'VV + 3.14 P = V21R is a maximum when R is a minimum; k = V2. What is the maximum power the resistors of Fig. 3-1 can absorb when connected across voltage V? the input current at maximum power condition. I 3.16 Fig. 3-1 If a voltage V is connected across the resistor combination of Prob. 3.13, find the condition for maximum power supplied from the source to the resistors. I 3.15 v From Probs. 3.13 and 3.14, 2V 2 k 2V2 V2 V2 P = ---;;- e + 2 = ---;;- 2 + 2 = V2a W Determine and Four resistors of ohmic values 5, 10, 15, and 200 are connected in series and a 100-V source is applied across the combination. How is this voltage divided among the various resistors? 18 0 CHAPTER 3 I Using the voltage division rule, we have Similarly, 3.17 VIO = 20 V, V, s = 30 V, V20 = 40 V Formulate the law of current division among three resistors R" R 2 , and R, connected in parallel, input current is i, I The common voltage across the resistors is V'= iR,.,., where IfR"I' = I fR, + I fRe + I fR,. Thc total Hence, , V Rep. 1 =-=-1 , 3.18 R, R, Determine the current through and the voltages acrm,s three resistors of ohmic values 5,7, and 8 n, connected in series and across a 100-V source. Total resistance = Re, = 5 + 7 + 8 = 20 n I ·· I V 100 C nCUlt current = = R = 20 = 5 A {'s Voltage across the 5-0 resistor = 51 = 25 V 8-0 resistor = 81 = 40 V 3.19 Voltage: across the 7-0 resistor = 7 1= 35 V Voltage across the Determine the voltage across and the currents throu:?;h three resistors of 5, 10, and 200, all connected in parallel and across a lOO-V source. There is 100 V lCross each resistor. I Current through the 5-n resistor = !¥' = 20 A. C'lrrent through the 10-0 resistor = \(~' = 10 A. Current through the 20-0 resistor =,*,' = 5 A. 3.20 Determine the current and power drawn from the source in the circuit of Prob. 3.19. I Total current from source = 20 + 10 + 5 = 35 A. Power supplied by source = VI = 100 x 35 = 3500 W. 3.21 Reduce the circuit between the terminals a and b, Fig. 3-2, to a single resistor. 20 a 10 30 d 6 n e 16 n b ~--~--~----~--~~----~---- c 60 80 Fig. 3-2 I From the law of parallel resistances, 1 I I 1 - = - + - -- or Red = I n Red 2 3 6 The series resistance between a and e is then I - I + 6 = 8 0, giving a net resistance Ra,. = (8)(8) = 4 n 8+8 3.22 R"h = 4 + 16 = 200 Calculate the resistances of 110-V light bulbs rated at 25, 60, 75, and 100 W. I From P = V2fR: R 2, W R how = ( 110)2 ---zs (110)2 = ~ = 484 0 , =201.670 R75 W = RlOo (l1W ~ = 161.30 W (110)" = 10() = 121 0 SERIES AND PARALLEL RESISTIVE CIRCUITS 3.23 D 19 An electric heating pad rated at 110 V and 55 W is to be used at a 220-V source. It is proposed to connect the heating pad in series with a series-parallel combination of light bulbs, each rated at 110 V; bulbs are available having ratings of 25,60,75, and 100 W. Obtain a possible scheme of the pad-bulb combinations. At what rate will heat be produced by the pad with this modification? • From Prob. 3.22 we know the resistances of the various light bulbs. The resistance of the heating pad is We must combine the bulbs to obtain a total resistance of 2200; then, by voltage division, the pad voltage will be the required 110 V. One possibility is a 100-W bulb in series with a parallel combination of two 60-W bulbs: Rb = R 100 + ~R60 = 121 + H201.67) = 221.83 0, which is on the safe side. Then 220 Rp + Rh = 220 + 221.83 = 441.830 Ip = 441.83 = 0.498 A Rp = (110)2/55 =2200. I!Rp = (0.498)2(220) = 54.56 W. and so the heat output of the pad is 3.24 Two resistors, made of different materials having temperature coefficients of resistance a l = 0.004 QC-I and a 2 = 0.005 QC- \ are connected in parallel and consume equal power at 10 QC. What is the ratio of power consumed in resistance R2 to that in R I at 60 QC? • At 10 QC, RI = R 2, which implies or ROI R02 1 + lOa 2 1 + lOa l Consequently, the power ratio at 60 QC is V2/R2 RI R'lI(1 + 60a l ) V2/RI = R2 = R 02 (1 + 60a 2) (1 + lOa 2 )(1 + 60a l ) (1 + lOa l )(l + 6Oa 2 ) Substituting the numerical values of a l and a 2 yields the value 0.963. 3.25 A 200-V source is connected across the circuit shown in Fig. 3-2. • From Prob. 3.21, Rab = 20 O. V 200 1= = = 10 A Rab 20 3.26 Calculate the voltage across the 8-0 resistor. Thus, Veb = ReJ = 16 x 10 = 160 V Vae = Vg n = V - Veb = 200 - 160 = 40 V In Prob. 3.25, determine the power dissipated in the 1-0 and 8-0 resistors. • From Prob. 3.21, 1= lOA Vgn =40V From Prob. 3-25, 1I 3.27 n = I - Ig II = 10 - 5 = 5 A Find the ratio of the currents 11/12 at 60 QC in the resistors of Prob. 3.24. • From Prob. 3.24, RI /R 2 = 0.963 = P2 / PI' RI If =0.963= 2 Notice that this result also follows from 3.28 I~ R2 12I IfI Since P2/ PI = I~R2/I~RI' or IIR I = 12R2 = V, A battery has internal resistance Ri and terminal voltage V,. cannot exceed V~/2Ri' • Let RI be the load resistance. Then and power taken by the load, we have since the two resistors are in parallel. Show that the power supplied to a resistive load 20 D CHAPTER 3 aPTiaR L = 0, For maximum power, which requires that (RL + Rf - RL[2(RL + R,)] = 0 Hence, 3.29 or (PTJrnax = V;/2R,. For the battery of Prob. 3.28, V = 96 V and R.= 50 mO. Discrete loads of 150, 100, 50, 30, and 20 mO are connected, one at a time, acr~ss the battery. Plot the curve of power supplied versus the ohmic value of the load. Hence verify that the maximum power IT.msfer occurs when Ri = R load = 50 mO. I PT. = V;' (RL :LR,)2 Substituting the given numerical values yields: _ 96 2 PIOO - . P IOO POD . 2 150 2 -_ 34._6 " kW (150 + 50) 100 . = 96 (100 + 50)2 = 40.96 kW = 96 2 2 30 P,o . = 96 (30 + 50) 2 = 43.20 kW P20 = 96 2 20 (20 + 50)2 = 37.62 kW 50 2 = 46.08 kW (50 + 50) which is plotted in Fig. 3-3 showing that (PTJrnax occurs at RL = 50 mO. 50 (P) = 46.08 kw ~o..)C _ i _ ---"" \ 45 40 ___ \ I 35' 30 so _ i i 7S Rc, .., 100 fL I~S i ISO Fig. 3-3 3.30 A battery has an internal resistance of 0.5 0, and has an open-circuit voltage of 20 V. The battery supplies a 2-0 load. Determine the power lost within the batl:ery and the terminal voltage on load. 20 2 I 1= 2 + 0.5 = 8 A V, = IRL = 8< 2 = 16 V Plos' = [2Ri = 8 (0.5) = 32 W 3.31 A resistor made of silver and another made of nickel, having temperature coefficients of resistance at 20 QC of 0.0038 QC-I and 0.006 QC-I, carry equal currents at 20 QC when connected across a voltage source. How will the total current be distributed if the temperature is raised to 150 QC? SERIES AND PARALLEL RESISTIVE CIRCUITS • 21 At 150 C, Q RSilve, = (R silve , )20[1 + 0.0038(150 - 20)] = 1.494(Rsilve, )20 Rnickel = (Rnickel)20[1 + 0.006(150 - Since the currents are equal at 20 QC, 3.5) at 150 QC, (R silve J2o 20)] = 1.78(RnickeJ20 = (Rnickel )20 = R. Therefore, by current division (sce Prob. 1.78R 3.32 D Isilver = 1.78R+l.494R 1=0.54371 or 54.37% Inickel = l.78R 1.494R + 1.494R 1 = 0.45631 or 45.63% Convert the delta-connected resistor bank of Fig. 3-4 into an equivalent wye-connected resistor bank. ~ ----------------~ b _ _ _...., c----------------~ Fig. 3-4 • For equivalence, the resistance between any two terminals (say, ab) for both the wye and the delta connections must be the same. Thus, equating them we get Ra(Ra + Ra) 2R y = R + R + R = a Hence, 3.33 a a 2 '3 Ra Ry = ~Ra' Three unequal resistors are connected in wye as shown in Fig. 3-5. resistor bank. • Obtain an equivalent delta-connected From Fig. 3-5 it follows that: (1) (2) (3) Solving RA' R B , and Rc yields: RA = 1 If (R t R 2 + R2 R 3 + R 3 R t ) 2 c__---. b Fig. 3-5 22 3.34 D CHAPTER 3 Suppose three resistors RA' R B' and Re are connected in delta as shown in Fig. 3-5. wye-connected resistor bank. • In this case also Eqs. (1), (2), and (3) of Prob. 3.33 are valid. obtain 3.35 Obtain an equivalent Therefore, we solve for RI' R 2 , and R3 to Convert the pi-connected resistors of Fig. 3-6a to an equivalent tee-connected set (Fig. 3-6b). • Notice that pi- and tee-connections are, respectivdy, the same as delta- and wye-connections. the results of Prob. 3.34, we obtain RARe RI = R A + R B + R e Thus, using 6x3 9 + 6 + 3 = 1.00 9x3 9 + 6 + 3 = 1.50 9x6 9+6+3 =3.00 >RI!> -_ (; L1Fig. 3-6 3.36 Verify that the converse of Prob. 3.35 is true; that is, show that if RI = 1.50, R2 = 1.00, are connected in wye, its equivalent delta will have the values shown in Fig. 3-6a. • Since RIR2 + R2R3 + R3R, = 1.5 x 1 + 1 x 3 + 3 >: 1.5 = 9, 3.37 Re=~=30 Determine the resistance across the terminals ab of the interconnected resistors of Fig. 3-7 a. (A.. _ _ _ _ _ __ I ('{2./ ,cz. < .... ~ ..... • _I'" "IV c~ ___ b _ _ _ __ R, - = 3.00 from the results of Prob. 3.33 we have: 9 R fl =-r:s=60 R A =t=90 and '1 b LL 'V,~ b 2. .(1. "sS- c ~:J1- b_____ 1- (b) (c) Fig. 3-7 SERIES AND PARALLEL RESISTIVE CIRCUITS 0 23 • First, we convert the upper delta to a wye to obtain the interconnection shown in Fig. 3-7 b which reduces to that given in Fig. 3-7c. Finally, Rab = 2 + (6 x 4) /(6 + 4) = 4.4 D. 3.38 What is the resistance across the terminals ab of the network shown in Fig. 3-8a? • By converting the delta-connected resistors to an equivalent wye we obtain the interconnection shown in Fig. 3-8b. Next combining the 3-D and 6-D resistors in parallel leads to the circuit shown in Fig. 3-8e. Hence, Rab = 2 + 2 = 4 D. b __~____L -_ _~_ h _ _ _----> c Cc) 3.39 Fig. 3-8 For the network shown in Fig. 3-8a, calculate the voltage across the terminals ae if a 36-V battery is connected across the terminals ab. • In this case, we convert the 6-D wye-connected resistors into an equivalent delta to obtain the circuit of Fig. 3-9a. Combining the 9-D and 18-D resistors in parallel gives the circuit of Fig. 3-9b from which we obtain the currents as follows: 18./7- 6.n. c 3&v Fig. 3-9 36 lac 3.40 or Calculate the power dissipated in the 9-D resistor connected across ab and in the 9-D resistor connected across be of the network of Fig. 3-8a when a 36-V source is connected across ab. • 2 From Fig. 3-9a we have, P9nab = V;b/Rab = 36 /9 = 144 W. From Problem 3.39, Vac = 18 V. Hence Vbc 3.41 = 6+6 =3 A = 36 - 18 = 18 V For the circuit shown in Fig. 3-lOa, determine R so that the power going into the terminals ab is maximum. Also calculate the maximum power. 24 0 CHAPTER 3 • The sequence of network reduction is shown source is given by ID Figs. 3-IOb-e. From Fig. 3-IOe the current drawn from the 12 1== 1+ 0.5R A and power is P For maximum power, Thus, aPlaR = 0 =I 2. Il-\I and 1.R "I- a. i '2.R. "l- = (1 + 0.5R)2 0.5( 1+ 0.5R)2 - 0.5R x 2(1 + 0.5R)0.5 = 0 requires that 12 =6A I = 1+ 0.5 x 2 1.Cl.. 144(0.5R) (O·:>ln R = 2 n. P max = 6\0.5 x 2) = 36 W ---7 P / or 1..!1.. I - ~ -\" ll-V !> ~ ?.R 2- b h (~ ) lC ") I /..I1- ~ o. sq .J1. Fig. 3-10 3.42 For the circuit in Fig. 3-lOa show by changing the delta-connected resistors to an equivalent wye that the maximu:n power entering the terminals ab is 36 W. • The network reduction is shown in Fig. .3-11.1--c. Therefore, from the results of Prob. 3.41, we havt: R=2n Notice that Fig. 3-11 C IS identical to Fig. 3-lOe. and Pmax = 36 W SERIES AND PARALLEL RESISTIVE CIRCUITS D 25 {'AV p ( 4.) /fI- II R 4" R 4 r r ~ g 4 b (6) ,tL- 1 • t't I'SR J Fig. 3-11 (c) 3.43 Four 60-W 11O-V bulbs are to be operated from a 230-V source. Determine the value of the resistance connected in series with the line so that the voltage across the bulbs does not exceed 110 V. 4x bO =240w Fig. 3-12 • For the circuit shown in Fig. 3-12 we have total power drawn from the source, P = 4 x 60 = 240 W Input current I P 240 = V = Do = 2.1818 A Voltage across the series resistor, VR 3.44 = 230 - 110 = 120 V = IR or R= VR T = 120 2.1818 =550 An alternate way of operating the bulbs of Prob. 3.43 is to connect them as shown in Fig. 3-13 with a series resistor. Calculate the value of the series resistance and state, giving reasons, which of the two methods is preferable. R 2?o I 4><{,o:;2fOW Fig. 3-13 26 D CHAPTER 3 • In this case, 1= P V 240 = 220 = 1.0909P., L VR = 230- 220= lOV= RI [11 Thus IJi909 = 9.167 0 R= 12R loss in method of Prob. 3.43 = 120 2 /55 = 261 82 W. 12R loss in method of the present problem = 10 2 /9.167 = 10.9 W. Second method is more efficknt but, if one bulb burns out, only two will function. 3.45 A 12-V battery is made of 36 cells each rated at 2 V c.nd 1.5 A for a given duty cycle. and power that may be drawn from the battery? • Since the battery is rated at 12 V and each ;;ell at 2 V, cells connected in series = parallel paths = ~ = 6. Rating for one parallel pa:h = 1.5 A (given). Line current = 6 x 1.5 = 9 A (Otherwise, 3.46 What is the line current ¥ = 6 cells. Number of Power = VI = 12 x 9 = 108 W power = 36 x 2 x 1.5 = 108 W.) For the circuit shown in Fig. 3-14, calculate R sucb that the power dissipated in the 3-0 resistor is 300 W. Fig. 3-14 • P3 fl = -tV2 = V, 30 :3 = :3 = 10 A I, n = V V I=1O+5=15A= I+R+(6X3)/(6+3) =3--~R 3.47 V3 = 30V V, 30 Ion = (; = (; = 5A or V=45 + 15R =90V or 90 - 45 R= - - - =30 15 For the value of R determined in Prob. 3.46, calcu late the power absorbed by each resistor. total power thus obtained is the same as that supplied by the source. 2 • Total power absorbed = 15 x 1 + 15 15 = 1350 W. 3.48 or :\00 W 2 X 2 2 3 + 30 /3 + 30 /6 = 1350 W. Verify that the Power supplied by the source = 90 x Calculate the voltage that must be connected across the terminals ab such that the voltage across the 2-0 resistor is 10 V (Fig. 3-15). • Input current to the circuit, Rpacallcl = 6 x 12 6 + 12 1= 1f = 5 A. = 40 Rtota' =, 2 + 1 + 4 = 70 V= Rtot"J= 7 x 5 = 35 V. Fig. 3-15 b 3.49 Refer to Fig. 3-15. Determine the voltage across the 6-0 resistor. verify that 1= 11 + 12, • V6 n = V - Vo !l .- Hence determine the currents 11 and 12 and 111 11 = 35 - 10 - 5 = 20 V 10 10 20 + 10 1+1 1 2 =-+-=--=5A=I 366 SERIES AND PARALLEL RESISTIVE CIRCUITS 3.50 D 27 Find the current in the 5-0 resistor in the interconnection of resistors shown in Fig. 3-16a. I By changing the delta-connected 3-0 resistors into an equivalent wye we obtain the circuit of Fig. 3-16b, which is reduced to the circuit of Fig. 3-16e. Thus, V 36 1= - = - - =12A R 2+ 1 VIO (Fig. 3-16b) = RI = 1 x 12 = 12 V V(I+S) 0 = V-VI = 36 - 12 = 24 V 0 24 Iso = 1 + 5 = 4 A Thus, c I V-:'?J'V .,. CA) 5..!L- I:> t).. '..fl..- " ,.fL I..£L 2.f7.'\.\ c I 36V (b) A. • ,..n.. IOI'V'" + 2.1l-- I"- • 4- -N' cl ~ (c) 3.51 3.f\. (d) Fig. 3-16 By adding the powers absorbed by the resistors of Fig. 3-16e, verify that the sum is equal to the power supplied by the 36-V source. I From the results of Prob. 3.50, Thus, Is 0 = 4 A and 120 = 8 A Vso=4 x 5=20V and V20 = 8 x 2 = 16 V Also the volt ages across the top and bottom of the 3-0 resistors become 16 and 20 V respectively as shown. The respective currents in these resistors are ¥ A and ~ A. Hence the current I (Fig. 3-16a) in the vertically I = ¥ - 4 = ~ A. drawn 3-0 resistor between be becomes Verification: 120 = ~ + ~ = 8A Pin 3.52 Find the resistance between the terminals ad for the interconnected resistors shown in Fig. 3-16a. I 3.53 = 36 x 12 = 432 W Using wye-delta transformations shown in Fig. 3-16b we obtain the circuit of Fig. 3-16d. A shunt is used to extend the range of an ammeter by connecting it across the ammeter as shown in Fig. 3-17. The ammeter has a resistance of 0.10 and gives a full-scale deflection of 2.5 A. Calculate the value of the shunt resistance to extend the range of the ammeter to 50 A. 28 D CHAPTER 3 I Refer to Fig. 3-17: Is = 50 - 2.5 = 47.5 A I=50A la =2.5 A or 2.5 x 0.1 47.5Rs = 2.5 x 0 1 Rs = - - - =5 .263mO 47.:> Hence, Fig. 3-17 3.54 The multiplying power of a shunt is defined as the ratio of the line current to the current through the ammeter. Obtain a general expression for the multiplying power. I From Fig. 3-17 we have: IaRa = (I - IJR s · I' Ra Mu Itip ymg power = TI = 1 + If or or a 3.55 s The resistance of a coil is measured experimentally by the voltmeter-ammeter method. Two possible arrangements of the meters are shown in Fig. 3-18. The resistance of the voltmeter is 10 kO and that of the ammeter is 0.1 O. For the setup of Fig. 3-18a the voltmeter reads 5 V and the ammeter reading is 25 A. What is the value of the resistance? I For Fig. 3-18a from Ohm's law: 5 = 25(0.1 + R) R= or fs - 0.1 = 0.1 0 I Ill. R (If) 3.56 (b) Fig. 3·18 If the ammeter reading in Fig. 3-18b is 25 A and the valiues of various resistances are the same as in Prob. 3.55, determine the voltmeter reading. I By the rule of current division we have: R 0.1 Iv = R + Rv 1= 0.1 + 10,000 (25 A) Voltmeter reading = RJv 3.57 == (HI,OOO) 0.1 +O~~,OOO (25) = 2.5 V Based on the results of Probs. 3.55 and 3.56, if the resistance is measured as the ratio of the voltmeter-toammeter readings, state which of the two connections of Fig. 3-18 is preferred for the measurement of (a) a low resistance and (b) a high resistance. ~T 5 I From Prob. 3.55: R= -i = B From Prob. 3.56: R= I V =0.20 2.5 = :~5 =0.10 Clearly, Fig. 3-18b is suitable for the measurement of a low resistance and Fig. 3-18a is preferred for measuring a high resistance. SERIES AND PARALLEL RESISTIVE CIRCUITS 3.58 D 29 Calculate the value of the shunt resistance to be used with a galvanometer having a resistance of 10 il if the current through the galvanometer is not to exceed S percent of the total current. I From Prob. 3.S4 we have: I Ra - =1+la Rs = ~ =0.S26il Hence, 3.59 or Rs An ammeter rated to read up to S A, having a resistance of O.S il, is to be converted into a IS0-V voltmeter by connecting a resistor in series with the ammeter. Calculate the value of this resistance. I Let R be the value of the series resistance. Then by Ohm's law, ammeter resistance. Substituting numerical values yields: ISO = S(R + O.S) 3.60 l~O R= or V = I(R + R.), where Ra IS the -0.S=29.Sil A 300-V voltmeter draws 2 mA current for full-scale deflection. This voltmeter is used to measure the voltage across a SO-kil resistor connected in series with a 2S-kil resistor, the combination of the resistors being connected across a IS0-V source. What is the voltmeter reading? I The voltmeter resistance Rv = 300/(2 x 10- 3) = ISO kil. From the circuit of Fig. 3-19, the total resistance across the IS0-V source is R = 2S + (ISO x SO) 1(ISO + SO) = 62.S kil. ISO 1= 62.S X 103 =2.4mA By current division rule, 3 I =SOX10 x2.4XlO- 3 =0.6mA (ISO + SO) x 10 3 IvRv = 0.6 x 10- 3 x ISO X 10 3 = 90 Y. v Thus the voltmeter reading is v I I + 3.61 50 k.£l.. ISO kn... V Fig. 3-19 If the voltmeter of Prob. 3.60 is now connected across the 2S-kil resistor, what is the voltage across the SO-kil resistor? I When the voltmeter is connected across the 2S-kil resistor, the value of the combined parallel resistance becomes Rp = (2S x ISO) 1(2S + ISO) = 21.428 kil. Or the total circuit resistance is SO + 21.428 = 71.428 kil. ISO x 10 - 3 = 2 .1 m A " CIrcUlt current I = 71.428 Voltage across SO kil = SO x 10 3 x 2.1 X 10- 3 = lOS V 3.62 A high-voltage dc transmission line supplies 900 MW of power over a distance of 800 km. This power is delivered at 600 kY. If the loop resistance of the line is 2 mil/km, determine the sending-end voltage. I The line is schematic ally represented in Fig. 3-20. and sending-end quantities. Then, PR = 900 MW (given) Subscripts Rand S are, respectively, used for receiving- I =I R 900 X 10 6 = I = 600 X 103 = 1.5 kA s R = 800 x 2 X 10- 3 = 1.6 il Vs = VR + IR = 600 x 10 3 + 1.S X 10 3 x 1.6 = 602.4 kV VR = 600 kV (given) 30 D CHAPTER 3 r v_:__ 3.63 R. --~--'VY-~~J L•• ' Fig. 3-20 Defining efficiency of transmission as the ratio of the power at the receiving end to the power at the sending end, calculate the efficiency of transmission of the line ,)j Prob. 3.62. Verify that the same result is obtained by line-loss calculation. I PR = p~ 900 MW (given) = VJ 3 ,= 602.4 x 10 x 1.5 X 3 10 = 903.6 MW 3 Efficiency of transmission is P RI p~ = 900/903.6 =' ~'9.6%. Line loss is 12R = (1.5 x 10 )2 x 1.6 = 3.6 MW. 6 p~ = PR + line loss = (900 + 3.6) x 10 = 903.6 MW, which is the same as previously calculated. 3.64 A dc generator has an open-circuit voltage of 123 V. When connected across a 4-0 resistor the generator delivers 3.6 kW of power. Calculate the internal power loss within the generator. I 3 36_X _ 10_ 12= _. 4 4+R g or Rg _ 123 - 120 _ n 30 - 0.1 .. - Ploss.. =('R g =30 2 xO.1=90W Hence, 3.65 123 1=30A='--- or A dc generator may be characterized by an ideal vol1:age source in series with a resistor. At the terminals of the generator, voltage and current measurements for two different operating conditions are V, = 115 V at 1= 10 A and V, = 105 A at 1= 15 A. Model the generator by a voltage source in series with a resistor. Fig. 3-21 I With the circuit model of the generator shown in Fig. 3-21, with the symbols defined, we have (for the two sets of data): \1;, = V, + IRg Solving for Vo and Rg yields 3.66 or Vo = 135 V \1;, = 115 + lOR g and and \1;, = 105 + 15R g Rg = 2.0n. A dependent voltage source is shown in Fig. 3-22. For the data given determine the power supplied by the 12 V-source and that by the 3/-dependent voltage source. + "1) Fig. 3-22 I From Ohm's law, 12 = 91 + 31 or 1= 1.0 A. Power delivered by the 12-V source is 12 x 1 = 12 W. Power delivered by the 3/-dependent voltage source is -1.0 x 3 x 1.0 = - 3 W. Negative sign is used since the current is going inlO the source. Thus the dependent source is absorbing (rather than delivering) power. SERIES AND PARALLEL RESISTIVE CIRCUITS 3.67 D 31 A 500-0 resistor is connected in parallel with a 250-0 resistor and the combination is fed by a 25-A current source. Calculate the power absorbed by each resistor. I The circuit is shown in Fig. 3-23. By current-division rule we have, 500 I1 = 250 + 500 25 = 16.67 A 250 = 250 + 500 25 = 8.33 A 12 Respective power losses are and li(250) = 69.472 kW I 1;(500) = 34.695 kW I 500£2.. Fig. 3-23 3.68 Determine the voltage across the resistors of Fig. 3-23. same as the total power dissipated in the resistors. I Voltage 104.187 kW. 3.69 Verify that the power supplied by the source is the P = V(II V=/I(250)=/2(500)=16.67x250=4167.5V. Power Power dissipated is 69.472 + 34.694 = 104.166 kW. + 12) = VI = 4167.5 x 25 = Figure 3-24 shows a circuit containing a 25-A independent current source and a 5-V dependent current source. Calculate V and R. I 1= 25 = 5 V. Thus, V = 5 V. Also, V = RI, 5 = R(25), or or R = 0.2 O. + V R l!iA Fig. 3-24 3.70 For the circuit of Fig. 3-24, determine the power absorbed by R. V2 I 3.71 PR = 52 0.2 = 125 W If the power supplied by the 25-A source of Fig. 3-24 is 150W and the dependent source absorbs power, calculate the voltages across the 25-A source and 5-V dependent current source. I V2S P25 A A = -1- '" 25 A 3.72 If '" 150 2s = 6 V V 5v = Ps v = 150 - 125 = V I 5 25 v 1 For the circuit in Fig. 3-25 the elements shown on the right are connected one at a time to the terminals ab. The control for the dependent sources is Ix' Determine the dependent parameter in each case. + 36 V _ 75 V + r:: I. 18 n a + Vu + V, ~ R I, b (a) (b) (c) (d) (e) Fig. 3-25 32 D CHAPTER 3 I Since the voltage across the 18-0 resistor is 36 \1. the current Ix must be 2 A. Then Vab + (2)(2) - 75 + Vab = 35 V. 36 = 0 or Vg = 35 V (b) Ig = Ix = 2 A kIx =35V (c) k =-1 (d) (a) VR (e) = I,R = 35 V R=17.50 k=17.50 A generator generating a ramp voltage, v = 6t V, i5. connected across a 25-0 resistor. dissipated by the resistor during the interval O;~ t;2 5 s. 3.73 I 2 2 p = v 1R = 36t /25 W. Instantaneous power Determine the energy Energy is A resistor supplied by a ramp generator producing a voltage v = lOt V is used as a heating element to boil a certain amount of water. It takes 2 kcal of heat energy to boil the water in 30 s. Calculate the value of the resistance. (1 kcal = 4.184 kJ.) 3.74 I Proceeding as in Prob. 3.73, (lOt)2 lOO (t 3 )30 - - dt= o R R 3 0 30 J W=2kcal =2 x 4184 = 8:·68J = 100 X 30 3 R = Tx8368 = 107.5 n or 3.75 For the data of Prob. 3.74, if I v = 10 V (constant), In this case, 30 W = 8368 J = 3.76 determine the value of the resistance. J o 102 3000 - dt ,: - R R 3000 R = 8368 = 0.3585 n For the circuit shown in Fig. 3-26a determine the current drawn from the 15-V battery. I First, we show the circuit reduction to a single equivalent resistor in Fig. 3-26b through d. Then, from Fig. 3-26d, V 1= R I 1£1.- tI , '" 15V 15 =6A 5/2 c: - I '? .. 1-.1.-- ~.a. ~ - r I ;1.f 1.fL (.!l 6)(2. ) _ ~ ( t:>/ +1 - -:2 + /SV \ T (~) I (c) 1ft :t.1 /5V £l- + I '2 (~) ~+6 ':. 2ft (b) (2 ·H.)-- I (l+t}~~fL '-(1_ fSV- :- Cd) Fig. 3-26 SERIES AND PARALLEL RESISTIVE CIRCUITS 3.77 D 33 Refer to Fig. 3-26a and calculate the power absorbed by the 2-0 resistor. I From Prob. 3.76, we have 1= 6 A. Now, applying the current division rule to the circuit of Fig. 3-26b, we obtain 12 = [2/(2 + 6)]6 = 1.5 A. Thus, P2 n = 1;(2) = (1.5)22 = 4.5 W. 3.78 What are the powers absorbed by the resistors of Fig. 3-26a? power drawn from the battery. I I1 = [6/(2 + 6)]6 = 4.5 A, From Fig. 3-26b and from Prob. 3.76 we have 13 6 = 3 + 6 4.5 = 3 A PI n 3 = 3 + 6 4.5 = 1.5 A 14 = (6)21 = 36 W P2 n = (1.5)22 = 4.5 W P4n =(1.5)24=9W Check: 3.79 Pbatte<y and from Fig. 3-26a we obtain = 1.5 A. Since P = 12R, we have 12 From Prob. 3.77, Verify that the sum of these powers equals the Ptotal = 90W = VI = 15 x 6 = 90 W . If the 3-0 resistor in the circuit of Fig. 3-26a is short-circuited, how much power will now be drawn from the battery? Also, determine the voltage across the 2-0 resistor. I From Fig. 3-26a it follows that short-circuiting the 3-0 resistor short-circuits the 6-0 resistor as well as the series combination of the 2-0 and 4-0 resistors. Hence, V2 n = 0 V. The current is limited by the 1-0 resistor, and we have 1= 1[- = 15 A. Power drawn from the battery is VI = 15 x 15 = 225 W. 3.80 How much power is drawn from the battery if the 6-0 resistor of the circuit of Fig. 3-26a is open-circuited? I In this case the equivalent resistance R = 1 + 3(2 + 4) = 3 0 3+2+4 Power drawn from the battery is 15 x 5 = 75 W. 3.81 A resistive circuit is shown in Fig. 3-27a. R = 15 = 5 A 3 Determine the equivalent resistance R. I The circuit reductions are shown in Fig. Is I=.!::': thus 3-27b through d, from which R = 1 + 2 = 3 O. 2.fl. I I. (.17... I /fl.- 14- VIfI ~ (,f2.. R..~ (~l b (a.) CC) I a 1>/.(L.. '?>(h - - ?>+£ - 2 1,.. (£I..NV' ]0-::'l..tL .f2- R~ r 6+?J ------------------------~ (ri) Fig. 3-27 34 3.82 D CHAPTER 3 Calculate the current through the 3-0 resistor and the voltage across the 1-0 resistor of the circuit of Fig. 3-27 a when 120 V direct current is applied across the terminals of the circuit. I From 3-27d: 120 6 12 = ~t3 40 = 26.67 A From Fig. 3-27c: VI n = 12 (1) = 26.67 x 1 = 26.67 V Thus, 13= 3!6 26 .67 =17.77A From Fig. 3-27a: 3.83 l/ R = -3 =40A 1= From the data of Prob. 3.82 verify that the voltage across the terminals ab is the sum of the volt ages across the terminals ac and cb. I From Prob. 3.82, I1 = 40 - 26.67 = 13.33 A Val> = I1 (6) = 13.33 x 6 = 79.98 V 12 = 26.67 A Vac = 12 (1) = 26.67 xl = 26.67 V 13 = 17.77 A Vel> = 13(3) = 17.77 x 3 = 53.31 V Vac + Veb = 26.67 -- 53.31 = 79.98 V = Vab Thus, 3.84 For the circuit of Fig. 3-27a determine 14 , Thus calculate the power loss in each resistor. Verify that the sum of the power losses is the same as the power delivered by the source. (If the two results are not identical, determine the percent round-off error.) 3 3 14 = 3 + 6 12 = 3 + 6 26.67 = 8.~;9 A I LP lOSS V= ! x 40 = 20 A = (2W2 + (20)22 + (13.33)26 + 126.67)2 1 + (17.77)23 + (8.89)26 = 4798.93 W PdeHve<ed 3.85 15 = 16 = = 120 x 40 = 4800 W Figure 3-28a shows a ladder network. % error = 4800 - 4798.93 4800 x 100 = 0.022% From the data, determine the voltage Vx and the current Ix' I The circuit reduction is shown in Fig. 3-28b thro~gh e. P From Fig. 3-28e: 1=, -...::..... =2A 4+2 From Fig. 3-28d: 6 I = ---- 2 = 1 333 A From Fig. 3-28b: Ix = 3! 3 1.:1:3 = 0.667 A = 13 From Fig. 3-28a: Vx = 13(3) = 0.667 x 3 = 2.0 V I 6+ 3 . I').V /w r ~(.~'~;f ,3 <... 4£l- ,wf '" Cc) rr.E] "f +(1. /,~.(t- 1~~ Fs~ b) I 2. V L_~.L1(q) Fig. 3-28 I ~ 3..n.. (e) 1<~ SERIES AND PARALLEL RESISTIVE CIRCUITS 3.86 D 35 Solve Prob. 3.85 without resorting to network reduction. I From Fig. 3-28a we have: or 3.87 Vx = lA3) = 3 and or, x 0.667 = 2.0 V Polarities of voltages across three resistors in series are shown in Fig. 3-29. points a, b, c, and d. J, /0 V 2" v 30 V e. Determine the potentials at the d -i- ~-------<=r=~-=-~-""~---'v\/v~--"'-1+1=------ v~---..... o-t -::- Fig. 3-29 I From Fig. 3-29, Va=-lOV 3.88 Vd = (30 + 20) = +50 V The conditions which determine the base bias of a transistor in an amplifier circuit are shown in Fig. 3-30. What are the potentials at points B, C, and E with respect to ground? What is the value of the base-to-emitter bias VBE ? c 8.5 V 10V -=- + 8 . - - - - - - VBE - - - - - + E Fig. 3-30 I The potential at point B is VB = +2.2 V. The potential at point C is Vc = + 10 Y. The potential at point E is VE = Vc - VCE = + 10 V - 8.5 V = + 1.5 Y. 2.2V-1.5V= +0.7Y. 3.89 The base-to-emitter voltage VBE is VBE = VBG + VGE = From the transistor amplifier circuit in Fig. 3-31 determine: (a) The voltage at the collector with respect to the emitter, VCE ; (b) the voltage at the base with respect to the emitter, VBE ; and (c) the voltage at the base with respect to the collector, VBC' _~ _ _ _--T"_ _-..,..12V C Col/EcToR +5'7v t p .." Fig. 3-31 36 0 CHAPTER 3 I From Fig. 3-31, VCE 3.90 = 5.7 - 1.2 = +4.5 V VHto 1 A - 1.2 = = +0.2 V VHC == 1.4 - 5.7 == -4.3 V A three-wire power line feeding a house is shown in Fig. 3-32. The loads on the line are as follows: PI = 1.2 kW, P2 = 3.6 kW, and P, == 9.6 kW. Calculate the currents la' I h , and In' la. ~ + t I, V,ol2oV V-~L/oV " I; ~ 1"" ~ .... p. I ;l V,.t2DV P:2.. " Jb Fig. 3-32 • I From the data: I == PI = 1.2 X 10' == 10 A I VI 120 la == 11 + I, = 10 + 40 = 50 A 3.91 12 == In = V;V 12 J.6 10' X = --~ - =30 - 11 ==30A 10 == 20 A A three-wire dc line supplying a resistive bank of loads is shown in Fig. 3·33. terminals a and e is 240 V, determine the voltage between a and b. I If the voltage between the Combining the resistors across ae as an equivaknt resistance, we have: (11) +- 20)60 Rac == 1.8 + 10+20 + 60 + 2.2 Thus (= By current division: I 240 'M- 60 = -----I 30 -I- 60 Voltage drop across 1.8-!! resistor is 1.8 x 10 = J ~ V Voltage across ab is 18 + 66.7 = 84.7 V. a, + I.BJl 'VV'v 24 n lOA 10 = 6 67 A . Voltage drop across 10-[2 resistor is 10 x 6.67 = 66.7 V. 10. ~ ~ 1/ = = 6 I.OA ---A/V'v--- C I j/ =----.tvvv----- + IO.fl.. ~ 12 ban 20fl 2.2.f'l.. 3.92 Refer to the circuit of Fig. 3-33. I Fig. 3-33 If 240 V is now applied across the lines ab, what is the voltage across be? In this case, the equivalent resistance Rab becomes 10(60 + 20) Rab = 1.8 + 10 + 60 + 20 + 1.0 ~ 11.689 n 240 la = 11.689 = 20.53 A By current division: = Voltage across 20-[2 resistor is 2.28 x 20 45.60V. Voltage across he is -(45.6 + 20.53) =: -66.13V. Voltage across 1.0-[2 resistor is 20.53 x I = 20.53 V. SERIES AND PARALLEL RESISTIVE CIRCUITS 3.93 37 In the circuit of Fig. 3-33, with line a open, determine the resistance between the terminals be. (10 + 60)20 I 3.94 D Rbc = 1.0 + 10 + 60 + 20 + 2.2 = 18.7560 A 12-V battery is used to supply resistive loads at 12, 9, and 6 V, the respective currents being 4, 2, and 1 A. Design a suitable voltage divider. I The voltage divider circuit is shown in Fig. 3-34. The total current supplied by the divider is 2 + 1 = 3 A. Let I, be 20 percent of the total divider current. Thus, Is + 16 = 11 = 12 + Is = 1.6 + 2 = 3.6 A I, = 0.2 x 3 = 0.6 A From Fig. 3-34, I3R, = 6 V 6 or R, = 0.6 = 100 3 I2R2 = 9 - 6 = 3 V or R2 = 1.6 = 1.8750 IIRI = 12 - 9 = 3 V or RI 3 3.6 = 0.8333 0 1" = 4A I" + I. ~ Iz + 12. V 2.. A + R5 + CJv R 12. V <4 "V ~ I LOA!) V.LT"jE Fig. 3-34 d;vidu 3.95 Calculate the total power dissipated in the voltage divider when supplying the load shown in Fig. 3-34. I From Prob. 3.94, 1=11 + 14 = 3.6 + 4 = 7.6 A. Power supplied from the battery is Pin = VI = 12 x 7.6 = 91.2 W. L output power is 1214 + 915 + 616 = 12 x 4 + 9 x 2 + 6 x 1 = 72 W. Plo" is 91.2 -72 = 19.2 W. 3.96 Verify the result of Prob. 3.95 by adding the losses in the resistors Rp R 2 , and R3 of the voltage divider of Fig. 3-34. I From Prob. 3.94, I~Rl = (3.6)20.8333 = 10.8 W I~R3 = (0.6?1O = 3.6 W which is the same as in Prob. 3.95. 3.97 For the voltage divider of Fig. 3-34, choose 13 to be 60 percent of the total divider current and calculate the power loss in the voltage divider. Compare the result with that of Prob. 3.96. I Proceeding as in Prob. 3.94, 12 = 13 + 16 = 1.8 + 1 = 2.8 A 13 = 0.6 x 3 = 1.8 A R3 = 6 1.8 = 3.333 0 P IOSS = (4.8)20.625 3 R2 = 2.8 = 1.071 0 11 = 12 + Is = 2.8 + 2 = 4.8 A 3 RI = 4.8 = 0.625 0 + (2.8)21.071 + (1.8)23.333 = 33.6 W which is much higher than the loss when 13 was 20 percent of the divider current. 38 D 3.98 CHAPTER 3 For the switch S open in Fig. 3-34, for the given values of R 4, R s' and R6 as in Prob. 3.94, and for the respective calculated values of RI' R 2, and R 3, find the volta;~e across R 3· I From the data of Prob. 3.94, R4 = ¥ =30 RI = 0.83330 Rs R2 = ~ =4.50 1.8750 = With S open, the total resistance across the battery becomes R = R 4{R 1 + [Rs(Rc + R 3.)]/(R 2 + R3 + Rs)} T R4 + RI + [Rs(R2 + R,)]/(R 2 + R3 + Rs) 3{0.8333 + [4.5( 1.87S- 1O)]/( 10 + 1.875 + 4.5)} = 1 73260 3 + 0.8333 + [4.5(1.875 + 10)]/(10 + 1.875 + 4.5) . 12 I = 1.7326 = 6.926 A J 1 = 1- 14 = 6.926 - 4 = 2.926 A By current division, Rs 12 = I) = 11 R2 + R3 + and 3.99 4.5 R,: = 2.926 1.875 + 10 + 4.5 = 0.804 A V, = 13RJ = 0.804 x 10 = 8.04 V With R4 = 3 0, Rs = 4.5 0, and R6 = 60 in Fig. 3-34, find the voltage across R3 with S open for the values of RI' R 2• and R3 calculated in Prob. 3.97. I Proceeding as in Prob. 3.98, we have R = 3{0.625 + [4.5( 1.071 + 3..3J3)]/( 4.5 + 1.071 + 3.333)} = T 3 + 0.625 + [4.5(1.071 + :1.333)]/(4.5 + 1.071 + 3.333) 1.4610 12 1= 1.461 = 8.21 A Jl = 1- 14 = 8.21 - 4 = 4.21 A By CUrlent division, 4.5 12 = 13 = 4.21 4.5 + 1.071 + 3.333 3.100 ,= 2.127 A V, = I)R) = 3.333 x 2.127 = 7.09 V Define: . (no· load voltage) - (voltage on load) Vio Itage regulatIon =: ----. voltage on load Calculate the voltage regulation from the results I Probs. 3.98 and 3.99. 0:' From the results of Prob. 3.98: . 8.04 - 6.0 3 rrt Voltage regulatiOu= 6.0 = 0.34 or 4'10 From the results of Prob. 3.99: . Voltage regulatIOn 3.101 7.09-6.0 = .. -~ = Summarize the results of Probs. 3.94 through 3.100. 0.182 or 18. 2rrt '10 What conclusions may be drawn from these results? bleeder current I b' % loss, W voltage regulation, % 20 19.2 34.0 60 33.6 18.2 Conclusion: Low bleeder current results in a low loss but a high voltage regulation whereas a high bleeder current results in a low voltage regulation but a high loss. 3.102 For the operation of a transistor amplifier, a power supply with a voltage divider shown in Fig. 3-35 is used. Calculate the values of R" R 2 , and Rv SERIES AND PARALLEL RESISTIVE CIRCUITS D 39 I R V. I Xl. bOV R.. A ----r--- -t 2S V J , - - - I - - -__ + + 2!;Y Vs. I -2SV Fig. 3-35 I From Fig. 3-35, and VI = 60 - (25 Let 13 be 10 percent of the maximum load current, which is 2 A. 3.103 1= 13 + 14 13 = 0.1 x 2 = 0.2 A RI = T = 2.2 = 4.545 0 VI 10 R2 Therefore, = 0.2 + 2 = 2.2 A 12 = 1- 11 = 2.2 - 1 = 1.2 A R = V3 = ~ = 125 00 3 13 0.2 . 25 V2 + 25) = 10 V = 12 = 1.2 = 20.833 0 A transistor base bias circuit and its equivalent are respectively shown in Fig. 3-36a and b. shown in Fig. 3-36, determine RI' Ca.) I For the values Fig. 3-36 (b) From Fig. 3-36b, 12 VBE 0.6 = R = 820 = 731.7 /-LA 11 = 12 + IB = 731.7 + 200 = 931.7 /-LA 2 VI 3.104 = Vcc - VBE 9.4 931.7 x 10 = 10 - 0.6 = 9.4 V _ 6 - 10.09 kO A dc generator having an internal resistance of 1 0 supplies a resistive load shown in Fig. 3-37 a. value of R x will the load draw the maximum power from the generator? I For what First we convert the delta-connected resistors to an equivalent wye, shown in Fig. 3-37 b, which is finally reduced to the circuit of Fig. 3-37c. Therefore, R load lOR x lOR x = R ab = 10 + R x + 10 + R x For maximum power transfer (see Prob. 3-28), or 20R x +R = 10 x 40 D CHAPTER 3 ~ tt l' R,. /).0 v b c- I~ (c) lo/%..(. (a.) ~ + r 2-0 v c (h) Fig. 3·37 Hence, 3.105 For the three cases Rx = 0, ~ 0, and voltage across the load is 120 V. I 3.106 00, calculate the power absorbed by the load of Fig. 3-37a. The For Rx = 0, Re = O. Thus, in the absence of .1 load, P10ad = O. For Rx=~' Re =Rg=[20(1O/19)]/(1O+ 10/19)= 10. Thus, Pload=1202/1=14.4kW. 2 For Rx = 00, Re = 20 0 and P10ad = 120 /20 = no w. A wattmeter measures power by measuring the product of the current through and the voltage across a circuit. Two possible connections are shown in Fig. 3-38a and b; find the power measured in each case. The voltage coil of the wattmeter has a resistance of 9 kO and the resistance of the current coil is 6 O. I For connection shown in Fig. 3-38a: V= 150V 150 P= VI= 144.23 W 1= 6 + 150= 0.9615 A For connection of Fig. 3-38b: 150 1= -6-+-"-[9-0-00-(-15-0)] /(9000 + 150) = 0.9769 A V= 150 - 0.9769 x 6 = 144.14 V I I v - /50 v V ISov (a.) 3.107 P = VI = 140.81 W {)'oL1- (b) Fig. 3-38 What is the actual power dissipated in the 150-0 resIstor for the two connections shown in Fig. 3-38a and b? I For Fig. 3-38a: For Fig. 3-38b: P 150 n = 12R P 150 n = = (0.9615)2150 = 138.68 W V2 ""if =, (144.14)2 --15()"" = 138.51 W SERIES AND PARALLEL RESISTIVE CIRCUITS 3.108 D 41 How much power is absorbed by the wattmeter in the two connections of Fig. 3-38? I For Fig. 3-38a: P w = 5.547 + 2.5 = 8.047 W For Fig. 3-38b: Pc = (0.9769)26 = 5.726 W 3.109 Pv = (144.14 )2 9000 = 2.308 W Pw = 5.726 + 2.308 = 8.034 W For the interconnected resistors shown in Fig. 3-39a find the net resistance between the terminals 1 and 3. 8.11- ~fL 3 -+ 4 (a.) (b) 6 (d) 4 CC) 6 (e) - (I) 6.0-. Z.LL ......----,~ 8 ~ g[t. --~--~------3 (.J.) I Fig. 3-39 First, we convert the two delta-connected resistors to equivalent wyes, as shown in Fig. 3-39b. reduction steps are shown in Fig. 3-39c through g. Finally, R l _ 3 = 8 O. 3.110 Find the total resistance between points 1 and 2 shown in Fig. 3-40a. Subsequent 42 D CHAPTER 3 2. (a.) b() ~ (L 40x 2.0 ,,,.11. 10 (I.. "0 4 .,/\/'of... J10 _ /3.33JL fo-t?D J'1- (b) ~~ ~ • .L2. 1<-. '~.L'l- ?c...fL. 6,,{l.... .'" bo x 1.7,s- 43 'l- 8.75':; ~ /?l.?> ~.(l. ,HIP ~7~ =-7. b4SL <I- (d) Fig. 3·40 Cc) I Notice that the 50-0 resistor is short-circuited and the 20-0 and 40-0 resistors are in parallel so the network reduces to that of Fig. 3-40b. Subsequent reductioll steps are shown in Fig. 3-40c and a from which 1.3.33 x 37.64 Rl _2 = 1333+ 37.64 = 9.844 0 /7 CHAPTER 4 UKirchhoff's Laws 4.1 Kirchhoff's voltage law (KVL) states that the algebraic sum of the voltages around any loop of a circuit is zero. Apply this law to the circuit shown in Fig. 4-1. I The direction of current I is arbitrarily chosen as indicated in Fig. 4-1. The voltage across each resistor is assigned a polarity; it is understood that Ohm's law yields V= IR if I enters the positive terminal of a resistor, and V= -IR otherwise. Application of KVL to Fig. 4-1 leads to VI + V2 + V3 - V= 0, where VI = IR I , V2 = IR 2 , and V3 = IR 3. VI + RI + V:a. - + t C R~ R2V, v~ R, V ~- v. R2 -1 + v_ C + R. V. v. _1+ Fig. 4-2 Write the Kirchhoffs voltage equations for the two indicated loops of the network of Fig. 4-2. polarities as marked. Also, express V2 in terms of II' 12 , and R 2 • I Assume Loop 1: Loop 2: - V{3 - V2 + V3 + V4 + V'Y Voltage across R 2 : 4.3 V2 V) R) + + 1- Fig. 4-1 4.2 + = 0 V2 = (11 - I 2 )R 2 Kirchhoff's current law (KCL) states that the algebraic sum of all currents leaving or of all currents entering any node of a circuit is zero. A node of a network is shown in Fig. 4-3, with currents flowing in the directions shown. Apply KCL to find the magnitude and direction of I. I 4A Fig. 4-3 I Assume the indicated direction for I: the node. 4.4 - 1- 1 + 3 + 5 - 4 - 2 - 6 = O. Hence, 1= - 5 A; Apply KVL to obtain an expression for the equivalent resistance formed by n resistances connected in series. i.e., 5 A into RI' R 2 , · •• , Rn I Let V be the voltage across the series circuit combination, I the current through it, and Res the equivalent resistance. Then, from Ohm's law, whence 43 44 4.5 D CHAPTER 4 Resistances RI' R 2 , by using KCL. •.• , Rn are connected in parallel. Obtain an expression for the equivalent resistance I Let I be the total current into the parallel combination, V the voltage across it, and Rep the equivalent resistance. Then, from KCL, 1= 4.6 V( ~ +~ RI + ... + -~-) R", 2 = ~ Rep or If the voltages, currents, and re si stances in a series circuit can be respectively transformed to currents, voltages, and conductances of a parallel circuit, such that the KVL equations of the series circuit go over into the KCL equations of the parallel circuit, then the circuits are said to be duals of each other. Construct the dual network for three resistors, RI' R 2 , and R 3 , connected in series with a voltage source V. I The network is shown in Fig. 4-4a. We draw the dual network, Fig. 4-4b, by replacing the series elements by parallel elements. (a) The equations for the two networks are: V=IR I +IR 2 +IR 3 V V I 2 V I=R+R+l~=VGI+VG2+VG3 (b) where conductances Gp G 2 , and G 3 are the reciprocals of the respective resistances. b v + a c (a) 1=1batter) d E Fig. 4·4 (b) 4.7 A bridge circuit is shown in Fig. 4-5. With the currents as marked, write (a) Kirchhoff's current law at the four nodes and (b) Kirchhoffs voltage law around the loops abda, bcdb, and adca. I (a) Node a: Node b: (b) 4.8 Fig. 4·5 1= i l + i3 i l = i2 + is Loop abda: ilR I + isRs Loop bcdb: isRs = = Node c: 12 = 1+ i4 Node d: I) i3 R 3 == i3 + i4 + is Loop adca: i3R3 - i4R4 = E i2R2 + i4R4 For the circuit of Prob. 4.5, consider the special case of the balanced bridge (is = 0). (a) If RI = 10 n, R2 = 20 n, and R3 = 30 n, determine R 4. (b) If E = 45 V, calculate the current supplied by the battery. I (a) Since is = 0, we have i l = i2 and 1:1 = - i4 • Also, nodes band d are at the same potential. From the above we obtain or ~ = R2 R 3 = (20)(30) = 60 n RI 10 J, Thus, KIRCHHOFF'S LAWS D 45 (b) The effective resistance Re across the battery becomes = (to + 20)(30 R + 60) . E and Ibattery SOV 100 + = 22 50 10+20+30+60 e = Re 45 22.5 = 2.0 A = SOV 20 il - i. + i, + i. 20 SO i, <D 100 V 30 Fig. 4-6 4.9 Determine the currents ix and iy in the network shown in Fig. 4-6. I On the basis of KCL, the currents in the remaining branches are also marked in Fig. 4-6. meshes 1, 2, and 3, By KVL for Solving these simultaneous equations yields ix = -3.S7 A, iy = 0.51 A. The negative sign on ix implies that the actual current flows in the direction opposite to that given in Fig. 4.6. 4.10 This problem relates to the concept of source transformation. Replace the voltage source v and its internal (series) resistance Rv in Fig. 4-7a by a current source i with internal (shunt) resistance Ri in Fig. 4-7b, such that the current through R remains unchanged. i" v R R I From Fig. Fig. 4-7 (b) (a) 4-7a: From Fig. 4-7b: V Then, for equivalence: which will hold if we take 4.11 Rv + R i = vi Rv and Ri = iRi Ri + R Rv. Find the current in, and the voltage across, the 2-0 resistor in Fig. 4-Sa. I Using the results of Prob. 4.10, we transform the 5-A current source to a voltage source; the circuit then becomes as shown in Fig. 4-Sb. For the two loops, KVL gives which when solved yield 25 - 15/1 - 10 + 312 12 = 5 A and - 3/1 = 0 20 - 3/2 + 3/1 Vab = (5)(2) = 10 V. - 212 = 0 46 D CHAPTER 4 20 30 SA d b 20 V (a I SO 100 c JOV C 2.5 V a -1 c 30 d (b) 4.12 Fig. 4-8 Determine I of Fig. 4-9 by mesh analysis. 60 c 10 V 20 a 2 + VI 10 30 c I 100 ·------------~--------------~~--~b 20V o Fig. 4-9 I In terms of the three mesh currents 11 , 12 , and I, (=1) indicated in Fig. 4-9, we have 7/ 1 - /2 =10 Solving for 13 yields 4.13 13 = 1= -1.68 A. Evaluate I of Fig. 4-9 by nodal analysis. I Figure 4-9 can be redrawn so that the two principal nodes labeled 0 coincide. Choosing this single node 0 as the reference, we have the nodal equations from which V2 = 3.2 V. 4.14 10 - VI VI VI - V - - - - - - - =2 0 612 Hence, VI - V2 V2 V2 - 20 ----=0 2 3 10 A resistive network with voltage and current sources is shown in Fig. 4-lOa. by mesh analysis. I Determine the currents I1 and 12 To apply mesh analysis, we first transform the to-A current source in parallel with the 5-0 resistor to a 10 x 5 = 50 V voltage source in series with a 5-0 resistor. Thus, we obtain the network of Fig. 4-lOb, for which mesh equations are 35/1 - 20/2 = 50 -20/1 + 5012 = -100 Solving, 4.15 I1 = 0.37 A and 12 = -1.85 A. Solve for I1 and 12 of the network of Fig. 4-lOa by nodal analysis. KIRCHHOFF'S LAWS IOn D 47 IOn sn + 2() lOA n o (a) (b) Fig. 4-10 I For nodal analysis, define the node volt ages VI and V2 as shown in Fig. 4-lOa. respectively, KCL gives At nodes 1 and 2, VI+ VI -V2 =1O 5 10 which have the solution V2 Thus, 4.16 V1 -V2 10 I1 = -1-0- = 27 =0.37 A 12 = ~V = V2 -100 -50 30 =T7=-1.85A For the network of Fig. 4-lOa, calculate the power supplied by the current source and by the voltage source. Verify that the sum of the powers from the two sources is the total power dissipated in all the resistances. I . 10(1300) Power supplied by the lO-A source = 10 x VI = 27 = 481.48 W Power supplied by the 100-V source = 100(-/2 ) = (100)(1.85) = 185 W Total power supplied by the two sources = 481.48 + 185 = 666.48 W Power dissipated in the 5-0 resistance = ~~ = ~ (1~~) 2 = 463.65 W Power dissipated in the 10-0 resistance = I~ x 10 = (0.37?(10) = 1.37 W Power dissipated in the 20-0 resistance = ~J = ;0 (4~r = 98.76 W Power dissipated in the 30-0 resistance = I~ x 30 = (1.85)2(30) = 102.67 W Total power dissipated = 463.65 + 1.37 + 98.76 + 102.67 = 666.45 W 4.17 Determine the current 1 shown in Fig. 4-11. I Without the values of the resistors, it is not possible to calculate the branch currents. However, the network within the shaded area may be viewed as a single node, at which KCL gives 2-3-1-4=0 or 1= -5A c b 11 In 4n sn v. 70 60 12n h Fig. 4-11 d ! 7.S mA 30 2n f g Fig. 4-12 e (I A) 48 4.18 D CHAPTER 4 For the ladder network shown in Fig. 4-12, find the source voltage Vs which results in a current of 7.5 mA in the 3-0 resistor. I A current of 1 A will be assumed. The voltage necessary to produce 1 A is in the same ratio to 1 A as Vs is to 7.5 mA because of the linearity of the network. Vc! = 1(1 + 3 + 2) = 6 V Then, by KCL, Vbg = 2.(4) + 6 = 14 V Again from KCL, lab = 2 + 2= 4A Vah and = 4(8) + 14 + 4(12) = 94 V Now, scaling down, Vah 1A V, V, = 0.705 V whence 7.5mA 90V a sn 20W h 4.19 13n d b 7!l Fig. 4-13 (a) terminals a and b and Determine the readings of an ideal voltmeter connected in Fig. 4-13 to terminals c and g. The average power in the 5-11 resistor is 20 W. I 1= o ~-5 = (b) -+2A The direction of I through the 5-0 resistor is determmed by noting that the polarity of the 90-V source requires that the current pass from d to c. Thus d is positive with respect to c and v'ie = (2)(5) = 10 V. (a) An ideal voltmeter indicates the voltage without drawing any current. infinite resistance. KVL applied to the clo~ed·lo()p acdba results in I) - 10 + 0 - VM = It may be considered as having an 0 VM= -lOV If the meter is of the digital type, it will indicate -10 V. A moving-coil galvanometer will try to go downscale, with the pointer stopping at the pin. If the leads are reversed, it will indicate 10 V. (And with its + lead at point b, it is known that hi:; 10 V positive with respect to a.) (b) KVL applied to the path cefgc gives 2(17) - 90+ 2(6) + VM=O VM=44V In this connection, the meter reads positive 44 V, indicating that point g is 44 V above point c. 4.20 Determine the current supplied by the 100-V battery in the circuit of Fig. 4-14. I With the currents marked in Fig. 4-14, the required current is loop 1241 we have 50 - lOll -- !o I 3 + 1012 = 11 + 12 , 0 Writing the loop currents, for the (1) For the loop 2342 we have: (2) KIRCHHOFF'S LAWS D 49 For the loop 1431 (containing the 100-V battery) we get: -1012 - 10(12 + 13) + 100 - 10(11 + 12) = 0 (3) Solving for 11 and 12 in Eq. (3) yields and Hence, 10--"-- + Fig. 4-14 loO" 4.21 lO-n resistor across the terminals 2 and 4 in Fig. 4-l4? How much power is consumed in the I Power = 1~(10) W. From Eqs. (1) and (2) of Prob. 4.20 we have: 11 - 12 + 13 = 5 Thus, 4.22 13 = 0 and and power = O. Determine the current I supplied by the battery to the resistive network shown in Fig. 4-15. Icv Fig. 4-15 I We apply KVL to the following loops: Loop 1231: Loop 2342: -1011 -3013 +5012 =0 -3013 - 10(12 + 13) + 50(11 - 13) = 0 Loop 1241: Rewriting the above equations yields - 11 + 512 - 313 = 0 Solving for the currents we obtain 11 Hence, = ~ A and 1 1 3 1= 11 + 12 = 5" + 10 = 10 = 0.3 A 50 4.23 D CHAPTER 4 For the network shown in Fig. 4-16, calculate the power supplied (or absorbed) by each voltage source. I Applying KVL to the loops, for the assumed directions of current flow we obtain 51 1 - 100 + 50 - 12 = 0 or 12 11 = 12.3077 A Thus == 11.538 A 11 + 12 = 23.8457 A P IOO V = 11 (100) = 12.3077 >: 100 = 1230.77 W (supplied) Pso v = 12 (50) = 11.538 >( 50 = 576.9 W (supplied) P200V = (/1 I -1 - , + 12 )200= 23.84S7 x 200= 4769.14 W (supplied) t I .l. L.-_lfV..n'V'--_L ...~...._~ Fig. 4-16 4.24 Determine the power absorbed by each resistor of the network of Fig. 4-16. Verify that the sum of the powers absorbed by the resistors is equal to the total power supplied by the sources. PIOn = (/1 + 12)210 = 23.8457 2 X 10 = 5686.17 W I Ps n = I~(5) = 12.3077 PI n = I~(1) = 11.538' 2: P From Prob. 4.23, 4.25 L P,uPPlied ab,orbed = 2 X X 5 = 757.40 W 1 = 133.12 W 6576.70 W = 6576.81 W. Solve for the current in the 2-0 resistor of the network of Fig. 4-17 by mesh analysis. I There are three meshes, and we define Ik as the current flowing in mesh Solving for 13 , the current in the 2-0 resistor, we 60 a + 20V g,~t \/, C SO C 2 () + L 4.26 Then, KVL gives 13 = 2.98 A. 30 IOV- k (k = 1,2,3). C V2 20 f3 By nodal analysis, obtain the current in the 2-0 resistor of the network of Fig. 4-17. Fig. 4-17 KIRCHHOFF'S LAWS D 51 I In Fig. 4-17 we identify the principal nodes 0, 1, and 2, and choose node 0 as the reference node. Next, we define the volt ages of nodes 1 and 2 with respect to node 0: VlO = VI' V20 = V2 . We now apply KCL at nodes 1 and 2 to obtain Solving for V2 , we get 20-VI V -VI VI VI-V; V2 +1O V; - - - + - 2- - -=0 ------=0 654 5 2 3 V2 = -4.046 V. Hence, the current in the 2-0 resistor is = I V2 3 and is directed from node 2 to node {3. 4.27 + 10 = -4.046 + 10 - 2 9 2 2 - . 8A This result is in agreement with that of Prob. 4.25. Using mesh analysis, find the currents I1 and 12 in the network of Fig. 4-18a. I Transforming the current sources to voltage sources yields the circuit of Fig. 4-18b, for which the mesh equations become 100 - 4/1 - lOll - 2(11 - 12 ) - 40 = 0 These equations simplify to -10/2 + 1013 = -20 Solving for I1 and 12 yields I1 = 5 A and 12 = 10 A. I~ lof/.. 1[2- (b) 4.28 Fig. 4-18 Apply nodal analysis to the network of Fig. 4-18b and solve for the currents I1 and 12 • I Let Vx be the voltage shown at node 1. At this node, we have V-lOO V-40 V-20 x 14 + _x-- + _x-- = 0 2 1 Thus, 22Vx = 660 1I 4.29 = 100-Vx 14 = 100-30 14 =5A or V -20 30-20 I =_X--=--=lOA 2 1 1 Calculate the power supplied by each source to the entire network of Fig. 4-18a. supply power. Determine if all sources 52 0 CHAPTER 4 I Since I) = 5 A (from Prob. 4.27), from Fig. 4-18a, 25 = 11 + Ix = 5 + Ix or P25 A = 25 x 80 = 2000 W V4 n = 20 x 4 = 80 V 20 = Iy - 11 + 12 = Iy -- 5 + 10 Similarly, V2n = 15 x 2 = 30 V or P20 A = 30 x 20 = 600 W Iz or and Iy = 15 A + 10 - 2= 0 P 2(1 V = -8 x 20 = -160 W or So the 20-V source absorbs power and the other sources supply power. 4.30 Calculate the power absorbed by each resistor of the network of Fig. 4-18a and determine the total power absorbed by all the resistors. I From the results of Probs. 4.28 and 4.29, P4fi = (Ij4 = (20)24 = 1600 W PlO n = (11 )210 = (5)210 = 250 W P 2n = (ly2 = (15)22 = 450 W p)n=(l2)21=(1O)21= l00W (20)2 Pwn=--W= 40W L P=2440W 4.31 Using the result of Prob. 4.29, verify the result of Prob. 4.30. I L PsuPPlied = L = Pabsorbed LP P25A + Pm. + P 20Y = ~ P or or 2000 + 600 - 160 = 2440 = which is the same as in Prob. 4.30. 4.32 Solve for the current in the 5-n resistor of the circuit shown in Fig. 4-19a. I Writing the mesh equations, we have 90 - 8/1 - 8(11 - 12 ) = 0 These equations simplify to 8/ 1 Solving for 13 yields - 4/2 = 45 -2/) + 512 - 2/3 = 0 -8/2 + 19/3 = 0 13 = 1.5 A. so ,.(2.. g.a. 60 8..a- 90V b f.n... 40 b (b) (a.) (t4 Ca.... "- 8"S r.;g - 8><9 1J+& 4.4 [ ::'4.1'\. ::. 'i?J),. 4-.0.. ::. "'f.sa- J. Cc) G.n.. (d.) Fig. 4-19 (e) LP 0 KIRCHHOFF'S LAWS 4.33 In the circuit of Fig. 4-19a find the voltage across ab if the 5-n resistor is removed and the terminals ab are open-circuited. I Writing the mesh equations we obtain 90 - 8/ 1 - 8(11 - 12 ) = 0 and 16/) - 812 = 90 or 12 = 2.8125 A Thus, 4.34 and Vab 8/2 + 4/2 + 8(12 - IJ = 0 -8/1 + 20/2 = 0 = 12(8) = 2.8125 x 8 = 22.5 V With the 5-n resistor removed from the circuit of Fig. 4-19a, the 90-V source is short-circuited. resistance that will be measured across ab. Determine the I With the voltage source short-circuited, the network reduction is shown in Fig. 4-19b through e. Rab = 10 n. 4.35 53 Hence, Refer to the results of Probs. 4.33 and 4.34. Using the open-circuit voltage Vab of Prob. 4.33 in series with the short-circuit resistance Rab of Prob. 4.34, connect the 5-n resistor across ab as shown in Fig. 4-20. Find the current in the 5-n resistor. Verify that the result agrees with that of Prob. 4.32. I 22.5 1= 10 + 5 = 1.5 A From Fig. 4-21 which is the same as in Prob. 4.32. The network to the left of ab is known as the Thevenin equivalent circuit (see Chap. 5). JOV Ca.} lo~ + I b Fig. 4-21 Fig. 4-20 4.36 Calculate the current in the 2-n resistor of the network shown in Fig. 4-21a. I First we transform the delta-connected' resistors to an equivalent wye to obtain the network of Fig. 4-21b. Writing the mesh equations, we get -1.5/) + 7.875/2 - 0.375/3 = 30 Solving for 13 , which is the current in the 2-n resistor, yields -0.51) - 0.375/2 + 2.875/3 = 10 13 = 5 A. 54 4.37 0 CHAPTER 4 Determine the power delivered by the 30-V source in the network of Fig. 4-21a. I Solving for 12 from the mesh equations of Prob. 4.36, we obtain 4.38 12 = 5 A. Thus, P30 v = 30 x 5 = 150 w. How much current flows through the 1-n resistor of the network of Fig. 4-21a? I From Fig. 4-2Ia, or 4.39 VI!l = 10 .- V 1!l = 10 - 5 x 2 = 0 or Find the value of R in Fig. 4-22 such that the power supplied by the 100-V source to the network is the same as the power supplied by the 5-A source. Fig. 4-22 5=.!:::: __ V-lOO R 10 I At node 1: V-lOO ------w-+1 At node 2: _ 100 _ 5 4 - For equal power: (1) 20 - (2) LOCI/ I = 5 V (3) From Eqs. (1) and (2) we obtain: \' V R +10 = 15 (4) V 14 +10 = 15 (5) V Thus, R J. = (6) Finally, Eqs. (3) and (6) yield 100 V =5V or R 4.40 R=20n Find the current in the lO-n resistor of the circuit shown in Fig. 4-23a. I By source transformation we obtain the circuit of Fig. 4-23h for which we have 50 = 30/) - 30/2 Hence, or -50 = 50/2 12 = lion = -1 A. ~.~ -r~~ V ~ ____ ~ _____ :;"f7... looV _ ~N Fig. 4-23 KIRCHHOFF'S LAWS 4.41 0 55 Solve Prob. 4.40 by using nodal analysis only. I With the voltages V2 and V3 and the current 1 defined in Fig. 4-23a, for node 1 we have· 50 50 - V2 1= 30 + -1-0- or 301 + 3V2 = 200 (1) For node 2 we have 50 - V2 V2 - V3 -1-0- + 5 = ---zo or 3V2 - (2) V3 = 200 Finally, for node 3 we obtain or V2 - 2V3 = 100 (3) Solving for V2 from Eqs. (2) and (3) yields V2 = 60 V. Hence, 4.42 110 n 50 - V 10 50 - 60 10 = ___2 = - - - = -1 A In the circuit of Fig. 4-23a calculate the current through the 50-V voltage source and the voltage across the 5-A current source. 1 = k(200 - 3V,) = k(200 - 3 x 60) = ~ A 50-V voltage source. From Eq. (3) 2V3 = V2 - 100 = 60 - 100 = -40 or V3 = -20 V. Voltage across the 5-A source = V2 - V3 = 60 - (-20) = 80 V. I From Eq. (1) of Prob. 4.41, we have 4.43 For the circuit of Fig. 4-23a verify that the total power dissipated in the resistors equals the total power supplied by the two sources. I Psupplied = P so V + Ps A = 50 x (50? Pdissipated 4.44 = current through the 2 = ~ + (1) 10 + ~ + 5 x 80 = 433.33 W (20)2 (80? -w + -w = 433.33 W = Psupplied Find the power dissipated in the 20-fl resistor of the circuit of Fig. 4-24a by using nodal analysis. I To use nodal analysis we convert to the lO-V voltage source to an equivalent current source to obtain the circuit of Fig. 4-24b. Defining the node volt ages VI and V2 as shown, we obtain: VI VI VI - V2 1 = 10 + "4 +---zo For node 1, V I -V2 ---zo For node 2, Solving for VI and V2 yields Thus, 120 n = VI - V2 20 VI = 1.11 V and +2 V2 = -11.11 V. = 1.11- (-11.11) = 0611 A 20 V2 ="8 . and /0./1.. Sf\.. lA (6) Fig. 4-24 56 4.45 0 CHAPTER 4 By applying only mesh equations to the circuit of Fi:~. 4-24a find the power dissipated in the 20-!l resistor. I Writing the mesh equations for the currents I) and 12 in Fig. 4-24a, we obtain () = -4/ 1 + 32/2 10 = 14/1 - 412 Solving for 12 yields 12 = 4.46 ¥. A or P,o n - 2(8) (33)2 = 54 20 = 7.469 W Solve for the power supplied by the 12-V source shown in Fig. 4-25a by network reduction, source transformation, and finally using KVL and KCL. I The network reduction is shown in Fig. 4-25b and c and the source transformation in Fig. 4-25d. From KVL, for the mesh current I we have 12 - 1(1) + 2 - 1(0.8) - 1(1.6) -- 1(0.8) = 0 Pl2 V and or 1= 1:'~ = 3.88 A = 121 = 12 x 3.88 = 46.56 W __ ~__~1~b~LL~~2__-V~V~V--+!___ -,(~~+~~~~ ~r-__~__~5 ~&~ : b£L[ lA ..f.f2.. 6 L2- ;1..(1. ~ , I 2..tL-. , " '-I _\'.... .... , '1,),1 l'"- fA ,..!:I.- . <-r lA t" , ;lL'lO·SA ~ (b) 7 ,.6.tl., t-: ~ [e) Fig. 4-25 4.47 Solve for the power supplied by the 12-V source of Fig. 4-25a using the following steps: First convert the two current sources to voltage sources. Then use network reduction and, finally, obtain the power by applying KVL and KCL. Verify that the result is the same as in Prob. 4.46. I The source transformations and the network reduction are shown in Fig. 4-26a. Thus we obtain the circuit shown in Fig. 4-26b. The mesh equations become -6 = 15/1 - 212 Solving for 12 gives 12 = 8 1!1 12 == 4 6712 - 2/1 - 2/3 = 3.88 A or P I2 v = 4 = 4/3 - 2/2 12 x 3.88 = 46.56 W. KIRCHHOFF'S LAWS 0 57 Fig. 4-26 4.48 Without using network reduction and source transformation, solve Prob. 4-46 by using only mesh equations. I First, we redraw the circuit (Fig. 4-27), identify the mesh currents, and write the corresponding equations. Notice from Fig. 4-27 that 11 = -1 A and 16 = 2 A. For the remaining currents, we have -6(-1)+18/2 -6/3 =0 -6/2 + 12/3 - 414 = 0 or -4/3 + 614 - 2/5 = 12 -2/4 + 415 = 2(2) Solving for 14 yields 8 14 = li1 = 3.882 A and or or or 3/2 - /3 =-1 (1) =0 (2) -2/3 + 314 - Is = 6 (3) -3/2 + 613 - 214 (4) - 14 + 2/5 = 2 P 12 v = 12 x 3.882 = 46.59 W. lA. Fig. 4-27 4.49 Find the power delivered (or absorbed) by each source shown in Fig. 4-27. I Solving for 12 and Is from Eqs. (1) through (4) of Prob. 4-48, we obtain 12 =0.117A and I s =2.94A. Voltage across the I-A source = (1 + 0.117)6 = 6.702 Y. PI A = 1 x 6.702 = 6.702 W = Psupplied. Voltage across the 2-A source = (2 - 2.94)2 = -1.88Y. P'A = 2(-1.88) = -3.76 W= Pabsorbed; P12 V = 45.59W= Psupplied (calculated earlier). 4.50 For the circuit of Fig. 4-27, verify that the net power supplied from the sources equals the total power dissipated in the resistors. Pdissipated = (12 - 11?6 + 1~(6) + (13 - 12?6 + 1~(2) + (14 - 13?4 + (15 - 14?2 + (Is - 16)22 = (0.117 + 1)26 + 6(0.117? + (1.35 - 0.117?6 + (1.35?2 + (3.88 - 1.35?4 + (2.94 - 3.88?2 + (2.94 - 2)22 = 49.47 W Psupplied 4.51 = 45.59 + 6.702 - 3.76 = 48.53 W; error = 0.94 W due to roundoff. A network excited only by current sources is shown in Fig. 4-28. resistor. I Defining the node volt ages VI and V2 as shown, we have: For node 1: or Determine the current through the 2-fl V V-V 10=.2 +5+ _ 1_ _2 4 2 3V1 -2V2 =20 (1) 58 D CHAPTER 4 VI -. v 5+----2 For node 2: 2 v V + -2 2 =- 8 16 -8VI + 11V2 = 80 or (2) Solving for VI and V2 yields VI = 22.353 V V2 23.529 V = 12 n = 1= VI ; V2 .= 22.353; 23.529 = -0.588 A . Sit + V loA 2.. Fig. 4-28 4.52 Verify the power balance for the network of Fig. 4·28. I Power balance implies that PsuPPlied from sources = Pabsorbed by resistors PlO A = lOVI = 10 X 22.353 = 223.53 W Vs A = V2 Ps A = L 4.53 VI - 5VS A P,uppHed = 23.529 - 22.353 = 1.176 V = 5 X 1.176 = 5.88 W = 223.53 + 5.88 = 229.41 W = ~~ = (22.!53? = 124.9141 W P2 n = 12(2) = (0.588?2 = 0.6915 W Pg n = ~~ = (23.~29)2 = 69.2017 W P I6 n L V~ = 16 = Pab,orbed (23.529)2 16 = 34.60 W = 229.41 W = L P,upplied For the network shown in Fig. 4-29, write a set of mesh equations to solve for the currents II' 12 , and 13 , I The mesh equations are: 3 4.54 P4 n \/----=-.). Fig. 4-29 In Fig. 4-29 if all the voltages are equal, each being :.0 V, and if all the resistances are also equal, each being 5 n, find the currents 11, 12, and 13, I From symmetry we have 11 = 12 = 13 = I. Thus, we may use only one of the three mesh equations of Prob. 4.53. or 10 = 51, or 1= 1f = 2 A. Hence, V= 3RI - RI - RI = RI, KIRCHHOFF'S LAWS 4.55 In Fig. 4-29, let the resistors R4 and Rs be short-circuited, What is the current in R6? R2 = R3 = S n, R6 = 10 n, and 0 59 V2 = 2V3 = 10 V. I Since R4 and Rs are short-circuited, the circuit reduces to that shown in Fig. 4-30. So, we use only the last two of the mesh equations of Prob. 4.S3. With the given numerical values we obtain 10 = (S + 10)/2 -10/3 = IS/2 -10/3 Current in R6 = I = 13 - 12 = ~ - S = -10/2 + (S + 10)/3 = -10/2 + IS13 ~ or 12 = ~ A and 13 = ~ A = -0.2 A. R, I R2.. Fig. 4-31 Fig. 4-30 4.56 The two ideal ammeters in the circuit of Fig. 4-31 read a current of 8 A each. resistors is 3200 W. Determine V, RI' and R 2. I The power absorbed by the Writing the nodal equation gives V V V 100 + 50 + If 2 = 8 or 1= 8+ 8 = 16 A Now, since P = VI = I: 12R, we have 3200 = V(8 + 8) V=200V or or V 200 R =-=-=25n I 8 8 Finally, or 200 _ 8 _ R - 200 _ lOO 200 SO- 2A or 2 4.57 Find the ratio 12/ Is for the circuit shown in Fig. 4-32. I Writing the nodal and mesh equations, we obtain, respectively, V Is =-+1 RI 2 V+~V=/2R2=V(I+~) (1) (2) or Substituting Eq. (2) into Eq. (1) yields Is=/2[RI(~:~) +1] Fig. 4-32 or 12 Is (1+~)RI R2+(I+~)RI Fig. 4-33 60 4.58 D CHAPTER 4 Determine the total power dissipated in the two resistors of the circuit of Fig. 4-33. I Since the voltage across the 5-0 resistor is VR , V R 1=-5 (1) Now writing the mesh equation we have 100 = 500(1 - 1) + VR (2) SVR - Substituting Eq. (1) into Eq. (2) yields VR 500 + 100 = 500 5 + VR - VR = ~ =6.25V or 5VR P soon = (1 -1)2500 = (1.25 -1)2500 = 3125 W P sn and I = 6.~5 = 1.25 A = (VR)2 = (6.25)2 = 78125 W 5 5 . Total power dissipated = 39.0625 W. 4.59 Verify that the power calculated in Prob. 4.58 is the same as the total power supplied by the two sources. I VI A = ( - I + 1)500 = ( -1.25 + 1)500 = -125 V P SVR = 5VR (1) = 5 x 6.25 x 1.25 = 39.0625 W P IOO v = 100(1) = 100 x 1.25 = 125 W PI A = VI A(l) = (-125)1 = -125 W I: P,uPPl;ed = -125 + 39.0625 + 125 = 39.0625 W, which is the same as in Problem 4.58. current source absorbs power. 4.60 Notice that the I-A If the power dissipated in the 12-0 resistor of Fig. 4-34 is 147 W, what is the value of the source voltage Vs? PI2 n = 147 W = n12), 1= v'¥l- = 3.5 A. Combining the 16-0 and 4-0 resistors in parallel and solving for I) yields I Since 4) 16 x Vs = ( 8.8 + 16 + 4 I, With 1= 3.5 A or I, V, Tz = Vx = 8.81, = and 8.8V --Us V (calculated above), we apply nodal analysis to node 1 to obtain Vx V 1 8.8Vs 7.3 = 3.5 + 14 ,= 7.3 --U = O. lVs Finally, for the mesh containing the 12-0 resistor, we have V 1(8.2 + 12 + 5.8 + 14) - 7.; 14 = 0 + or v " 8. g.t1... i0 4.61 4.0- ".g.. 3.5(40) = 14Vx D + { /0.(1... ~<~ v· v+ ·_1 \ 73 / ( = 14 O.lV,) or Vs = 100 V 2.2.0- 1 1"i-.sL IZ.{L 5.9.(1.. Fig. 4·34 In the circuit of Fig. 4-34, calculate the current through the voltage source and the voltage across the controlled-current source. I As determined in Prob. 4.60, the current through the voltage source is 11 = Vs 100 12 ~= 12- = 8.33 A The voltage across the controlled-current source is V, = Vlon + V14n = 10 ; ; + 14( ; ; 4.62 I) = 1O(0.1V,)'" L4(0.lV, - 1) = 10 x 0.1 x 100 + 14(0.1 x 100 - 3.5) = 191 V Verify that the total power supplied by the two sources in Fig. 4-34 equals the total power absorbed by the resistors. KIRCHHOFF'S LAWS D 61 Fig. 4-35 I From the results of Prob. 4.60 and 4.61, PsuPPlied = VJI V + Vi 7.3 = 100 x 8.33 + 191 x 0.1 x 100 = 2743 W To determine the power absorbed by the resistors we reduce the network shown in Fig. 4-35a and b. and V)7.3 = 0.1 Vs = 10 A, Since Vs = 100 V Pabsorbed = (loW 12 + (10) 2 (10 + 9.1) = 2743 W which is the same as the power supplied. 4.63 Determine the power supplied (or absorbed) by the controlled-voltage source in the network of Fig. 4-36. I Defining the node voltage VI (Fig. 4-36), the nodal equation becomes: VI -4 VI - 3VR 2 + -2- + 5 =0 Solving for VI yields VI = -24 V 1= 3VR VR = - VR = VI - 4 and Now, VI = 3(VI -4) - VI = 3(-24- 4) - (-24) = -12A - 5 24 - 4 = - 28 V 5 5 PWR = 3VR I = 3(VI - 4)1 = 3( -24 - 4)( -12) = 1008 W Fig. 4-36 4.64 Solve Prob. 4.63 by using mesh analysis. I Let the mesh current be 11 , 4 + VR - Then, the corresponding mesh equation becomes 511 - 3VR = 0 4.65 511 = 4 - 2VR 511 = 4 + 4(11 + 2) Thus, Hence, or 1= -12 A or VR = -2(12 + 2) = -28 V and but V R = -2(11 + 2) 11 = 12 = - I P 3VR = 3( -28)( -12) = 1008 W Determine the open-circuit voltage (across the terminals ab) with the polarities as marked in the circuit of Fig. 4-37. I For the mesh we have 1= 3 A, -t.(1.'\J'>~ V4 n = 41 = 12 V, and Vab = 20 - V4 n = 20 - 12 = 8 V V ".n. Gt --,---.,\,\.1----;, , • V" Fig. 4-37 62 4.66 D CHAPTER 4 What is the voltage across the 3-A current source in Fig. 4-37? I 4.67 - V3 A V3A or = 16V In the circuit of Fig. 4-37 if the terminals ab are short-circuited, determine the current through the short circuit. I With the short circuit, we write the nodal equatIOn as 20-V =3+ ~ 4 6 4.68 = Vab - Vg n = 8 -- 3 x 8 = -16 or V= ¥v and l,hort-circuit = V "6 = 24 5 x 6 = 0.8 A Solve for the current I) shown in the network of Fig. 4-38a. I, (c) (6) Fig. 4-38 I First, we change the current source to a voltage source and transform the 6-.0 delta-connected resistors to an equivalent wye, shown in Fig. 4-38b, which is then reduced to the network of Fig. 4-38c. For this network, the mesh equations become 12 - 61) - 141) + 1412 - 24 =, 0 and 24 - 1412 - 1412 + 1411 = 0 From Eqs. (1) and (2), 4.69 or -12 = 201) - 1412 24 = -141) + 2812 (1) (2) I) = 0 A. The circuit of Fig. 4-38a is redrawn in Fig. 4-39. I or Using only nodal equations, find the current 12, At node 1: 12 - V) VI - V3 VI - V~ + ---466 --- = --- (1) or At node 2: At node 3: (2) V 1 -V3 V2 -V3 "'3 --6- + 2 + --6- = 12 Solving for V2 from Eqs. (1), (2), and (3) yields -2V) - 2V2 + 5V3 = 24 or V! = ?f V Hence, (3) KIRCHHOFF'S LAWS D 63 + Il·V Fig. 4-39 4.70 Determine the total power dissipated by all the resistors of Fig. 4-39 and show that the entire power is supplied by the 2-A current source. I From Eqs. (1), (2), and (3) of Prob. 4.69 we have and VI = 12V Since VI = 12 V, 11 = 0 A, P4 n = O. Power in the remaining seven resistors becomes: L Pd;,,;pated = H(VI - V3)2 + (VI - V2? + (V2 _ V3)2] + (~{ + (~~2 = H(12 - ~)2 + (12- ?f)2 + (?f - ~)2] + b:(?f)2 + b:(~)2 = 27.428W Voltage across the 2-A source, V3=~V which is the same as the total dissipated power. 4.71 P2A=~x2=27.428W Since 11 = 0, PI2 V = 12 X 11 = O. Since 11 in Fig. 4-39 is zero, the 12-V source may be disconnected (by open-circuiting). reduction that 12 = ~ A. Verify by network lAcbFig. 4-40 I Network reduction is shown in Fig. 4-40a and b. ~ ~ 7 With the node voltage defined in Fig. 4-40b, we have 2 = 12 + 12 + 4 = 48 VI Hence, 4.72 or VI = % T V VI 96 6 12 = 4 + 12 = 7 x 16 = "7 A Using nodal analysis, find the voltage across the lO-A current source in the circuit of Fig. 4-41. I Let the node volt ages be VI' V2, and V3, as shown in Fig. 4-41. lO-VI VI -V3 V1 -V2 or - 5 - = -1-0- + -1-0- At the three respective nodes we have: (1) 64 D CHAPTER 4 (2) VI -V3 V2 -V3 -1-0- + -1-0- = V3 5 + 10 Solving for V3 from Eqs. (1), (2), and (3) gives or VI 2f V, '\1 3 == - + V2 - (3) 4V3 = 100 which is the required voltage. lOA fo.{l- ... + v 10 . .('1.· 1 '" Fig. 4·41 4.73 In the circuit of Fig. 4-41, transform the lO-V source to a current source and the wye-connected lO-n resistors to an equivalent delta. Thus calculate the voltage across the lO-A source. IOf"L VI !.fA Y3 lOA- ~Jl- 7 (}J Lt<. ) Fig. 4·42 I The source transformation and the equivalent circuits are shown in Fig. 4-42a and b. From Fig. 4-42b we obtain: For node 1: For node 2: Solving for V3 yields 4.74 V3 = - 2f V, which is consistent with the result obtained earlier. For the circuit of Fig. 4-41, verify that the total po\\er absorbed by all the resistors is the same as the net power supplied by the two sources. I From the results of Prob. 4.72, we have LP = [10 - (-36/7W + [-36/7 - (-204/7W + ~ .L... 4.75 to 5 absorbed by resistors [-36/7-(-80/7W 10 (_B0/7)2 + "-10- + 204 PsuppHed by sources = PlO A + PlO V = 10 X 7 + [-80/7-(-204/7W 10 10[10 - (- 36/7)] 5 + (-204/7)2 = 321.71 W Find the current in each of the resistors of the circuit shown in Fig. 4-43. I For node 1: 9 = VI - V2 + VI + VI -4 5 vi. - 32 2 or 19V - 15V = 500 I 2 5 = 321. 71 W D KIRCHHOFF'S LAWS For node 2: Solving for VI and V2 yields VI = 50 V, I = VI = 50 I I 5 = 5 V2 = 30 V. 10 A 1= 2 -t - 17V2 = 240 Hence, VI - V2 - 32 2 = 50 - 30 - 32 =-6A 2 V2 and V, 15VI or 30 14 = 10 = 10 = 3 A 1 ~ t- lo.fi- r-~/vv--~ ~ Fig. 4·43 Fig. 4·44 Transform the two current sources in the circuit of Fig. 4-43 to voltage sources and find the current in the 4-.0 resistor. I The transformed circuit is shown in Fig. 4-44, for which we have: 45 - 32 - 40 = 17/1 - 212 and or 12 = 5 A. Solving for 12 yields 4.77 __AAr---~ ~f2... o /oJL 4.76 In the circuit of Fig. 4-45, determine the node volt ages VI' V2 , and V3 • I The node equations yield 9+ v -V2 +....!. V =0 1 50 _1_ _ Solving for the node voltages VI' V2 , and V3 yields VI I fl. Iz 2.(1.. Fig. 4·45 4.78 65 Solve for the power balance in the network of Fig. 4-45. I With the currents marked in Fig. 4-45 and using the results of Prob. 4-77, we have: VI 50 11 = 50 = 50 = 1 A 12 V2 - VI 60 - 50 = - 1 - = - 1 - = 10 A 13 V2 I = 7 L Pres;s,ors V2 60 = 15 = 15 = 4 A 170 - V3 4 60 14 = 20 = 20 = 170 - 94 4 = = 19A = 50(/1)2 + 1(12)2 + 15(13)2 + 20(14)2 + 2(15)2 + 47(16)2 + 4(17)2 = 50( 1)2 + 1(10)2 + 15(4)2 + 20(3)2 + 2(17)2 + 47(2)2 + 4(19)2 = 2780 W L Psources = 170(/7) + 9( - VJ = 170 x 19 - 9 x 50 = 2780 W Notice that the 9-A current source absorbs power. 3A 66 4.79 D CHAPTER 4 Calculate the current in each resistor of the network of Fig. 4-46. I lJ 120 n = 13 = 20 - 50-10 20 ,£I- I'2. = -2 A n = 12 = 20-10-9 I = IA 15 :1 =14 =lQ=2A 5 -'-E}l M.il- 1, r--;~ Is 5J11~ Fig. 4-46 4.80 Determine the currents through each voltage source of the circuit of Fig. 4-46. I 4.81 Which of the sources supply power and which absorb power in the circuit of Fig. 4-46'1 power absorbed by the resistors. I P20 V = 20(1.) = 20(1) = 20 W (supplied) P9v =9(-/2) =9(-1) =-9 W (absorbed) Determine the total Pso V = 50( - IJ = 50(2) = lOO W (supplied) PlO v = lO(lb) = 10(8) = 80 W (supplied) Total power absorbed by the resistors = 20 + 100·- 9 + 80 = 191 W. 4.82 An operational amplifier is schematically represented in Fig. 4-47a, and Fig. 4-47b shows the corresponding equivalent circuit. For the ideal model assume 1+ = 1_ = 0 and V+ = V_. Draw the equivalent circuit of the inverting operational amplifier shown in Fig. 4-47c. l-z, + + (a.) (c) &~\-: l' I:: 0 ,"---"._-- (d) (0) Fig. 4-47 KIRCHHOFF'S LAWS I D 67 For the circuit of Fig. 4-47c we have (1) or Now V_ = V+ = 0 (since V+ is grounded). v, + Vo RI R2 Thus Eq. (1) becomes: = 0 or Hence, we obtain the equivalent circuit of Fig. 4-47d. 'v.f> l-t V_ 'v0 L + R... ! ....------\; V~ () R, ." (b) (4.) 4.83 Fig. 4-48 Draw the equivalent circuit of the ideal noninverting operational amplifier shown in Fig. 4-48a. I Since V_ = V+ = Vs and L = 0, we have v,+Vs-Vo=O RI Solving for Vo yields Vo = (1 + R)RI)V,. R2 Hence, we obtain the equivalent circuit of Fig. 4-48b. Fig. 4-49 4.84 The circuit shown in Fig. 4-49 responds to the difference of two input signals. hence, show that this circuit is a differential amplifier. I Find an expression for Vo and, Since the operational amplifier is ideal, we have and, by voltage division, but v- =V+ Thus, 4.85 For the operational amplifier circuit of Fig. 4-50 determine Vo. All volt ages shown are with respect to ground. 68 D CHAPTER 4 r----4.-- V' " Fig. 4-50 I Since 1_ = 0 and V_ = V+ =0, the nodal equation becomes V I +V2 +VO=0 R 4.86 R R3 The equivalent circuit of an operational amplifier circuit is shown in Fig. 4-51. Determine its input resistance. Av. I Fig. 4-51 I Writing the loop equations we have: VI = V; + RJ(lJ --/J = (R; + RI)/I (1) - RJ2 or I2 = __R_J!....-_A_R..Li - RI +R2 + Ro I I (2) Substituting Eq. (2) into Eq. (1) yields or the input resistance is 4.87 In the circuit of Fig. 4-52, find the currents through each resistor. I Let the required currents be II' 12 , and 13 as shJwn. 4 = 6/J + I J + 12 = 6 VI VI The nodal equations may then be written as: VI - 20 "2 + -:;;- + - - 3 - V or J =~ 23 V Hence, II ,=~1 =~ _ 2 I = VI -:..~~ 2 3 E.. 23 x 2 - 23 = A (64/23) - 20 = _ !E. A 3 23 ->---, 5"1 I fA Fig. 4-52 Fig. 4-53 KIRCHHOFF'S LAWS 4.88 D 69 Evaluate the currents through the resistors of the circuit shown in Fig. 4-53. I Writing the nodal equations we obtain V V-V 5=1 +1 =--.!+_1__2 121 2 which, respectively, simplify to 3VI - V2 = 10 Thus, and 45VI = -7V2 VI = 1.06 V ~I = 1.~ = 1.06 A I1 = 12 V2 = -6.82 V = VI; V2 = 1.06 - ~-6.82) = 3.94A 1= V2 = -6.82=_1.364A 5 5 4.89 Find k in the circuit of Fig. 4-54 such that the power dissipated in the 2-0 resistor does not exceed 50 W. I Since P2 n = 2(11)2;;;: 50 W, the nodal equations are: or we have 11 ;;;: VW = 5 A and VI = 2/1 ;;;: 10 V. 16 -_ 5k_- _10 1=-IA= 16 - klI - V1= _ 2 4 4 With the critical values, -4 = 6 -5k or k=2 'v 411\1 ~---~'~~---<tL'-'--~~~~~~----~ t1. t~ ->.. 2.!L (,A: Rn I, Fig. 4-54 ~--~------------,,----. 4.90 For the value of k determined in Prob. 4.89, determine the power supplied by each source of the circuit of Fig. 4-54. I 4.91 Calculate the power dissipated in the resistors of the circuit of Fig. 4-54 and verify the power balance in the circuit. I P2 n = (11)22 = (5)22 = 50 W Ps n = (13)28 = (2)28 = 32 W From Prob. 4.90, 4.92 E Psupplied 2: Pdissipated = 86 W = 60 + 10 + 16 = 86 W. Determine the node volt ages VI' V2 , V3 , and V4 shown in the circuit of Fig. 4-55. 2.11. Fig. 4-55 I Writing the nodal equations we have VI V -V - + -I- - 3+ 2 1 VI+40-V3 =0 6 or (1) 70 D CHAPTER 4 or V V3 - v~ 6+ - 4 = 10+ - - - 3 5 and (2) or (3) Solving for VI' V3 , and V4 from Eqs. (1), (2), and (3) yields V4 = 15V VI = lOV Finally, 4.93 or 120 = ~I 1~ =5A I1 + 10 = 5 A 1 = 160= -5 = VI 0 ~ V, =: = V3 -~~= 5 50 10 ~ 20 = -10 A 20 - 15 5 =1A Verify the power balance for the circuit of Fig. 4-55 L I L 4.95 or V2 = -20 V. Calculate the currents in the resistors of the circuit of Fig. 4-55. I 4.94 V2 + 40 = 20 = (5)22 + (10)2[ + (5)26 + (1)25 + (5)23 = 380 W PdiSSipated PsuPPlied = 40(5) + 6(V3 - Solve for 1 in the circuit of Fig. 4-56. V4 ) + iOV4 = 200 + 6 x 5 + 10 x 15 = 380 W Also determine the current in the 3-0 resistor. I Writing the nodal equation yields 12-V V - - +5=- --41 4 3 12 - V 1=-4 Solving for 1 yields 1 1= 0.6A 30 = ~ =~ 3 !:! - 41 = 12 - 4(0.6) = 32 A 3 . 3 ~f1--. 1) I 41 4fl.. 1 5.!l- 3) I() It lo.{lFig. 4-56 4.96 Fig. 4-57 Solve for the four mesh currents shown in the network of Fig. 4-57. I Let the voltage across the 5-A source be VI. Then 50 = 20/ 1 + VI - 15/3 = 20/1 + VI - 150, since = 10 A. 12 - I1 = 5. -30 = 35/4 - 512 - 20/3 = 35/, - 512 - 200, since 13 = 10 A. VI = 15/2 - 514 . 13 The above equations simplify to 20/1 + VI ,= 200 35/4 - 512 = 170 12 - I1 =5 VI + 5/4 - 15/2 = 0 Solving these yields I1 = 4.46 A 12 = 9.46 A 14 =6.22A VI = 110.8 V KIRCHHOFF'S LAWS 4.97 Verify that Kirchhoff's current law is satisfied at the node 0 of the circuit of Fig. 4-57. I KCL at node 0 reads or 4.98 D 71 5 - 12 + I1 = 5 - 9.46 + 4.46 = 0 How much power is supplied by each source in the circuit of Fig. 4-57? I P sov = 50/1 = 50 x = 223 W 4.46 P SA = 5VI = 5 x 110.8 = 554 W P30V = 30(-1) = 30(-6.22) = -186.6 W (absorbed) P IOA = lOV2 = [15(13 - /1) + 20(13 - /4 )]10= 10[15(10 -4.46) + 20(10- 6.22)] = 1587W = 223 + 554 - Net power delivered 4.99 186.6 + 1587 = 2178 W Verify the power balance in the network of Fig. 4-57. I PlO n = 10(14)2 = 10(6.22)2 = 386.884 W PIS n = 15(11 - 13)2 = 15(4.46 - 10)2 = 460.374 W Ps n = 5(12 - 14)2 = 5(9.46 - 6.22)2 = 52.488 W P20n = 20(l.-IY = 20(6.22 - 10)2 = 285.768 W E = 2179.9 W, which is approximately the same as the power calculated in Prob. 4.98. is due to round off error. 4.100 A difference of 1.9 W For the circuit of Fig. 4-58, find the value of R such that it absorbs 5 W of power. -+ V x R Fig. 4-58 I From Fig. 4-58: (VI)2 2 PR=/I R = 2+R R=5 (1) The nodal equation may be written as Vx -VI 20-V1 --4- + --4- = I1 = VI 2+ R Combining these two equations yields VI Vx -4 - or VI VI 4 + 5 - 4 = -2+-R = v(!+ I 2 1 )=5 2(2 + R) or From Eqs. (1) and (2) we obtain 102R [ (3+R)2 Solving for R yields R = 13.325 n or 0.675 n J2 =5 VI V, 2(2 + R) 2 V = 10(2 + R) I 3+R + 5 (2) ~ CHAPTER 5 Network Theorems~ 5.1 Thevenin's theorem may be stated as follows: At the terminals 12 in Fig. 5-1a, the arbitrary linear network A, containing resistances and energy sources, can be replaced by an equivalent circuit consisting of a voltage source Vrh in series with a resistance R Th . The voltage V~h is the open-circuit voltage across 12, and RTh is the ratio of Rn. Linear network A with sources and resistances I Network Network B B 2 2 (b) (a) Fig. 5-1 the open-circuit voltage to the short-circuit current. Alternatively, RTh is the equivalent resistance at the terminals 12 when all independent sources are suppressed. Figure 5-1a shows the arbitrary network, and its Thevenin equivalent is pictured in Fig. 5-1b. Replace the network of Fig. 5-2a to the left of terminals ab by its Thevenin equivalent circuit. Hence, determine 1. 60 20 C 10 V 10 2 a + C' 12 V2 30 ---" 1 C 0 100 b 20 V (a) 60 20 1.46 0 a 30 -19.27 V 1 a 100 b b (b) (c) Fig. 5-2 To determine R Th , short-circuit the voltage sources to obtain the network of Fig. 5-2b. parallel are equivalent to i~2!!2 = 6+1 ~0 7 which, in series with the 20, is equivalent to Then, the parallel combination of (2017) 0 and 30 gives (2017)(3) 60 RTh = Rab = flal7) +3 = 41 = 1.460 72 The 60 and lOin NETWORK THEOREMS To find VTh , open-circuit the terminals ab. D 73 The lO-V battery then sees a resistance 6 + (1)(2 + 3) = 41 0 1+2+3 6 and so a current of 10 60 41/6 = 41 = 1.46 A is drawn from it, of which 1 1 + 2 + 3 (1.46) = 0.243 A passes through the 3-0 resistor. Then, Vn = (0.243)(3) - 20 = -19.27 V Thevenin's equivalent circuit becomes as shown in Fig. 5-2c, whence -19.27 1= 1.46 + 10 = -1.68 A The negative sign indicates that the current actually flows in the 10 0 from b to a. 5.2 Calculate the Thevenin voltage and resistance at the terminals ab of the circuit of Fig. 5-3a. so 60 a so 90V b 40 (et) er) Fig. 5-3 I The Thevenin voltage is the open-circuit voltage appearing across the terminals ab, which is the same as the voltage across cd. For this open-circuited network we need not consider the 6-0 resistor. resistance seen by the 90 V is given by Re = 8(8 + 4) 8 + 8 + 8 + 4 = 12.80 The current I, shown in Fig. 5-3b, becomes 90 1= 12.8 = 7.03 A Thus, the voltage Vco =- 90 - 81 =- 90 - 7.03(8) =- 33.75 V Current I Hence, co = 33.75 =4 22A 8 . and led = 7.03 - 4.22 = 2.81 A Vcd = Vn = 2.81 x 8 = 22.48 V To find the Thevenin resistance, we refer to Fig. 5.3c from which we obtain Rn = Rab = (4 + 4)8 6+ 4+4+ 8 = 10 0 The equivalent 74 5.3 D CHAPTER 5 Reduce the delta-connected resistors of Fig. 5-3a to an equivalent wye. Find the Thevenin voltage and resistance for the modified circuit. Verify that the results agree with those of Prob. 5.2. sa "'- left c 6.f2- !!.n5 SA b ~n.s (t:{) I The network reduction is shown in Fig. 5-4a. the 16/50 resistor and is given by --- ~a (}~ 5 b Fig. 5-4 The Thevenin voltage is across en, which is the voltage across 90 VTh = Ven = 8 + Cl.. 16 16/5+ 8/5 5 90 = 4 = 22.5 V The circuit for the Thevenin resistance is shown in Fig. 5-4b from which (8+8/5)16/5 RTh 5.4 = 8+8/5~~16/5 + 8 5 +6=100 Determine the current ix in the 5-0 resistor of the circuit shown in Fig. 4-6, by using Thevenin's theorem. I The circuit to determine the Thevenin voltage is shown in Fig. 5-5a, from which we have (for the two meshes) Solving for 11 yields 11 Thus, = 12.88 A VTh = 100 - 1(10) = 100 - 12.88(10) = -28.8 V Figure 5-5b shows the circuit to determine the Th~venin resistance. 10(2 + 1.2) 10+2+1.2 Vrh -28.8 RTh + 5= 2.42 + 5 Therefore, = - - - - - = 2.42 0 R Th Hence, Ix = cb) 5.5 = -3.87 A Fig. 5·5 Fig. 5·6 For the circuit shown in Fig. 4-5, we have RI =, 10 n, R2 = 200, R3 = 300, and R4 = 60 O. Using the Thevenin equivalent circuit show that the current in Rs is zero and is independent of the values of the battery voltage E and Rs' NETWORK THEOREMS D 75 I The circuit, shown in Fig. 5-6, is drawn to determine the Thevenin voltage. Thus, we obtain E 11 And E E 10 + 20 = 30 A E 12 = 30 + 60 = 90 A Hence the current in 5.6 = Rs = 0 A. Using Thevenin's theorem, determine the current in the 2-0 resistor of the network shown in Fig. 4-8a. Fig. 5·7 I First, we change the 5-A current source to an equivalent voltage source. Consequently, we have the Thevenin equivalent circuit to determine VTh as shown in Fig. 5-7a. Writing the mesh equation yields 25 - 10 = (5 + 10 + 3)1 Hence, 20 = - 3( ~) + V Th or or VTh = 20 + 2.5 = 22.5 V The circuit to determine RTh is shown in Fig. 5·7 b from which (10 + 5)3 RTh = 10 + 5 + 3 = 2.5 0 VTh 22.5 12 n = RTh + 2 = 2.5 + 2 = 5 A Thus, 5.7 Find the current 1 in the 10-0 resistor of the circuit of Fig. 4-9. /PV Use Tbevenin's theorem. J~IL_'\~;~ ).0 \t Fig. 5·8 76 D CHAPTER 5 I The circuit to determine the Thevenin voltage is shown in Fig. 5-8a, from which 10 11 = 6 + (5 ;(1):'(5 + 1) = 1.46 A By current division, 1 1= (1.46) :~~t5 = 0.244 A Thus, V3 n = 3(0.244) = 0.732 V and VTh = -20 + 0.732 = -19.268 V From Fig. 5-8b, (2 -+- 6/7)3 RTh = 2 :-6~' + 3 VTh 1= RTh + 10 Finally, 5.S 1.4630 = -19.268 ,= 1.463 + 10 = -1.68 A By applying Thevenin's theorem to the circuit of Fig. 4-lOa determine the current 11 in the 10-0 resistor. I After converting the current source to a voltage source, the Thevenin equivalent circuits to determine VTh and RTh are shown in Fig. 5-9a and b, respectively. Cb) (tL) Fig. 5-9 From Fig. 5-9a we obtain 100 V20 n = 2(20) = 40 V and /= 30+20 =2A Also, VTh or = 50 - 40 = 10 V From Fig. 5-9b we have 30 x 20 = 5 + 30 + 20 = 170 RTh Hence, 5.9 11 = VTh 10 + 10 = 17 + 10 = 0.37 A RTh Find the current in the 30-0 resistor of the circuit of Fig. 4-lOa by Thevenin's theorem. 5Jt /o..{l..- V ..... 'Th ~ ')fL :-~l ~o o,~ 2o.R ----- -=-1 n20~'~J (4.) I (6) From the circuit of the Fig. 5-lOa we have 1=, 51) 35 = 1.43 A 100 + Vn = 20( 1.43) or "fl.- VTh =-71.43 V Fig. 5-10 NETWORK THEOREMS D 77 From Fig. 5-lOb, 20(5 + 10) RTh V lh Hence, 5.10 = 20 + 5 + 10 = 8.57 0 -71.43 130 n == 30 + RTh = 30 + 8.57 == -1.85 A Determine the current supplied by the 100-V source to the circuit of Fig. 4-14 by applying Thevenin's theorem. - \lTt, Yf-/.-.v----' Fig. 5-11 Ca.) I The Thevenin equivalent circuits are drawn in Fig. 5-11a and b. From Fig. 5-11a, for the two meshes we obtain Hence, For the mesh containing VTh we have or 100 - VTh + 10(11) + 10(12) == 0 VTh == 100 + 10(2.5) + 10(2.5) == 150 V From Fig. 5-11b we get R Th 10 1(10+10) ==100 ==-+3 2 3 Therefore the required current is VTh 150 1100 v == RTh + 10 == 20 = 7.5 A 5.11 By applying Thevenin's theorem, determine the current supplied by the battery of the circuit of Fig. 4-15. (0.f1- RTh __-N~_ _ _ _ _ _~~~_ _~ 'v V (b) (a.) lOA IOfL- L_Th _ _-V\IIv ~.n. q (c) Fig. 5-12 0 78 D CHAPTER 5 I The Thevenin equivalent circuits are drawn in Fig. 5-12a and b. From Fig. 5-12a it is clear that no current fio\\'s through the resistors. Therefore, VTh = lOV To obtain RTh we reduce the circuit of Fig. 5-12b to that shown in Fig. 5-12c from which (10 + 50/3)(50 + 10/3) 0 _ 100 0 10 + 501:\ t 50 + 10/3 + 1 - 3 VTh 10 1= R : =. 100/3 = 0.3 A _ 50 RTh - 9 + Hence, T 5.12 Using Thevenin's theorem find the current in the t-n resistor of the circuit of Fig. 4-16. lo.r1- IOrL (6) (a.) I From the corresponding Thevenin equivalent circuits drawn in Fig. 5-13a and b we have VTh = 50V Hence, 5.13 Fig. 5-13 11 n = R VTh Th lOx5 and +1 RTh 10 = 10 + 5 = '3 0 50 150 li)l3 + 1 = 13 = 11.538 A = Solve for the current in the 2-0 resistor o{ the network of Fig. 4-17, by Thevenin's theorem. (;JL J.ov f) ".fL '5.('"/.. S> 4Sl- in ofl- s.n... ."- ~ 3JL J.'V'- 1loV 1 V RTh T~ .- (a.) I (6) The Thevenin's equivalent circuits are shown in Fig. 5-14a and b. Fig. 5-14 Thus, we obtain 20 = 1011 - 4/2 0= --4/1 Or, V,n 30 VTh = 10 + 13 = 12.31 V + 1212 =3C~) = ~~ V {[6 x 4)/(6 + 4)] + 5}3 R Th = 3 + 5 + [(6 x 4)/(6 + 4)] = 2.1340 12.31 2.134 + 2 = 2.98 A 5.14 Find the current in the 1-0 resistor of the circuit of Fig. 4-18a by Thevenin's theorem. NETWORK THEOREMS {~ft- + It n,---r-----. Ion RTI, ~ 2.t"L 411- Ion.. 1 ~ (a.) I (6) Fig. 5-15 Then, we define VTh and First, we transform the current sources to voltage sources as shown in Fig. 4-18b. shown in Fig. 5-15a and b, respectively. From Fig. 5-15a, RTh _ 100 - 40 _ 15 A /- 10+4+2 - 4 VTh = 100 -/(10 + 4) - 20 = 100 - 415 (14) - 20 = 255 V From Fig. 5-15b, RTh = (10+4)2 7 10 + 4 + 2 = 4 0 =~=lOA Hence, 5.15 1 + 7/4 By Thevenin's theorem, determine the current in the 4-0 resistor of the circuit shown in Fig. 4-18b. loA ID- lofl _.L(4.) I Fig. 5-16 (h) By nodal analysis, from Fig. 5-16a, we have _ 40 -20 _ 20 A /oA-~-3 VAO = 40 - 20) 2( 3 = 380 2(1) 32 VTh V = 100 - 80 3 = 220 3 V From Fig. 5-16b, RTh = 10 + 2 + 1 = 3 0 220/3 4 + 32/3 = = Hence 5.16 In Fig. 4-22, given R = 5 0, 5A find the current in the 20-0 resistor by Thevenin's theorem. lO.tL t RTh Fig. 5-17 I From Fig. 5-17 a and b we have VTh = 100 V and D 79 80 D CHAPTER 5 VII, 100 120 n = 20 +l~~ = 20 + 0 = 5 A Hence, 5.17 Calculate the current in the 10-0 resistor of the circuit shown in Fig. 4-23a by Thevenin's theorem. Fig. 5-18 I From Fig. 5-18a, Vs A 50 - VTh = 5 x 20 = lOC V - 100 = 0 Or From Fig. 5-18b, RTh Hence 5.18 lIon = 20 + 20 = 40 0 =~~h --=~=-lA 10+ Rn 10+40 Find the current in the 20-0 resistor of the circuit of Fig. 4-24a by Thevenin's theorem. I Fig. 5-19 (6) (a.) From Fig. 5-19a (with polarities as shown) V8 n = 2 x 8 = 16 V 10 20 V4n = 1O+4(4)=7 V Thus, From Fig. 5-19b, 10 >: 4 RTh VTh Hence, 5.19 120 n = 20 + 76 = 10~:-:t + 8 = 7 0 RTh 13217 132 =2() + 7617 = 216 = 0.611 A Find the current in the 20-0 resistor of the circuit of Fig. 4-24b by Thevenin's theorem. -t V, ... 5 Cl- V Th lA- lA- (a.) ,.nj i+~f~ ~ \ on (D) Fig. 5-20 NETWORK THEOREMS I D 81 From Fig. 5-20a solving for VI and V2 yields VI VI 10+4=1 Thus, or 20 Vl=--;:;-V or ~ - VTh and + 16=0 V2 = -16V or From Fig. 5-20b 10 x 4 RTh VTh Hence, 5.20 = 10 + 4 +8= 76 --;:;- 0 13217 132 20 + 7617 = 216 = 0.611 A 120n = 20+ R Th Determine the current through the 12-V source of Fig. 4-25a by applying Thevenin's theorem and nodal analysis. 6.{L ~, ... 'l.f1.V vvv.. ').. ,{1.. lA &{l.. ;. V1, + 'v - Th- "f-.n- 2A M- O (a.) Fig. 5-21 I From Fig. 5-21a we have V V-V 1+-t+~=0 or Vl -V2 V2 V2 --6-=6 + 2+4 Thus V2 6=-2Vl +V2 or 3V2 =Vl 6 5V =- - or and From Fig. 5-21b, 4x6 17 RTh = 1 + 4 + 6 = 5 0 112 Y 5.21 = VTh = 66/5 = 3.88 A RTh 17/5 Repeat Prob. 5.20 using the circuit of Fig. 4-25d. I In this case, by inspection, we have VTh = 12 + 2 - 1(0.8) = 13.2 V Hence = 1 + 0.8 + 1.6 = 3.4 0 VTh 13.2 112y = -R = - = 3.88 A Th 5.22 RTh 3.4 Find the current in the 2-0 resistor of the circuit of Fig. 4-28 by Thevenin's theorem. 82 D CHAPTER 5 SA 2. v, + V I Th ----,~R v.... - Th I 4Jl- toA. $ (a.) , %.cV -:; .....(1- (6) Fig. 5-22 From Fig. 5-22a we obtain v I} = 10-5 = 5 A 2 = 5(16)(8) = 80 V 16 + 8 3 Thus, From Fig. 5-22b, 16 x 8 RTb 5.23 R2 = R3 = 5 n, In Fig. 4-30, given theorem. -20/3 28 = 4 + 16 + 8 = 3 n = - 0.588 A 22 + 28/3 R6 = 10 n, and V2 = 2V3 = 10 V, find the current in R6 by Thevenin' 5f1.-. (12.) , From Fig. 5-23a and b we obtain 5+10=1(5+5) 15 1= 10 = 1.5 A or VTh = 5 - 5(1.5) = -2.5 V Thus, 5x5 RTb = 5 + 5 = 2.5 n and = -2.5 I R6 =~_ 10 R 10 + 2.5 = -0.2 A Hence 5.24 Fig. 5-23 (b) + Th Determine the current in the 5-n resistor of the circuit of Fig. 4-33, by Thevenin's theorem. f( TI( ::- 50o.t\.. I (a.) , (6) Fig. 5-24 From Fig. 5-24a, Vl = 1(500) = 500 V Also, and 100 + 500 - VTh = 0 RTb =, 500n or VTh = 600 V NETWORK THEOREMS 0 83 From Fig. 5-24b, 600 = 500( ~R ) Thus, 5.25 + VR - VR = 6.25 Y or 5VR 1= 6.25 5 and = 1 25 A . Determine the current through the short-circuited terminals ab in the circuit of Fig. 4-37 by Thevenin's theorem. , By open-circuiting ab in Fig. 4-37, we obtain VTh = 20 - 4(3) = 8 Y Also from Fig. 4-37, RTh Thus, lab = = 6 + 4 = 10 0 VTh 8 = 10 =0.8A R Th 5.26 Find the current in the 4-0 resistor of the circuit shown in Fig. 4-38a without any network reduction. Thevenin's theorem. Fig. 5-25 , To determine VTh we redraw the circuit as shown in Fig. 5-25 from which we obtain VI 12 V1 -V2 V1 -V2 + --6- + 6+6 = 2 or 4V1 - 3V2 = 24 V 1 -V2 --6- and V 1 -V2 + 6+6 = V2 12 3Vl -4V2 =0 Or V=96 Y I 7 Thus, V=72Y 2 7 and and Hence, 5.27 By applying Thevenin's theorem to the circuit of Fig. 4-38b determine the current I in the 12-0 resistor connected to node 2. 'lSL 2a 2{1.. ~~-?~--~~~ If ~4-J1... + I~[l... ," _\ 2A 'V R rh (0) (a.) , From Fig. 5-26a we have I = I 24 -12 =~A 12 + 2 + 2 + 4 5 Fig. 5-26 Use 84 0 CHAPTER 5 3 VTh =24-11 (12+2)=24-"5 (14)= Thus, 78 5 V From Fig. 5-26b, RTh _'" 1.4(6) _ 31 -.<.+ 14+6 - 5 0 78/5 12+31/5 Hence 5.28 6 = '7 A Find the current I in the 5-0 resistor of the circuit of Fig. 4-41 by Thevenin's theorem. roJi- ,.0- ,os.?- IOn. + + V'L I ef) -_r Vn loft A- (0 (6) (a.) RTh Fig. 5-27 (e) , From Fig. 5-27a we have or or - V 1 + 3V2 VTh = - V1 I) - V1 + V2 or -- T Th V2 - VTh 10 + 10 - = 10 2VTh = 100 Solving for VTh yields Vrh 1020 '" -13 V Now, referring to Fig. 5-27b, we transform it to an equivalent delta shown in Fig. 5-27c, which may be reduced to that shown in Fig. 5-27c. Therefore, from Fig. 5-27d we obtain R - (165/14)30 _ 110 0 165.'14 + 30 - 13 Th - Hence, 5.29 = I 5!l VTh == ---(1020/13) = _ 204 A 5 + RTh 5 + (110/13) 35 Find the current in the 5-0 resistor of the circuit of Fig. 4-43 by Thevenin's theorem. NETWORK THEOREMS JL 1 :&4,"++ ".fl..l~._f 'A v" (J) (tJ..) , Fig. 5·28 From Fig. 5-28a we obtain 9 = VTh - 32 - VI + VTh - VI 2 4 or 15VTh and v Hence, Th 17VI = 240 - = 490 V 3 From Fig. 5-28b, =10 R (4)(2)=34 0 3 + 4+2 Th 490/3 Thus, 5.30 5 + 3413 = lOA Determine the current in the 2-0 resistor of the circuit of Fig. 4-43, by Thevenin's theorem. Fig. 5·29 , The Thevenin equivalent circuits are shown in Fig. 5-29a and b from which we obtain or and - 7V2 = -80 810 V2=19 V Thus, v Since, Th RTh 5.31 5VI 60 (10 + 5)4 == 10 + 5 + 4 = 19 0 = 830 _ 810 _ 32 = _ 588 V 19 19 19 = -588119 2 + 60119 = -6 A Evaluate the current in the 50-0 resistor of the circuit of Fig. 4-45 by Thevenin's theorem. Fig. 5·30 0 85 86 0 CHAPTER 5 , From Fig. 5-30a we have V2 -VI --1= 9 V3 - V 2 --2- 170 - V3 4 V2 or \12 = 9 + 20 + is V3 - \1 V3 = 47 + --2--2 V2 VI = 9 - or -37V2 + 30V3 = 540 or -94V2 + 145V3 = 7990 Solving for V2 yields V2 = 63.42 V VTh Hence, = VL = V2 -- 9 = 63.42 - 9 = 54.42 V From Fig. 5-30b we obtain R Th 4.420 VTh Thus, 5.32 = 150 n 54.42 50 + 4.42 = 50 + RTh = 1.0 A Calculate the current 13 in the circuit of Fig. 4-46 by Thevenin's theorem. If).. V R Th- 1h /d.IL L 5'n..~ 212-~ Fig. 5-31 , From the Thevenin equivalent circuits of Fig. 5-31a and b we have VTh + 50 - 20 + 10 = 0 or VTh = --40 V Hence, 5.33 I, VTh = 20 + R· Tt = - RTh = 00 40 20 + 0 = -2A Find the current 12 in Fig. 4-46 by Thevenin's theorem. , By inspection, 20 -- V-rh - 9 - 10 = 0 or Vn 1 I = ---- = - - = 1A Hence 2 5.34 1- RTh 1+0 Find the current in the 3-0 resistor of the circuit of Fig. 4-55 by Thevenin's theorem. , The Thevenin equivalent circuits are shown in Fig. 5-32a and b. I 2n = lOA or Also, VI V V - -2 - -L + I = 10 x 2 = 20 V V -20--40 2 _ = 10 6 or From Fig. 5-32a, NETWORK THEOREMS ~A If\. 0 87 ,.(1... VI SA + 1-.f'2- 1(a.) Fig. 5-32 or Finally, VTh = 20 + V2 = 20 + 240 380 "7 = "7 V From Fig. 5-32b, RTh = 5 + ~ + 2 = ?f 0 = Hence, 5.35 380(7 =5A 3+55(7 By Thevenin's theorem, calculate the current through the galvanometer, having a resistance of 200, in the Wheatstone bridge shown in Fig. 5-33a. (D) (4) , (c) From Fig. 5-33b we have 36 11 = 10 + 30 = 0.9 A 36 12 = 40 + 50 = 0.4 A VlOn = 11(10) = 0.9 x 10 = 9V V40n = 12(40) = 0.4 x 40 = 16 V or Also, VTh = V40 n - VlO n = 16 - 9 = 7 V From Fig. 5-33c, RTh 10 x 30 40 x 50 = 10+30 + 40+50 =29.720 VTh Hence, 5.36 IG = 20+ R Th 7 20 + 29.72 = 0.14 A Determine the current in the 10-0 resistor of the circuit of Fig. 5-34a. , Fig. 5-33 From Fig. 5-34a, 36 11 36 = 40+[50(30+20)]/(50+30+20) = 65 A 1 12 = 2 11 = 18 65 A 88 0 CHAPTER 5 3bV '---\t--(a.) (b) Fig. 5-34 And VTh = V20n + V40n = 12 (20) Or 18 + /[(40) = 65 36 20 + 65 (40) x 360 = 13 V From Fig. 5-34b, 360(13 360 10 + 228(13 = 358 Hence 5.37 = 1.0 A Find the current in the 3-0 resistor of the circuit of Fig. 5-35a by Thevenin's theorem. 6A (a..) (b) , (c) Fig. 5-35 From Fig. 5-35b we have or VTh =36V From Fig. 5-35c we obtain 12 x 6 RTh = -1'-6 =40 -+ Hence, VTh 13 n = 3 + R Th 5.38 36 = :3 . 36 + 4 = 7" = 5.143 A By Thevenin's theorem, find the current in the 19.2-n resistor of the circuit shown in Fig. 5-36a. NETWORK THEOREMS 0 89 O.IJ\. -.-~ 0·117..- (4.) (D) o. 1/1.- o.o'lca.Jl. o.()DI~ Q.OqSJl... o. lA S.7"J\. (tI) (c) I From Fig. RT~ Fig. 5-36 5-36b, 5.96/1 Or - -0.111 + 9.8/2 = 120 0.112 = 120 11 = 20.34 A and 12 = 12.45 A VTh = 9.6/2 + 5.76/1 = 9.6 x 12.45 + 5.76 x 20.34 = 236.7 V Hence, From Fig. 5-36c and d RTh 5.39 (0.098 + 5.76)(0.001 + 0.1) = 0.098 + 0.098 + 5.76 + 0.001 + 0.1 = 0.197 0 236.7 119 . 2 n = 19.2 + 0.197 = 12.2 A Find the current in the 9-0 resistor of the circuit of Fig. 5-37a by Thevenin's theorem. (a.) (6) Fig. 5-37 (c) I Since the circuit contains a dependent voltage source, we obtain RTh = V ae ( lse' where Vac and Ise are, respectively, defined in Fig. 5-37b and c. From Fig. 5-37b, 20+6Ix =4Ix +6Ix or Ix =5A and Vae=VTh=5x6=30V 90 0 CHAPTER 5 From Fig. 5-37c, or 20 = 4/se R Hence, 5.40 Th .- = Vue = 30 = 6 n /se 5 /se = 5 A VTh and /9n = 9 + R 30 = - - =2A 9+6 Th Find the current in the lO-n resistor of the circuit of Fig. 5-38a. IO..D-~ ~T,-=------------. tr + -t Ip.v V, +~~v"-~ ... ,oA ~.f1..- Vx ~ SA (ooV )( -L__----'--------' - - I (b) (a.) (c) I Proceeding as in Prob. 10 A Fig. 5-38 5.39, from Fig. 5-38b we have 100- Vac + IOVx - Vx=O and Vx = IOx5=50V or Vac::: VTh =550V From Fig. 5-38c we obtain / Also, se + 10 = 100~~ _ 5 V=_lOOv or /se 9 x =- 550 = -45 -550/45 Thus, Hence, 5.41 or / 10 n 550 10 - 45 = --- 550 4s A n = -15.7A Calculate the current in the 6-n resistor of the circuit of Fig. 5-39a by Thevenin's theorem. 1ft --I/V'v-..,...--<.. - 'v-,. i (C) Fig. 5-39 NETWORK THEOREMS I First, we calculate Vac from Fig. 5-39b. D 91 Thus, 18 + Vx + 2Vx - Vac = 0 V=3xl=3V x But or Vac = VTh = 27 V From Fig. 5-39c, 18+Vx +2Vx =0 or 18+3Vx =0 Vx - = 3 - Isc or Isc = 9 A I R = Vac = 27 = 3 0 Th Isc 9 But Thus, 27 Hence 5.42 = 3 +6 =3 A I6f1 Determine the current in the 2-0 resistor of the circuit of Fig. 4-55 by Thevenin's theorem. lit _ _.AAJV------r---- .... t - - - - - ; <S €;A + 6(t I loA I Fig. 5-40 I From Fig. 5-40a, or V5 n = 6 x 5 = 30 V 11 40 =--;; A V3 n = 10 x 3 = 30 V 40) 240 V6n =6 ( --;; =-::;-V 240 380 VTh = -::;- - 40 + 30 + 30 = -::;- V From Fig. 5-40b, 6 xI RTh 5.43 62 = = 5 + 3 + 6 + 1 = --;; 0 380(7 = 5 A 2 + 62(7 Find the current in the 24-0 resistor of the circuit of Fig. 5-41a by Thevenin's theorem. I To determine VTh , we remove the 24-0 resistor. Then, VTh = Vab = -lOIx (12) = -120Ix = Vx But Ix = Thus, To determine VTh = -120 RTh 48 - 3Vx 48 - 3VTh 1000 = 1000 (48 - 3VTh ) 1000 we short-circuit the terminals ab. or The circuit reduces to that shown in Fig. 5-41b, since 92 D CHAPTER 5 t 14 ft 12Jl. (a.) f>_I>< --,---12It---l (6) Vx =0. Thus, , _ and 5.44 Fig. 5-41 Vac Hence, RTh Finally, 124 48 or 'x - 1000 = I~~ == 9 0.48 Isc = -0.48 = 18.750 9 n = 24+18.75 = 0.21 A Find the current in the 14-0 resistor of the circuit of Fig. 5-42 by Thevenin's theorem. SJl. ().. + " IC C). I V" ,4.f'l. Fig. 5-42 b I To find VTh = Vac' we open-circuit ab. Thus, Vac = VTh = 10 - (-0.lVx)5 = 10 + 0.5Vac Or Voc = 20V =V Th Next we short-circuit ab resulting in Vx = 0 and 0.1 Vx 10 10 Isc = 5 + 8 = 13 A 5.45 Vac or RTh = o. Thus, 20 = Is,~ == 10(13 = 260 and 114 n = 20 14 + 26 = 0.5 A Determine the current in the 16-0 resistor of the circuit of Fig. 5-43 by Thevenin's theorem. I Open-circuiting ab in Fig. 5-43 yields Vac = VTh ='40V Short-circuiting ab implies that Isc = Ix. smce And 40- V Isc + 0. 8/sc = ~ NETWORK THEOREMS But D 93 VI = 6/se 40 Ise = 24 A Therefore, Hence, / 16 0.8]. Vae and n RTh = 40 = -1- = 40(24 = 240 40 16 + 24 se = 1.0 A It.fl. I bn Fig. 5-43 5.46 Fig. 5-44 Repeat Prob. 5.45 by converting the 40-V voltage source to an equivalent current source. I The new circuit is shown in Fig. 5-44 from which the open-circuit voltage is Voc = iO x 4 = 40V = V Th To find lse' we short-circuit ab and use the nodal equation to obtain VI 4 = 10 + 0.8/x + Ix and Ix VI = "6 = Ise 40 Ise = 24 A Hence, which is identical to the result of Prob. 5.45. 5.47 Determine the current in the 8-0 resistor of the circuit of Fig. 4-54, for I By inspection, with the k = 2, by Thevenin's theorem. 8-0 resistor removed, Vae = VTh = 16 V When the 8-0 resistor is short-circuited, Ise = Thus, 5.48 00 16 and 16 RTh = ~ = 00 Iso = 0 + 8 = 2.0A Find the current I in the circuit of Fig. 4-56 by Thevenin's theorem. v, 4fl Fig. 5-45 I First, we remove the 4-0 resistor to obtain the circuit shown in Fig. 5-45 to determine Vac- Thus, and VI = 5 x 3 = 15V and To find Ise we short-circuit ab through which Ise = I flows. 12 - VI + 5 = ~ + 4(12-V I) 434 or, and V, = 9.6V I - sc- 12 - V, = 06 4 . A R Th,= -3= 50 0.6 or Then, Vae = 15 - 12 = 3 V 1=0 94 5.49 D CHAPTER 5 Obtain the Thevenin equivalent circuit at the terminals ab of the circuit shown in Fig. 5.46a. I 3A ~v + -----: 31 E -~ ',,_ J oJ'L ( I If ~ 1.- - I_---.l...-.--J - - - .: (a.) (b) I Note that this circuit does not contain any independent sources. apply a voltage source of 1 V at ab. Therefore, + 1 - 31 or I:: ! A 6 3 4 The Thevenin equivalent circuit is shown in Fig. 5-46b. 1= 5.50 ! To obtain the Thevenin equivalent we 1 R Th =1(4=40 and The equivalent circuit of an operational amplifie r is shown in Fig. 5-47. resistance at the output terminals. Assume R, ~ Ri' ...R. . < Fig. 5-46 Find the Thevenin equivalent tl. -......,.------ - -, 1.. 3) R I '+ ,..J..." " IV) '-)- )- __ J Fig. 5-47 !. I As in Prob. 5.49, we apply a voltage source of 1 V across ab. We further observe that the parallel combination of RI and Ri is approximately RI' sin(:e RI ~ Ri' Then, for the resulting circuit we have II(R I + R2 + Ro) - 12RO + AVi = 0 Solving for 12 yields Hence, 5.51 Apply Thevenin's theorem to the circuit to the left and to the right of node 1 of the network shown in Fig. 5-2a. Hence, determine the current in the 1-0 resistor. loV Fig. 5-48 I To the left of node 1, the Thevenin equivalent is simply the lO-V source in series with the 6-0 resistor. the right of node 1, we have VTh = V3 n = 20 60 13 x 3 = 13 V and RTh = 2+ 10 x 3 56 -u = 13 0 V and Finally, we obtain the circuit of Fig. 5-48, from which 10- VI + (60/13)- VI 6 (56/13) ..!:.l. =0 I or VI = 1.96 _V, I III - -1- = 1.96 A To NETWORK THEOREMS Linear network A with sources and 0 95 1 Network 8 Network 8 2 2 ~sis'lJ"ces (b) (a) Fig. 5-49 5.52 Norton's equivalent circuit is the dual of Thevenin's equivalent circuit and may be obtained from the latter by source transformation. Explicitly, Norton's theorem may be stated as follows: At the terminals 12 in Fig. 5-49a, the arbitrary linear network A, containing resistances and energy sources, can be replaced by an equivalent circuit consisting of a current source IN in parallel with a conductance G N, shown in Fig. 5-49b. The current IN is the short-circuit current (through 12, when these terminals are short-circuited) and G N is the ratio of the short-circuit current to the open-circuit voltage. Note that G N = 1fR Th , where RTh is the Thevenin equivalent resistance. Find the current in the 10-0 resistor of the circuit of Fig. 5-2a by Norton's theorem. 6f1-. loJ i) 2.f1. In. V, ~fI- J) RTh=1.460, IN loA I (b) 'oV (4) I FromProb.5.1, J) Fig. 5-50 or 1 1 G N = RTh = 1.46 = 0.685 S To determine IN we refer to the circuit of Fig. 5-50a from which Solving for 13 yields 13 = -IN = 13.167 A Therefore, from Fig. 5-50b, VI VI -13.167= 1.46 + 10 Hence, 5.53 lIOn = - or 16.8 10' = -1.68 A In Prob. 5.52 we have shown the details of calculating IN' GN = VI = -16.8V However, from Figs. 5-1 and 5-49 it follows that 1 R (1) Th b (a.) (6) Fig. 5-51 96 D CHAPTER 5 V = , .Th - I and N (2) RTh Apply these results to the circuit of Fig. 5-3 to obtain directly the current through the 5-n resistor. I From Prob. 5.2, 1 R Th =-=1Of! GN VTh = 22.48 V and Thus, Hence we obtain the circuit of Fig. 5-51b from which we have 2.248 5.54 = VI 10 + 5VI = O.03V or I 7.5 and VI =7.5V ISfl =""5 = 1.5 A Obtain a Norton equivalent circuit to determine the current in the 5-n resistor of the circuit of Fig. 4-6. I From Prob. 5.4 we have RTh = 2.42 VTh Hence, n VTh = -28.8V md 1 28.8 IN = RTh = - 2.42 = -11.9 A 1 G N = RTh = 2.42 = 0.413 S Thus, we obtain the desired circuit shown in Fig. 5-52. 0.413 S Fig. 5-52 5.55 Determine the current in the 2.4 A lO-n resistor of the circuit of Fig. 5-53a by Norton's theorem. /0£7-. 3ft (a.) S (6) I A 0.2", --L(c) 1~" Fig. 5-53 To determine IN we short-circuit ab to obtain the circuit of Fig. 5-53b from which, by current division, I",=8A With the source and short-circuit removed from Fig. 5-53b we obtain RN 1 9(6 + 3) = G = 9 + 6 + 3 = 4.5 n or 1 G N = 4.5 = 0.222 S N Hence we obtain the Norton equivalent circuit of Fig. 5-53c which yields V]CO.222 + 0.1) = 8 or v] = 24.84 V 1]0 n = 10.] S = 2.484 A NETWORK THEOREMS 5.56 D 97 Obtain the Norton equivalent circuit for the network of Fig. 5-54a to determine the current in the 50-0 resistor. (a.) (h) I Fig. 5-54 With the 50-0 resistor short-circuited, the total current will flow through the short circuit. Thus, IN = 0.75 + 0.75 = 1.5 A Next, with the sources and the 50-0 resistor removed, the two 100-0 resistors become in parallel resulting in a net resistance of 50 O. Therefore, 1 G N = 50 = 0.02 S Consequently, the Norton circuit of Fig. 5-55b follows, from which Ison = 0.75 A 5.57 The network shown in Fig. 5-55a is an equivalent circuit of a transistor switching circuit. resistance is 1250, calculate the current through it by Norton's theorem. I"2. s.n. 1.'71 ""\1' 1.2Z ....S If the base-to-emitter (a.) 14-.11 - 11.1. = 2.SI_A Fig. 5-55 (h) I The Norton equivalent of the circuit of Fig. 5-55a is shown in Fig. 5-55b from which IB = IN G BE 5.58 3 G BE +G = 2.51 x 10 N -3[ ( 8 x 10) 8 + 2.691 10 ] 3 = 1.878 x 10 Find the Norton equivalent at the terminals ab of the circuit of Fig. 5-56a. -3 = 1.878 mA 98 D CHAPTER 5 I~ 3fi .~~.-~----~-----.~ (a.) I First, we short-circuit ab. (b) Fig. 5-56 Then I" = 1, + 2Ix = 3Ix Ix = But 12 "4 = 3A Isc = 3 x 3 = 9 A Next, we determine the open-circuit voltage Voc a1: ab. Ix = 12 - Voc 3 and + 21 = Voc I x 4 x or 9 45 - - == -- Isc G N = -V Hence, . Jx = 48/5 oc 48 From Fig. 5-56a with ab open-circuited, Voc 12 or Voc = 12 - Voc 12 3 or and Thus, we obtain the circuit shown in Fig. 5-56b. 5.59 The circuit in Fig. 5-57 contains two types of independent sources. Obtain the Norton equivalent circuit at ab. r--------,---------, A. f.n.. ,_ _ _-'--_.&-----.b Fig. 5-57 (a.) I By short-circuiting ab we observe that 35 Isc· . = 1./ . = -3A Removing the sources and the short-circuit yields R == ~'i~+ 1) = 2 0 N 3+5+1 Hence we obtain the Norton circuit of Fig. 5-57h. 5.60 Converting the current source of the circuit of Fig. 4·lOa results in the circuit of Fig. 4-lOb. the 10-0 resistor by applying Norton's theorem. v. 30.n. 17.J1- (<2.) I /t>fl-. (b) Short-circuiting the 10-0 resistor results in the circuit of Fig. 5-58a, from which 50 - V1 100- V --5- + 30 I Find the current in F1 20 or V==800 y 1 17 Fig. 5-58 NETWORK THEOREMS 50 - VI 50 - (800/17) 10 lsc = --5- = 5 = 17 A And Removing the short-circuit and the sources yields 30 x 20 RN = 5 + 30 + 20 = 17 n Thus we obtain the Norton circuit of Fig. 5-58b which yields 10 V2 V2 17 = 17 + 10 5.61 V2 100 V2 = or 27 V and lID 100 n = 10 = 10 x 27 = 0.37 A Obtain the Norton equivalent of the circuit shown in Fig. 5-59a. R, ,... Vo o "to (t.) (4.) Fig. 5-59 (c) I We short-circuit AB (Fig. 5-59b) to obtain -J.LVgk lsc = Rb + Rk where Vgk = -lscRk - V. The Norton equivalent becomes as shown in Fig. 5-59c. 5.62 The circuit of Fig. 5-60a contains only a dependent source. Obtain its Norton equivalent. +n- ~~ ,,: f 4n A- I l"sc ~~.a. I 51l of? (4.) 1- 'l- I oS (b) le, \ T~:30J2 16_fl- (c) I A Fig. 5-60 (tI.) First we short-circuit AB as shown in Fig. 5-60b, from which v 10 + 21 V + "4 = 0 or To find RN we refer to Fig. 5-60c where or V=O and lsc = 0 D 99 100 D CHAPTER 5 or And at node 1, At node 2, VD I = o 5 VD - V 5 'VD +- = -VD + -VD - -4 5 4 12 2Vo 60 Vo ID R =-=300 or = - N Hence we obtain the Norton circuit of Fig. 5-60d. 5.63 Find the current in the 5-0 resistor of the circuit of Fig. 5-61 by Norton's theorem. 4Jl. ~.fl.. A 2n.' - tI + V, SJ"t. ,Isc. Fig. 5-61 lJ I At node A, after short-circuiting the 5-0 resistor, we obtain VI 4 - V I /2 -2- - Isc ,= 0 Hence no current will flow through the 5-0 resistor, 5.64 or Ise = 0 Is n = O. Determine the Norton equivalent of the circuit of Fig. 5-61. Fig. 5·62 I From Prob. 5.63, Iso: = IN = 0 To find RN we use the circuit of Fig. 5-62a from which VD-VI Vo+VI/2 3 ID = --4- + --2-- = 4 VD or VD 4 -=R =-0 ID N 3 Thus we obtain the Norton circuit of Fig. 5-62b. 5.65 Find the current in the 1-0 resistor of the circuit of Fig. 5-63a by Norton's theorem. In IJl (b) (a.) I Short-circuiting the 1-0 resistor yields =0 5 + I se _ 20 -10 5 By inspection, or Fig. 5-63 NETWORK THEOREMS D 101 Hence we obtain the Norton circuit of Fig. 5-63b from which v V 5 - + - =3 1 5.66 or V= 15 6 and =2.5V Two networks are shown in Fig. 5-64; the ammeters in both circuits have negligible resistances. Calculate the ammeter readings in the two networks and comment on the results. Note that in the two networks the positions of the ammeter and the voltage source are interchanged. I. so 100 I" + 90V + 300 90V (a) I In Fig. Fig. 5-64 (b) 5-64a, the resistance seen by the voltage source is R' = 5 + (10 + 5)(30) = 15 n e 10+5+30 so that 90 15 ]'=-=6A Hence, by current division, Ammeter reading = la = !~ (6) = 4 A Similarly, for Fig. 5-64b, R" = 15 e (30)(5) = 135 +30+57 n and r= 90 14 13517 = 3 A By current division, 30(14) . = I b = 35 Ammeter readmg 3 = 4A The equality of the ammeter readings constitutes a special case of the reciprocity theorem: In a linear bilateral network, if a voltage V in some branch produces a current I in some other branch, then the same voltage V in the second branch will produce the same current I in the first branch. 5.67 Solve for the current I of Fig. 5-65a by applying the superposition theorem. 20 60 iOV c + VI 10 c 2 a c + lOO L-----------------~------~--------~------------~r---b o 20 V (a) 60 20 I, lOO ~------~------~-----1I---b (p) Fig. 5-65 102 D CHAPTER 5 I According to the superposition theorem, we determine the current due to each source (in the absence of all other sources). The net current is the sum of all these currents. So let us eliminate the 20-V source by replacing it by a short circuit. The network of Fig. 5-65a then becomes as shown in Fig_ 5-65b. Combining resistances as in Prob. 5.2, we find I, = 0.0636 A Next, we eliminate the lO-V source to obtain the circuit of Fig. 5-65c. Proceeding as before, we determine I; ,= -1. 744 A Consequently, I = I, + 12 = 0.0636 - 1. 744 = -1.68 A 5.68 Determine the current I, in the circuit of Fig. 4-lOa by superposition. I,' r," IOn. ~ loA ~o!<- 1.. IOIl. {24i1- ( 01.) Fig. 5-66 I For superposition, we first excite the circuit by the::urrent source only as shown in Fig. 5-66a from which we observe that the lO-A current from the source is dividl!d into a 5-0 resistor and a 10 + [(20 x 30)/(20 + 30)] = 22-0 resistor. Hence, by current division, , 'i ) I; = lOj\) -=-.:_- = 1.852 A + 22 Next, removing the current source and keeping only the voltage source, from Fig. 5-66b we obtain 12 = 30 + ([20(1O 4- 10(1 5 )]-/(-20-+-1O-+-5)~} = 2.593 A By current division, I~ = 2.593 20-=;3.10 + 5 = 1.481 A 0 I[ = I; - Hence, 5.69 I~ ,= 1.1<52 - 1.481 = 0.37 A Calculate the current I in the circuit of Fig. 4-16 by superposition. 5n. 51l. \.M I' 1, 1" I"' I j'l. lIt _/\,I\f\ Ion lOll. + ( D) (t'.) I (r:") We use one source at a time as shown in the circuits of Fig. 5-67a to c. Fig. 5-67 From Figure 5-67a we have 201) I' = 10 + [( 5-;(-1 ) /=(5=-+--:"1""-:-)] = 18.46 A Fig. 5-67 b yields - ·9') I , -- 5 + [(10 x 100 1)/(10 + 1)] - Ih ..• i\ and 1 I" = 16.92 10 + 1 = 1.54 A and 5 I'" = 11.54 5 + 10 = 3.85 A Similarly, from Fig. 5-67c we obtain 50 12 = 1 + [(5 x 10)/(5 + 10)] = 11.54 A Hence, 5.70 I = I' + I" + I'" = l~;.46 + 1.54 + 3.85 = 23.85 A Find the current in the 2-0 resistor of the circUIt (If Fig. 4-17 by superposition. NETWORK THEOREMS D 103 2.0" II I" 1J1- Fig. 5-68 , From Fig. 5-68a, with the 20-V source removed, we obtain ]' == 2 + (3{5 + [(6 x 4) 1(6 + 4)1~~1 {3 + 5 + [(6 x 4)/(6 + 4)]} == 2.42 A From Fig. 5-68b we have 20 I1 == 6 + {4(5 + [(3 x 2)/(3 + 2)])} 1{4 + 5 + [(3 x 2)/(3 + 2)]} == 2.37 A 4 12 =2.37 {4+5+[(3X2)/(3+2)]} =0.93A I" == 0.93 3! 2 = 0.56 A 12 n == ]' + I" = 2.42 + 0.56 = 2.98 A 5.71 Find an expression for the output (v o ) of the amplifier circuit of Fig. 5-69. What mathematical operation does the circuit perform? , The principle of superposition is applicable to this linear circuit. With appearing at the noninverting terminal is found by voltage division to be R v2 = R + R Let vol be the value of Similarly, with V SI Vo with V S2 = O. Assume an ideal basic op amp. V S2 = 0 (shorted), the voltage V SI V SI = ""2 Now, = 0, By superposition, the total output is The circuit is seen to be a noninverting adder. R VS2 _-..J"" V. I 64 $I g.n.. r--.--_-~ Fig. 5-69 Fig. 5-70 104 5.72 D CHAPTER 5 Calculate the current in the 4-0 resistor of the circJit of Fig. 5-70 by superposition. , Assuming that all currents in the 4-0 resistor are in clockwise direction, with the 6-A and 6-V sources removed, due to the 12-V source we get 12 I' ,= - - - = 1.0 A 4+8 With the 6-A and 12-V sources removed, for the 6-V source we have 6 J" = -- ---- = -0.5 A L. + 8 Finally, with the two voltage sources, we obtam for the 6-A source, by current division: J'" = -6( -4.0 A 14 n = I' + J" + I'" =, l.0 - 0.5 - 4.0 = - 3.5 A Hence, 5.73 -~) = 4+8 Calculate the current in the 4-0 resistor of the circuit of Fig. 5-71a by superposition. lOA- 2..fl. S /.!1- (0.) 2.0 3£1.- /..11- r---~'--~~~-------LI' 4n (c) (h) In (d) 2fL In. 4-.ll..: 1---";"+-( I"' 4 V 1---------1 (/J Fig. 5-71 NETWORK THEOREMS D 105 , Considering one source at a time, from Fig. 5-71b we note that the 2-0. and the 4-0. resistors are in parallel, so that the circuit may be redrawn in Fig. 5-71c. By current division we have 1 11 == -5 1 + 3 + [(2 x 4)/(2 + 4)] = -0.9375 A and I' = -0.9375 2 ~ 4 = -0.3125 A Similarly, we redraw the circuit of Fig. 5-71d as that in Fig. 5-71e, from which 3 12 and = -10 3 + 1 + [(2 x 4)/(2 + 4)] = -5.625 A -i"= -5.625 2~4 = -1.875 A or i"= 1.875 A Finally, from Fig. 5-71f we obtain I Hence, 5.74 m 3~A W = 4+{[2(1+3)]/(2+1+3)} = . I = I' + i" + l'" = -0.3125 + 1.875 + 3.75 = 5.3125 A Determine the indicated voltage VI2 in the circuit of Fig. 5-72 by superposition. I, \ I __ I Fig. 5-72 , Instead of redrawing the circuit, we remove the voltage sources by closing S2 or SI. closed. Then, Let SI be open and S2 20- VI VI VI -4-=10+4 Or, I == 20 - VI = 20 - 25/3 = 35 A I = _ VI = _ 25/3 = _ 25 A I 4 4 12 2 4 4 12 6 V~ = 6(11 - 12 ) = 12 (35 + 25) = 30 V Next, let S2 be open and S2 closed. Then, or 6 (25 + 35) = -15 V V "12 = 6(11 - 12 ) == - 24 or 5.75 Find the current through the dependent voltage source, 2VI , of the circuit shown in Fig. 5-73 by superposition. Fig. 5-73 , First we remove the 3-A current source. Then, 24 - 2F + 2 VI + 4 VI =0 and 106 D CHAPTER 5 Thus, Next, we remove the 24-V voltage source. 1ben . from nodal analysis, /" =, - 36 A 1 = I' Hence, 5.76 e 6 + /" "". - 14 ~424 ) = -0.857 A For the operational amplifier circuit of Fig. 5-74, f:nd Vo by superposition. Note that V+ = O. R, + , V R].. -t V 2- T V R~ 0 --N\f\ "'_'.I v ,.. , Fig. 5·74 Removing the voJtages V 2 and V3 by short-circuiting them we have Similarly, Hence, 5.77 Find Vo for the operational amplifier circuit shown V+ = V_. III Fig. 5-75 by superposition. Given 1+ = 1_ = 0 and 1<'11<, + VI R., V. I. -+ V2. I , Sl \t. R. . . -+ V D Fig. 5·75 Removing V2 and applying VI we have V· ,,: - R2 V o Removing VI and applying V2 yields (with 1+ = I RI I = 0 and V+ = V_) \1'''== R2 V o RI 2 Hence, Vo = V ~ + V;. = R (V2 - VI) / 1 5.78 By superposition, calculate the current 1 in the circuit shown in Fig. 5·76a. , First, we remove the 70-V source to obtain the circuit of Fig. 5-76b. in parallel and then write the mesh equations as (~~ + 4 )/; + 2/; - ~~ I' = () We combine the 2-0 and 20·0 resistors (~~ + lO)I' - ~~ I; = 50 + 2/; NETWORK THEOREMS D 107 If I 1oS"L- + 1-----'v\'fV----.-<2.1 l.fl.- I IOf/... (a..) (h) /o.n.. Fig. 5-76 (c) Solving for ]' yields ]' = 4.575 A Next, with the 50-V source removed, the mesh equations for the circuit of Fig. 5-76c become 22/2 - 21" - 20/~ = 70 2/~ 121" - 2/2 = 24/~ - 20/2 = -2/~ Solving for I" gives I" = 3.425 A 1 = ]' + I" = 4.575 + 3.425 = 8.0 A Hence, 5.79 Determine the Thevenin voltage at the terminals ab of the network shown in Fig. 5-77a by superposition. {/2... 2A /2{1- 4.f1- w~ 20~ + /-z.v 1 b (h) (a..) Fig. 5-77 , To obtain the Thevenin voltage VTh , we open-circuit ab, convert the delta-connected resistors to an equivalent wye, and transform the 2-A current source to a voltage source. Hence we obtain the circuit shown in Fig. 5-26a, which is redrawn in Fig. 5-77b. Now, with the 12-V source removed, we have I' = 12 24 +2+2+4 = 2 1. A 108 D CHAPTER 5 Similarly, with the 24-V source removed, we obtain lr, /" = ----:~-- = 0 6 A 12 + 2 +2 + 4 Hence, ] = ]' - /" = 1.2 - 0.6 = 0.6 A . VTh == 24 - 0.6(12 + 2) and = 15.6 V Verification: 24 - 12 ] = ]' - /" = 5.80 -I" +2+2+4 L, = 0.6 A Find the Thevenin voltage VTh shown in the circuit 1)1 Fig. 5-27a by superposition. Removing the lO-V source, converting the lO-U wye-connected resistors, and combining the series-parallel resistors yields the circuit of Fig. 5-78a. See also Fig. 5-27 c and d. 7.sn 1 I" I' (b) (a.) , Fig. 5-78 From Fig. 5-78a we obtain , (7.5 -1 4.286 ) ] = 10 30 +-7-.'5 + 4.286 = 2.82 A Next, we remove the lO-A source and insert the lO-V source in series with the 5-0 resistor to obtain the circuit of Fig. 5-78b. Then, [0 ] = -5+---C-O{[--=30:-C(7=-.5=-+-----,----c30W(:1O + 7.5 + 30]} = 0.4615 A And 130 n VTh 5.81 = V,o n = -2.615 = /" - ]' = 0.205 - 2.82 = -2.615 A x 30 = -78.45 V Determine the Thevenin voltage shown in the circuit I)f Fig. 5-28a by superposition. 4.n. -"',....- - • vv L'" .. I,' I,' Ion Ion.. (a.) , ~2fl.- 1)2.. 1~' 2(L I' 4-fl- IDJ2. (6) (c) Fig. 5-79 With the notation of Fig. 5-28a, we have VTh 32 + V2 - - VI = 0 Therefore, to find VTh we must determine VI and V;, wilich requires the calculation of the currents in the 2-0 and 10-0 resistors. We apply one source at a time as shown in Fig. 5-79a to c. From Fig. 5-79a, ]; =OA From Fig. 5-79b, ]~ =9A ]" =_9(_4_) = -6 A 2 - 2 +4 NETWORK THEOREMS D From Fig. 5-79c, 1';' = 4A Thus, I I = I; + I'; + 12 = I~ + r~ + r~ = ~ - 6 + 0 = - ~ A = 0 + 9 + 4 = 13 A VTh = 32 + VI - V2 = 32 + 130 + Hence, 5.82 I'~ I'~=OA 1= 163.33 V Obtain the Th6venin voltage shown in the circuit of Fig. 5-29a by superposition. 4-11- r"':'ro=', J ~5-:;;' 5"11 (/1.) , 1111 I 5,n I," (6) (c) Fig. 5-80 From Fig. 5-29a, 32 + VTh + V2 where VI = Is 0(5) and - VI = 0 V2 = 110 0(10) From Fig. 5-80a, I; = 0 I~ = 0 From Fig. 5-80b, 11 (10+4) 11 =9 10+4+5 =6.63A From Fig. 5-80c, I'; = 4( 10 +1~ + 4) = 2.11 A Iso = 0+ 6.63 + 2.11 = 8.74A VI = 8.74 x 5 = 43.7 V Or V2 = 4.26 x 10 = 42.6 V or Finally, 5-83 1 100 = 0 + 2.37 + 1.89 = 4.26A VTh = V1 - V2 -32=43.7-42.6-32=-30.9 V Determine the VTh of the circuit of Fig. 5-32a by superposition. /fL t ____-L(a.) (b) In. (.0- I:' j"' '+ (c) Fig. 5-81 109 110 0 CHAPTER 5 , From Fig. 5-32a, VTh - Vs n -- 40 + V6 n - V2 n = 0 To find the volt ages across various resistors we mu~.t determine the currents through them. we draw the circuits of Figs. 5-81a to c each having one source, others being removed. From Fig. 5-81a, ]' =, 40 A From Fig. 5-81b, 1'; = 0 For superposition 7 1 From Fig. 5-81c, ]11' 1 = 10(_1_) = 10 A 1+6 7 ] 6 n , = ]1 /I' - ] 1 ,= ]~ From Fig. 5-81a, l~ From Fig. 5-81b, From Fig. 5-81c, 1'; = 10 A ]sn 10 _ 30 40 -'7 - "7 - "7 A =OA =, -5 A = ]; + 1';= --5 + 10 = 5 A Vs n = ] s n(5) = 5 x 5 = 25 V From Fig. 5-81a and b, From Fig. 5-81c, ] 2 VTh Hence, 5.84 ~'2 n = 1'; = 10 A - 25 - 40 + 1~0 - n = ] 2 n(2) = 10 x 2 = 20 V ~:O = 0 or VTh = 59.28 V Determine the current through the 10-0 resistor of the circuit of Fig. 5-38a by superposition. (a.) , (~) We remove the current source to obtain the circuit of Fig. 5-82a. 100 = 101' -- lOVx + 51' Fig. 5-82 Thus, Vx = 51' ],=_100 A 35 Next, from Fig. 5-82b, where we have removed the voltage source, we obtain Thus, 100 = 101' - 501' + 51' or 10 = Vx +]" 5 or 5.85 I" = 450 A 35 or ] = I" - I' ,= ~~~ + 100 35 35 = 1571 A . (from right to left) Find the current in the 6-0 resistor of the circuit of Fig. 5-39a by superposition. '~~~~Jt'6.f1- I" ~--------------- ca.) (6) Fig. 5-83 NETWORK THEOREMS , Retaining one source at a time we obtain the circuits of Fig. 5-83a and b. v, = or 0 111 From Fig. 5-83a, (I) f f= 2A From Fig. 5-83b, 3 = Vx + Vx + 2Vx 1 6 Or 6 6 16 Cl = I' + I" = 2 + 1=3 A Hence, 5.86 1"= Vx +2Vx =2+4=IA Vx =2 Find the current in the 4-0 resistor of the circuit of Fig. 5-84a by superposition. I ,./2... SA (a.) ..l' "-~l'r /rL Z;.n 41' 54.f'L '_____--"- -- - J ~[t- (b) r In ~ /~r \. t ·;l->V--1 V sA I'fi 3JL. 4I" ...p- '-----------!._~ __. L ____ ___ j Fig. 5-84 Cc) , From Fig. 5-84b we have or And I" = 15 A Or 5.87 I' = 12 -13.5 = _ ~ A 1 2 8 5 + I" = 41" + V= 13.5 V ~ 3 1"= - ~ 1 3 15) =15A I 4Cl =4(I'+I")=4 ( --+2 8 . Verify, by nodal analysis, that the current I is one-fourth the current in the 4-0 resistor of the circuit of Fig. 5-84a. , From Fig. 5-84a, 12- V 1=-1 or V= 12- I 112 0 CHAPTER 5 V 1+5 = 3 and 5.88 12 - I + 41 or I + 5 = - 3 - + 41 31 + 15 = 12 - I JJ' 3 1 + 121 or I = "8 A = 4 14 Cl Find the current in the 3-0 resistor of Fig. 5-84a, by superposition. , From Fig. 5-84b, I' 3 Cl = I' - 4]' = - 3I' = .- 3 (~ .- 2~) = ~2 A (from Prob. 5.86) From Fig. 5-84b, I;Cl 5.89 = ]" + 5 - 41" = 5 - 31":= 5 -- 3( ¥) = - ~ A (from Prob. 5.86) Apply superposition to the network of Fig. 5-85a to find the current in the 1-0 resistor. In. 20A ''''A '~,~:rc~ 1'1 5bA (a.) , I; = 20(:. + ~ + From Fig. 5-85b, I"=16( 1 . From Fig. 5-85c, J Fig. 5-85 = 10 A 3 + 21+2 ) =16/3A 11 = 11 Cl = I; - I~ = 10 - 16/3 = Hence, 5.90 (c.) (IJ) .!!. A 3 Find the voltage across the 2-0 resistor of the circuit of Fig. 5-85a by superposition. , From Fig. 5-85b, (from Prob. 5.89) "_ /1 1 + 3 ) _ 32 12 - 16 \ -I + 3 + 2 - 3' A From Fig. 5-85c or 5.91 12 ~ ~ = I~ + I; = 10 + 3' = 3'- A V2fi ~ I~ = 12(2) = 3' (2) = 3' = 41.33 V In a linear resistive circuit, whereas superpositior. of currents is valid, show that superposition of powers is not valid. , Let currents I' and]" flow through a resistanc:e R due to one source at a time in the network (with an other sources removed). Then the total current I = I' +]" (assuming two sources only for simplicity). Power dissipated in R is 5.92 Refer to the circuit of Fig. 4-5. Given R, = 20, R2 = 30, R3 = 20, R4 = 60, and R5 = 10. If E= I V, calculate the current 15 . Then, apply superposition and reciprocity to calculate the current through Eif a 2-V source is connected in series with R5 , with the negative polarity connected to b. , To calculate 15 we redraw the circuit in Fig. 5-86a. For the three mesh currents we have -21 1 - 612 + 813 = 1 NETWORK THEOREMS 0 113 Fig. 5-86 (0.) Solving for I), 12 , and 13 yields I) = 0.1757 A 15 = and I) - 12 = -0.0405 12 = 0.2162 A 13 = 0.331 = current through E Now, by reciprocity, if the I-V source is transferred to the branch bd as shown in Fig. 5-86b, then the current 15= -0.0405 A will flow through eta. By superposition, a 2-V source will produce -2 x 0.0405 = -O.OBl-A current through eta. Hence, the new current through eta with a 2-V source in the network is I 5.93 Refer to Fig. 5-86b. , bd = 0.331 - 0.081 = 0.25 A By mesh analysis verify the result of Prob. 5.92. For the three mesh currents we obtain 51) - 12 - 213 = 2 Solving for the three currents yields 13 = 0.25 A I) = 0.5 A 5.94 For the circuit of Fig. 5-86b show by superposition that no current flows through the 3-0 resistor. c -'-"''''-- I" ~ o.Sn. 11. (a.) ,fl- c:t -1 2V , 0.40 IV (b) The circuits to which we apply superposition are shown in Fig. 5-87 a and b. I = I 2 _~ 1 + [(2 x 6)/(2 + 6)] + [(2 x 3)/(2 + 3)] - 3.7 A Fig. 5-87 From Fig. 5-87 a, , 2 ( 2 ) I = 3.7 2+3 =0.2162A In Fig. 5-87 b we show the abd delta-connected resistors (with the voltage source removed) transformed to an equivalent wye. Then I _ 2 - Hence, 1 _ 0.8 + {[(3 + 0.4)(6 + 0.4))/[3 + 0.4 + 6 + 0.4))} - 0.331 A 6 + 0.4 I II = 033 . 11 3 + 0.4 + 6 + 0.4 = 0.2162 A 13 n = l' - ]" = 0.2162 - 0.2162 = 0 114 5.95 D CHAPTER 5 By superposition, find the current in the 1-0 resIstor of the current-excited bridge circuit shown in Fig. 5-88. , With the 5-A current source removed, 1,=lOA and With the IO-A current source removed, 6+3 I'; = 5 6 + 3 + 1 = 4.5 A 1 = 1'; - I; = 4.5 - 6 = -1.5 A b ~----~~r-----~ lOA 5.96 Fig. 5-88 Fig. 5-89 Calculate the current in the 3-0 resistor of the circUit of Fig. 5-89. From this result, by reciprocity theorem, determine the current in the 2-0 resistor if the 24-V source is removed and a 6-V source is connected in series with the 3-0 resistor, with the positive polarity connected to a. , From Fig. 5-89, 24 1= 2+[(3X6)/(3+6)[ =6A 6 1 30 =6 3 +6 =4A Thus, a 24-V source in series with the 2-0 resistor produces a 4-A current in the 3-0 resistor. By reciprocity, a 24-V source in series with the 3-0 resistor will produce a 4-A current in the 2-0 resistor. By linearity, a 6-V source will produce I-A current in the 2-0 resistor. Because of the specified polarity, the current will flow from right to left. 5.97 Calculate the current 1 I in the circuit of Fig. 4-21 a by superposition. 4Q / 7 - -__ ~'v'\ I,'·'·/~6n. 4~ \~{l-~ In _____ _ 2Q Ca.) (h) (c) Fig. 5-90 , From Fig. 5-90a, I' = , 4+ 20 = 25 4[(6 x 3)/(6+3)+(2 x 1)/(2+ I)] 7 4 + [(6 x 3)/ (6 + .3)] + [(2 x 1)/ (2 + I)] = 3.571 A NETWORK THEOREMS 0 115 From Fig. 5·90b, 30 255 I~ = 6 + [(8/3 x 3)/(8/3 + 3») = 63 I~ = ~ (4.0476 3 +38 /3) = 4.0476 A = 1.071 A Finally from Fig. 5·90c, I'" = 3 10 50 3 571 A 2+{[1(2+2»)/(1+2+2)} = 14 = . 1I Hence, 5.98 = I; + I';' = ~ [3.571( 1 + ~ + 2)] = 0.357 A I'; + I';' = 3.571 + 1.071 + 0.357 = 5.0 A Find the Thevenin voltage VTh shown in the circuit of Fig. 5·91 by superposition. , From Fig. 5-91, With the 6-A source removed, With the lO-A source removed, I~ = 6C + ~ + J = 3 A 5.99 I~ 1I = or - I; = 3 - 1.67 = 1.33 A or VTh = 1.33(2) = 2.67 V Verify the result of Prob. 5.98 by nodal analysis. , For node 1, v V-V I Th 1 6 + 10 = .....! + 3 or For node 2, 10 + VTh = VI - VTh 2 1 or Solving for VTh yields VTh = 2.67 V f--:-+--V'VV_ '). . I," I, 2.n 1.' GA + V Tt< Fig. 5-91 5.100 Fig. 5-92 Transform the current sources of the circuit of Fig. 5·91 to equivalent voltage sources, and hence obtain the voltage VTh • , The new circuit is shown in Fig. 5-92 from which 18 -10 1= 3+1+2 8 =6 8 A 8 VTh = 1(2) = -6 x 2 = -3 = 2'67 V CHAPTER 6\\ Capacitors ~ 6.1 The circuit element C, measured in farads (F) ancl shown in Fig. 6-1, is called the capacitor for which the voltage-current relationships are + v ~ v= , 6.2 If the voltage across a 50-ILF capacitor is i= C . !I Lt idt= o 1 6 100 x 10- , dq = --- what is the current through the capacitor? A 6 X l' i dt 100 sin 200t = 5000 sin 200t is switched through c. 100-ILF capacitor at t = O. Determine the rms voltage Lt lOcos377tdt=- 10 sin 377t . 6 = 265.25 sm 377t V 100 >:377 x 10- 0 V= 265.25 ';-;:;2 =187.6V VL. Find the charge associated with the capacitor and the current of Prob. 6.3. {t q= Jo {t 10 i dt = 10 Jo cos 377t dt ,= 377 sin 377 t = 26.525 sin 377t mC Determine the instantaneous and average powers in the capacitor of Prob. 6.3. 26~2.5 p = vi = (265.25 sin 377t)(1O cos 377tl ,= With sin 2 x 377t = 1326.25 sin 754t W T= (1/60) s the period of the current wave, Pav ,= - LT p dt = 0 ~~ "'r J (0 Obtain an expression for the energy stored in a capacitor C (F) charged to voltage V (V). , In a time dt, energy dW is given by dW = p dt = vi dt = vC 6.7 q= cit q = Cv = 50 >: 10 or C A current i = 10 cos 377t A across the capacitor. , 6.6 C -dv dt v= 101) sin 200t V, 1 v= 1 'v=-C 6.5 1= = 50 x 10- 6 (100 x 200 cos 200t) = cos 200 t Since and 6.4 . i dt + k Since current is defined as the rate of change of charge, we may write i = dq/dt, where q is the charge in coulombs (C). From the data of Prob. 6.1, find an expression for the charge on the capacitor. , 6.3 ~~ f Fig. 6-1 ~~ dt or W= C v 1 Lo v dv = -2 CV 2 J Determine the energy stored in the capacitor of Prc.b. 6.1. , Since W= ~CV2, we have, from Prob. 6.1, wet) = ~ C[ v(tW = ~ x 50 x 1O- 6 ( 101) sin 200t)2 = ~ x 50 x 1O- 6 (loo? sin 2 200t = 25 x 10- 2 x 6.8 116 i (1 - cos 2 x 200t) = 125(1 - cos 400t) mJ What are the time-average and peak values of the tnergy stored in the capacitor of Probs. 6.1 and 6.7? CAPACITORS , D From Prob. 6.7, wet) = 125 - 125 cos 400t The time-average value of the second term is zero. ml Hence w(t)av = 125 ml The peak energy occurs at 4OOt= TT. Or, at t=(TTI400)s, we obtain w(t)lmax = 125 + 125 = 250 ml 6.9 A voltage pulse given by o t~O v(t)= 2tV {4e-(t-2) V 20~t~2 ~ t , is applied across a 10-JLF capacitor. Sketch the voltage across and the current through the capacitor. i= C dv = 0 t<O dt = 10 x 10- 6 (2) = 20 JLA 0<t<2 = 10 x 1O- 6 [ _4e-(t-2)] = -40 e-(t-2) 2<t JLA Hence the sketches shown in Fig. 6-2. 4 +---+---_~>--_---=+=~~f (s) ~ o,.., )..CJ + ____--, Iv Fig. 6-2 6.10 , Sketch the power and energy curves from the data of Prob. 6.9. p= vi t~O =0 0~t<2 = 2t(20) = 40t 2<t w= !Cv = ! 2 x 10 X 10- 6 X 0= 0 t~0 =!xlOxlO- 6 (2t)2=20t 2 JLl 0~t~2 6 = ! x 10 x 1O- [4e-(t-2)j2 = 80e -2(t-2) JLl See Fig. 6-3. 2~ t 117 118 D CHAPTER 6 2 6.11 Fig. 6-3 Obtain the equivalent capacitance for two capacitances C, and Cz connected Generalize a and b to the case of n capacitances. I (a) v= ~ es f i dt = v, + V z = ~ 1 f f i dt + (~ 2 i dt (a) in series and (b) in parallel. whence n capacitances in series: . (b) 1 = C ep dv . . C dv dv dt = I, + 12 = 1 dt + C2 dt whence n n capacitances in parallel: Cel' ,= L Ck k'"1 6.12 A 40-JLF capacitor is charged to store 0.21 of energy An uncharged 60-JLF capacitor is then connected in parallel with the first one through perfectly conducting leads. What is the final energy of the system? I The initial charge on the 40-JLF capacitor is obtained from 0.2 = or Q2 2(40 x 10 6) W= Q2/2C (Prob. 6.21); thus, Q = 4 X 1O- 3C When the capacitors are connected in parallel, the common voltage V is given by V = total Q total C 4 x 10 - 3 (40+60)10 = 40 V == _ 6 Then Final energy in 40-JLF capacitor = H40x 10- 6 )(40)2,=0.0321 Final energy in 6O-JLF capacitor = H60 x 10- 6 )(40)2 := 0.048 J Final total energy = 0.032+ 0.048=0.081. The energy lost, 0.2-0.08=0.121, the charges on one another in spreading out over the, two capacitors. 6.13 represents work done by What is the equivalent capacitance between the terminals ab of the capacitive system shown in Fig. 6-4a? I Using the results of Prob. 6.11, we obtain Fig. 6-4b to d from which C"q =', 14 C CAPACITORS 3C a 10 IOC C3"'H,)C~b~ b 119 c ~~I C D 1:1-6 12.( l.1.; IOC --I \- ----I fA.. ~ --- [ ----11 ----L b bX/1.,\C,-:.-1C ( 6.14 (lO+4-)C ~ 14 C. _r-----i{ r----~ j (rh -r:;I1. ) Fig. 6-4 A 40-JLF capacitor is connected in parallel with a 60-JLF capacitor and across a time-varying voltage source. At a certain instant, the total current supplied by the source is 10 A. Determine the instantaneous currents through the individual capacitors. I Since the capacitors are in parallel, the voltage v across them is related to the currents i l and i2 by . II = C dv I dt ~ or Cl C2 i2 But il . 12 = C dv 2 dt 40 60 + i2 = 10 A Hence, 6.15 A 50-JLF capacitor is charged to 300 JLc. An uncharged 1oo-JLF capacitor is then connected in parallel with the first capacitor. Evaluate the charge transferred to the 1oo-JLF capacitor. I For two capacitors Cl and C2 , connected in parallel and charged to a voltage V, we have CIV=QI Cl or C2 C2 V=Q2 QI Q2 From the data, 50 100 Hence, 6.16 and A combination of four capacitors is shown in Fig. 6-5. of 0.5 JLF. (transferred to the 1oo-JLF capacitor) Find the value of C to obtain an equivalent capacitance Fig. 6-5 120 D CHAPTER 6 I Applying the rules for series-parallel capacitors to Fig. 6-5 we obtain = (C + 0.6)(0.2 + 0.8>. C eq = 0 5 C + 0.6 + 0.2 + 0 . 8 ' or C + 0.6 = 0 5 C + 1.6 . Hence 6.17 Find the equivalent capacitance of the combination of capacitors shown in Fig. 6-6a. c CC) Fig. 6-6 I The network reduction procedure is shown in Fig 6-6b to e from which Ceg 6.18 = 2.85 p,F A series combination of two capacitors Cl = 20 p,F and 100 V. How is this voltage divided across the capacirors? Cz = 40 p,F is connected across a dc source of I Since the capacitors are in series CIVI = CzV, VI = 66.67 V Hence, 6.19 or Vz = 33.33 V Express the two series-connected capacitors of Prob. 6.18 as an equivalent capacitor. equivalent capacitor? Also, find the charges on Cl and Cz' I CI C2 C eq = Cl QI + Cz = CIVI = = 20 x 40 20 + 40 20 x 66.67 X = 13.33 p,F 10- 6 = 1333 p,C What is the charge in the Q = CV = 13.33 x 100 x 10- 6 = 1333 p,C CAPACITORS 6.20 ! Wl = ~ Cl V~ = Wz = ~ x 40 x 20 6 1O- (66.67)Z = 44.446 mJ X 1O- 6 (33.33)z = 22.22 mJ X Wl + Wz = 44.44 + 22.22 = 66.66 m] Q=CV X 13.33(100)z X 10-6 = 66.66 mJ W= 2: V=Q C or 1 Hence, C (Q)2 C Q2 = 2C An uncharged 50-JLF capacitor is connected to a 40-mA constant current source. the capacitor after (a) 5 JLS and (b) 10 ms. I Since Q= f Determine the voltage across Q = It = CV i = I (a constant) and i dt I V= - ( C or (a) V= (b) V= 40 X 10- 3 6 50 x 10 40 X 5 x 10 X lOx 1O- 3 =8V 10- 3 50 x 10 6 -6 X = 4 mV The voltage across a 50-JLF capacitor rises at a constant rate of 10 V/ms. Calculate the current through the capacitor. What is the increase in the charge at the capacitor plates as the voltage increases by 60 V? I 6.24 ! From Prob. 6.6, But 6.23 We = Obtain an expression for the energy stored in a capacitor C in terms of the charge Q on it. I 6.22 121 What is the energy stored in each capacitor of Prob. 6.18? Verify that the total energy stored in the two capacitors equals the energy stored in the equivalent capacitor of Prob. 6.19. I 6.21 D i= C~~ 6 =50xlO- x 1~~3 =500mA ilQ = C ilV= 50 X 10- 6 X A 200-JLF capacitor is charged to store 9O-mJ energy by a constant current of 0.1 A. the capacitor? I 60 = 3000 JLC What is the voltage across Since 1 2 W=- CV 2 = V ~ 2W C = ( 2 x 90 x 10 - 3) 112 = 30 V 200 X 10- 6 6.25 How long does it take to charge the capacitor of Prob. 6.24, and what is the value of the charge on the capacitor? Q 6x 10- 3 (= - = =60ms and I 0.1 6.26 A 30-JLF capacitor is charged by a voltage source having a saw tooth waveform shown in Fig. 6-7. the charging current. v /0 --+-----~~----~------~------~·r ? 7nS Fig. 6-7 Determine 122 D CHAPTER 6 I dv 10 , = ---~- = 0.5 x 10- Vis dt 20 x le 3 - From Fig. 6-7, dv i = C dt = 30 >( 0.5 x 10- 6 x 10 3 = 15 mA Thus, which remains constant. 6.27 Refer to the data of Prob. 6.18. After the series·connected capacitors are fully charged, the dc source is removed and the capacitors are then connected in parallel. Calculate the final charge on each capacitor. I From Prob. 6.19, Q = 1333 f..tC. When connected in parallel, Q V= - Thus Hence, 6.28 C QI = CIV= 20 x 22.217 X 10- 6 + Cz = (20 + 40) x 10- 6 = 60 l333 60 22.217 V /LC Q2 = C2 V= 40 x 22.217 = ----- = = 444.33 C = Cl X 10- 6 = f..tF. 888.66 f..tC The voltages across two capacitors Cl and C2 conrlected in series are 4 V and 2 V, respectively, when the combination is connected across a 6-V source. A 2-/LF capacitor is now connected across Cl and the voltage now measured across C 2 is 3 V. Evaluate Cl and C2 + 4v - + It! + Fig. 6-8 I From Fig. 6-8a, 4C, From Fig. 6-8b, (Cl = 2C2 + 2)3 = 3C2 Solving for Cl and C 2 yields 6.29 A 100-f..tF capacitor is charged to 100 V. It is then connected to a 400-f..tF uncharged capacitor. energy is dissipated in the connecting leads? I Wi = ~CIV2 Initially energy =\ x 100 x 10- 6 (100)2 = 0.5 J 6 When connected in parallel, the initial charge Qi = Cl V= 100 X 10- X 100 = 10 mC parallel combination C = Cl + Cz = (100 + 400) f..tF capacitor. The common voltage then becomes V= Q. = C is redistributed to the 3 lOx 10- 500 x 10 =20V 6 W2 = ! C2V2 = ! Wf = WI Final energy: W. - Energy dissipated: 6.30 How much >< 400 X 1O- 6 (20)z = 0.08 J + W2 =, 0.02 + 0.08 = 0.1 J "'~ = 0.5 - 0.1 = 0.4 J A 20-f..tF capacitor is connected in parallel with a 30-f..tF capacitor. At t = 0 the combination is connected across a voltage source. At a certain instant the \'oltage begins to increase and the current drawn from the source is 5 mA. Determine the current division be1:ween the capacitors. Also calculate the rate of change of voltage across the capacitors. I . 1 I dl' = C -dv = 20 x 10 -6 --1 dt dl . 12 = C2 dv dt = 30 x 10 -6 dv dt CAPACITORS i, =2mA Hence 6.31 123 3 X 10- 3 30 x 10 6 = 100 Vis 2 X 10- 3 dt = 20 x 10 6 dv and i2 = 3mA D A 30-JLF capacitor is sequentially charged by a 12-V battery and then discharged through a resistor R by switching between positions 1 and 2 (Fig. 6-9) at a rate of 100 operations per second. Determine the average current through the resistor. I ~v Fig. 6-9 With the switch at position 1, Q = CV = 30 X 10- 6 x 12 = 360 JLC Provided R is small enough, essentially all this charge passes through the resistor during discharge. Thus, Q through R in 100 operations = 360 x 10- 6 x 100 = 36 mC, which takes 1 s. Hence the average current = 36 mCI 1 s = 36 mA. 6.32 How large can R be in Prob. 6.31 and still allow a 99% discharge of the capacitor in the time available (10- 2 s)? , By KVL, the discharge process is governed by Thus, the critical value of R satisfies 100 = e3331R or 6.33 -R dqldt = qlC, or which integrates to q = Q e- tlRC• R=~""'72n In 100 For the circuit shown in Fig. 6-10, determine the voltages v, and v 2 ' with S open . . L -t , '\T ; loo $;"'" (oaot Fig. 6-10 With S open, Ceq (1 + 2)6 = 1 + 2 + 6 = 2 JLF v2 = 6 x VI ~o = i = i3 = Ceq 6 f i3 dt = 6 x V - V2 ~~ = (2 x 10- 6 )(100)(1000 cos 1000t) = 0.2 cos 1000t ~O 6 f 0.2 cos 1000t dt = 33.33 sin 1000t = (sin 1000t)( 100 - 33.33) = 66.67 sin 1000t V Fig. 6-11 V A 124 6.34 D CHAPTER 6 For the circuit of Prob. 6.33, with S closed, find C sllch that the current from the source has a peak value of 0.25A. I From Fig. 6-10, with S closed, Ceq . I 6.35 dv = Ceq dt = 18+3C -6 9 + C (100 x 1000 cos 1000t) x 10 = 0.25 cos 1000t The capacitors may be combined as shown in Fig. 6-11. VI = Cl I v2 _ I C Note that With I VI 2 f f i dt = 1 3 x 10 i dt -- 15 x 110 6 6 f f VI Since A or C=9 p,F and v, from Fig. 6-10. i = 0.25 cos 1000t A 0.25 cos 1000t dt = 83.33 sin 1000t V 0.25 cos 1000t dt = 16.67 sin 1000t V from Prob. 6.34, + v2 = V = 100 sin lOOOt V. C = 9 p,F and S closed, determine the curre:l1 lhrough each capacitor of the circuit of Fig. 6-10. With the currents shown in Fig. 6-10 and the values il = i2 VI and v 2 calculated in Prob. 6.35, we have 2 X 10- 6 ~ (83.33 sin 11)OOt) = 166.66 cos 1000t mA 1 X 10- 6 ~ (83.33 sin IOOOt) = 83.33 cos 1000t mA = i,- = 6 X 10- 6 -ddt (16.67 sin WOOt) = 100 cos 1000t mA ~ (16.67 sin WOOt) = mA i4 6.37 (1+2)(6+C) 18+3C 1 + 2+6+C = 9 + C For the value of C determined in Prob. 6.34, with S closed, find I 6.36 = = 9 X 10- 6 150 cos 1000t Find the capacitance at the terminals of the wye-connel;ted circuit shown in Fig. 6-12, taking any two terminals at a time. I 6.38 The capacitors of Prob. 6.37 are reconnected in del1a as shown in Fig. 6-13. terminals taking any two terminals at a time. I Cab = C2 C I C3 + C +C I C 3 Fig. 6-12 6.39 = ca Determine the capacitance al the C I + CC 2 3 C 2 + C3 .b _ _ _ _- - - . J Fig. 6-13 Let three capacitors, each having a capacitance Cy , be connected in wye (Fig. 6-12). If three capacitors, each having a capacitance Cd' are connected in delta (Fig. 6-13), determine the equivalent relationships between the two connections. CAPACITORS I D 125 From Prob. 6-37, From Prob. 6.38, For the two to be equivalent we must have Cy =3Cd or 6.40 and Conductor-to-conductor and conductor-to-sheath capacitances of a three-core cable are shown in Fig. 6-14a. Determine the per-phase capacitance (defined as the net capacitance between a conductor and sheath and ground). I Figure 6-14b shows all the capacitances and their interconnections. The delta-connected capacitances Cc are then converted to an equivalent wye, shown in Fig. 6-14c, using the results of Prob. 6.39. Hence Cn = Cs + 3Cc shown in Fig. 6-14d. \ C b S c c, c:c+;.cC h ~ Fig. 6-14 126 6.41 D CHAPTER 6 In a test on a three-conductor cable, the three conductors are bunched together and the capacitance measured between the bunched conductors and sheath is Cl' [n a second test, two conductors are bunched with the sheath, and the capacitance measured between thes;: and the third conductor is C 2 • Determine C, and Cc, shown in Fig. 6-14a. I C= 3Cs From the first test we obtain Cz = Cs + 2Cc From the second test we get Hence, 6.42 In a certain test, the capacitance C3 is measured betwel~n two conductors, with the third conductor connected to the sheath. Determine Cn shown in Fig. 6-14d. I For the connection indicated we have C 3 = C c + ~Cc+ ~Cs= H3C C, Hence, 6.43 + C.)= = ~Cn from Prob. 6.40 2C3 Find the capacitance between the terminals ab of the circuit shown in Fig. 6-14b. I First the delta-connected capacitors are transformed to an equivalent wye, as shown in Fig. 6-14c, from which by symmetry Hence, 6.44 Convert the wye-connected capacitors in Fig. 6-14b to an equivalent delta and verify the result of Prob. 6.43. I From wye-delta equivalence we have and the circuit becomes that shown in Fig. 6-15 from which Cab = Cc + ~Cs + HCc + ~CJ =HCc + 2Cc + ~Cs + ~C,) =: H3Cc + C,) ,--~~--- ..!.C ~ 6.45 s Fig. 6-15 The capacitances per kilometer of a three-wire cable are 0.90 p,F between the three wires bunched and the sheath and 0.40 p,F between one wire and the other two connected to the sheath. Determine the line-toground capacitance for a 20-km-long cable. I From Prob. 6.41 Cl = 0.9 p,F and C2 = =: 3Cs 0.4 = Cs + 2Cc = 0.3 + 2C, Cs = 0.3 p,F /km or Cc = HO.4 - 0.3) = 0.05 p,F /km From Prob. 6.40 Cn = C, + 3Cc = 0.3 + 3 x 0.05 = 0.45 ,uP/km (Cn )20km = 20 x 0.45 = 9.0 p,F CAPACITORS 6.46 From Prob. 6.42, Cn = 2 x 0.6 = 1.2 p,F Ikm or Find the capacitance between any two conductors of the cable of Prob. 6.45. I From Prob. 6.43, Cab = H3Cc + C,) = H3 x 0.05 + 0.3) 6.48 X 20 = 4.5 p,F In a three-conductor cable a short circuit occurs between the conductor a and the sheath. What is the capacitance between the conductors a and b, if Cs = Cc = C? Compare the result with that of an unfaulted cable. Fig. 6-16 , The circuit of Fig. 6-14b changes to that of Fig. 6-16 from which C(C + C) 8 C (C + C ) C =C+C+ cc s =C+C+ =-C ab c s Cc + Cc + Cs C+C+C 3 For the unfaulted cable, from Prob. 6.43, Cab = H3Cc + CS> = H3C + C) = 2C 6.49 127 The capacitance between any two conductors of a three-core cable, with the third conductor grounded, is 0.6 p,F Ikm. Calculate the line-to-ground capacitance for a 25-km-long cable. , 6.47 D Find the capacitance at the terminals ab of the circuit shown in Fig. 6-17a. (b) I Fig. 6-17 By series-parallel combination we obtain the circuit of Fig. 6-17b from which Cab 3x9 = 9 + 3 + 9 = 11.25 p,F 128 6.50 D CHAPTER 6 What is the equivalent capacitance at the terminals ah of the circuit shown in Fig. 6-18a? (a..) a. _-__....( b----------+-~~ ---------~ (I» ISy+ IS·tf H~,I J. "tf ----+-.-I~__ _ 1 J:- t;;)Nf (c) , The circuit reduction steps are shown in Fig. 6-ISb and c, from which Cab 15[(15 x 6)/(15 + 6) + 15] = 15 + [(15 >: 6:1/(15 + 6) + 15] = 8.44 p,F Fig. 6-18 /1CHAPTER 7 L/ 7.1 Inductors The circuit element shown in Fig. 7-1 is known as an inductor, and has the voltage-current relationship ±J di or v=L i= v dt+ k dt where L is the inductance in henries (H). A 50-mH inductor carries a 5-A current which reverses in 25 ms. What is the average induced voltage across the inductor? i. A I -2S' I Fig. 7-1 Fig. 7-2 , The induced voltage v is zero except at the instant of current reversal, when it is infinite. interval T = 25 ms (Fig. 7-2) we have i(to + T) - i(to) = 7.2 1 J,'O+T T L,o v dt = L vov or v =L ov i(t + T) - i(t ) 0 0 T = 50 X 10- 3 However, over an 5 - (-5) 25 x 10 3 = 20 V Plot the voltage across an inductor L if the current through it is as shown in Fig. 7-3a. , Since v = L dildt, we obtain the voltage as shown in Fig. 7-3b. (a) v I Fig. 7-3 (b) 7.3 Determine the equivalent inductance of n inductors , Lp L 2 , ••• , in series. With common current i, we have v = Les di di di di dt = Ll dt + L2 dt + ... = (Ll + L2 + ...) dt 129 130 D CHAPTER 7 n Hence 7.4 Les What is the equivalent inductance of n inductors , f v dt f = -.l LJ v dt + _L J' L2 Two inductors, ductance? L J = 30 mH and L z = 60 mH, , , f + ... = (-.l + -.l + ... ) LJ L2 v dt 2: -Lk1 b'J art' connected in parallel. What is the equivalent in- 2 The current through a 60-mH inductor is given by v=L ~ = 60 X 1== 15 sin 377t A. Determine the associated power. = vi = (339.3 cos 377t)(15 sin 377t) = 2544.75 sin 754t ~ v 3 10- x 15 >< 377 cos 377t = 339.3 cos 377t Determine the energy stored in an inductor L having =L .1 , Find the energy stored in the inductor of Prob. 7.7. 3 wet) = ~ Li 2 = ~ x 60 X 10- (15)2 sin 2 377t Energy = W = = W current I. ~~ = vi = Li Power = p = 3.375(1 - cos 754t) Determine the = L J + L}= 20 + 10 = 30 mH Les Voltage = v 7.10 dt The combination of inductors of Prob. 7.5 is connected in series with a lO-mH inductor. equivalent inductance. p 7.9 IJ L J L2 (30)(60) Lep = L + L- '" . 30 + 60 = 20 mH J 7.8 in parallel? L], L z , ... , L ..p or 7.7 Lk J 1 '~ - - == Hence, 7.6 2: <=, With common voltage v we have i = _1_ Lep 7.5 == ~ x 60 X 10- 3 X f: p dt = L 225 x f i di = ~U 2 HI - cos 2 x 377t) J What are the time-average and peak values of the energy stored in the inductor of Prob. 7.7? , From Prob. 7.9, wet) = 3.375(1 -- cos 754t) The time-average value of the second term is zero w(t)I.,. The peak energy occurs at 754t= 71. Or, J Hence, == 3.375 J t== 71/754s, w(t)lm.x = 3.375 + 3.375 = 6.75 J 7.11 Flux linkage A is defined by A = Ne/> = Li, where L is inductance, i is the current through the inductor, N is the number of turns making the inductor, and e/> is the flux "linking" the N turns (see Fig. 7-1). Determine the flux linking a lOO-turn coil having a 50 cos 377t voltage connected across it. , Since A = Li = Ne/> v= then Thus, Note: 1 e/> = IV f dA dt =N and dq. ~li- v = L r!!:. dt = (50 cos 377t) 1 (' 50 sin 377t v dt = 100 Jo 50 cos 377t dt = 100 377 = 1.326 sin 377t The unit of flux is the weber (Wb). mWb INDUCTORS 7.12 D 131 The voltage across a 50-mH inductor is given by v(t) = 0 t~O t~O Sketch the voltage across and the current through the inductor. v-, V +0 ~.o / / / 100 Fig. 7-4 , The plot of vet) is shown in Fig. 7-4a . 3 X 10 [ - e -'(t . 1 J v d t = 50 x 110 3JlOte- t dt= 10 50 I = L = 200(1- e- I - te-I) 3 + 1)]t0 = lOX10 50 [ - e -I( t + 1) + 1] A which is sketched in Fig. 7-4b. 7.13 The current through a 200-mH inductor is given by i = 2e- t - 2e- 2t What is the energy stored after 1 s? , At v 7.14 dt t = 1 s, Also determine the voltage across the inductor after 1 s. 10- 3 ~ dt (2e- ' - 2e- 2') = 200 x 10- 3 x 2(-e- t + 2e- 2') v=200xlO- 3 x2(-e- 1 +2e- z )=-38.87mV t=1s, Energy, At = L ~ = 200 X A W= !Li 2 = ! x 200 x 1O-\2e -I - 2e- zl )z = ! x 200 X 10- 3 x 4(e- ZI + e- 41 - 2e -31) W= ! x 200 X 10- 3 x 4(e- 2 + e- 4 - 2e- 3) = 21.63 mJ Three inductors are connected as shown in Fig. 7-5. equivalent inductance is 0.7 H. Given L1 = 2L z , find L1 and L z such that the 132 D CHAPTER 7 Fig. 7-5 , From Fig. 7-5, or L2 Hence 7.15 = L1 =0.6H 0.3 H Three inductances, 0.6 mH, 0.12 mH, and L mH, Ire connected in parallel. Find L for the equivalent inductance to be maximum. What is the maximum value of the equivalent inductance? , The 0.6-mH and 0.12-mH inductances in parallel yield L = Q.~-":~-~ 12 = 0.1 mH P 0.6-rO.12 0.1 1 + (O.IIL) mH Thus, which is maximum as 7.16 L--HfJ. and the maximum Leq = 0.1 mHo A current of the waveform shown in Fig. 7-6a passes through a 2-H inductor. waveform. Sketch the corresponding voltage ----/L' /, , \ f----~-_4_---.-~---2. ' :'45(; '7 i s ' (a. ) tT, V + _I -::I---r.-,-r--_ t 7 l) rb) , i=0 i = ~t -- Fig. 7-6 di v'=Ldt=O 0~t<2s lj! i= --5t+30 v =<~( -- 5) = -10 V Hence the sketch shown in Fig. 7-6b. 7.17 ,~ How much power is associated with the inductor of Prob. 7.16? 5< t < 6s INDUCTORS I 0~t~2s = ~0t - ~) = (~t - I~n 2~ t < 5s W W =-1O(-5t+30)=(50t-300) 5<t<6s A 40-mH inductor is connected in parallel with a 60-mH inductor and across a time-varying voltage. At a certain instant, the total current supplied by a source is 10 A. Determine the instantaneous currents through the individual inductors. I Since the inductors are in parallel, for the common voltage v we have di l di2 Tt = L2 Tt v = LI . = L1 Thus, 12 f v dt 2 ~ = L2 = 60 And i2 Hence, 40 LI + i z = 10 il Also 7.19 i 60mH = 4A i40 mH = 6 A Find the currents il' i z , and i3 shown in the circuit of Fig. 7-7 with the switch S open. 5 /2 'M H i, , -\+ /; "IT~ +- c.~ loCI 1000 . , = 11=1 3 Leq I Then, f 6 x 12 Leq = 6 + 6 + 12 = 10 mH where .. 1 1 = 13 = 10 x 10 3 f 100 cos lO00t dt = 10 t Fig. 7·7 v dt v = 100 cos lO00t 3 100 sin lO00t . 10 x 1000 = 10 Sill lO00t X From Fig. 7-7, il = il L2 ~ I f VI dt 6 ~ = LI = 12 = 0.5 Or i l = 3.33 sin lOOOt 7.20 133 p = vi =0 7.18 D A i2 = 6.67 sin lOOOt A Determine the voltages VI and v 2 shown in Fig. 7-7, with S open. I VI di, -3 d . ) = Lld(= 12 x 10 dt (3.33 Sill lOOOt = 40 cos lOOOt v2 = v Note that v 2 = 6 X 10- 3 di3/ dt. VI = cos 1000t(100 - 40) = 60 cos lOOOt v V A 134 7.21 D CHAPTER 7 For the circuit of Fig. 7-7, with S closed, find L such that the current from the source has a peak value of 12.5 A. , From Fig. 7-7 with S closed, we have 6X12 6L) 24+lOL Leq = ( 6 + 12 + 6~:-L mH = 6 + L mH . 1= L1 f f (6 + L)103 v dt = 24 + lOL 100 cos 1000t dt = ( 6 + L) . 24 + lOL 100 Sill 1000t eq or 7.22 12.5 = (6 + L)l00 24 + lOL Since i J +i 2 =12.5 i3 v J = 12 X 10- 3 i2 = 8.333 sin 1000t A ~ (4.167 sin 1000t) = 50 cos 1000t V= v 2 Find the inductance at the terminals of the wye-connected circuit shown in Fig. 7-8, taking any two terminals at a time. , From Fig. 7-8, e.. c b ! ~ r L ... -------c~~~ I 3 ~----------------~ Fig. 7-8 The inductors of Prob. 7.24 are reconnected in deha as shown in Fig. 7-9. terminals, taking any two terminals at a time. , 7.26 = Find v J and v 2 ' with S closed, with L obtained in Prob. 7.21. , 7.25 12mH i = 12.5 sin 1000t A and 7.24 = For the value of L determined in Prob. 7.21, with S closed, find ip i2, i 3, and i4 from Fig. 7-7. , 7.23 L from which Fig. 7-9 Determine the inductance at the From Fig. 7-9, Let the three inductors, each having an inductance L y ' be connected in wye (Fig. 7-8). If three inductors, each having an inductance L d , are connected in delta determine the equivalent relationship between the two connections. From Prob. 7.24, From Prob. 7.25, From equivalence we must have, or and INDUCTORS 7.27 D 135 Find the equivalent inductance at the terminals ab for the circuit shown in Fig. 7-lOa by converting the I-H wye-connected inductors to an equivalent delta. a. ----,.---' 31+ I.ff (a.) (h) 1.1+ 6---~----~ (eL) (c) , FiR. 7-10 The steps in circuit reduction are shown in Fig. 7-lOb to d from which 2x4 Lab = 2 + 4 = 1.33 H 7.28 Verify that the result of Prob. 7.27 is correct by transforming the delta-connected inductors to an equivalent wye. (a..) a. h (6) Fig. 7-11 136 D CHAPTER 7 , In this case, the steps in circuit reduction are shown in Fig. 7-lla and b from which Lab = 7.29 n + 0 = 1.33 H A 40-mH inductor is charged to store 2 J of energy. An uncharged 60-mH inductor is then connected in parallel to the first one through perfectly conducting leads. How much energy is transferred to the second inductor? , In terms of the flux linkage A (which is conserved), we have A2 2=---2x40xlO or W= A2/2L. Thus, A = 0.4 Wb 3 Since the voltage, or the flux linkage A, remains un,;hanged, we have or . A L 1 +L 2 1='·----= 0.4 (40+60)XIO 3 =4A Hence, 7.30 How is a 300-mWb flux divided between a SO-mH and a lOO-mH inductor connected in series? , Let the series current be i. Then or But Hence A] = lOOmWb A2 = 200 mWb CHAPTER 8 AC Sources, Waveforms, and Circuit Relationships 8.1 A time-varying voltage is given by v = Vm sin wt. Sketch this waveform. maximum values, the frequency, and the period of the voltage. Identify the instantaneous and , The instantaneous and maximum values are denoted by v and Vm , respectively, in Fig. 8-1. defined by the condition vCt + T) = vCt) for all t; thus, . d T = -27T s peno frequency w f = 1 T= The period Tis w 27T Hz L, A -t--4---+:--\-:----f-:---_ i Fig. 8-1 8.2 8.3 Fig. 8-2 , T= T = 2 ms = 2 x 10- 3 From Fig. 8-2, 1 f= -T = 1 2x1Q3 S = 500Hz What is the angular frequency of a waveform having a period of 2 ms? , 8.6 71 = 601 = 16.7ms Find the frequency of the waveform shown in Fig. 8-2. Thus, 8.5 m'S" What is the period of a 6O-Hz voltage wave? , 8.4 I What is , T 27T =- w (a) the maximum value and or w 27T T =- = 27T 2x10 3 = 3141.6 rad/s Cb) the period of a waveform given by (a) Vm = 100 V v = 100 sin 377t V? 27T 27T (b) T= ---;;; = 377 = 16.7 ms A voltage wave and two current waves, shown in Fig. 8-3, have respective maximum values Vm The waveforms are sinusoidal. Write mathematical expressions for these waves. , Imp and 1m2 , Fig. 8-3 137 138 D CHAPTER 8 I v(t) = Vrn sin wt Note that i 1 lags behind v by cPl and i2 leads v by 42 8.7 These angles-cPl and cP2-are known as phase angles. Find the period, frequency, and the amplitude of the current waveform shown in Fig. 8-4 . .1.,/\ : -!C---------'L..---7-~~~ t ~, I 8.8 "?lo 20 /0 1 1 f= - = - - - - -3 = 100Hz T lC >( 10- T= lOms What is the period of a sinusoidal waveform having a 100-MHz frequency? I 8.9 or 10 ns Find the frequency in hertz and the angular frequency in rad/s of a waveform whose period is 5 ms. 1 1 f= - = = 200Hz T 5X103 I 8.10 Fig. 8-4 I = lJ) 2TTf= 2TT(200) = 1256.64 rad/s v = 200 sin 377 t V? What are the amplitude, frequency, and period of a voltage wave I v'" = 200 V Since w = 2TTf = 377 rad/s 8.11 A voltage wave is given by cycle? I ~ 8.12 v = 100 sin 314t. T = 2TT/W = 2TT/314 = 20 ms, 60Hz T= 71 = 601 = 16.67ms Ho'>' long does it take this waveform to complete one-fourth which is the time for 1 cycle. Thus, the time for one-quarter cycle = x 20 = 5 ms. (a) v = 100 sin (wt + 30 ) i = 10 sin (wt + 60°) (b) v = 100 sin (wt + 30 ) i = 10 sin (wt - 30°) (c) v = 100 sin (wt - 60 ) i = 10 sin (wt - 90°) C C (a) (b) (c) C i leads v by (60° - 30°) = 300. i lags v by [30° - (- 30°)] = 60°. i lags v by [-60°-(-90°)]=30°. Sketch the waveforms of the voltages and currents of Prob. 8.12 and verify the results obtained in Prob. 8.7. I The waveforms are shown in Fig. 8.14 = What are the phase relationships between the following waveforms? I 8.13 377 f= --. 271 8-5a to c. Determine the average value of the current waveform period. I (a) Since I avg over (a) a period and (b) a half- w = 2TT/T, lavg (b) i = Irn sin wt 1 l'o+T . 2TT Irn 2TT ]'O+T = T '0 Irn SIn -y t dt = - 2TT cos T t '0 = 0 21'0+TI2 . 2TT Irn 2TT ~1'0+T/2 Irn ( 2 TT to 2TTto) 2Irn = I SIn - t dt = - - cos -- t = - -cos - - - cos - - = cos wt T '0 rn T TT' T ~ '0 TT T T TT 0 Thus, the average value depends on the initial pont to" For to = 0, lavg = 2Irn/TT =0.6366Irn' AC SOURCES, WAVEFORMS, AND CIRCUIT RELATIONSHIPS D 139 VIt' v (c) Fig. 8-5 8.15 Determine the instantaneous and average powers dissipated in a resistor R connected across an ac voltage v = Vrn sin wt. v 2 V2 V2 I P = - = --'!!. sin 2 wt = --'!!. (1- cos 2wt) R R 2R 8.16 Determine the dc voltage that must be connected across a resistor R such that the power dissipated is the same as the average power determined in Prob. 8.15. = V~e = V~ p I de R from Prob. 8.15 2R Or This value is known as the effective, Or root-mean-square, value. 8.17 By rms we imply "the square root of the mean value of the square." waveform shown in Fig. 8.6. Determine the rms value of the current I i 4 I 10 T-:.g ~ I 'l.. - 4 l't s , I g t,s Fig. 8-6 140 8.18 0 CHAPTER 8 Find the ratio rms value/average value of an ac sinu_oidal voltage, where the average is taken over the first half-cycle. I From Prob. 8.16, From Prob. 8.14, This ratio is known as the form factor of the waveform. 8.19 Determine the rms value, average value, and form factor of the current waveform shown in Fig. 8-4. I From Fig. 8-4, rms= llT ~ -T 2 i dt= ~ (J 1 10 x 10 average = 2 A 8.20 (0 < t < 10 ms) i = 400t A 3 3 l'OO<'() (400t)2dt= ° (400)2 , 10 x 10 . [t-3 JIOXIO- _ -2.31A (J 2.31 form factor = -2- (linear function) 3 3 = 1.155 Determine the rms value, average value, and form factor of the current waveform shown in Fig. 8-7. I The wave has period T = 3 s. rns value 8.16 form factor = -------- = - - = l. 22 average value 6.67 1.-, A. J- (0 A, I 8.21 2. ~ ~ 5 7 t, I() $ Fig. 8-7 v = Vm sin wt. Obtain the voltage-current relationship for a resistor R connected across an ac voltage I Since v = Ri and i= Or V J?'!2 5 in w t = I rn sin w t where V and I are rms values. Note from (1) anc 12) that the current and the voltage are in the same phase. Obtain the voltage-current relationships for an I Since mdu~tor L carrying a current i = Irn sin wt. di dt Im sin wt l'='L - and Or (2) V=RI and we obtain (1) v = Vrn sin wt Vm sm wt= Ri Or 8.22 'J i = v= L ~ (Irn sin wt) = wLI", cos wt = Vrn cos wt = Vrn sin (wt + 90 and (1) 0 (2) ) V= wLI=XJ where XI. = wL is defined as the inductive reactance, and is measured in ohms. Note from (1) and (2) that the current through the inductor lags the voltage across the induct or by 90 0 • 0 141 AC SOURCES, WAVEFORMS, AND CIRCUIT RELATIONSHlPS 8.23 What are the various forms of voltage-current relationships for a capacitor C connected across a voltage source, v = Vm sin wt? I Since i = C dvldt we obtain and i = wCVm cos wt = Im cos wt = Im sin (wt Or Xc = l/wC + 90 0 is defined as the capacitive reactance in ohms. From (1) and (2) it follows that the current through the capacitor leads the voltage across it by 90 8.24 An inductor draws 5 A of current at 110 V 60 Hz. cally. I (2) ) I V=--=-XI wC c and where (1) v = Vm sin wt • Express the instantaneous voltage and current mathemati- First, we note that (unless stated otherwise) the values are all rms. Im = V2(5) = 7.07 A And 0 Thus, w = 27Tf= 27T(60) = 377 rad/s Vm = V2(110) = 155.56 V i = 7.07 sin 377t A 0 Since the current lags the voltage by 90 in an inductive circuit (see Prob. 8.22), we have v = 155.56 sin (wt 8.25 0 ) Determine the inductive reactance and the inductance from the data of Prob. 8.24. V I X L =wL=337L 22 L = 377 = 58.36 mH Hence, A capacitor draws 2 A of current at 120 V 50 Hz. cally. I 110 X /, = -I = - 5 =220 Now 8.26 + 90 = 155.56 cos wt V Express the instantaneous voltage and current mathemati- Proceeding as in Prob. 8.24, Im = V2(2) = 2.828 A w = 27T(50) = 314 rad/s Vm = V2(120) i = 2.828 sin 314t A 169.7 V = Since the current leads the voltage by 90 in a capacitive circuit (see Prob. 8.23), we have 0 v = 169.7 sin (wt - 90 8.27 = -169.7 cos wt V V 120 X =-=-=600 c I 2 1 Now, I Xc = wC = 314C I C= 314x60 =53.0JLF Hence, The voltage across a 0.5-H induct or is I v = 200 sin lOOt V. What is the instantaneous current? XL = wL = 100 x 0.5 = 500 Since i lags v by 90 0 , we finally have i = Im sin (wt - 90 8.29 ) Determine the capacitive reactance and the capacitance from the data of Prob. 8.26. I 8.28 0 The current through a 50-/-tF capacitor is 0 ) = 4 sin (lOOt - 90 i = 2 sin 1000t A. 0 ) A What is the instantaneous voltage? 142 D CHAPTER 8 1 1 Xc = wC = 1000 x 50 x 10 I 6 =: 20 0 Since the voltage lags the current by 90°, we have v = Vrn sin(wt - 90 C ) = 40 sin (lOOOt - 90°) V 8.30 The voltage and current through a circuit element are 70°) A. Identify the element and find its value. v = 100 sin (377t + 20°) V and i = 4 sin (377t- I The voltage leads the current by 20 - (-70) = 90°. Hence, the element is an inductor. X = Vrn L Irn Now, 8.31 = 100 = 25 0 =, wL 4 The voltage and current for a circuit element = 377 L L = 66.3mH or art~ v v = 200 sin (314t - 10°) i = 20 sin (314t - 10°) A Identify the element and find its value. I Since the voltage and the current are in the same phase, the element is a resistor, the ohmic value of which is given by 8.32 The voltage and current for a circuit element are v v = 6 cos (lOOOt - 80°) i = 3 cos (lOOOt + 10°) A Identify the element and find its value. I Since the current is ahead of the voltage by 10- (-80) = 90°, the element is capacitive and 1 wC Vrn 6 X =-=-=20 c Irn 3 8.33 1 = TiiOoc or 1 C = 2 x 1000 = 500 /-t F The voltage across and the current through an ac circuit are given by v = Vrn sin wt Find the instantaneous power. I The instantaneous power is given by p = vi = Vrn sin wtlrn sin (wt - c/J) =, V,Jrn1 cos [(wt - wt + c/J) - cos (wt + wt - c/J)] = Wrnlrn[cos c/J - cos (2wt - c/J)] 8.34 Determine the average power for the circuit of Prob. 8.33, in terms of rms values of voltage and current. I Since the time-average value of cos (2wt - c/J) is zero, from the result of Prob. 8.33 we have Pay == p =, ~ Vrnlrn cos c/J Now 8.35 V= Vrn1v2 and 1= Irn1v2. Hence, P ,= VI cos c/J, where cos c/J is known as the power factor. An ac circuit is purely resistive, having an equivaknt resistance of 15 n at the terminals. A 110-V 60-Hz ac source is connected across the terminals. Determ:ne the input current, power, and power factor. I Hence, 1= .!:": = 110 = 733 A R 15 . P = 12R = (7.33)215 = 806.67W= VI cos c/J 806.67 cosc/J'= 110 x 7.33 =1.0 Otherwise, since the current through and the voltage across a resistor are in phase, its power factor is cos 0° = 1.0. 8.36 How much power is dissipated in the circuit element of Prob. 8.32 and what is its power factor? AC SOURCES, WAVEFORMS, AND CIRCUIT RELATIONSHIPS D 143 I Power factor, cos cP = cos 90° = 0 (leading, since the current leads the voltage. This is a convention.) Power = VI cos 90° = 0 W 8.37 Repeat Prob. 8.36 for the data of Prob. 8.30. I Power factor, cos cP = cos 90° = 0 (lagging since the current lags the voltage): Power = VI cos 90° = 0 W 8.38 An ac circuit draws 5-A current at 220 V and consumes 1000 W power. I Since P = VI cos cP, where P is the power in watts, we have P 1000 cos cP = VI = 5 x 220 8.39 What is its power factor? =0.91 The voltage and the current in an ac circuit are v = 200 sin (377t + 30°) i = 10 sin (377t + 60°) Determine the average power. I From the data, V=200/V2V, 1= lO/V2A, Average power = 8.40 and (~) (~) cos 30° = 866 W The voltage and current in ac circuit are v = 100 sin (377t - 30°) 8.41 cP =60°-30°=30°. Hence, V What is the power factor? Is it leading or lagging? I cP = 30 - (- 30) = 60° i = 6 sin (377t + 30°) A cos cP = cos 60° = 0.5 leading A circuit draws 3 A current at 50 V and consumes 150 W power. the circuit. Determine the power factor and the nature of P 150 cos cP = VI = (50)(3) = 1 I The circuit is resistive. 8.42 A 5-0 resistor, a 100-/-tF capacitor, and a 100-mH inductor are connected in parallel across a 100-V 60-Hz source. Determine the average power drawn by the circuit. I Since the capacitor and the inductor do not draw any power (Probs 8.36 and 8.37), the power is dissipated only in the resistor. Hence, P= V2 = (100)2 =2kW R 8.43 What is the reactance of 50-mH inductor at I (a) (b) At dc, XL = wL = (0)(50 x 10At 60Hz, w = 27Tf= 27T 8.44 ) (a) dc and = 0 O. (b) 60 Hz ac? Thus, an inductor acts as short circuit across dc. 60 = 377 rad/s XL = wL = 377 x 50 x 10- 3 = 18.850 Determine the inductance of a coil having a 50-0 reactance at 60 Hz. I 8.45 X 3 5 XL XL 50 L = -;;; = 27Tf = 27T(60) = 132.63 mH Determine the frequency at which a 50-mH inductor has a 50-0 reactance. I XL 50 = L 50 x 10 = 1000 rad/s = 27Tf w = - 3 144 0 CHAPTER 8 Thus, 8.46 f= 1000 2;- == 159.15 Hz What is the reactance of a 500-/-tr capacitor at I (a) (a) d,; and (b) 100 Hz? X =_1_= 1 =00[1 , wC (O)(:iOO x 10 6) At dc: Thus, a capacitor acts as an open circuit across dc. (b) 8.47 At 100 Hz: Xc = 21T X Determine the capacitance of a capacitor having Cl 50-0 reactance at 60 Hz. 1 1 w = - - = - - - - - . - - = 100 rad/s = 21Tf CX, 100 x 10- (, x 100 101) f= --- = 15.9Hz Thus, 2'r The power factor of a circuit is 0.866 lagging. If the inJ:·ut power is 600 W at a voltage v = 100) V, what is the instantaneous current? I cos cfJ = 0.866 lagging Given: or p= VI cos cfJ 600 or 1 = 110(0.866) = 6.3 A Now 8.50 n = 3.18 At what frequency will a 100-/-tF capacitor have a 11)(1-[1 reactance? I 8.49 6 111 C = wXc = 21TfX, = ::1T·-x----,6-=-0-x--=5:c:-0 = 53.05 /-t F I 8.48 100 x 500 x 10 cl> = 300 cfJ = COS-I Im = 6.3V2 = 8.9 A and lagging implies that i lags v by 300. Power factor angle, (0.866) = 300 lagging 600 = (110)/(0.866) 01 Determine the power factor and the input power for a ,;ircuit with v I .J2 x 110 sin (3771 + cfJ = 20 - 10 == 100 Hence i = 8.9 sin (3771 - 20 0) A. =50 sin (I + 10°) V and i = 2 sin (I + 20°) A. cos cfJ = cos 100 = 0.9848 leading 50)( ~V2 2 ) cos 10° = 49.24 W Power = ( V2. CHAPTER 9 Complex Numbers and Phasors 9.1 A complex number represents a point on a plane with reference to two perpendicular axes. The horizontal axis is called the real axis and the vertical axis the imaginary axis. The symbol j is used to denote the unit along the imaginary axis. Locate the following complex numbers in a complex plane: A = 3+ j4 B= -j5 C = -6- j6 D=4 - j5 -- I The locations of these points are shown in Fig. 9-1. If we join these points to the origin, with arrowheads directed toward the points, we obtain directed line segments such as OA, OB, etc., which are called phasors. ---- A- .... -3+J ~ "'t I I c ::r - (, -j' Fig. 9-1 9.2 Fig. 9-2 With the definition of a phasor just given, locate the following phasors: A= 2/4SO B = 1/120 0 C=3/-30° I The locations of the phasors are shown in Fig. 9-2. This form of representation is known as polar form, in contrast to the form shown in Fig. 9-1, which is known as the rectangular form. 9.3 Diagrams such as those of Figs. 9-1 and 9-2 are known as phasor diagrams. By means of a general phasor diagram obtain the relationships to convert polar forms to rectangular forms, and vice versa. o I - Fig. 9-3 Consider the phasor DC (or C) shown in Fig. 9-3. In rectangular form we may write C = a + jb Or ICI = Va + b 2 2 (1) 145 146 D CHAPTER 9 Also tan fi = -J b a (2) Equations (1) and (2) give the conversion from rectangular to polar fom. is given by a = ICI cos () b 9.4 A = 2 + j2. Given: IAI = ',12 2 + 22 = VS Icl sin () or A =,/8 /4SO A = -3 + j4, what is the corresponding po,ar form? IAI = \fI-3)2 + 4 I (4) tan () = ~ = 1 Hence, Given (3) Express A in its polar form. I 9.5 = Polar to rectangular form conversion 2 =5 Referring to Fig. 9-4, we have a = tan- J 1 = 53.13° () = 180° - a = 126.87° Of Hence, 2o~ -J b Fig. 9-5 Fig. 9-4 9.6 Convert A = 4 / -30° I Let A = a to its rectangular form. + jb. Then, a = 4 cos (-30°) A Hence 9.7 Convert A = 5 /200° I Let A = -a - jb (see Fig. 9.5). From Fig. 9 ..5, a = 200 -180 = 20°. Then, -b = -5 sin 20° = -1.71 A ,= -'4.7 - j1.71 The imaginary quantity j is defined by may be drawn from this diagram? Since :,.464 - j2.0 = -5 cos 20° = -4.7 Hence I = to its rectangular form. -a 9.8 b = 4 sin (-30°) and j= j = V-I. Lucate /, /, and t on a phasor diagram. What conclusion V-I, we have /= -1 t=(/)j=-/=1 These are shown as phasors in Fig. 9-6. We may conclude that j may be treated as an operator which when applied to a phasor rotates it by 90° in the counterclockwise direction. 9.9 Find the reciprocals of the following phasors: (a)}; (b) 3 + j4; (c) 6 /30°. and imaginary parts. For b plot the phasor and its complex conjugate. I (a) Reciprocal of In each case separate the real D COMPLEX NUMBERS AND PHASORS 147 t ... 1.-. J __________~__~~------~~ R~ \. Fig. 9-6 (b) Fig. 9-7 . Reciprocal of 3 + J4 = 3 + j4 Multiplying the numerator and the denominator by (3 + j4) (c) IAI2. 9.10 (the complex conjugate of 3 + j4), we obtain and its complex conjugate are shown in Fig. 9-7. From Fig. 9-7 we infer that the phasors Thus, 6 3 - j4 3 - j4 3 - j4 1 . (3 + j4)(3 - j4) = 9 + 16 = 25 (3 - J4) 1 3 + j4 The phasor 1 IAlil and IAI /-() /~Oo = (6 ~~ L=lQ:) = ~ /-30° = ~ (cos 30° - are complex conjugates, with product j sin 30°) = 0.144 - jO.083 To add (or subtract) two or more phasors we add (or subtract) the respective real and imaginary parts. According to this rule obtain C = A + B, where A = 4 + j2 and B = 2 + j4. Show A, B, and C graphically. C = A + B = (4 + j2) + (2 + j4) = (4 + 2) + j(2 + 4) = 6 + j6 I The phasors are plotted in Fig. 9-8. 1.. c = A+/3 4 / / = ID +A Re. Fig. 9-8 9.11 Subtract I A = 3 + j3 Fig. 9-9 from B = 1 - j2. Show the operation graphically. C = B - A = (1- j2) - (3 + j3) = (1- 3) - j(2 + 3) = -2 - j5 which is shown in Fig. 9-9. .. 4 148 9.12 D CHAPTER 9 Multiply two phasors A = 2 + j3 in rectangular and polar forms. I 9.13 and B = 1 + j2. Separate real and imaginary parts. Express the results A· B = (2 + j3)(1 + j2) = 2 + j4 + j3 - 6 = -4 + j7 = 8.06/119.74° Express A and B of Prob. 9.12 first in polar form. mUltiplying the magnitudes and adding the angles. I A = 3.605 ili.31~ Verify that A· B in polar form may be obtained by B = 2.236/63.43° A· B = (3.605)(2.236) /(56.31 + 63.43) = 8.06/119.74° which agrees with the result of Prob. 9.12. 9.14 The procedure for Prob. 9.13 can be extended to <iivision where we divide the magnitudes and subtract the angles. Hence evaluate A/B, where A and Bare thl! same as in Prob. 9.12. I From Prob. 9.13, A = 3.605 ili.31~ A 3.605 / ) / ° B = 2.236 (56.~-= 63.43 = 1.612 -7.12 Hence, 9.15 B = 2.236/63.43° Verify the result of Prob. 9.14 in rectangular form. ~ = 2 + j3 = (2 + j3)(1 - j2) = (2 - j4~~fI + 6) = I B 1 + j2 (1 + j2)(1 - j2) 5 ! 5 (8 _ '1) = 1 6 _ '02= 1612/-712° ] . J.. . which agrees with result of Prob. 9.14. 9.16 Calculate (a) 10/45°+10/-45°; 2/60° + (5 - j1.232). I (a) (b) (c) (b) 10 /45° = lO(cos 45° + j sin 45°) (6.8+J3.2}+(5-j1.2); 2/60°+3/-30°-4/45°; (c) and (d) 10 L-4:~ = lO(cos 45° - j sin 45°) k = lO(cos 45° + cos 45°) = 14.14 (6.8 + j3.2) + (5 - j1.2) = (6.8 + 5) + j(3.2 - 1.2) = 11.8 + j2 2/60° = 2(cos 60 + j sin 60) = 1 + j1. 732 3/- 30° = 3(cos 30 - j sin 30) = 2.598 - j1.5 -4/45° = -4(cos 45 + j sin 45) = -2.828 - j2.828 k = 0.77 - j2.596 (d) 9.17 2/60° = 2(cos 60 + j sin 60) = 1 + j1.732 Perform the following calculations in polar form: (5/45°)(4/-20°); and (d) (5!45°)(4!200)(3l-~~~). I 3 + j3 = 4.243 Lt~ (a) (a) k = 6 + jO.5 (3 + j3)(5 + j8); (b) (5/45°)(4/20°); (c) 5 + j8 = 9.434/58° (3+j3)(5+j8)=' (4.243)(9.434) (45°+58°) =40/103° 9.18 (b) (5/45°)(4/20°) = (5)(4) /(45° + 20°) = 20 /65° (c) (5/45°)(4/-20°) = (5)(4) /(45° - 20°) = 20 /25° (d) (5/45°)(4 /20°)(3/-75°) = (5)(4)(3) /(45° + 20° -75°) = 60 /-10° Perform the following calculations in polar form: (4/-20°); (d) (5/45°)(4/-20°)/(3/75°). I (a) (b) 3 + j3 = 4.243/45° (b) (5/45°)/(4/20°); la) (3 + j3)/(5 + j8); 5 + j8 = 9.434 ,~i8° (c) (5/45°)/ 3 + j3 4.243 ° 5 + j8 = 9.434/(45 - 58) = 0.45 L..=.rr 5 /45° _ 5 /. )_ /" .. 4 /20° - 4 ' (45 - 20 - 1.25 L1i. 0 COMPLEX NUMBERS AND PHASORS 5 /45° (5 /45°)(4 / -20°\) 3 /75° -----=---=-=~c=-= = (d) I (a) Thus 1= ° (5)(4) /(45 - 20 -75) = 667 /-50° 3 . 1= 141.4 = 100 v'2 or Im = 141.4 100~. (b) 1= ~ /60° = 7.07 /60° (c) 1= ~ /90° = 1.414 /90° Express the following current phasors as instantaneous currents, all at a frequency w: 25/-30°; and (c) 6/-90°. I 9.21 149 Phasors are used to denote sinusoidal alternating quantities. For instance, v = Vm sin (wt - 4» is written as V = V 1- 4>, where V = Vm IV2 is the rms value. Similarly, i = I m sin (wt + 0) is written as I = I ~. Express the following currents as phasors: (a) 141.4 sin wt, (b) 10 sin (wt+600), and (c) 2coswt. 9.19 9.20 5 4 1(45+ 20) = 1.25 M 4~ = (c) D (a) i = v'2(10) sin wt = 14.14 sin wt (b) i = v'2(25) sin (wt - 30°) = 35.35 sin (wt - 30°) (c) i = v'2(6) sin (wt - 90°) = -8.484 cos wt (a) 1O~; (b) The instantaneous voltage and current for an ac circuit are v v = 155.6 sin 377t i = 7.07 sin (377t - 30°) A i = 7.07 e i377t -(w/6) A Represent these as complex exponentials. I v = 155.6e i377t v and Note that radian measure must be used in the argument of a complex exponential. 9.22 For the voltage and current given in Prob. 9.21, determine seconds). (a) the frequency (in hertz), I f= w = 377 = 27Tf (a) (b) 9.23 T= 1 f 377 = 60Hz 27T 1 = 60 = 0.0167 s The volt ages across two series-connected circuit elements are VI = 50 sin wt V and v 2 = 30 sin (wt - 30°) V. What is the rms value of the applied voltage? Also determine the phase angle of this voltage with respect to v l' I In phasor notation, VI = V = VI + V2 ~~=35.36+jO = (35.36 Vz = ~ / - 30° = 18.37 - jlO.6 + 18.37) - jlO.6 = 53.73 - jlO.6 = 54.75 / -11. W Ivl = 54.75 V Hence 9.24 or (b) the period (in V 4> = -11.W Two circuit elements are connected in parallel. The current through one of them is i 1 = 3 sin (wt - 60°) A and the total line current drawn by the circuit is i = 10 sin (wt + 90°). Determine the rms value of the current through the second element. I In phasor notation, 11 -= Jz (cos 60° - j sin 60°) = 1.06 - j1.832 1= ~ (cos 90° + j sin 90°) = 0 + j7.0n 150 D CHAPTER 9 Since I"" 11 + 12, we have I z = I - I1 =- j7.0n - l.06 + j1.832 =, -1.06 + j8.904 9.25 A.. - 'I' - 8.904) -- 180 - 8273 tan -1 ( -l.06 . -- n~' "7° ':11 .. _ iz = 12.68 sin (wt + 97.27°) A In a parallel circuit consisting of two elements, we haY!! I1 = 40 /20° A, 12 = 30 / -65° A, and V= 100 ~ V. Determine the rms line current and the power factor. I 1= I1 + 12 = 40 /20° + 30 / -65° = 4(1(l:os 20 + j sin 20) + 30(cos 65 - j sin 65) And = 37.588 + j13.681 + 12.678 - j27.1~;9 = 50.266 - j13.508 13.508 ° tan 4> = - 50.26(i or 4> = -15 1I 1 = 52.05 A And power factor cos 4> = 0.966 lagging. Determine the input power to the circuit of Prob. 9.26. I 9.28 8.967 A 12m = v'2(8.967) = 12.68 A From Prob. 9.23, and 9.27 Ilzl = What is the instantaneous current i z in Prob. 9.24'1 I 9.26 and P = VI cos 4> = (100)( 52.05) cos 15° = 5027.6 W Determine the input power to each circuit element of Prob. 9.26. in Prob. 9.27 is the sum of the powers for the two elements. I Verify that the total input power calculated PI = VII cos 4>1 = (100)(40) cos 20° = 3758.8 W P z = VI2 cos 4>z = (100)(30) cos 6:5 = 1267.8 W 0 PI + P2 = 5026.6 W which is approximately equal to the power calculated in Prob. 9.27. 9.29 If the current through the circuit of Prob. 9.23 is 1= 5 ~ A, I = P From Prob. 9.21, 9.30 calculate the input power. III cos 4> P = (54.76)(5) cos (-Il.16") = 268.62 W Find the power dissipated in each element of the circuit of Prob. 9.23 for 1= 5 ~ (as in Prob. 9.29). that the sum of the powers is the same as the total power calculated in Prob. 9.29. I Verify From Prob. 9.23, PI:::: VJ cos 4>1 = (35.36)(5) cos (0°) = 176.80 W PI + P; ,= 268.67 W which is the same as calculated in Prob. 9.29. 9.31 For the circuit shown in Fig. 9-10, we have Find i3 at 60 Hz. 1~ - t', I, + tr I Fig. 9.10 I Or, I, :::: I, - 12 = 6 /30° - 2 /20° = 6(cos 30° + j sin 3(1°) - 2(cos 20° + j sin 20°) = 5.196 + j3 - l.879 - jO.684 = 3.317 + j2.316 = 4.045 /34.9° COMPLEX NUMBERS AND PHASORS Thus 13m = 0(4.045) = 5.72 A 9.32 2 I=-A and Apparent power = Hence, v'2 150 2 v'2 v'2 = 150 VA. Just as P = VI cos cf> denotes the true, or active, power in an ac circuit, the quantity VI sin cf> is defined as the reactive power; it is measured in volt amperes reactive (var). Find the reactive power for the circuit of Prob. 9.32. I In this case cf> = 30° - (- 30°) = 60°. V= Hence, 9.34 + 34.9°) In ac circuits the product of rms voltage and rms current is defined as apparent power, measured in voltamperes (VA). If v = 150 sin (wt + 30°) V and i = 2 sin (wt - 30°) A, what is the apparent power? I 9.33 151 w = 27Tf= 27T(60) = 377 i3 = 5.72 sin (377t Hence, D 150 v'2 2 I=-A and V Reactive power = VI sin cf> = v'2 (~)( ~) sin 60° = 129.9var Given v = 200 sin 377t V and i = 8 sin (377t - 30°) A for an ac circuit, determine (b) true power, (c) apparent power, and (d) reactive power. I (a) (b) The current lags the voltage by () = 30°. From the data, V= (200/V2) V and 1= (8/V2) A. True power = VI cos () = (c) (d) v'2 v'2 (0.866) = 692.8 W Apparent power = VI Reactive power = VI . Therefore, 200 8 200 8 = v'2 v'2 = 800 VA Sill () = 200 8 v'2 v'2 (0.5) = 400 var (a) the power factor, CHAPTER 10 \\ AC Circuits Under Steady State~ 10.1 From Chap. 9 we know that the voltage across, and the current through, a resistor R are in the same phase. Express this as a phasor relationship. I 10.2 or Draw a phasor diagram for a resistor showing the VI relationship. waveforms in the time domain. Also show the vi relationship between the V' RIR f1.' It? (a.) (b) I Fig. 10-1 The phasor diagram is shown in Fig. 1O-1a assuming the common phase angle to be 0°. in the time domain is given in Fig. 1O-1h. 10.3 The vi relationship Repeat Prob. 10.1 for an inductor L operating at angular frequency w. In an inductor the current lags the voltage by 90°. Also IlL 1= IVLI IwL (see Chap. 9). Thus, if VL = VL !.St., then I L = I L / - 90° = VL 1w L(B~J.::I= VL 1jw L, since 11 j corresponds to a rotation by 90° in the clockwise direction (see Fig. 9-6). I 10.4 Repeat Prob. 10.2 for the inductor of Prob. 10.3. ,/ VI..-_ jwLIL ~ I / I / I~ Ca.) (b) Fig. 10-2 I The phasor diagram and the vi relationship are shown in Fig. 1O-2a and h. 10.5 Repeat Prob. 10.1 for a capacitor C operating at angular frequency w. I In a capacitor the current leads the voltage by 90° and Ilel = Ivcl 1(1 IwC) (see Chap. 9). So, if = le /90° = [Vcl(1 Iw C)lC~rQ:) = jw CVe, since j corresponds to a rotation by 90° in the counterclockwise direction (see Fig. 9-6). Vc = Ve!.St., then le 10.6 152 Repeat Prob. 10.2 for the capacitor of Prob. 10.5. AC CIRCUITS UNDER STEADY STATE D 153 ~ V c Ca..) (D) I The phasor diagram and the 10.7 vi relationship are shown in Fig. 1O-3a and b. From Probs. 10.1, 10.3, and 10.5, express resistance, and inductive and capacitive reactances, as complex numbers in rectangular and polar forms. R = R + jO = R !.St.. I 10.8 Fig. 10-3 = 0 + jwL = wL /90 XL X 0 c =0- ---.L = _1_ /- 90 wC wC 0 Impedance Z of a circuit is defined as the ratio of the voltage V across the circuit to the current I through the circuit. Find the impedance of a circuit having a resistance R in series with an inductance L operating at angular frequency w. I The circuit is drawn in Fig. 10-4, from which V=VR + VI. = RI + jwLI = (R + jwL)1 V L = jwLI Thus Z= ~ = R + jwL = R + jX L R j'-" L. p- I~ ... 10.9 V V t - Fig. 10-5 Proceeding as in Prob. 10.8, = (- ---.L)I wC Vc Hence, Z = -VI = R - What is the impedance of an LC series circuit? I V L = jwLI V c For Z=O, wemusthave wL=l/wC, V=VR + Vc = - j wC =R - . JX (R - ---.L)I wC c At what frequencies will the impedance be zero or infinite? =~I wC Z=~=j(WL- or 10.11 f Fig. 10-4 Find the impedance of the RC series circuit shown in Fig. 10-5, at angular frequency w. I 10.10 ~~ Ft ~ I I - -j/wC or 1 w c)=j(XL -Xc ) w=l/vTZ: and Z=oo at w=O. Determine the impedance of an RLC series circuit (Fig. 1O-6a). I In this case, for the circuit current I, the terminal voltage V may be written as 1C)]1 V=[R+j(WL- w or Z=~=R+j(wL- 1 w C)=R+j(XL -Xc ) (1) (2) 154 D CHAPTER 10 -j v- 1- + ~- I ~ Vo - L- y + /wc- 11 jwl.- R. _ (a..) VL .,VA ..... VL 1" VI:. - - ~ 10.12 ~ , , ~V (,5) Fig. 10-6 At what frequency will the input current and the terminal votage of an RLC series circuit be in phase with each other? I For V and I to be in phase, the circuit must act like a purely resistive circuit. Thus, the imaginary part of the impedance must be zero. Hence, from Eq. (1) or (2) of Prob. 10.11, we must have wL = 1/wC or w = 1/v'YC. 10.13 Draw a phasor diagram for the RLC series circuit (If Fig. 1O-6a. I The phasor diagram is shown in Fig. 1O-6b, where V=_-.L 1 V L = jwLI V=VR+VL+Vc c wC and cf> is power factor angle. 10.14 Admittance Y is defined as the reciprocal of imped.lI1ce Z; that is, Y = I/Z. Using this definition, find the admittance of a parallel RL circuit, shown in Fig. 10.. 7, under sinusoidal steady state. L x.. R 4- vl<- V I Fig. 10-7 From Fig. 10-7, and where 10.15 Y = G - jB = admittance, G = 1/ R = condu.;tance, and BL = I I wL =1 I XL =inductive susceptance. Find the admittance of an RC parallel circuit. I Proceeding as in Prob. 10.14, and where Y = G + jBc = admittance and I=IR+lc= Bc = (oC ~ +jwCV=(G+jBJV=YV = 1 /Xc = capacitive susceptance. AC CIRCUITS UNDER STEADY STATE 10.16 D 155 Repeat Prob. 10.15 for the general RLC parallel circuit shown in Fig. 10-8. I In this case we have Hence, ~~----~--------~~v I I v Fig. 10-8 10.17 Fig. 10-9 Draw a phasor diagram for the circuit of Fig. 10-8. I The phasor diagram is shown in Fig. 10-9 where 10.18 cf> is the power factor angle. A 4-0 resistor in series with a 7.96-mH inductor is connected across a 11O-V 60-Hz source. Determine (a) the impedance, (b) input current, and (c) the voJtages across the resistor and the inductor. (d) Draw a phasor diagram showing the current and the voJtages. XL = wL = 27TfL = 27T(60)(7.96 x 10- 3 ) = 3 0 I (a) Z = R + jXL = 4 + j3 = 5 /36.87° 0 (c) V R = RI = 4(22)/-36.87° = 88 /-36.87°V (b) 1= V Z = 11O~ 5/36.87° = 22/-36.87° A V L = jXLI = j3(22) /-36.8r = 66/90° - 36.87° = 66 /53.13° V (d) See Fig. 10-10. " .... ..... :1I0·L.E" V " ~-7'-----------~ / / / / V :88L-3 6• S7 0 V R 22 10-19 L:~t·gt A Find the power factor and the power input to the circuit of Prob. 10.18. dissipated in the resistor. I From Fig. 10-10, cf> = -36.87°. Fig. 10-10 Verify that the total input power is Thus, Power factor = cos (- 36.87°) Input power Pin = VI COS cf> = 0.8 lagging = (110)(22)(0.8) = 1936 W Power dissipated in R = 12R = (22?4 = 1936 W 156 10.20 D CHAPTER 10 Given v = 200 sin 377t V and i = 8 sin (377t - 30°: A for an ac circuit. (b) true power, (c) apparent power, and (d) reactive power. I 10.21 (a) The current lags the voltage by () = 30°. (b) From the data, V= (200 1V2) V (200/V2)(8/V2)(0.866) = 692.8 W. (c) (d) Apparent power = VI = (200/V2)(8/V2) = E:OO VA. Reactive power = VI sin () = (200 1Y2)(8 N2)( 0.5) = 400 var. (a) the power factor, Power factor = cos 30° = 0.866 lagging. l = (8/V2) A. and true power = VI cos () = Therefore, A coil has a resistance of 10 n and draws a current of 5 A when connected across a 100-V 60-Hz source. Determine (a) the inductance of the coil, (b) the power factor of the circuit, and (c) the reactive power. I (a) Z= 100 =VR2+(wL)2= V'lO'+(377L)2 5 (b) R 1 Z 2 or L = 45.94mH or cos(}=-'=-~ Reactive power = VI sill ,9 = (100)(5)(sin 60°) = 433 var (c) 10.22 Determine A series RLC circuit is excited by a 100-V 79.6-Hz S'JUfCe and has the following data: R = 100 n, C = 5 JLF. Calculate (a) the input current and Cb) the voltages across the elements. I (a) w = 27Tf= 27T(79.6) =500radls L = 1 H, XL = wL = (500)(1) = 500n 6 1 10 Xc = wC =, (:iOO)(5) = 400 n Z = R + j(XI. - X(J = 100 + j(5()(I-- 400) = 100 + jlOO = 141.4 /45° n V 100 ~)C ] =z= 141.421~ = 0.707 /-45° A V R = RI = (100)(0.707 L.= 45°) = 70.7 / -45° V (b) VI. = jXLI = (500 L2Q~)(0.707 /-45°) = 353.5 /45° V Vc = - jXcI = (400 L-9Q':)(0.707 / -45°) = 282.8 / -13SC V ...... " ...... ........ " I" / 10 0 / V\ I I 12 1\ / ........ ........ " ....../ /1 173.2/SX. V 200 -jl1.55 n / V = 173.2/SX. V / 0 1\ (a) (b) Fig. 10-11 10.23 For the circuit shown in Fig. lO-lla, evaluate the current through, and the voltage across, each element. draw a phasor diagram showing all the volt ages and currents. I Applying nodal analysis at node 1, with VI = VH " VI - 173.2 10 VI VI + 20 + -jll.55~ =0 whence Then AC CIRCUITS UNDER STEADY STATE From this, VID n = 173.2~ D 157 - VI = 100 !300V VI 100 /-30° 0 11 =20= 20 =5/-30 A It can be readily verified that 10.24 1 = I1 + 12 , The phasor diagram of Fig. lO-11b shows all voltages and currents. For the circuit shown in Fig. 10-12, calculate the current supplied by the voltage source and the voltage across the current source. 2.83~ lOOL!r V 500 A Fig. 10-12 I At node 1 we have, with VI = VlO = voltage across current source, Current in = current out from which 10.25 VI = 25 + j75 V. Then the current from the voltage source is 1 = (100 -Vl)/50 = 1.5 - j1.5 A. For the circuit of Fig. 10-12, by using the relationship P = Re (VI*), determine the power delivered by (a) the voltage source and (b) the current source. (c) Verify that the sum of these two powers is the power dissipated in the two resistances. I (a) (b) Since, from Prob. 10.24, 1 = 1.5 - j1.5 A, Again, from Prob. 10.24, VI = 25 + j75 V PI = Re [(25 (c) 10.26 P v = Re [(100 + jO)(1.5 + j1.5)] = 150 W. 2.83! 45° = 2 + j2 A. and + j75)(2 - j2)] = 50 Thus, + 150 = 200 W Power dissipated in the two resistances, A voltage source V ~ having an internal impedance Zs = Rs + jXs supplies a load having impedance + jXL · If RL and XL are individually variable, show that maximum power is transferred to the load when RL = Rs and XL = - Xs; that is, when ZL = Z~. (This is known as the maximum power transfer condition.) What is the power transferred to the load under this condition? ZL = RL I Because Z sand Z L are in series, we have the load current [ V = L 2 1. / y(Rs+RL) +(XS+XL) 2 Power transferred to the load is P= R = L L (Rs [2 V2R L + RL)2 + (Xs + X L )2 V2R L ~ = _V2 [ 1 _ (R S -R L - (Rs + RL)2 4Rs Rs + RL Equality in the first line is attained for XL = - Xs; absolute maximum, V 2/4R s , for ZL = Z~. 10.27 in the second, for )2] RL = Rs' V2 ~ _ - 4Rs Hence, P attains its A 20-n resistance is connected in series with a parallel combination of a capacitance e and a 15-mH pure inductance. At angular frequency w = 1000 radl s, find C such that the line current is 45° out of phase with the line voltage. I For V and 1 to be 45° out of phase, the net reactance of the parallel R = 20 n. Hence, 1 ±j20 j(1000)~0.015) + j1000 e or e= For the smaller (larger) capacitance, 1 lags (leads) V by 45°. 1010 Us + Le combination must be ±20 n, since fo) = 16.67 p,F, 116.7 p,F 158 10.28 D CHAPTER 10 A 46-mH inductive coil has a resistance of 10 n. (a;' How much current will it draw if connected across a 100-V 6O-Hz source? (b) What is the power factor of the coil? I (a) 3 wL=(27Tx60)(46xlO- )=17.34n and ZL=1O+j17.34=20.0/60°. Then, _ (100 L.<o~ ) _ ° IL - (20.0 -i6a') - 5.0 /-60 A L~L. (b) 10.29 Power factor = cos 60° = 0.5 lagging. A capacitor is connected across the coil of Prob. Determine the value of the capacitance. 1(1.:~8 to make the power factor of the overall circuit unity. I The admittance of the parallel combination will be l/Z L +jwC=1/(1O+j17.34)+j377C=i..+ j(377C - 0.0434). For unity power factor, the imaginary part must ,'anish, yielding 10.30 C=0.0434/377= 115 JLF. A 20-n resistor is connected in parallel to a 26.52·mH inductor. The circuit operates at 60 Hz. (a) the input impedance and (b) the input admittan';e of the circuit? What is I (a) 1 (b) 10.31 Y= Z= 1 8.95.z~~r =0.1117/-63AoS Three circuit elements R = 2.5 n, XL = 4 n, and Xc = 10 n are connected in parallel, the reactances being at 60 Hz. Determine the admittance of each element and, hence, obtain the input admittance. I YR = G = ~ = OA + jO Y L = -jBL = ._; 1 X =0- jO.25 L Hence, 10.32 Y=YR+Y L +Y c =OA-jO.15S If the circuit of Prob. 10.31 is connected across each branch and (b) the input current. I (a) IR =VY R = (10 L!r)(OA L!r) = 4 L!r A le (b) 10.33 lO-V 60-Hz ac source, determine (a) the currents through IL =VY L = (10 L!r)(0.25/-900) = 2.5/-90° A =VY c = (10 lO° 1(0.10 /90°) = 1.0 /90° A I=VY = (10 L!r)(OA - jO.15) = (1O~r)(OA272/-20.56°) = 4.272/-20.56° A Verify that the power input to the circuit of Prob. 10 32 is all dissipated in the resistor. I 10.34 Cl Pin = VI COS cf> = (10)( 4.272)(cos w'56) = 40 W At what frequency would the circuit of Prob. 10.31 be in resonance? I The condition w = 1/v'LC (Prob. 10.12) holds also for parallel resonance. Thus, from the given reactances at 60 Hz, w= 10.35 1 377\fIO == - - - rad/s \"(4/377)(1/377 x 10) 2 or about 95 Hz A 6-n resistor is connected in parallel with a 300- /J.F capacitor, and the circuit is supplied by a lO-A 60-Hz current source. Find the instantaneous currents through the resistor and the capacitor. I 1 1 Xc = wC = 27T(60)(300 x 10 6) = 8.84 n Voltage across the circuit, "= R(-jXc) = (6)(8.84) ~ = 4965/-3417° n ... R - jXc 6 - j8.84 . . AC CIRCUITS UNDER STEADY STATE D 159 V= ZI = (4.965 / -34.17°)(1O~) = 49.65 /-34.17° V V 4965 IR=R=-6-/-34.17°=8.275/-34.17°A iR = IRm sin (wt + cfJR) = V2(8.275) sin (377t - 34.17°) = 11.7 sin (377t - 34.17°) A ie = V2(5.617) sin (377t + 55.83°) = 9.42 sin (377t + 55.83°) A Similarly, 10.36 V 49.65/-34.17° le= -jXe = 8.84/-90° =5.617/55.83°A For Prob. 10.35 verify that the input current I is the phasor sum of IR and le I From Prob. 10.35, le = 5.617 /55.83° = 3.154 + j4.647 IR = 8.275/- 34.17° = 6.846 - j4.648 Hence, 10.37 IR An inductive coil consumes 500 W of power at 10 A and 110 V and 60 Hz. inductance of the coil. I Since at 12R=500W l=lOA, or 377L=v'% since VR 2 +(wL) 2 W=27Tf=27T(60)=377. Determine the resistance and the Now, R=500/(1O)2=5n. V 110 Z=-=-=11!l= I 10 10.38 + le = 10 + jO = 10 L!r = I or Hence, L=v'%/377=25.99mH. A lossy capacitor dissipates 11 W of power while taking 0.3 A of current at 110 V and 60 Hz. Represent this capacitor by a circuit having an ideal capacitor in parallel with a resistor and find the numerical values of the circuit elements. -t V If( Ic C I/oon ,"g~F Fig. 10-13 I The circuit is shown in Fig. 10-13, for which we have to evaluate Rand C. P=V 2IR, R= V2 P (110)2 = -1-1- = 1100 n 1= 0.3 = V I~ + I~ An RL parallel circuit is shown in Fig. 1O-14a. same current and power at a given voltage. le = V(0.3)2 - (0.1)2 = 0.2828 A = wCV Obtain its series equivalent such that the series circuit draws the jX~L.~ j I. 9zn Rs :./.44fl t J"3 " .n... I V + (a.) I 110 1100 = 0.1 A R= le 0.2828 C = w V = (377)(110) = 6.82 JLF Hence, 10.39 V IR = From the data, since (6) The equivalent series circuit is shown in Fig. 1O-14h. Fig. 10-14 From Fig. 1O-14a, R(jXL) (4 L!r)( 3 /90°) 12/90° . . Z= R+jX = 4+j3 = 5~ =2.4/53.13°= 1.44+J1.92=R s + JX Ls L 10.40 Generalize the result of Prob. 10.39 to the circuit of Fig. 1O-15a. 160 D CHAPTER 10 z. -... +'x -J p p Fig. 10-15 I Thus, 10.41 Solve Prob. 10.39 in reverse; that is, obtain the circuit of Fig. 1O-14a as the parallel equivalent of the circuit of Fig. 1O-14b. Zs = 1.44 + j1.92 I 1.44 - j1.92 1 or (1.44 + i1.92)(1.44 - j1.92) 1.44 + j1.92 1 1 1 1 Rp = G = 025 = 4 n Xp = 13 = 0 333 = 3 n Thus, p 10.42 Zs=Rs±jXs, Since Y 1 Zs s= = +, +, 1 R+X Rs -' Xs s J s G - 'B R ± ,'Xs = (R ± ,'X )(R += ,X ) = R2 + X2 R2 + X2s = P p s ss Ss ss ~ R~ 1 + X;' R.. R =-=--_. Thus, p Gp Obtain the series equivalent of the circuit shown in Fig. 1O-16a. (a.) I FromFig.1O-16a, (6) Rp=4n, jXp =j3-j6=--j3. 4(3)2 36 (4)2 + (3)2 = 25 Hence 10.44 . Generalize the result of Prob. 10.41; that is, given an RsXs series circuit, find its equivalent parallel circuit. I 10.43 p • Z, = 1.44 - j1.92 = From the resultsofProb. 10.40, 1.44n + (4)23 48 .. (4)2 + (W = 25 = 1.92n (capaCitive) shown in Fig. 1O-16b. The equivalent circuit of a transformer is shown in Fig. 62.5/-36.8°. Calculate VI and 11 , I, Fig. 10-16 10-17. Given: Ve = 2400 L!r and O.2Sl RI V I Fig. 10-17 le = AC CIRCUITS UNDER STEADY STATE D 161 I With the numerical values and the symbols shown in Fig. 10-17 we obtain: le = 62.5/- 36.8° = 50 - j37.5 A El = (2400 + jO) + (50 - j37.5)(0.2 + j0.45) = 2427 + j15 = 2427 /0.35° Y 0 _ • 2427 /0.35° 1550 = 1.56 /-89.65 - 0.0095 - } 1.56 A m Im = I = 2427 + j15 e Therefore, 10 = le VI 10.45 + Im = 0.25 = 10 "=' 10 3 X 0.2427 + '0 ] A 11 = 10 + le = 50.25 - j39.06 = 63.65/-37.85° A j1.56 A (2427 + j15) + (50.25 - j39.06)(0.2 + j0.45) = 2455 + j30 = 2455/0.7° Calculate the power dissipated in the circuit of Fig. 10-17 for the data of Prob. 10.44; also determine the input power. I Power dissipated = (l1)2R I + (lyRe + (le)2R2 (63.65)\0.2) + (0.2427)\10 000) + (62.5)\0.2) = 2.18 kW = Input power = output power + dissipated power = (Ve1e cos cP)1O- 3 + 2.18 kW = (2400)( 62.5)(cos 36.8)10- 10.46 3 + 2.18 kW = 120.1 + 2.18 = 122.28 kW Repeat Prob. 10.44 for the modified circuit shown in Fig. 10-18. f>., S1- I, + VI Fig. 10-18 I From Fig. 10-18, R = 0.2 + 0.2 = 0.4 n From Prob. 10.44, Hence, le = 62.5/-36.8° VI = (2400 + jO) + (50 - le = I + V R. .. 2453 iJlZ ° Im = 1550 /90° = 1.58/-89.3 A 11 = 50.25 - j39.08 = 63.66/-37.9° A XL = 12 n, L.. and Xc = 6 n. If 1= 1O~, find VR , Vv 1 'V'(I'---~~--...., '-\1_ R +v/. c.1" _____----JT-~ I jO.9 n + jO.9) = 2453 + j30 = 2453/0.7° Y 2453 /0.7° 10 X 10 3 = 0.2453/0.7° A For the circuit shown in Fig. 10-19, R = 8 n, Vc, and V in rectangular and polar forms. + 0.45) = Ve = 2400 ~ and j37.5)(0.4 10 = 0.2453 - j1.58 A 10.47 jX = j(0.45 Fig. 10-19 = IR = (10 ~)8 = 80 ~ = 80 + jO° Y V L = I(jXL ) = (10 ~)(12/900) = 120 /90° = 0 + j120 Y Vc = 1(-jXc) = (1OL!L)(6/-900) = 60 /-90° = 0 - j60 Y V=VR + V L + Vc = 80 + j60 = 100 /36.87°Y VR c 162 10.48 D CHAPTER 10 In the bridge circuit shown in Fig. 10-20, calculate the current through the inductor, the capacitor, and the three resistors. Fig. 10-20 I Choosing 0 as the reference node, writing the node ,equations for nodes 1 and 2 (after multiplying through by 10), we obtain: V2 =Vk , = 10(3 - j1) Thus V 1.= -J-=I-j3A 'J jlO I ke =1+j3A led = 1- jl A 10.49 le! = Igh = j2A 1 + j1 A Draw a phasor diagram showing all the voltages of the circuit of Fig. 10-20. I Choosing Vab as the reference phasor, the phasor diagram is shown in Fig. 10-21. ,- - ..... -~ Vef -.. .... ... ' ........' , ,, ,, Fig. 10-22 Fig. 10-21 10.50 Draw a phasor diagram showing all the currents in the circuit of Fig. 10-20. I 10.51 The phasor diagram is shown in Fig. 10-22. How much power is supplied by the voltage source to the circuit of Fig. 1O-20? dissipated entirely in the resistors. Verify that this power is Since Iba = led + le! we have, from the results of Prob. 10.48, Iba = (1 - j1) + (1 + j1) = 2 L!r A. Input power = Vab lab cos cP = (40)(2) cos 0° = 80 W. Power dissipated in the resistors = 1O( 11 - jl2 + 11 + j112 + Ij212) = 10(2 + 2 + 4) = 80 W (verified). I 10.52 For the circuit shown in Fig. 10-23, we have vab= 141.4 sin 4000t V mine the currents through and the volt ages across ,~ach element. and if! = 4 sin (4000t + 45°) A. Deter- AC CIRCUITS UNDER STEADY STATE e £;O{"L of 'l~ D 163 ~ ~ s-ofl- ~ IO/'F f ~ t 1 -. JL J Fig. 10-23 I Choosing node 1 as the reference node, we have: j) = 1 1 V3 ( 50 + 50 + 25 Igh . or V3 =Ve! =Vgh =Vij = 25(3 - jl) = 79/-18.4°V = 1.5 - jO.5 = 1.58/-18.4° A V,d = Vab 10-53 100 50 + (2 + J2) - le! V3 = 25(1 + jl) = 35.3/45° V = 1 + j3 = 3.16/71.6° A led = 0.5 + jO.5 = 0.71/45° A Draw a phasor diagram showing all voltages and currents in the circuit of Fig. 10-23. I The phasor diagram is shown in Fig. 10-24. V<lb Fig. 10-24 10.54 Determine the powers supplied by the voltage source and the current source in the circuit of Fig. 10-23. I Power supplied by the voltage source: iVab Il/bal cos CPu = iVabll/edl cos CPu = 141.4 V2 (0.71) cos 45° = 50 W Power supplied by the current source: 4 lV;jll/jil cos CPi = \V,II~il cos CPi = 79 V2 cos (45 + 18.4) = 100 W 10.55 Calculate the total power absorbed by the two resistors of the circuit of Fig. 10-23. equals the sum of the powers supplied by the two sources. Verify that this power I Power absorbed by the resistors: (led)250 + (lgS50 = (0.707)250 + (1.58)250 = 25 + 125 = 150 W Sum of the powers from the two sources (from Prob. 10.54) is 10.56 50 + 100 = 150 W (verified). For the circuit shown in Fig. 10-25, find the power delivered by the source. 164 D CHAPTER 10 0.1 Lolr Fig. 10·25 I By using nodal analysis (after multiplying through by 1500), we obtain (8 + j6)V2 - 5V3 = 15(1 V2 = VOb = 25 /- 36.87° V. Solving for V2 yields -5V2 + (5 - j3.75) V, =0 l-Ience, power supplied is Voblba cos cjJ = (2S)(O.1) cos 36.87° = 2 W 10.57 How much power is dissipated in the 300-0 resistor of the circuit of Fig. 1O-25? V~ I (25)2 500'= 500 = 1.25 W p}OO l! == 2 - 1.25 = 0.75 W. Otherwise, P500 fl = From Prob. 10.56, Pin = 2 W. Hence, V} 20 L.[: 1 1300 n = j400= ~mo /90° = 20 /-90° A where V3 is obtained from Prob. 10.56. 10.58 Hence, 1':100 n = (1/20)2300 = 0.75 W. Calculate the total power delivered to the circuit of Fig. 10-26. • J.o.n. J Fig. 10·26 I The input impedance is 1 Zi = 20 Thus, 10.59 + 1I(-j40) + j 140 + 11(20 + j20) = 400 200 L.[: Vi li = Zi = 40 L.[: = 5 L.[:A A (15 + j20).0 impedance is supplied by a 125-'1 60-Hz source. impedance during one cycle of the applied voltage~' I Let V = 125 L.[:. How much energy is delivered to the Then 125 L.[: 1= 15 + j20 = 5/-53.1': Energy 10.60 Pi = VJi cos cjJ = (200)(5)(cosOO) = 1 kW and = P = (125)(5) cos (53.1°) = 375 W power >~ time = (375 J Is)( io s) = 6.25 J The terminal voltage and current for a series circuit are v = 141.4 sin 2000t V and i = 7.07 sin (2000t + 36.87°) A. Obtain the simplest two-element circuit which would have the above vi relationship. I In terms of phasors, AC CIRCUITS UNDER STEADY STATE V= 1~~2.4 10° = 100 10° Y v L. ~ ~ 1= /36.87° = 5 /36.87° A Since Specify the V= 100 Y and 1= 10 A, \Z\ = 11~ = 100 P 600 cos () = VI = (100)(10) = 0.6 Hence, Z=1O(cos53.1°±jsin53.1°)=(6±j8)0 and mH induct or or a 125-JLF capacitor. 10.62 Z = R + ,'X = 100 &' - 20 /-3687° - 16 ,'12 5 /36.87° . - A two-element series circuit draws 600 W of power and 10 A of current at 100 Y and 5OOI7T Hz. values of these circuit elements. I 165 1 1 C = wXc = (2000)(12) = 41.67 JLF R= 160 Thus, 10.61 :'~27 V L. D R=60, () = ±53.1° or X=±j80, which requires either an 8- An inductive coil having a 30-0 resistance and unknown inductance is connected in parallel with a 100-0 resistor. The combination is connected across a 100-Y 5017T-Hz source. If the power delivered by the source is 400 W, find the value of the inductance. + IOD.!'). L Fig. 10-27 I From Fig. 10-27, (100)2 P IOO n = 10() = 100 W 2 P30 n = 400 - 100 = 300 W = (/30 n) 30 (300 or 130n ='-J30=3.162A or or XL = 3.162 = 100 10 L = 27T(50/7T) = 100 mH Hence, 10.63 31.62 or The resistance R and the inductance L of a coil are to be determined experimentally. The available equipment are a voltmeter and an 8-0 resistor. The 8-0 resistor is connected in series with the coil and the combination across a 120-Y 60-Hz source. If the voltmeter reads 32 Y across the resistor and 104 Y across the coil, determine Rand L. I + ?1- V v - R + R Vc.<l , L o , ~"Z..II ,/,.V ' ~~--~------------~~a' (h) Fig. 10-28 166 D CHAPTER 10 I The circuit and the phasor diagram for the experiment are shown in Fig. 1O-28a and b respectively, from which 32 ~ + 104 ~ = 1201.['. The law of cmines gives cos 82 = (104)2 + (120)2 - (3::)2 2(104)(120)- = 0.9692 81 = -53.1° Similarly, Icoil Hence VR or 32 / VR =32/-53.1°Y = I = R = 8 - 53 . 10 = 4 /-5~r R = 10 n, or _ Veoil A Zeoil- Ieoil _ - 104~ _ 0 '2 4/-53.10 -1 +] 4 L = 24/377 = 63.66 mHo Fig. 10·29 10.64 In the circuit shown in Fig. 10-29, X is purely reactive. With the switch closed, the circuit draws 1 A of current at 100 Y and consumes 80 W of power. With th,~ switch open, the current remains unchanged at 1 A. Determine X and XcP I R = -/2 80 == - - (1)2 = 80 n With the switch closed: X= ±60n or With the switch open: Ix - xci = Ixl = 60 or The unique solution with positive Xc is 10.65 Xc = 120 n and X = +60 n. A circuit draws 2 A at 120 Y and 60 Hz and conSUlles 120 W of power. A capacitor of unknown value is connected in series with the given circuit and the combination put across the same voltage source. It is found that the magnitude of the input current is increased. Determine the values of the original circuit elements. li I Let Z = Z n. Now, since the current increases when a capacitor is connected in series, the original circuit is inductive and cjJ is positive. Or, cjJ = Thus COS-I.!..... VI = __ I~_- = 60° COS-I Z = 60 /60° = 30 + j51.96 wL = 51.96 n 10.66 Izl = ~ = I (12(')(2) or L= 120 = 60 n 2 R=30n 5~~~6 = 137.83 mH For the circuit shown in Fig. 10-30, calculate C such that the input current is 45° out of phase with the input voltage at a frequency w = 2000 rad/s. ~D.n. ---'V\I'~ _ __ nT :/~H ________'l~ C Fig. 10·30 AC CIRCUITS UNDER STEADY STATE D 167 I For the input current to be 45° out of phase with the input voltage, the net reactance of the parallel portion of the circuit must be ±20fl. Therefore, ±jI20= -jlwL + jwC, or 1 (1 1) C = ~ wL ± 20 10.67 1 ( = 2000 1 2000 x 15 X 1) 10- 3 ± 20 = 10- 3 120 (2 ± 3) = 41.66 JLF An inductive load takes 480 W of power at 0.8 lagging power factor while operating at 120 V and 60 Hz. It is desired to make the power factor of the input current (with respect to the input voltage) unity by connecting a capacitor in parallel with the load. What is the value of the capacitance? I·, Fig. 10·31 I The circuit configuration is shown in Fig. 10-31. 480 IL = (120)(0.8) Thus, IL = 5 /-36.87° = 4 - j3. Let and 5A Ic=jb. Then 3 C = (377)(120) Hence, 10.68 = Then cjJ = cos- 1 0.8 = -36.87° Ii=IL+lc=4+j(b-3) = 66.3 or Ic=j3=jwCV. JLF A variable capacitor C is connected in parallel with an inductive coil of inductance 0.2 H and resistance 10 n. The combination is connected across a lO-V 50-Hz voltage source. Find the value of C which will result in a minimum current from the source. 1 I .. + v v II (a..) (1,) Fig. 10·32 I The circuit and the phasor diagram are shown in Fig. 1O-32a and b, where R = 10 n, j62.8 n. The current I will be minimum when the circuit operates at unity power factor. XL = Xc = 62.8 = l/wC. Hence, jXL = j314(0.2) = For this condition 1 C 10.69 = (314)(62.8) = 50.7 JLF The circuit of Fig. 1O-33a shows an ac generator with internal impedance Rg + jXg delivering power to a load. (a) Obtain an expression for the magnitude of the current in terms of the voltage source and the elements of the circuit. (b) Use the expression for current to obtain an expression for the power dissipated in Rx' (c) Show that, if Rx and Xx are considered as the variables (with P as a parameter), the relationship of part 2 (b) can be written in the form of the equation of a circle the center of which is at the point (V /2P - Rg, - Xg) 2 and the radius of which is (V2/2P)(1-4R gPIV )1/2. (d) Let V= 100 V and Rg and Xg be 10 and 20n, respectively. Draw the circles for various convenient values of P between zero and the value at which the radius of the circle is zero. Label each circle with the value of power to which it corresponds. 168 D CHAPTER 10 Xx Ge.rI. , Loa.d. I - ___ .J (a..) \ 0) I (a) 1= V . V(R x + Rg)2 + (Xx Fig. 10-33 (b) + Xg)2 (c) The relationship of part (b) may be rewritten m,: [R x V2)]2 + (X + X )2 = -V42 (1 - -4Rg2P) + (Rg - -2P x g 4p V which is the equation of a circle as described. 10.70 An ac bridge circuit is shown in Fig. 10-34. detector D. (d) The circles are shown in Fig. 1O-33b. Obtain the condition for balance when no current flows through the 1----4-2. Fig. 10-34 I Since the current through D is zero, nodes 1 and 2 are at the same potential. Thus, I1 flows through ZI and Z3 whereas 12 flows through Z2 and Z4' Under balance condition we have IIZI = 12Z 2 and II Z 3 = 12Z4· Hence, the required condition is 10.71 Z I /Z 3 = Zzl Z 4' As a special case for the bridge of Fig. 10-34, we have ZI = RI' Z2 = Rx + jwLx, Z4 = R4 + l/jwC4· The bridge is balanced. Solve for the unknowns Rx and Lx' I From Prob. 10.70 we have Z3 = IljwC 3 , and AC CIRCUITS UNDER STEADY STATE RI(R4 + or ~C) = JW ~C3 (Rx + jwLJ JW 0 169 or 4 Equating the real parts and solving for Lx yields Lx = R I R 4 C3 (1) Equating the imaginary parts and solving for R x gives (2) 10.72 Verify that (a) the operation of the bridge of Prob. 10.71 is independent of frequency and be balanced by adjusting resistances only. (b) the bridge can • (a) Since the frequency W does not enter in the balance conditions (1) and (2) of Prob. 10.71, the operation is independent of frequency. (b) Condition (2) of Prob. 10.71 can be satisfied by adjusting R I only (for a given C3 /C4 . Then condition (1) can be satisfied by adjusting R4 only. 10.73 For the circuit shown in Fig. 10-35 determine the condition such that the input impedance is purely resistive at all frequencies. R v jwL Fig. 10-35 • The voltage source "sees" the impedances impedance is R + jwL and (R + jwL)[R - j(l/wC)] 2R + j[wL - (l/wC)] where X = wL - l/wc. 10.74 Zin R2 + LlC + jRX 2R+jX (1) _ R __ LlC- R2 2R+jX (2) = R for all w if LlC = R2. L = 100 mH, For the circuit of Fig. 10-35 we have input power. • in parallel; hence, the input Subtracting R from both sides of Eq. (1), Zin and Eq. (2) shows that R - j(l/wC) C = 10 ILF, R = 100 n, and V = 100 V. Calculate the From the data we may verify that 3 L 100 x 10- = 104 = (100)2 = R2 C-lOx10 6 Thus, according to Prob. 10.73, the input impedance is purely resistive, or Zin = R = 10.75 lOOn Power = VI cos cjJ = (100)(1)(1) = 100 W The bridge of Fig. 10-34 is used to measure the resistance and inductance of a coil. The arms of the bridge are: ZI = RI/(l + jwCR I ), Z2 = R 2, Z3 = R 3 , and Z4 = Rx + jwLx· For balanced conditions we have RI = 10 kn, C = 0.5 J.LF, R z = 400 n, and R3 = 600 n. (a) Derive the general conditions for balance. (b) Determine Rx and Lx. • (a) For balance we must have or or Equating the real and imaginary parts, the conditions for balance are 170 D CHAPTER 10 = R2 R 3 (1) L, ,= CR 2 R 3 (2) RIRX (b) For the given numerical values, from Eqs. (1) and (2) we obtain R = (400)(600) = 24 n x 10,000 10.76 Lx = (0.5)10-\400)(600) = 120mH Two impedances ZI = 10 / - 53.13° nand Z2 == 20 /36.87° n are connected in parallel. The combination draws 2 + j1 A current from a voltage source. Delermine the complex power for each branch. • Total admittance is Y -- Y 1 + Y 2 -- ~ Z +~ Z -- 1 L10 15'1 l4° + L20 J~"-_ /-3687° • 2 = (0.06 + jO.08) + (0.04 - jO.031 = 0.1 + jO.05 = 0.1118 /26S S hr.. ,,0 1= 2 + j1 =2.236~A 11 = VY I = (20 L!L)( fa /53.13°) = 2 /53.13° A ,I 2.236 ~ !ro0 \ = Y = 0.1118 /26S = 20&' V 12 = VY 2 = (20 L!L)( 10 / - 36.87°) = 1 / - 36.87° A SI =VIr = (20L!L)(203.J~) = 40 /-53.13° = 24 - j32 VA S2 = VI; = (20 L!L)(1 D6.!Q~) = 20 /36.87° = 16 + j12 VA 10.77 For the circuit of Prob. 10.76 verify that the total compilex power is the sum of the two complex powers obtained in Prob. 10.76. • From Prob. 10.76, SI = 24 - j32 Thus, S = SI S2 = 16 + j12 + S2 = 40 - j20 VA. Otherwise, S = VI* = (20 L!L)(2.236 / -22';~ 1'= 44.72 / -26S = 40 - j19.95 VA The slight error is due to roundoff. 10.78 A parallel circuit consisting of two impedances is shown in Fig. 10-36. and, hence, calculate the input current I. • Z = ~ = (8 + j6)(lC·- j5) 8 + j6 + 10 + j5 ZI + Z2 Since Replace it by an equivalent series circuit = 50 + jlOO = 53 /32° n 18 + jll . V= 200 L!L V, V 200,L!lo 1= - = ---=. = 37 74/-32° A Z 5.3il£: . I Fig. 10·36 10.79 Determine the power factor of the circuit of Fig. 1(1-36 by calculating the power dissipated in the equivalent resistor. Verify that the result is consistent with that of Prob. 10.78. • From Prob. 10.78, Z = 5.3 /32° = (4.495 + j2.80B) n. P = [2R = (37.74)2(4.495) = 6.402 kW From Prob. 10.78, Thus, R = 4.495 n. P 6402 cos cjJ = V[ = (200)(37.74) = 0.848 lagging cos cjJ = cos (- 32°) = 0.848 Jagging. D AC CIRCUITS UNDER STEADY STATE 10.80 A series-parallel circuit is shown in Fig. 10-37. 171 Determine the currents I, 11' and 12, j SA 311. 1; SA V.: loo Fig. 10-37 I The input impedance Zi is given by = 2 + '3 + (3 + j5)(5 - j6) = 2 + '3 + (5.82 M)(7.81/ -50.2°) 3+j5+5-j6 J Z, 8.06/-7.120 J = 2 + j3 + 5.65/15.92° = 7.43 + j4.55 = 8.71/31.48° n _ V_lOO jJl _ ° 1- Zi - 8.71~ -11.48/-31.48 A The voltage across the parallel branch is V23 = IZ 23 = (11.48/-31.48°)(5.65/15.92°) = 64.86/-15.56° V V23 64.86/-15.56° / 11 = ~ = 5.83/59° =11.12 -74.56° A 10.81 12 = V23 Z; = 64.86/-15.56° /'lA t:'l0 7.81/-50.19° = 8.3 Ll±&l.A Draw a phasor diagram for the circuit of Fig. 10-37 showing V12' V23 , V, I, 11' and 12, I From Prob. 10.80, V12 = 1(2 + j3) = (11.48 /-31.48°)(3.6/53.3°) = 41.39/24.83° V V=100jJlV 1= l1.48/-31.48°A V23 = 64.86/-15.56° V I 1 =11.12/-74.56°A 12 = 8.3 /34.63° A Hence we draw the phasor diagram of Fig. 10-38. I - -- ----- -- ,.. \ \ - /" ,,/ - C ~ I /' /' V::. IOo'f ,/ LD• I, I, Fig. 10-38 10.82 Fig. 10-39 The circuit of Fig. 10-39 operates at a frequency of 50 Hz. voltage V and the input current I are in the same phase. Determine the value of C in order that the input I Input impedance: Z= - i. + wC i. i. (6)(j12) = _ + (j72)(6 - j12) = _ + '24+ 48 6 + j12 wC 36 + 144 wC J. . For V and I to be in the same place, the imaginary part of Z must be zero. Hence, 1 C = (2.4)(314) = 1326.9 J1. F w = 27T(50) = 314. Thus, l/wC = 2.4. At 50 Hz, 172 10.83 D CHAPTER 10 For the C obtained in Prob. 10.82 calculate the power input to the circuit of Fig. 10-39 by: (a) VI cos cI>; (b) I2R, where R is the resistive portion of Z (obtained in Prob. 10.82); and (c) (II?6, where 11 is the current through the 6-0 resistor. Verify that the results are identical., • (a) Since Z = 4.80, 1= (100/4.8) L!r = 20.83 ,~!J~A. Hence, (b) I2R = (20.83)2(4.8) = 2083 W. (c) By current division rule, Z2 j12 I, = I ZI + Z2 = (20.83 L!r) 6 + rl2 VI cos cl> = (100)(20.83)(cosOO) = 2083 W (20.83 L!r)(12/900) 13.42 ~ = 18.63/26.56° A = - I~R = (18.631\6) = 2083 W 10.84 V,= 10 L!r V. For the circuit shown in Fig. 10-40, we have Find 11' 12, and 13, _ 10+ jO _ _1~~~': __ 77/° I1 - 3 + j2 - 3.60 L~,.690 - 2. 33.69 A • Z3 ( / 0) 9 - j6 (:~.77 /-33·n(10.82/-33.7°) 57/ 73 ° 12 =IIZ +Z = 2.77-33.7 19+j2'= 19.1& = 1 . - .4A 2 3 _ Z2 _ ° 10 + j8_ (2.77 /-33.7°)(12.8 ~) _ ° 13 - I1 Z2+ Z 3 - (2.77/-33.7) 19+}"219.1& -1.86L.=.rA ~ j III.n an. I I 3.f2 ;;&, j1Jl + ..,. v.,-- IDU" - '1ft v Fig. 10-40 10.85 Determine the terminal voltage V for the circuit of Fig. 10-40. • V = VI + V2. Since V, = 10 L!r = 10 + jO V (given) and, from Prob. 10.84, V2 = 13Z3 = (1.86/-1°)(10.82 L-3:ht> = 20.13/-34.7° = 16.55 - jl1.46 V V= (10 + jO) + (16.55 - j11.46) = 26.55 - jl1.46 = 28.9/-23.3° V we obtain 10.86 Obtain the complex power in each of the three impedances of the circuit of Fig. 10-40. the three complex powers is the same as the input complex power. • Verify that the sum of SI = VII~ = (10 L!r)(2. 77 L11.n =, 27.7 /33.7° = 23.05 + j15.37 VA S2 =V21; = (20.13/-34.7°)(1.57 L~'.4°) = 31.6/38.7° = 24.66 + j19.76 VA S3 =V31; = (20.13/-34.7°)(1.86 LD = 37.44/-33.7° = 31.15 - j20.77 VA S = SI + S2 + S3 = 78.86 + j14.36 VA =VI~ = (28.9 /-23.3°)(2.77/33.7°) = 80 /10.4° = 78.68 + j14.44 VA which are approximately equal. 10.87 Repeat Prob. 10.86 for the circuit of Fig. 10-37. • From Prob. 10.81, SI =V'21* = (41.39/24.83°)(11.48 LU.AB~) = 475.16/56.31° = 263.57 + j395.36 VA S2 = V231~ = (64.86/-15.56°)(11.12 L]~.56°) = 721.24/59° = 371.46 + j618.22 VA S3 = V231; = (64.86/-15.56°)(8.3 84./~~) = 538.34/- 50.19° = 344.67 - j413.54 VA S = S, + S2 + S3 = 979.7 + j600 =VI* = (100 L~)(11.48/31.48°) = 1148/31.48° = 979.04 + j599.49 VA Hence AC CIRCUITS UNDER STEADY STATE 10.88 D 173 Three circuit elements R = 10 n, L = 0.1 H, and C = 600 /LF are connected in parallel and the combination is placed across a 11O-V 60-Hz source. Determine the input current and input power. Also calculate the power factor angle. • YI=~= 1 j Y 2 = jwL = 0 - (377)(0.1) = 0 - jO.0265 11 + jO =0.1+jO 0 1 6 Y 3 = ----;---/ C = jWC = j377(600 x 10- ) = 0 + jO.2262 -J w Thus, Y=Y +Y 2 +Y 3 =0.1+jO.1997=0.223/63.4°S. Let V=110L!L. Input current I=VY= (110 L!L)(0.223/63.4°) = 24.57 /63.4° A = 11 + j21.97 A. Input power = VI cos cjJ = (110)(24.57) cos 63.4° = 1210 W. Power factor angle cjJ = 63.4°. j 10.89 10.90 Determine the power dissipated in the lO-n resistor of the circuit of Prob. 10.88. found to be 63.4°, calculate the input current for the circuit. If the power factor angle is • PR = V2/ R = (110)2/10 = 1210 W. 10.88. as Hence 1= 1210/ (110 cos 63.4°) = 24.57 A in Prob. We have two circuits: Circuit 1 draws no direct current whereas circuit 2 takes 5 A at 50 V dc. When connected across 50-V 60-Hz ac, circuit 1 takes a 2-A current and circuit 2 draws 3 A. Identify the circuit elements and obtain their numerical values. • Since circuit 1 is an open circuit to dc, it is a capacitor. takes a 2-A current. Thus, Vc =50=ljwC=2/377C Now, at 50-V dc, circuit 2 takes a 5-A current. R = ~ = 10 n. At 50-V 60-Hz ac, V 50 1=-=-=3 Z wL Hence 10.91 determined = V(16.67)2 - Z When connected across a 50-V 60-Hz source, it or Hence, the circuit is inductive and its resistive component is or (10)2 = 13.33 n Z = lj! = 16.67 n 13.33 L = 377 = 35.37 mH or A 11O-V 60-Hz inductive load draws (500 - j500)-VA complex power. A capacitor C is connected across the load to bring the overall power factor to 0.866 lagging. Determine the value of C, and new value of complex power of the load/capacitor combination. j &2, VAN. I /. VA P, W j 50 0 -: k..l/c .... ~ 1C VA "'-«..= 500 - VA.k..t. Fig. 10-41 • First, we show the complex power for the load in Fig. 10-41, which shows a lagging power factor angle of 45°. From Fig. 10-41, it follows that to change the power factor to 0.866, we have: Reactive power supplied by Cis500tan300=211.3var or Ic=211.3/110==1.921A=wCV==(377)C(110). Hence, C= 46.325/LF. New VA = (500)2 + (211.3)2 == 542.8 VA and the corresponding complex power S = P + jQ == 500 j211.3 VA. V 174 10.92 D CHAPTER 10 Determine the reading of a voltmeter connected between points 1 and 2 of the circuit of Fig. 10-42. T 1... IJl o '10 V1'1 I.! Fig. 10-42 • The impedances are: Z2 = 1 + jl = 1.414 /45° n ZI = 1- jl = 1.414 /-4~U _ ~ _ (1.414 L45~)(1.414 ~) _ 2L!r - I L!r Z - ZI + Z2 (I - jl) + (I + jl) - 2 L!r 0 n V IlJL!r 1= - == - - - - =1 1OL!r A Thus IL!r Z By current division, Similarly The voltmeter reading Vl2 is given by VI n + VI2 - VI n or = 0 (77. P /45°)(1) + VI2 - (77.77 / -45°)(1) = 0 55 + j55 + V 12 -55 + j55 = 0 or - V 12 = j55 + j55 = jllO V O)r Hence the voltmeter reading is 110 V. 10.93 Obtain the active and reactive powers of the circuit of Fig. 10-42. • S = VI' = (110 L!rXllOL!r) = 12100 + jO. Taking the current from Prob. 10.92, Hence P = 12100 W, Q = Ovar. 10.94 Check Prob. 10.93 by adding the complex powers in the two branches. For ZI we have SI = VI~ = (110 L!r) (77.77 L:-4:~)= 6050 - j6050 VA. For Z2 we obtain (110 L!r) (77.77 / +45°) = 6050 + j6050 VA. Hence SI + S2 = 12100 + jO S. I 10.95 = Find the current through the capacitor of the circuit of Fig. 10-43. v I -j 2.fl_ 1 , vt ~_ .... 11.. I ~ I..{L. 511- 5 LoO;., z• I '4 ' 2.Sl ~j 111- 1 b • Fig. 10-43 Impedance presented between ab: . 2.828~ . h k C;:k O . 2(1 + jl) Z2 = -J2 + 2+ 1 + jl = -J2 f 3.162 /18.43° == -J2 + 0.894~ = - j2 + 0.8 + jO.4 = 0.8 - jl 6 = 1.79 / -63.4° n . ° 5L!r ° 12 = current through capacitor = 5 L!r 5 + 0.8 _ j1.6 = 4.16 ~ A S2 = VI; = AC CIRCUITS UNDER STEADY STATE 10.96 D 175 Calculate the current in the 15-0 resistor of the circuit shown in Fig. 10-44. . "t- I ... T, J~ 5.n, I.f I ~? I.:'~o~ t V j SA -5 '0 Isn j ~.".n. Fig. 10-44 • First we determine the input admittance: Y = YI + Y2+ Y3+ Y4= or Z= 1 Y= 1 1 1 0.221/-58° =4.53/58°0 13 = Y 3V = 10.97 1 jO.187 = 0.221/-58° S V=ZI=(4.53/58°)(33/-13°)= 149.5/45°V (is L!r)(149.5/45°) = 9.97 /45° A Determine the complex power drawn from the current source of the circuit of Fig. 10-44. • From Prob. 10.96. Z = 4.53/58°. So the circuit is inductive. + jQ =VI* = (149.5)(33) /(45 + 13) = 4933.5/58° = 2614 + j4184 VA S= P 10.98 1 J5 + 5 + j8.66 + 15 + -jlO = 0.117 - Calculate the input active and reactive powers in the circuit elements of Fig. 10-44. complex power. Verify that the result agrees with that of Prob. 10.97. • Now. Hence determine the V2 (149.5)2 Q = - = = 4470var I XI 5 1 12 = Y 2V= 5 + j8.66 (149.5/45°) and 1121 = 14.95 A Q2 = (/2)2X2 = (14.95)28.66 = 1935.5 var P3 = (/3)2 15 = (9.97)2 15 = 1490 W S2 = Pz + jQ2 = 1117.5 + j1935.5 VA S3 = 1490 + jO V A Q3 = 0 2 Q4 := V = (149.5)2 = 2235 X 10 var S4 = 0 - j2235 V A 4 S = SI + S2 + S3 + S4 =j4470 + 1117.5 + j1935.5 + 1490 - j2235 = 2607.5 + j4170.5 which is approximately the same as obtained in Prob. 10.97. 10.99 VA The error of about 0.3 percent is due to roundoff. Determine L in the circuit shown in Fig. 10-45. o T: II.f/S· I 2. 220 L! <> f40 H • -j L 15.52. i Fig. 10-45 V 220 L!r 12 = Z2 = 15 - j15 220 L!r _ °_ . 21.2~ - 10.35 ~ -7.34 + J7.34 A _ _ . _ . _ ° _ V _ 220 L!r I1 - I - 12 -11.81/ -7.12 - (7.34 + J7.34) - 4.38 - J8.8 - 9.83/-63.6 A - ZI - 10 + jX L from which 10.100 XL = 200 = wL = 377 L Determine Zx in the circuit shown in Fig. 10-46. or L = 53.05 mH 176 0 CHAPTER 10 ~0:J1.'L J 3.n. Fig. 10-46 --------~--------~ I Y= 31.5 M "0 • I 1 50 /60° = 0.63 L::..:;,-Q_ ,= 0.51 - ,0.37 = Y x + 10 + 4 + j3 1 V= or Thus, 10.101 Zx = 2 + j2 n Find the voltage V AB in the circuit of Fig. 10-47. y -jsa. • ).,0 L.?JiJ So ~IfSD S..Il.. I Fig. 10-47 Notice that the inductor jXL is irrelevant became no current flows through it. 20m , 11 = 10 + jlO = 1.414 L=.!i A Now, V AX = 11 (5) = (1.414/-15°)5 = 7.07 L=.l:r V Hence 10.102 V AB =VAX 12 50~ = 5-j5 Thus, V XY = O. =7.07~A V YR = -1 2 (5) = -(7.07 ~)5 = - 35.35 ~ Y + V XY + V YR = 7.07 /-15·: + 0 - 35.35 LiL = -28.6 - j1.83 = 28.7 /183.7°Y Determine V AB in the circuit of Fig. 10-48. I By current division we obtain 11 = 1 Z 1 ~2 Z = 18/45° 30 2 ! j ,·8 = 4.64/120. 1..:: A Z1 _ 18 / A <:0 30 _ 7 0 /3()O 1 ° 12 - 1 Zl + Z2 & 30 + j8 - 1 .4 ~ A For the path AB we have V AB = V AX + VXB = 11 (20) - l!( j6) ,= (4.64/120.1°)20 - (17.40 /30.1 0)j6 = 92.8/120.1° + 104.4 1.=59.2~ = 11.6/-59.9° Y t- sS\.- -1-.0.. V A VA8 - 8 .~ .~L, S19 -J .if.J2. J )< Fig. 10-48 10.103 In the circuit of Fig. 10-49, we have I Fig. 10-49 V AB = 48.3 LK':. Evaluate V. By voltage division, -j4 J V AX = --4 ·4 V= -1---: V , -- I j8.66 VBX = 5 + j8.66 V D AC CIRCUITS UNDER STEADY STATE v and _ VBX - (_1_ _5 +j8.66 ) _ 1+j j8.66 V - V -0.268 + j1 Determine Zx in the circuit of Fig. 10-50. I Y= I V= 27.9~ 50 /30° = 0.558 /27.8° S = 0.494 + jO.26 = Yx + Y x = 0.174 + jO.1 = 0.2 /29.9° S Thus 10.105 _ AX V= (-0.268 + j1)VAB = (1.035 /105°)(48.3 /30°) = 50 /135° V Hence, 10.104 =V AB 177 or 1 1 5 + 3 _ j4 = Yx + 0.2 + 0.12 + jO.16 Zx = 5 /-29.9° 0 Calculate the current I in the circuit of Fig. 10-51. Fig. 10-51 I First, we determine Y ab : Y ab = or ~+ k + 3! j4 = 0.2 Zab = 0.~67 jO.5 + 0.12 + jO.16 = 0.32 - j0.34 = 0.467 /-46.7° S /46.7° = 2.14 /46.7° = 1.47 + j1.56 Z = 2 + j5 + 1.47 + j1.56 = 3.47 + j6.56 = 7.42 /62° 100 / _ {:'' 2 00 A 1= 7.42 /62° = 13.48 ~ Thus 10.106 How much reactive power is taken by the circuit of Fig. 1O-51? I From Prob. 10.105 the complex power is given by S =VI* = (100 L!L) (13.48 /62°) = 1348 /62° = 633 + }1190. Thus, reactive power is 1190 var. 10.107 Several impedances are connected in series. Given: Z, = 15 /30° 0 across Z, is V, = 60 /15° V, find Vs, the voltage across Zs' I 10.108 V, 60 L..!1.:: ° I, = 12 = ... = Is = Z, = 15 = 4 L=.rrA m and = 10 LQ: O. If the voltage Vs = IsZs = (4 /-15°)(10 L!L) = 40 /-15° V Several impedances are connected in parallel. Given: Y, = 0.015 L!L Sand rent through Y, is I, = 2.24 /31.16°, determine Is, the current through Ys' I Zs Ys = 0.05 /30° S. If the cur- I, 2.24~ /'211{:.0 V, =V2 = ... =Vs = Y, = 0.015 L!L = 149.3 Ll!J.2.. V Is =Vs Ys = (149.3 /31.16°)(0.05 /30°) = 7.465 /60.16° A 10.109 A 5-0 resistor is connected in series with a 3-0 capacitive reactance, and the combination across a 4-0 resistor. Is the power factor dependent on the input current (or voltage)? If the total input current is 40 L!L A, determine the power factor of the circuit. I From the data, 178 D CHAPTER 10 (5-j3)4 Z = 5 _ j3 + 4 = 20-j12. ° 9 _ jf-= 2.4 - ,0.53 = 2.46 /-12.45 n Power factor = cos (-12.45°) = 0.976 leading, 10.110 which is independent of the input. Determine the complex power for the circuit of Prob. 10.109. P=/ 2R=(40)22.4=3840W, I From Prob.1O.109, R=2.4n and X=0.53!l. 848 var. Hence, S = P - jQ = 3840 - j848 VA. 10.111 An impedance, (3 + j4) n, is connected in parallel with a lO-n resistor. 1100 W, what is the power in each resistor? I Let Z\ = 3 + j4, Z2 = 10 + jO. Then IjZ\ 12Z 2 , = Q=/ 2X=(40)20.53= If the input power to this circuit is or l!J=l!~1=_ 1121 P, P Also, 2 10.112 = 1I/JR, = (2)2l\0 12R 12 10 =2 V3 2 + 4 2 Iz,1 = 1 ') ,,- and P, + P2 = 1100W 2 600 Wand Hence, PI Two impedances, Zj = 2 + j4 and factor of the circuit is 0.9 lagging. Z2 = R + jO, = P 2 = 500 W are connected in parallel. Determine R so that the power I Equivalent admittance is given by Ye=1/(2+J4)+lIR=(1I1O+lIR)-j/5S. factor, the angle of the admittance must be cos-' 0.9= - 25 .84°. Thus, l/l~~lIR 10.113 = tan 25.84c 0,0.484 R=3.2n A 250-VA 0.5-lagging-power-factor load is connected in parallel to a 180-W 0.8-leading-power-factor load, and the combination to 300-VA 100-var inductive load. Determine the complex power for the combination of the three loads. I S, = 250 /cos -, 0.5 = 250 /60° = 250(cos 60 + j sin 60) = 125 + j216.5 VA S2 = 180 - j180 tan (cos -, 0.8) = 180 - j135.0 Vi\. Thus, 10.114 or For 0.9 lagging power S3 = y' (300)2 - (100)2 + jlOO = 282.8 + j100 VA S = S1 + S2 + S,= 587.8 + j181.5 VA Two impedances, Z, = (2 + j3) nand Z2 = (3 + j6) n, are connected in parallel. for the second branch is 1490 VA. Determine the total complex power. I Since IS21 = (/2)2IZ21, The apparent power we obtain (I) 2 2 = 1490 --==2 = 222 A +6 v:f 2 By current division, or S2 = (/2)2 Z2 = 222(3 + j6) = 666 + j1332, 10.115 S = S\ + S2 = 2202 + j3636 An impedance Z\ = (4 + j4) n is connected in parallel with an impedance reactive power is 2500 var (lagging), what is the total active power? I Y 3 =Y,+Y 2 = Then 10.116 and 4:j4+12~j6=0.2488/-39.57°S P = 2500 cot 39.57° = 3025 W. Find the power consumed in each branch of the circuit of Prob. 10.115. I From Prob. 10.115, Z2 = (12 + j6) n. If the input D AC CIRCUITS UNDER STEADY STATE 1 Y I = 4+ j4 =0.177/-45° But 10.117 PI 11 YI 12 Y2 + P 2 = 3025 W; hence, Y2 179 = 12 ~ j6 = 0.0745/-26.6° 0.177 0.0745 PI = 1975 Wand A general two-loop network is shown in Fig. 10-52. formally solve for the mesh currents. I P 2 = 1050 W. Write the mesh equation in complex (or phasor) form and ZIIII + Z12I2 =VI By KVL we have, in general, matrix form as [ ZIl Z21 Z21I1 + Z22I2 =V2, and which may be written in ZI2] [11] = [VI] Z22 12 0 where ZIl = ZA + ZB = self-impedance of mesh 1 through which 11 flows; Z22 = ZB + Zc is the self-impedance of mesh 2 through which 12 flows; and Z12 = Z21 = ZB which is common to 11 and 12, Using Cramer's rule, the formal solutions are: I =IVI ZI21 I 0 Z22 and .:l where .:l = determinant of the Z matrix. Fig. 10-52 10.118 Fig. 10-53 Using the procedure described in Prob. 10.117, solve for the current 11 in the circuit of Fig. 10-53. I Identifying the various impedances we have j15) + (10 + j20). Hence, .:l = Now, since V = 220 LQ: V, 1 10 + j20 -(10 + j20) 10.119 = ZI2 = Z21 = -(10 + j20), and Z22 = (15 - -(10 + j20) 1 = 450 + ,'150 = 4743/184° 25 + j5 . . we obtain 220 + jO 11 ZII = 10 + j20, 1 o -(10 + j20) 25 + j5 474.3/18.4° 1 = Extend the result of Prob. 10.117 to a three-mesh circuit. 10-54. 220(25.5 ~) 474.3/18.4° ° = 11.82 ~ In particular, solve for 13 shown in the circuit of Fig. SI..2 'v 0 /0J1. SI/.. 10.12- Fig. 10·54 180 D CHAPTER 10 I For the mesh currents shown, in matrix form the VI relationship is: [ Il 7+ j5 j3 5 7 + j3 j5 5 1 = -- (2 - [2) = 17-,21 3 j5 12 + j3 -(2-j2) 10 L~ 5 n~ 5 0 13 = ----Il,-------'--7 + j3 j5 Thus, j5 12 + j3 -2 + j2 1 10.120 5 J[II~ J [~0/3~: L!L] 12 j5 + j3 -(2-j2) 5 0 I -(2 - j2) = 1534.5/25° 17-j2 I For the network shown in Fig. 10-55, what V make:; 667.96/-169° 1534.5m =0.435/-194° A = 12 = O? Fig. 10·55 I Proceeding as in Prob. 10.119, 5 + j5 -j5 12 Solving for V yields 10.121 1 o 30 L!L 0 I 6 10 0 V = ----Il---- = 0 or (5 + j5)(-6)V + 30 L!L(j5)1O = 0 V = 35.4/45° V. Express the result of a nodal analysis of a general network in matrix form. I For an admittance Y across which we have a voltage V, the current I is given by YV = I. For a system of currents and voltages, this result is generalized to [Y][V] = [I], where the [Y] is the admittance matrix and is given by In the symmetric [Y] matrix Y;; is the self-admittant:c of node i (the sum of all the admittances connected to node i) and Y;j is the coupling admittance between nodes i and j. 10.122 Use Prob. 10.121 to solve Prob. 10.120. I Defining the node voltages as V1 and V2 (Fig. For 12 = 0, we must have V1 = 0, or le-55), the matrix equation becomes AC CIRCUITS UNDER STEADY STATE 30 which yields 10.123 m:5 D 181 -0.2 + ]0.2 =0 V = 35.4 /45° V. Find the current I in the circuit of Fig. 10-56 by nodal analysis. Fig. 10-56 l I !+~+! 5 ]2 4 1 4 .:l = 1 0.45 - jO.5 -0.25 -0.25 1 / ° 0.75 + jO.5 = 0.546-15.94 10 -0.25 1 j25 0.75 + jO.5 13.52/56.31 ° VI = 0.546/-15.94° = 0.546/-15.94° =24.76/72.25°Y 1 Hence, 10.124 1= (24.76/72.25°)/(2/90°) = 12.38/-17.75° A. Determine the input impedance Z;t = VII I for the network shown in Fig. 10-57. - Vz. 3~J) v S'n... I :z..n. JJ -j'l..n... Fig. 10-57 I The input impedance is given by Z;t = M.:l \ I ' where.:l is the determinant of the impedance matrix and .:l11 is the cofactor of Zw Hence, 8 Z;] = 1 j2 -3 -3 8 + j5 0 -5 8 + j5 1 10.125 -5 01 -5 7 - j2 = 315.5 ~ = 6.98/-8.7° 0 -5 I 45.2/24.9 7 - j2 For the circuit of Fig. 10-57, determine V, which results in a voltage 5 I m: Y across the 5-0 resistor. Notice that the voltage across the 5-0 resistor is the same as the node voltage V2 • we obtain: Thus, by nodal analysis 182 0 CHAPTER 10 1 5 - j2 l 1 V + j + J5 5 - j2 --1 j5 o - - - - - =V(0.134/-61°) -1 j5 -1 j5 1 1 1 J5 + "5 + 2 - j2 Thus, 10.126 Determine I, in the circuit of Fig. 10-58 by mesh analysis. I The mesh equations in matrix form may be written as 4 + j2 [ -4 ][1,] I, 4 - ,1 -4 [2&:] = -6 L!r Solving for I" we obtain I ,= 1 2 + jO -6+ jO -4 1 4 - ;1 -4 1 4 + j2 1 -4 = -16 - j2 ° 2 + j4 = 3.6/123.7 A 4 - jl Fig. 10-58 10.127 Find the power supplied (or absorbed) by each source of the circuit of Fig. 10-58. I To find the power we must know I, and I,. From Prob. 10.126, I, = 3.6/123.7" A and 4 + j2 21 -4 -6 16 + j12 20 L-143.13° ° ° 12 =----_-4-1 =- 2+,·4 = 447/63.4° =4.47/-206.53 =4.47/153.47 A 4+ j2 -4 4 - jl 1 1 Thus P, = VJ, cos <p, = (2)(3.6) cos 123.7 = -4 W (absorbed) P 2 = V2 I 2 cos 10.128 <P2 and = (6)(4.47) cos (180 - 153.47) = 24 W (supplied) Determine the power dissipated in the 4-D resistor of the circuit of Prob. 10.127. Verify the power balance. I From Prob. 10.127 10.129 2: P,ou<Ce = -4 + 24 = 20 W. which is the power absorbed by the resistor. Write a set of nodal equations in matrix form for the ~ircuit of Fig. 10-59. Then evaluate VI from the following data: Y I = 0.25 L!r S, Y 2 = 0.2/-90° S, Y 3 = o.s L90° S, 11 = 6 L!r A, and 12 = 4 &: A. Y1 V, Y, I, ) Il. Fig. 10-59 AC CIRCUITS UNDER STEADY STATE D 183 I The general nodal equations may be written as Substituting the numerical values and solving for VI yields: VI = 6 - JO.2 I ~ 4 jO.3 I I -----'----'--------'- 0.25 - jO.2 1 10.130 jO.2 jO.21 jO.3 - jl.8 + jO.8 jO.075 + 0.06 + 0.04 -jl.O 0.1 + jO.075 1.0i=2!L 0.125 /36.87° 8.0/-126.87°Y Convert the current source of the circuit of Fig. 1O-60a to a voltage source and find the current through the capacitor by mesh analysis. I If the current through the capacitor is 12 , the mesh equations may be written as 8 + j4 [ -(3 + j4) -(3+ j4)] 8 - j1 Solving for 12 yields 8+ j4 12 = I 100 -43.3 - j25 -3- j4 1 8+ j4 I -3 - j4 8 - j1 I -3 - j4 = 53.6 + j26.8 = 59.92 ~ = 080 /2656° A 75+jO 75!Sr. .. -J &"Jt. fi.fl~{L. .:;fL loo~· j41l V (a.) , _j~.n.. Q.ft ,.n.. ?J sfl... i)3 (b) 10.131 S0L.: V Fig. 10-60 Convert the voltage source of the circuit of Fig. 1O-60a to a current source and find the current through the capacitor by nodal analysis. V, '5'[2. Fig. 10-61 I The circuit is shown in Fig. 10-61. The current through the capacitor is (1/ - j5)(VI - V2 ). The node equations may be written as 184 D CHAPTER 10 [ 0.32 + jO.04 -jO.2 .- ;0.2 ] [ Vj ] 0.2 + jl).2 V2 = [ 20 ] 8.66 + jS Solving for Vj and V2 yields: 20 = V j I8.66 + jS _10.32 + jO.04 Do-jO.2 where I 1 V2 = 0.32 + jO.04 - jO.2 Do 20 8.66 + jS I +, ('>j ) I = 0.096 + jo.on = 0.12 /36.87° - jO .... O. 2 .~ + jS.732 = ~~~1'62.37° = 48.66 + '23.2 V 0.12 /36.87° 0.12 /36.87° ] = 3.0 Thus, Vj + jS.9464 = 6.478 ~ = 4686 + '268 V 0.12 /36.87° 0.12 /36.87° . J. = 2.S7 and V2 Vj - V2 = 1.8- j 3.6 = 4.02 / -63.43° V or Capacitor current is 10.132 Do -jO.2 0.2 + jO.2 (4.02 / -63.43°)/(S / -90°) = 0.3 /26.S7° A, which agrees with the result of Prob. 10.130. Solve for the current 12 in the network of Fig. 10-62. Fig. 10-62 I The mesh equations may be written as 14 + j1 -8 [ -j4 -8 9- j3 jS Solving by Cramer's rule, Do = and 10.133 14 + j1 -8 1 -j4 12 = 14 + j1 -8 -j4 1 SO /30° SO /-30° 0 From Prob. 10.132, j4 -jS 7 - jl --------'-Do--- Determine the complex power drawn by the I -·8 9 -- j':. jS 12 = 2.88 A. (l j4 -jS 1 =676/-28.16° 7 - j1 1 = 19S0 /-148.S3° 676 / -28. W = 2.88 / -120.37° A + }2)-0 impedance of the circuit of Fig. 10-62. Then, S =VI; = Z121; = I~Z == (2 ..38)2 (I + j2) = 8.29 + j16.S8 VA 10.134 Find the current in the 2-0 resistor of the circuit of Fig. 10-63. AC CIRCUITS UNDER STEADY STATE 1- - j ~.f1. -ro6' - - - - : -j 2- r~ ~ - - - - - - ""', 2. 2fl-- 0 185 I~ II_----tI---..-~ • j : :U\ fooL"6D V -to /00 .J 1·4 ~ v L90 -j I.2, ."=_--::_-:_:-::_:-::_:--_-_-_...J.._-_-_-_---_....J~, I In Fig. 10-63 the loops are chosen such that the 2-ll resistor lies on only one of them. -j2 [ 12 = 1 -~j2 0 2 - j4 o o 0 j4 - jl j4 ][11] 12 -oj2 0 2 - j4 j4 0 j4 - jl = 13 (100 /90° - 100 /30°) -100 /90° lOOm lo 10.135 Fig. 10-63 0 [lOOM-lOOm] -100 /90° 100 /30° I j4 -jl = I 721.1 /-163.9° 24.3 /-99.5 = ° 29.7 /-64.4 A Convert the voltage sources of the circuit of Fig. lO-64a to current sources and write a set of nodal equations in matrix form for the new circuit. I The circuit with current sources is shown in Fig. lO-64b for which 50/30° 50m I1 = ~ = 2/-90° = 25/120° A 12 = 75 /20° ]4 = 75 /20° 0 4/90° = 18.75/-70 A Choosing node 5 as the reference node for nodes 1 through 4, we have V V-V + _ I_._2 + 3 -J2 ......!. V 5 .....2 + v-v _ 1 _ ._4 -J5 V = -25/120° .....2 + 2 V -V V-V + _3_ _2 = -18 75/-70° j4 j3 . V-V V-V + _2_._3 -J2 J3 _ 2 _ ._I = 25 /120° V V-V V-V + _4_ _3 + _ 4_ _1 = 18 75/-70° 4 j4 - j5 . ~ _3_ _ 4 which may be written as 1 1 1 --+--j2 3 j5 1 1 1 1 2-72+13 1 j2 1 10.136 1 0 1 1 1 -+ - + 5 j3 j4 1 j3 1 0 j5 j5 j3 1 0 1 0 j2 j4 j4 n[ V2 V3 = V4 -25~ 25/120° -18.75/-70° 18.75/-70° 1 1 1 4+]4-]5 Find the active power supplied by the current source to the circuit of Fig. 10-65. I First we determine the input admittance: 1 1 1 10+]5+2 Y 1 I = - 1 2 111 1 2 2 + 3 + j4 - jlO = --------'--------'--- = VIII 1 2 + 3 + j4 + - jlO 0.31/-50° S j 186 0 CHAPTER 10 .~ J So l}oo V S5L. -j + SA (a.) (b) 2.n. i 'IVy + VI Fig. 10-64 lOA j 5.r.~ 2-t 2>{l. V 2. T-5 1 .SI. Fig. 10-65 I 5 LO" 0 V, = Y = o.312~500 = 16.13/50 y Thus, i Input power or 10.137 P = VI cos cP = (16.13)(5) cos 50° = 51.84 W Without solving for V2 in the circuit of Fig. 10-65, Iktermine the current through the 2-ll resistor. I Since V, = 16.13/50° (from Prob. 10.136), 110 n = 16.13/50° 10 LQ:' and Thus, = 1.613/50° = 1.037 + j1.236 A Ij5 n = 16.13/50° ° . 5 = 3.226/-40 = 2.471 - J2.074 A m 1= 5 /.S}~ = S + jO A (given) 12 n = I - (I,D n + Ij5 {]) =, l.492 + jO.838 = 1. 71/29.3° A AC CIRCUITS UNDER STEADY STATE 10.138 D 187 Evaluate the voltage V2 in the circuit of Fig. 10-65. I From Prob. 10.136, Vj = 16.13/50° V. From Prob. 10.137, 12 n = 1.71/29.3° A. Hence, V2 =Vj - 12n2= 16.13/50°- (1.71/29.3°)2= 10.37 + jI2.36- 2.984- j1.676=7.386+ jlO.684 = 12.98 /55.34° V 10.139 Calculate the total power dissipated in the resistors of the circuit of Fig. 10-65. same as the input power obtained in Prob. 10.136. Verify that this power is the I From Prob. 10.138, V2 = 12.98 /55.34° V. Thus, 13 n = 12.98~ 3 + j4 = 2.596/2.2° A 12 n = 1.71/29.3° A (from Prob. 10.137) lIOn = 1.613/50° A (from Prob. 10.137) 2: Pdissipated = (2.596)23 + (1.71)22 + (1.613)210 = 52 W , which agrees with the power (51.84 W) calculated in Prob. 10.136. 10.140 Determine the reactive power for each element of the circuit of Fig. 10-65. Sum these reactive powers and obtain the complex power by using the result of Prob. 10.139. Verify that the complex power thus obtained is the same as the power obtained from the relationship S = VI*. I To find the reactive powers, we determine the currents through the reactive elements. From Prob. 10.137, I jsn =3.226/-40°A. From Prob. 10.139, V2 Lj1Q n = _ jlO = 12.98~ 10 /-90° L Q<eactive = (3.226)25 + (2.596)24 From Prob. 10.139, P = 52 W. Thus, S =Vjl* = and = 1.298/145.34° A (1.298)210 = 62.14 var S = 52 + j62.14 VA. (16.13/500)(5~) I j4n =2.596/2.2°A, Now, to verify: = 51.85 + j61.78 VA (The two are approximately equal.) 10.141 Obtain the Thevenin equivalent at the terminals ab of the circuit shown in Fig. 10-66. SA a. b L -________- L_ _ _ _ _ _ _ __ _ _ I 10.142 . ZTh 5(3+j4) . = J5 + 5 + 3 + j4 = 2.5 + J6.25 0 Fig. 10-66 1O~ VTh = 8 + j4 (3 + j4) = 5.59/26.56° V Determine the current through the 1-0 resistor of the circuit of Fig. 10-67 by Thevenin's theorem. r I, I J?. b Fig. 10-67 188 0 CHAPTER 10 • From Fig. 10-67, . _ ° _ (10 - j2.65)j7.65 _ ZTh - 10 + 10 _ j2.65 + j7.S4 - 14.59 + J5.3 -15.52 /19.56 0 100 LP'. VTh = -1O---j2-.-65=+ j7.54 J7.54 ° 67.74'&£ V 5".74.&£ ° 14.59 + j5.3 + 1 = 4.11 MA Thus, 10.143 = Knowing the current in the 1-0 resistor, without writing mesh equations, find the output complex power from the voltage source of the circuit of Fig. 10-67. • Since 11 n = 4.11 /45° = 2.905 + j2.905 A 12 1= 11 0 = '"2 J7.54 = 45.21 /45° 7.54/..2S!.. = . 6 / -45° = 4.242 - J4.242 A + 12 = (2.905 + j2.905) + (4.242 - J4.242) A = 7.147 - j1.337 = 7.27 / -10.6° A S =VI* = (100 iJ!..)(7.27 LE)i~) 0= 727 /10.6° = 714.6 + j133.7 VA 10.144 By calculating the sum of the powers dissipated in the resistors of the circuit of Fig. 10-67, verify that the result is consistent with that obtained from Re S in Prob. 11).143. • 10.145 Obtain the Norton equivalent (phasor version) of tbe circuit of Fig. 10-66. • To find IN we short-circuit the terminals ab. Thus, by current division, 10 LQ: 3 + j4 ° IN = 5 + [j5(3 + j4)]/(3 + j9) 3 + j9 = 0.83 Ljl&A 10.146 YN = 1 1 ZTh = 2.5 + j6.25 = 0.148 / -68.2° S Obtain the Thevenin equivalent (phasor version) at t he terminals ab of the network shown in Fig. 10-68. 50fL t Q. V -0;----+ Th I> i- Fig. 10·68 c • By voltage division, Vac V Th =V =V ab ac = 12 + j24 ~ . )' 33 + j24 LO L1_ - Vbc 10.147 _ 30 + j60 0 80 + j60 20 ~ - =(Q~J~~_30+j60) ° 33 + J24 80 + j60 20 L!r - 0.326 /169.4 _ 21(12 + j24) 33 + j24 ZTh - V bc 0_ 5 )(30 + j60) 80 + j60 = 47.35 /26.8° 0 1 + - Replace the network of Fig. 10-69 at the terminals ,,ob by its Thevenin equivalent. AC CIRCUITS UNDER STEADY STATE 0 189 _iron.... loA ~~N.---~----~r--~ () (o!.! 1/ '---___~_____ b • By voltage division, VTh 10.148 Fig. 10-69 = (3 + j4)(1O LQ::) 13 + j4 = 3.68 /36.03 V . 1O(3+j4) ZTh=-'lO+ 1O+3+j4 =8.38/-69.23 Obtain the Norton equivalent of the network of Fig. 10-69 at the terminals. • Verify that ° n IN = VTh/ ZTh. By current division, 1OLQ:: 3+"4 IN= 10+ [(-jlO)(3+j4)]/(3-j6) 3-;6 =0.44/105.26°A Z~h = 8.38 / ~69.230 = 0.119 /69.23° S YN = Verification: From Prob. 10.147, VTh ZTh 3.68~ _ ° 8.38 /-69.23° - 0.44 /105.26 A which agrees with IN calculated otherwise. 10.149 Three impedances, Zl' Z2' and Z3' are connected in wye, whereas three other impedances, ZA' ZB' and Zco are connected in delta, as shown in Fig. 1O-70a and b, respectively. Obtain the conditions for their equivalence. (b) b Fig. 10-70 • For equivalence we must have the same impedance at a given pair of terminals in both connections Therefore, Similarly, the delta-to-wye transformation is 10.150 (D == ZA + ZB + Ze): If the three impedances in a delta connection are equal to one another, obtain the ratio Z/>/Zy. • Let ZA = ZB = Ze = Z/>. Then Zl = Z2 = Z3 = (Z/»2/ 3Z/> = Zy. Thus, Z/>/Zy = 3. 190 0 10.151 CHAPTER 10 Given: Zl = 5 0, Z2 = jlO 0, equivalent delta impedances. • = ZA and ZB Given: ZA = (3 - j2) 0, impedances. 10 0 = eonnected in wye, as shown in Fig. 1O-70a. 15 - j5 n ZB = (2 + j3) 0, • and = 5 Find the '150 +] Zc = 10+ j300 Zc = (2 + j16) 0 in Fig. 1O-70b. Find the equivalent wye _ ZAZB = (3-j2)(2+j3) =05- '050 Zl - ZA + ZB + Zc 7 + ]'17 . J. Z2 Similarly, 10.153 = Z l Z2 + ZZ Z 3 + Z3 Z 1 = 5(j1O) + 5(10) + (j1O)1O Z3 10 Similarly, 10.152 Z3 = 3 - j1 0 and Z3 = 1 + j30 Find the input impedance of the network of Fig. 10·71. sJt b------- Fig. 10-71 • First convert the delta-connected reactances to an equivalent wye as shown by dashed lines in Fig. 10-71. Thus, = '1 670 Z 1 = (j5)(j1O) j30 ] . Z2 = L~)(j15) = '2 50 j30 ] . Z = 3 (j1O)(j15) j30 = '50 ] Notice that Z3 and the 5-0 resistor are in series. Similarly, Z2 and the capacitor are in series, and the two series combinations are in parallel. Therefore, Z2 - j22.5 = j2.5 - j22.5 = -j20 Z3 + 5 = 5 + j5 Zi=ZI + Finally, 10.154 (5 + j5)( - j20) ., .. ° 5+j5-j20 ='JI.h7+8+J4=8+J5.67=9.8/35.20 In the circuit of Fig. 10-71, if 100 L!r V is applied a'; ab, what is the input power? same as that absorbed by the 5-0 resistor. • Since Zi = 9.8/35.2° 0, from Prob. 10.153, V 1= Zi Input power vision, 10.155 = 100~ ° 9.8L~,.20 =10.2/-35.2 A P=Vlcos q,=(100)(1O.2)cos(-35.2°)=833.8W. Iso:::: (10.2/-35.2°) 5 + or Verify that this power is the g~ j~:::~22.5 0= Now, P 5n =(/5n)2 5. By current di- (10.2/-35.2°)(1.256/18.43°) = 12.91/-16.77° A P5 n = (12.91)25 = 833.3 W Find the input impedance at the terminals ab of the network shown in Fig. 10-72. AC CIRCUITS UNDER STEADY STATE D 191 "ii b Fig. 10-72 • Converting the lower delta-connected impedances into an equivalent wye, as shown by dashed lines, we obtain (5/20°)( 10 /90°) Zl ° = 5 flQ~+ 10 m + 8 L!r = 2.9/67.32 n Z3 = Z2 = (10 /90°)(8 L!r) 17.27 ~ = 4.63/47.32° n (8 L!r)(5 /20°) ° 17.27 ~ = 2.32/-22.68 n Zl + 6/-45° = (1.12 - J2.67) + (4.24 - j4.24) = 5.59/-16.34° n Z2 + 2 L!r = 5.14 + j3.4 = 6.16/33.48° n Hence, . Zi = 2.14 - ,0.89 = 10.156 (5.59/-16.34°)(6.16~) + (5.59/-16.34°) + (6.16/33.48°) . . = 2.14 - ,0.89 + 3.2 + ,0.41 5.34 - j0.48 = 5.36/-5.14° n By converting the wye-connected impedances of the circuit of Fig. 10-73 to an equivalent delta (shown by dashed lines) obtain the input impedance at the terminals ab. Fig. 10-73 • ZB = 33.62/117.49° ° 6/90° = 5.6/27.49 n Zc = 33.62/117.49° 2/30° = 16.81/87.49 Now the parallel combinations (Zc and 10 /-60° n) and (Z A and 7 /50° n) are in series. be Zse. Then (16.81/87.49°)(10 /-60°) (11.21/77.49°)(7 /50°) Zse = (16.81~) + (10 ~) + (11.21~) + (7 M) = 17.19 - j3.9 Finally, 10.157 ° n Let this impedance ° ° 16.89/-27.32 + 4.43/60.51 = = 17.62/-12.7JD n (5.6 ~)(17.62/-12.7n (5.6/27.49°) + (17.62! -12.770) = 4.45/18.1° n Determine the voltage Vx in the circuit of Fig. 1O-74a. Then transpose the current source, as shown in Fig. lO-72b, and determine Vx • Hence verify the reciprocity theorem. 192 0 CHAPTER 10 ;- v x (b) • Fig. 10-74 From Fig. 1O-74a, I= Thus, 5 -- ;5 7---:-,3 ° 5~ V = 5 + ~5 (5 L90')( --,'2) = 9.53/21.8° V 7+ ,3 x - , From Fig. 1O-74b, (- '2)5/90° Vx = (5 + j5)1 = (5 + j5) - '7 + j3 10.158 = 9.53/21.8° V Verify the reciprocity theorem by determining Ix in the circuit of Fig. 1O-75a. Then insert the voltage source in the branch where Ix flowed and find the current in .he branch which formerly contained the source (see Fig. 10-75 b) III --jzn _j.. n... (6) • Fig. 10-75 From Fig. 1O-75a, Z = 2 + (3 , Thus, + 1~~_~- j2) = 5 -- /2 24 + j6 5 + j2 n 50 /90° 3 +)4 50 ~(3 + j4) Ix = -Z-;- 5:t~2 = 24 + j6 A From Fig. 1O-75b, Z = , (2 _ '2) + ~j3 + j4) , 5 + j4 = 24 + j6 5 + j4 n Thus, 10.159 A (6 + jlO)-n coil is supplied by a (30 sin lOt)-V ac source connected in series with a 12-V battery, as shown in Fig. 10-76. The internal impedance of the generator is (0.1 + j2) n and that of the battery is negligible. Determine the current i by superposition. AC CIRCUITS UNDER STEADY STATE • 0 193 With the battery replaced by its internal impedance, v (30/\f2)~ ° 1= ZT = (0.1+j2)+(6+jlO) =1.58/-63 A iac = 1.58 v'2 sin (lOt - 63°) = 2.23 sin (lOt - 63°) A or With the generator now replaced by its internal impedance, 12 Idc = RT i = iac + Idc Hence, 10.160 Determine the current in the 4-n 12 6 + 0.1 = = 1.97 A 2.23 sin (lOt - 63°) + 1.97 A = inductor of the circuit of Fig. 10-77 by superposition. Fig. 10-77 • First remove V2 and replace it by a short circuit. Then, owing to V, alone, ,_ 5/60° _ ° 1 - -:---4 ·6 -2.5~A J -J Next, remove V, and replace it by a short circuit. 1"= Thus, 10.161 Ij4 n Then, owing to V2 alone, ~O /3?0 = 5/120° A J4 - J6 =1' + 1"= 2.5/150° + 5/120° = 7.27 /130° A Repeat Prob. 10.160 by solving mesh equations. • We define the mesh currents as shown in Fig. 10-77. [ 10 -10 o Then M] 12-10 - j2 -2 - 0][1,] 2 12 = [5 0 2 13 10 /30° Solving for 12 yields 1 2 = 10 5/600 -10 0 0 10 /30° 0 -2 2 -10 12 - j2 -2 -2 2 10 1-10 0 01 ~ ~ 290.9 /40° 4D / -90' ~ 7.27 A 130 which agrees with the previous result. 10.162 In the circuit of Fig. 10-78 we have Z5 = 6/25° n, and Z6 = 15/40° n. • (a) (b) I, = 20 /60° A, 12 == 50 /30° A, 13 = 10 /210° A, 14 = 70 /100° A, Determine (a) the current through Z3 and (b) the voltage across Z5. 13 - 12 = 10 /210° - 50 /30° = (-8.66 - j5) - (43.3 + j25) 12Z 5 = (50 /30°)(6 /25 0 ) = -51.96 - j30 = 300 /55° V = 60 /210 0 A 194 0 CHAPTER 10 -;s\--- !J ~ t, !;) tl ~ G;' z" t-J 10.163 Calculate the apparent power, true power, and reactive power taken by Z5 of Prob. 10.162. • s= Thus, 10.164 Fig. 10-78 P =, l::'.S94 kW IS.000 kVA Q = 6.339 kvar Solve for the currents I) and 12 in the circuit of Fig. 0-79a. O.O"J...J1.. ~:;~ loo Doe V (a) sOoo/.2° A t~~~· s_L30 A O.O:l.JL J:~'11.. D a02. J? &------'---- (b) • Fig. 10-79 The mesh equations may be written as 1.02 + j3 [ 1 + j3 .l = 1+j3 1.02 I 1.02 + j3 1 + j3 + j3 ][11] I, [100~] = 100 /30° J + j3 I 1.1)2 + j3 = 0.127 /71.4° 100~ 1 100 m 1 = Thus 1 12 = 10.165 1+/l I 1.02 + j3 16S.7 ~ - = 0.127/71.4° 0.127/71.4° = ° 130S.S/-74.S A 1.02 + j3 100 ~ I +j3 100 flQ~, 162.0 /176.2° ° 0.127 L11£ - =, -0.127 L11£ = 1279.9/104.S A Determine the true power supplied by each source of the circuit of Fig. 1O-79a. powers supplied equals the sum of the powers dissipated in the resistors. • Owing to 100 ~-V source we have PI = 100(U()S.S) cos 74.So = 34.31S kW. have P2 = 100(1279.9) cos (104.S - 30°) = 33.SSR kW, or P , + P2 2: PdiSSipated = Verify that the sum of the Owing to 100 /30° source we 67.S73 kW = (130S.S)2(0.02) + (127 9.9:\0.02) + 1(1308.8/-74.So + 1279.9/104.8)12 1 = (34.26 + 32.76 + 0.916) =, 67.93 kW D AC CIRCUITS UNDER STEADY STATE 10.166 195 In the circuit of Fig. 10-80, determine the complex power supplied by the 40 i!r-V source. j 3J1.. IlL l. .12.. '----..,---"IV\ -j /./1.. ~-H-: 'i c 4cL! v Fig. 10-80 • The mesh equations may be written as 40i!r + 60 /90°= 40+ j60 = (8 + j8)11 + (-5 - j6)12 _I -58 -+ j6j8 a1 Thus I1 and SI = VI~ 10.167 = = 40+j60 -5-j61 80 - j60 6 + j5 33.84/55.840 (40 i!r)(30.1/9.24°) - 5 - j6/_ ° 6 + j5 - 33.84/55.84 = 1018.63 ~ 33.84/55.840 = 30.1/-9.24° A 1204/9.24° = 1188.38 + j193.33 VA = What is the voltage aross the capacitor of the circuit of Fig. 1O-80? • Vc = (-jl)1 2 . The voltage across the capacitor is 8 + j8 / -5 - j6 a 12 = Hence 10.168 80i!r - 60 /90° = 80- j60= (-5 - j6)11 + (6+ j5)12 40 + j60 / 80 - j60 12 is found from the mesh equations of Prob. 10.166: 1188.1 ~ = 33.84/55.840 = 35.1/-19.74° A Vc = (1/-90°)(35.1/-19.74°) = 35.1/-109.74° V Determine the sum of the complex powers supplied by the three sources of the circuit of Fig. 10-80. From Probs. 10.166 and 10.167 we have • SI = 1188.38 + j193.33 VA I1 Thus = 12 = 35.1/-19.74° = 33.04 - j11.85 A 30.1/-9.24° = 29.71- j4.83 A 13 = I1 - 12 = -3.33 + j7.02 = 7.77 /115.44° A S2 =V21i = (80i!r)(35.1/19.74°) = 2808/19.74° = 2643 + j948.4 VA and Hence, 10.169 S3 =V31; = (60 /90°)(7.77/-115.44°) = 466.2/-25.44° = 421- j200.3 VA 2: S = SI + S2 + S3 = (1188.38 + jI93.33) + (2643 + j948.4) + (421 - j200.3) = 4252.38 + j941.43 VA Find the sum of the complex powers drawn by all the passive elements of the circuit of Fig. 10-80. the result is the same as that obtained in Prob. 10.168. • For the nomenclature of Fig. 10-80 and by using the results of the preceding problems, we obtain Vx Thus, Similarly, = (3 + j2)11 = (3.6/33.7°)(30.1/-9.24°) = 108.36/24.46° V Sx = V)~ = (108.36/24.46°)(30.1 /9.24°) == 3261.64 /33.7° Vy = =2713.5 +j1809.7 VA (1 - jl)12 = (1.414/-45°)(35.1/-19.74°) = 49.63/-64.74° V Sy =Vyli = (49.63/-64.74°)(35.1/19.74°) = 1742/-45° = 1231.78 - j1231.78 VA and Verify that Vz = (5 + j6)13 = (7.81/50.2°)(7.77 /115.44°) = 60.69/165.63° V Sz =V); = (60.69/165.63°)(7.77/-115.44°) = 471.56/50.19° = 301.91 + j362.24 VA 196 0 CHAPTER 10 2: S = Sx + Sy + Sz = (2713.53 + j1S09. 7) + (1231. 7B- j1231.7S) + (301. 91 + j362.24) = 4247.22 + j940.16 VA which is very close to the result of Prob. 10.169. 10.170 A circuit having a voltage source and a current wurce is shown in Fig. 1O-S1a. Combine the current source with a lO-n inductor, convert it to a voltage source, and draw the resulting circuit. , V2 = (100 LQ::)(j10) = 1000 /90 V. 0 Hence the circuit of Fig. 1O-S1b. j S'.IL 2ft.. .I +.Jl J'OJl};' 4oJl. 1 (a.) 2-.n. j SA , 'OOO/:!, V j/'J'Z. ~ Fig. 10-81 10.171 In the circuit of Fig. 1O-S1b calculate the current in the 4-n resistor. , To determine the desired current we choose th,! mesh currents 11,12' and 13 as shown, so that we have to solve for only. The mesh equations may be written as 11 S.7 + j1.71 4.7 + j1.71 [ o a= S.7 + j 1.71 4.7 + j1.71 I 0 40 /30 40 /30 0 1000 /90 0 Thus, 10.172 Determine 11 = 14 n = 4.7 + j1.7t 46.7 + j6.i1 -40 o 0 -40 40 + jlO 4.7+j1.71 46.7 -" ;6.71 -40 4.7 + j1.71 46.7 + j6.71 -40 ------a-- ][11] [40 /30 12 = 40 /30 0 -40 0 1000 /90 13 1 0 0 I = 590S/91.1° 40+jlO 0 --40 40 + jlO = 175,453 L=.Z.r 590S/91.1 =29.7/-162.1°A 0 11 in the circuit of Fig. 1O-S2. o 'o~ v ~ 2/1- '-----'_ _ _ _ _..J j "'I-/l.. Fig. 10-82 AC CIRCUITS UNDER STEADY STATE , 0 197 The mesh equations may be written as = [~O/~] = [;.~: j~] [104 -~j2j2 6-~3j6 94~l2l[~:l + j2 13 0 0 a= Thus, 10 + jO 5.2 + j3 0 1 and 10.173 11 = 10 - j2 0 1 4 - j2 0 6-j6 -3 4 - j21 -3 9+j2 0 6 - j6 -3 4 - j21 -3 =627.7/-35.67° 9 + j2 -----a-----'-- = 648.2/-40.95" ° 627.7/-35.67° = 1.03/-5.27 A , How much complex power is supplied by each voltage source of the circuit of Fig. 1O-82? To determine S2 we must know 12, which is obtained from Prob. 10.172 as 10 -0 j2 1 12 4 - j2 4 - j21 -3 9 + j2 _ 472.6 ill..:J.Z _ ° - 627.7/-35.67° - 0.75 /93.4 A S2 = V2 1i = (6/30°)(0.75/-93.4°) = 4.5/-63.4° VA. Evaluate the voltage Vx in the circuit of Fig. 10-82. , Vx = (2 + j4)13' where 10 0- j2 1 4 - j2 13 = 0 6 - j6 -3 10 5.2 + j3 0 1 a 422.3 /82.66° ° = 627.7/-35.67° = 0.67/188.33 A Vx = (2 + j4)(O.67 /118.33°) = (4.472 /63.43°)(0.67 /118.33°) = 3/181.76° V Hence, 10.175 0 a = Thus, 10.174 10 5.2+}3 How much total power is dissipated in the resistors of the circuit of Fig. 1O-82? same as the sum of true powers supplied by the two sources. , 2: Verify that the result is the 2 2 (11 11) 6 + 111 + 13124 + 112 - 131 23 + (11 21) 3 = (\.03)26 + 10.363 - }0.1921 2 4 + 10.619 + }0.8461 23 + (0.75)23 = 12.02 W Pd;,,;pated = From Prob. 10.173, 2: PsuppHed = 4.5 cos 63.4° + 10.3 cos 5.27° = 12.27 W The results are close and the error of (12.27 - 12.02)/12.27 = 2.0 percent is due to round off. 10.176 Determine the current 11 in the circuit of Fig. 10-83. o 10 ~;ZV V Fig. 10-83 198 0 CHAPTER 10 , As shown in Fig. 10-83, we choose the mesh currmts so that we have to solve for only one current. the mesh equations become -jl 21 1- jl -j2 o -j22 77.96/-7.37° 12 -jl 21 =, 2~ =38.98/-7.37°A -jl 1 - jl -j2 2 -j2 2 10 10 /120° 1 11 = 10.177 Calculate the power drawn by the 12-D resistor of tilt circuit of Fig. 10-84. Fig. 10-84 , For the meshes shown we have [ 20 + j4 -2- j4 -2- j4 -6 j5 2-.i1 ~6l[lll-[60~l J5 Iz 0 9 - j5 13 0 Solving for 11 yields: -2 - j4 -6 1 2 - jl j5 10 j5 9 - .is 2549.1/-26.57° 1 20 + j4 -2-j4 -61= 617.2/-26.57" -2-j4 2-jl j:i -6 j5 I) - j5 60 o 1 I Hence, 10.178 = =4.13~A P IZll = (4.13)212 = 204.7 W. What is the voltage across the 2-D resistor of the circuit of Fig. 1O-85? Fig. 10-85 Thus, AC CIRCUITS UNDER STEADY STATE , We choose the mesh currents so that we have to solve for 11 only. 15 + j6 10. [ - 3 - J6 17.32+jlO -27.68 - j67.94 I o 15 + j6 10 10 18 - j6 1 -3-j63-j6 1[ - 3 - j61 3 - j6 6+j6 12,395/86.7° 1414 ~ 17.32 + jlO -27.68 - j67.94 0 = 1 8.77 /25.8° A V2fi = 2(8.77/25.8°) = 17.54/25.8°Y. and 10.179 -3- j6 1 3 - j6 6 + j6 10 18 - j6 3 - j6 199 Thus, the mesh equations are: -3 - j6][I1] [ 20 flQ: 3 - ~6 12 = 20 /30° - 90 /60° = 6 + J6 13 0 10 18 - ~6 3 - J6 D The detector of the bridge circuit of Fig. 10-86 has 5-0 resistance. conditions shown. Determine the current through it for the Fig. 10-86 , Again, we choose the mesh currents so that we have to solve for 11 only. 7 + j3 -j3 -j3 7 + j3 2 6 0 -j3 100 7 + j3 I100 6 2 6 8 - j4 [ Hence 10.180 11 13 100 I 2 I 6 8 - j4 -j3 7 + j3 6 2 ][11] 12 = [0 100 1 6 8 - j4 = --------- 7 + j3 -j3 1 2 The mesh equations become 1000 &: ° 284.78/-10.5" =3.51LI0.5 A What is the current in the 3-D resistor of the circuit of Fig. 1O-87? ".,11- -)2./L --\~---,-----4 So[}:V + ~ 4-.n.. Fig. 10-87 200 0 CHAPTER 10 , With the mesh currents shown, we have 5 - j2 -2 [ o -2 7 + j3 -5 43.3 + j25 I70.48 ~ j25.65 7 [I,] [ o ·-5 12 9 + j'l. I, ~~3 ~5 43.3+j25 0 70.48 + j25.65 = 1 I - 5 I) + j 4 3319.4 ~ I, = --15---j2----2---0-1 -- = 290.6/45.4° = 11.42/36.7° A -2 o 10.181 7 + j3 -5 -5 9+j4 Determine the complex power supplied by each sourx of the circuit of Fig. 10-87. , Since I, = 11.42/36.7° A, from Prob. 10.180. S, = V,Ii = (50/30°)( 11. L·2 L-36.7°) = 571 /-6.7° YA To determine S2 we must solve the mesh equation; of Prob. 10.180 for 13: 5 - j2 -2 1 o 13 = ° 11.26/-22.6 A Determine the sum of active and reactive powers abso rbed by the passive circuit elements of the circuit of Fig. 10-87. Check the sum against S, + S2 of Prob. 1 J 181. , First, we obtain 12 from Prob. 10.180: 5 - j2 43.3 + j25 ~2 ~ () ~~4 I 70.48 j25.65 9 1 12 = ------:2=-=9-=-0.-:-6---;/-,.,45=-.4=0- ---IZll or I2l! Thus, = P 11.46 A and S=P = 12 - I, = 9.29/--22.~~ - 11.2/36.7° = -1.08 - j11.41 ISll = 12 - I, = (8.07 -- ;~·.6) SI - (10.39 - j4.32) or I," = 2.34 A + (I2)z3 + (I,)2 4 = (11.42 )2( -2) + (9.29)23 + (11.26)24 = 505.23 var + jQ = 1188.44 + j505.23 YA. From Prcb. 10.181, S, = 571/-6.7° = 567.1 - j66.62 Thus, ~ 2700 ° 290.6/45.4° = 9.29/-29.6 A = = (I,)23 + (I2ll)2 2 + (Is n)25 + (I,) '4 .= (11.42)23 + (11.46)22 + (2.34)2 5 + (11.26)24 = 1188.44 W Q = (I, )2( -2) 10.183 = S2 =V2I~ = (75 /20°)(11.26/22.6°) = 844.5/42.6°YA Hence, 10.182 -2 43.3 + }25 I 7+j3 0 -5 70.48 + j25.6~ 3271.5 /22.8° 290.6 ill.£ .- = 290.6 ill.£ + S2 = 1188.40 + j504.66 YA, S2 = 844/42.6° = 621.3 + j571.28 in agree1l1enl with S above. By nodal analysis, determine the voltage V2 of the cilcuit of Fig. 10-88, :In j 5J? r----.-------~r---~~A 4Jl. - V.. -jIOJ/. Fig. 10-88 AC CIRCUITS UNDER STEADY STATE I D 201 For node 1: For node 2: ° 1 (1 1 1) -100 ~ = - 2 + j5 VI + 2 + j5 + -jlO + 20 Vz 7.88 + j1.39 = (0.5069 - jO.2408)VI - (0.0690 - jO.I724)Vz or -100 + jO = (0.0690 - jO.I724)V1 + (0.1190 - jO.0724)Vz Solving by determinants, we find 54.88/155.42° ° Vz = 0.0796/- 31.490 = 689.5/186.91 Y 10.184 By nodal analysis, determine the current through the 1-11 resistor of the circuit of Fig. 1O-79a. I First, we convert the voltage sources to current sources and obtain the circuit of Fig. 1O-79b, from which we obtain: 5000 ~ + 5000 /30° = (0.~2 + 1: j3 + 0.~2) V V or V=96.5/15.17°Y 96.5 /15.17° II n = 1 + j3 = 3.16!J..12:.. = 30.52/-56.4° A jSfl- 1 Fig. 10-89 10.185 Convert the voltage source of the circuit of Fig. 1O-81a to a current source, write a set of node equations, and determine the voltage across the 4-11 resistor. I From the equivalent circuit, Fig. 10-89, we have: For node 1: For node 2: 1) 1 1 1 8 /10° = ( 5/20° + 4 + 2 + j5 VI - 2 + j5 Vz o 1 (1 1 1) -100 ~ = - 2 + j5 VI + 2 + j5 + jlO + 40 Vz 7.88 + j1.39 = or (0.5069 - jO.2408)VI + (-0.0690 + jO.I724)Vz -100 + jO = (-0.0690 + jO.I724)VI + (0.0940 - jO.2724)Vz Solving by determinants, 16.285/110.79° VI = 0.1371/-87.07" = 118.78/197.86° V 10.186 For the circuit of Fig. 10-89, determine the current through the (2 + j5)-11 impedance. I From the node equations of Prob. 10.185, 54.875/155.43° Vz = 0.1371/-87.07° = 400.26/242.50° Y The voltage across the (2 + j5)-11 impedance is V = VI - Vz = 118.2/197.86 - 400.26/242.50 == (-113.05 - j36.43) - (-184.83 - j355.02) = 326.85 /77.28° Y and 202 0 10.187 CHAPTER 10 Convert the voltage sources of the circuit of Fig. 10-82 to current sources and determine the current through the resistor by nodal analysis. 4-n Fig. 10-90 I The new circuit is shown in Fig. 10-90 for which the nodal equations become ° (1 1 1) 1 -1.6667 LQ:: = (; +4-=,2 + 2 + j4 VI - 2 + j4 V2 1 (1 1 1) -0.894/93.43° = - 2 +74 '\ + 2 + j4 + :3 + 3 - j6 V2 -1.6667 + jO = or 0.0535 - jO.8924 (0.4667- jO.l000)V1 + (-0.1000 + jO.2000)V2 = (-O.1OOel + fO.2000)VI + (0.5000 - jO.0667)V2 Solving by determinants, 1.0067 /179.36° ° VI = 0.260 /-9.1040 = 3.87 /188.46 V 10.188 Determine the voltage Vx shown in Fig. 10-90. I From the node equations of Prob. 10.187, 0.2473 L-1~1.05° V2= 0.260L-91040 and 10.189 3.87 /188.46° _ ° 4.47 /-26.56° - 0.865 /215.02 A whence 095/ = .-1 4995° V . V x = VI - V2 = (-3.83 - jO.57) - (-0.82 - j0.48) = 3.01/-178.22° V How much complex power is supplied by each of the two current sources of the circuit of Fig. 1O-90? I SI = VI I~ = (3.87 D88~6°)(1.667 LQ::) = 6.45 /188.46° VA S2 =V2I~ = (0.95/-149.95~)(O.894/-93.43°) 10.190 = 0.8493/-243.38° VA By using nodal analysis, find the voltage across tt'.e I-n resistor of the circuit of Fig. 10-83. Fig. 10-91 I First we convert the voltage sources to current sources to obtain the circuit of Fig. 10-91. The nodal equations may then be written as: For node 1: 5 LQ:: - 5/210° = (!2 + .!.jl + _1_)V - (_l_)v - ! V - j2 j2 2 I For node 2: - 2 3 ;')v - .!. V j2 5/210° - 5/-210° = -. _..!:-. VI + (_1_. +! + -,2 - J2 1 J2 2 3 AC CIRCUITS UNDER STEADY STATE For node 3: 5/-2100-5~=-~V,-kV2+G+ -~1 D 203 +k)V3 9.33 + j2.5 = (0.5 - jO.5)V, + (0 - jO.5)V2 + (-0.5 + jO)V3 0- j5 = (0 - jO.5)V, + (1 + jO)V2 + (0 + jO.5)V3 -9.33 + j2.5 = (-0.5 + jO)V, + (0 + jO.5)V2 + (0.5 + jO.5)V3 Solving by determinants, 2.5~ V2 = 0.25 ~ = 10.0 /-90° V 10.191 and Determine the currents through the 1-0 inductive reactance and the 1-0 capacitive reactance of the wyeconnected impedances in Fig. 10-91. Verify that the sum of the currents reaching node 0 is zero. I From the node equations of Prob. 10.190 we obtain 9.33 + j2.5 -j5 1-9.33 + j2.5 - jO.5 1 -0.5 1 jO.5 0.5 + jO.5 jO.5 9.745 ~ V, = -----o-.2-5~~-00---- = 0.25 ~ = 39/97.36 V 0 0.5 - jO.5 - jO.5 9.33 + j2.51 -jO.5 1 -j5 1 _-_0_.5_ _J_·0_.5~---9-.3-3-+-j2-.-5- = 9.745 mM = 39/ 2640V V = _ 3 0.25~ 0.25~ 8 . The three currents are V, 39~ /'7 ,l.t~o . I, = 1/90° = 1/90° = 39 ~ = 38.68 + J5 12 = 10 /-90° = - jlO A (from Prob. 10.190) V3 39mM ° . 13 = 1 ~ = 1/-90° = 39/172.64 = -38.68 + J5 A Thus, 10.192 1\ + 12 + 13 = (38.68 + j5) - jlO + (- 38.68 + j5) = 0 (verified) Calculate the current through the 6-0 resistor of the circuit of Fig. 10-84 by nodal analysis. 1'-.12- VI 1 - - - - - - / -+-~-----I ~ 6.12. -j S.Jl. 3Jl. Fig. 10-92 I First, we convert the voltage source to a current source to obtain the circuit of Fig. 10-92 for which the node equations are: Node 1: 1) (1 1) 1 1 1 -50~= ( 12+2+j4+ -jS+3 V1 - 12+3 V2 Node 2: or (currents and admittances being real here) -50 = 0.5166V\ - 0.4166V2 Solving, V2=39.13=39.13~V, and 50 = -0.4166V\ + 0.5833V2 I=V)R=39.13~/6~=6.52~A. 204 D 10.193 CHAPTER 10 The circuit of Fig. 10-93 has a voltage source and a dependent current source. the 25-0 resistor. I Determine the current through 'ut. Fig. 10-93 I By KCL, 1 10° - V iI~l=- j100 11 = 501 = 50 Solving, 10.194 11 = 1/(27 - j2) = 36.94/4.2° mA. What is the voltage across the l-kO resistor of the circuit of Fig. 1O-93? I By using the results of Prob. 10.193, we have Vx = (-491)(1000) 10.195 1 - 251 1 = 2 _ j2 = ( - ~~ 11 )(1000) --%0/1 = (980 /180°)(0.03694/4.2°) = 36.2/184.2° V = How much complex power is delivered by the 5 LlQ':-A current source to the circuit of Fig. 1O-94? I SI =V1(5/-300)VA. To solve for VI' we use nodal analysis: V ~ J2 V-V + _1_._2 =5 D!f -Jl .- which may be written as ~ VI From these, VI -- (2 + jl)Vl jV2 = 5/30° + 0 + jl)V2 = 0 = 3.07 /167.48° V and SI = (3.07 D67.48°)(5/-300) = 15.35/137.48° VA. Fig. 10-94 10.196 Determine the complex power supplied by the dependent current source of the circuit of Fig. 10-94. I In this case, S2 =V22Vi. From Prob. 10.195, Vi = 3.07/-167.48°V, and the node equations give 11.82 jj6.57~ 6 5/13062° V V2 = 1.82 L=.74.1~~ =.1 . Hence, 10.197 S2 = (6.15/130.62°)2(3.07 L:_167.48°) = 37.78/-36.86° VA. From the results of Probs. 10.195 and 10.196, verif) the complex-power balance for the circuit of Fig. 10-94. I Complex power supplied by the two sources: SI + S2 = 15.35/137.48° + 31. 7 8/-36.86° = 18.91 - j12.29 VA Complex power absorbed: p = p Thus, 2n = (V2)2 2 S= P = + jQ (6.15)2 = 18 91 W 2' = 18.91 - j12.29 VA, Q= Q j2 n __ Q -jl n = (Vl)2 _ (1V1 - V21)2 = (3.07)2 _ 17 = -12 29 2 1 2 1 . var which is the same as the total power supplied. AC CIRCUITS UNDER STEADY STATE 10.198 The circuit of Fig. 10-95 contains an independent voltage source and a dependent voltage source. power is delivered by the dependent source? D How much Fig. 10-95 I Since Ix = 11 , the mesh equations become [~2-!:1 /~ jJ[~:J [~J = 12 I = 1 Hence, I j1 -2+j1 Vx 10.199 = 21x = 5.46 /7.66° V Re Sx or = 48 + j12 18 + j2 I= 49.48 M ° = 18.11 /6.34° = 2.73 /7.66 A = Ix 4+j 12 - 11 = 1.48 / -32.9° - 2.73 /7.66° = -1.463 - j1.168 = 1.872 / -141.4° A Re [Vx(12 - 11)*) = Re [(5.46 /7.66°)(1.872 /141.4°») = -8.77W (absorbed) Verify the active-power balance in the circuit of Fig. 10-95. I From the mesh equations of Prob. 10.198, Pabsorbed 12 = 1.48 /-32.9° A. = 4(12)2 + 2(11)2 + 8.77 (from Prob. 10.198) = 4(1.48)2 + 2(2.73)2 + 8.77 = 32.44 W Psupplied 10.200 1 0 4+j 4 - j1 j1 1 = Re [(12 ~)In = Re [(12 ~)(2.73 / -7.66°») = 32.46 W (verified) Obtain the Thevenin equivalent of the circuit of Fig. 10-96 at the terminals ab . .)"t..fl- I. '7J2 ~--~--~A---~~ VT- j2S1 Solil V '-------'-------b I VTh = (-j2)1= (2/-90°) 3 _ ZTh - 10.201 1.67 + 50~ "4 +J Fig. 10-96 '2 =(2/-90°)(13.85/-33.69°)=27.7 /-123.69° V - J (-j2)(3+j4) _ . _ . _ ° _ j2 + 3 + j4 - 1.67 + 0.92 - J2.61 - 2.59 - J2.61 - 3.68 / -45.2 n Represent the circuit of Fig. 10-97 at the terminals ab by its Thevenin equivalent. I Thus, j5(-j4). Zp = j5 _ j4 = - J20 n VTh = 1(10) = 50 ~ V 100 ~ 1= 10 + j20 _ j20 + 10 = 5 ~ 10(10 + j20 - j20) ZTh = 7 + 10 + 10 + j20 _ j20 = 12 n 205 206 D CHAPTER 10 11 loo '-0 V IOJL ~ r 10.202 Fig. 10-97 If the terminals ab of the circuit of Fig. 10-97 are 5t.ort-circuited, determine the current through the capacitor. I By the Thevenin equivalent circuit developed in Prob. 10.201, the current delivered by the generator is VTh ~O LQ:: Ig = ZT~= 12LQ:: = 4.167 .&: A This current is divided in Le-parallel circuit. Hfllce, by current division, _ j5 _ 1(10 j5 _ In o le - I, -:---5'4- - 4.l67 & -:-1 - 20.83 &A 'J - J ] 10.203 Determine the load impedance Z/ across ab of the circuit of Fig. 10-97 which will draw maximum power from the source. Also calculate the maximum power. I For maximum power, using the Thevenin eqUIvalent, we must have Z/. 10.204 = ZTh = 12.&: VIh ZTh+ Z /. 50" and = 12+L~=·:·()8A P max = (Ij12 = (2.08)212 = 51. 92 W Obtain the Thevenin equivalent at the terminals al. of the circuit shown in Fig. 10-98. I Yp,,<alICI = ~ + ~ + ~ = 0.589/-25.1° S or Zpa<aIlel = 0.589)-25.1° = 1.698/25.1° ZTh = 1.698 fl2.J~ + (- j3) Thus, To determine VTh we combine the 3-0 and 5-n Hence, re~.i,.tors = n 2.75/-56° n in parallel to obtain Rp = [(5)(3)1/ (5 + 3) = 1.875 O. 1 ''75 ') VTh = 50 /30° ( :--8'7':':<~-;-4 = 21.22/- 34.9° V L . +] . ·~3J1.. rF J 1-11- ":oJ/- ~ ~. 2' 10.205 --T-lI-' ----'tt S-fl- ')OL.?;~_Vl _______ _ b Fig. 10-98 If a (6 + j9) 0 load is connected at ab of the cirCUit of Fig. 10-98, determine the complex power drawn by the load. I Using the Thevenin equivalent developed in PTb. 10.204, we determine the load current, 1/., as given by 21.22/-34.9° ° 1/. = 2.75/-56° + (6 + j9) = 2.1 L.=:]§J?.. A VI. = 1/(6 + j9) = 22.7 /-20.3° V Hence, 10.206 Determine by Thevenin's theorem the load that must be connected across the terminals ab of the circuit of Fig. 10-99 to draw the maximum power from the source Also find the maximum power. AC CIRCUITS UNDER STEADY STATE .----r--------. D 207 ~ Fig. 10-99 I For the Thevenin equivalent, we must have =450~ 3:6 = 150~Y VTh For maximum power transfer, ZL = Z~h = 5.66 /- 45° = 4 - j4. I == L and maximum power is 10.207 6x3 ZTh=(2+j4)+ 6+3 =5.66/45°0 ZL VTh + ZTh = 1508~ Thus =18.75 L.T: A Pmax = (IL)2 RL = (18.75)2 (4) = 1406 W. Obtain the Thevenin equivalent of the circuit to the left of the terminals ab in Fig. 10-100. currents through the 14-0 resistor and the jlO 0 reactor. Determine the 3Jl.. J• Ion I ;' 1ft I From Fig. 10-100, with Th Fig. 10-100 I ab open, VTh = 30 /20° (1 + j3) + Z I (~~ ~~) + (6 _ j5) = 23.43 /-19.81°Y = [(3 + j2) + (1 + j3)](6 - j5) = 5 /11 55° = 49 + '10 3 + j2 + 1 + j3 + 6 - j5 . . ] 14(+jlO) '6 62 = 8.1 4 /54 .5°.0. ZL = 14 jlO = 4.73 + J. VTh IL=ZTh+ZL 23.43/-19.81° 1 91 / (4.9+j1) + (4.73 +j6.62) = . - 58 ° .17 A V L = ILZ L = (1.91 /-58.17°)(8.14 /54.5") = 15.55/-3.67° Y Hence, 10.208 I 14n = 15.55/-3.67° 14 = 1.11/-3.67° A I j10 n = The Thevenin equivalent values of a section of a circuit are mine the parameters of the Norton equivalent circuit. 15.55/-3.69° ° 10 /90° = 1.55 / -93.69 A ZTh = 5/20° 11 and VTh = 100 /60° V. I 10.209 By Thevenin's theorem, determine the current through the 4-0 resistor of the circuit of Fig. 10-95. Deter- 208 D CHAPTER 10 (a.) I (6) Fig. 10-101 First, we determine VTh from Fig. 1O-101a from which or and VTh = Ix( - jl) + 21x = Ix(2 - 12~ Ix = 4 _ jl = 2.91 /14° A jl) =, (::.91 /14")(2.24 / -26.56°) = 6.51 / -12.56° V From Fig. 1O-101b, which may also be written as (4 - jl)lx 12~ _ VTh and ZTh - Hence, 10.210 + (jl)lsc = _ ""C" - (-2 + jl)I, 6.51/-12.56° 4.24/-98.13° == + (jl)lsc = 0 or Isc = 4.24 / -98.13° ° . 1.53 /85.57 0 = 0.118 + J1.52 0 VTh 6.51L-12.56° 148/3282°A 14 n = ZTh + 4 = -4.39 /20.26° = . - . What is the voltage across the 3-0 resistor of the circuit of Fig. 1O-102? Fig. 10-102 I By mesh analysis we obtain 61 1 - j3(11 - 12) = 9 ~ and The above equations may be combined to yield j311 + (1- j3)12 = 0 or -j9 12 = 2 - j7 Hence, 10.211 An impedance ZL = 2.47 /16° is connected at the terminals ab of the circuit of Fig. 10-102. current through ZL by Thevenin's theorem. I Determine the Notice that we have already determined VTh in Prob. 10.210. Thus, VTh = VJ n = 2.78 / -16° V. To determine ZTh' we short-circuit the 9 ~-V voltage source and apply a voltage V at ab such that a current I flows, as shown by dashed lines in Fig. 10-102. Then, ZTh =V/I: Now, from Fig. 10-202 (as modified by the dashed lines), we have or Hence Thus, ZTh = VTh I L = ZTh + Z L ~ = 7 ~8j2 and V I VI VI 7 + j2 = "3 + 6" + - j3 = V 1:"8 ,= 2.47 / - 16" 0 = 2.377 - jO.679 0 2.78 L--16° 2.78 /-16° 2.47 / -16° + ::.47 /16° = 2.377 + 2.377 = 0.584 / -16 A 0 AC CIRCUITS UNDER STEADY STATE 10.212 Find the Norton equivalent at the terminals ab of the circuit of Fig. 1O-103a. (b) ~j 4 ±-j~n ,fl h (c) I First, we find IN from Fig. 1O-103b. Fig. 10-103 Thus, 0 6(l00 1~(l0 12 /30 111") AO IN = 21 d - /1 = 2L2Q. - -j-l- = 13.1 L..!!bi.A Next, we obtain ZTh from Fig. 1O-103c, from which Z Th 10.213 = (jl)(-j2) = '2n ]'1'2 -] or ] YN = f- = Th ~ =0.5 /-900S ] Calculate the current in the 5-n resistor of the circuit of Fig. 1O-104a. j "ft-. ,o.I2.. yo..n.. . _j S"1l. loo /!.Iy 5.52.. b (CoL) ID J1-- _j'ClJ2. -\\ I V, a. loo ,=,80' b (6) Fig. 10-104 D 209 210 D CHAPTER 10 I First, we replace the circuit to the left of ab by its Thevenin equivalent to obtain the circuit of Fig. 1O-104b, for which VTh = 100& j1~~j5 = 100B80' V = ZTh (- j5)(j1O) '5 -) '10 +) . + 10 = 10 - JlO By using nodal analysis at node a, we obtain: -100 -VI 10 - jlO + 2Ix Solving for VI yields VI = Ix -100 - VI 10 - jlO or 2V I -]5 VI VI = - j5 + 5 VI = 24.27 / -104.04° V. Is n = Hence, 10.214 and +5 ~I = 24~27 L=..!!.11.04° = 4.85 / -104.04° A Determine the true power supplied by each source of the circuit of Fig. 1O-104b. I To find the power Pv supplied by the voltage source, we must know I. Thus, -100 - VI -100 - 24.27 / -104.04° ° 1= 10 _ jlO = 10 __ jlO = 6.86 /210.95 A Pv = Re VI* = Re [(100 L-1~r)(6.86 /-210.95°)] = 588.28 W Similarly, 10.215 PI = Re VI 21; = 2 Re [(24.27/-104.04°)( 24 27 5 /14.04°)] = 0 W Verify that in the circuit of Fig. 10-104, the power dissipated in the resistors is entirely supplied by the voltage source. I Pdi"ipated 2 (VI)2 = I 10 + -5- )2 (24.27)2 (6.86 10 + 5 = 588.4 W = which is the true power supplied by the voltage sour;;e, the power supplied from the current source being zero (from Prob. 10.214). 10.216 Verify the reactive-power balance for the circuit of Fig. 10-104. I Total reactive power absorbed = -(10)(1)2 _. (5)(IJ2 = -[(6.86)210 + (4.85)25] = -588.2 var From Prob. 10.214, Qv = Im (VI*) = Im [(100.1 - !J~00)(6.86 /210.95°)] = - 352.8 var Qv 10.217 + QI = -588.4var Solve for the current I in the circuit of Fig. 1O-105a by superposition. 6JL Ca..) "'SI.. [----"W\)v~· o IOOL.! (h) I Since the impedances (5 + j8) nand (4 1O-105b and c. Hence, (c) j3> n Fig. 10-105 have no effect on I, we use only the circuits of Fig. AC CIRCUITS UNDER STEADY STATE Similarly, ,100~ ° I, = 6+3 = 11.11 L!LA and I' = 25/30° I", = 6+3 and 1"= 211 = 5.56 L!L A 2.78/30° 2 = 1.39/30° A I = I' + I" = 5.56 ~ + 1.39/30° = 6.8/5.87" A Hence, 10.218 = 2.78/30° A 11.1~ ~ D By superposition, find the current in the capacitor of the circuit of Fig. lO-l06a. 'l.f'L (lL) (h) (c) Fig. 10-106 I The circuits with one source at a time are shown in Fig. 1O-106b and c. From Fig. 1O-106b it follows that I~ = 100 ~ / - jlO = 10 /90° A. From Fig. 1O-106c it is clear that I~ = 0 A. Hence, le = I~ + I~ = 10 /90° A. 10.219 Determine the current in the induct or of the circuit of Fig. 1O-106a by superposition. Again, from Fig. 1O-106b we have I~ = 0 A, 2.5 /-90° A. Hence, IL = I~ + I~ = 2.5 /-90° A. I 10.220 and from Fig. 1O-106c we obtain I~ = 25 ~/jlO = By superposition determine the current in the 3-0 resistor of the circuit of Fig. 1O-106a. I Figure 1O-106b yields I~ = 100 ~/(3 + 2) = 20 ~ A. Hence, IR = I~ - I~ = 20~- 5 ~= 15~A. Figure 1O-106c gives I~ = 25 ~/(3 + 2) = 5 ~ A. _j'll 4-5" L(,t V + ~ ~ n. CD lo.fL { '-----...........~ rn"' ~Sn. I '00 J 90" V L!!. I (4.) _j'ft. _j bJ'1. foo/..ioV jSJl (c) Fig. 10-107 212 10.221 D CHAPTER 10 Find the current in the inductor of the circuit of Fig. 10-107 a by superposition. I The component circuits, Fig. 1O-107b and c, give [ ' = 45 M = 45 /150° A L -j6+ j5 Thus, 10.222 mr I" = 100 = 100 /170° A L -j6 + j5 ~.nd IL = I~ - I~ = 45 /150° - 100 /170° = 59.73 ~.J13° A. Using superposition, determine the steady-state current through the 12-V battery of the circuit of Fig. 1O-108a in the time domain. h .. \I (a.) • 11 t. 2..n. . -J /.'19Jl.. + 30fjV /lV (b) (e) Fig. 10-108 I For superposition, we use the circuits of Fig. lO-]08b and c. In Fig. 1O-108b the capacitor acts as an open circuit and the inductor as a short circuit under steady state. Hence, i' = ¥ = 6 A. In Fig. 1O-108c we have shown the ohmic valLles of inductance and capacitance at 1 Hz, since 27T(1)(0.1)=0.630 and Xc = !7TJ(1)(0.08)=' 1.990. Thus, XL = 3°iJr - - - = 20.96 /82.78° A - }1.99 + [2(j0.63)(2 + }0.63)] Ic = and, by current division, I" = (20.96 /82.7[:) ') jO .. ~363 = 6.3 /155.29° A ~ which, in the time domain, becomes 155.29°) A. 10.223 +] . i" = V2(6.3) sin (27Tt + 155.29°). Thus i = i' - i" = 6 - 8.91 sin (6.28t + A 6-V dc generator has a 0.5-V 7200-Hz ripple in the generated voltage. A load having a 2-mH inductance and a 100-0 resistance is connected across the generator terminals. If the internal resistance and inductance of the generator are 30 and 1 mH respectively, determin(! the steady-state time-domain current. I The generator may be represented by two voltag'~ sources in series and the circuit for the generator and load becomes as shown in Fig. 1O-109a. We may then apply superposition by using the circuits of Fig. 1O-109b and c. From Fig. 1O-109b, under steady state, j' =,61103 = 58.3 mA. From Fig. 1O-109c we obtain 0.5iJr I" = -103'1-3<;-:-5 = 2.9 / - 52.8° mA + ] . ~.b Hence, i = 58.3 + V2(2.9) sin (27T7200t - 52.8°) = SR.3 + 4.1 sin (45,239t - 52.8°) mA. AC CIRCUITS UNDER STEADY STATE f.- 0 213 ).- ~ ~ G~-.\ ,---------, 100ft {).S !JV ,_J ./' (a.) t if 6v (6) 10.224 (e) Fig. 10-109 Using superposition determine the current through the 4-0 resistor of the circuit of Fig. 1O-110a. I 4Jl , l.L/v I' 4.(t.. "'f.n.. (l» r'/ Fig. 10-110 (c) I In applying superposition, using Fig. 1O-110b and e, it is clear that the inductive and capacitive branches have no effect on I' and 1". Hence, from Fig. lO-110b, 1'=-0 12&/4=3&A, and from Fig. lO-110e, I" = 30 &/4 = 7.5 & A. Thus, 1 = I' - I" = 3 & -7.5 & = 4.5 /180° A. 10.225 What complex power is absorbed by the inductive branch of the circuit of Fig. lO-110a? 12& I Thus, 10.226 SL = VL VL = 12 & (given) I1 = (12 &)(5.72 /17.44°) = 68.69 /17.44° = 65.53 + j20.59 VA Determine the value of C in the circuit of Fig. 1O-110a if C takes 5 var. is 60Hz. I Hence, 10.227 ° IL= 2+jO.628 =5.72/-17.44 A The operating frequency of the circuit or C = 14.74 p.F. What are the total active and reactive powers supplied by the two sources of the circuit of Fig. lO-110a? value of C is that found in Prob. 10.226. The 214 0 CHAPTER 10 I From Probs. 10.225 and 10.226, the active power supplied by the sources is the same as the powers absorbed by the resistors: p 10.228 Q = 20.59 - 5.0 = 15.59 var = (4.5)24 + 65.53 = 146.53 W Determine the ammeter reading in the circuit of Fig. 10-111. sL!°.{L Fig. 10-111 _ I ° 0 (5 LQ::)(6 /20°) __ ') ° 120 LQ:: ° 1= 8.32 M = 14.42 /-30 A Zin - 2L2Q::+4 M+ 5LQ::+6flQ: --D-MO By current division, 5LQ:: o le! ° = (14.42 /-30 ) 5L:Q~+ 6 /20° = 6.67 /-41 A Ammeter reading is 6.67 A. 10.229 Find the transfer impedance between branch abe and branch ef of the circuit of Fig. 10-111. 10.230 = Vin Z I transfer lout ~E.£ I,I 120 LQ:: 6.67 L..=.1L = 17.9 ffi.:: 0 0 Calculate the current in branch abe (Fig. 10-111) if tre generator is replaced with an ammeter and the ammeter in branch ef is replaced with an 80 /200-V general or. as shown in Fig. 10-112. I V ~ Z"an'fc, = ~ = I out e 0 = 17.9 /41 (from Prcb. 10.229) or labe ~m = 17.9 ill = 4.47 /-21° A abe I • lojJ.oA 3S2 Fig. 10·112 10.231 Solve for I in the circuit of Fig. 10-112 by superposition. I Owing to the current source only (by current di"bion), '4 I' = (20 /60°) 3 ~ j4 = 16 ~~6.8T A = -1.91 + j15.88 A Owing to the voltage source only, 1-2'\ 13°A-276- '118A 1"- 15/30° 3 + j4 -3 - ,---'-'-. J. Thus, 10.232 I = I' + I" = 0.85 + j14.7 = 14.72 /86.7° A. Convert the current source to a voltage source in the circuit of Fig. 10-112 and solve for I. AC CIRCUITS UNDER STEADY STATE D 215 Fig. 10-113 I The modified circuit becomes as shown in Fig. 10-113, from which I 10.233 = 80 ~ +]5 3 + j4 M = -56.29 + j47.5 = 14 73/86 7° A 3 + j4 . . The circuit of Fig. 1O-114a is excited by sources of different frequencies. I We solve the problem by superposition. Thus, from Fig. 2/30° _ ° 1'= 4 + ]'6 + 1 - ]'1 -0.28~A or Determine the current i(t). 1O-114b, in terms of maximum values, i'(t) = 0.28 sin (3t - 15°) A and from Fig. 1O-114c, I" = -(3/10°) 4 + Hence, i(t) = jl~ :jll~ j3/5 = -3.03/16.21° A or i"(t) = -3.03 cos (5t + 16.21°) A i'(t) + i"(t) = 0.28 sin (3t - 15°) - 3.03 cos (5t + 16.21°) A. i (,..e) J-fL .LF "">;, (a) j6.n.. ~ l'-~/A T -J (h) fA . J 10J1. Cc) 10.234 Fig. 10·114 Determine the average power delivered to the 1-0 resistor of the circuit of Fig. 1O-114a. I From the results of Prob. 10.233, the power due to the voltage source is P' = HO.28)2 1 = 39.3 mW. The power due to the current source is P" = H3.03)2] = 4.59 W. Thus, PI n = P' + P" = 0.0393 + 4.59 = 4.63 W. 216 10.235 0 CHAPTER 10 In the circuit of Fig. 1O-114a, if the current source has the same frequency source, show that we cannot use superposition of average power. I From Prob. 10.233, I' = 0.28 / -15° A. (w = 3 rad/s) as the voltage Owing to the current source, 4 + j<i I" = -3 /10° 4 + j6+1-=- jl = -3.06 /21.31° A i'(t) = 0.28 sin (3t - 15°) A Thus, i"(t) = - 3.06 cos (3t + 21.31°) = -3.06 sin (:It + 90° + 21.31 0) = -3.06 sin (3t + 111.31°) A Now we have both currents due to sinusoidal excitations, and I = I' + 1"(1 /90°) = 0.28 Thus, PI n c.li':- 3.06 /111.31° = 3.23 / -64.69° A = ! (3.23)2 1 = 5.23 W By superposition, P 10.236 = P' + PI' = 0.0393 + ~ (3.06)2 1 = 4.72 W "'" PI n If a current through a resistor R is composed by superposition of two parts owing to sinusoidal sources of different frequencies such that i(t) = Ilm sin W I t + 12m sin w2 t, find its rms value and show that the rms value is not the sum of the rms values of the two compon~nts. I Average power dissipated in R is P = H~mR + ;!I~mR = (I~ + I~)R (1) If 1 is the rms value of i(t) then where I1 and 12 are rms values of the two components. 2 P='/ R 1= Comparing Eq. (1) with Eq. (2), we observe that values. 10.237 (2) VI~ + I~, which is obviously not the sum of rms Solve for i(t) for the circuit of Fig. 1O-115a by appl.ying Thevenin's theorem. values. Base phasors on maximum iF r,r ~___JtH Ul i et.) 2.H ----_--1 _)1..11- lA \---11-----1 I", v,"" :? Lt; 0• Fj' Y ~- j" .n. {L. _ _ _ _ L ________- - - - Fig. 10-U5 I At w = 3 rad/s, for the given voltage, the circuit of Fig. 1O-115a is redrawn as that of Fig. 1O-115b from which '1 VTh = 1 ~ jl 3 /50° = 2.12 /95' V 1= Thus, i(t) = 0.38 cos (3t + 10.19°) A. 2.12L95" . 0.5 - jO.S + j6 ZTh = 0.71 /45° - jl = 0.5 - jO.5 = 0.38 /10.19° A n AC CIRCUITS UNDER STEADY STATE 10.238 Determine i(t) in the circuit of Fig. 1O-116a. I In terms of maximum values from Fig. I' D 217 Notice that the sources are of different frequencies. 1O-116b and 1O-116c, we obtain, by superposition, = 3/50° = 1.34 /- 13.43° A 1 + j2 I" = _1_ 2 /100° I+jl i' = 1.34 sin (2t -13.43°) A = 1.41 /55° A i" = 1.41 cos (t + 55°) A i = i' + i" = 1.34 sin (2t - 13.43°) + 1.41 cos (t + 55°) A IJl. r ~---'t-tiH-(t-)-~-. 1 3""('t+~\~ a co< (f+'''') A (a.) I JL. Ul • 2.Jl J w:'- (,.0:;1 (D) 10.239 Fig. 10-116 (c) Change the frequency of the voltage source in Fig. 1O-116a from 2 rad/s to 1 rad/s and repeat Prob. 10.238. Obtain your solution (a) by using superposition, and (b) with both sources acting simultaneously. (c) Would method (b) be valid for the original circuit of Fig. 1O-116a? I (a) In terms of maximum values, 3 /50° I' = - - = 2 12 /5° A 1 + j1 . L:!.. Thus, i'(t) = 2.12 sin (t + 5°) A i" = 1.41 cos (t + 55°) A i(t) = 2.12 sin (t + SO) + 1.41 cos (t + 55°) = 2. 12(sin t cos 5° + cos t sin 5°) + 1.41(cos t cos 55° - sin t sin 55°) = 0.95 sin t + 0.99 cos t A (b) 1= 3 /5~10 + - 11 '1 (-2 /10°) I+}· which, together with (c) 10.240 w +] = 2.12 ~ -1.41 / -35° = 0.95 + jO.99 = 1 rad/ s, implies that i(t) = 0.95 sin t + 0.99 cos t. No, because the sources are of different frequencies. In the circuit of Fig. 10-117 determine the power delivered by the sources. dissipated in the resistor. _ If) Cos Verify that this is the average power 2. t V Fig. 10-117 I _ 10 ~ - 5 /45° _ 7.37 / -28.68° _ ° 12 + j2 _ j - 2.24 /26.57° - 3.30 / - 55.24 A where 3.30 A represents a maximum value. 218 D CHAPTER 10 Ps v = - ~ Re [(5 /45°)(3.3 /55.24°)] = 1.47 W Thus, 10.241 PlO V = ~ Re [(10 L!r)(3.3 /55.24°)] = 9.41 W 1.47 W + 9.41 W = 10.88 W. In the circuit of Fig. 10-118, determine the avera:se power and complex power delivered by the source at w = 1 rad/s. I H Fig. 10-118 I 1= 10.242 2~ 2L!r j1+[(j2)(-j1)]/(j2-j1) = -j1 =2~~A S = ~ [(2 L!r)(2 / -90°)] = 2/ -90° VA P= ReS =0 w = 2 rad/s. (b) Could this circuit be li so, give the values of these elements. (a) Determine the power factor of the circuit shown in Fig. 10-119 at replaced by a resistor in series with an inductor" Fig. 10-119 I (j2)(2 - j 2 1 . ° = 4 + J2 = 4.41 /26.57 n Zeq = 2 + - - 2 ___ (a) n Power factor = cos 26.57° = 0.89 lagging. (b) Yes; from (a), Zeq=R+jwL, 2. Cos with R=4!1, wL=2!1 or L=lH. z.t.v Fig. 10-120 10.243 In the circuit of Fig. 10-120, find Z L such that maximum average power is delivered to it. average power. I Basing phasors on maximum values, we have ~ VTh = 2: j2 2 = 1.41 /45° V ZL = Z;h = 1 + jO n 10.244 IL = 1.412~?': ZTh = - j1 = 0.705 /45° A Determine the voltage v(t) in the circuit of Fig. 10·121. ~ + 2: j2 = = 1 + jO n P= ~(IL)21 = 0.25 W Calculate this D AC CIRCUITS UNDER STEADY STATE 219 j Ifl- ~I.n. 2si",ztv Fig. 10-121 I From Fig. 10-121, V =V I - V2 • To solve by nodal analysis we replace the 2 cos 2t voltage source by j2 and the 2 sin 2t source by 2. Hence, for the two nodes we have - jV2 =-2 V = 2.83 /135° and R -jVI = j2 v(t) = 2.83 sin (2t + 135°) V or _j ID..fl- + If' i" + 1- ~ - i" 1rN Fig. 10-122 10.245 For the circuit of Fig. 10-122 we have I VI = 10 sin (lOOOt + 60°) V and v2 = 5 sin (1000t - 45°) V. Find Zn' In the frequency domain we have Thus, 10.246 The circuit of Fig. 1O-123a operates at 1 MHz. Solve for v(t). (a.) _3/ ,oS!-. r-~~~~--~~----~ 0 .n, ( IJ) Fig. 10-123 I First, we convert the current source to a voltage source to obtain the circuit of Fig. 1O-123b in which Zp = [5(-jlO)]/(5 - jlO). Then writing the node equation for node 1 we have 220 D CHAPTER 10 (IO - V)(5 - jllll V 'v - )100 -~,=-+--- j50 j5 IO - jIO Solving for V yields V= 10.247 -5+)4.8 0.25 - jO.05 6.93 /136.2° 0.255/-11.30 = 27.2 6 or U:!LX'V Determine the ammeter reading in the circuit of Fig. 10-124. V(t) = 27.2 sin (27T x 10 t + 147.so) V The meter reads rms values. Fig. 10-124 I In terms of its maximum value become V = 2 LQ:: Y Zp and (2.8 - )0.1 )1, - (I - )0.5)1, = 2 UI'- = j2 /(1 + j2) = 0.8 + jO.4 n. The mesh equations - (I - 0.67 - )0.5)1, + (2 - )0.5)1, = 0 Solving for 12 yields 0.66 - )1.0 4 9° A 5.47-)0.935 =0.216/- 6. I 2- 10.248 or . 0.216 Ammeter readmg = 7- = 152.8 mA Draw a phasor diagram showing the voltages and currents indicated in Fig. 1O-125a. + V + 1ft I --~IL -- :; ~/0" Fig. 10-125 I II = If. + IR2 = 5 + jIO A Vc 'VIII = 1(5 + jIO) = 11.18 /63.4° Y = (-jl)(5 + jlll) ,= 10 - j5 = 11.18 /-26.6°Y VI =V2 + VIII + Vc = jlO + 5 -1 /10 + 10 - j5 Hence the phasor diagram of Fig. 1O-125b. 10.249 Find vl(t) in the circuit of Fig. 10-126 if v2=sin(l.~tV. = 15 + j15 = 21.2 /45 0 AC CIRCUITS UNDER STEADY STATE I If' r D 221 2 H ~..t"L t I Fig. 10-126 I In terms of maximum value, V2 = 1 L[: Y. The admittance of the RC-parallel branch is 1 + jO.25 S. The series impedance is ZI = 3 + jO.5(2) = 3 + jl.O D. I L[: . 12 = -1- = 1 +JO VI =V2 I} = (1 L[:)(j0.25) = 0 + jO.25 + II (3 + j1.0) = (1 + jO) + (1 + jO.25)(3 + j1.0) = 3.75 + j1.75 = 4.14 /25° Y or 10.250 Y 2 = 1 + jO.5! = VI (t) For the circuit of Fig. 10-127, v 2 =2sin2tY. I, Ifl = 4.14 sin (0.5t + 25") Y Find vl(t) and v}(t). '.J1.. 1~ ~ v;.l-~ j 1.rL ~.l. _j I.f<' Fig. 10-127 I In terms of maximum values, or II = 12 VI = 1(1 10.251 V2 2L[: 2L[:. + j3) + V2 = 1 + j3 + 2=3 + j3 =4.24/45°Y In the bridge circuit of Fig. 10-128, V2 • I V2 + 13 = -·1 + --J 1 ·1 = - - . + - 1 ·1 -J -J -J R I Cl = R2 C 2 == T, = J2 v}(t) = 1.414 sin (2t - 45°) Y . . + 1 + J1 = 1 + J3 A or VI = VI L[: Y, vl(t) =4.24 sin (2t+ 45°) Y and / cp. jwRICI ) V2 = V2 By current and voltage division we obtain 1 + jwRIC I Thus or VI Ivll -=-= V2 11 + jwTI Iv 1 11- jwTI 2 Fig. 10-128 =1 Show that VI = 222 D 10.252 CHAPTER 10 Determine the phase angle 4> for the circuit of Prob. 10.251. I From Prob. 10.251 we have (VI LQ::)(1-jwT)=(V2 li)(1+jwT). Hence, tan'l(wT) or -tan-\wT)=4>+tan- l (wT). J'h!nce 4>=-2tan- l (wT). 10.253 In the circuit of Fig. 10-129, RI = R2 = tan-I (-wT)= 4>+ vi LlC. Show that Vi. =VR2 at all frequencies. L RI IIR • + 72 \: ~ R. ... ..- VR• Vc v)= + VL = RI since RI = vi LI C. Vi. = VR2 V 1 + (l/jwVLC) V R2 Thus, + jwL ;~)L = Similarly, V 10.254 Fig. 10-129 V I - V = ------ R = - - - - , = = R2 + 1/ju.C 2 1 + l/jwVLC at all frequencies. For the data of Prob. 10.253, determine I1 and 12 , I From Fig. 10-129 and Prob. 10.253, VRI =V-VL =V(1 - Consequently, I1 = :-t-l/j~VLC) = IT VRI If; = VT V RI 1 + j:VLC VCTL = 1 + jwVLC V Similarly, 10.255 Ammeters are connected in the two branches of the circuit of Fig. 10-129. the frequency at which the ammeters will show equal readings. For RI = R2 = vi LlC, determine I For the ammeters to show equal readings we mu,;t have 1I11 = 1121, which holds if and only if 11 + jwVLCI = 10.256 1 1 - j wJrc 1 01 1 wVLC= wVLC or 1 w=vrc Find the total power dissipated in the resistors of the circuit of Fig. 10-130. j lo.llV, SA _j'ofL 'ofL Fig. 10-130 AC CIRCUITS UNDER STEADY STATE I D 223 By nodal analysis we obtain: Vj 5 + Vj -j10 Vj - V2 Vj - V2 _ 1 V2 -Vj - j5 +~ +ji(J- + + V2 V2 -Vj jlO j5 V2 = -(-'05) ] . + 10 which simplify to (0.2 + jO.2)V1 - jO.1 V2 = -jO.1Vj 1 + (0.1- jO.1)V2 = jO.5 Using Cramer's rule and solving for Vj and V2 yields = V1 0.1 - jO.1 - 0.05 = 1 _ '2 = 2.24 / -63.4° V 0.02 - jO.02 + jO.02 + 0.02 + 0.01 ] ~~- V = -0.1 + jO.1 + jO.1 = -2 + j4 2 0.02 - jO.02 + jO.02 + 0.02 + 0.01 P 5 10.257 n = (V1)2 = (2.24)2 =1 OW 5 5 . Total power loss = 3.0 W How much active power is supplied by each of the two sources of the circuit of Fig. 1O-130? I P j A = VI, cos ct>j = (2.24)(1) cos (-63.4°) = 1 W P 05 10.258 4.47 /116.6° V = A = V 2 I 2 cos ct>2 = (4.47)( -0.5) cos (116.6° + 90°) = 2 W What is the total reactive power absorbed by all the elements of the circuit of Fig. 1O-130? I Total reactive power absorbed equals total reactive power supplied. Complex power supplied, S = Sj + S2 =Vjl~ + V21; = (2.24 / -63.4°)(1 LQ::) + (4.47 /116.6°)(-0.5 /90°) = (2.24 / -63.4°) - (2.235 /206.6°) = 3 - j1 or, reactive power absorbed = 1 var. 10.259 In the circuit of Fig. 10-131, calculate the complex power supplied by each source. _j~..n. '?Jt , :l- j+JL LfJ V yI -- 10 - 11- I-I , ':t I _ I - y~ Fig. 10·131 I To solve for the power, we must know the currents I and 12 , Writing the mesh equations yields j or and j4(12 - I j) - j212 14 + j8 I j = -1-3- Hence, Thus, = + 21 j = 0 or ° 1.24 /29.7 A 12 = = V,l~= (10 LQ::)(1.24 / -29.7°) = S, (3 + j4)lj - j412 = 10 (2 - j4)lj 20 + j30 13 = + j212 = 0 ° 2.77 /56.3 A 12.4/ -29.7° = 10.77 - j6.14 VA S, = - V,I; = -21 j I; = -(2)( 1.24 /29.5)(2.77 / - 56.3°) = ~6.87 / -26.6° = -6.14 + j3.08 VA 10.260 Calculate the total complex power absorbed by the elements of the circuit of Fig. 10-131. result of Prob. 10.259 is correct. I Hence verify that the P3 n = (lj)23 = (1.24)23 = 4.61 W Qj4fl =jl (I, - 12f4 =jl (1.24 /29.7°) - (2.77 /56.3°W4 = j( 1. 75)24 = j12.27 var or From Prob. 10.259, above result. S = Sj + S2 = 10.77 - j6.14 - 6.14 S = 4.61 + j12.27 - + j3.08 = 4.63 - j15.35 = 4.61 - j3.08 VA j3.06 VA, which agrees with the 224 0 10.261 CHAPTER 10 Determine VI of the circuit of Fig. 10-130 by using superposition . . V. ~----~1-4-;:~~~~~'~~----~ -J V L -_ _ IO~ ~j4fL ~_ _ _ _ _ _~r _____~ Fig. 10·132 I First we reduce the circuit of Fig. 10-130 to lhat shown in Fig. 10-132 (by combining all the parallel elements), from which: 4 V, = 1 10° ( - j2)( _. j10 + 2 + j4) = 2 - '2 V 1 ~ 4-j2.-jlO+2+j4 J Due to 11 : V~=(-0.5/-900)4 '2 '102 + '4=-lV +J V = V; + V'; = 2 - j2 - 1 = 1 - j2 V Hence 10.262 (2 + j4)(4 - j2) -] -J Rework Prob. 10.261 by using Thevenin's theorem. ZTh a. c:::r--;-1L"'b I T- j, 0.11. ' - -_ _ __iO_ b Fig. 10·133 I For this problem also we use the reduced circui: of Fig. shown in Fig. 10-133. 10-132 and obtain the Thevenin equivalent at ab, Thus, ZTh = 4 - j2 + j4 + 2 = 6 + j2 n VTh = (1 LQ:)( 4 - j2) + (0.5 /-90°)(2 + ;4) = 6 - j3 V 6- j3 and Hence, Finally, 10.263 . 6 + J'2 - J'10 = 0.6 + JO.3 1= 1 LQ: - lab = 1 -·1).6 - jO.3 = 0.4 - jO.3 A VI = I( 4 - j2) = (0.4·- jO.3)(4 - j2) =1- j2 V Draw a phasor diagram for the currents shown in the circuit of Fig. 1O-134a. I The phasor diagram is shown in Fig. 1O-134b. - .-c 4 -- '~ r- - . - ~- -1 ,ok. - ~ '- -j I2.SJ2. L .. IT . .~ : - -. ((1,) : Ji. ~- J.. - 1 - c~ r.\ r- f--- . -f--- ~ (~) Fig. 10·134 AC CIRCUITS UNDER STEADY STATE 10.264 If 13 = 2 L!L A 0 225 in the circuit of Fig. 1O-135a, find all other currents and voltages and draw a phasor diagram. . v -+- ~- - - --l'Ws -[- il tv,. V t-- V ~ ~ .', I1 1-,]ILV t- v, t-j, I- r (a. ) - :--.. ,-c- ., ~~ -- I- t-r- '- -f-r- 1/ .~~ "~ _.- -t- I~ - ~ - -- . -+ t. .Y'1" ..... t-. ~--- t-- hA. t Fig. 10-135 (h) I 12 =- 2 + jl I) = jl A VI = 6+ j8 V The phasor diagram is shown in Fig. 1O-135b. 10.265 In Fig. 1O-136a I V = 1 L!L v. Draw a phasor diagram showing all currents. The phasor diagram is shown in Fig. 1O-136b. _. , .7 ~ -- iLJ- - 1': 1: c :, .J '-J ~ ~ ..~ ~- 1/ J -j 'SA I - t- [~I d' 1'- I- ~,l'. 1/ -I~ r--. t;:r- t- t- . c- 1- - t - 1- I"'/ I : 0' / 3.7 -~ t-- (a.) - -~7, r-I- -r-- 0·lA. ,- (6) 10.266 Fig. 10-136 Two impedances Z) = 9.8 / -78° nand Z2 = 18.5 /21.8° n are connected in parallel and the combination in series with an impedance Z3 = 5 /53.1 ° n. If this circuit is connected across a 100-V source, how much true power will be supplied by the source? I 1 Yp = Y I 1 + Y 2 = 9.8 /-78° + 18.5 /21.8° = 0.107 /48.14" S or Zp = Y1 . = 6.236 - J6.961 n p = 5 /53.1° + Zp (3 + j4) + (6.236 - 100 L!L ° 1 = 9.7/ -17.780 = 10.3 /17.78 A P = VI cos Zin = j6.961) (J = 9.7 / -17.78° n = (100)(1O.3)(cos 17.78°) = 980 W 226 0 10.267 CHAPTER 10 The circuit of Prob. 10.266 operates at 60 Hz. Determine the value of inductance which must be connected in series with the circuit so that the power factor becom;:s unity. I From Prob. 10.266, Zin = 9.236 - j2.961. With the inductor in series we have Z;n = 9.236 - j2.961 + For the unity power factor, Xl. = wL = 2.961 or L = 2.961/21T60 = 7.85 mHo jXl.. 10.268 Find the current 1\ through the 6 /40°-0 impedance of the circuit of Fig. 10-137. Fig. 10-137 I For the parallel branch we have Yp = 6 /~Oo + 5 ~; +;;:/20 = 0.8585 / -26.14° S 0 By current division, the required current is 1\ = I ~\ = 20 /30° (6 p 10.269 /40C)((1.l~5~5 /-26.14°) How much complex power is absorbed by the 6 L§!.t:-H impedance of the circuit of Fig. 1O-137? I The voltage across the given impedance is (120 /90°)(20 / - 30°) = 2400 /60° VA. 10.270 = 3.88 /16.14° A V = ZI = (6 /60°)(20 /30°) = 120 /90° V. Thus S = VI* = If the circuit of Fig. 10-137 operates at 60 Hz, find the current 12 in the time domain. I As in Prob. 10.268, 12=1~: =20/30° (5/300)«(1.~5185/-26.140) =4.66/26.14°A Hence 10.271 i2 = v2( 4.66) sin (21T60t -+ 26. J40) = 6.59 sin (377t + 26.14°) A From the results of Probs. 10.268 and 10.270, solve 'or the current 13 shown in the circuit of Fig. 10.137. I By KCL, 13 = I - (1\ + 12)' Now, 1= 20 /30° = 17.32 + jlO (given) 1\ '=3.88/16.14°=3.73+j1.08 (from Prob. 10.268) 12 = 4.66 /26.14° = 4.18 + ;2.05 (from Prob. 10.270) Thus, 10.272 Determine the voltage V, across the current source of the circuit of Fig. 10-137. I 10.273 13 = 9.41 + j6.87 = 11.65 /36.13° A Vs = I( 4 /50° + 6 /60°) + 1\ (6 /40°) = (20 L2.!r)(9.96 /56°) + 23.3 /56.14° = 218.52 /83° V How much total power is absorbed by all the resistive components of the impedances of the circuit of Fig. 1O-137? I PR = Re V,I* = Re [(218.52 /83°)(21) / - 30°) 1= Re (4370.4 /53°) = 2630 W AC CIRCUITS UNDER STEADY STATE 10.274 0 227 A certain series-parallel circuit contains an impedance Zl = 1 - jl 0 in series with a parallel combination of Z2 = 1 + jl 0 and Z3 = 1 - jl n. The sinusoidal driving voltage is 240 /30° V. Sketch the circuit and determine (a) the driving-point impedance and (b) the driving-point admittance. I The circuit is shown in Fig. 10-138, from which Z = (1 + jl)(l- jl) = 1.00 P (1 + jl) + (I - jl) Zin = Zl + Zp = (1- jl) + 1.0 = 2 - jl = 2.2359/-26.57° 0 (a) (b) 10.275 Yin = L = 2.236/ ~26.57° = 0.4472/26.57° S Find the voltage across Z 1 in the circuit of Fig. 10-138. I 1= VY = (240 /30°)(0.4472/26.57°) = 107.34/56.57° A VZ1 = IZ 1 = (107.34/56.57°)(1 - jl) = 151.78/11.57° A 1 1~ Fig. 10-139 Fig. 10-138 10.276 Determine the current supplied by the voltage source of the circuit of Fig. 10-139, where Zl = (3 + j2) = 3.606/33.69° 0 Z4 = (3 + j6) = 6.708/63.435° 0 I Zp(I.2) = 1.952/24.53° 0 Zp(3,4) Z2 = (4 + jl) = 4.123/14.036° 0 Zs = (4 + j7) = 8.062/60.255° 0 Zp(3.4) = 7.0827 /-16.06° 0 Z3 = (2 - j5) = 5.385/-68.199° 0 Z6 = (2 + j3) = 3.606/56.310° 0 Zp(5,6) = 2.4932/57.53° 0 + Zp(5,6) = ZA = 7.0827 /-16.06° + 2.4932/57.53° = (8.1448 + jO.1441) = 8.1460 /1.0133° 0 Zin = (1.952~)(8.1460 /1.0133°) ° (1.7758 + jO.81041) + (8.1448 + jO.1441) = 1.5957 /20.04 0 V 240 L!r ° 1= Zin = 1.5957 /20.040 = 150.41/-20.04 A 10.277 Find the current through Z3 of the circuit of Fig. 10-139. I The current 13 is determined by the division of the current Ix' 240 L!r ° 8.146~ =29.46/-1.01 A Hence, 10.278 IJ 0.1857 /68.2° 0.1412~ = 38.74/51.13° A How much total reactive power is absorbed by the impedances of the circuit of Fig. 10-139. Q =ImS =lm VI* 10.279 = 29.46/-1.01° =Im [(240 L!r)(150.41/20.04°)] = Im (36,098.4/20.04°) = 12.37 kvar The voltage across a certain parallel section of a series-parallel circuit is 120 /60° V. The parallel section consists of a (2.0 + j3.0)-0 branch in parallel with a (4.0 - j8.0)-0 branch. Determine (a) the current in each branch and (b) the total current to the parallel section. 228 0 CHAPTER 10 I Let the branches be labeled A and B. Then ZA~' 2 + j3 = 3.61/56.31° 0, ZB = 4 - j8 = 8.94/-63.43° O. 120i.§!r lA = 3.61/56.31° = 33.24 /3.69° A (a) (b) 10.280 I = lA 120 /60° IB = 8.94 /-63.43° = 13.42 /123.43° A + IB = (33.17 + j2.14) + (-7.39 + jl1.2) = 29.02 /27.36° A (a) the input impedance and For the circuit shown in Fig. 10-140, determine j (b) the input current. (.{L - + ,\It... - j?>SL. lb;l (I.. .1\1\"..--\ V:.4-50VLo~ S V -t !} 25'H,? Fig. 10-140 I ZA = 4 + j3 = 5 /36.87° 0 (a) Z = p ZAZB =3.00/-19.44° ZA+ZB 1= Zin= (2.83-jI.00) -z.- =, 2];3 on 10.281 +j1.0=2.83~0 j·50~ V (b) ZB = 2 - j3 = 3.61 /-56.31° 0 100 ~ = 159.01 A Solve for IB and VA in the circuit of Fig. 10-140. YB 0.277 /56.31° IB = I Y p = 159.01 0.333-/19.44;;- = 132.5 /36.87° A I VA = IZp = (159.01)(3.00 / -19.44°) = 477 / -19.44° V 10.282 Determine the current lA in the circuit of Fig. 10-140 in the time domain. YA 0.20 .I - 36.87° / ° lA ~ I Y p = 159.01 -0.333 /19.44° = 95.53 -56.31 A I i A = 95.53 Y2 Slll (501Tt - 56.31°) A 10.283 Draw the phasor diagram showing all the voltages and currents of the circuit of Fig. 10-140. I See Fig. 10-141. ......... 10.284 V,.,.. V" Fig. 10-141 If 450-V dc is applied across the circuit of Fig. 10-140, determine I, lA' and I B under steady state. Since C acts as an open circuit and L as a short circuit to dc (under steady state), we have I = lA = 4~O = 112.5 A. 10.285 What is the voltage across the 10-0 resistor of the circuit of Fig. 10-142? , Since 1=5 /20° A, VI = RI = 10(5 /20°) = 50 DO° V. IB = 0 and AC CIRCUITS UNDER STEADY STATE lOSt- 1 + - VI + V r-------I-. J - - - - - - - ' 5" l...J.oo A Fig. 10-142 Solve for the current 12 in the circuit of Fig. 10-142. 10.286 I The admittance of the parallel branch is Yp 1 1 = 3 + j4 + 4 - j6 ° 1 + J3 = 0.4262 /-62.48 S By current division, Y2 ° 1I(4-j6) ° 12 = 1 Yp = 5 flQ: 0.4262/-62.48° = 1.63/138.8 A 10.287 Determine V2 , the voltage across the parallel branch, in the circuit of Fig. 10-142. I 10.288 Find V, the voltage across the source in Fig. 10-142, from the results of Probs. 10.285 and 10.287. I 10.289 V = VI + V2 = 50 /20° + 11.75 /82.5" = 48.52 + j28.75 = 56.4/30.65" V What is the input impedance of the circuit of Fig. 1O-142? I 10.290 10 + Zp = 10 + 0.4262 )-62.48° = 11.08 + j2.08 = 11.27 /10.64° n V = Zinl = (11.27 /10.64°)(5/20°) = 56.35/30.64° V, S =VI* = (56.4/30.64°)(5 /-20°) = 282/10.64° VA Solve for the currents I1 and 13 in the circuit of Fig. 10-142. I I1 = 5/20° (3 + 13 = 5 /20° 10.293 which agrees with the result of Prob. 10.288. How much complex power is delivered by the source of the circuit of Fig. 1O-142? I 10.292 = Recalculate V (Prob. 10.288) from the result of Prob. 10.289. I 10.291 Zin j4)(0.42~2/-62.480) (j3)(0.426~ /-62.48°) = 2.35/29.35° A = 3.91/-7.52° V Find the voltages across the various elements of the circuit of Fig. 10-142. I Vj3 n =V2 = 11.75/82.5"V V_ j6n = (1.63/138.8°)(6/-90°) = 9.78/48.8°V Vj4 Il = (2.35/29.35°)(4/90°) = 9.4/119.35" V V3 n = 3(2.35/29.35°) = 7.05 /29.35° V V4 n = (1.63/138.8°)4 = 6.52/138.8° V 10.294 Draw a phasor diagram showing all the currents and volt ages in the circuit of Fig. 10-142. I See Fig. 10-143. 0 229 230 0 CHAPTER 10 Fig. 10-143 10.295 Verify that the true power supplied by the source powers dissipated in the resistors. Pwu<Ce Pdissipated = 3(11)2 or the circuit of Fig. 10-142 is the same as the sum of the = Re S = Rc, (282/10.64°) = 277.15 W + 4(12)2 + 10(1)2 = 3(2.35)2 + 4(1.63)2 + 10(5)2 = 277.19 W CHAPTER 11 L/ Magnetically Coupled Circuits ~ 11.1 Figure 11-1 illustrates magnetic coupling of two circuits, which is measured by the mutual inductance M. Like L, M carries the units H. Including coupling, write the equation of either circuit in terms of instantaneous values and in terms of sinusoidal steady-state values (phasors). M RI L'2) C LI VI N, V2 <D Fig. 11-1 The required equations are: . RIll . Rh 11.2 + (1/,. ([) I R2 di l + LI Tt ± di2 + L2 Tt di2 M Tt ± M Ttl di = VI = v2 or (1) or (2) The mutual coupling between the circuits of Fig. 11-1 is shown in more detail in Fig. 11-2, where the two coils are shown on a common core which channels the magnetic flux <p. To determine the proper signs on the voltages of mutual inductance, apply the right-hand rule to each coil: If the fingers wrap around in the direction of the assumed current, the thumb points in the direction of the flux. Resulting positive directions for <PI and <P2 are shown on the figure. If fluxes <PI and <P2 aid one another, then the signs on the volt ages of mutual inductance are the same as the signs on the voltages of self-inductance. Apply this rule to find the sign for M in Eqs. (1) of Prob. 11.1. Fig. 11-2 I 11.3 In Fig. 11-2, <PI and <P2 oppose each other; consequently, Eqs. (1) would be written with the minus sign. The circuit of Fig. 11-3(a) represents a transformer in which source VI drives a current i 1 , with a corresponding flux <PI as shown. Now, Lenz's law implies that the polarity of the induced voltage in the second circuit is such that, if the circuit is completed, a current will pass through the second coil in such a direction as to create a flux opposing the main flux established by i l . The right-hand rule, with the thumb pointing in the direction of <P2' provides the direction of the natural current i 2 • The induced voltage is the driving voltage for the second circuit, as suggested in Fig. 11-3b. This voltage is present whether or not the circuit is closed. When the switch is closed, current i2 is established, with a positive direction as shown. Suppose the switch in the passive loop is closed at t = 0 when i I = O. Write the KVL equations for the two loops. 231 232 0 CHAPTER 11 VI dil dl M (b) I Fig. 11-3 The equation of the passive loop is R . I 2 2 di di z - M - l =0 + L 2 --dr dt whereas that of the active loop is 11.4 Write the KVL equations for the magnetically coui='led circuits of Fig. 11-4 operating under steady-state ac. The circuit also shows the dot convention stated as follows: (a) When the assumed currents both enter or both leave a pair of coupled coils by the dotted terminals, the signs on the M terms will be the same as the signs on the L terms; but (b) if one current enters by a dctted terminal while the other leaves by a dotted terminal, the signs on the M terms will be opposite to the Sigls on the L terms. I The KVL equations in matrix form may be written as [ R, -iwM R; + jwL 2 ][1,] [VI] 12 0 = R, R2 0J ~. V, RI +jwLI -jwM jwL, V, jwL 2 J R2 J J Fig. 11-4 11.5 jw(L,- M) jw(L 1 - M) jwM Fig. 11-5 Represent the circuit of Fig. 11-4 by a conductively coupled circuit. I Let the desired circuit be as shown in Fig. 11-5, wnere an inductive reactance X M = wM, carries the two mesh currents in opposite directions, whence Z'2 =, Z21 = - jwM in the Z matrix. If now an inductance LI - M be placed in the first loop, the mesh current equation for this loop will be (RI + jwL,)I, - jwMI 2 = Similarly, L2 - M in the second loop result~. in the same mesh current equation as for the coupled-coil circuit. Thus, the circuit of Fig. 11-5 is equivalent 10 that of Fig. 11-4. VI' 11.6 In Fig. 11-1, we have shown two fluxes-<I>lI linking with coil 1, and <1>\2 linking with coils 1 and 2 both. This latter flux gives rise to the mutual coupling between the two coils and is known as the mutual flux. The flux <1>'1 is known as the leakage flux. With the use of the definition of inductance as flux linkage per ampere (see Prob. 7.11) and defining coupling coefficient k = <1>12/<1>1 = <1>21/<1>2' obtain an expression for k in terms of self- and mutual inductances. Coils 1 and 2 in Fig. 11-1 have NI and N2 turns, respectively. I From Prob. 7.11, since A = N<I> = Li, L di di we haw = N d<l> dt . .= e (mduced voltage) From Fig. 11-1, induced voltage e 2 in coil 2 is given by MAGNETICALLY COUPLED CIRCUITS e2 = M di , di 0 233 d = N2 di <P12 or (1) Similarly, e I in coil 1 is given by M = N d<P21 or I (2) di2 Multiplying Eqs. (1) and (2) yields (3) (4) Now (5) (6) and Substituting Eqs. (4) through (6) into (3) gives M yL I L 2 k=-- Hence, 11.7 The device shown in Fig. 11-6 is a transformer. parameters. Represent this by an equivalent circuit. Identify all the circuit Fig. 11-6 , Referring to Fig. 11-6, the primary winding of NI turns is connected to the source voltage VI and the secondary winding of N2 turns is connected to the load impedance ZL. The coil resistances are shown by lumped parameters R I and R 2 • Current 12 produces flux <P2 = <P21 + <P22 and II produces <PI = <P12 + <Pll" In terms of the coupling coefficient k, From these flux relationships, leakage inductances Lll and L22 can be related to the self-inductances LI and L2 by = (1- k)LI LII The corresponding leakage reactances are XII = (1- k)X I The flux common to both windings in Fig. 11-6 is the mutual flux emfs by Faraday's law. <Pm = <P12 - <P21. This flux induces the coil 234 0 CHAPTER 11 dcP," e = N ---I I dt Defining the turns ratio, a = N I /N2 , we obtain from these the basic equation of the transformer, e l /e 2 = a. In terms of rms values, E/E2 = a. The relationship between the mutual inductance can be developed by analysis of the secondary induced emf, as follows: d<P m d<P12 d<P 21 d<P 12 d(k<p2) e 2 = N2 Tt = N2 Tt- N2 Tt = N2 Tt - N2 --;{f Using the results of Prob. 11.6, we have e2 = M di l Tt - di2 kL2 df di l M di2 Tt - -;; Tt = M Now, defining the magnetizing current i.p by the equation or di q• e =M-2 dt we have or Hence, in terms of the coil emfs and various reactances and resistances, we obtain the equivalent circuits of Fig. 11-7a and b. R, jX22 jX" R, R2 a + V, ~ E, + ~ ~ E2 V, ZL R2 jXM + ~ jX , V, (a) jX2 ZL V2 (b) I'ig. 11-7 11.8 From the circuit of Fig. 11-7b determine the input impedance of the transformer. I The governing equations are Eliminating 12 gives or 11.9 An ideal transformer is represented by Fig. 11-8. [t is characterized by the fact that it is lossless (RI = R2 = 0), has no leakages (XII = X 22 = 0), and has an infinitely permeable core (XM = 00). Under these conditions determine its input impedance for a load Z L' + - I, V, a + E, + 11 E2 - + 12 ZL v2 Fig. 11-8 MAGNETICALLY COUPLED CIRCUITS I From Fig. 11-8, VI = El = aE2 = aV2 • 0 235 Since the transformer is lossless, or Hence, 11.10 For an ideal transformer the ampere-turn dot rule states that the algebraic sum of the ampere turns for a transformer is zero. Now refer to the three-winding transformer shown in Fig. 11-9. It has turns NI = 20, N2 = N3 = 10. Find II given that 12 = 10.0 /-53.13° A, 13 = 10.0 /-45° A. Fig. 11·9 I With the dots and current directions as shown, 10(10.0 /-53.13°) + 10(10.0 /-45°) 11.11 When one coil of a magnetically coupled pair has a current 5.0 A, the resulting fluxes <PII and <P12 are 0.2 and 0.4 mWb, respectively. If the turns are NI = 500 and N2 = 1500, find Lp L 2, M, and the coefficient of coupling k. I + <P12 = 0.6 mWb. From Prob. 11.6, <PI = <P11 L = NI <PI = 500(0.6) = 60 mH I Then, from 11.12 from which NIII - N 212 - N313 = 0, or II = 6.54 - j7.54 = 9.98/-49.06° A. II M= N2<p12 = 1500(0.4) = 120mH II 5.0 5.0 M=kyL I L 2, k= ;12= 0.667 L 2 =540mH. Two coupled coils have self-inductances L, = 50 mR and L2 = 200 mH, and a coefficient of coupling k = 0.50. If coil 2 has 1000 turns, and i l = 5.0 sin 400t A, find the voltage at coil 2 and the flux <PI' I M = hI LI L2 = 0.50Y (50)(200) = 50 mH di v 2 = M d: d = 0.05 dt . (5.0 Sill 400t) = 100 cos 400t Assuming a linear magnetic circuit, M = N2<p12 = N 2(k<Pl ) iI iI 11.13 or M 4 <PI = N k i l = 5.0 X 10- sin 400t Wb 2 Write the mesh equations in terms of instantaneous values for the circuit of Fig. 11-10. R v M Co Fig. 11·10 I Examination of the winding sense shows that the signs on the M terms are opposite to the signs on the terms. . di di 1 RI + L - - M - + I dt dt C I. + I dt di di L - - M - = 2 dt dt V L (1) 236 0 CHAPTER 11 di + -1 Ri + L' -dt C or where 11.14 L' = L1 + L2 - 2M. Because i dt = v M;;;,. 'ILIL~;;;" (L, + L 2 )/2, (2) L' is nonnegative. Rewrite the mesh equations of the circuit of Fig. 1l-lO for sinusoidal steady condition. , These equations follow directly from Eqs. (1) and (2) of Prob. 11.13. (R + jwLI - jwM + j~C + jwL where 11.15 I L' = L, + L2 - 2 -- ;lJJI'I1) 1 =V or Thus, (R + jwL' + j~C)1 =V 2M. In a series aiding connection, two coupled coil~ have an equivalent inductance LA; in a series opposing connection, LB' Obtain an expression for M in te:-ms of LA and LB' , As in Prob. 11.13, LI + L2 + 2M = LA and LI + L2 - 2M = L B, which give This problem suggests a method by which M can be determined experimentally. 11.16 M = HLA - L B ). Write the mesh current equations for the coupled coils with currents i, and i2 shown in Fig. 11-11. v Fig. 11-11 , The winding sense and selected current directiom, result in signs on the M terms as follows: 11.17 , 11.18 Obtain the dotted equivalent circuit for the coupled circuit shown in Fig. 11-12a, and use it to find the voltage V across the 10-0 capacitive reactance. Repeat Prob. 11.16 if i2 is given by the dashed line in Fig. 11-11. , To place the dots on the circuit, consider only tne coils and their winding sense. Drive a current into the top of the left coil and place a dot at this terminal. The corresponding flux is upward. By Lenz's law, the flux at the right coil must be upward-directed to oppose the first flux. Then the natural current leaves this winding by the upper terminal, which is marked with a dot. See Fig. 11-12b for the complete dotted equivalent circuit, with currents 11 and 12 chosen for calculation of V. 5 - j5 [ 5 + j3 and 5 + j3 ] [11] [10 L!r -I 10 + j6 12 = 10 - jlO _ V= 1,(-jlO) = 10.15 /23.96° V MAGNETICALLY COUPLED CIRCUITS 0 237 so IOM' v + + v IO~ V -jlO 0 (Cl.) SO. jS 0 IOM' ~ jS 0 • S0 -jIO 0 v IO~ V (h) Fig. 11·12 11.19 Obtain the dotted equivalent for the circuit shown in Fig. l1·13a and use the equivalent to find the equivalent inductive reactance. j40 (£~ jS O • • j60 • j30 Ch) Fig. 11·13 I Drive a current into the first coil and place a dot where this current enters. The natural current in both of the other windings establishes an opposing flux to that set up by the driven current. Place dots where the natural current leaves the windings. (Some confusion is eliminated if the series connections are ignored while determining the locations of the dots.) The result is shown in Fig. l1·13b. Z = j3 + j5 + j6 - + 2(j4) - 2(j3) = j12 il 2(j2) Le., an inductive reactance of 12 il. 11.20 Compute the voltage V for the coupled circuit shown in Fig. 11·14. ·S 0 .' SOM' V J k = 0.8 '10 0 ~' . 30\ 12 SO + V -j40J Fig. 11·14 238 0 CHAPTER 11 I X M = (0.8y5(1O)= 5.66!1, and so the Z matnx is 3 + jl [ -3 - j1.66 _ 12 - Then 50 o ~z 1 =8.62/-24.79 A 0 Repeat Prob. 11.20 with the polarity of one coil reversed. I In this case, the Z matrix is as follows: 3 + '1 Z= [ ] -3 + j9.66 and 11.22 3 + jl -3 - j1.66 V= 12(5) = (8.62i--:~.79°)(5) = 43.1/-24.79°Y and 11.21 1 --: -- j1.66] 8 + j6 -3 + j9.66] (2 8+ j6 3 + jl 50 I -3 + j9.66 0 I-_-'---c~-z--=; 3.82/-112.12° A = V= 12 (5) = 19.1/-112.12° Y. Obtain the equivalent inductance of the parallel-connected, coupled coils shown in Fig. 11-15. I Currents I1 and 12 are selected as shown; then jwO.3 Z = [ jwO.043 or Leq jwO.043 ] jw0.414 Zin =V,II,. (jwO.3)(jw0.414) - (jwO.043)2 _. 02 6 jw0.414 - JW . 9 and = 0.296 H. +~ 12 ) 0.8 H 0.3 H VI 50~ k = 20 50 ~ ~ Y j50 0.7 Fig. 11-16 Fig. 11-15 11.23 jIO 0 For the coupled circuit shown in Fig. 11-16, show that dots are not needed so long as the second loop is passive. I Currents I, and 12 are selected as shown. 5~ ± j4 I 11 = ---,_ _ _ 5 _+...!..j_lO---,-_ 1 2 + j5 ± j4 1 ± j4 5 + jlO 2 + j5 250 + j500 _ 1.° -24 + j45 - 10.96 L...::.:~·64 A 501 12 = _±--,,-j_4_ _0_ = 3.92/-118.07 + 90° A ~z 1 The value of ~z is unaffected by the sign on M. Since the numerator determinant for I, does not involve the coupling impedance, I, is also unaffected. The expression for 12 shows that a change in the coupling polarity results in a 180° phase shift. With no other phasor voltage present in the second loop, this change in phase is of no consequence. 11.24 For the coupled circuit shown in Fig. 11-17, find the ratio V2 /V, which results in zero current 11, V, I = 0= I Then , VI (2 + j2) - V2 ( j2) = 0, from which lV 2 j2 2 + j2 ~z V21VI ,= 1 - jl. I MAGNETICALLY COUPLED CIRCUITS 2a sa j4 Ja J c • :;o~ v + a jS I, -j8 a Fig. 11-18 Fig. 11-17 11.25 239 a jJ{~ e 0 In the circuit of Fig. 11-18, find the voltage across the 5-0 reactance with the polarity shown. I For the indicated choice of mesh currents, 50 /0450 1 -;3 _150~ '81 _ ° 11 = ------'J'-.8-1- 109 -J'9 -1.37 /-40.28 A 3 + j15 j8 -j3 1 Similarly, 12 = 3.66 / -40.28° A. The voltage across the j5 is partly conductive, from the currents 11 and 12 , and partly mutual, from current 11 in the 4-0 reactance: V = (1 1 + I 2 )j5 + 11 (j3) = 29.97 /49.12° v. Of course, the same voltage must exist across the capacitor: V= -I2(-j8) =29.27 /49.12° V. 11.26 Obtain Thevenin and Norton equivalent circuits at terminals ab for the coupled circuit shown in Fig. 11-19a. 40 lOll!' jlO 0 30 (;\ v jS 0 40 a J \ \ I' I / I b (a.) 8.62/48.79" 0 ,.....-£=::J--oQ a 4.82/- 34.60° V ,.....----,----------0 O.559! - 83.39° A _ ------Ob 8.62/48.79" a 0 -----......--------ob (b) (c) Fig. 11-19 I In open circuit, a single clockwise loop current I is driven by the voltage source: 1= (10 L!r)/(8 + j3) = 1.17 /-20.56° A. Then VTh = I(j5 + 4) - I(j6) = 4.82 /-34.60° V. To find the short-circuit current 1', two clockwise mesh currents are assumed, with I 1'= I 8 + j3 -4+ j1 8 + j3 -4 + j1 1~ I = 0.559 / -83.39° A and -4 + j11 7 + j5 The equivalent circuits are shown in Fig. 11-19b and c. ZTh = VTh I' = 12 = 1'. 4.82/-34.60° 0.559 /-83.39° = 8.62 /48.79° 0 240 11.27 0 CHAPTER 11 Obtain a conductively coupled equivalent circuit for the magnetically coupled circuit shown in Fig. 11-20a. • jsn SOL![ V j6 n ~jIOn. 3nJ J I, 12 -j4 z" so SOL\L V n Fig. 11-20 I Select mesh currents I, and 12 as shown, and write the KVL equations in matrix form. 3 + jl [ -3 - j2 The impedances in Fig. 11-20b are chosen to give l;he identical Z matrix. Thus, since I, and 12 pass through the common impedance Zh in opposite directions, Z'1 in the matrix is -Zb. Then, Zb = 3 + j2 o. Since ZII is to include all impedances through which I, p.u,ses, 3 + jl = Z" + (3 + j2), from which Z" = - jl o. Similarly, Z22 = 8 + j6 = Zb + Zc and Zc = 5 +)4 o. 11.28 For the transformer circuit of Fig. 11-7b, k"O.96, R,=1.20, R2=0.30, Xl=200, X2=50, ZL = 5.0 /36.87° n, and V2 = 100 ~ V. Obtain th,~ coil emfs E, and E 2, and the magnetizing current I</>. I X'I =(1-k)X, =(1-0.96)(20) =0.80 a=~=2 X X 22 = (1- k)X2 = 0.2 0 X M = kyX IX 2 =9.60 2 Now a circuit of the form Fig. 11-7a can be constructed, starting from the phasor voltage-current relationship at the load and working back through E2 to E,. 12 = ZV 2 L 100 L'!r ° = 5.0-i)687" L __ = 20 /-36.87 A E2 = I 2(R 2 + jX22 ) + V2 = (20 b19..:..::E~)(0.3 + jO.2) + 100 ~ = 107.2 - jO.4 V E, = aE 2 = 214.4 - jO.8 V 11.29 E2 . I</> = .X = -0.042 - Jl1.17 A J M For the linear transformer of Prob. 11.28, calculate the input impedance at the terminals where V, is applied. I Completing the construction begun in Prob. 11.28, 1 I, = I</> + - 12 = (-0.042 - j11.l7) + 10 /-36.87° = 18.93/-65.13° A a V, = I,(R, + jXII ) + E, = (18.93/-65 . .L~)(1.2 + jO.8) + (214.4 - jO.8) = 238.2/-3.62° V Therefore, 11.30 In Fig. 11-21 three identical transformers are primary wye-connected and secondary delta-connected. A single load impedance carries current IL=30~A. G,ven Ih2=20~A and I"2=I c2 = iO~A, with N, = lON2 = 100, find the primary currents I"" I b " (cl. MAGNETICALLY COUPLED CIRCUITS 0 241 • Iul- NI 11 IL =i) • I bl - ZL NI I 11 " Fig. 11-21 The ampere-turn dot rule is applied to each transformer. + N21a2 = 0 or NIlbl - N21b2 = 0 or = -l~(lO~) = -1 ~A Ibl = l~ (20 ~) = 2 ~ A + N21e2 = 0 or lel = - II~ (10 ~) = -1 ~) = -1 ~ A NIlal l NI c1 lal The sum of the primary currents provides a check: 11.31 lal + Ihl + lel = O. A pair of coupled coils is shown in Fig. 11-22a. Redraw the figure by eliminating the geometrical information and replacing it by the dot notation. If the, coil resistances are negligible, write the voltage equations for the circuit. IH r-h c '\'l.H tI - f--- 511 Fig. 11·22 I The required circuit is shown in Fig. 11.32 In the circuit of Fig. 11-22b I Vab 11-22b for which diab 1 Tt - 2 Tt died (1) di dt di dt (2) Vab = V = -2 --""- +5 ~ cd = 100 sin 1000t Vand iab = 5 sin(lOOOt + 30°) A. Determine v ed . From Eq. (1) of Prob. 11.31, died = ! (di ab _ V ) 2 dt ab dt Thus Vcd = - 2 diab 5 (diab Tt + 2" Tt - Substituting the given expressions for Vab and iab yields Vab ) ved = 2385 sin (1000t + 114.8°) V. 242 11.33 0 CHAPTER 11 Terminals band c of the coils of Fig. 11-22a are CI)nnected and a voltage across a and d. Determine vab and v cd' vad = 400 cos lOOOt V is applied I The circuit becomes as shown in Fig. 11-23, which gives Vab diad diad diad = 1 Tt - 2 Tt = -1 Tt Hence, diad . ,diad -dt = 5 Tt -- ,~ giving diadldt=200coslOOOtA/s, . diad = 3 Tt Vad vab= -200coslOOOtV ~ diad = 400 cos lOOOt = Vab + Vcd = 2 Tt and vcd=600coslOOOtY. L~c.. i"b -:. Ltu!. a.. :) IH 5H Fig. 11-24 Fig. 11-23 cl. 11.34 v cd vac = 400 cos lOOOt Y. Repeat Prob. 11.33, assuming that band d are now connected and I The circuit becomes as shown in Fig. 11-24, frem which di clc vab = 1 --dl Thus, 11.35 di d;c = 40 cos lOOOt A/s so that diac dt +2 - Vab = 120 cos lOOOt V and Vcd = -280 cos lOOOt V Obtain expressions for currents iab and ied in Prob. [1.31. I Solving Eqs. (1) and (2) and integrating yields 11.36 Three coupled coils are wound on a core as shown in Fig. 11-25a. Redraw it by using the dot notation. Use circles to relate coil 1 to the other coils and squares to relate coils 2 and 3. Neglecting coil resistances write the voltage equations. ~ '-, '1. _ _+--4. (J) T---+---f I....-;"--P--l Fig. 11·25 I The required circuit is shown in Fig. 11-25b. The voltage equations are Vab diab = L1 Tt + M12 died _ M ~i!...1 dt 13 d' Vej = -M13 11.37 diab -di- - Vcd M 23 diab = M12 Tt + L2 died Tt + L3 died - M ~ dt 23 dt ~ dt For the coupled coils shown in Fig. 11-26, Lp L 2 , and M are equal to 9, 16, and 12 mH, respectively. currents iab and ied are identical and equal to 3 sin WOOt A. Find the voltages Vab and vcd' The MAGNETICALLY COUPLED CIRCUITS 0------. I 11.38 Vab diab • L, L2 L..----_d died = Lt Tt + M Tt = 63 cos lOool V Vcd = M Fig. 11-26 diab died Tt + L2 Tt = 84 cos WOOl V Find the total instantaneous power absorbed by the magnetic field of the coils of Fig. 11-26 (Prob. 11.37). What is the maximum energy stored in the magnetic field? I P = Vabiab At 1=0, + Ve)cd = 220.5 sin 20001 W stored energy W is zero because both iab and ied are zero. w = 11.39 r-----... c • IJ-------' 0 243 Thus, J: P dr = 220.5 J: sin 2000r dr = 0.11(1 - cos 20001) J W max =0.22J The current iab in the coupled coils of Prob. 11.31 is held constant at 1 A. The current ied has the form shown in Fig. 11-27. Plot the voltages vab and ved for one period (8 time units). 10 - f 10 4 Fig. 11-27 _of Fig. 11-28 I Vab = 1 d~;b - 2 d~;d = -2 d~;d Thus, and The plots are shown in Fig. 11-28. 11.40 Sketch the power absorbed by each winding over 0 $; 1$;8, I Pt = vabiab = 1vab Fig. 11-29. (see Fig. 11-28) and P2 = ve)ed using the data of Prob. 11.39. (see Figs. 11-28 and 11-27). /0 -+t The plots are shown in 244 11.41 0 CHAPTER 11 Sketch the total energy absorbed in the magnetic field of the coils of Prob. 11.39, over 0:::; t:::; 8. w = !Lli~b + !L2i~d - Miabied' Inserting the known numerical values gives w = ~ + ~i~d ~(iCd - ~)2 J. Referring to Fig. 11-27, we see that the w curve will be flat between t = 1 and - 2ied = fa + t = 3, and between t = 5 and t = 7; it will be parabolic over the rest of the cycle. We shall always have w ~0.1 J, with equality at certain times only if Im ~ 0.4 A. Figure 11-30 graphs w in the representative cases Im = 0.2, 0.4, and 0.8 A. I 0.8 {WIw,y, .i 'cil CI{ • O. if A w::O./ J _ _ 1_ _ _ _ _ _ _ _ o.t 0.2 0 "1.0 r- ---, , I I 1...- I"", '" O.iA 3.0 IP2.d I.~ 0./ 2 3 of ~ ~ 7 8 'i t 11.42 10 Fig. 11.30 An electronic amplifier is modeled by the circuit shown in Fig. 11-3l. Determine the turns ratio of the ideal coupling transformer for maximum power transfer to the load resistor R. I Power absorbed by R is Combining the above four equations, The expression above is a maximum when J N, ~ 1"1-' ,4.v~~ of • • NI/N2 ='YrpIR. r- ;J e R RJ;! ... la d "1Z:jL':;b Fig. 11-31 I jwC Fig. 11-32 MAGNETICALLY COUPLED CIRCUITS 11.43 Figure 11-32 shows the equivalent circuit of an amplifier. G =VcdIVab . I 0 245 Obtain an expression for the voltage gain Establishing mesh currents counterclockwise and clockwise in the left and right meshes respectively, [(rp + RI) + jWL 11 ]1 1 + jwMI 2 = jJ.V ah jwMl 1 + [ R2 + j( wL 22 1 w c) ]12 = 0 - Solve for 12 and then obtain G= IhwC = Vab 11.44 -f-LMIC [(rp + RI) + jWL ll ][ R2 + j( wL 22 In the amplifier of Prob. 11.43, if amplifier. I rp "pY(Rl)2 + (WLlI)2, - 1 w C)] + w 2M2 obtain a simplified expression for the gain of the We may write (rp + RI) + jwL l l = rp ( 1 + ( (R2+ 2L2 )1/2 R +. L ) 1 :w 11 = rp 1 + 1 W r 11 p for the given condition. 11.45 li] = rp p Therefore An equivalent circuit of a transformer-coupled amplifier is shown in Fig. 11-33. G=VcdIVab , assuming that rp "p1/wC 1. Determine the voltage gain c I,2. R, Rz LII Lzz C, I, Cz It. d _ fLVcb _ I, z- --;;;- - gm Vc/} I Fig. 11-33 Choosing mesh currents upward through both of the inductances, jwMl 1 + [ R2 + j(WL 22 - w~) ]12 = 0 12 = jwC2V cd Elimination of 11 and 12 between the three equations yields G= VCd = Vab (RI 11.46 -jgm(MlwC1C2) + j(wL l l -1/wC1)][R 2 + j(WL 22 -1/wC2)] + w 2M2 In a two-winding unloaded transformer, the flux linking the primary and the mutual flux are 10 mWb and 8 mWb, respectively. What is (a) the primary leakage flux, and (b) the coefficient of coupling? I (a) Leakage flux = 10 - 8 = 2 mWb. (b) Coefficient of coupling, k = fa = 0.8. 246 11.47 0 CHAPTER 11 If the fluxes in the transformer of Prob. 11.46 are produced by a 2.4-A primary current, and and N2 = 160 turns, I LI = N, Hence, 11.48 4>1 I; 0.01 = 100 2.4 = 0.417 H L2 = 1.064 H. NI = 250, N2 = 300, k = 0.S2. The reluctance seen by the two What is the mutual inductance between the two windings? For a certain two-winding transformer, windings is 420 ampere-turns per weber. M = k NIN2 = 082 ~ . I 11.49 NI = 100 turns determine the self- and mutual inductances of the two windings. !250 )(300) 420 = 146 H A transformer has a coefficient of coupling of 0.7 and primary and secondary turns of 270 and 540, respectively. The primary has a resistance of 10 n and draws 2 A at 120 V 60 Hz. What is the no-load secondary voltage? I Representing the transformer by the circuit of Fig. 11-7b (with v, = (R, + jX,)I, or flRI = 2(10) = 20 V (V,)2 ZL = 00), we have = (J,RY + (J,X,)2 VI = 120 V (given) Thus, Thus, 11.50 E2 = A 60-Hz 240-V generator connected to the primar) 01 a transformer causes 20 /10° A in a load connected to the secondary. The coefficient of coupling is 0.75, and th,~ parameters of the transformer (measured at 60 Hz) are RI = 2.3 n, XI = SO n, R2 = 4.S n, X 2 = 100 n. Determine (a) the magnitude and phase angle of the primary current and (b) the primary current if the load is removed. I Magnetizing reactance is (b) With Xm = kyX I X 2 = 0.75\'(SO)(100) =67.1 n. _ Egen + j X m l 2 _ 240 LQ': + ;67.1(20 M) _ /1 '10 I1 R 'X ~I 3 ·SO -16.5 Ll,lA I+J I •... +J (a) 11.51 O·n*" 118.32 = 165.65 V 12 = 0, I, = (240 L!t) 1(2.3 + JSO) = 3 L.:J~~~ A. The equivalent resistance and equivalent reactance (Ieferred to the high-voltage side) for a certain 25-kVA 2400/600-V 60-Hz transformer are 2.S and 6.0 n, rt:sp,~ctively. If a load impedance of 10 /20° n is connected to the low side, determine (a) the equivalent input impedance of the combined transformer and load; (b) the primary current if 2400 V is applied to the high-voltc.ge side. (a) Using the procedure of Prob. 11.9, . Zin = (R e • Hs + JXe ,lls) + Z'oad,HS " . 0(2400)2 (2.S + J6.0) + 10 fl!r 600 =, (2.S + j6.0) + (150.35 + j54.72) =, (153.15 + j60.72) = 164.75/21.63° n (b) From Ohm's law, I p . HS = 11.52 V HS Z -= ,".HS 2400 L!t / ° 164:]"5-2l.63 L_ _ = 14.57 -21.63 A 0 A 50-kVA 4160/600-V 60-Hz transformer has an equivalent impedance referred to the high side of 9.5 /50° n. Determine the input impedance of the combined transf,)rmer and load if the secondary is supplying rated kVA at 600 V to an O.SO-pf lagging load. MAGNETICALLY COUPLED CIRCUITS I The power factor angle is (J = COS-I 0.80 = 36.86° lagging. The phasor power supplied by the secondary is S2 = 50,000 /36.86° VA 50,000 /36.86 = (600 L!r)I; 0 247 S2 =V21; I; = 83.33 /36.86° 12 = 83.33 / -36.86° A From Ohm's law, Z'oad = V2 I2 Zin.HS 11.53 600 L!r 83.33/-36.86° = 7.20 /36.86° 0 Z,oad.HS = 7.20 /36.86°(~ )2 = 346.11 /36.86° 0 = R e . HS + jXe • HS + Z'oad.HS = 9.50 /50° + 346.11 /36.86° Zin.HS = 355.37 /37.21° 0 Determine the required turns ratio for an impedance-matching transformer that will provide maximum power transfer to a 6-0 loudspeaker from an amplifier with an output impedance of 50 kO resistive. The circuit is shown in Fig. 11-34a. Amplifier (a) (b) Fig. 11-34 I The output impedance of an amplifier is its Thevenin impedance. Thus the amplifier acts as a generator whose Thevenin impedance (resistance, in this case) is 50 kO, as shown in Fig. 11-34b. For maximum power transfer, the amplifier must "see" a load the resistance of which is equal to its own resistance. Placing a transformer between the load and the amplifier causes the amplifier to "see" a resistance equal to Zin = (N I IN2 )2Z!Oad' Thus, for maximum power transfer, Zin must equal 500000. NI -N =913 . 2 11.54 An impedance-matching transformer is used to couple a 20-0 resistive load to an amplifier whose Thevenin equivalent is given in Fig. 11-35a. Determine (a) the turns ratio required for maximum power transfer; (b) the primary current; (c) the secondary current; and (d) the secondary voltage. Vn" (h) (e) Fig. 11-35 248 0 CHAPTER 11 I (a) For a perfect match the Zin of the combined transformer and load must equal Z;h' Thus, N, N = 7.07 2 (b) The equivalent circuit for perfect matching is shown in Fig. 11-35b. VTh 1, = ZTh + Zin 25 L!r =, 1"':)("10 + 1000 = 0.012 _ _1_2 __ . = 7.07 0.0125 LQ: 1 (c) Applying Ohm's law, 5 L!r A 12 = 0.0884L!r A (d) Referring to Fig. 11-35a, V2 11.55 " 201 2 = 20(0.0884) L!r = 1.768 V For the circuit and data of Prob. 11.54, calculate (a) the power delivered to the load if the transformer is not used and (b) the percent increase in delivered pow,~r when the transformer is used. I (a) The circuit becomes as shown in Fig. 11-.35/". 1= 25 iQ: _ 1000 + 20 = 0.0245 A VTh ZTh + 20 P = l~oadR'oad = (0.0245)2(20) = 12 mW (b) From Prob. 11.54, P = (0.0884)2(20) = 156.3 mW (with the transformer). Percent increase is [(0.15630.0120)/0.0120](100%) = 1202.5%. The addition cof an impedance-matching transformer resulted in a twelvefold increase in power delivered to the load. 11.56 Two coupled coils, having a coefficient of couplillg 0.9, have a IO-H mutual inductance. Other data are NI = 100 and N2 = 300. Determine the primary current to produce 0.8-Wb flux to link with the secondary coil. I or 0.8 300 = 100 -11 Thus, 11.57 11 = 24 A. From the data of Prob. 11.56, find the self-inductances of the coils. I eP12 = keP, 0.8= 0.9ePl or Ll = NI eP 1 = 100 0.89 = 3.7 H I, 24 Thus 11.58 eP, = 0.89Wb or M" = ky L1L2 10 = 0.9'113.7 L2 or L2 = 33.37 H. A 30-A current in the primary of a transformer res1jlts in a primary flux of 0.5 Wb. and k = 0.9. Find Lp L 2 , and M. I M = N eP'2 = N. 2 L, = N, eP, T = 200 I, 0.5 NI = 2N2 = 200 ~,~. = 100 0.9 x 0.5 = 15 H '1 , 30 = 3.33 H Given: 30' M= kyL ,L 2 or 1.5 = 0.9Y3.33L2 I Thus, 11.59 L2 = 0.83 H. A 120-V primary voltage produces a 15 /- 60'~ A secondary current in a transformer. Other data are: k = 0.86, RI = 1.8 n, R2 = 4.6 n, X, = 57 n, and X 2 = 150 n. Calculate the primary current. I Proceeding as in Prob. 11.50, we have Xm = h/X1 X 2 =,0.86'11(57)(150) = 79.5 n Thus, 11 = 120iQ:+(15i=60,:)(79.5L2Q::) =228/-60 1.8 + ]57 . 0 .8 A MAGNETICALLY COUPLED CIRCUITS 11.60 0 and = 120L.Q':+(:~8~i?9.5~) Other data are: R1 = 19.1/58.7°A = 3.90, X1 = 65 0, XM = In this case, we have 4/25° = 60 LQ:: + 12 (j48) 3.9+j65 Hence, 11.62 (b) 11 = A transformer draws a 4/25°-A primary current of 60 V. 48 O. Determine the secondary current. I 249 (b) 12 = 15/60° A. 120 L.Q': (a) 11 = 1.8 + j57 = 2.1/ -88.7° A I 11.61 (a) 12 Repeat Prob. 11.59 for 0 60 LQ:: + 12 (j48) 65.11/86.6° = or 12 = 6/32.7° A. A 4160/208-V transformer has a load impedance of 0.01/30° 0 connected to the low-voltage side. The equivalent impedance referred to the high-voltage side is 2/70° O. What is the input impedance of the transformer and the load? I Referred to the high-voltage side, _ (4160)2 ° _ 0_ . 208 (0.01 M) - 4 M - 3.46 + J2 0 Zload - Thus, 11.63 Z;n = 4/30° + 2/70° = 3.46 + j2 + 0.68 + j1.88 = 4.14 + j3.88 = A 2300/120-V transformer is connected to a 0.04/10°-0 load on the low-voltage side. impedance referred to the high-voltage side is 4/70° 0, calculate the primary current. I = (2:~00) 2 (0.04/10°) = 14.69/10° 0 2300 _ /_ 21 76° A 11 = 17.02/21.76° - 135.14 . Thus, The primary and secondary impedances of a 2.2-kV /600 - V transformer are ZI = (1.4 + j3.2) 0 and Zz = (0.11 + jO.25) O. If the secondary current is 41.67 /- 36.87° A at 600 LQ:: V, determine the magnitude of the primary input voltage. I Zload Z2 = 41.6~0~ ~.870 = 14.4/36.87° = ) (14.64/37.39°) 2~gg z Z;n = = 196.81/37.39° 0 (referred to high-voltage side) (1.4 + j3.2) + 196.81/37.39° = 199.87 /37.88° 0 or Hence, 11.52 + j8.64 0 + Ztoad = (0.11 + jO.25) + (11.52 + j8.64) = 14.64/37.39° 0 (referred to low-voltage side) =( 11.65 If the equivalent Referred to the high-voltage side Zload 11.64 5.68/43.1° 0 600 It = 2200 41.67 = 11.36 A VI = 11 (Z;n) = (11.36)(199.87) = 2271.4 V. Find the input current to the two magnetically coupled coils of Fig. 11-36. j + IOn. j 301l. SA + loo Ll V Fig. 11-36 250 0 CHAPTER 11 I The circuit equations may be written as: (2+)IO+)147)l j +)14.71 2 =100&: ) 14.71, + (5 + )3') + )14.7)1 2 20 /90° = Solving for I, yields I, = 5.03 /-86.44° A 11.66 For the coupled circuit of Fig. 11-37, we have R ,= 70 n, IxM I = 94.25 n. Determine 1 if VT = 120 &: V. I Xj = 62.83 n, R2 = 80 n, X 2 = 251.33 n, and • T V -= 12Q~·v''--------Fig. 11-37 I The dots show that X M is negative (see Prob. L.4). V T = 1[(R j + R 2 : + f(Xj + X 2 - 21xMI)] 120 &: = 1{(70 + 80) + j[62.83 + 251.33 - 2(94.25)]} 120 L~ ° 1 = 195.68 L12~~~ = 0.61 /-39.95 A 11.67 Repeat Prob. 11.66 if the connections to coil 2 are interchanged. I Now X M is positive. V T = 1[(R j + R z ) - j(X; + X 2 + 21xMI)] i20 &: = 1{(70 + 80) + j[62.83 + 251.33 + 2(94.25)]} = 1(524.56 /73.38°) 1 = \}.23/-73.38° A 11.68 For each sketch in Fig. 11-38, determine the equivalent inductance of the series-connected coupled coils, if Ll and L2 are 2 and 8 H, respectively, and k = 1. IMI = ky L j L 2 = 1'11(2)(8) = 4 H. For Fig. 11-3:~a and c an assumed direction of current through the series connection will show current entering one dot and leaving the other dot. Hence M is negative, and Leq = Ll + L2 - 21MI = 2 + 8 - 2(4) = 2 H. For Fig. 11-38b and d an assumed direction of current through the series connection will show the current entering both dots or leaving both dots. Hence M is positive, and Leq = L + L2 + 21 MI = 2 + 8 + 2( 4) = 18 H . I j • T • ~ (a) (h) I * (d 11.69 • • (d) Fig. 11·38 For the coupled coils shown in Fig. 11-39, determine I k - 0.7 • (3 + j2) (a) IxMI; (b) Rcq; • (5 + j4) (c) Xeq; Fig. 11-39 (d) Zeq. (a) IXMI= kyX X 2 = 0.7'11(2)(4) = 1.98n (b) R eq ==3+5=8n (c) From Fig. 11-39 we see thatXM is negative. Thus, Xeq=Xj +X2 -2IxM I=2+4-2(1.98)=2.04n. (d) Zeq = Req + jXeq == 8 + j2.04 = 8.26 /14.31': n j MAGNETICALLY COUPLED CIRCUITS 11.70 0 251 In the coupled circuit of Fig. 11-40 determine the ammeter reading. k =, 4SL P" + /ov 0 LE.. Fig. 11-40 V I The ammeter reads 1111, to solve for which we use mesh analysis. First, however, the mutual reactances between the coils AB and CD must be determined: X MAB = ky XAXB = 1y(1)(9) = 3 n X MCD = ky XCXD = 1Y(2)(8) = 4 n Now the mesh equations become (5 + j1)11 - j312 =0 - j31, + (7 + jll)/2 - j413 Solving for 1\ yields 11.71 1\ = 4.23/32.87° A. = lOO M + (11 + j8)13 = 0 -j41 2 Ammeter reads 4.23 A. Three coils (A, B, and C), with k = 1, are wound on a common core so that they are mutually coupled and may be represented by the circuit of Fig. 11-41. Determine 1\. loo L!/V •:J-of(Z~j+)A ~C Fig. 11-41 LA_ _ _ _ _ _ _ _.-Lh_ _ _ _ _ _ _ _....l c I The mutual reactances are X MAB = 1y(1)(4)=2n X MBC = 1y(4)(9) = 6 n X MCA = 1y(9)(1) = 3 n The mesh equations may now be written as [(1 + j1) + (2 + j4) + 2(j2)]11 + [(2 + j4) + j2 + j3 - j6]12 = 100 M [(2 + j4) + j2 + j3 - j6]11 + [(2 + j4) + (3 + j9) - 2(j6)]12 = 100 /90° or Solving for 1\ yields 11. 72 = 100 M (2 + j3)1\ + (5 + j1 )12 = lOO /90° 1\ = 21.42/- 80.16 A. Repeat Prob. 11.71 assuming that coils are wound on separate cores. I In this case all XM's = 0, [(1 Solving for 1\ yields 11.73 (3 + j9)11 + (2 + j3)12 and the mesh equations become = lOO LQ:: + j1) + (2 + j4)]11 + (2 + j4)12 (2 + j4)1\ + [(2 + j4) + (3 + j9)]12 = lOO /90° 1\ = 23.22/-77.65° A. Determine the complex power supplied by each generator of the circuit of Prob. 11.72. I Solving for 12 in Prob. 11.72 yields 12 = 11.55/-300.25° A. Thus, SI =V\I~ = (lOO LQ::)(23.22/77.65°) = 496.6 + j2268.3 VA S2 =V21; = (100 /90°)(11.55/300.25°) = 997.7 + j581.9 VA 252 11.74 0 CHAPTER 11 The coupled circuit of Fig. 11-42a is to be represented by an equivalent T circuit shown in Fig. 11-42b. Determine the parameters of the T circuit. Aa I •• JI.lL. 3 f" -2~ ~ Xl -j2.fl- 4 2 -j3 (a) In- j r-----1-.-vv.-.--1 Z} _ '--_~I(---..... _---_+_---__4.__.-------J n - 13 2 4 X M = kVJ[;.r; I Fig. 11-42 (iI) = 1y(4)(2) = 2.83 n For the magnetically coupled circuit, the mesh equations are 100 LQ:: = (j1 0= + 4 + j4 -- (3 + 0)1\ + (0 - j2.83)1 2 (0 - j283)I J + (2 + j2 + 10 /30° + 0)12 100 = (4 + j2)I J or + (- j2.83)1 2 + (2 + j2 + 10 /30°)1 2 0= (-j2.83)1, For the conductively coupled circuit, the mesh equajons are 100 LQ:: = (j1 + ZA -+ Zc - j3)I J + (-ZC)12 0= (-Zc)IJ + (Zc + ZB + 10 /30°)1 2 For the two circuits to be equivalent, we must have 4 + j2 = Z A + Zc - - j2.83 = --Zc j2 Solving the three simultaneous equations yields ZlI= 2 - jO.83 ZA=4+j1.17n lOO k - 1 4 3 A (3 + j9) 118 Zc = j2.83 n Ler \' k - I • • (I+jl) .!1-. n • ;~~1 ~ n.. C • (6;L8) ~ 512... /) (a) 100 ~v 4 .n.. (h) Fig. 11-43 MAGNETICALLY COUPLED CIRCUITS 11.75 0 253 Replace the transformers in Fig. 11-43a by equivalent T sections. I Figure 11-43b represents the circuit with the equivalent T sections whose parameters are to be determined. The mesh equations for the magnetically coupled circuit of Fig. 11-43a were obtained in Prob. 11.70 as 0= (5 + jl)ll + (- j3)12 100= (-j3)11 0=(0)1 1 + (0)13 + (7 + jll)I 2+ (- j4)I3 + (-j4)12 + (11 + j8)13 The mesh equations for the conductively coupled equivalent circuit in Fig. 11-43b are 0= (4 + ZA + Zc)I, + (- ZJI2 lOO = (- Zc)I, + (Z8 + Zc + ZD + ZF) 0=(- ZF)1 2 + (ZF+ ZE+ 5) I J =12 + (- ZF)I J Equating the respective coefficients, -j3 = -Zc 5 + jl = 4 + ZA + Zc 7 + j11 = Zc + ZB + ZD + ZF -j4=-ZF Solving the five simultaneous equations yields ZA = 1- j2n Referring to Fig. 11-34b, ZB is in series with ZD and is, therefore, represented as one impedance 11. 76 (ZB + ZD)' Assign polarity markings to the coils of Fig. 11-44a . ~\ . . • A (a.) Fig. 11-44 I The markings are shown in Fig. 11-44b where the dots are for coils 1 and 2, triangles for coils 1 and 3, and squares for coils 2 and 3. 11.77 Determine the input impedance for the circuit in Fig. 11-45. ZI=(5+j9)n and Z2=(3+j4)n. I The impedances of the respective coils are Zin = 3 + j4 + 3 + j4 = 6 + j8 = 10 /53.13° n. Fig. 11-45 11.78 Determine the input impedance for the circuit shown in Fig. 11-46. ZI=(5+j9)n and Z2=(3+j4)n. I 11.79 Fig. 11-46 The impedances of the respective coils are Zin = 5 + j9 + 3 + j4 - 2(j6) = 8 + jl = 8.06/7.13° n For each connection shown in Fig. 11-47, determine the overall inductance of the coupled coils in series if L2 = 5 H, k = 0.60. LI = 3 H, I (a) (c) M = 0.6Y(3)(5) = 2.32 H. L = 3 + 5 - 2(2.32) = 3.36 H L = 3 + 5 + 2(2.32) = 12.64 H (b) L=3+5-2(2.32)=3.36H (d) L = 3 + 5 + 2(2.32) = 12.64 H 254 0 CHAPTER 11 Fig. 11-48 Fig. 11-47 11.80 The coil impedances in Fig. 11-48 are ZI = (3 + j8) n, Z2 = (1 + j2) V, If k = 1.0, determine the impedance measured at rerminals TI T 2 . I 11.81 MI2 = l"y!(8)(2) = 4 n Z = (3 Z3 = (5 + jl) n, Z4 = (6 + j3) n. -+ j8) + (1 + j2) - 2(j4) + (6 + j3) = (10 + j5) n For the circuit of Fig. 11-49 calculate the mutual rea·;tance and mutual inductance. k - I Fig. 11-49 I Thus, 11.82 M = 5 127T( 60) = 13.26 mHo In the circuit of Fig. 11-49 determine the ammeter reading. I The mesh equations become [(1 + j2.5) + (5 + j2)]11 + (j5)12 (j5)11 Solving for 12 yields 11.83 12 = 1.12 / -128.61 ° A. = 50 /30° + [(2 + j5) + (3 + jlO)]12 = 20 /60° Amrnder reads 1.12 A. Determine the ammeter readings in the circuit of Fig. 11-50. X,.,.A - 4/L x,.,.- 2SL xolle - 3.11- Fig. 11-50 I The mesh equations are [(2 + j3) + j2]I J + (j2 + jJ)12 =30iQ:: (j2 + j3)I, + [j4 +";5 + (5 + j7) + 2(j4)]12 = 50 /30° Solving these equations yields A2 reads 1. 09 A. 11.84 I1 = 5.12 /-78.0£/\ Determine the complex power drawn by the (5 I S = VI*. 12 = 1.09 /-31.16° A. Thus, Al reads 5.12 A and -+ )7)-n impedance of the circuit of Fig. 11-50. From Prob. 11.83, 1* Hence, and = I; = 1.09 /31.16° A S = (9.37 /23.36°)(1.09 /31.16°) and V= (5 = 5.94 + j8.31 + j7)(1.09 / -31.16°) = 9.37 /23.36° V VA. MAGNETICALLY COUPLED CIRCUITS 11.85 0 255 Determine the ammeter readings in the coupled circuit of Fig. 11-51. loo ~'V .... (011 Gen I • LP~V .... Gen 2 • x., ... -/3J:1... x.,. -i6 J'l- • x.,. -i2.tL 8 Fig. 11·51 I The mesh equations may be written as: [1 + jl + 5 + j4 + 2(j2)]lj + [-(5 + j4) + (j6 - )3 - j2)]12 = 100 LQ:: [-(5 + j4) + (j6 - j3 - j2)]11 + [5 + j4 + 7 + j9 - 2(j6)]12 Hence, 11=7.56/-76.72°A and 12 =6.32/-160.23°A. 17.56/-76.72° - 6.32/-160.23"1, or 9.29 A. 11.86 Al reads 7.56A; A3 reads 6.32A; A2 reads Determine the complex power supplied by each source of the circuit of Fig. 11-51. I SI = Vjlr = (100 LQ::)(7.56/76.72°) = 173.7 + j735.8 VA S2 = - V22 1* 11.87 = -100 LQ:: =(- 100 LQ::)(6.32/160.23°) = 594.8 - j213.77 VA What is the ammeter reading in the circuit of Fig. 11-52? (2 +j4) r::; £1.. (3+j5).a. •.... x.,. -i4 {L x.,. f1., x.,. -i2 f2-. ....• A /"A (4 + j6 .. 0.) -j3 -/3 /1... Fig. 11-52 I The mesh equations are [(2+ j4) + (4+ j6) + 2(j2)]ll + [-(4 + j6) - j2- j4- j3]12 = 100 LQ:: [-(4 + j6) - j2 - j4 - j3]ll + [(4 + j6) + (3 + j5) - j3 + 2(j3)]12 = 0 Hence 11 = 18.89/15.01° A. 11.88 Calculate the coefficient of coupling between coils A and B of the circuit of Fig. 11-52. I 11.89 The ammeter reading is 18.89 A. X A =5n and X B =6n. Also X M =3n (given). k=X M IYX A X B =3;y'(5)(6)=0.548. Determine the number of turns of coil B of the circuit of Fig. 11-52 if coil A has 300 turns. I Thus, 11.90 Hence, or NB = 329 turns. Find the ammeter reading in the circuit of Fig. 11-53. 256 0 CHAPTER 11 (3 • ~j l).f'L. A .... --j 3 60llz 1I r:'- .~ 100 ~ V ~j4).fl- (2 + f::\ C • .... -+___ L -_ _ _ _ _ 150 ~V "01-\1 (6 +j8)f"L X., .. ~j2n. A/--_--' X",.· j5..n.. X",e· j1 I Jl Fig. 11-53 The mesh equations are [- j3 + (3 + jl) + (6 + j8) + 2(j2)]11 + [-j3 + (6 + j8) + j2 - j5 + j1]12 = 100 &: [-j3 + (6+ j8) + j2- j5 + j1]11 + [-j3 + (6+ j8) + (2+ j4) - 2(j5)]12 = 150&: Solving for 12 we obtain 11.91 12 = 21.71/25.53° A, c.nd the ammeter reads 21.71 A. Determine the voltage across the capacitance of the circuit of Fig. 11-53. I Xc = 3 0 (from Fig. 11-53) 12 = 21.71/25.53° A (from Prob. 11.90) The mesh equations of Prob. 11.90 may be solved to yield Vc 11.92 What is the inductance of coil B of the circuit of Fig. 11-53? 4 LB = 2'7T(60) = 10.61 mH Determine the ammeter reading in the circuit of Fig. 11-54. ,,-.--- • ( (3 +jl) I ..fl.. ( K= .1 (2 +j4) ( • -.1L \ I , 11, I I .. l' _.- -- ( lOO ~ V '"'v +t \ '- I , --+~ i) I We choose the mesh currents as shown in Fig. 11.-54. + '"'v 150 !!J§:. V --- ,.. --~ ( '/ [-(2 + j4) + (-j2)]11 + [(2 + j4) - j1]12 which give = 100 &: -150 /90° = 150 /90° Thus, the ammeter reads 61.7 A. Determine the polarity markings for the coupled coils of Fig. 11-55a and draw the corresponding circuit. I 11.95 12 = 61.7 /74.42° A. Fig. 11-54 Then we have to solve for 12 only. [(3 + j1) + (2 + j4) + 2(j2)]lj + [-(2 + j4) + (-j2)]12 11.94 Hence, = - j3(8.57 /-132.77° + 21.71/25.53°:, .= (3/-90°)(14.11/12.53°) = 42.33/-77 .47° V I 11.93 I1 = 8.57 /-132.77° A. See Fig. 11-55b. For the polarities determined in Prob. 11.94, find the mutual reactances between the coils if I Referring to Fig. 11-55b, X Me = 1y(1)(4) = 20 X MD = 1\/(4)(9) =60 k = 1. D 257 From the mesh equations of Prob. 11.96, we obtain IJ = MAGNETICALLY COUPLED CIRCUITS 0 lOO /-110 0 V lOO /70 V (3 + il) (2 + i4)'I--_+- .fL. .fZ.. (5 + i9) Jl. Lz (b) (a.) Fig. 11-55 11.96 Determine the reading of the ammeter A3 in the circuit of Fig. 11-55a. I The mesh equations are [(2 + j4) + (3 + j1) + 2(j2)]IJ + [(2 + j4) + j2 - j6 + j3]12 = 100 /70 0 [(2 + j4) + j2 - j6 + j3]I J + [(2 + j4) + (5 + j9) - 2(j6)]12 = -100 / -110 Solving for 12 yields 11.97 12 = 10.48 /72.4 0 A. 0 Hence, A3 reads 10.48 A. How much power is dissipated in the resistor of the coil L2 of the circuit of Fig. 11-55a? I Power dissipated, P = 111 + 1212(2) W. Thus, 8.41 /-12.83 0 A. I J + 12 = 8.41 /-12.83 + 10.48 /72.4 Hence, 0 = 8.2 - j1.87 + 3.17 + j9.99 = 13.98 /35.37 A P = (13.98)2(2) = 390.88 W. 50 L1!r V lOO~V Fig. 11-56 11.98 Find the ammeter reading in the circuit of Fig. 11-56, assuming I First, we determine the mutual reactances: The mesh equations may now be written as k = 1. 0 258 D CHAPTER 11 [j2 + (2 + j8)]I\ -j2.451\ + (--/2.45)12 + (-j2.83)1 3 = 100 LQ: + [j3 + (5 11.99 13 = 6.13 /49.44° A. Hence the ammeter reads 6.13 A. How much reactive power is taken by loop 3 of the ;;ircuit of Fig. 11-56? Q = 1~(XL - Xc) I 11.100 + j3.4613 = 50 /20° + j3.461 2 + (j4 - j5)13 = 0 -j2.831) Solving for 13 yields t j7)]12 =, (6.13)\4 - 5) = -37.58 var Calculate the complex power supplied by the voltage :;ource in the circuit of Fig. 11-57, given k = 1. (5+jlO).lL o j6 l~- .~ (2+12)~ ~ '---------+-.- I XM=1V(2)(8)=4n. [(5 4.n. Fig. 11-57 Themeshequationsare + j10) - j6 + (2 + j2)]I) + [-( - j6) - j4]K2 [-( - j6) - j4]1\ [-(2 + j2) Solving for I] yields (7+j8) + [-(2 + j2) + j4]13 = 100 LQ: + [- j6 + (7 + j8)J1 2 + [-(7 + j8) + j4]13 = 0 + j4]IJ + [-(7 + j8) + j4JI 2 + [(2 + j2) + (7 + j8) + 4 - 2(j4)]13 = 0 1\ = 8.71 /-50.79° A. Hen,x, S =VI; = (100 LQ:)(8.71 /50.79°) = 871 /50.79° = 550.73 + j674.78 VA 11.101 If the lOO-Vac source in the circuit of Fig. 11-57 is rtplaced by a 100-V battery, what will the ammeter read? I The steady-state circuit for dc source is as shown in Fig. 11-58. +G-_~·A - ' ==.rOOV Fig. 11-58 4(2 + 7) Zp = 4 + (2 + 7) = 2.77 n 11.102 Find the voltage across the capacitor and the I Zi"= 5 + 2.77 = 7.77 n ener~,y 100 IT = 7.77 = 12.87 A stored in the capacitor, from the data of Prob. 11.101. The voltage across the capacitor may be obtained by describing the path shown with the dotted line in Fig. 11-58: MAGNETICALLY COUPLED CIRCUITS VAB = L driving volts - Ik = IT L volt drops = 100 - [2/2 1 Yk Y YT = Z T 0.1111 IK = 12.87 0.3611 D 259 + S(12.87)] 1 = 277 = 0.3611 S p' 1 Y K = 7 + 2 = 0.1111 S = 3.96 A = 12 V AB = 100 - [2(3.96) + S(12.87)] = 27.73 V 1 Now Xc = 27TfC Thus 1 6 = 27T(2SC) or 2 W = (1 12)CV = (1/2)(1061 x 10- 6 )(27.73)2 C= 1061 J.LF 20 = 408 mJ ~v (3 + i9) 11- "" Fig. 11-59 11.103 What is the ammeter reading in the circuit of Fig. 11-S9? I Using the outer loop for 13 reduces the number of impedance terms. The mutual reactances are 0= [(1 + j3) + (S + j7) + (2 + j4) - 2j4.S8]lt + [-(1 + j3) + j4.S8 + j6]12 0= [-(1 + j3) + j4.S8 + j6]lt 20 LQ: = [(S + j7) - j4.S8]lt Solving for 13 yields 11.104 13 = 1.6 /19° A. + [(S + j7) - j4.S8]13 + [(1 + j3) + (3 + j9) - j10]12 + (- j10 + j4.S8)13 +[-jlO+(S+j7)]13 + (- j10 + j4.S8)12 Hence the ammeter reads 1.6 A. How much total complex power is drawn by all the passive circuit elements of the circuit of Fig. 11-S9? I The complex power drawn by the passive element equals complex power supplied by the source: S = VI* = (20 LQ:)(1.6 / -19°) = 32 / -19° = 30.26 - jlO.42 VA 11.105 Determine the parameters of the equivalent Tsection that can be used to replace the transformer in Fig. 11-60a. -i4 IL V N 2..1t. _j 4-..(/.•. i6.t1.. ~ i5(2.~ • • k - 0.52 ;LiL (a..) Fig. 11-60 (b) 260 D CHAPTER 11 I The T equivalent is shown in Fig. 11-60b. For Fig. 11-60a we have v = (-j4 + j3 + 2)1, + (- j2.01)1 2 X M = 0.52v'3X5 = 2.01 0 0= (-j2.01)1 1 + (j5 + j6 + Zload)1 2 For the T section of Fig. 11-60b the mesh equations are V= (-j4 + ZA 0= (-ZC)ll Zc -I- + 2)11 + (-ZC)12 + (Zli + Zc + j6 + Zload)1 2 Equating coefficients of II and 12 , ZA + Zc + 2 - j4 = 2 - j1 = Z A = jO.99 0. Solving, 11.106 Zc Zc + ZB + j6 + Zload = j11 + Zload j2.01 Z" :: j2.99 0. Zc = j2.0 I 0. Determine the parameters of an equivalent T sectioll that can be used to replace the transformer in Fig. 11-61. it .Cl.- I:) ,< ·~l{:·j2)0 I I Fig. 11-61 k- 1 -i3 fL Mutualreactanceis X M =l'V(4)(2)=2.83n. Mesh equations for the circuit of Fig. 11-61 are 100 L!r = [j1- j3 + (4 + 4j111, + j2.8312 + [(2 + j2) + 10 /30°112 0= j2.831 1 Mesh equations for the T section are: 100 L!r = U I - j3 + Z4 + Z, )1, + (-ZC)12 + (Zc + ZB + 10 /30°)1 2 0= -Zcll Thus, 4 + j2 = j1 - j3 + Z A + Zc 2 + j2 + 10 /30° = Zc + ZB + 10 /30° -Zc:= l2.83 Solving for the impedances, ZB :: (2 + j4.83) 0 11.107 Obtain the T equivalent for the coupled circuit (2 + i ~hown Zc = - j2.83 0. in Fig. 11-62. 1)..n. Fig. 11-62 • - 0.75 I The mutual reactance is X M = 0.75v'2X8 = 3.00. 100 L!r = (2 + j1 For the coupled circuit, + : + j2)1, + (- j3)12 + (5 + j8 + Z,oad)1 2 0= (-j3)1, For the T equivalent, Solving for the impedances, l.B = (5 + j5) 0 Zc = j30 MAGNETICALLY COUPLED CIRCUITS 11.108 D 261 Determine the Thevenin equivalent to the left of the terminals T1 and T2 in Fig. 11-62. Fig. 11-63 I To determine the Thevenin equivalent we use the T equivalent obtained in Prob. 11.107 and shown in Fig. 11-63. Thus, VTh = 100 is!.. (2 + j1) + (2+ j1 +5 - j1)(j3) = 276/668°= (1086+,·2535)0 Z p = (2+j1+5-j1)+(j3) . . . . Now, ZTh 11.109 {~_ j1) + j3 = 39.39/66.8° V = (1.086 + j2.535) + (5 + j5) = (6.086 + j7.535) = 9.69/ 51.07° 0 Determine the Zload in Fig. 11-62 for maximum power transfer and calculate the corresponding active and reactive powers. I For this problem, we may use the Thevenin equivalent of Prob. 11.108. transfer, Zload = Z~h = 9.69 /- 5107° fl. Now, Thus, for maximum power 39. 39 ~ h.t:. QO 9.69/-51.07° + 9.69/51.07° == 3.24 ~ A 11.110 Q= p = [2R = (3.24)2(6.086) = 63.89 W Thus, [2Xc = (3.24)\7.535) = 79.10 var leading Repeat Prob. 11.107 for the circuit of Fig. 11-64a. /oo/./ V • '11 (.6) Fig. 11·64 I For Fig. 11-64a we have X M = 1Y(7)(7) = 7 O. -100 is!.. == [(6 + j7) + (3 + j6)]I1 + [-(3 + j6) + j7]I2 0== [-(3 + j6) + j7]Il + [(6 + j7) + (8 - j4) + (3 + j6)]I2 For the T equivalent of Fig. 11-64b the mesh equations are -100 is!.. = [ZA + Zc + (3 + j6)]I1 + [-Zc - (3 + j6)]I2 0= [-Zc - (3 + j6)]I1 Comparing coefficients yields: + [ZB + Zc + (3 + j6) + (8 - j4)]I2 262 D CHAPTER 11 ZA -\- Zc -\- (3 -\- j6) = (6 -\- j7) -\- (3 -- ;6) Z" -\- Zc -\- (3 -\- j6) -\- (8 -- j4:' 11.111 (6 -\- j7) -\- (8 - j4) -\- (3 -\- j6) Z,,~'(6-\-j14)fl ZA = (6 -\- j14) fl Hence, == -Zc - (3 -\- j6) = -(3 -\- j6) -\- j7 Repeat Prob. 11.108 for the circuit of Fig. 11-64a. I Proceeding as before we have i1 3 _. VTh = 100 (6 -\- j1~) + (3 _ j1) = 19.99 j-73.73°Y (6-\-j14)(3-jl) . Zp = (6 ·14) (' ·--:---1) = (3.03 - J.37) fl and -\- J + ~'-1 ZTh = (3.03 - j.37) -t- (5 + j14) = (9.03 -\- j13.63) fl 11.112 Find the current in the (8 - j4)-fl impedance of the circuit of Fig. 11-64a. I By using the Thevenin equivalent obtained in Pro:J. 11.111, we obtain _ 19.991..::.:]).73' _ ° 1 - (9.03 -\- j13.63) -\- (8 _ j4) - 1.02 j -103.36 A 11.113 Repeat Prob. 11.107 for the circuit of Fig. 11.65a. k -0.41 Tf • • (1 + j3) !.:)n (3+j2)(L -j6J1- 4.Q. Fig . 11-65 I For Fig. 11-65a we have X M = 0.41Y(3)(2) 1.0 fl. = -60 LQ: = [(1 -\- j3) t 4]1\ -\- (-4 - j1)12 0=(-4-jl)1 -\-(3-\-j2-j6-\-4)lz For the T network shown in Fig. 11-65b we have: -60 LQ: = (ZA -\- Zc + 4)1\ -\- (-Zc - 4)12 0= (-Zc - 4)1\ Hence and 11.114 ZA -\- Zc -\- 4 = (1 -\- j3) -\- 4 -\- (Z" -\- Zc - j6-\- 4)l z -Zc - 4 = --4 - j1 ZA=(1-\-j2)fl ZB -\- Zc - j6 -\- 4 = (3 -\- j2) - j6 -\- 4 ZB=(3-\-j1)fl Repeat Prob. 11.108 for the circuit of Fig. 11-65a. Zc=j1fl MAGNETICALLY COUPLED CIRCUITS D 263 (b) ca.) Fig. 11-66 I From Fig. 11-66a we obtain VTh ° 4 + j1 = 60 LQ: (1 + j2) + (4 + j1) = 42.39 /-16.92° Y and from Fig. 11-66b we get (1 + j2)(4 + j1) . Zp= (1+j2)+(4+jl) = (1.09 +J1.15) 0 ZTh 11.115 = (1.09 Determine the current through and the voltage across the capacitor of the circuit of Fig. 11-65a. I Using the Thevenin equivalent of Prob. 11.114, we have I z = le = (7.55 /26.35°)(6 /-90°) = 45.3 /-63.65°Y What is the maximum energy stored in the capacitor of the circuit of Fig. 11-65a? 1 I 1 C = 27rfX c or = 27r(60)(6) = Vc = 45.3 Y (from Prob. 11.115) 442.1p,F Vmax = V2Vc = V2( 45.3) = 64.06 Y Hence, 11.117 42.39 /-16.92° (4.09 + j2.15) _ j6 = 7.55 /26.3SO A = Vc =lc(-jXc) 11.116 + j1.15) + (3 + j1) = 4.62 /27.73° 0 Wmax = !(442.1 X 1O- 6 )(64.06)z = 907 mJ. Find the voltage ve(t) across the capacitor of the circuit shown in Fig. 11-67. 11-/ /M'O.Z, I H l.vr(.o~ t V .~ ~~--:~ \ I V = 2 LQ:, and since w = 1 radls, .J Fig. 11-67 we have XI = X z = 10 X M =0.250 and The mesh equations become j1.251 2 and Thus, V, 11.118 =- 1,(-jX) =-(-jI.6)(-jl) = 1.6 V or = 2 j1.251 1 = 0 I z = -j1.6A (Ve)max = 1.6(V2) and Ve(t) = 2.26 cos t Y The mutually coupled circuit shown in Fig. 11-68 is that of an autotransformer. For an ideal autotransformer the ratio of the voltages VIIVz = (NI + Nz) I Nz = a + 1, where a = NI I Nz (as may be inferred from Fig. 11-68). Determine the current transformation ratio ILl lab' 264 0 CHAPTER 11 -- <p, I, + a --- + ( N, 1-1-) N, ( 1-1-) ( V, ( ......- - I- -) N2 ( I-) CI- + V2 In - - V, + V2 N2 -It ZL C <1>2 (a) I From Fig. (b) Fig. 11-68 11-68a and b the apparent-power balanc~ is expressed by or 11.119 Obtain an expression for the output complex power orm ideal autotransformer (Fig. 11-68) in terms of the current through the winding having NI turns. I S2 =V2 1r =V2 (I:h + I:b ) =V2 1:b + V2 1:b (b - But (1 ) N, =a lV, = --- lab (2) or From Eqs. (1) and (2), S2 = V2 1:b + a(V2 1:b ) = Se + Si where the first term corresponds to power transferred by conduction and the second term represents that by induction or by transformer action. 11.120 For the ideal autotransformer shown in Fig. 11-69, find V2 , leb' and the input current 11 . I, a + llob NI "'40 VI = 150Lft V --It b t N 2 ", 80 Icb ZL = 10&r Cl V2 Fig. 11-69 C I V = 2 v _1_ a +1 = 100~V lab = 11.121 a; 1 = 6.67 / -60° A In Prob. 11.120, find the apparent power delivered to the load by transformer action and that supplied by conduction. I Scond = V2 1:b = (100 ~)(6.67 /60°) == 667 /60° VA MAGNETICALLY COUPLED CIRCUITS 11.122 0 265 A lO-kVA 440/110-V two-winding transformer is reconnected as a stepdown 550/440-V autotransformer. Compare the voltampere rating of the autotransformer with that of the original two-winding transformer and calculate Si and Se· I Refer to Fig. 11-68b. The rated current in the 11O-V winding (or in ab) is 10,000 1\ = ~ =90.91 A Current in the 440-V winding (or in be) is 10,000 ---:t4O = (b = IL - 1\ = which is the rated current of the winding be. IL Check: For the autotransformer = The load current is 1\ + (b = 90.91 + 22.73 = 113.64 A a = ~g = 1.25 IL = al\ which agrees with IL calculated above. 22.73 A and 1.25 = 10,000 ---uo== 113.64 A Hence the rating of the autotransformer is Sauto = VJ\ = 550 10,000 ---uo= 50 kVA Inductively supplied apparent power is Si a-I = VZ(IL - 1\) = -a- S= 1.25 - 1 1.25 50= lOkVA which is the voltampere rating of the two-winding transformer. Se 11.123 The conductively supplied power is S = 50 ~ = 1.25 =40kVA Repeat Prob. 11.122 for a 440/550-V step-up connection. + + Fig. 11-70 I The step-up connection is shown in Fig. 11-70. The rating of winding ab is 110 V and the load current I L flows through ab. Hence I = 10,000 = 90 91 A L 110 . The output voltage is Vz = 550 V. Thus the voltampere rating of the autotransformer is VZI L = 550(10,000/110) = 50 kVA, which is the same as in Prob. 11.122. Power transferred inductively = 50 - 40 = 10 kVA. Consequently, a two-winding transformer connected as an autotransformer will have a voltampere rating a/(a - 1) times its rating as a two-winding transformer. 11.124 A lO-kVA 4160/450-V 25-Hz ideal autotransformer is used for step-down operation and is delivering rated kVA at rated voltage and 0.85 pf lagging. Determine (a) the current to the load and (b) the current supplied by the generator. I (a) Refer to Fig. 11-68b. (} = cos-\ 0.85 == 31.79° 10,000 /31.79° = (450 L!r)I~ 82 = 10,000 /31.79° VA I; = 22.22/31.79° I/- =22.22/-31.79° A 266 D CHAPTER 11 (b) IL/I\=NI/N2 where N\/Nz =4160/450=a. 22.22/-31.79° = ~!!~I 1\ 450 11.125 In an ideal step-up autotransformer (Fig. 11-70) primary, what is the voltage across the load? I and Nz = 400. If 100 V is applied across the 600 200 Vz 100 Vz = 300 V. For the autotransformer of Prob. 11.125, determine the input current if a 30-0 resistor is connected as a load on the secondary. V 300 I = - 2 = _ . ,= 10 A L Z2 30 I 1\ Thus, 11.127 NI = 200 NI + Nz = 200 + 400 = 600 Thus, 11.126 1\ = 2.4/- 31.79° A = ~~ (10) = 30 A. A 280-turn ideal autotransformer has a secondary lapped at 14 turns. If the input voltage is 240 V and the secondary load is a 6-mO resistor, find the current in the section of the winding supplying the load. I Since Vz/V\ = Nz/Ni' 14 Vz = 280 240 = 12 V or 11.128 1\ = z~o(2000) = 100 A. From Fig. 11-68b, Icb = IL - 1\ = 2000 - 100 ,= 1900 A. The input apparent power to an ideal step-down aU':otransformer is 20 kVA at 400 V. 11-68) = 30 A, determine the load voltage. I 1\ = 20,000/400= 50 A. 11.130 = VzI L = 80Vz V ~~ ~Q:0OO = 250 V 2 80 Thus, In the autotransformer of Prob. 11.128, how much power is transferred to the load by induction? I (b (Fig. IL ~~50+30=80A, From Fig. 11-68a, S\ = 20,000 = Sz 11.129 If the current (a) Pc = VzI\ = (250)(50) = 12.5 kVA (b) (a) conduction and (b) Pi = V2(b = (250)(30) = 7.5 kVA An ideal transformer is used to couple a 50-n resistor load to a generator whose internal impedance is 400 L!r O. If the coil connected to the generator has 100 turns, determine the required number of secondary turns so that maximum power will be transferred to the load. I Zgen Zin = = 400 L!r O. 400 L!r 0, For maximum power transfer, the generator must "see" a 400 L!r-O load. 400 =(~)Z50 Thus, Nz = 35.4 turns z 11.131 The primary and secondary turns of an impedan~e-matching transformer are 200 and 600, respectively. Determine (a) the input impedance at the primary terminals if a 20 /30°-0 load is connected to the 600-turn secondary, and (b) the rms primary current if thl! source voltage is 50 sin 300t V. Zin (b) V\ = 50 v'2 = 35.36 V = (~~)2(20 DO') = 2.22/30° 0 _ 35.36 L!r _ < ( _ ° 1, - 2.22LJ!L - L.13 L=l!rA MAGNETICALLY COUPLED CIRCUITS 11.132 For the operating condition stated in Prob. 11.131, calculate (a) the secondary current (rms), secondary voltage (rms), and (c) the active power drawn by the load. 200(15.93) = 600/2 35.36 T Thus (c) D 267 (b) the I z = 5.31 A or 200 = 600 Vz = 106.08 Y. For an ideal transformer, the active power drawn by the load is equal to the power input to the transformer. Thus, SI =VII; = (35.36 L!r)(15.93 /30°) = 563.28 /30° = 487.81 + j281.64 VA PI =487.81 W 11.133 A l00-Hz generator supplies a 2.0-0 resistor through an ideal transformer whose primary to secondary turns ratio is 200/600. The rms voltage across the load is 120Y. Determine (a) the primary current and (b) the input impedance of the circuit. 11.134 Find the input impedance of the circuit shown in Fig. 11-71. Fig. 11-71 I Writing the mesh equations we have (6 + j5)11 - jlz =60L!r - jl, + (10 - ) 2 + } 3 + } 4)12 = 0 Hence, 11.135 60 L!r Z =--=6+j5+ 'I j2 1O-}2+}3+}4 =6+j5+ I 1O+J·5 =6.1+j50 A 200-p,F capacitor is connected in series with the primary winding of a 60-Hz ideal transformer. The primary winding impedance is (2 + jlO) 0 and the secondary circuit impedance is (0.4 + jO.2) O. If the turns ratio is 2, determine the input impedance. 1 1 X = = = 13.260 c wC (377)(200)10- 6 I Zseconda,y <et. to pdma,y Zp,;ma,y = 2z(0.4 + jO.2) = 1.6 + jO.8 0 = 2 + jlO - j13.26 = 2 - j3.26 0 Z, = 1.6 + jO.8 + 2 - j3.26 = 3.6 - j2.46 0 CHAPTER 12~ Resonance ~ 12.1 Resonance is the condition that exists in ac circuits und:r steady state when the input current is in phase with the input voltage. Using this definition, find the frequency at which a series RLC circuit will be in resonance. I For resonance, we must have Z=R + j(X[ - Xc) = R + jO er or 1 wrL= -C Wr Hence 12.2 Wr = 1 vrc or fr = l 2-:;;~rtc = resonance frequency (Hz) Draw a phasor diagram showing the current and all the voltages in a series resonant circuit. I The circuit and the phasor diagram are shown in Fig. 12-l. Notice that VL + Vc = 0 I----~~--.. and V=VR = RI. I VI(= RL I Fig. 12-1 12.3 Plot XL' X 0 graphically. and R as functions of frequency I See Fig. 12-2, which shows that, at f= fr' -6 12.4 268 Thus show how the resonance frequency can be obtained XL = Xc t Fig. 12-2 IIz IIz For a series RLC circuit, plot impedance and current as functions of frequency. frequency. I 12.5 f See Fig. 12-3, from which Z= R and I = Im at Fig. 12-3 Hence obtain the resonance f = fr. Show graphically how the power-factor angle of a stones RLC circuit varies with frequency. the resonance frequency. Hence determine RESONANCE <;b - 'xt Y-. ~ 0 269 \ -I Hz 0 t------\:-------~ Lauingpf -900 - - - - - - - - - - - - Fig. 12-4 I See Fig. 12-4, which shows that at fr' (} = O. 12.6 A coil with inductance and resistance of 1.0 mH and 2.00, respectively, is connected in series with a capacitor and a 120-V 5-kHz supply. Determine the value of capacitance that will cause the system to be in resonance. I The circuit is similar to that shown in Fig. 12-1a, for which 1 Solving, 12.7 C = 1.01 X 1 27TV 0.001 C 10- 6 F = 1.01 p,F. From the data of Prob. 12.6 determine (a) the current at the resonance frequency and (b) the maximum instantaneous energy stored in the magnetic field of the inductance at the resonance frequency. I (a) At series resonance, Z;n = R. ~ 1= (b) 12.8 5000 = or t,. = 27TVLC Wmax = ! Ll~ax = = R 120~ =60~A 2 Ll~ms = (1.0 x 10- )(60)2 = 3.6 J 3 The quality factor Q of an ac circuit is defined by Q= 27T (maximum energy stored per cycle) energy dissipated per cycle (1) Using this definition, obtain Q in various forms for a resonant series RLC circuit. I From Eq. (1), (2) where 1= rms current at resonance and and Tr = lit,. = 27T/Wr in Eq. (2) yields Tr = period at resonance frequency. wrL Wr = I/VLC, Eq. (3) may also be written as 1 Qs = wrRC = 12.9 Im = V21 (3) QS=R Since Substituting 1 [I R VC (4) A series RLC circuit has a variCl.ble resistance. If the frequency of the input voltage is varied, sketch the currents as functions of frequency for two arbitrary values of the resistance. I See Fig. 12-5. 270 0 CHAPTER 12 1 1, R, L..Ra. 0.1011, Y!.!.'~ ___ ___ _ I • 12.10 I = L = 25.4 mH, 1 1 27rvTC fr = Q, 12.11 R = 2.42 n, In the circuit of Fig. 12-1a, (4) of Prob. 12.8. ---r======;= = 27rYO.0254(52 x 10 -6) 1 1 wrRC = 27r(138.6)(2.42)(52 I I; li X For the circuit shown in Fig. 12-6, Rp R 2 , and R, C = 52 p,F. and 133.6 Hz 10- 6 ) = ar,~ = wrL Qs Qs 9.13 Fig. 12-5 = R = Evaluate Q, as given by (3) and 27r(138.6)(0.0254) = 9 3 2.42 .1 1 ILl I 0.0254 R 'VC = 2.42 'V 52 x 10 6 =9.13 0.51, 1.3, and 0.24 n, respectively; Cl and C z are 25 and Determine the resonance frequency. 62 p,F, respectively; LI and L2 are 32 and 15 mH, rl~spectively. I Before the resonance frequency formula may be used, the circuit must be reduced to the form shown in Fig. 12-1a. Req = RI + R2 + R3 = Leq = LI + L2 = 0.032 + 0.015 = 0.047 H Ceq = 17.8 12.12 X = S (c) _1_ Req & Ceq = wrL Z Qeoil2 R2 1 1 = Ceq Cl - = _1_ 2.05 ~ (a) Qs' 0.047 ="'i 17.8 x 10- 6 ~. (b) Qeoil l ' ( b) ""'" 1 + 25 x 10 6 and (c) Qeoil 2' Q . = wrL I = 27r(174)(0.032) eo.11 RI 0.51 = 686 . = 27r(174)(0.015) = 26 1.3 1 . 0.1011 A, R3 1 1 = -----;;6 Cz 62 x 10 +- 10- 6 F For the resonant circuit of Prob. 12.11 find I (a) Q + 1.3 + 0.24 = 2.05 n 0.51 : LI,R I C2 Fig. 12-6 - --.- -- Cl :J L 2 .R 2 Coil 2 0 1.1 I, 12 "Bandwidth" Fig. 12-7 f RESONANCE 12.13 0 271 Fig. 12.7 shows a general form of resonance curve. The range of frequencies within which the variable does not drop below I/V2 times its maximum value is called the passband or bandwidth. Thus, the bandwidth is (1) Determine the powers dissipated at the frequencies fl and f2 in a series RLC circuit and compare the results with the power dissipated at resonance frequency. , At f=F },I , PI=I2IR=«~I)2R=H2R v2 At f= f2' P =I2R=(_I_I)2R=!I2R At f= fr' Because 12.14 r 2 PI = P 2 = ! Pr' v2 2 r 2 r r the frequencies fl and f2 are known as half-power frequencies. Obtain a relationship between the resonance frequency fr and the half-power frequencies fl and f2 for a series RLC circuit. , From Fig. 12-7, the current at fl and f2 drops to I/V2 of its resonant value. Thus the impedance must be equal to v2 times the resonant value, which is R; i.e., YR 2 +(XL -Xc)2=v2R, which implies that XL-Xc=±R, or 1 wL- wC=±R For XL < Xc which corresponds to f, or wp Eq. (1) yields W Similarly, for (1) XL> Xc , = - R + 2:1 VI(R)2 4 L + LC (2) 2L corresponding to f2 or W 2' Eq. (1) gives (3) Multiplying Eqs. (2) and (3) yields 1 [(R)2 wI W 2 =4 L + 4] LC -41 (R)2 L = fr = Hence, 12.15 1 2 or LC=(wr) Wr = y'W I W 2 (4) VfJ2 Find the bandwidth BW of a series RLC circuit in terms of its parameters. , From Eqs. (2) and (3) of Prob. 12.14, we have (1) 12.16 Express the bandwdith of a series RLC circuit in terms of Q s and , From Eq. (3) of Prob. 12.8 we have Prob. 12.15 yields 12.17 RI L = w)Qs and fr. 1/27T = f)w r . Substituting these into Eq. (1) of With reference to Figs. 12-5 and 12-7, comment on the relationship among f;, f2' fr, and BW. , As indicated in the figures, the resonance frequency is not centrally located with respect to the two half-power frequencies, especially when BW is large (Qs is small). Indeed, as the geometric mean of fl and f2 (Prob. 12.14), fr must always be below their arithmetic mean. However, if Q ~ 10, the resonance frequency is sufficiently centered with respect to the half-power frequencies to allow us to write (1) 272 12.18 D CHAPTER 12 A series circuit has a resonance frequency of 150 kHz and a bandwidth of 75 kHz. frequencies. Determine the half-power I Since Qs = 150175 = 2 < 10 approximations (J) of Prob. 12.17 cannot be used. The exact relations are BW = f2 - f1 and f, = Vf1 f2' and we may work in kilohertz. Thus, we eliminate f2 between 75 = f2 - fl and 150 = vr:Tz to obtain f~ + 75f1 - 22,500= O. Solving, fl = 117.1 kHz and f2 = 75 + f1 = 192.1 kHz. 12.19 Determine the parameters of a series RLC circuit that will resonate at 10 kHz, have a bandwidth of 1 kHz, and draw 15.3 W from a 200-V generator operating at th,~ resonance frequency of the circuit. I For a series circuit operating at resonance, P = V R = V= 200 V V' Q =~=1O=1O BW 1 Qs = s 15.3 = -~, R R 2'Ti"IL --8.-- 10= (2~0)2 2'Ti"(1O,000)L 2610 1 A series RLC circuit has a Q s of 5.1 at its resonance frequency of 100 kHz. Assuming the power dissipation of the circuit is 100 W when drawing a current of 0.8(1 A, determine the circuit parameters. I R 5 1 = 2'Ti"(100,000)L . 156 = L = 1.26mH 1 f..= 1 2'Ti"vTC C = 2.01 nF What is the bandwidth of the circuit of Prob. 12.20? f.. Qs= BW I 12.22 _ 2'Ti"f,L QS--R- 1560 100 000 = - - = = = = , 2'Ti"W.2f) >( lO-')C 12.21 L=416mH C =61OpF 10,00(1 ,= 2'Ti"V 0.416C 12.20 R = 2.61 kO BW= 19.6 kHz Determine the half-power frequencies of the circuil of Prob. 12.20. I Proceed as in Prob. 12.18. 19.6= f2 - fl Substituting, f~ and 12.23 + 19.6f1 - 10 4 = 0 f1 = -19.6 + Y (19.6)2 + 4(10 4 ) 2 = 90.7 kHz f2 = 19.6 + f1 = 110.3 kHz. A 125-V ac source supplies a series circuit conslstillg of a 20.5-p,F capacitor and a coil whose resistance and inductance are 1.060 and 25.4 mH, respectively. The generator frequency is the resonance frequency of the circuit. Determine (a) the resonance frequency and (b) the current. 1 I (a) (b) 2'Ti"YO.0254(20.5 x 10 '") At resonance, Z, = R; ,= 220.6 Hz hence V 125 L!r 1= -Z, = ----= 117 .9 L!rA 1.06 RESONANCE 12.24 From the data of Prob. 12.23, calculate the voltage across I (a) (a) the capacitor and 0 273 (b) the coil. Xc = XL = 27rfrL = 27r(220.6)(0.0254) = 35.21 0 Vc = 1Xc = (117.9)(35.21) = 4151 V (b) Vco ;! = IR + IjX L = 117.9 ~(1.06 + j35.21) = (117.9 ~)(35.23 /88.3°) 12.25 = 4154 /88.3° V or For Prob. 12.23, determine the resistance that must be connected in series with the circuit to limit the capacitor voltage to 300 V. I 1= 300 = 1(35.21) * =8.52= 1~5 1= 8.520A R = 14.670 Hence the required additional series resistance is 14.67 = 1.06 + Rx 12.26 Rx = 13.610 In a series RLC circuit under resonance, express the voltages across the capacitor and the inductor in terms of Q s and the applied voltage. I 12.27 A coil with resistance and inductance of 400 and 50 mH, respectively, is connected in series with a 450-pF capacitor and a generator. Determine (a) the resonance frequency and (b) the circuit impedance at the resonance frequency. 1 1 27rV (0.050)( 450 x 10 I (a) fr = 27rvTC (b) 12.28 A 60-V source having an internal resistance of 10 0 is connected to the circuit of Prob. 12.27. the circuit current and (b) the voltage across the capacitor. 1 = -- = C (b) 27rfC 1 27r(33,553)(450 x 10- 12 ) = Determine (a) 1O.54kO V= 1Xc = (1.2 A)(1O.54 kO) = 12.65 kV A 25-J.LF capacitor is connected in series with a coil whose inductance is 5.0 mHo Determine (a) the resonance frequency; (b) the resistance of the coil if a 40-V source operating at the resonance frequency causes a circuit current of 3.6 mA; and (c) the Q of the coil. 1 27rV (0.005)(25 x 10 6) 12.30 = 33,553 Hz = 33.553 kHz Zr=R=400 X 12.29 12) ~ ~ fr = (b) 1= (c) . = wrL = 27r( 450.2)(0.005) = 1.28 x 10-3 Qco.\ R 11.1 X 10 3 0.0036= R R = 450.2 Hz 11.1 kO A coil with inductance and resistance of 3.0 mH and 200, respectively, is connected in series with a capacitor and a 12-V 5.0-kHz source. Determine (a) the value of capacitance that will cause the system to be in resonance and (b) the circuit current at the resonance frequency. 1 I (a) fr = 27rvTC (b) Zr =200 5000 = or 27ry'~.003C V 12 Ir = Zr = 20 = 0.6 A C=338nF 274 12.31 0 CHAPTER 12 What is the maximum stored energy in the capacitor of Prob. 12.30? I 12.32 12.33 . VC(max) = 56.5v2 = 79.9 V A coil having a 2-0 resistance is connected in series with a 50-p,F capacitor. What is the inductance of the coil? The circuit resonates at 100 Hz. I L = 50.66mH. At resonance, XL = Xc or 27T(100)L = 1I27T(lOO)50(1O- 6 ). Thus, If the circuit of Prob. 12.32 is connected across a 100- v 100-Hz ac source, determine the power dissipated in the coil. I 12.34 1 9) =, 56.5 V . . 27T(5000)(338 x 10 Vc = 1Xc = 06 At resonance, [= VIR = I~ = 50 A and P'dl = [2R = (50)22 = 5 kW. Calculate the voltages across the capacitor and the cOli of Prob. 12.33. I XL 12.35 wL = 27T(100)(50.66 x 10- 3) = 3U3 0 VcOil = 50Y(2)2 + (31.83)2 = 1594.6 V A 120-V 20-Hz source supplies a series circuit consisting of a 5.0-0 capacitive reactance, a 1.6-0 resistor, and a coil with resistance and inductive reactance of 3.0 and l.20, respectively. Determine (a) the input impedance and (b) the circuit current. I (a) Zin = - (a) _ Zin - = 120 LQ:: 5.97/-39.56° 5.97 /-39.56': 0 1= 20.11 m-=-~~ A (a) the voltage across the coil and Calculate I (3 + j1.2) + 1.6 - j5 ~ (b). I _ 12.36 = Vcoil = IcoilZcoil = (20.11/39.56°)(3 + jl.2) ,= 64.98/61.36° V Xc L 9.5mH = (b) the resonance frequency in the circuit of Prob. 12.35. = 1 27TfC 1.2 = 27T(20IL 1 5 = 27T(20)C C= 1592p,F 1 - = 40.93 Hz 27TY (9.5 x 10- 3)( 1592 x 10 -") 12.37 Determine the Q of a coil whose resistance and inductance are 10 0 and 0.04 H, respectively, if the coil is at resonance at 2 kHz. . = wrL = ~~T(2000)(0.04) = 50 2 Qcool RIO· 7 I 12.38 A coil whose resistance and inductance are 5.00 and 32 mH, respectively, is connected in series with a 796-pF capacitor. Determine (a) the resonance frequent:} of the circuit; (b) the quality factor; and (c) the bandwidth. I (a) fr (b) Qs Thus, = = 1 27TY (32 x 10- 3)(796 x 10- 12) W~L = f.. ==.31.53 kHz 27T(3l.53 x 1~3)(32 x 10 -3)= 1268 BW= 24.9 Hz. (c) Qs = !w or 1268 = 31,530 BW RESONANCE 12.39 The circuit of Prob. 12.38 is connected to a 120·Y source and operates at resonance frequency. the input current and (b) the voltage across the capacitor. I (a) D 275 Calculate (a) VT 120 I=-=-=24A Zr 5 27T(31.53 x 10 3 )(796 x 10 12) = 6341 n 12.40 A 400-Y 200-Hz ac source is connected in series with a capacitor and a coil whose resistance and inductance are 20 mn and 6 mH, respectively. If the circuit is in resonance at 200 Hz, determine (a) the capacitance of the capacitor and (b) the circuit current. I 12.41 1 200= 27TVO.006C !,. = 27TvT:C (a) 3 1 Vc = IXc = (20 x 10 ) 27T(200)(105.54 x 10 6) = 151 kY (b) Wmax = ~ LI~ax = HO.006)(20,000v2)2 = 2.4 MJ Q = w~L = 27T(2~~6;~·006) = 377 Since Q> 10, BW (b) the maximum and BW= !,. Q= 200 377 = 0.53 Hz = 200 + 0.265 = 200.265 Hz BW 11 = Ir - 2 = 200 - 0.265 = 199.74 Hz A series resonant circuit has a resistance of 1 kn and half-power frequencies of 20 and 100 kHz. (a) the bandwidth and (b) the resonance frequency. I (a) (b) BW=/,-f, =100-20=80kHz !,. = Vh/1 = V( 100)(20) = 44.72 kHz Calculate the inductance and the capacitance of the circuit of Prob. 12.43. Q = ~ = 44.72 = 0 56 I BW L=2.0mH 12.45 V 400 Ir = Zr = 0.020 = 20 kA Prob. 12.17 gives 12 = !,. + 2 12.44 (b) Calculate the half-power frequencies for the circuit of Prob. 12.40. , 12.43 C= 105.54 J-tF (a) the voltage across the capacitor and In the circuit of Prob. 12.40, determine instantaneous energy stored in the coil. I 12.42 1 (a) 80 . 1 wrL= -C wr Q= wrL R O. 56 = 27T(44,720)(L) 1000 1 27T( 44,720)(0.002) = 27T( 44,720)C C=6.3nF For the circuit of Fig. 12-1a sketch the current and the various voltages as functions of frequency. I See Fig. 12-8. I, V Fig. 12-8 Determine 276 12.46 0 CHAPTER 12 Evaluate the frequencies I Wc and WL of Fig. 12-8. From Fig. 12-1a, V 1--- -Z and V~ Squaring both sides, = V2X~/Z2 V~ and = V;X~/Z2. Expressing Xc, XL' and Z in terms of w, To find the frequency Wc for which Vc is largest, we lllillimize the denominator of the expression for V~. differentiating with respect to w, equating the result to zero, and solving, we find Wc = Thus, ~ L1C ~~~~: = w,~l- 2~~ By a similar procedure, W Observe the interesting relation 12.47 L WCWI_ ;-2--w, = 'J 2LC -"~R2Ci = VI - 1 12Q; = w; = (cf. Prob. 12.14). W1W" From Fig. 12-9, determine (a) the resonance frequency, width, and (d) the quality factor. (b) the cutoff half-power frequencies, (c) band- A 50 Fig. 12-9 I f, = l.0 kHz. (b) At A = 100/V2 =71 (d) Q = f,lBW = l.010.9 = 1.1. (a) 0.9 kHz. 12.48 A resonant series RLC circuit having half-power frequency. V R I 12.49 h = 1.6 kHz. ::>perates from a 20-V source. (c) 20 5 Qs L = 15.91 mH; and, from = ~ BW = 1O(~~ = 100 Q, = l/w,RC, 0 = w,L 1 R = Since Qs > 10, fl = f, - 27T(1000)L 10 Q, = 60 and f, = 12 kHz. we have (Prob. 12.17), f.. BW 2 If the circuit C = 1I27T(1000)(1O) (10) = 1.59 J-LF. Find the half-power frequencies of a series RLC circuit having I BW = f2 - fl = Determine the power at The bandwidth of a series resonance circuit is 100 Hz and the resonance frequency is 1000 Hz. resistance is 10 ,0" determine Land C. Thus, 12.51 and I=-=-=4A I 12.50 R = 5 ,0, fl = 0.7 kHz = f, - 2Qs = 12,000 - 12,00) 2(60'1 = 11,900 Hz f2 = 12,000 + 1000 = 12,100 Hz If we define wlw, == a (a dimensionless relative frequency), obtain an expression for the admittance of a series RLC circuit in terms of R, Q" and a. I y = 1 = _1_1R _ _ _-,-R + j(wL - 1/wC) 1 + ,(alu,LlR - 1Iaw,CR) R[1 + jQs(a - 1 la)] RESONANCE 12.52 Express the half-power frequencies in terms of a == w/wr and Qs' and show that, for values of a at half-power points are approximately given by a = 1 ± l!(2Q,}. I For 12.54 Qs = = V(wj + (RI2L)2 ± :L 10, the radical becomes v'1.0025 ~1 + (2:Lr ± 2:L = VI +(1/2 Q,)2 ± 2~s a= = 1. Plot the magnitude of the admittance of the circuit Also indicate the corresponding values of Q. See Fig. 12-10. From the graph of Fig. 12-10 find the bandwidths. Wr I BW = - - = 27TQ l/VLC 1000 Hz = --::--27TQ 27TQ 10 n 15.9 Hz for R = 39.7 Hz = 159 Hz for R = 25 n for R = 12.55 :r = or In a series RLC circuit, L = 100 mH and C = 10 j.tF. as a function of a == w/wr for R = 10, 25, and 100 n. I the two From Prob. 12.14, half-power frequencies are given by W 12.53 Qs> 10, D 277 = = lOOn Show that the condition for resonance in a parallel RLC circuit (Fig. 12-11) is the same as that in a series RLC circuit. I For resonance, the circuit impedance (or admittance) must be purely resistive. y = .!. + j(wc __1_) R wL Resonance occurs when 1 wC--=O wL which is the same as that of Prob. 12.1. or 1 w= vrc or f= 1 27TVLC For Fig. 12-11 we have 278 D CHAPTER 12 v 1------.--- V I Fig. 12-11 12.56 Draw a phasor diagram showing the voltage and all the currents in a parallel resonant circuit. I 12.57 Fig. 12-12 See Fig. 12-12. Notice that IL +lc=O and I'=IR=VIR. Using the definition of Q given by Eq. (1) of Prob 12.8, obtain an expression for Qp for a parallel resonant circuit. I From Eq. (1) of Prob. 12.8, + wC)max P T 2,T(W L Qp = - We let i = Im cos wrt. where we have used Then R v = Ri = RIm cos w/. _ 1 WL - 2 w; = 1 I Le. Thus, 2: .IoL V d)t 2 -_ L·2 _ 1 ((' 1 IL - (1) r 1 '2 12 R2c . 2 Sill m (2) WJ Furthermore, Wc = !Cv 2 = H~R2Ccos2 (3) wrt From Eqs. (2) and (3), (conservation of energy) 1 Tr = 1, (4) (5) (6) and Substituting Eqs. (4) through (6) into Eq. (1) yields Qp 12.58 = 27Tf.RC, r R ~C L = R = w RC wrL r (7) = - A coil having resistance Reoil and inductance Leoil is ,;onnected in parallel with a capacitor C and across a voltage source V (Fig. 12-13). Determine the equivalent conductance and susceptance of the circuit at angular frequency w. Rc«.L C<><'l To-n1e I -t c~ {:. V,- ----- Fig. 12-13 ~ I The input admittance for the circuit in Fig. 12·13 is y= 1 R eoil + jXL __ + ---coil --jXc = 1 Reoil + jwLeoil 1 + ----j(lIwC) 0 RESONANCE 279 Rationalizing the respective denominators, and then separating the real and imaginary terms, y = Reoil - jwLeo;\ +. C _ R2coil + w 2 L coil 2 'W R e2o,.\ - i i B G 12.59 Find the resonance frequency for the circuit of Prob. 12.58. , The condition for resonance is B =0 w ~ 0); this gives (and w = 27TJ: = r r 12.60 1_1_ _ R~oil If R~o;\ > LeojC, Wr as given by Eq. (1) of Prob. 12.59 is imaginary, and so resonance cannot occur. For the circuit of Prob. 12.58 if , If R eo ;\ = 0, R eo ;\ = 0, show that the circuit will draw no current at resonance. then y=_._l_++=~ but, at resonance, XL coil = Xc; ,Xc, In the circuit of Prob. 12.58 (Fig. 12-13), L eo ;\ = 2 mH, cally, the variation in the resonance frequency. , (_1___1) Xc so Y = 0 and no current is drawn. ,XL coil 12.62 (1) L~o;\ \j LeoilC When is resonance impossible in the circuit of Prob. 12.58? , 12.61 R eo ;\ + w2Le2o,.\ XL coil C = 7.75 /-LF, and Reoil is variable. Show, graphi- See Fig. 12-14. 0.13 0.12 0.11 <Il - 0.09 ~J 008 t . ~ 0.07 ·s 0.06 "i 0.05 ·3 0.04 l: 0.03 0.02 (:i 0.01 o owu~~wu~~~-. 3 5 7 9 11 1315 Coil ,.liItanc., n 12.63 2 4 6 8 10 12 14 16 18 20 W)t. Fig. 12-14 If Reo;\ = 10 n, L eo ;\ = 2 mH, and nitude as a function of frequency. F,equency, C = 7.75 /-LF Fig. 12-15 in the circuit of Prob. 12.58, plot the admittance mag- , See Fig. 12-15. Note that the frequency at which resonance occurs and the frequency at which Y is a minimum are not the same; coincidence can occur only for the ideal condition Reoil = 0 (by Prob. 12.61). The equation relating admittance magnitude to radian frequency is 280 12.64 0 CHAPTER 12 For the data of Prob. 12.63 plot the input impedance and the input current as functions of frequency. , See Fig. 12-16. I I IL >Ie : le >IL 90· - - - , - - - - - - - - - - I I I Le.din. pr Or------+----------------Frequency, Hz o 2 4 10 6 12 W,-. 14 16 18 20 -90 A&tI/~ hequency, 0 Fig. 12-17 Fig. 12-16 12.65 Sketch the phase angle of the circuit of Prob. 12.63 as a function of frequency. , 12.66 See Fig. 12-17. A 60-V source supplies a parallel circuit consisting of a 2.5-/-LF capacitor and a coil whose resistance and inductance are 260 mH and 15 n, respectively. Determine the resonance frequency. , By Prob. 12.59, f. 1 r = 27T 12.67 ~ r- 2 2 1 Reoil 1 J 1 _ (15) =197Hz LeoilC - L~O'l = 27T ~ (0260)(2.5 x 10 6) (0.260)2 The circuit of Fig. 12-13 is modified to that ShOWl in Fig. 12-18. independent of R and is still given by Eq. (1) of Prob. 12.59. Show that the resonance frequency is R I v , From Fig. 12-18, we have 1 . C Y= R + JW Setting 12.68 Fig. 12-18 + 1 ReoB + jwLeoil Im (Y) = 0 and solving for Wr 1 -_ R +- 2 Rcoil R 2 2 + W Leoil . ( + JW C- again gives Eq. (1) of Prob. 12.59. Find the Q p at resonance for the circuit shown in Fig. 12-18. Leo'l) 2 2 2 Rcoil + W Lcoil RESONANCE lC) , The given circuit is reduced to that of Fig. 12-19c by the steps indicated in Fig. 12-19. and the resistor R, shown in Fig. 12-19b, we have: Similarly, Xp R2 + X = sX 2 s R2 X S S s s = X X + Xs = Xs s 0 281 Fig. 12-19 Combining the coil (1 ) 1 + Q2 s Now, we combine Rp and R of Fig. 12-19b to obtain Re of Fig. 12-19c such that Hence, Re = RpR/(Rp + R). (1) (2) as in Prob. 12.16. 12.69 Simplify the results of Prob. 12.59 when , Qeoil ~ 10. From Eq. (1) of Prob. 12.59, R~Oil 1 2 wr=LC-Ucoil coil Rearranging terms, (1) Q but = wrL coil (2) Reoil coil Solving Eq. (2) for R eoiP substituting into Eq. (1), and solving for Wr yields or 2( 1) cOil If 1 cOil Q ~ 10, and 1 w=~~= r 12.70 or Wr 1+-Q2 =LC VLeoilC A parallel circuit consisting of a 65-pF capacitor and a coil whose inductance and resistance are 56 /-LH and 600, respectively, is connected to the output of a transistor, as shown in Fig. 12-20a. The transistor acting as a current source has a source resistance of 37 kO. Determine (a) the resonance frequency and (b) Qeoil at the resonance frequency. 282 0 CHAPTER 12 -, rI I (0.) h61'1I 60n I I U.our~ I y- ":R 65 pF .J Transistor ~--.. ~ 37 kn (b) Xp Rp :: ::: 65 pF ::r 65 pF (c) Fig. 12-20 12.71 What is the Q p of the circuit of Fig. 12-20a at resonance? , Reduce the given circuit to that of Fig. 12-20c: thus, as in Prob. 12.68, Re = R 12.72 RRp 37,000 x 14,300 + R p = 37000 = 10.3 kfl , + 14300 , Determine the bandwidth of the circuit of Fig. 12-20a. , From Prob. 12.71 we have BW=239kHz 12.73 Determine the capacitance of the circuit of Fig. 12-21a so that it will resonate at 22.3 kHz and have a bandwidth of 4.05 kHz, when including a coil whose resistance and inductance are 56 fl and 3.2 mH respectively; assume the circuit is supplied by a current source whose source resistance is 8081 fl. 8081 (} 56 (} 3.2mH 15.7 nF c (hI }'ig. ]2-21 I From Fig. 12-21, Q coil = 3 wrLcoil R. = 27"(22.3 x 10 )(0.0032) = 56 8 COIl Qcoil < 10; hence high-Q formulas cannot be used. Instead, as in Prob. 12.68, RESONANCE fr R = RsaurceRp 8081 x 3640 e R R = 8081 + 3640 = 2509.58 fl source p Qp = wrCR e (Prob. 12.68). Thus, 5.51 = 27T(22.3 Also 12.74 + X 10 3)( C)2509.58 X X and 283 10 3 103 = 5.51 C = 15.7 nF. How much additional parallel resistance is required to change the bandwdith of the circuit of Fig. 12-21a to 6kHz? , From Fig. 12-21b, f, Qp = BW = 22.3 X 10 3 6 X 103 = 3.72 = w,CR e 1 Re 1 1 1691 = 8081 1 1 Rp 1 Rx -=--+-+- Re = 1691 fl 12.75 22.3 Qp = BW = 4.05 0 1 Rsaurce 1 + 3640 + R x A circuit consisting of a capacitor in parallel with a coil whose inductance and resistance are 1.05 mH and 100 fl, respectively, is driven at its resonance frequency of 600 kHz from a constant-current source. The source consists of a 2.30-mA 600-kHz constant-current generator in parallel with a 60-kfl source resistance. Determine (a) Qcail and (b) capacitance. I (a) The circuit is similar to that shown in Fig. 12-20a. 3 = W,L cail = 27T( 600 X 10 )(0.00105) = 9 Qcail R. 100 3 .58 COIl (b) Since Qcail > 10, 03 1 600 xl = 27TVO.00105C 12.76 Determine , (a) (a) Qp and Rp = R cail (l + C=67pF (b) the bandwidth of the circuit of Prob.12.75. Q~a.J = 100[1 + (39.58)2] = 156.8 kfl R = RpRsaurce = (156.8)(60) = 43.39 kfl e Rp + Rsaurce 156.8 + 60 11.0= 12.77 600 X 103 BW BW=54.5kHz What is the maximum energy stored in the capacitor of the circuit of Prob. 12.75? Also determine the power dissipated in the resistor. , The voltage across the capacitor is the voltage across the parallel circuit. resonance, Zin = Re. Therefore, V= IRe = 0.0023(43,390) = 99.8 V P = I2Re 12.78 = 12 Wc max = H67 x 1O- )(99.8v'2Y = 667 nl (0.0023)\43,390) = 230 mW Determine all currents in the circuit shown in Fig. 12-22. , Since XL = Xc, the circuit is in resonance. 120 lJt. I R = -10- = 12 D!.... lo° A Referring to Fig. 12-20c, at Thus, _ 120 lJt. _ ° IL - 4/90° -30/-90 A 284 0 CHAPTER 12 1, I + Ion Fig. 12·23 Fig. 12·22 12.79 Determine I and 1\ in the circuit of Fig. 12-23. , Notice that the circuit is in resonance. 120~ 1= - - = 12 ~ A 10 12.80 1\ = le + IR = 30 /90° + 12 ~ = 32.31 /68.2° A A parallel RLC circuit has R = 0.5 0, L = 10 mH, and the corresponding input currents at 24 V. , fr and C = 2.5 mF. 1 1 = 27TVLC = 27TY(iJ.0=1=)(=2=.5=X=I=0=3;=) = 31.83 Hz A 0.5-/-LF capacitor is connected in parallel with a coil whose resistance and inductance are 1.00 and 2.0 H, respectively. The parallel circuit is supplied by a 100-V sinusoidal generator that is operating at the resonance frequency of the circuit. Determine (a) the resonance frequency; (b) the input impedance; and (c) the current. , (a) By Prob. 12.59, f, = ~ ~_1_ - LR~o;/ = -~ J 1 27T Lca;/C 2'TI· 2(0.5 x 10 (2~ 6) - co;/ (b) At the resonance frequency, Y;n = G + jC·, 1= - V Z;n = 1 Z;n = G = 0.25 X 10-6 =4MO = 0.25 /-LS 100 = 25 /-LA 4 x 10 --6 A parallel section consisting of a 5.4-/-LF capacitor in parallel with a coil whose resistance and inductance are 18.80 and 8.0 mH, respectively, is driven at it:; resonance frequency by a 240-V sinusoidal generator. Determine the resonance frequency. ~-R~2 ( 18. 8 )2 6 , 1 1 caH 1 , II 106 fr 12.83 = 159.15 Hz 1 1 (c) f where I +(27TX 159.(5)2(2)2 12.82 Determine the resonance frequency I=VIR =24/0.5 =48A. At resonance, 12.81 Hence, = 27T ~ Lca;/C - L~a;/ = 2-:;'~ V (0.008)(5.4) - = 68 Hz 0.008 A 480-V 60-Hz source supplies energy to a parallel circuit consisting of a 25 /30°-0 branch and a 12 /-40°-0 branch (Fig. 12-24). Determine the impedance of .1 circuit element that, if connected in series with the source, will cause the system to be in resonance. , Fig. 12·24 (25m)(12~) . ° Zp = 25 /30° + 12 / -400 = 9.61 L=18.82 0 Z;n = Zp +Z = For the system to be in resonance, Z;n must have no imaginary term. XL = 27TfL, 3.1 = 1207TL or L = 8.23 mHo (9.10 - j3.1) Hence +Z Z=+j3.1=jXL • Since RESONANCE 12.84 D 285 Find the input current and input power for the circuit of Prob. 12.83. I From Prob.12.83, at resonance, 480 1= 9.10 = 52.75 A 12.85 P = (480)(52.75) = 25.32 kW A series-parallel circuit consisting of a (5 - j3)-fl impedance in series with a parallel section consisting of (4 + j2)-fl and (2 + j3)-fl impedances is supplied by a 490-V 30-Hz generator. Sketch the circuit and determine (a) the input impedance and (b) the input current. , The circuit is shown in Fig. 12-25. Z = (4+j2)(2+j3) =206/4307°fl p 4 + j2 + 2 + j3 . . (a) Z;n = Zp + (5 - j3) = (1.51 + j1.41) + (5 - j3) = 6.70 /-13.73° fl (b) V 490 LQ:: ° 1= Z;n = 6.70/-13.73° = 73.12 /13.73 A I + Fig. 12-25 12.86 (a) For the circuit of Prob. 12.85 determine the current in the parallel circuit operating at its resonance frequency? Explain. , (a) (4+ j2)-fl impedance. (b) Is the series- The voltage across the parallel branch is Vp = IZp = (73.12 /13.73°)(2.06 /43.0r) = 150.63 /56.80° V Vp 150.63 /56.8° ° 11 = Z1 = 4+ j2 =33.7 /30.23 A (b) 12.87 If the circuit were at resonance the phase angle between V and I would be zero. No: A tank circuit is supplied by a current source whose source resistance is 56 kfl. The tank circuit is composed of a 56-nF capacitor in parallel with a coil whose inductance and resistance are 35 mH and 80 fl, respectively. Determine (a) fr and (b) QCOB" , (a) By Prob. 12.59, 1 fr = 27T (b) Qcoil ~ I Looi'C - R~OH L~oil 1 = 27T ~ 80 1 (0.035)(56 x 10 9) - ( 0.035 = wrL coB = 27T(3578)(0.035) = R. 80 9.83 COIl 12.88 Determine , (a) (a) Zin and (b) Qp of the circuit of Prob. 12.87. Rp = Rco;l(l + Q~Oil) = 80[1 + (9.83)2] = 7.81 kfl Z = R = RpRs = (7.81)(56) _ ;n e Rp + Rs 7.81 + 56 - 6.854 kfl (b) Qp = wrCR e = 27T(3578)(56 x 10- 9 )(6.854 x 10 3 ) = 8.62 )2 = 3578 Hz 286 12.89 0 CHAPTER 12 Find , (a) the bandwidth and (b) the half-power frequencies of the circuit of Prob. 12.87. (a) Qp (b) From Prob. 12.87(b), fr 862 . = 3578 BW or = BW < 10. Qca;l fr = vr;t; BW= f~ - fl 3578 = f2 = 415 + fl f~ 12.90 BW = 415 or vr;t; 12.80 6 + 415f1 - 12.8 X 10 = O. Hence f1 = 33'16 Hz, 415 = f2 - f1 10" = (415 + fJf1 X f2 = 3791 Hz. Determine the capacitance required for a tank circuit that uses a 35-/-LH 60-!l coil and resonates at 1.65 MHz. The tank circuit is to be fed from a 1.65-MHz 2.0-mA constant-current generator whose source resistance is 6Ok!l. , The circuit is shown in Fig. 12-26. 6 = w,L = 27T( 1.65 x 10 )(35 x 10- QcaH R 6 0 6< 1 = ) 60 so we use Eq. (1) of Prob. 12.59: 1.65 10 6 = X ~~ 27T 1_____ ( 35 X 60 35 x 10 to-'e c= 259pF r~. ~ S<>-.&. - - - - r -... Fig. 12-27 Fig. 12-26 12.91 Calculate the current through the capacitor of the circuit of Prob. 12.90. , The circuit of Fig. 12-26 is modified to that shown in Fig. 12-27 from which (cf. Prob. 12.68) Rp = R ca ;/(l + Q~OH) = 60[1 + (6)2J R = R = RsRp = (60)(2.22) ,= 2 141 kfl e on Rs + Rp 60 + 2.22 . 2.22 k!l = V= IR;n (0.002)(2141) = 4.28 V = V le 12.92 Repeat Prob. 12.90 if the coil is replaced by one having , 373 = 11.5 mA LcaH =: 350 /-LH. ) _ - 90.7 we may write 1 27T fr = 12.93 4.28 = 6 40 R Xe and = w,L = 27T(1.65 x 10 )(350 x 10- caH QcaH> 10, 40 fl 6 Q Since RcaH =: =: I 1 'J L7; C=26.6pF COil Repeat Prob. 12.91 for the data of Prob. 12.92. , Rp Re = R;n =: R~R~ R s =: p = R ;l(l + Q;:a;/I= 40[1 + (90.7)2J = 329.1 kfl Ca ~g~~~~) =, 5(1,75 kfl V= IR;n = (0.002)(50,750) = 101.5 V le = Xe • Xc = 3626fl V 101.5 = 3626 = 28 mA RESONANCE 12.94 0 287 A tank circuit consisting of a coil whose resistance and inductance are 40.6 fl and 21.5 mH, respectively, is connected in parallel with a capacitor and supplied by a 1000-Hz 125-V generator of negligible impedance. Assuming the circuit is at resonance, determine its bandwidth. I = Qcoil W,L eoil R. coll = 271'(1000)(0.0215) = 3 40 6 3. 3 < 10 • Then, from Prob. 12.68, 12.95 A current source consisting of a sinusoidal 2.6-mA constant-current generator in parallel with a 60-kfl resistor supplies a tank circuit whose parameters are C = 105 nF, Leoil = 10.5 mH, and Reoil = 106 fl. Determine (a) f, and (b) QeOil at the resonance frequency. I (a) ( b) - fr Q ~ 1_1__ R~Oil 271' 'V CL COil L~oil ~ I 271' 'V (105 x 1 10 9)(0.0105) _(~)2 -41 0.0105 R. 106 - 2.81 Using a circuit similar to that of Fig. 12-27, R = RsRp e Rs + Rp Qp = w,CR e = 271'(4518)(105 x 277 = 4518 . BW (60,000)(943) = 928 4 fl 60,000 + 943 . 1O~9)(928.4) = 2.77 = :w BW= 1631 Hz V= IZ in = (0.0026 LQ:)(928.4 LQ:) = 2.41 LQ: V XC"il Zeoil = 2rrJ,Lcoil = 2rr(45 18)(0.0105) = 298 = R + jXL = (106 + j298) 2.41LQ: 316.29 L1.QA£: = = n 316.29 /70.42° fl 7.63 / -70.42 mA For the amplifier represented by the circuit of Fig. 11-32, find the gain at the resonance frequency of the secondary. I 12.99 = Under resonance, determine the voltage across the capacitor and the current through the coil of the circuit of Prob. 12.95. I 12.98 z 5 8H What is the bandwidth of the circuit of Prob. 12.957 I 12.97 - = w,L eoil = 271'(4518)(0.0105) _ coil coli 12.96 _ - Substituting W = W,= l/y L 22 C in the result of Prob. 11.43, we obtain In the amplifier circuit of Fig. 11-33, if the primary and the secondary have the same resonance frequency, obtain an expression for the gain in terms of Q, a SE wlw, and the coefficient of coupling k. I From Prob. 11.45, we have Letting M W;LllL22' = ky LII L22 and w, = l/y LII Cl = l/y L22 C 2 , and dividing numerator and denominator by 288 D CHAPTER 12 Letting Q = w,LI R = lIw,RC a = wlwr , and - j(g,)u.,;..jC I C2)(kla) 22 [1IQ I + j(a - lIa)1I1IQ2 + j(a - 1Ia)] + k a G=--------=-~ -j( gm Iw,...;c;c;)(k la) [lIQ\Q2 + ea 2 -- (a -- 1Ia)2] + j(a -lla)(l/Q\ + 1IQ2) -=--~--~ 12.100 Define 0 = a-I terms of o. I Substitute a = 1). (with Q\ = Q2 =, Q If in the amplifier of Prob. 12.99, express the gain in a = 1 and 1 a- a in Eq. (1) of Prob. 12.99 (with G= 12.101 (1) a2 -- a+1 (a - 1) = 20 a 1 = - - - ,,= - - a Q\ = Q 2 = Q): - j( gmlw;V Cl C2)(k/1) - j( gm/w,Y Cl C2)kQ2 --- = ::--------=---7--'-"----'--'-------'c~------"-(11 Q2 + k 2 - 4( 2) + j20(21 Q) [\ +(kQ)2 - (2Qo )2] + j( 4Qo) For the amplifier of Prob. 12.99, we have w, ,= 11)6 rad/s, gm = 1500 p,S, C\ = C 2 = 1 nF, and QI = Q2 = 40. Plot IGI (obtained in Prob. 12.100) as a function of Qo, with -2 < Qo < 2, for k = 2/Q, lIQ, and 1/2Q. I For the given values we have G=--- l.5kQ2 [1 + (kQ)2 -- (2QO)2] + j(4Qo) Letting y = Qo, 80 (5 - 4/) + j4y G= 40 (2 - 4/) + j4y 20 [1.25 - 4/ + j4y -::----- for k =2/Q for 1 k=Q for 1 k = 2Q the magnitudes of which are plotted in Fig. 12-28. I~I 20 \ /0 o ~--+--+--- -l -/ 0 ~ _) ' F i g . 12-28 RESONANCE 12.102 1JJI V I V=[R+j(wL- w Thus 12.103 Q = wLl R For a series RLC circuit, obtain an expression for V c/V in terms of Vc V =~ = lIR[l + jQ(a Ye 289 a = w/w,. ~~ c =- 1 -lIa)] j(alQ)[l jwC + jQ(a - (a 2 - I) + j(a/ Q) -lIa)] Q = 1,4, and 10. Plot IVc/vl, as obtained in Prob. 12.102, as a function of a for I and D See Fig. 12-29. lYe/vi -j/2W .2i. + .j .... o·s '·0 ~ Z.... 2.0 J Fig. 12-30 Fig. 12-29 12.104 Vc Determine the frequency at which the circuit of Fig. 12-30 will be in resonance. I For resonance, we must have Im (Z;n) = O. Now, Z ='w + - jl2w ='w _ _,_._ = 'w _ j(2w + j} 1 - jl2w' 2w - j ' 1 + 4w 2 on' Im (Z;n) = w 12.105 2w 1 +4w =0 or w = 0.5 rad/s If the capacitor and the inductor of the circuit of Fig. 12-30 are interchanged, what is the resonance frequency? I Proceeding as in Prob. 12.104, - j jw j Z =-+--=--+ m 2w 1 + jw 2w Im (Z ) = - m 12.106 2 - 1 2w jw( 1 - jw) 1 + w2 w + - -2 1+ w = 0 20 IS!.. -, ° = -'1- = 20 flQ: A IL = 20 /90° + 20 IS!.. = 20V2 /45° A W 2 Vc = 20V1" sin O.St V. Find the instantaneous Solving the problem in the frequency domain we have (since le jw w = 1 rad/s or The voltage across the capacitor of the circuit of Fig. 12-30 is current through the inductor. I - j =-+-+ - -2 2w 1 + w2 1+ w w = w, = 0.5 rad/s and Vc = 20 IS!.. V), 20 IS!.. IR = 1 IS!.. = 20 IS!.. A or iL = 40 sin (0.5t + 45°) A 290 12.107 D CHAPTER 12 What are the instantaneous energies stored in the capadtor and inductor of the circuit of Fig. 12-30 for the data of Prob. 12.106? I Since w = 0.5 rad/s, L = 1.0 Hand we have C = 2 F. Thus, WL = ~ L(iL)2 = Hl)[40 sin (O.St + 45°)]2 = 800 sin 2 (0.5t + 45°)J 2 Wc = !C(V(J2 = H2)(20V'2sinO.5t)2 = 800sin 0.5tJ 12.108 Determine the Q of the circuit of Fig. 12-30. I From Eq. (1) of Prob. 12.57, Q = ~J~'L + WC)max PR From Prob. 12.107 and standard trigonometrical idemities, _ 80C ( . sin LOt - cos LOt) 11-+ 2 J WL+W C - Maximizing by calculus, we find (WL Hence 12.109 + WC)max ( 2V2) J = 800 1+ PR=I 2 R=(20) 2 1=400W w, = 0.5 rad/s V2 _ 07 _ (0.5)(800)(1 + V2/2) _ Q400 - 1 + 2 - 1.7 If the resistor and the inductor of the circuit are interchanged, what is the resonance frequency? I The resistor has no effect on the frequency. Since L = 1.0 Hand f, = 1 21TvTC 1 27TY (1)(2) ---;== = 0.1125 Hz C = 2 F, CHAPTER 13 uFrequency Response and Filters ~ 13.1 For the RC circuit of Fig. 13-1a, sketch Vou • as a function of frequency and show the cutoff frequency fe' I By KVL we have (1) which is qualitatively plotted in Fig. 13-1h, where Vou • = 0.707 V;n f= fe' at 0.707 V.. ,,", " , , R 0 """- Vil"'l 0 0 er Passband-+l V out " I O"l. 8",,""~ 0 " I I c.l , ,I o le Frequency. Hz (bl 13.2 Relate R, C and !c for the circuit of Fig. 13-1a. I At f=!C, VOU1 = 0.707V;n' Substituting in Eq. (1) of Prob. 13.1 yields 2 X~ (0.707) = R2 + X~ 13.3 The circuit of Fig. 13-1a is a low-pass filter. I 13.4 Let R = SOO fl. Then C = 1I21T!cR = 1/21T(SOO)(SOO) = 636.7 nF. A low-pass RC filter has a 800-Hz cutoff frequency. If R = 2 kO, 1 800 = 21T(2000)C In a low-pass RC filter, having a cutoff frequency of 2 kHz, I 13.6 1 R = Xc = 21Tfc C or Design one having a cutoff frequency of SOO Hz. I 13.5 Fig. 13-1 2000= 21TR(80 x \0 9) The RL circuit of Fig. 13-2 is also a low-pass filter. what is the value of C? C=99.S nF C = 80 nF. Determine R. R =99S 0 Determine its cutoff frequency fe' I Thus, R2 2 (0.707) = Hence, 2 2 R +XL or fe = RI21TL Hz. 291 292 D CHAPTER 13 R V out Fig. 13-2 0-----13.7 Repeat Prob. 13.3 for an RL circuit. I Let R = 500 fl. Then R 500 L = 21Tlc = ~;i500) = 159.15 mH 13.8 For the circuit of Prob. 13.7, determine v I VOU! = 100 out at (I Hz, 100 Hz, 500 Hz, 1 kHz, and 100 kHz for a 100-V input. 500 Y(500)2 +=(==2=1TI===x=1=59=.=15=X=1O~37";)2 The following table gives the desired result. f,Hz o 100.0 98.0 70.7 44.7 0.5 100 500 1000 100,000 13.9 13.10 Find L in a low-pass RL filter having R = 1.5 kfl I R f. c A low-pass RL filter has L = Determine R. and a cutoff frequency of 4 kHz. The circuit of Fig. 13-3a acts as a high-pass filter. frequency. I L = 149 mH R 4000 = 21T(0.025) I 13.11 1600 == 1500 21TL = 21TL 25 mH and a cutoff frequency of 1600 Hz. Plot R = 628fl VOU! as a function of frequency and determine the cutoff From KVL, at cutoff VOU! = O. 707V;n' Thus, 2 (0.707) Hence, le = R2 = or -2R,,;T +."c R = Xc = 2 1 f. C 1Tc 1I21TRC Hz. Vin i c > =o~o~~~ / I / o~-----;II~--~~--~ I (a) Vout / o / R I I I I ,.I.----Puaband----.. I .-..I oK I IJhl/~ I I f, Frequency, Bz (h) Fig. 13-3 FREQUENCY RESPONSE AND FILTERS 13.12 The RL circuit of Fig. 13-4 also acts as a high-pass filter. 0 293 Determine its cutoff frequency. I At cutoff, Thus, fe = RI21TL Hz. Fig. 13-4 13.13 Design an RL high-pass filter having a cutoff frequency of 2 kHz. I 13.14 Let R = 2 kO, Then, For the circuit of Prob. 13.13, determine VOUI at 0 Hz, 100 Hz, 500 Hz, 2 kHz, and 100 kHz for a 100-V input. I V = 100 oul 3 21Tf(159.15 x 10- ) V(200W + (21Tf159.15 x 10 3)2 The desired result is given below: 13.15 A high-pass RC filter has 13.16 A high-pass RC filter has Vou " V 0 100 500 2000 100,000 0.00 5.00 24.25 70.70 99.98 R = 12500 and a cutoff frequency of 1600 Hz. 1 1 fe = 21TRC I f,Hz 1600 = 21T(1.25 C= 0.65 nF X 10 3 )C 13.17 An RL high-pass filter has R = 1.6 kO and a cutoff frequency of 10 kHz. R 1600 fe = 21TL 10,000 = 21TL The circuit of Fig. 13-5a may be used as a bandpass filter. show its frequency response. The circuit of Fig. 13-5a shows the load on the filter. resistance, we obtain the circuit of Fig. 13-5b from which VOUI =V m ,/ RT V R~ RT = ROUIRLI(RoUI + RL)' + (XL - XC)2 =V m Find L. L = 25.5 mH Obtain an expression for its output voltage, and I where Determine R. R=27.2kO I 13.18 C= 79.6nF and a cutoff frequency of 9 kHz. I Determine C. ,/ Combining the load resistance with the filter V R~ RT + [21TfL - 1I(21TfCW A typical frequency response plot is shown in Fig. 13-5c. 294 0 CHAPTER 13 -_- O>----~'-----ll{-[ ';. _r;- - - ! "0.' R_. 0--------- R ••• - - - _IL 0___ _ (a) ----V In ]~.:b.nd o2°2-vln -, ,-- :J __ 1 ~ ,g I , I \ \ I \ , ... I .,,"/ I / \, " ' .... ..... _-- Frequency. Hz (h) (e) Fig. 13-5 13.19 State quantitatively the criteria which may be used t·) characterize a bandpass filter. I 13.20 The criteria are Ir = 1I27TVLC; selectivity Q, = wrLlR; bandwidth BW= I)Qs' A parallel circuit yielding a bandpass characteristic similar to that of Fig. 13-5e is shown in Fig. 13-6. that the load draws negligible current, obtain an expfl~ssion for Vout ' V I =V out in z where ab R 1 + Z ab Z ab Assuming = (RCOil+jXcoil)(-jXC) (Rcoil + jXcoiJ + (-jXC) 9-- - - 0 a R 10ad c 0---- 13.21 Assuming the circuit parameters in Fig. 13-5a are L,o;1 = 50 mH, determine (a) the resonance frequency and (b) the bandwidth. I (a) (b) 1 !,. = 27TVLC Fig. 13-6 C = 127 nF, Rout = 63 n, 11 = 2000 BW BW= 182 1 =2kHz 27TY(0.05)(127 x 10 9) ~-r==========~ R = RoutRload = 63(600) =57n TRout + R 10ad 63 + 600 Qs 13.22 0 = wrL = 27T(2000)(0.05) _ RT 5 7 - 11 Ir Qs= BW For the circuit of Prob. 13.21 determine the cutoff frequencies. I Since Qs > 10, BW 182 12 = Ir + -2- = 2000 + T = 2091 Hz BW 182 II = Ir - -2- = 2000 - T = 1909 Hz R 10ad = 600 n, FREQUENCY RESPONSE AND FILTERS 13.23 0 295 If the input voltage to the circuit of Prob. 13.21 is 30 V, calculate the output voltage at f,; fi' f2' and lOf,. v I =V out At the resonance frequency XL = XC' on RT YR~ + (XL - XC)2 Hence, f1 = 1909 Hz, At XL = 21TfL = 21T(1909)(0.05) = 6000 1 1 Xc= 21TfC = 21T(1909)(127 x 10-9) =6560 V = 30 out At 57 Y(57)2 + (600 - 656)2 21.4 V f2 = 2091 Hz, V XL =6560, 57 =21.4V Y(57)2+(656-600)2 =30 out At =' f =' lOf, = 10(2000) = 20,000, XL = 21T(20,OOO)(0.05) = 6.28 x 103 0 1 Xc= 21T(20,000)(127 x 10-9) =62.70 V = 30 out 13.24 57 = 0.28 V Y (57)2 + (6280 - 62.7)2 For the filter circuit of Fig. 13-5a we have R L = 50 kO, filter having cutoff frequencies of 25 and 23 kHz. I Leoi' = 45 mH, fr 1 + 2~ 21TYO.045C Q = ~ = 24,000 = 12 = w,L s BW 12 = 21T(24,000)(0.045) RT RT 2000 RT = 565 0 1 1 1 -=-+-565 Rout 50,000 ~ = _1_ + _1_ RT Rout R 10ad Rout = 571 0 Design a series-resonance-type bandpass filter that has cutoff frequencies of 15 and 35 kHz. resistance is 60 kO, and the coil has an inductance of 50 mH and negligible resistance. 35 X 3 10 = 1 Hence The load BW = (35 - 15)10 3 = 20 kHz I 13.26 f, Design a bandpass 1 = 24,000 Hz = 21TvTC C = 978 pF Reoil = O. 25,000 = BW = f2 - f1 = 25,000 - 23,000 = 2 kHz or 13.25 and f, + 20 X 2 10 3 or f, = 25 kHz 1 25000 = 21TVO.050C C= 811 pF. Assume the circuit parameters for the series-resonance bandpass filter in Fig. 13-5a are C =' 1.8 pF, Leoi' = 25 mH, Rout = 52 0, and R 10ad = 9000 O. Determine (a) the resonance frequency and (b) the bandwidth. 296 0 CHAPTER 13 I 1 1 (a) f, = 2 1 T V I T = 750,644 Hz (b) Rau,Rlaad = 52(9000) = 51 7 n RT - R out + RI oa d 52 + 9000 . 21TV (0.025)(1.8 x 10 12) _L_ _ w,L _ 21T(750,644)(0.025) = 2280 0= Qs - RT 51.67 BW Thus, 13.27 BW = 329 Hz. f2 BW 329 = f, + -2- = 750,644 + T = 750,809 Hl f1 = BW 329 f, - -2- = 750,644 - T = 750,480 Hz Obtain an expression for the output voltage as a function of frequency from the circuit of Prob. 13.26, for a 60-V input voltage. I v = V au, 13.29 BW For the filter of Prob. 13.26, determine the cutoff frequencies. I 13.28 2280 = 750,644 or m = 60 RT VR~ + (XL - xcf 51.7 r- V (5l.7)2 + [6.28f(0.025) -1I(6.28f1.8 x V 10 12)] From the result of Prob. 13.28, determine the output voltage at resonance frequency, cutoff frequencies, and at lOf,· I The result is tabulated below: 13.30 f Vout 750,644 750,834 750,49:; 750,440 60.00 39.31 44.47 0.00266 The circuit of Fig. 13-7 acts as a bandstop filter. Its resonance frequency and bandwidth are shown in Fig. 13-8. Obtain a relationship between the input and output voltages. I v out =: V Zab ~ 10 cue Hence Stop band IJ C>----JVlI'v-----Q- - - - ~ - - - - 0 I I V out r______ _ o_ _ _ _ _ _ _ _ _ c 13.31 o o 11 12 b Frequency. H. Fig. 13-7 Fig. 13-8 Sketch the frequency of the circuit of Fig. 13-7. I See Fig'. 13.32 I I I I 13-8. The parallel circuit of Fig. 13-9 also acts as a bandstop filter. Relate its output voltage to the input voltage. FREQUENCY RESPONSE AND FILTERS V I out Z where = ab Reoih =Vin R 0 297 Rout + Z ab out (Reail+jXLeail)(-jXc) (Reail+jXLeail)+(-jXc) Lcon IJ c 0-----------_ - - - - 13.33 RI = 10000, In the circuit of Fig. 13-7, bandwidth. I Thus, 13.34 1 = 21TVLC Ir Qs Leoil Reail = 2 0, = 160 mH, and C = 396 pF. Determine its 1 12 = 20kHz 21TVO.160(396 x 10 ) = wrL = 21T(20,000)(0.160) = 2 = ~ Rs 2 + 1000 0 BW or 20 = 20,000 BW BW = 1000 Hz. Since Q > 10, BW 1000 12 = Ir + -2- = 20,000 + -2- = 20,SOO Hz 1 I = Jrf - 1000 BW 2 = 20 ' 000 - - 2 = 19 ' SOO Hz - Find the output voltage of the circuit of Prob. 13.33 as a fraction of the input voltage at resonance frequency and at cutoff frequencies. I At resonance, XL = Xc; hence, =V V out At m ~ 'f(R . + R )2 = V coli V in 2 0002V 1002 =. in 1 I1 = 19,5OO Hz, 1 Xc = 21T(19,SOO)(396 XL = 21T(19,SOO)(0.160) = 19.6 kO XL - Xc = -10000 V 12 = 20,SOO Hz Xc= =V out 10 12) = 20.6 kO on V (2 + 1000)2 + (-1000)2 = 0.706V on XL = 21T(20,SOO)(0.160) = 20.6 kO 1 21T(20,SOO)(396 V X V(2)2 + (-1000)2 =V out 13.36 Fig. 13-9 What are the cutoff frequencies of the circuit of Prob. 13.33? I 13.35 0 m X 10 12 =19.6kO ) XL-XC=10000 V(2)2 + (1000? = 0.706V y(1000)2 + (100W m Determine the capacitance required for a series resonance bandstop filter that will block 8S kHz. The inductance and resistance of the coil are 60mH and ISO, respectively, RI =20000, and R laad = 1.4 MO. I Referring to Fig. 13-7, Ir = 1 21TVLC 1 8S,000 = 21Tv'D.06C c= S8pF 298 13.37 0 CHAPTER 13 What is the bandwidth of the circuit of Prob. 13.36? = wrL = 27T(85,000)(0.06)= 15 9 = ~ Q, Thus, 13.38 15 + 2000 R . 159= 85,000 . BW or BW BW = 5.34 kHz. The circuit of Fig. 13-10 is a double-resonant filter. The two resonance frequencies are for the tank (parallel) circuit and for the series circuit. If Cl = 3.5 nF. jetermine the remaining parameters required in order that the filter will reject a 100-kHz signal but accept 50 kHz. I The resonance frequency of the stopband is determined by the tank circuit. Assuming Qeoil ~ 10, f. = r 1 27TVLC 1 100 000 = --- c=====:;= ' L J (3.5 x 10 '9) L J = 724 ILH 2,T'! The impedance of the tank circuit at 50 kHz is 1 XCI = 27T(50,000)(3.5 x 10 9) = 910 n Ztaok X L1 X C1 - X Cl ) LI = J'(X (227)(910) =' 02 n j(227 - 910) J3 Thus, at 50 kHz the tank circuit behaves as a pure il1ductive reactance of 302 n. 50 kHz, capacitor C2 must have a capacitive reactance of 302 n. OI Thus, To cause series resonance at 1 302 = 27T(50,000)C C = 10.5 nF. C2 ~\I---'" ~---- - . ., vnul. ------------_.( ---- C>--------------() - - -- Fig. 13-11 Fig. 13-10 13.39 The circuit of Fig. 13-11 is also a double-resonant filter. ters to reject 150 kHz and accept 200 kHz. If LI = 2.5 mH, determine the remaining parame- I The resonance frequency of the passband is determined by LI and Cl: f. r 1 27TY L I C 1 = --== 1 200,00J = 27TYO.025C I Cl = 25pF At 150 kHz, the impedance of the series LC branch is XL ,= 27T(150,000)(0.025) = 23.6 kn 1 XCI = 27T(150,000)(25 x 10 12) = 42.5 tn Z,er = j23,600 - j42,500 = -jI8,900 n Thus the series branch is in effect a capacitive react,mce of 18,900 n at 150 kHz. the parallel branch must have an inductive reactance of 18,900 n. 18,900 = 27T(150,000)L2 13.40 For tank resonance to occur L2 = 20mH Assume the circuit parameters for the bandstop filter shown in Fig. 13-7 are RI = 1500 n, Reoi' = 1.5 n, and C = 300 pF. Determine (a) f, and (b) the bandwidth. Leoil = 140 mH, FREQUENCY RESPONSE AND FILTERS I 1 21TvTC 1 == 24,558 Hz 21TY(0.14)(300 x 10 12) -=--r======~ (a) fr = (b) Assuming the load draws insignificant current, Q = wrL = 21T(24,558)(0.140) = 14 39 R s Hence, 13.41 1500 + 1.5 fl 1707 = 24,558 - -2- = 23,705 Hz 12W 21Tf What is the output voltage for a 30-V input at resonance frequency in the circuit of Prob. 13.40? Vou• = 30 II!;:II where Zab Hence, at = ~(1.5)2 + [21Tf(0.14) - fr = 24.558 kHz, 21Tf(300IX 10 12) r 0 Vou • = 0.444 V. Determine the required capacitance for a series resonance bandstop filter that will block 65 kHz. The load resistance is 50 kO, RI is 30000, and the coil inductance and resistance are 55 mH and 10 0, respectively. 1 I or fr = 21TvTC Hence, 65 000 = _-=1== , 21TVO.055C C = 109 pF. What is the voltage across the load in Prob. 13.44 at resonance frequency? I At resonance Zab = 10 0 Given: 10 VOU! = 80 10 + 3000 = 0.27 V Assume capacitance Cl for the double-resonant filter shown in Fig. 13-11 is 2.1 nF. parameters that will block 90 kHz and accept 100 kHz. J: I V;n = 80 V. resistive, 10 x 50,000 _ R T = 10+50,000 =100 13.46 1707 f2 = 24,558 + -2- = 25,412 Hz Z;n = y(1501.5)2 + [21Tf(0.14) - _1_ (300 x 10 and Z;n is found in Prob. 13.42. 13.45 2~~8 For the circuit of Prob. 13.40, determine the input impedance as a function of frequency. I 13.44 14.39 = For the circuit of Prob. 13.40, find the cutoff frequencies. I 13.43 or . BW = 1707 Hz. I 13.42 0 299 r = 1 21TV LIC I Determine the remaining or Thus, Ll = 1.207 mHo At 90kHz, XL! = 21T(90,000)(1.207 x 10- 3 ) = 682.20 Z,", = jXL1 - XC! = 1 9) = 842.520 21T90,000(2.1 x 10 jXcl = j(682.2 - 842.52) = - j160.32 0 For parallel resonance to occur at 90 kHz, jXL2 == + j160.32 Hence, 13.47 21T(90,000)L2 = 160.32 L2 = 284 ILH. Assume capacitor Cl in Fig. 13-10 has a capacitance of 6.5 nF. block 75 kHz and accept 20 kHz. I Determine the remaining parameters that will For stopband, 75000= 1 , 21TY LI (6.5 x 10 9) At 20kHz, 300 0 CHAPTER 13 XLI = 27T(20,000)(693 X 10- 6 ) = 87.00 X Cl - 1 27T(20,000)(6.5 X 10- 9 ) -122490 - • The impedance of the parallel section at 20 kHz is 87/90° X 1224.9/-90° . '87 _ '1 ';;-4-9-- = J93.65 0 Zp = J J .-•.. For series resonance to occur at 20 kHz, XC2 Hence 13.48 = I 27T(20,OOO) C2 or 93.65 93.65 C2 = 85 nF For the circuit shown in Fig. 13-12, determine the ratio V2 /VI • I By mesh analysis we have 141.411 + jlO0I 2 = 100 iQ~ is!.. A (0.471)(141.4) = 33.4 is!.. V I1 = 0.471 Thus, V, = 100 - + 141.412 = 0 jlOOI, 12 = 0.333/-900 A 0 V, = (0.333/-90°)(141.4) =47.\0 /-90 y V 47.10 ",-=90° . - 2 = - - - - . - - = -]141 V, 33.4 LQ~ . fooL!!" jJ··~~~·v~ f~l. V 4.rt L-------OL 13.49 In the circuit of Fig. 13-12 let L = 100 mH and frequency) and plot IV/VII as a function of a. I Fig. 13-12 C = 10 /-tF. Define a = wlwr (w r being the resonance Using nodal analysis, after some manipulation it may be shown that V2! 1 !VI 0= ,r[=.=_=4=a72=+=4=a"7 (1) 6 where a = wlwr and wr = v21 LC = 10 3 \12. Hen:e Eq. (1) is plotted as shown in Fig. 13-13. u 13.50 Fig. 13-13 A filter circuit using an operational amplifier is sho"n in Fig. 13-14a. It may be represented by its equivalent circuit shown in Fig. 13-14b. If Z, = RI and Z2 corresponds to an RC parallel circuit with elements R2 and C, obtain the ratio Vo/Vi • I From Fig. 13-14b we have V)Vi=-Z2/ZI''NOW, ZI=RI and R2(1/j.'j)C) R2 Z = - - - - - = ----''--=2 R2 + l.'jl'j)C 1 + jwCR 2 Thus, Vo =- R2/ RI Vi 1+ jwCR 2 (1) FREQUENCY RESPONSE AND FILTERS + + 'v.t \;. t VD We may substitute R = - R2! RI and Vo = in Eq. (1) of Prob. 13.50 to obtain CR 2 = L R Vi l+jwL which is identical to the result of Prob. 13.6. R V" = Vi or YR2+X~ Hence the circuit acts as a low-pass filter. What is the cutoff frequency of the filter of Prob. 13.50? I From Prob. 13.6, le = R/27TL. R = R2/RI From Prob. 13.51, le = 13.53 Fig. 13-14 Obtain the filter characteristic of the circuit of Prob. 13.50. I 13.52 301 + (D) 13.51 0 and L = R 2C. Hence R2/R] 1 27TR 2C = 27TR I C The operational amplifier circuit of Fig. 13-15a may be represented by its equivalent circuit shown in Fig. 13-15b. If Zl = l/jwCI . and Z2 consists of an RC parallel circuit, with elements Rand C2, determine V)Vi' Vo = 1 + Z2 Vi Zl I Z = _1_ I and jwC = R(lIjwC2) + lIjwC2 jw(RC I + RC2) + 1 = ---:---"::"-:::---:-jwRC2 + 1 R 2 Vo R/(l + jwC2 R) = 1+ Vi lIjwC] Hence, Z - R 1 + jwC2 R (1) V.l (a.) 13.54 (6) Fig. 13-15 Approximate, by straight lines, the frequency response of the circuit of Fig. 13-15a. I From Eq. (1) of Prob. 13.53, Vo _ jw(RCI + RC2) + 1 _ jwT I V, jwRC2 + 1 - jw T2 where TI = R(C I + C2) is shown in Fig. 13-16. and T2 +1 +1 = RC2. Notice that or Ivol _ Y(wT I )2 + 1 lV.T - y(wT2)2 +1 and the approximate frequency response 302 D CHAPTER 13 Fig. 13-16 13.55 Show that the circuit of Fig. 13-17 acts as a low-pa:;s filter. I By nodal analysis we have V, -Vc 6 Vc V, +lIjw 6 5Vc =V2 - Vc . ,V, VI - V2 = (1 + Jw6)- __6 Consequently, VI -V2 = (1 or --- = -- or 6Vc =V2 or 0 which is similar to the result of Prob. 13.6. + jw6)Vc Henct:, the circuit is a low-pass filter. 6ft V, Fig. 13-17 13.56 What is the cutoff frequency of the filter of the circuit of Fig. 13-17? I From Prob. 13.55 we have V,=Y' 2 At cutoff V2 = 0.707V, = ~--2 + (6W)2 or 49 + (6(u,J Thus, Wc = 2'TTJc = 0.8 /=0.13Hz Obtain a relationship between Vo and Vi shown in the circuit of Fig. 13-18. Vi - Vo Vo 1= 1000 = 1000/i~ I Vo V, Hence, 13.58 6 Y49 or (0.707)2 13.57 I ( jw Vo 1) + i + jw =Vo 1000 + 1 + jw 1 + jw (1- + jW)2 + 1000 What is the input impedance of the circuit of Fig. 13-18 at resonance? ..I y. I "I E~T ,,,," F V 0 IH LJ~ Fig. 13-18 FREQUENCY RESPONSE AND FILTERS w- I Thus, 13.59 1 _ Zpa<alle' = 31.6rad/s = (Q, )2R = (31.6)21 = 1000 il. Hence, Q=(31.6)(1)=316>1O sI' Zin = 1000 + 1000 = 2 kil. Determine the cutoff frequencies of the circuit of Fig. 13-18. I From Prob. 13.58, 1,= w, 31.6 2'TT = 2'TT =5.03Hz B W= I, Q= 5.03 31.6 =0.16Hz 12 = 5.03 + 0.~6 = 5.11 0.16 4. 95 H z I1 = 5.03 --2-= Thus, 13.60 1 ,- VIT - y(l)(lO 3) 0 303 The circuit of Fig. 13-19 is a double-tuned filter. If C = 100 pF, to be rejected and a frequency of 800 kHz accepted. , Hz find LI and L2 if a frequency of 400 kHz is For series resonance, f= s 3 1 =800xlO 2'TTV LIC or L = (271"800 x 103 )2(100 x 10 I (2) At 400 kHz, XI = wL, = 271"f L, = (271")(400 X 10 3 )(0.396 X 10- 3 ) = 995 X . = _1_ = 1 = 3979 il ( wC 2'TT(400 x 10 3 )(100 x 10- 12 ) j(XL - Xc) = j(995 - 3979) = - j2984 At resonance at 400 kHz, XL = Xc or Xc L 2 =W- 2984 = 1.19 mH 271"(400 x 103) + + V, RL. V, Fig. 13-19 n = 0.396 mH ~ CHAPTER 14 Three-Phase Circuits\~~ 14.1 Three voltage waveforms are shown in Fig. 14-1. I Write mathematical expressions for these voltages. We may mathematically express this system of voltages as Va'a = Vm sin wt v -+--\--!---L-_ _ t-Vt Fig. 14·1 14.2 Express the voltages of Prob. 14.1 as phasors. I These voltages may be written in phasor notation as Va'a = V !.Jt.. = V(l + jO) "b'b = V /-120° = V(-0.5 - jO.866) Vc'c=V /-240°= V(--O.S +jO.866) 14.3 Draw a phasor diagram showing the three volt ages ·)f Prob. 14.2. these three voltages. where Show three voltage sources corresponding to I Figure 14-2a shows the phasor diagram, whereas Fig. 14-2b shows the three voltage sources corresponding to the three equations. Consequently, we may define a three-phase (voltage) source having three equal volt ages which are 120° out of phase with one another. In particular, we call this system a three-phase balanced system-in contrast to an unbalanced system, in which the magnitudes may be unequal and/or the phase displacements may not be 120°. c ~ # + ) - - - - - . V•.• = vm - + Vb ·b = VL=lN (a) 14.4 c·.!'J(;\~ b~ Vb ' b b (b) Fig. 14-2 What is the phasor sum of the three volt ages in a balanced three-phase system? I From Prob. 14.2, we have: 2: V = 304 vc' c Va'a + Vb'b + Vc'c = V + (-0.5 - jO.866)V + (-0.5 + jO.866)V = 0 THREE-PHASE CIRCUITS 14.5 The order in which the three voltages of Fig. 14-1 attain their maxima is known as the phase sequence. definition, determine the phase sequence of the volt ages of Fig. 14-1. 0 305 By this I We abbreviate va'a' Vb'h' and vc'c as Va' Vb' and Vc respectively. Now referring to Figs. 14-1 and 14-2a, we observe that the voltages attain their maximum values in the order V 0 ' V h' and vc' This order is known as the phase sequence abc. A reverse phase sequence will be acb, in which case the volt ages Vc and Vb lag va by 120 and 240 respectively. 0 , 14.6 A way to interconnect the voltage sources of Fig. 14-2b is shown in Fig. 14-3, and is known as the wye connection. Obtain a relationship between the line voltages and the phase voltages. I The line voltages are related to the phase voltages such that Voa + Vab = Yob or (1) Similarly, V bc =Voc -Yob jO.866) - V(l + jO) and Vca =Voa -Voc' From Eq. (1) and Fig. 14-2 we may write In polar form, V ah = V(-0.5- = V(-1.5 - jO.866). (2) Similar relationships are valid for the phasors V bc and Vca' Because Vab is the voltage across the lines a and b and V is the magnitude of the voltage across the phase, we may generalize Eq. (2) to VI = v3Vp where VI is the voltage across any two lines and Vp is the phase voltage. I. r-------~~------oA 0 b 14.7 'b 8 'e c Fig. 14-3 Draw a phasor diagram showing the phase and line voltage relationships of Prob. 14.6. I See Fig. 14-4. / v.. =-v j---- \ I / ~ Vbo = -Vob I / '--~---~ vco =-v QC 14.8 vc. Fig. 14-4 What is the relationship between the phase and line currents of a wye-connected system? corresponding phasor diagram. Draw the I For the wye connection it is clear from Fig. 14-3 that the line currents I, and phase currents Ip are the same. Thus, we may write I, = Ip. The mutual phase relationships of the currents are given in Fig. 14-5. 306 D CHAPTER 14 Fig. 14-5 14.9 Repeat Prob. 14.6 for the connection shown in Fig. 14-6, which is known as the delta connection. corresponding phasor diagram. I From Fig. 14-6 it follows that V, =Vp' Draw the See Fig. 14-7 for the phasor diagram. I.A .---------------~~------~A 'b8 ~----~--------~8 'cC ......- - - - - - - - - - -__' - - - - - - 0 C 14.10 Fig. 14-f, Fig. 14-7 I We show in Fig. 14-6 the phase currents and line currents for the delta-connected system. currents and line currents are related to each other by The phase Repeat Prob. 14.8 for the delta connection. and These relationships may be generalized to 14.11 I, = v'3/, , where I, is the line current and Ip is the phase current. Draw a phasor diagram showing the currents of Prcb. 14.10. I See Fig. 14-8. Fig. 14-8 14.12 The circuit of Fig. 14-9a is supplied by a 240-V three-phase source. , From Fig. 14-9b, the voltage Vca is applied to th;: 3 ~-n load. Determine le' So, for the given direction of la' we have THREE-PHASE CIRCUITS 0 307 -Vea -240~ la = 3 ~ = 3 &: = -80 /120° A Ib = Similarly, Now, since la 240 / -120° 4M == 60 /-180° A + Ib + le = 0, we have V e"" IQ. 4 - - - -......- - - 1 VILb \ :fir> \ \ re. c ____--__ Vbc ------------~ (a.) 14.13 (h) Fig. 14-9 If the circuit of Fig. 14-9a operates at 60 Hz, write the phase currents in the time domain. I Since w = 27Tf= 27T(60) = 377 rad/s, ia = -80\12 sin (377t + 120°) A 14.14 \. - VC4 -:: Va.c. ib = 60\12 sin (377t - 180°) A The circuit of Fig. 14-lOa has Zl = 5/10° n, three-phase 450-V source. Solve for la' I Z2 = 9 /30° n, and ic = -72.11\12 sin (377t -106.1°) A Z3 = 10 /80° n, and is supplied by a From Fig. 14-lOa and b, la = -z: + -z: Vab Vac 450 ~ = 5/10° + -450 ~ 0 / 36 9 /30° = 11 .29 - .52° A I~ b _ _ _ _~ C - _ _ _._~--l 3 (a.) 14.15 Fig. 14-10 The circuit of Fig. 14-11 is supplied by a 240-V three-phase four-wire system. currents. I Referring to Fig. 14-4 and choosing Vab as the reference, we have Vab =240~V Similarly, V ho = 240 Vao = -/-30° = 138 .56/-30° V V3 138.56/-150° V and Vco == 138.56 /90° V Determine the three line 308 0 CHAPTER 14 Hence, Similarly, 1:... 4- ~ b ~ 1'10 1e. ..1:1'1 c IV 14.16 k Neutral line ~ Fig. 14-11 What is the current IN in the circuit of Fig. 14-11? I By KCL, la+lb+lc+IN=O. Using the re,ults of Prob. 14.15, we have jI7.32) + (27.71 + jO) + IN = O. Thus IN = -3R.l:; L-8.71° A. 14.17 Determine the three ammeter readings for the circuit shown in Fig. 14-12. three-phase 240-Y 60-Hz source. I (40.00-j23.1O)+(-30+ The circuit is supplied by a The two mesh equations may be written as Vab = (3 LQ: + 4 /60°)1] or Vcb = (4 /600)1] + (5 /90° + 4 /60°)1 2 + (4 &Q:') [~ 240 LQ: = 6.08 /34.72°1] + 4 /60'1 2 - 240 / -120° = 4 /60°1 + 8.70 /76.71°1 2 I Solving for I] and 12 yields 12 = 23.29 /26.18° A Referring to Fig. 14-12, we have la = I] = 41.4 ~"!!.~~ A Ib = -(I] + 12) I, = 12 = 23.3 /26.2° A = -(41.4 ,~=2!).40 + 23.3 /26.2°) = -50 / -28,91° A Therefore, ammeter A] reads 41.4 A; ammeter.4 2 reads 50.0A; and ammeter A3 reads 23.3 A. Fig. 14-12 14.18 Solve Prob. 14.17 by converting the wye-connected impedances to an equivalent delta. I The equivalent delta-connected impedancesue shown in Fig. 14-13, for which we have, by wye-delta transformation, ZB = 9.29 /47.74° n Similarly, and Zc = 12.39 /107.74° n The three line currents to the equivalent delta are: la Vab Vac 240 L!:~ -(240 /120°) = -Z A + -Z B = --;---._+ 9.29 ~ =41.4 /-56.4° A 7.43 LJ1J.4:~ THREE-PHASE CIRCUITS Similarly, 0 309 - (240~) 240 ~ Ib = 7.43 /17.74° + 12.39 /107.74° = 50.0 ill1..1:: = -50 / -28.91° A I _ 240 LllQ: ( - 9.29 /47.74° and -(240 ~) _ ° + 12.39 /107.74° - 23.3 /26.2 A As expected, the three line currents obtained through the application of the wye-delta transformation are identical to those obtained by using loop analysis. I, ~ b - e. - Al I. A2 I, 14.19 A3 Fig. 14-13 An unbalanced three-phase load supplied by a three-phase four-wire system is shown in Fig. 14-14. The currents in phases A and Bare 10 and 8 A, respectively, and phase C is open. The load power factor angle for phase A is 30° and for phase B is 60°, lagging in both cases. Determine the current in the neutral. I With VA as the reference phasor: 18 = 8 / -180° = -8.0 + jO A lA = 10 /-30° = 8.66 - j5 A Hence, IN = lA + 18 = 0.66 - j5 Fig. 14-14 circuit 14.20 = 5.04 /- 82S A Draw a phasor diagram for the circuit of Prob. 14.19. I See Fig. 14-15. Fig. 14-15 14.21 A balanced three-phase load is wye-connected, and has an impedance Zp = (4 - j3) n in each phase. the line current if this load is connected across a 220-V three-phase source. I v = 220 = 127 V Pv3 Zp = 4 - j3 = 5 /- 36.87° 127 J[ = Jp = 5 /_ 36.870 = 25.4 / 36.87° A n Find 310 14.22 0 CHAPTER 14 Draw a phasor diagram for the currents and voltagt s of the circuit of Prob. 14.21. I See Fig. 14-16, which also shows the circuit diagram. Fig. 14-16 14.23 Calculate the phase currents for the delta-connected load shown in Fig. 14-17a. I We show the voltages in the phasor diagram of Fi:~. 14-17b. Vab 11. 2 Since the load is balanced, we may write = 10 /600 =, 450 L.Q:: ii~/600 12-) = 45 L:.!.§O° A Thus, = 45 / -600 A and 13 _1 = 45 /600 A, which are shown in Fig. 14-17c. I. 3-~ -tSO- V S~ - I, (a) E.. liI-l I. 1200 1200 E•• 1200 (b) 14.24 I, (c) Determine the line currents in the circuit of Prob. 14.23. I From Fig. 14-17, la Similarly, = 11_2 Ib = 13] 45 /600 = 77.9 / -900 A and le = 77.9 /300 A = 45 L-6(~ - 77.9 /150 0 A Fig. 14-17 THREE-PHASE CIRCUITS 14.25 0 311 A wye-connected load has a 5 /20°-0 impedance per phase and is connected across a 120-Y three-phase source. Calculate the line current and the phase current. • Van 120/V3/-30° = 69.28/-30 0 y, = where Vab=120LQ::. Thus, V bn =69.28/-1500y and V,n = 69.28 /90° V. I = Hence, a Ib = 13.86/-170° A Similarly, rents. 14.26 IN = le = 13.86 /70° A. The line currents are the same as the phase cur- (from Prob. 14.25) = (8.91- jlO.62) + (-13.65 - j2.41) + (4.74 + j13.02) = OA A three-phase delta-connected load having a (3 + j4 )-0 impedance per phase is connected across a 220-Y three-phase source. Calculate the magnitude of the line current. V line = Vphase lphase 14.29 69.28 tl2:. = 13.86/- 50° A 5 /20° -(la +lb +I,) = 13.86/-50° + 13.86/-170° + 13.86/70° • 14.28 5 /20° If the neutral of the load of Prob. 14.25 is connected to the neutral of the source, determine the neutral current. • 14.27 and ~= = 220 Y = 2~O = 44 A Zphase lline = = 3 + j4 = 5/53.2° 0 v3 x 44 = 76.21 A A 220-Y three-phase source supplies a three-phase wye-connected load having an impedance of (3 + j4) 0 per phase. Calculate the phase voltage across each phase of the load. • • Vphase = 220 v3 = 127 Y Zphase = 3 + j4 = 5/53.2° 0 lphase = 1~7 = 25.4 A = lline Calculate the line current lA in the delta-connected system shown in Fig. 14-18. 240 LQ:: 240 /120° IA=IAB-IcA= 3/20° - 4L!Q'. =127.16/-41.19°A Q. ~""o-v 3-p looV ?J-r b b S-..\cc. _ _ _ _ _ _ ~ -------------~c Fig. 14-18 Fig. 14-19 14.30 Determine the current la for the three-phase system shown in Fig. 14-19. 14.31 • • 14.32 _ _ 100 LQ:: 100 LQ:: 100 L..!lQ: _ ° la -I+I AB -I cA - 2LQ:: + 4LQ:: - 5/60° -67.27/-14.92 A In the circuit of Fig. 14-19, verify that Ic is unaffected by the 2LQ::-0 impedance. Ie=ICA-IBc= 100 L..!lQ: 100 L.=..Wr 5/60° - 2/-30° =68.05/81.55°A In a three-phase four-wire system the currents in two phases are 10 /- 36.87° A third phase is open-circuited. Determine the current through the neutral. • Also calculate le. IN = -(10 /-36.87° + 6/-53.13°) = - [(8 - j6) + (3.6 - j4.8)] = and 6/- 53.13° A -15.85/-42.9SO A and the 312 14.33 0 CHAPTER 14 If v = V", sin wt is the voltage across and i == I" sin wt is the current through a resistor, determine the instantaneous and average powers, and show that the instantaneous power pulsates at a frequency 2w. • p = (Vm sin wt)(Im sin wt) == V Jm sin 2 wt = W,Jm(1 - cos 2wt) I P = '2 V:Jn,= Vm Im v'2 v'2 = VI where V and 1 are rms values. 14.34 Extend the result of Prob. 14.33 to a three-phase balanced system. Verify that the instantaneous power has no pulsating component. Obtain an expression for the total average power. • p = v)a + v,,ih + vJc = (V", sin wt)(l" sin wt) + [Vm sin (wt - 120 )][1m sin (wt - 120°)] 0 + [Vm sin (wt + 120°)][1", sin (wt -~ 1200)] = VmIm(sin2 wt + ~ sin 2 wt + ~ cos 2 wt + ~ sin 2 wt + ~ cos 2 wt) = ~ VmIm(sin2 wt + cos 2 wt) = ~ VmIn, = 3 ~ ~ = 3V/ p where VI' and I" are rms values of phase voltage and current. Since the instantaneous power is constant, the average power is the same as the total instantaneous power; that is, PT = 3VI'Ip' 14.35 What is the total power in a three-phase circuit having a balanced load with a phase power factor angle Op? • 14.36 Power per phase = VpIp cos 01' Total power PT Express Eq. (1) of Prob. 14.35 in terms of line values. the same. = 3VI'Ip cos 01' (1) Verify that the results for wye and delta connections are • For wye connection, 11' = 11' VI' = V;tV3, so that Eq. (1) of Prob. 14.35 becomes PT = v1V,I/ cos 01" For delta connection, 11' == I/V3, VI' = 11' ancl Eq. (1) of Prob. 14.35 yields PT == v1V;I/ cos 01" 14.37 Depict graphically the instantaneous and average po\\<ers in a three-phase balanced system. • A graphical representation of the instantaneous ~ower in a three-phase system is given in Fig. 14-20. It is seen that the total instantaneous power is constant and is equal to three times the average power. This feature is of great value in the operation of three-phase motors where the constant instantaneous power implies an absence of torque pulsations and consequent vibrations. Total instantaneous power lor three phases Fig. 14-20 14.38 Find the power delivered to the load of Prob. 14.27 • Since Zp == 3 + j4 = 5 /53.2° n, Op = 53.2°. Thm, PT = v1(220)(76.21) cos 53.2° = 17.4 kW. THREE-PHASE CIRCUITS 14.39 From Prob. 14.28, we have ~=220V Hence, PT are connected in wye and are sup- The currents are: (480/v3)~ 6 /20° = 46.19 / - 500 A Determine the power drawn by each impedance of the circuit of Prob. 14.40 . • la=46.19/-500A Hence, Pb Pa = PZl = (277 .13)( 46.19) cos 200 = 12.03 kW. = PZ2 = (277.13)(34.64) cos 40° = 7.35 kW Similarly, and Pc = PZ3 = (277.13)(27.71) cos 0° = 7.68 kW Find the reactive power of the circuit of Prob. 14.40. • 14.43 and = V3(220)(25.4) cos 53.2° = 5.8 kW. la = 14.42 l[ = 25.4 A Three impedances, Zl = 6 /20° n, Z2 = 8 /40° n, and Z3 = 10 LQ: n, plied by a 480-V three-phase source. Solve for the line currents. • 14.41 313 Determine the power consumed by the load of Prob. 14.28. • 14.40 D = Q Z1 = (277.13)(46.19) sin 20° = 4.38 kvar Q b = Q Z2 = (277.13)(34.64) sin 400 = 6.17 kvar Qc = QZ3 = (277.13)(27.71) sin 0° = 0 Q = Q a + Q b + Qc = 10.55 kvar Qa What is the overall power factor of the circuit of Prob. 14.40? • From Prob. 14.41 we have P = Pa + Pb + Pc = 27.06 kW and from Prob. 14.42, Q = 10.55 kvar. Thus, S = P + jQ = 27.06 + jlO.55 kVA = 29.05 kVA Power factor cos 14.44 = P S= 27.06 . 29.05 = 0.93 laggmg For the load shown in Fig. 14-16, calculate the power consumed by the load. • 220 V = -=127V PV3 l[ = lp Hence, 14.45 (J Zp =4 - j3 = 5 / - 36.87° n = 5 / ~~:.87 = 25.4 /36.87° A P = V3 x 220 x 25.4 cos 36.87 = 7.74 kW. A three-phase balanced load has a lO-n resistance in each of its phases. The load is supplied by a 220-V three-phase source. Calculate the power absorbed by the load if it is connected in wye; calculate the same if it is connected in delta. • In the wye connection, 220 V = PV3 = 127V Hence, P = V3V)l cos (Jp In the delta connection, lp = 127 10 = 12.7 A = = V3 x 220 x 12.7 xl = ~ =220V l[ cos (Jp =1 (load being purely resistive) 4.84 kW. l[ = V3 x 22 = 38. 1 A 314 0 CHAPTER 14 Hence, P = v3VJ[ cos Op = v3 x 220 x 38.1 Xl=' 14.52 kW. Notice that the power consumed in the delta cc·nnection is three times that of the wye connection. 14.46 A three-phase 450- V 25-Hz source supplies power to a balanced three-phase resistive load. If the line current is 100 A and, determine the active and reactive powers drawn by the load. • The power factor of a resistor load is 1.0. P = v3( 450)( 100)( 1) = 77 .94 kW 14.47 A three-phase 600-V 25-Hz source supplies power ':0 a balanced three-phase motor load. If the line current is 40 A and the power factor of the motor is 0.80, clett:rmine the active and reactive powers drawn by the motor. • p = v3(600)( 40)(0.8) = 33.26 kW cos Op = 0.8, Since 14.48 14.49 sin Op = 0.6; hence, Q = \13(600)(40)(0.6) = 24.945 kvar. A 25-hp induction motor is operating at rated load from a three-phase 450-V 60-Hz system. The efficiency and power factor of the motor are 87 and 90 percent, respectively. Determine (a) the active power in kW and (b) the apparent power in kVA. 1 • A three-phase motor is a balanced load, and the phase angle for the given motor load is 0 = cos- 0.90 = 25.84°. (a) P = hp(746) = 25(746) = 21 436 78 W eff. 0.87 " (b) Pf=~ 5 0.9=21 44 P" 21.44 kW S=23.82kVA Determine the reactive power and the line current for the motor of Prob. 14.48. • Q=S(sinOp) Op=cos-1(0.9),=25.84° Q = 23.82 (sin 25.84°) = 10.38 kvar 14.50 Q = v3( 450)( 100) sin 0° = 0 S=23.82kVA P ,= v'3VJ[ cos Op or (from Prob. 14.47) I[ = Jj~~~;)·~.9 = 30.56 A The load on a three-phase wye-connected 220-V system consists of three 6-0 resistances connected in wye and in parallel with three 9-0 resistances connected in delta. Calculate the magnitude of the line current. • Converting the delta load to a wye, Rw = ~ x') = 3 O. This resistance combined with the wye-connected load of 60 per phase gives a per-phase resistance Rp as Rp = (6 x 3)/(6 + 3) = 2 O. The phase voltage is 220/\13 = 127 V. Hence, Ip = I[ = 1~7 = 63.5 A. 14.51 A wye-connected load has a (5 + jlO) fl impedancel plw.se. Also, a (2 + j6) fl impedance is connected between lines a and b. Determine the current la' • Vab = 240 L.Q:: V I Thus, 14.52 138. SI) / - 30° =- + ----.--= 49 .68 / -76 .9° A 2 + j6 5" jlO Solve for I, in the circuit of Prob. 14.51, and verify that it is independent of the (2 + j6)-fl impedance. • 14.53 240 L.Q:: a ven = ~~ /90° v3 /90° = 138.56 ~- If the neutral of the circuit of Prob. 14.51 is conne,tee • IN = -(la + Ib + IJ le = 10 Hence, the neutral of the source, determine the neutral current. la = 49.68 /-76.9° A le = 12.39 /26.21' A Ib 13~.~~~ = 12.39 /26.57° A (from Prob. 14.51) (from Prob. 14.52) = 138.56 L=.1lQ0 _ ~O L.Q:: = 4032 /11754° A 5 + jlO 2 + j6 . . IN = -(49.68 / -76.9° + 48.32 /117.54° + 12.39 /26.57°) = 0 A. THREE-PHASE CIRCUITS 14.54 Thus, Q = 3(100)(20) sin 53.13° = 4.8 kvar S = P + jQ = 3.6 + j4.8 kVA. A balanced delta-connected load draws 10 A of line current and 3 kW at 220 V. resistance and reactance of each phase of the load. • 220 Zp = 5.77 = 38.1 0 Thus, Rp = Zp cos (Jp = (38.1 )(0.7873) = Determine the values of P = v3VJ[ cos (Jp 3000 = v3(220)(1O) cos (Jp or 14.56 315 The power factor of each phase of a balanced wye-connected 5-0 impedance is 0.6 lagging. This impedance is connected to a source having a 100-V phase voltage. Determine the apparent power drawn by the circuit. 100 (Jp = cos- 1 0.6 = 53.13° I Vp = 100 V (given) Jp = 5 =20A P = 3(100)(20)(0.6) = 3.6 kW 14.55 D or cos (Jp = 0.7873 30.0 0 The connection shown in Fig. 14-21 is known as an open-delta connection. Calculate the power absorbed by the resistors (with the values as shown). The load is connected to a 220-V three-phase source. • The magnitudes of the currents in the resistors are 19.36 kW. IA--...,. ,t ---,-___ T '21o V Is -+ l.zov S.JI.. PJ Ie 14.57 Fig. 14-21 In wye-connected resistive load, shown in Fig. 14-22, the resistor of phase B burns out. determine the power absorbed by the resistors. • In this case, the circuit no longer acts as a three-phase circuit, and P = (22)\5 + 5) = 4.84 kW. l.t ----I~-_--. + For the values shown, JA = Jc = 220 1(5 + 5) = 22 A. Thus, A 2.20V --_--J/ SJ'l <: Fig. 14-22 14.58 The apparent power input to a balanced wye-connected load is 30 kVA, and the corresponding true power is 15 kW at 50 A. Calculate the phase and line voltages. • Hence, kW 15 cos (Jp = kVA = 30 = 0.5 P = 15,000 = v3~J[ cos (J = v3~(50)(0.5) Vp = 346.42/V3 = 200 V. J[ = 50 A or (given) ~ = 346.42 V = v3Vp 316 14.59 14.60 0 CHAPTER 14 Determine the kvar drawn by the load of Prob. 14.5i!. • s = P + jQ Q= or y(-sy'= (p)2 = V(30)2 - Calculate the input power to a three-phase load formed by a delta-connection of the load of Prob. 14.58, at the value of the line voltage determined in Prob. 14.58. • From Prob. 14.58, Z = I' 2lK) 50 = 4n Z 4 p P = ,nV1I1cos 01' . = V3(346.42)(150)(0.5) = 45 kW Find the ohmic values of the resistance and the reactance of the load in Prob. 14.60. From Prob. 14.60, • Rp = Zp cos 01' 14.62 1 = VI' = 346.42 = 86 6 A ? 11 = V3lp = V3(86.6) = 150 A 14.61 (lW = 25.98 kvar 4(0.5) = =: 2n XI' = Zp sin 01' = 4(0.866) 3.46 n = For the three-phase four-wire system shown in Fig. 14-23, we have the loads as shown. currents. • VI' 230 = -V3~ 10° = , 30,000 L=1Q: ° lA = V3(230)(0.866) = 86.96/-30 A 132 .8m: V I~ = 86.96/-15Q~ A I" A = IB In e = 20,000 /120° 132.8 = 150 .6/120° A = 86.96/-150° + 112.9 L~-120° = 164.04/-126.58° A le = 86.96/90° + 150.6/120° -= 229.46/109° A I~ 1,. I.' B 230-V 86.96/900 A = I~ + I~ = 86.96 L::.o1Q: + 75.3 m: = 156.75/-16.1° A lA A I~ = I"B = 15,000 1-120'= 112.95/-120° A 132.8 L..2. __ 10,000 10° = 75 3 10° A 132.8 ~ . ~ Determine the line x" ~-f is kw e~ 4-Wk It c S~ N 14.63 Fig. 14-23 Draw a phasor diagram showing the line currents determined in Prob. 14.62 and, hence, determine the current I N through the neutral. • 14.64 IN See Fig. 14-24, from which IN = 47 ~ A. Verify analytically the result of Prob. 14.63. • From Prob. 14.62, lA = 150.6 - j43.47 IB = -97.".'5 -- j131.73 IN = lA + IB + le = - 21.)\6 + j41.76 le =-74.70 + j216.96 =47.14 (I 17.63° A which agrees with the result obtained graphically. 14.65 A 460-V three-phase 60-Hz source supplies energy to the following three-phase balanced loads: a 200-hp induction-motor load operating at 94 percent efficienc:t and 0.88 pf lagging, a 50-kW resistance-heating load, and a combination of miscellaneous loads totaling 40 kW 11 a 0.70 lagging power factor. Determine (a) the total kW supplied; (b) the total kvar supplied; (c) the total apparent power. THREE-PHASE CIRCUITS / / / / / / \ \ \ \ \ \ \ \ \ \ \ Fig. 14-24 • Active power: (a) For motor, = hp(746) = 200(746) = 58 72 kW p M eff. 0.94(1000) 1 . For heating load, PH = 50 kW. For miscellaneous load, Pmise = 40 kW. Total kW = PT = 158.72 + 50 + 40 = 248.72 kW. (b) Reactive power: For motor, tan(}= Q pM tan 28.36° = 1~;2 M QM = 85.68 kvar For heating load, QH = 0 kvar For miscellaneous load, (}mise = cos - 1 0.70 = 45.57° t an () Qmisc P misc =-- tan 45.57° = Q4~se Q T = 85.68 + 0 + 40.81 = 126.49 kvar (c) Apparent power, S3f,bal 14.66 = P3f,bal + jQ3f,bal = 248.72 + j126.49 = 279 /27.0° kVA Calculate the total line current for the circuit of Prob. 14.65. • Thus, s = V3~/, I, = 279,000N3(460) = 350 A. or 279,000,= V3( 460)/, D 317 318 0 CHAPTER 14 ~~~~----------------- ~-4r---~~-----------__ c_~~r-____r-__~~______ 480 V E Thr~~ :I~.se "'\F«'A,';0'~6 I --5 E /_~ \~..........E••. ,.30" c' Zl - 10 L30° 14.67 ------I " '/fr. n Y Z2 - 20 L-50° Fig. 14-25 n A balanced wye-connected load and a balanced delta·connected load are supplied by a three-phase 480-Y 50-Hz generator, as shown in Fig. 14-25. The branch impedances of the wye and delta loads are 10 /30" nand 20 / - 50° n, respectively. Determine the active and r;!active powers drawn by each three-phase load. Determine the phasor voltage and phasor current for anyone branch of each three-phase load, and then substitute into the power equation for balanced three-phase lo}(ls. • The branches selected for this example are drawn with heavy lines in Fig. 14-25. The branch currents are determined by using the standardized phasor diagram and Ohm's law. Thus, for the selected branch of the wye bank, E aa , 480/V3~ I hey = ~ = 10 /30° I br , y = 21.71 / -60° A Vbr'y=E aa , 0 =277.13 /-30 y For the selected branch of the delta band, Ib',~ = 480~ Ea/> -z; = 20 / -50° Thus, Using the phasor diagram, measure the angle from the branch-current phasor to the respective branch-voltage phasor. The direction is shown with heavy arrows ire 1he phasor diagram. For the selected branch of the wye bank, P Y,bal = 3(277.13 )(27.71) cos (+ 30°) = 19,951.2 W QY,bal = 3(277.13)(27.71) sin (+30°) = 11,518.9 var For the selected branch of the delta bank, P~,bal = 3(480)(24) cos (-50°) = 22,214.7 W Q~,cal = 3( 480)(24) sin (-50°) = -26,474.5 var Thus 14.68 Determine the total apparent power and the overall p,)wer factor of the circuit of Prob. 14.67. PT = P y + P~ = 19.95 + 22.21 = 42.16 kW • Q T = Qy + Q~ = 11.52 + (-26.47) = -14.95 kvar ST = PT + jQr =42.16- j14.95 =44.74 /-19.5" kYA Power factor is cos (- 19.5°) = 0.94 leading. 14.69 In the circuit of Fig. 14-26, determine the line current lA for the impedance values shown. • From Fig. 14-26, Vab = 450 ~ Y V bc V aa , = Thus, V__ a' + I = __ A ZI Vab Z2 + Vac Z3 = 259.8 = 450 L - 120 y 0 Vca == 450 /120° Y ~ L::- 3(~ = 259.8 / -30° Y L=lQ 13 /20° 4:iO ~ -450 /120° - 189 6 / 7965" A +l:~/45° + 3 iN: . - . THREE-PHASE CIRCUITS IN ------~~~ 0 319 -----------~---------+-------- IA, ------j~~ _ _ _.." ... L ----I--___._--+---i----.--I. Thre;';h,;se- -.....-=--; 450 V 60Hz Z\-13L20° n Z2 - 15 L45° n Z3 - 3 L30° n Z4 - 10 L60° n Z~ - (3 + /4) n 14.70 , Fig. 14-26 What is the current, IN' in the neutral of the three-phase four-wire system shown in Fig. 14-26? IN = i;' Thus, 14.71 IN = but Vcc' 259.8/90° 3 + j4 = =~ /90° = 259.8/90° Y ° -51.96/36.87 A A four-wire three-phase 450-Y 60- Hz system supplies power to the following loads: three impedances (13 /20° each) connected in wye; an unbalanced delta load consisting of 15/4SO n between lines a and b, 3/30° n between lines a and c, and 10 /60° n between lines band c; a single-phase load of (3 + j4) n between line c and the neutral line. Draw the corresponding circuit and calculate the current in line a. , The circuit is shown in Fig. 14-27, from which 10 /60° n, and Zs = 3 + j4 = 5 /53.13° n. Zl = 13 /20° n, Z2 = 15/45° n, Z3 = 3/30° n, Vaa' = 450 ~ = 259.8/-30 y 0 = 259.8/-30° 450iQ:: -450/120° ° 13m +15/4SO+ 3M =189.6/-79.65A IN AI 450 V ~~~__~I~.~~______________________~~ ?J-f 4-Wh S~ le _e~~____+-__~________-+______+--r__~ Fig. 14-27 14.72 14.73 Determine the current IN in the circuit of Fig. 14-27, , IN = Vcc' Z; = 259.8L=2Q': 5/53,13° = / ° 51.96 -143.13 A Caiculate the line current Ic in the circuit of Fig. 14-27. , _ Vca VCb Vce' Vce' _ 450 ~ - 450 ~ 259,8 L2Q:: 259,8 L2Q:: = 220 7 /64 96° A le - Z3 + Z4 + Z, + Zs - 3/30° + 10 /60° + 13/20° + 5/53,13° . . Z4 = 320 14.74 0 CHAPTER 14 From the results of Probs. 14.71 through 14.73 determine lb' , Since la + Ib + Ic + IN == 0, Ib == -(la + le + IN) == -(189.6/-79.65" + 220.7 /64.96° + 51.96/-143.13°) = 87.7 /168.3° A 14.75 A wye-connected load and a delta-connected load are supplied by a four-wire 400-V three-phase 50-Hz system. The connections for the wye load are 2 /20° n to line a, 30 /50° n to line c, and 6/75° n to line b. The connections for the delta load are 50 /30° n between lines a and b, 25/-60° n between lines band c, and 17.3 /90° n between lines a and c. Sketch the circuit and calculate the current lb' 11 == 50 /30° n, , The circuit is shown in Fig. 14-28 from which 2 /20° n, Z5 == 6/75° n, and Z6 = 30 /50° n. Vbb' = Vbb' Ib = Zs + 400. V3 "2 L~~fL = Z2 == 25/-60° n, 1_ 1 30.94~ ,(l0 Z3 == 17.3/90° n, V V 230.94 ~l50: -400 ~ 400 L..=.m ° z:; +-----z;= 6/75" + 50 /30° + 25/-60° = 31.38/146.4 A Vbc be A --__---,~------------- b le c~~-;----+---~--·----------~--~ Fig. 14-28 14.76 , In the circuit of Fig. 14-29 find the current 11 , 450 iQ~ ° 11 == 38Z~- = 11.84 ~ A Io.. 450-V ~----.---~-----~---------~-- 3-cj> b------~--~----------~~---+~-- ~ C --~--~----+_----~r_----~-----+----- 12; Fig. 14-29 14.77 14.78 , , Determine the current 12 in the circuit of Fig. 14-29. 450 Vaa' == -V3 / - 30° == 259 .8 /- 30° V Vaa' 12 == 40 ~ = 259.8 L=lQ': 40 ~ == 6.5 /- 30° A What is the current I} in the circuit of Fig. 14-29? Vab I 3 == 20/30° 450 LO" Vac + 20/30° = 20~~Q:: + -450 ~ 20 /30° = 38.97 /-60° A Z4 == THREE-PHASE CIRCUITS 14.79 la = I, + 12 + 13 = 11.84 ~ + 6.5 /-300 + 38.97 /-60 0 = 52.3 /-45 0 A Determine the current la in the circuit shown in Fig. 14-30. I 1= a V aa , Z, + V ,=208 /-30 0 = 120/-300 y Vab+ Vac Z2 The delta consisting of Z2'S is closed. Z2 V3 aa 0 208 flL' Y and Vac = - 208 /120 Y 120 / - 30 0 208 flL' - 208 /120 0 _ 0 la = 6M + 4/250 + 4 -108.52 /-59,5 A Vab = m Thus, N~ 20ft V __ -=====~ ____ ~ __________________ 9 ~--------------~----;-------------------r- ~ c~~---+------r-----------.---~----~- 3- b------~------r-~.-----------_r--_r-- ~ L2St. .n.. / 14.81 ~ Fig. 14-30 With the delta closed determine IN in the circuit of Fig, 14-30, I Notice that delta-connected Z2 's have no effect on IN' Vb'b IN = Z3 = 14.82 321 From the results of the preceding three problems obtain la' I 14.80 D Thus -120~ 1_1t;(l0 2/100 = -60~A If the Z2 connected between the lines band c in the circuit of Fig. 14-30 is opened, find Ic' I 14.83 Repeat Prob. 14.82 for I b , I 14.84 Ib = z: + ---z; + Z; V bb , V bb , 120~ V ba = 2 /10 0 + 120~ 6 /500 -208~ + 4 /250 = 122.58 /176.120 A In the circuit of Fig. 14-31, Z1 = 10 /20 0 n, Z2 = 10 /50 0 n, Z3 = 10 /800 n, Z4 = 10 /-10 0 n, 10 / -40 0n. The circuit is supplied by a 450-Y three-phase four-wire source, Determine IN' I IN =- V cc' ZI _ (450/ J3) L2!r. 10 ~ -25.98 /70 0 A et b C .lA '" 7N Fig. 14-31 and Z5 = 322 14.85 D CHAPTER 14 Determine the currents I, and Is shown in the circuit of Fig. 14-31. , 14.86 14.87 11 = , Vah 450 &: --z; = 10 /80° = / _Q(l0 45 ~ A == V" = (450/Y3) L2.Q.:: = 25 98 /130° A Z, 10 / -40° . L Solve for the current Ib in the circuit of Fig. 14-31. V ha Vb, Vw -450 &: 450 / -120° 259.8 / -150° Ih = Z1 + Z2 + Z, = 10 /80° + --to /50° + 10 ~ = 62.2 /168.8° r\. thre\~-phase The circuit of Fig. 14-32 is energized by a 450-V , 60-Hz source. Determine la' At 60Hz = wL I = (377)(7.96 Xlt X L2 = wL 2 = (377)(15.92 I Thus, a X X I 10 ') = 3 n 10-') = 6 n Vab Vaa , V ah = 24 &: + 5 / -90° + 6/90 I = wC t = (377)(530.5 X 10 b) = 5 n 1 1 X =-= =8n C2 wC2 (377)(331.6 X 10 6) XCI Vac 0 +- 8i~90° = 450 &: 259.8 ~ + 451~L~ + -450 ~ = 9346 / -I ISO 24 &: + 5 /-90° 6 ,~~I~ 8 /-90° . . A 10.. Vl ~ b 530.5 ~'I"fl C, 7.96 mH Lt Iz C N ~ ~F 2.00 ~ L ./,j''.9-? " 't,. "'1- C ~~- Fig. 14-32 331. 6/lt' 14.88 14.89 14.90 , Solve for IN in the circuit of Fig. 14-32. , V" Vb'b V"" -259.8flil" -·259.8~ -259.8/-30° ° I N =2&:+3/900+5/-900= 2&:·t 3/90° + 5~ =250.5/-86A How much active power is supplied by the source '" 2 the circuit of Fig. 14-32? 10 2 P'UPPtiCd=~P"b""bed=(Jt)24+(J2)2+(J,)4= 2 (450)2 24 (259.8)2 (450)2 24+ -2- 2+ 4 4=92.81kW In the three-phase system shown in Fig. 14-33a phase b gets open-circuited owing to a fault. remaining wye-connected load to an equivalent delta. , Similarly, _ j3(5.82~) + (5.82 ~0)41~4(j3) _ 40.43/106.22° _ ZA Zn= 5.82/75.96° 40.43/106.22°, 4&: =1O.1l/106.22,H where ZA' ZR' and Zc are shown in Fig. 14-33b. Zc= /'1(1 ,,)1.0 5.82/75.96° -6.95~n 40.43 /106.22° j3 ° =13.48/16.22 n Convert the THREE-PHASE CIRCUITS IQ.. S. g2. US. 9' ~ 0 323 I ... ~ 1./1.. b c '1ft 1,; ...J\/'..r ~ .l3A IN N Fig. 14-33 lA.) 14.91 If the circuit of Fig. 14-33a is connected across a 450-V three-phase source, calculate the current IN' I We use the circuit of Fig. 14-33b to solve for the currents in the various impedances, the voltages across which are vac = -vca = -450 /120° V Thus, la = VaN = 450 V "v'3 /-30° = 2598/-30° . 259.8 L=N: -450 ~ 10.11/106° + 13.48 /16.2° 450~ le Hence, 14.92 259.8~ = -10.54 - and V eN = 259.8/90° V . j50.27 A . = 13.48/16.2° + 6.45/30.26° = 10.89 + j64.71 A IN = -(Ia + IJ = 10.54 + j50.27 - 10.89 - j64.71 = -0.35 - j14.44 = 14.44/-91.4° A. In the circuit of Fig. 14-34 phase b gets open-circuited while the circuit is connected to a 450-V three-phase source. Determine IN' I Proceeding as in Prob. 14.91, la = -450 ~ 259.8 L=N: . 8/-90° + 5/-90° = 74.69 + j73.13 A 450 ~ 259.8 ~ . 2~ = -48.71 + jlO1.79 A le = 8/-90° + IN = -(la + le) = -74.69 - j73.13 + 48.71- j101.79 = -25.98 - j174.92 = 176.84/- 98.45° A c:t - .....- - - - t " - - - - l ,n. ~ z~;----j-8n--~+_--.NV2~~-- C __ I~ N __~--------__------__~ 14.93 Fig. 14-34 Calculate the power dissipated in the 6-0 resistor of the circuit of Fig. 14-35. I Power dissipated = (/N)26, where V bN = 480 ~ = 277 /-150° V Thus, and Power dissipated = (46.17)26 = 12.79 kW 324 D CHAPTER 14 3.ft ;1.J1.. j3.s./L ~ 4to-v 3-~ Ib b Ir s~ ~J2. 1, c. 2.il j3.S.n. ja.s.n I .. -j4Jl - - - - ' V V '.....- - - e,ft 14.94 Fig. 14-35 (2 t j3.5) n Determine the apparent power taken by the 14-35. , Zh = 2 + j3.5 = 4.03 /60.26° n impedance in phase b of the circuit shown in Fig. V 277 /-150° -z;: = 4.03 ~ = 68.73 / -210.26° A bN I~ S = (I)2(Zh) = (68.73)2(2 + j3.5) = 9.45 + j16.53 kVA 14.95 Verify that the result of Prob. 14.94 can also be obtained from , From Prob. 14.94, V= Vb" = 277 / -15Q~ V Thus, 14.96 14.97 , , S =VI*. 1* = 68.73 /210.26° A S =VI* = (277 /-150°)(68.73 /210.26°) = 9.4·l t j16.53 kVA, which is the same as in Prob. 14.94. Determine the current It in the circuit of Fig. 14·35. V V/m 277 / - 150° 277 / -150° ° ° ° I t = ~n+2+j3.5= 6 + [email protected],=46.17/-150 +68.73/-210.26 =100/-173.37 A How much power is supplied to phase b of the circuit of Fig. 14-35? _ _ ° --"80 LQ" 480 / -120° Ih - 11 + 12 + 13 -100 /-173.37 + --3--:--4 + '4 ... J 4- J = 100 / -173.37° - 96 /53.13° + 848 L -75° = -135 - j147.2 = 199.74 t 133° A Op = -l,iO .- (.- 133) = - 17° Thus, 14.98 Ph = (277)(199.74) cos (-17°) = 52.91 kW. Three impedances, Zab = 8 /20° n, Zhc = 15l!!SO 0, and Zca = 10 LQ" n, across a 300-V three-phase source. Find the line current la' , la = Vab Z ab 14.99 + Vac Zae 300 = 8/20° + ---300/120° 10 LQ" = are connected in delta and ° 63.48 / -37.68 A Repeat Prob. 14.98 if the phase sequence of the source is reversed. , With reversed phase sequence we have _ Vac Vab _ - 300 L121.!: 300 L!r _ ° la - Z + Z 16 /65 0 - + 8/20° - 37.66 / -50.87 A ae ab ~_ THREE-PHASE CIRCUITS 14.100 Determine the three ammeter readings in the circuit of Fig. 14-36, where Z3 = 15 L!r n, and the three-phase applied voltage is 300 V. , Vab Vac - 300 ~ 15 L!r 300 L!r = z; + Z; = 10 /25 + la 0 Zt = 10 /25 0 n, D 325 Z2 = 20 /60 0 n, 0 = 47.78 / -38.89 A So Al reads 47.8 A. _ V ba V bc _ -300 L!r 300 ~ _ Ib - Z; + Z; - 10 /25 + 20 /60 44.05 /163.27 A 0 0 0 - Ammeter A2 reads 44.1 A. Vca Ic VCb 300~ = z; + Z; = 15 L!r + -300~ 20 /60 0 0 = 18.03 /73.90 A Thus, A3 reads 18.0 A. - I. &l " I. b - I, e=. 14.101 Fig. 14-36 Repeat Prob. 14.100 for a reversed phase sequence and compare the results. , In this case we obtain the circuit of Fig. 14-37, from which Vab la Vac 300 L!r = Z; + z; = 10 /25 + 0 - 300 ~ 20 /60 0 0 = 32.35 /-52.51 A Al reads 32.4 A. Ib = 300/-120 z: + Z; = -300L!r 10 m + 15 L!r V ba V bc 0 0 = 37.48 /-172.89 A A2 reads 37.5 A. Vca _ - 300 ~ 300 ~ _ 3 0 11:.(\000 I - Vcb c Z3 + Z2 15 L!r + 20 /60 5. ~ A 0 - A3 reads 35.0 A. The following tabulation emphasizes the effect of phase sequence on the line currents to an unbalanced three-phase load. Amperes Line a b c Sequence abc Sequence cba 47.8 44.1 18.0 32.4 37.5 35.0 Fig. 14-37 326 D 14.102 CHAPTER 14 Power is measured in a three-phase circuit by two wattmeters, as shown in Fig. 14-38a. showing the volt ages across and the currents through the two wattmeters. , Draw a phasor diagram The phasor diagram is shown in Fig. 14-38b. Fig. 14-38 14.103 Determine the angles 01' O2 , and 03 shown in Fig. 14-38b. Also calculate the currents la and Ic if 0, and the circuit is connected to a 450-V three-phase ZI = 10 /20° n, Z2 = 30 /10° n, and Z, = 5 source. DO° , _ Vab Vac _ 450 L!r ---450 Lmr ° la -I10 /20° -I- 30 /10° = 55.84/-31.88 A z: le = Thus, 14.104 01 = 31.88°, Vca Vcb 450 mOoz; -I- Z3 = 30 L!.Q~- + - 03 = 39.06°, , z; - and O2 = 120" 450 ~ 5 /300 - 0 1 -- 60° = 120° - = ° 93.78/39.06 A 39.06° - 60° =20.94°. From the results of Probs. 14.102 and 14.103 determine the wattmeter readings in the circuit of Fig. 14-38a. Wl = Vab1a cos 01 = (450)(55.84) cos 31.88° = 21.3 kW W2 = VcJc cos O2 == (45 1))(93.78) cos 20.94 = 39.4 kW 14.105 Show that the sum of the two wattmeter readings gives the total instantaneous power input to the circuit of Fig. 14-39. , Current through Wl = ia; voltage across Wl = vac = va - Vc; instantaneous power read by Wl = (V a - VJ i a' Current through W2 = i b; voltage across W2 = v bc = Vb - Vc; instantaneous power read by W2 = (Vb - vc)ib' Total instantaneous power read by Wl and W2, (1) but ia -I- ia -I- ic phase power. = 0 so that Eg. (1) becomes a. p = L)a -I- ill.. Vbib -I- vcic' which is the total instantaneous three- £l, -----, Vvi b h • z,c, b W2 Fig. 14-39 C 14.106 If the impedances in the circuit of Fig. 14-39 are id~l\tical, that is, Za = Zb = Zc angle is 0 (lagging), draw a phasor diagram showing all voltages and current. , See Fig. 14-40, where Va = Van' Vb = V bn , and Vc = V cn ' = Z, and the power factor THREE-PHASE CIRCUITS 14.107 0 327 With a balanced load, using the phasor diagram of Fig. 14-40, show that the sum of the two wattmeter readings equals the power consumed by the circuit of Fig. 14-39. , Current through Wl = la; voltage across Wl = Vac =Va - Vc' From Fig. 14-40, Vac is behind la by 30° - 0, so that reading of Wl = VaJa cos (30° - 0). Similarly, from Figs. 14-39 and 14-40, it follows that the reading of W2 = VbJb cos (30° + 0). Now, in terms of line quantities, we have la = Ib = I, and Vac = Vbc = "". Thus, WI + W2 = ""I, cos (0 - 30°) + ""I, cos (0 + 30°) = ""1,(2 cos 30° cos 0) = V3V/I, cos 0 = total three-phase power Fig. 14-40 14.108 , A 220-V three-phase motor takes 21,437 W at 0.71 lagging power factor. or Hence, 14.109 What is the line current? 21,437 = V3(220)I,(0.71) I, = 79.2 A. Let the motor of Prob. 14.108 be wye-connected and balanced. If wattmeters are connected in lines a and c to measure the input power to this motor, determine the individual wattmeter readings. , cos 0 = 0.71 W1 or 0= cos- 1 0.71 = 44.77° = VabIa cos (44.7r + 30°) = (220)(79.24) cos 74.77° = 4580 W W2 = VcbIc cos (90° - 44.7r - 60°) = (220)(79.24) cos 14.77 = 16,857 W 14.110 14.111 , Use the approach of Prob. 14.107 to find the wattmeter readings of Prob. 14.109. = V/I cos (0 + 30°) = (220)(79.24) cos (44.77 + 30°) = 4580W W2 = (220)(79.24) cos (44.77 - 30°) = 16,857 W The line current to a lightly loaded three-phase motor, operating from a 450-V three-phase system, is 24 A. The power factor of the motor is 0.47 lagging. If the two-wattmeter method is used to measure the three-phase power supplied to the motor, what would each wattmeter read? , 14.112 Wl 0= COS-I = ""I, cos W + 30°) = (450)(24) cos (61.966° + 30°) = -371 W2 = ""I, cos W - 30°) = (450)(24) cos (61.966° - 30°) = 9162 0.47 = 61.966° Wl Obtain an expression for the power factor angle 0 in terms of the wattmeter readings of Prob. 14.107. , From Prob. 14-107, Wl + W2 = V3""I, cos 0 (1) 328 D CHAPTER 14 WI - W2 = V,I,lcos (0 But -I- 3)') - cos (0 - 30°)] = - V/I sin 0 (2) Thus, Eqs. (1) and (2) yield tan 0 14.113 W2 WI "i; W2 - Wl P=W1-1-W2=60-l-40=100kW and the power factor is 14.115 - W2- WI 3 c-ccc=---I---c-C:- Two wattmeters, connected to measure three-phase pl>wer, read 60 and 40 kW (for WI and W2 respectively). Determine the total power and the power factor. , 14.114 = \I tanO=v- W2 -1-WI • M 40 - 60 =v3 -l=-0.3464 40 60 or 0=-19.1° cos (-19.1) = 0.945 lagging. , S cos 0 = WI , Total Calculate the total apparent power from the data of Prob. 14.113. -I- W2 60 -I- 40 or S = 0.945 = 105.8 kVA What is the per-phase reactive power in the circuit of Prob. 14.113'1 Q = YS2 - p 2 =\~105.8)2 - (100)2 = 34.6 kvar or per-phase reactive power = ~ Q = ~ (34.6) = 11.53 har. 14.116 A balanced wye-connected load draws a line current tlf 60 A from a 450-V 60-Hz three-phase system. The power factor of the load is 0.70 lagging. Determine the active power, apparent power, and reactive power drawn by the load. , P = V3V/, cos 0 = v'3( tSO)(60)(0.7) = 32.736 kW S = V3V/, = v:~( 4SIl)(60) = 46.765 kVA 2 ,r;-;-;- 2 Q= vS~-P == V(46.76:Jr-(32.736) =33.396kvar ,/ 14.117 0 A three-phase motor draws a line current of 30 A when supplied from a 450-V three-phase 25-Hz source. The motor efficiency and power factor are 90 and 75 percent, respectively. Determine the active power drawn by the motor. , 14.118 0 P = V3V/, cos 0 = V:~(4Sil)(30)(0. 75) = 17.537 kW , Calculate the motor output and the total input reactive power for the motor of Prob. 14.117. Pout = (efficiency)(PiJ = (0.9)(17.537) = 15.783 kW Q = P tan 0 = (17.537) tan (cos -1 0.'75) 14.119 = or 21.16hp (17 .537)(tan 41.41°) = 15.47 kvar A four-wire 208-V three-phase 60-Hz system is used to supply power to a three-phase 5-hp induction motor and a single-phase 6-kW heater connected between line c and the neutral line. The operating efficiency and power factor of the motor are 81 and 71 percent, respectively. Draw the corresponding circuit diagram and calculate the current in line c. , The circuit is shown in Fig. 14-4l. P= 5 x 746 ---0.81 = 4604.94 W Powerfactor=0.7I=cosO Vc,) Rhca,ec where Rhca,e, or For motor load, = V3~I, cos o= (J ,= V'3(208)I,(0.71) 47.77° lagging. = 6000/(208/V3)2 = 0.416 Thus, n. or I cc.=18/-47.7rA. Hence (2081'/3) /90 0 le = 18/-47.77° + --(~:1I-6-- = 275.62 /87.6° A Thus, THREE-PHASE CIRCUITS D 329 Io. .... ~------------------~--~ , I bb-; b _________ c le. Ice' ------.~-J~-~~r~ ------{I~6~I<.~wOI Fig. 14-41 H~ 14.120 Draw a phasor diagram showing all the currents and voltages of the circuit of Prob. 14.119. , To draw the phasor diagram we must know la = laa' and Ib = I bb " These are obtained from the results of Prob, 14,119, Hence la=l aa .=18/-77,77°A and I b =l w =18/-197,77°A, and the phasor diagram becomes as shown in Fig, 14-42, / I / H~ , I I Ic.'C' Fig. 14-42 14.121 A 450-V three-phase three-wire 60-Hz feeder supplies power to a 25-hp induction motor and a 30-hp induction motor. The 25-hp motor is operating at full load, is 87 percent efficient, and has a power factor of 90 percent. The 30-hp motor is operating at one-half of its rated horsepower; at this load, it is 88 percent efficient and has a power factor of 74 percent. Sketch a one-line diagram for the system and determine the feeder active power. 4Sov ?- 4> ~ 1111 ~k 2.S-H P 0.87 ~. 0.'10 t+ 1111 ~oHP f:lkJ. o. u .c.fb. 0.7+1'1 Fig. 14-43 330 D CHAPTER 14 , The one-line diagram is shown in Fig. 14-43. PM1 = 14.122 For the two motors we have: H30)(746) PM2 = - -0.88 - - - - = 12 , 715 . 91 W 25 x 746 0.87 =21,436.78W P= PM1 + PM2 = 34,152.69W Calculate the total reactive power supplied by the keder of Prob. 14.121. , The power factor angles are: (Jl = QMl cos- 1 (0.9) = (J2 25.84' = cos- 1 (0.74) = 42.27° = PI tan (Jl = (21.436.78) tan 25.84° = 10,381.41 var QM2 = P z tan (J2 = (12,715.91) tan 42.2r = 11,558.42 var Q = Q M1 + Q M2 = 10,381.41 + 11 ,558.42 = 21,939.83 var 14.123 What is the overall power factor of the system of motors of Prob. 14.122? , From Prob. 14.121, P = 34,152.69 W. Q = 21,939.83 var Q tan (J = f> :~1 ,939.83 :M,152.69 = 0.642 = (J == 31.72° Calculate the line current through any of the feeders of Prob. 14.121. , Apparent power, S = P + jQ = 34,152.69 + j21,939.83 = 40,5<)2.64 /32.72° VA Hence, 14.125 or cos (J = cos 32.72° = 0.84 lasging. and the power factor is 14.124 From Prob. 14.21, I, 5 = V3""I, = 40,592.64 or = 40,592.641\1'3(450) = 52.08 A. A 440-V three-phase source supplies a 1O-kVA n.8-pf-lagging load, which is connected in wye, and a delta-connected 1O-kVA unity power factor load. Calculate the total apparent power input to the two loads. , P == P y + Pt>. = (10)(0.8) + (10)(1) == 18 kW Q = Qy + Qt>. = (10)(0.6) + 0 = 6 kvar S = P + jQ = 18 + j6 = 18.97 /18.43° kVA 14.126 What is the overall power factor of the two leads of Prob. 14.125? true power calculations. , cos (J = cos 18.43° = 0.95 Verify that the same result is obtained from , __ 5 __ 18.97 x 10 3 _ ,,- V3"" - V3( 440) - 24.89 A 18,000 cos 14.127 14.128 , f,1 = V3( 440)(24.89) = 0.949 = 0.95 Calculate the line currents to each of the loads of Prob. 14.125. or 10,000 ly= V3(440) =13.121\ or 10,000 It>. = V3(440) = 13.12 A A three-phase three-wire 500-V 60-Hz source supplies Cl three-phase induction motor, a wye-connected capacitor bank that draws 2 kvar per phase, and a balanced three-phase heater that draws a total of 10 kW. The induction motor is operating at its rated 75-hp load and has an efficiency and power factor of 90.5 and 89.5 percent, respectively. Sketch a one-line diagram of the system and determine (a) the system kW; (b) the system kvars; (c) the system kVA. , (a) A one-line diagram is shown in Fig. 14-44. PM = 75 x 746 0.905 = 61.82 kW From the data given: PH ,= 10 kW (JM = cos -1 0.895 = 26.49° THREE-PHASE CIRCUITS QM = PM tan ()M = (61.82) tan 26.49° = 30.81 kvar (b) Qc = -(2 + 2 + 2) = -6 kvar Thus, (c) (leading) S = P + jQ Finally, (lagging) QH = 0 = 71.82 + j24.81 = e ® 75.98 /19.06° kVA loJc.k.J 3-~ -tfi. B9,Sff Cfo.57. (a) the line current and 14.130 ~ Fig. 14-44 (b) the overall power factor of the system of load of Prob. 14.1287 or (b) 1 H M 7~HP What is 331 Q = Q M + Qc + Q H = 30.81 - 6.0 = 24.81 var. Soov ?>-~ 14.129 D 75,980 I[ = V3(500) = 87.73 A Power factor = cos () = cos 19.06° = 0.95 lagging. Determine the line currents drawn by each load of the system of Prob. 14.128. - 61,820 -7 7 9. 6A IM - V3(500)(0.895) - 6000 or 14.131 le = V3(500) = 6.93 A PH = V3~/H(I) 10,000 or IH= V3(500) =11.55A A balanced delta-connected load whose impedance is 45/70° n per branch, a three-phase motor that draws a total of 10 kVA at 0.65 pf lagging, and a wye-connected load whose impedance is 10 n (resistance) per branch are supplied from a three-phase three-wire 208-V 60-Hz source. Sketch the circuit and determine the line current to each three-phase load. , The circuit is drawn in Fig. 14-45. For the Y load, lay = 120~ 10 L!r = 12 /- 30° A For the A load, For the motor, 10,000 = V3(208)laM or laM =27.76A 2.og V b 3-~---~----------~~--~ c IOLE-".n/~ Fig. 14-45 332 D 14.132 CHAPTER 14 Determine the active and reactive powers taken by each load of the circuit of Prob. 14.131. , For the Y load, P y = V3V;I{ cos Opy = V3(208)(12) cos 0° = 4320 W For the A load, Q I>. = V3(208)(8) sin 70° = 2709 var Pt> = V3V// cos 0pl>. = V3(208)(8) cos 700 =' 986 W For the motor, PM = S M cos OM = (10,000)(0.65) = 6500 W Now, cos OM = 0.65. Thus, OM = 14.133 0.65 COS-I 49.46° = QM = PM tan OM = 6500 tan 49.46° = 7600 var What is the overall power factor of the system of Pfl)b. 14.131? S = P + jQ = (4320 + 986 + 6500) + j(O + 2709 + 7600) = 11,806 + jlO,309 tan 14.134 (J = 0.873 Power factor =cos (J = 0.753 lagging For the circuit shown in Fig. 14-46, determine the currents la and le' and Z 3 = 14/-45° n , and a 208-V three-phase applied voltage. , Vab la Vea Given: ZI = 20 L!r n, Z2 = 14/45° n, 208 L!;~ - 208 ~ 20 10" ~- 14/45° = 15.78/-65.46° A ~_ 208 D20~ -208/-120° 14 L45'O- + 14 ~ Veb z; + Z; = .... ,zoS Vac = Z I + Z2 = le = = 28.70 /90° A I,. ~C;V V 3-t/> b c 14.135 = 41.13 0 (J 11. l .. ~ ""~ Fig. 14-46 Find the wattmeter readings in the circuit of Fig. lL.-46. x: Vab = (2 )H)(15.78) cos 65.46° = Vab1a cos ~ Wl 1 = 1363.21 W la :>('Veb W2 = Veb1e cos ~ =, (208)(28.7) cos (-30°) = 5169.83 W le 14.136 Calculate Ib in the circuit of Fig. 14-46. Explain the significance of your result. , Ib lab = z: Vab = z: + Z; V ba V be 208 L!r DetermilH rhe phasor sum of the three currents in the three phases. = -208 L!!' 208/-120° 20L!r- -I- 14~ 1000 = 20 L!;!' = 10.4 ~ A I 10 = lab + Ibc + lea ca [be = Vbe Z; = = 15.78/-114.54 ° A 208 ~ 1_ 7,0 14/-45° = 14.86 c:1..2. A = Vea = ~~~i.!.20° = 14 86/75° A Zz 14 L45° . = 10.4 L!r + 14.86 L-7r + 14.86/75° = The current lu is a circulating current in the closed delta. 18.09 + jO A = 18.09 L!r A cjo 0 THREE-PHASE CIRCUITS 14.137 333 Verify that the sum of the line currents in a delta-connected load is always zero. , Referring to Fig. 14-46, we have or 14.138 0 la + Ib + le = 0 (verified). Use the result of Prob. 14.137 to find Ib and verify that the result is consistent with that obtained in Prob. 14.136. , From Prob. 14.137, Ib = -la -le = -15.78/-65.46°-28.7/90° (from Prob. 14.134) = -6.55 + j14.35 - j28.7 = -6.55 - j14.35 = 15.77 /-115.6° A which is close to the result of Prob. 14.136. 14.139 Draw a phasor diagram showing the voltages and currents which affect the readings of the wattmeters Wl and W2 in the circuit of Fig. 14-46. I See Fig. 14-47. I / J r : - . - - - - - -.. ~ + "'5.4,,-0 Fig. 14-47 14.140 The two-wattmeter method is used to measure the power drawn by a balanced delta-connected load. Each branch of the load draws 2.5 kW at a power factor of 0.80 lagging. The supply voltage is 120 V, three phase, 25 Hz. Sketch the circuit showing wattmeters in lines a and c, and determine the respective wattmeter readings. I The circuit is similar to that shown in Fig. 14-46, except that in the present problem the load is balanced. For the load we have a rating of 2.5 kW/phase with 0.8 pf lagging. Thus, 0 = cos -10.8 = 36.87°. Threephase power = 3(2.5) = 7.5 kW = V3VJ, cos o. Thus, If = 7500/"\13(120)(0.8) = 45.11 A and W1 = ~I, cos (0 + 30°) = (120)(45.11) cos (36.87° + 30°) = 2126 W W2 = VJ, cos (0 - 30°) = (120)(45.11) cos (36.87 - 30°) = 5734 W 14.141 Three parallel-connected three-phase loads are supplied by a three-phase 240-V 60-Hz source. The loads are three-phase, 10 kVA, 0.80 power factor lagging; three-phase, 10 kVA, unity power factor; three-phase, 10 kVA, 0.80 power factor leading. Sketch the circuit showing wattmeters in lines a and c, and determine (a) the system active power; (b) the system reactive power; (c) the system apparent power. I The system is represented by the circuit of Fig. 14-48 where the loads are: Load A-lO kVA, 0.8 pf lagging; load B-lO kVA, unity pf; load C-lO kVA, 0.8 pf leading. (a) P = PA + P B + Pc = 10(0.8) + 10 + 10(0.8) = 26 kW (b) Q= Q A + QB+ Qc=8tan36.87+0-8tan36.87=Okvar (c) S = P + jQ = 26 + jO = 26 kVA 334 D CHAPTER 14 ,,-----I' ~---~-----T4-- I I c ,----] l 8 J I J 14.142 Determine the overall power factor of the loads of Prob. 14-141 and calculate the line current. P 26 Power factor = - = S 26 I 14.143 Since cos fJ = 1 and W2 = VJ/ cos (fJ 26,000 I( = V3V( = V3(240) = 62.55 A + 3{f) = 240(62.55) cos 30° = 13,000 W - 30°) = 240(62.55) cos (-30°) = A 100-kVA balanced three-phase load operates at 0.65 pf lagging at 450 V. powers of the load. I S cos fJ = P fJ = COS-I 0.65 = 49.46° 13,000 W Determine the active and reactive P = 100(0.65) = 65 kW or Q = S(sin fJ) = 100(sin 49.46°) = 75.99 kvar If the load of Prob. 14.144 is connected in delta, what is the phase impedance? S I 14.146 S 1 fJ = 0°, W1 = VJ( cos (fJ 14.145 ~= What are the readings of the wattmeters of Prob. 14.141? I 14.144 Fig. 14-48 100,000 I( = V3V( = V3( 450) = 128.3 A I p = 128.3 V3 = 74 .08/-49 .46° A Determine the power dissipated in the resistive components of the load of Prob. 14.144. total power and verify the result of Prob. 14.144. I Rp = Zp cos fJ = 6.07 cos 49.46° = 3.95 n Pp Hence determine the = (Ip)2Rp = (74.08)2(3.95) = 21.676 kW P = 3Pp = 3(21.676) = 65 kW 14.147 Repeat Prob. 14.146 for reactive power. I Xp = Zp sin fJ = 6.07 sin 49.46° = 4.61 n Q p = (Ip)2Xp = (74.08)2(4.61) = 25.31 kvar Q = 3Q p == 3(25.31) = 75.93 kvar 14.148 If power is measured by two wattmeters in the circuit of Prob. 14.144, determine the wattmeter readings. I W1 = VJ/ cos (fJ + 30°) = 450( 128.3) cos (49.46° + 30°) = 10,561 W W2 = VJ( cos (fJ - 30°) = 450(128.3) cos (49.46 - 30°) = 54,437 W 14.149 What equivalent load connected in wye will produce n:sults identical to those of Probs. 14-144 through 14-148? I By wye-delta transformation, Per phase Zy = ~Za =, ~ (6.07 /49.46°) = 2.023/49.46° n THREE-PHASE CIRCUITS 14.150 0 335 Determine the kVA rating of a bank of delta-connected capacitors required to improve the power factor of the system of Prob. 14.144 to 0.9 lagging. I Since cos Of = 0.9, Of = cos- 1 0.9 = 25.84°, Reactive kVA of the load and capacitor, Qf = P tan Of = 65 tan 25.84° = 31.48 kvar. Reactive kVA of the load without capacitor, Qj = P tan OJ = 75.99 kvar, from Prob. 14.144. Reactive kVA of capacitor bank, Qc = Qj - Qf = 75.99 - 3l.48 = 44.51 kvar, which is also the kVA rating since the capacitors are considered lossless. 14.151 If the system of Prob. 14.150 operates at 60 Hz, determine the capacitance of each capacitor in the bank. 4451 I Thus, Hence, 14.152 V 2 = C = lIh(60)(13.65) = 194.32 MF. A 240-V 25-Hz three-phase system supplies a lOO-kW 0.6-pf-Iagging balanced load. Determine the capacitance in each phase of a wye-connected capacitor bank to adjust the power factor to 0.95 lagging. I = cos -1 0.6 = 53.13° Final powerfactor angle Of = cos -1 0.95 = 18.19° Initial reactive kVA = 100 tan 53.13° = 133.33 kvar Final reactive kVA = 100 tan 18.19° = 32.86 kvar Initial power factor angle OJ -x:-- = 3(V)2 . kvar of capacitor band = 133.33 - 32.86 = 100.47 kvar = c Thus, 14.153 1 ,1 Xc = hfC = 0.57 n or and h(25)C = 0.57 3(240/V3)2 X c C = 11.12 mF/phase Repeat Prob. 14.152 for a delta-connected capacity bank. I Proceeding as in Prob. 14.152, we have: 100,470 3 Hence, 14.154 (450)2 14.84 kvar = ~ = - Xc Xc (450)2 1 1 Xc = 14,840 = 13.65 n = 27rfC = 27r(60)C kvar/capacitor = -3'- (240)2 Xc 1 Xc = 1.72 = 27r(25)C or C = 3.70 mF. For the instrumentation shown in Fig. 14-49 we have the following readings: W1 = 2000 W, Al = 18 A, VI = 440 V; W2 = 8000 W, A2 = 18 A, and V2 = 440 V. Find the power factor of the system. I S= P = W1 + W2 = 2000 + 8000 = 10,000 W V3VJI = V3(440)(18) = 13717.4 VA Power factor = and P S= 10,000 13717.4 = 0.729 fhb-wr S,~~--______~____~______ To~ Fig. 14·49 14.155 Determine the kVA rating of a delta-connected capacity bank needed to improve the power factor of the system of Prob. 14.154 to 0.8 lagging. I Q, cos Of = 0.8 P tan OJ = 10,000 tan 43.2° = 9390.6 var j Qf or QcapacilOr = Qj - Qf which is also the kVA rating. = =9390.6 - 7500 = 10,000 tan 36.87° = 7500 var =1890.6 var = 1.89 kvar 336 14.156 D CHAPTER 14 Repeat Prob. 14.155 to determine the capacitance (If a wye-connected bank to adjust the power factor to 0.8 lagging. The system operates at 60 Hz. I From Prob. 14.155, Qc = 1890.6 var or Per phase Qc , = ,( 1890.6) Thus, 14.157 = (Vcf Xc or c= 25.90 /-LF and A three-phase 230-V l00-kVA load operates at 0.8 lagging power factor. The power factor is to be improved to 0.8 leading by a capacitor bank. However, the capacitors are lossy and take 50-kW power when connected t6 a 230-V three-phase source. Determine the kVA tatinls of the capacitor bank. I Initial kvar = 100(0.6) = 60 kvar Final kvar = 100 ( -0.6) = -60 kvar kvar of capacitor ba:1k = 60 - ( -60) = 120 kvar kVA rating of the capacitor bank =, 14.158 'V p2 + Q2 = V(50)2 + (120)2 = 130 kVA Calculate the three line currents in the circuit of Fig. 14-50. I .. __•__~_'___________,l.~~,.... ~,- _______~l-h- b ~ __~_~~ - ______1.~~ ... 2400 V Three phase 25 Hz 14.159 Using the two-wattmeter method, -/ Vab p = VabIa cos ~ la + Vcb ( cos (From Prob. 14.158 it follows that )(" Veb ~ = 2400<44.37) cos 59.45° + 2400(51.96) cos 0° = 178.3 kW le Vcb and le are in the same phase.) What is the total reactive power taken by the load of Prob. 14.158? I 14.161 Fig. 14-50 Determine the total active power drawn by the load of Prob. 14.158. I 14.160 ZI ·100 L20· n Z2· 80 L30· n Za· 80L30· n Q = 2400( 44.37) siu 59.45° + 0 = 91. 7 kvar Find the overall power factor of the load of the circuir shown in Fig. 14-50. I From Prob. 14.159, tan Q (J = p P = 178.3 kW; 91.7 = 178.3 = 0.5143 from Prob. 14.160, or (J Q = 91.7 kvar. = tan'! (0.5143) = 27.22° and cos (J = 0.889 THREE-PHASE CIRCUITS 14.162 How much apparent power is supplied to the load of the circuit of Fig. 14-50? from apparent power calculation. s = yp2 + Q2 = Y(178.3)2 + (91.7)2 = 200.5 kVA I 14.163 (J = 337 Verify the result of Prob. 14.161 P cos 0 S= 178.3 200.5 = 0.889 Determine the active power, reactive power, and the power factor of each load of the circuit shown in Fig. 14-51. VaN (600/V3) /-30° ° 11 = ~ = 50 M = 6.93 /-40 A I 13 == V _ VbN 12 - Z; (600/V3)~ _ 6 /_1"5,,00 40~ -8. 6~A _ - (600 /V3) /90° 15 L=lQ: = 23.09/120° A eN Z; = S1 =VaN1i = (346.41/-30°)(6.93/40°) = 2400.62/10° = 2364.15 + j416.86 VA S2 = VbNI~ = (346.41/-150°)(8.66/155°) = 2999.91 ~ = 2988.49 + j261.46 VA S3 =VeNI~ = (346.41/90°)(23.09/-120°) = 7998.61/-30° = 6927 - j3999.30 VA = 2364.15 W Q1 == 416.86 var cos P2 = 2988.49 W Q2 = 261.46 var cos P1 Q3 = - 3999.30 var P3 = 6927 W cos = cos 10° = 0.980 (J1 (J3 (J2 = cos 5" = 0.996 == cos 30° = 0.866 ....!....!' ....!...~'-j-'--=-'---.,.. - : : ; - - - - ...!....!"'-+--+---=-~...., N - ....... 600 v IIIH. 14.164 Fig. 14-51 Obtain the system's active and reactive powers for the circuit of Fig. 14-51. I S = S1 + S2 + S3 = 2400.62/10° + 2999.91 ~ + 7998.61/-30° = 12,279.64 - j3320.98 P= 12.28 kW 14.165 What is the overall power factor of the system of loads shown in Fig. 14-51? I From Prob. 14.164, -3.32 Q tan 14.166 (J = p = 12.28 or = -0.27 (J = -15.13° and cos (J = 0.965 Determine the parameters of equivalent series-connected circuit elements to represent each impedance of the circuit of Fig. 14-51. I Z1 = R1 R1 =49.240 Similarly, and 14.167 Q = -3.32 kvar + jwL 1 = 50 /10° = 49.24 + j8.68 0 w = 21T(60) = 377 8.68 L1 = 377 = 23 mH + j3.49 0 yields R2 = 39.85 0 L2 =9.25 mH Z3 = 15/-30° = 12.99 - j7.5 0 gives R3 = 12.990 C 3 = 353.67 MF. Z2 = 40 ~ = 39.85 Determine the parameters of equivalent parallel-connected circuit elements to represent the impedances of the circuit of Fig. 14-51. 1 1 Y 1 = Z1 = 50/10° = 0.02/-10° = 0.0197 - jO.0035 S I R1 = Thus, L1 = 0.764 H. 0.0~97 = 50.77 0 1 jX1 = 0.0035 L:::2Sr. = 287.94/90° = j27T(60)L1 Similarly, for the other impedances we have R2 = 40.15 0 L2 = 1.22 H R3 = 17.320 and 338 D CHAPTER 14 Notice that the components of the equivalent parallel circuit are relatively larger compared to the equivalent series circuit parameters obtained in Prob. 14.165. 14.168 For the circuit shown in Fig. 14-52a use a delta-to-wye transformation to obtain the impedance between the terminals 3 and 4. I The steps are indicated in Fig. 14-52b and c, frem which Z34 = 20 + jO = 20 LQ: n. I 3 loJL ~----~'V~------~l 10..Q. 4 Cc.) 14.169 If a 40 LQ:-Y source is connected across the circuit of Fig. 14-52a, determine the voltage V31 • I From Fig. 14-52c, 134 V34 131 14.170 = 10 + JIO 40 LQ: ° = 20 LQ: = 2~!... A 20 LQ: _ 10 + JIO - v.'.." L-·45° V31 = 101 31 = 14.14/-45°Y -- A Obtain an expression for the reactive power in a thre,~·phase system in terms of two wattmeter readings if power in the system is measured by two wattmeters (which read W1 and W2). I From Prob. 14.107, W1 = VJ( cos (0 or W1 - W2 But 14.171 Fig. 14-52 + 300) W2 = V, I, cos (0 - 30°) = VJ([cos (0 + .3(0) - cos (0 - 30°)] = VJI sin 0 Q = V3VJ( sin O. Hence, the total reactive power is given by Q = V3(WI - W2) var. The current coil of a wattmeter is connected in phase a and the voltage coil is connected between phases band c of a wye-connected three-phase load, as shown in Fig. I4-53a. Draw a phasor diagram showing the wattmeter current and voltage with respect to the phase voltage~;. I The wattmeter current is la and the voltage is V,,,, as shown in the phasor diagram of Fig. 14-53b. _ _....L.!"'__+-____ To..R..,.L (6) Fig. 14-53 THREE-PHASE CIRCUITS 14.172 0 339 From Fig. 14-53a and b show that the single wattmeter may be used to measure reactive power in a balanced three-phase circuit. I From Fig. 14-53a the wattmeter reading is ..( V bc W = VbJa cos ' \ la From Fig. 14-53b the angle between V bc and la is W= VbJa cos (90° - 14.173 = ~I{ sin Thus, cos (J = 0.6 or (J _ V3(W1 - W2) _ 0_ tan(JW1+W2 -tan 53.13 -1.33 or W1- W2 = ~ (Wl + W2) = (1.33)(20) W1- W2 = 15.4 and W1 + W2 = 20. = 17.7 kW W1 Hence, Let W2 = O. = 2.3 kW. ~ 50 = 25 W. Also, tan (J = At what power factor 0 or (J = O. Hence, Then (J W1-0 Wl+0 = V3 - - - = V3 or (J = tan -1 V3 = 60° and cos (J = cos 60° = 0.5 The input power to a three-phase load, measured by two wattmeters, is 100 kW, the wattmeter readings being equal. If the power factor of the load is changed to 0.866 leading and the load still draws 100 kW, determine the readings of the wattmeters. I or cos (J = 0.866 W1 - W2 = 33.33 and or W1 + W2 = 100.00. = 30° t Hence, W1 (J an (J - - J3cWI - W2) Wl + W2 = 66.67 kW The three phases of a delta-connected load consist of 10 /45° il impedances. 200-V three-phase source. How much power is consumed by the load? I Zp Hence, and W2 = 33.33 kW. The load is connected across a = Rp + jXp = 7.07 + j7.07 P = 3Pp = 3(20)27.07 = 8484 W. If power input to the load of Prob. 14.177 is measured by two wattmeters, find the reading of each wattmeter. I From Prob. 14.177: W1 + W2=8484 Hence, 14.179 W2 and At what power factor will one of the wattmeter readings be zero in a three-phase system. tan 14.178 J3 The total input power in a three-phase circuit, as measured by two wattmeters, is 50 W. will the wattmeter readings be equal? Determine this reading. I 14.177 = 53.13° J3 If the wattmeters readings are equal, then each reads the power factor is cos (J = 1. 14.176 Q = V3VJI sin (J = V3 W or (J I 14.175 Hence (J. A three-phase load operates at 440 V, 0.6 pf lagging. Two wattmeters used to measure the input power read a total of 20 kW. Find the individual readings of the wattmeters. I 14.174 (J) 90° - tan (J = tan 45° = 1 = v'3 =W-:-:C Wl-+---: - W2 =2 W2= 1792.8 and 1 W or Wl - W2 == 4898.38 Wl = 6691.2 W. A three-phase wye-connected load is balanced and has a (4 + j3)-il impedance per phase. The load is connected to a 220-V three-phase source. If power input to the load is measured by two wattmeters, determine the wattmeter readings. 340 D CHAPTER 14 Vp 127/0° I = I = - =----:- = 25.4/-36.87° 220 V = = 127V I PV3 1 p 4 -- ,3 Zp cos Op P == V3VJI cos Op = V:3(221l)25.4(0.8) = 7742.7 = W1 W1-- W2 tan 0 = 0 75 = V3 - - - - - . . W1 + W2 Thus, 14.180 W1 = 5547.75 Wand = cos 36.87° = 0.8 + W2 W1 - W2 = 3352.8 W or W2 = 2194.95 W. Repeat Prob. 14.179 if the impedances are connected in delta. I Ip 220 ""5 = 11 = 44 A = V3lp = V3(44) = 76.208 A P = V3(220)(76.20SI(0.8) = 23,230.6 = W1 + W2 W1-- W2 tan 0 = 0 75 = V3 -----. W1.,..W2 Hence 14.181 or 10,059.4 = W1 - W2 W1 = 16,645 Wand . W2 = 6585 W. A three-phase balanced load is connected across a 220-V three-phase source. A wattmeter reads 600 W when its current coil is connected in line a and its voltage coil across lines a and b. Next, the voltage coil is connected between lines band c, but the current coil remains in line a. The wattmeter again reads 600 W. Calculate the load power factor. I VJ( cos (30° + /I) W1 = ..,( V bc W2 = VbJa cos -'\ la As seen from Fig. 14-53b, the angle between Vb, alId la is 90° - O. VJ( cos (30° + 0) = VJ( cos (90° - 0) Thus, 14.182 and 0 = 30°. Hence, the power factor is cos 0 W1 = W2, we finally have 30° + 0 = 90° - 0 = cos 30° = 0.866. Determine the line current in the load of Prob. 14.181. I 14.183 20 = 60° or Since P = W1 = 600 = (220)(11) cos (30° + 30°). Hence, I1 = 5.45 A. The ratio of the wattmeter readings in a three-pha~e inductive system is 2: 1, when power is measured by the two-wattmeter method. Determine the power fact)r of the load. I Let W2 = x and W1 = 2x. Then, W1 -- W2 2x - x 1 tan 0 = V3 ----.-- = V3 - - = W1 + W2 2x + x V3 or 14.184 0 = 30° and cos 0 = cos 30° = 0.866. How much active and reactive powers are supplied by the source shown in the circuit of Fig. 14-54? la. a. z.z.o-v s-4 5 ............... ~ 1, ~ C • I "-... /oJL .-~ Fig. 14-54 THREE-PHASE CIRCUITS I 0 341 or or lea = :0 or 212~ = 22 A lea = P = (/ab)24 + (/bY3 + (/ejlO = (44)24 + (44)23 + (22)210 = 18,392 W Q = (/ab)2( -3) + (/bj4 + 0 = (44)2( -3) + (44)24 = 1936 var 14.185 Determine the apparent power and the overall power factor of the load of Prob. 14.184. I s= 14.186 P + jQ = 18,392 + j1936 = 18,494/6.0°. Power factor is cos 6.0° = 0.995. Determine the line currents la' I b , and le in the circuit of Fig. 14-54. I 220 L!r ° lab = 5/-36.87° = 44/36.87 A Ibe 220~ ° = 5/53.13° = 44/-173.13 A lea = 220 L.!1!r 10 L!r = 22/120° la = lab - lea = 44/36.87" - 22/120° = 46.78!'!£" A Ib = Ibe - lab = 44/-173.13° - 44/36.87" = 85/-158° A Ic = lea - Ibe = 22/120° - 44/-173.13° = 40.73/36.6° A 14.187 In the circuit of Fig. 14-54, verify that the phasor sum of the line currents is zero. I From Prob. 14.186, la = 46.78!'!£" = 46.2 + j7.35 A le 14.188 Ib = 85/-158° = -78.88 - j31.66 A = 40.73/36.6° = 32.68 + j24.31 A L = 0 + jO Draw a phasor diagram showing the line currents and voltages pertaining to the circuit of Fig. 14-54. I See Fig. 14-55. \ -158° \ "-V Fig. 14-55 CA. 14.189 The current coil of a wattmeter is connected in line a and the voltage coil across the lines a and c of the circuit of Fig. 14-54. Determine the wattmeter reading. I From Fig. 14-54 and Prob. 14.187 the angle between Vac and la is (220)(46.78) cos 69° = 3688.18 W. 14.190 (60+9)=69°. Find the circulating current in the delta-connected load shown in Fig. 14-54. I I circ = lab + Ibc + lea = 44/36.87" + 44/-173.13° + 22 /120 0 = 44.66/115.86° A Hence, W= 342 14.191 D CHAPTER 14 Replace the delta-connected load of Fig. 14-54 by an equivalent wye-connected load. I The wye-connected load is shown in Fig. 14-56, for which we have Za = 1O( 4 - j3) = 2936 / -4024" n 10 + 4 - j3 + 3 + j 4 ' '-. Z Z" = (4 - j3)(3 + j4) = 1.468 /12.9° n 10 + 4 - j3 + 3 +j4 10(3 + ;4) __ = 2936 /498° n , = 10 + 4 - j3 ~ 3+ j 4 ' . Whereas we have obtained an equivalent wye, it may be readily seen that by using this circuit it is more difficult to obtain the results of Probs. 14.184 through 14.189. The reason for this difficulty is that we do not know the volt ages Van' V bn , and V en · l~ .to l2o-V Ib • , h ~b {:::I ?J-f te C 14.192 Fig. 14·56 III The above-mentioned difficulty may be circumvented by using Millman's theorem which enables us to determine Van' V bn , etc. According to Millman's theorem, the voltage of the load neutral point n with respect to the neutral of the source n' is given by (1) Apply Millman's theorem to obtain the voltages V,In' V bn , and Ve~ of the circuit of Fig. 14-56. I From Prob. 14.191, Ya 2.936/40.24° == 0.34 /40.24° = 0.26 + jO.22 S 1 2 ° 0 .664-,0.152S . Y b =1.468/12.9 o =0.68/-1.9 = 1 Y c = 2.936 /49.8° =~ 034 / -49.8° = 0.22 - jO.26 S L Y = 1.144 -. jO.192 = 1.16 / -9.52° S V bn , = 127 /-1200y, and V cn '= 127 /120° V. Thus from Eq. (1) above, 1 V nn , = 1.16 /_ 9.520 [(127 L!r)(0.34 /40.24°) + (127 / -120°)(0.68 / -12.9°) + (127 /120°)(0.34 / -49.8°)] Now, Van' = 127 L!rY, = 10.86 /164.9° Y Van = Van' - V nn , = 127 L!r - 10.86 /164.9° = 137.5 / -1.18° Y V bn = V bn , - V nn , = 127 / -120° - 10.86 /164.9' ,= 124.65 / -115.17" Y Vcn = V cn ' - V nn , = 127 /120° - 10.86 /164.9° 14.193 = 119.55 /116.32° Y Repeat Prob. 14.186 for the circuit of Fig. 14-56. I la = z: = 137.5 L::.l.l~ 2.936 L-=:-;jQ.240 = 46.8 /39.06° A Ib = V bn z: 124.65 elIS.17° Ie= Ze Van = ~8ZL~.90 Ven = = 84.9 /-128.07° A 119.55 1ll6.32° 2.936-'4~8') L___ _ = 40.72 /66.52° A As expected, the results are in close agreement. THREE-PHASE CIRCUITS 14.194 D 343 Repeat Prob. 14.184 for the circuit of Fig. 14-56. I Za = 2.936/-40.24° = 2.24 - j1.89 n Zb = 1.468/12.9° = 1.43 + jO.33 n = 2.936/49.8° = 1.89 + j2.24 n P = (Ij2.24 + (IS 1.43 + (IJ21.89 = (46.8)2(2.24) + (84.9)2(1.43) + (40.72)\1.89) = 18,347 W Ze Q = (46.8)\-1.89) + (84.9)2(0.33) + (40.72)2(2.24) = 1953 var Considering roundoff errors the results are in good agreement. 14.195 Repeat Prob. 14.185 for the circuit of Fig. 14-56. s = P + jQ = 18,347 + j1953 = 18,450 /6.11° I 14.196 cos 6.11° = 0.994 Draw a phasor diagram showing all the voltages and currents pertaining to the circuit of Fig. 14-56 and compare the result with the phasor diagram of Fig. 14-55. I See Fig. 14-57. \ \ \ I Fig. 14-57 14.197 Calculate the line current, in the wye-connected purely resistive circuit of Fig. 14-58a, if the line voltage is 220-V three-phase balanced. I Since a direct solution is cumbersome, we convert the wye-connected load to a delta-connected load shown in Fig. 14-58b, for which we have obtained the values of the results from Rab = RaRb + RbRe + Rc Ra _ 18 + 54 + 27 _ Rc - I = I - I = Vab _ Vea = 220 LQ: a ab ca 11 16.5 11 Now, 9 - 11 _ 220 L!1!r 16.5 n , etc. = 29.06/-23.42° A A.--~---, C ----t- C --t--t'---"V (a.) (b) Fig. 14-58 344 D 14.198 CHAPTER 14 Because of certain incorrect connections, such as incmrect polarities, a three-phase circuit takes the form shown in Fig. 14-59. Solve for la and Ib with the polarities as shown. I The mesh equations become 11.5I a - Ib - 101 = 127 LQ: -la + 2l.51/.- 201 = 127 LJL -101 a 201 b - + (60 + j30)1 = 0 Solving for la and Ib yields la = (127 LQ:)(1150 t j675) 7625 + j7387 Ib = (127 LQ:)(850 .- ;375) ° 7625 + J7387 = 11.11/-20.3 A = ° 15.95/-13.7 A 0.511 ::> (o.f1- 10 ___________, ~. ;l."jL. ~~~'L O.s.n. 14.199 Fig. 14-59 How much apparent power is supplied by each source to the circuit of Prob. 14.198? Sa =Val: = (127 LQ:)(15.95 L!3.7:) = 2025.65/13.7° = 1968 + j479.75 VA I Sb = Vb I:; = (127 LQ:)(11.11/20.3°; c= 1410.97/20.3° = 1323.23 + j489.52 VA 14.200 Determine the power dissipated in the 30-n resistor of the circuit of Fig. 14-59. I Solving for I from the kVL equations of Prob. 14.198 we obtain " _ 127 LQ:(475) _ 1- 7625 + j7387 - 5.69/-44·LA 14.201 Calculate the currents in the 1-n, lO-n, and 20-f! resistors of the circuit of Fig. 14-59. I 110 n = la - 1= 15.95/-13.7° - 5.69/-44.1° = 11.41 + jO.18 = 11.41/0.9° A 120 n = Ib - 1= 1l.11 L='20~>: - 5.69/-44.1° = 6.33/0.9° A 11 n 14.202 = Ib - la = 1l.11/-20.3° - Verify the active power balance in the circuit of Fi€:. 14-59. I Psupplied Pdissipated 14.203 = (15.95)2(0.5) = Pa + Pb ,,= 1323.23 + 1968.0 = 329l.23 W + (11.11)2(0.5) + (11.41 )2(10) + (6.33)\20) + (5.077)2(1) + 970 = 3287.95 W Verify the reactive power balance in the circuit of Fig. 14-59. I 14.204 15.95/::'13.7° = -5.07 - jO.08 = 5.077 /-179.01° A QsuPPlied = Qa + Q b = 479.75 + 489.52 = 969.27 ,ar A method of solving a three-phase unbalanced circuct is the method of symmetrical components. According to this method, a set of three-phase unbalanced phasors (volt ages or currents) can be resolved into three sets of symmetrical components, which are termed the positive-sequence, negative-sequence, and zero-sequence components. The phasors of a set of positive-sequence components have a counterclockwise phase rotation (or phase sequence), abc; the negative-sequence component:; have the reverse phase sequence, acb; and the zerosequence components are all in phase with each other. These sequence components are represented geometrically in Fig. 14-60a through c. Graphically combine these components to obtain an unbalanced system. THREE-PHASE CIRCUITS D 345 }---+--- VDI (a) (b) (c) Vd) r-- Vc I I V,ll I 1\ \ \ ,, , V,<I\ (d) Fig. 14·60 I The phasors of Fig. 14-60a through c are combined to obtain the phasor diagram of Fig. 14-60d. I'onversely, we may say that the voltages Va , Vb' and Vc can be resolved into their sYIQ.metrical components of Figs. 14-60a through c. 14.205 Express, mathematically, the voltages Vo, Vb' and Vc as phasor sums of V aD , VoP and Vo2 ' the symmetrical components of Va' From Fig. 14-60 notice that the positive-sequence component is designated with a subscript 1. The subscripts 2 and 0 are used for the negative- and zero-sequence components, respectively. I According to the method of symmetrical components, as illustrated in Fig. 14-60, we have (1) Vb = V bD + Vb! + Vb2 (2) Vc = V CD + Vel + Vc2 (3) We now introduce an operator a such that it causes a rotation of 120° in the counterc1ockwise direction (just as the j operator produces a 90° rotation), such that a = 1/120° = 1 x e j !2D a 3 = -0.5 + jO.866 a2 = 1/360° = 1 ~ 1+a +a = 0 = 1 /240° = -0.5 - jO.866 = a* 2 Using the above-mentioned properties of the a operator, we may write the components of a given sequence in terms of any chosen component. Expressed mathematically, we have, from Fig. 14-60, 346 D CHAPTER 14 Consequently, Eqs. (1) through (3) become (in terms of components of phase a), Va =VaO + Val + Va2 (4) Vb = VaO + a2Val + aVa2 (5) (6) 14.206 Express mathematically the symmetrical components of Va in terms of Va , Vb' and Vc' I Solving for the sequence components from Eqs. (4) through (6) of Prob. 14.205: VaO= l (Va Val + Vb + VJ (1 ) 2 ~ (Va + aVb + a VJ = Va2 ,= ~ (Va (2) + a2Vb + aVJ (3) In deriving Eqs. (1) through (3), properties of a such as those given in Prob. 14.205 have been used. Relationships similar to those of Eqs. (1) through (3) are valid for phase and sequence currents also; that is, for current relationships, we simply replace the V's in Eqs. (1) through (3) by I's. 14.207 A three-phase wye-connected load is connected across a three-phase balanced supply system. Obtain a set of equations giving the relationships between the symmetrical components of line and phase voltages. I The symmetrical system, the assumed directions (If voltages, and the nomenclature are shown in Fig. 14-61, from which we have Because Vab + V be + Vca = 0, we get V abo = "bcO =, VeaO = O. We choose Vab as the reference phasor. the positive-sequence component, we have 2 V abl = HVab + aVbc + a Vea )= [(Va - VI» + ~,(Vh - V,) + a 2(V, - VJ] 2 2 2 = ~ [(Va + aVb + a VJ - (a2Va + Vb + aV)] = ~ [(Va + aVb + a VJ - a\Va + aVb + a Vc )] = H(l- a 2)(Va + aVb + a 2VJ] = (1- a 2lVal = V3Val e i3 0" Similarly, for the negative-sequence component, V ab2 = = ~ (Vab W~ For (1) obtain + a2Vbe + aVea ) = Heva - Vb) t a\Vb - VJ + a(Ve - Va)] 2 Heva + a2Vb + aVJ - (aVa + Vb + a V,.)] = Heva + a2Vb + aVJ - a(Va = H1- a)(Va + a2Vb + aVJ = (1 - a)V"2 = V3Va2e-i30° + a2Vb + aVJ] (2) In Eqs. (1) and (2) above, Val and Va2 are, respectively, the positive- and negative-sequence components of the phase voltage Va' '.. vca VQb b Ib Vb" C I,. Fig. 14-61 THREE-PHASE CIRCUITS 14.208 D 347 The line voltages across a three-phase wye-connected load, consisting of a 10-0 resistance in each phase, are unbalanced such that Vab = 220 /131] V, Vbe = 252 LQ: V, and Vea = 195/-122.6° V. Determine the sequence phase voltages. I Since line voltages are given, we determine the sequence components of line voltages. and (3) of Prob. 14.206, 2 Vbc1 = HVbe + aVea + a Vab ) Thus, from Eqs. (2) = H252LQ: + 1/120° x 195/-122.6°+ 1/-120° x 220 /131.7°) = 221 + j12 V = ! (252 LQ: + 1/-120° x 195/-122.6° + 1/120° x 220 /131. 7°) = 31 - j11. 9 From Eq. (1) of Prob. 14.206, V beO = Hvbc + Vea + V ab ) = Sequence components of phase voltages are H252 LQ: + 195 /-122.6° + 220 /131.7°) = 0 V VaO = O. Since 221 + j12 V3(-j) 31 - j11.9 V3( j) 14.209 = = Vbc1 = - jV3Val and V bc2 = jV3Va2 , . -6.9+ j127.5V -6 9 - '179 V . j. Determine the volt ages across the 10-0 resistances and the currents through them. I From Eqs. (4) through (6) of Prob. 14.205 we obtain Va = -6.9 + j127.5 - 6.9 - j17.9 = -13.8 + j109.6 V Vb = a2Val + aVa2 (upon substitution and simplification). la = V R V Ib = ; Since la + Ib + le 13.1 1-155.3° A. 14.210 = 0, = 132.8 - j54.8 V The line currents are given by = 10 (-13.8 + j109.6) = -1.38 + j1O.96 = 11.05 A /97.2° = 10 (132.8 - we finally get j54.8) = 13.28 - j5.48 = 14.37 A /-22.4° le = - la - Ib = - (-1.38 + jlO.96 + 13.28 - j5.48) =-11.9 - j5.48 = Corresponding to sequence currents, we may define sequence impedances. Thus the positive-sequence impedance corresponds to an impedance through which only positive-sequence currents flow. Similarly, when only negative-sequence currents flow, the impedance is known as the negative-sequence impedance, and when zero-sequence currents alone are present, the impedance is called the zero-sequence impedance. A three-phase synchronous generator, grounded through an impedance Zn' is shown in Fig. 14-62. The generator is not supplying any load, but because of a fault at the generator terminals, currents la' I b, and le flow through the phases a, b, and c, respectively. Obtain sequence networks (or equivalent circuits) for the positive-, negative-, and zero-sequence currents. I Let the generator-induced volt ages be Ea' E b , and Ee in the three phases (Fig. 14-62a). The induced volt ages in the generator are balanced. Therefore, these volt ages are of positive sequence only. For the positive-sequence (phase) voltage, we have (1) where Ial Zl is the positive-sequence voltage drop in the positive-sequence impedance (of the generator) Zl' If Z2 is the negative-sequence impedance of the generator, the negative-sequence voltage at the terminal of a phase is simply (2) since there is no negative-sequence generated voltage. The generator zero-sequence currents flow through Zn as well as through ZgO' the generator zero-sequence impedance. The total zero-sequence current through Zn is 348 D CHAPTER 14 laO + Ino + leO = 31aO' but the current through Z,o is laW written as Hence, Vao = - laoZ.o - 31"oZ", which is also (3) .•, + 3Z". Sequence networks corresponding to Eqs. (I) through (3) are shown in Fig. l4-62b Z\} -- Z'" h were through d. r--------------'0 h c (a..) '01 Cb) 14.211 Vo2 L2 V'II - Rer. bus Rd. bus Rd. bu, + '"2 u VoU L0 +u - __ Cc) '00 (d) A line-to-ground fault occurs on phase a of the generator of Fig. 14-62a. representation of this condition and determine the current in phase a. + a Fig. 14-62 Derive a sequence network I The constraints corresponding to the fault are In = le = 0 (lines being open-circuited) and Va = 0 (line-to-ground short-circuit). Consequently, the symmetrical components of the current in phase a are given by Hence laU = lal = la2 = ~Ia' Consequently, the sequence networks must be connected in series, as shown in Fig. 14-63. The sequence voltages appear across the respective sequence networks. To determine the current, we have from Fig. 14·63, VaO + Val + Va2 = Ea - lalZI - lalZZ - lalZu' But Va = VaU + Val + Va2 = O. COilsequently, and THREE-PHASE CIRCUITS D 349 + + + Fig. 14-63 14.212 The line currents in a three-phase four-wire system are la = (300 + j400) A, Ib = (200 + j200) A, le = (-400 - j200) A. Determine the positive-, negative-, and zero-sequence components. I la! = ~ (la + aI, + a 21J = ~ (300 + j400 + 73.2 - la2 + a21b + alJ = H(300 + j400) + 1;-120°(200 + j200) + 1;120°(-400 - = Hla = H300 + j400 j273.2 + 372.9 - and j246.2) = 249 - j119.4 = 276 /25.6° A + 282.8~ + 447.2 ~)= ~ (300 + j400 + 73.2 - j200)) j273.2 + 373.3 - j246.2) = 248.8 - j119.4 = 275.97;25.6° A laD = Hla 14.213 + Ib + IJ = H300 + j400 + 200 + j200 - 400 - j200) = 33.3 + j133.3 = 137.4 /76° A The line currents in a delta-connected load are la = 5 ~, Ih = 7.07 ;225°, and l = 5~ Calculate the positive-, negative-, and zero-sequence components of currents for phase a. Also determine the positive- and negative-sequence components of the current lab and hence calculate lab' I From the given data, laO = H5 LQ: + 7.07 /225° + 5 /90°) = 0 la! = ~ [5 ~ + I ~(7.07 ~) + I L=.l.2O.:'(5 L..2lt:)] = H5 LQ: + 7.07 /345° + 5 /-30) = 5.38 - j1.44 = 5.57 /-15° &0 + 1;-120°(7.07 ;225°) + I ;120°(5 ;90°)] ~ (5 ~ + 7.07 L~ + 5 L2.1..O.:') = -0.387 + jl.44 = 1.49 L..l5!t. la2 = H5 = _ lab! - 1 la! ~ /'l1l0 _ 1 v'3 - v'3 1 lab2 lab 14.214 X 1 = v'3 la2 / - 30° = v'3 = lab! 5.57 ;-15 ° + 30 ° -- 3. I + j U.8 3 + lab2 + Ibo X 1.49 /-30° + 150° =0.223 + jO.831 =3.1 + jU.83 + 0.223 + jO.83 I =3.323 + jl.661 =3.715;26.56° A A three-phase unbalanced delta load draws 100 A of line current from a balanced three-phase supply. An open-circuit fault occurs on one of the lines. Determine the sequence components of the currents in the unfaulted lines. 350 0 CHAPTER 14 I Let the fault occur in phase c. Then, for a delta-connected load, we have le = I,-a - Ibe = 0 and I ca =1 bl or la = -Ib la = 100 LQ: Hence, and laO = H100 LQ: + 100 l.HQ~ + 0) = 0 14.215 la! = ~ [100 LQ: la2 = HlOO LQ: + 1 L=l~r:(lOO /180°)] = 50 + j28.86 + 1 mOol100 /180°)] = 50 - j28.86 Balanced three-phase system problems are often solved on a per-phase basis. A three-phase wye-connected load consists of equal resistances of 100 n / phase. The load is connected across a 220-Y three-phase source. Determine the total power drawn by the load by cckulating the per-phase power. I On a per-phase basis we have Zp = 127 LQ: Ip = 100~ = 1.27 A 220 _ 100 + jon Vp = -v'3 L:::.._ 10'= 127 ~ (ooy P = 3Pp 14.216 3(161.33) = 484 W Repeat Prob. 14.215 if the resistors are connected in delta. I Zp = 100 LQ: n = 220 LQ: Y Vp Pp = (2.2)2100 = 484 W 14.217 = Ip ° = 220 100 LQ: = 2.2 A P = 3(484) = 1452 W A balanced wye-connected load consisting of Zy = (3 + j4) n/phase and a balanced delta-connected load having Z:,. = (9 + j12) n/phase are connected to a 220-Y three-phase source. On a per-phase basis, calculate the line current la' I First, we convert Z:,. into an equivalent ZYe Thus, = ~Z:,. Thus, = 1(9 t j12) = 3 + j4 f1Iphase ° _ (3 + j4)(3 + j4) __ 25/106.26° _ 3 + j4 + 3 + j4 -- 10 /53.13° - 2.5/53.13 V Zp - I Hence, ZYe' p = I = Vp a Zp but la = 127 ~ 12.5 /53.13° = 50.8/-53.13° A. V p = 220 LQ: v'3 = 127 LQ: Y 17 CHAPTER 15 UTransients in DC Circuits 15.1 The voltage across a capacitor C in an electric circuit at if the current through the capacitor is i(t)? I t = to is v(to). What is the voltage at some instant t The v-i relationship for a capacitor is which becomes, for the problem at hand, ve(t) = v(to) 15.2 1 (' + C )'0 i(u) du (1) Obtain a relationship similar to Eq. (1) of Prob. 15.1 for an inductor L if the current through it is i(to) at t = to' I For the inductor, we have Thus, 15.3 A capacitor C initially charged to a voltage Vo is suddenly connected across a resistor R. I The circuit is shown in Fig. 15-1. The voltage vet) = Vc dV e Find vdt). satisfies Vc CTt+/i=O from which we obtain (1) since Vc = Vo at t = O. Fig. 15·1 15.4 The quantity RC in Eq. (1) of Prob. 15.3 has the dimension of second (s), and is known as the time constant. R = 10 kO and C = 50 JLF, what is the time constant r? I 15.5 r = RC = (10 x 10 3 )(50 x 10- 6 ) = 500 ms (or 0.5 s) Given R = 2MO in an RC circuit of the form shown in Fig. 15-1, find C if we want a time constant of lOs. I 15.6 If r = 10 = RC = (2 X 10 6 )C or C= 5 X 10- 6 = 5 JLF To what value will the voltage Vo of the capacitor of Prob. 15.3 decay over a period of one time constant? I Substituting t= r = RC in Eq. (1) of Prob. 15.3 we obtain 1 ve(r) = Voe- = 0.368Vo V or 36.8 percent of its initial value. 351 352 15.7 0 CHAPTER 15 Determine the time constant of the circuit shown in Fig. 15-2a. I The given circuit may be reduced to that of Fi~;. 15-2b, from which = T R C e where R == R e e l R + R2R2+ R3 3 and R, (bj 15.8 Fig. 15-2 Determine the energy stored in the capacitor of Prob. 15.3. I wcCt) = !C(V C)2. Vc = ~,e-lIRC. From Prob. 15.3, Hence, (1) where 15.9 Wo = ~ CV~ Obtain an expression for the transfer of energy from the capacitor to the resistor of the circuit of Prob. 15.3. I 15.10 is the initial stored energy. WR(t) = Wo - Wc = Wo - Woe - 2. me = Wo(1- e -:!',RC) J A 50- JLF capacitor is discharged through a 100-k!1 resistor. If the capacitor was initially charged to 400 V, determine its initial energy and the energy stored after 600 ms. I 15.11 (from Prob. 15.8) T = RC = (100 x 10 3 )(50 x 10- 6 ) = 5 s How long will it take the capacitor of Prob. 15.10 to discharge to O.072-J stored energy? I Substituting the numbers in Eq. (1) of Prob. 15.8 we obtain 0.072 = 4e- Hence, 2 /S • or -2t 0.072 -5- = In -4- = -4.017 t = 10 s. v 0.018 Vc O.007V.. (00) = 0 ----- L---~--~--~~~~--~+-~rl Fig. 15-3 TRANSIENTS IN DC CIRCUITS 15.12 Sketch the decay of voltage of a charged capacitor C being discharged through a resistor R as a function of time. I Vc = Voe- tiT • The voltage is given by Eq. (1) of Prob. 15.3, which may be written as Fig. 15-3. 15.13 0 353 The plot is shown in A portion of the decay of voltage of the capacitor of Prob. 15.12 is shown in Fig. 15-4. If Vc = VI at and Vc = v 2 at t = t2 are the readings from Fig. 15-4, determine the time constant of the circuit. t = tl I Hence, v; I \rz. --LI I 15.14 In the circuit of Fig. 15-1 we have constant of the circuit? I VI = 600 V ic = C at t2 = 4 s. What is the time 4-1 = In 600 _ In 300 = 4.328 s Thus, Vc = 10.75 - 1.5e-IOOOt V. What is the dt = (20 x 10- )(-1.5)( -1000)e -IOOOt = 0.03e -IOOOt A dv In the circuit of Fig. 15-1, we have I 15.17 v 2 = 300 V tl = 1 sand The voltage across a 20-JLF capacitor varies with time and is given by current through the capacitor? I 15.16 at From Eq. (1) of Prob. 15.13, T 15.15 Fig. 15·4 12 11 6 C = 40 JLF and RC = (40 x 10- 6 )(400) = 16 ms Obtain the current transient if Vo = 100 V. R = 400 il. 1 _ RC i= 1000 _ 16 -62.5s -I VO=100V i; = 0.25e- 625t A How does the charge vary as a function of time in the circuit of Prob. 15.16? I 15.18 In the circuit of Fig. 15-5 the switch is closed at t = 0 when the 6-JLF capacitor has charge Obtain the expression for the transient voltage vR" I Qo = 300 JLc. The two parallel capacitors have an equivalent capacitance of 3 JLF. Then this capacitance is in series with the 6JLF. Thus, T=RCeq =40JLs. At t=O+, KVL gives v R =300/6=50V; and, as t~oo, t140 tlT VR~O (since i~O). Therefore, v R = 50e- = 50eV, in which t is measured in JLS. 354 0 CHAPTER 15 Fig. 15-5 15.19 An RC transient identical to that in Prob. 15.3 has a power transient charge Qo, if R = 10 O. I PR=Poe-2t1RC Then, 15.20 }C=10 5 wR(oo) = 3.6 mJ = Q~/2C, or from which C=2J.1,F 184 WR=L'PRdt=3.6(1-e·-tlo.ooOOl)mJ Q'J= 120 J.l,c. we may write Eq. (1) of Prob. 15.3 as Since q = Cvc or Qo = CVo, X 10- 0 = 500 x 10- 6 e- tl (SOXlOOxIO 6). Solving for t yields t = 5.0 ms. From the graph of Fig. 15-3, find the time taken by I th,~ For the circuit of Fig. 15-1 we have circuit 2 s after the switch is closed. I 15.23 ic = C d~tC =- C = 1 mF, R = R~ Voe-tlRe = - Determine the Thus, capacitor of Prob. 15.20 to discharge to a level of 25 J.l,c. 6 T = RC = (50)(100 x 10- ) = 5 ms. 25 J.l,C corresponds to 0.05Q o (since at 0.05Qo' t = 3T. Hence the required time is t = :1(5 x 10- 3 ) = 15 ms. 15.22 Obtain the initial A 100-J.I,F capacitor, carrying an initial charge of 500 ,!.le, is discharged through a 50-0 resistor. time it takes the capacitor to discharge to 184-J.I,C charge. I 15.21 or PR = 360e-tIOOOOOI W. 2 kO, and Q o = 500 J.l,C). ""J = 120 V. From Fig. 15-3, Determine the current in the 3 (1 X 10 - )120e- 2/2 = -0.0221 A 2 A 100-J.I,F capacitor charged to 24 V is connected in ~.eries with a 200-J.I,F uncharged capacitor, a 1-kO resistor, and a switch as shown in Fig. 15-6. Find the current 0.1 s after the switch is closed. I By kVL we have 10 6 200 I 1(6 i dt + 100 which upon differentiating with respect to t becomes J' i dt + 1000i = 0 i Si + (di/ dt) = 0 or i= Ae- ISt or (i)too+ Thus, From Eqs. (1) and (2), -0.00536 A. A=-0.024 and 0+ 24 + (1) 1000(i)t~o+ --0.024 A i==-O.024e- 15t. Finally, at =0 = Fig. 15·6 t=O.ls, (2) 15 i=-0.024e- (OI)= 0 TRANSIENTS IN DC CIRCUITS 15.24 What is the voltage across the 200-J.I,F capacitor of the circuit of Prob. 15.23 at t = 0.1 s? 106 (01 (01 4 15t I VC2 = 200 Jo i dt = 0.5 x 10 Jo (-0.024e- ) dt = -6.215 V 15.25 In the circuit of Fig. 15-1, R = 1 kn and C = 100 J.l,F. the initial current when the switch is closed? I 15.26 By kVL, Vc + v R = 0 or 24 + 1000i = O. The capacitor is initially charged to 24 V. Since RC = (1000)( 100 x 10- 6 ) = 0.1 s, Ri + i(O+) = -0.024 A. Thus, the solution to bJ i dt = 0 or d' 1 R...!.+-i=O dt C becomes i = Ae- tIRC = Ae- 10t. From Prob. 15.25, i(O+) = -0.024 A yields i"" -0.024e- 10t. At t = 0.02 s we obtain i = _0.024e- 10 (002) = -0.0196 A. What is the voltage across the capacitor of Prob. 15.25 at I By KVL, vR+VC=O or v c =-1000i. Vc = (-1000)( -0.0196) = 19.6 V. 15.28 What is Determine the circuit current 0.02 s after the switch is closed, for the circuit of Prob. 15.25. I 15.27 355 At A = -0.024. Thus, t = 0.02 s? t=0.02s, i=-0.0196A (from Prob. 15.26). Thus, Sketch ic and Vc for the circuit of Prob. 15.25. I See Fig. 15-7. 1• ...:1. . . , ; .. . ::;...... ..• •· ,.·• •· ·• ··r • . • •. .• l":~~~L ~ o I ~.JJ ~ t, s I J. . . . : : . . .;:. . . . . . . ..L ..............:................[.............. . -O·02.V . , : : I 15.29 Fig. 15-7 The circuit of Fig. 15-8 was under steady state before the switch was opened. If RI = 1.0 n, C=0.167F, and the battery voltage is 24 V, determine vc(O-) and vdO+). Also find i(O+). I Since the capacitor voltage cannot change instantaneously, we have switch is opened, at t = 0+, Vc + VR1 + VR2 = 0 or 24 + i(l + 2) = O. R2 = 2.0 n, dO +) = V c(O-) = 24 V. Hence, i(O+) "" -8 A. V After the + Fig. 15-8 15.30 Determine i in the circuit of Prob. 15.29 1 s after the switch is opened. I RC = (1 + 2)(0.167) = 0.5 s. Thus, the current is of the form i = Ae -tIRC = Ae -2t. t=O+, weobtain i=-8e- 2t. At t=l, i=-8e- 2(1)=-1.08A. Since i = -8 A at 356 0 CHAPTER 15 15.31 The circuit parameters in Fig. 15-9 are C = 2.4 F and R = 5.0 n. The battery voltage is 100 V. the circuit is at steady state, determine the current in the resistor 10 s after the switch is opened. Assuming I or, at (=0+, 5i + 100 = 0 tl12 • Thus, i = AeAt i = _20-10/1Z = -8.69 A. (= and 0+, i i(O+) ,= - = -20 A I~ = -20 A A :y'ields RC = 5(2.4) = 12 s = -20. i = -20e- tI12• Thus, At (= 10 s, c ~--1(-R F b ., ~-......-----111 + 15.32 Fig. 15-9 What is the energy stored in the capacitor of Prob. 15.31 (10 s after the switch is opened)? I At (= 10 s, Vc Thus, 15.33 - W= ~C(vc)z + (-8.69)5 == I) or Vc = 43.45 V = H2.4)(43.45)2 = 2265.5 1. Determine the current at (= 0+ in the circuit of Fig. 15-10. and the capacitor is charged to 100 V initially. Given: C = 0.2 F, A=-lO and RI = 3 n, R z = 7 n, I 100 + (3 + 7)i ;: () or 15.34 What is the current at (= or i(Q+) = -lOA 6 s in the circuit of Fig. 15-1O? I Since RC=(3+7)(0.2)=2, yields i = -0.5 A at ( = 6 s. i=Ae-· 'IZ • Since i(O+)=-lO, i=-lOe- 051, which c Fig. 15-10 15.35 The initial current in the inductor L of the circuit of Fig. 15-11 with S open is 10 , is closed. L Fig. 15-11 Determine the current after S TRANSIENTS IN DC CIRCUITS I By KVL we have L(dildt) + Ri = 0, 15.36 i = 10 at The quantity LlR in Eq. (1) of Prob. 15.35 has the dimension of second (s) and is known as the time constant. What is the time constant of a coil having R = 10 nand. L = 100 mH? L T L T Hence, = 10 ms = - = 2 ms = R 100 X 10- 3 10+ Rx =2 x 10 -3 Rx=40n. To what value will the current 10 through the inductor of Prob. 15.35 decay over a period of one time constant. I Substituting its initial value. t = T = LI R i L(T) in Eq. (1) of Prob. 15.35 yields = loe - 1 = 0.368/0 A or 36.8 percent of Sketch the decay of current through the inductor of Prob. 15.35. I 15.40 100 x 10- 3 10 Determine the resistance of the resistor. I 15.39 =R= It is desired to decrease the time constant of the coil of Prob. 15.36 to 2 ms by connecting a resistor in series with the coil. 15.38 (1) t = O. I 15.37 357 from which we obtain iL = 10e-(RIL)t since 0 The sketch is identical to that of Fig. 15-3, except that for Vo we substitute 10 , What is the time constant of the circuit shown in Fig. 15-12a? I The given circuit may be reduced to that shown in Fig. IS-12b, for which we have RIR2 R e-- R + R 1 2 Hence, T and = Le/ Re s. Fig. 15-12 15.41 The decay of current in a coil is recorded. It is found that at t = 6 ms. Determine the time constant of the coil. I From Eq. (1) of Prob. 3 15.35 we obtain for the given data: t = 6 ms, 3.68 = loe -6Xl0- /T mA or T 15.42 vL=L = 10 mA at at t = 2 ms, t = 2 ms and iL = 3.68 mA At (6 - 2) x 10- 3 = In 10 -In 3.68 = 4 ms The current through a 50-mH inductor is given by I iL i L = 5 - 2e -lOt A. What is the voltage across the inductor? d~:=(50XlO-3)(_2X_1O)e-lOt=e-l0[V 358 0 CHAPTER 15 15.43 i= 6e~tl".2 Wo = !Ll~ (~L.r~)e~(2Rlqt = Woe~(2R!L)t is the initial stored energy. Obtain an expression for the transfer of energy from the inductor to the resistor of the circuit of Fig. 15-11. I WR(t) = Wo - WL = W,,(1- (from Prob. 15.44) J R = 4n If the initial current is 4 A through the inductor, what is Wo = !Ll~ = !(2)(4f = 161 r=LlR=2/4=0.5s From Prob. 15.46, 0.8 = 16e~4t. t = 0.7S ~,. Thus How does the flux linkage with the inductor of the coil of Prob. 15.43 vary with time? I 15.49 -- l-I~Je -(2RiL)t How long will it take the inductor of Prob. 15.46 to discharge to 0.8-1 stored energy? I 15.48 = W" e~(2RIL)t) In the circuit of Fig. 15-11, L = 2 Hand the energy stored at t = 0.25 s? I 15.47 =0.2s Determine the energy stored in the inductor of the ·;ircuit of Fig. 15-1l. where 15.46 Determine the current transient. = 6e~5t A. W/,(t) = !L(i/Y = 15.45 0.4 R= T I" = 2 + 2 = 6 A Hence, 15.44 L 24 I L = 0.4 H. RI = R = 2 nand In the circuit of Fig. 15-11, we have Flux linkage A == Li or In the circuit of Fig. 15-13, the switch is closed at Find the voltage across the resistor. I I = 0 Le R L =2+ (3)(6) =4H 10 = 10 A. 4 4 r=-=-=ls 3+6 e when the 2-H induct or has a current 6H Fig. 15·13 15.50 The power transient in an RL circuit of the type shown in Fig. 15-11 is given by the initial current if R = 2 n. L 2 10 L 2 = 72 J: e~10t dt = 7.2(1- e- r=-=-=-- R WR = Maximum energy stored is J: PR dt 7.2 = !Ll~ = HO.4)I~:. or Thus L = O.4H 10 = 6 A. 101 ) 1 PR = 72e~10[ W. Determine TRANSIENTS IN DC CIRCUITS 15.51 In the circuit of Fig. 15-11 we have R = 1 nand 10 A, determine the current after 3 s. I 15.52 L = 1 n. i(3) == lOe~3 "" 0.4978 A or At (== 0+, The switch is closed at ("" O. we have, by KVL, Le Thus, 359 If the initial current through the inductor is The 3-H inductor in the circuit of Fig. IS-14 carries a lO-A initial current. Solve for i. I 0 ~ + Ri =0 where i""Ae-(4/2)t=Ae~2t=lOe~2t since L ",,(3)(6)=2H i=lOA at R=4n and 3+6 e (=0. Fig. 15-14 15.53 A 240-V dc generator supplies current to a parallel circuit consisting of a resistor and a coil as shown in Fig. IS-ISa. The system is at steady state. Determine the current in the coil one second after the breaker is tripped. I The new circuit is shown in Fig. IS-ISh, from which we have iL(O+)""iL(O~)= ~6Z =0.8A. By KVL (600 + 300)i + 200dil d( = O. Thus, i = Ae ~45t. Since i(O+) = 0.8, A = 0.8 and i = 0.8e ~45t A. Hence, at (= 1 s, i "" 0.8e~45(1) = 0.0089 A. i ...----::;'\ ilL 200H "R2 300n 600n I;;' (0+) (a) Fig. 15·15 15.54 What is the voltage induced in the coil and the voltage across the coil 1 s after the breaker is tripped in the circuit of Fig. IS-ISa? I By KVL, from Fig. IS-ISh, or At (= 1 s, i = 0.0089 A from Prob. IS.S3. Thus, v L(1 s) = -0.0089(600 + 300) = -8.01 V or . ~.(2.. V coil = -V R1 = -0.0089(600) = -S.34 V 10 H t Fig. 15·16 360 0 CHAPTER 15 15.55 The circuit in Fig. lS-16 is at steady state. t,=O At Determine i(O+) and i(O-). the switch is opened. I 15.56 Find the current in the 2-n resistor at t = 0- and at t = 0+ in the circuit of Fig. lS-16. I 15.57 What is the current through the inductor of the circuit of Fig. lS-16 at I = 1.S s? By KVL, 10 Since 15.58 t i =SA at t = 0, A ~ + (3 + 2)i = = Sand i = Se-os, i = Ae- O. 5 ' or () t = 1.S s, At i = Se- O. 5 (1.5) = 3.7S A. Sketch i(t) for the circuit of Fig. lS-16 after the swtch has been opened. I See Fig. lS-17. i, A . . . . . . . . !. . . . . . . . . '('- ................ !..............., , .............. ................ . ············· ..]············.····1·· .. ······· .. •··· 4 2. +.,.... _.......,....", o 15.59 Fig. 15·17 In Fig. IS-1Sa, V = 100 V, RI = SO n, and the coil parameters are 100 Hand 200 opened. What is the coil current at t = 1.S s? I i(O-) = i(O+) = ~gg = O.S A. Since i = O.S at t = 0, A n. At t =0 the switch is With the switch open, by KVL, 100 15.60 /0 2. -itd' + (200 + SO)i =0 = O.S. Thus, i = i = Ae -25, A or O.5e -25, A. At t = 1.S s, i = O.Se -(25)(1.5) = 0.012 A. Determine the time constant of the circuit of Fig. lS .. lSa and sketch i(t). I T = LI R = 100/(200 + SO) = 0.4 s. For i(t) see Fig. lS-lSb . .t L, '" Coil ..........:. ...............;...............1".............. 0.5' 6 i ~ .............!.................;.................!................ ! 'l..5. ..."...... ···:·... ···.·.· ..···4·· .. ·•··••·· .. ···;··.. ·· ..·• .. •..··l.··· ............ . ··:.t'=:$t-~t-·2~ S t, Fig. 15·18 15.61 In Fig. lS-19a a field discharge resistor is connected in parallel with the motor field winding to discharge the energy in the magnetic field when the switch is open.!d. This allows for a gradual discharge of the energy and thus avoids damage to the switch and to the coil when the switch is opened. The energy stored in the magnetic field is dissipated as heat energy in RD and in the resistance of the field windings. Assuming RD is 1000 n, the generator is operating at 120 V, the parameters of the fIeld winding are 100 Hand 94 n, and the circuit current is at steady state, determine (a) the current in the di5charge resistor at t = (0-), and (b) the current in the discharge resistor at t = (0+). TRANSIENTS IN DC CIRCUITS 0 361 i,v '.,~~ ..- 4----+~~~~~==~__~t, s 17: Fig. 15-19 I The circuit at t=(O+) iL(O-) = 1';40 = 1.28 A. 15.62 is shown in Fig. 15-19b. iD(O-) = 1~=0.12A In the circuit of Prob. 15.61, determine the field current 0.5 s after the switch is opened. I From Fig. 15-19b, 100 At t = 0, 5.39mA. 15.63 (a) i = 1.28 A. Thus, di dt + (94 + lOOO)i = 0 A = 1.28 and or i = Ae-1094t i = 1.28e-1094t A. At What is the voltage across the field winding in the circuit of Prob. 15.61 at t = 0.5 s, i = 1.28e(-lO 94)(0 5) = t = O+? I 15.64 In the circuit of Prob. 15.61, determine the time elapsed for the voltage across the field winding to decay to 40 V. I At the instant the voltage across the field winding is 40 V, the voltage across Ra is 40 V, and iD = V R= But iD = 1.28e-10.94t, sothat 0.04=1.28e-10.94t. 15.65 40 1000 =0.04A Hence, t=0.317s. Sketch i(t) and Ufie1d(t) for the circuit of Fig. 15-19. I See Fig. 15-19c. Fig. 15-20 362 15.66 0 CHAPTER 15 In the circuit of Fig. 15-20 a charged coil is discharged through a diode of negligible resistance by opening the switch at t = O. Determine the coil current 0.2 s after the switch is opened. I i(O+) = i(O-) = ~~g = 0.714 A. By KVL, 200 Since i = 0.714 A 0.503 A. 15.67 15.68 at t = 0, A = 0.714. Thus, What is voltage across the coil at (a) t = (0-) and I (b) (a) v(O-) = 250 V = Vbattery v(O+) or i = Ae- L7S1 A i'~ 0.714e- L7S1 A. (b) t = (0+) At t = 0.2 s, i = 0.714e(-175)(02) = in the circuit of Fig. 15-20? = vdiode = 0.714(0) = 0 V What is the instantaneous power in the circuit of Prob. 15.66? I 15.69 -itd' + 350i == 0 = i 2 R = (350)(0.714i(e 2 (-L75)1) = 178.43e- 351 W p How much energy is dissipated in the resistor of the circuit of Fig. 15-20 in 0.2 s? I W = f.0.2 (i2 R) dt = 178.43 (')2 e - 351 dt (from Prob. 15.68) o .'(1 = 25.661 15.70 In Fig. 15-21 the two paralleled field windings are protected by a common discharge resistor. The parameters of winding 1 are 300 Hand 200 n, the parameters of winding 2 are 100 Hand 200 n, RD is 600 n, and the generator voltage is 240 V. Determine (a) the steady-state current in each winding and in the resistor before the switch is opened and (b) the voltage across each winding at t = (0 -). I (a) iWl (b) VW1 (0-) (0-) = ~~ = 1.2 A i W2 (0-) = ~~ == 1 2 A = VW2 (0-) = VRD(O-) = 240 V Fig. 15-21 15.71 What are the voltages across the circuit elements in the circuit of Fig. 15-21 at I Because of the inductances the currents cannot change instantaneously. iRD(O+) t = O+? Thus, = 1.2 + 1.2 = 2.4 A (the windings being in parallel). 15.72 Determine the value RD in the circuit of Prob. 15.70 to limit the voltage across the windings to 240 V when the switch is opened. I Thus 15.73 or 240 = 2.4R;' R;' = 240/2.4 = 100 n. When the switch is closed, a current i flows in the circuit of Fig. 15-22 because of an initial charge on the capacitor. Write the equation governing i and the form of the solution to the equation. TRANSIENTS IN DC CIRCUITS L I 0 363 Fig. 15-22 By KVL, di 1 Ldi+c I. or Idt=O which has the characteristic roots given by vrc - - JW _+_J_'_=+. pp P2 - - where w == 1/VLC. Hence the solution is of the form (1) 15.74 Using Euler's identity, express Eq. (1) of Prob. 15.73 as a sum of sine and cosine functions. I Since, e jwI = cos wt + j sin wt and i = At(cos wt + j sin wt) e- jwI = cos wt - j sin wt, Eq. (1) of Prob. 15.73 may be written as + A 2(cos wt - j sin wt) = (At + A 2) cos wt + j(A t - A 2) sin wt = BI cos wt + B2 sin wt 15.75 (1) Show that B t and B2 in Eq. (1) of Prob. 15.73 are real numbers and At and A2 are complex conjugates. I Since i is real, the quantity on the right-hand side must be real. Thus B t = At + A 2 and B2 = j(A t - A 2) must be real. Consequently, At = a + jb and A2 = a - jb must be complex conjugates. 15.76 If the initial voltage on the capacitor of the circuit of Fig. 15-22 is Vo' determine the constants B t and B2 in Eq. (1) of Prob. 15.73. I At t = 0, i = O. Thus, 0 = B t + B 2(0) or Bt di L-+V,=O dt 0 Since Bt=O, B2 = -(VolwL). 15.77 dildt=B 2wcoswt. At = O. At t = 0+, or t=O+, we have -(VoIL)=B2W. Hence, Bt=O and Sketch the current for the circuit of Prob. 15.76. I From Prob. 15.76, i = (VolwL) cos wt. Hence, the sketch is as shown in Fig. 15-23. i,A \--_+-_-\--_-+__... wo(, ~ Fig. 15-23 15.78 In the circuit of Fig. 15-24, L = 1.0 H, R = 6.0 n, and C = 0.2 F. The capacitor is initially charged to 24 V and the switch is closed at t = O. Determine i one second after the switch is closed. I By KVL (and after differentiation), d 2i dt 2 di 1. + 6 di + 0.2 I =0 The characteristic roots of this equation are - 5 and -1. Thus, the solution is of the form i = Ate-SI + A 2e- ' ; i(O+)=O yields A t +A 2 =0 and dildt+0+24=0 yields dildt=-24=-5A t -A 2. Hence, A I =6 and A 2 =-6; i=6(e- sl -e- I). At t=ls, i=6(e- s -e- t )=-2.17A. 364 0 CHAPTER 15 , t , A- .. , -2 + IT.. L R 15.79 Fig. 15-25 Fig. 15·24 See Fig. 15-25. In the circuit of Fig. 15-22, e = 2.55 /LF and is closed at t = O. Calculate i at t = OA s. , From Prob. 15.74, L =, 1 - vrc -- \1(200 x 10 i(O+)=i(O')=O yields B2 = -0.214, - Vo L The capacitor is charged to 60 V The switch I '=======;;= = 1400 rad I s ')(2.55 x 10 6) = -3')0 ,= wB cos wt= wB 2 At 2 = 1400B 2 i = -0.214 sin (1400 x OA) = -0.153 A. t = 0.4, Determine vc(t) from the data of Prob. 15.80. Vc or Vc + 0.2 ~ ('-0.214 sin 1400t) = 0 = -(0.2)(0.214)(1400) IX'S 1400t = -59.92 cos 1400t A From the data of Prob. 15.81 determine the first zero crossing of the voltage across the inductor. 'vL=O 15.83 = -60 0.2 i = - 0.214 sin 1400t A. , 15.82 mHo B 1 =0. t!!. (0+) = dt Thus, :~OO i = B, cos wt + B2 sin wt. w - 15.81 ~. Sketch the current in the circuit of Fig. 15-24. , 15.80 - when cos1400t=0 or 1400t=7T/2. In the circuit of Fig. 15-22, e = 100 /LF. capacitor occurs at t = 10 ms. , Vc = 0 when cos wt = 0 or Hence, t=1.122ms. Find L so that the first zero crossing of the voltage across the wt = 7T/2. For t = 10 ms, w = (7T/2) X 10 2 = I/VLC, from which L = 0.405 H. 15.84 For the circuit of Fig. 15-24, write the equation governing the current i and obtain a general form of solution to the equation. , By KVL, 1 di R'1+-L -+ dt e 2 d i dt 2 or L - di tit J.dt=O 1 1. + R '- + - e 1= 0 (1) The characteristic roots are PI' Pz = and the form of the solution is i = Al e P1 ' R :f V!(2RL)2 2/= + A 2eP" A. Le TRANSIENTS IN DC CIRCUITS 15.85 Write the form of solution to Eq. (1) of Prob. 15.84 when it has two distinct real roots. between R, L, and C for this case. I For distinct real roots, we must have (R!2L)2> 1ILe. form of the solution becomes i = A] e -al + A 2 e -bl A. 15.86 (R!2L)2 Let -a and -b be the two real roots. = 11 Le. The roots are 365 Obtain the relationship What is the form of solution to Eq. (1) of Prob. 15.84 when it has two real but coincidental roots? relationship between R, L, and C for this particular case? I For the roots to be coincident we must have If we write the solution as D p] Then the What is the = P2 = -(RI2L) = -a. we have only one unknown constant. For a second-order differential equation we must have two arbitrary constants. Therefore, the correct form of solution is 15.87 Obtain the form of solution to Eq. (1) of Prob. 15.84 when it has a pair of complex conjugate roots. relationship between R, L, and C for this case. I For a pair of complex conjugate roots we must have (RI2L)2<1ILe. Let the roots be where 0'= R!2L and w = Y(RI2L)2 -11Le. Then, the solution becomes Find the -O'±jw, (1) 15.88 Simplify Eq. (1) of Prob. 15.87 to obtain the solution in terms of sin wt and cos wt. I Proceeding as in Prob. 15.74, using Euler's identity, we may write Eq. (1) of Prob. 15.87 as i = e -al(B] cos wt + B2 sin wt) 15.89 Sketch the form of Eq. (1) of Prob. 15.88 for B] = 2, B3 = 3, w (1) = 1, and 0'=0.0796. I See Fig. 15-26. Fig. 15-26 In Eq. (1) of Prob. 15.88, a is known as the damping coefficient and w is called the damped frequency of oscillation. 15.90 In the circuit of Fig. 15-24, C = 14.28 mF, R = 45 n, and L = 5 H. The capacitor is charged to 50 V The switch is closed at t = O. List all conditions on voltages and currents at t = 0- and at t = 0+. I Because of inductance, (1) Because of capacitance, VR(O+) = O. But vc(0+)=vc(0-)=50V. KVL at t=O+ implies that VL(O+)+Vc(O+)+ 366 0 CHAPTER 15 di v =LL dt and We have i(O + ) = O. since Consequently, (2) 15.91 For the circuit of Prob. 15.90, obtain the general solution for the current i. I Since the characteristic roots are -7, - 2, (1) 15.92 Apply the initial conditions obtained in Prob. 15.90 to Eq. (1) of Prob. 15.91. I Hence evaluate AI and A 2 • From Eq. (1) of Prob. 15.90 we have or From Eq. (2) of Prob. 15.90 and Eq. (1) of Prob. 15.91 we obtain ~ (0+)=' -10= -7A I Solving for A 1 and A 2 yields 15.93 AI = 2 and A2 i = 2(e- 7! From Probs. 15.91 and 15.92 we have -0.675 A. 15.94 t '= 0.5 s? e- 2') A. At vL = L t=0.5s, vL ~ 7 = (5)(2)(-7e- ' R = 10 n, In the circuit of Fig. 15-24, C = 0.04 F, switch is closed at t = O. Find i(t). i=A 1e- s'+A 2te- s,. ~ i(O+)=O requires that ~ = A 2(e- s, - 5te- s,) At t = 1 S, VI. i = 2(e(-7)(OS) _ e(-2)(OS» = t = 0.5 s? + 2e- 2') V and A1=0. L = 1 H. The capacitor is charged to 20 V and the PI = P2 = P = -5. The solution takes the form Thus, i=A 2te- S " (0+) =,-20 = A2 and Determine the voltage across the inductor of the circuit of Prob. 15.95 at v 15.97 t = 0.5 s, (0.5)=1O(-7e- 3s +2e- I )=5.24V. I The characteristic roots are, from Prob. 15.84, 15.96 At What is the voltage across the inductor of the circUlt of Prob. 15.90 at I 15.95 = - 2. What is the current in the circuit of Prob. 15.90 at I -2A 2 L i = -20te- S, A t = 1 s. =L~=-20(e-S'-5te-S') dt = -20(e- s - 5e- s ) = 0.54 V. In the circuit of Fig. 15-24, C = 76.92 mF, R = 4 n, and L = 1 H. The capacitor is charged to 100 V and the switch is closed at t = O. Determine i for O.l:; after the switch is closed. I In this case the roots are -2 ± j3 and the s·Jlution takes the form of i = e -2'(B I cos 3t + B2 sin 3t). Since i(O+)=O, B1=0 and i=B 2e- z'sin3t. (aildt)(0+)=-100 requires that -100=B2(3) or B 2 = -33.33. Thus, i=-33.33e- L 'sin3tA. At t=(I.ls, i=(-33.33)e- Z (01)sin[3(0.1)]=-8.06A. 15.98 Determine the damping ratio and the damped frequency of oscillation of the current in the circuit of Prob. 15.97. I Thus, Damping ratio f= 3/27T = 0.48 Hz. Cl' = --2 w = 27Tf = 3 rad/s TRANSIENTS IN DC CIRCUITS 15.99 0 367 Sketch i(t) as obtained in Prob. 15.97. I See Fig. 15-27. 1.78 - - - - - - - - - - - 0.25 0.50 0.75 - - -:::ao-_ _ 1.0 1.25 1.50 1.75 2.0 Time,. < 1 -14.4 15.100 Fig. 15-27 In this problem we present a technique of plotting i(t) for the circuit of Fig. 15-24. The procedure is as follows: (a) Determine the period of the sinusoidal function (or portion) of the solution, mark off time scale corresponding to time constants, and plot the sine wave. (b) Plot the exponential and its mirror image. (c) circle all points on the curves of (b) that correspond to the positive and negative peak values, respectively, of the sine wave. (d) Draw a smooth curve joining all the circled points. Using steps (a) through (d) plot i = 6e- 05t sin 4.186t A. I The procedure is illustrated in Fig. IT 15-28a from which we obtain Fig. 15-28b. 2T 3T 4T 6T 5T Periods 1.0 co x ( o~~~~~~~~~~~~__~~~____ ~ 8 2r 9 10 I I 4r 5r 11 12 Seconds Time. canatan ts (a) 1.0 co x ! !. 0 ~ Seconds -1.0 (b) 15.101 Sketch the curve i = 6e- I See Fig. 15-29, where 05 Fig. 15-28 sin 3.142t A. w=3.14 or f=O.5Hz and r=2s. 368 0 CHAPTER 15 l.0 - ." ~\ o '33 0. 3 3 0,4' -I .~- 15.102 Sketch f\ ~ I' ~~ /V- V \ /' ,(" I -- ~ /\ \ I ----~ ~. --= \ ) ,0 'f ,">e'" V J ,-I i r. J V J Fig. 15·29 i = 10e- 251 sin (12.56t + 30°) A. I See Fig. 15-30, where f = 12.56/27T = 2 Hz, T = 0.5 s, and T = 12.5 = 0.4 s. /. O---k:-7......-!---+--~ ----.+0;0----+---+J I I Fi~:. 15.103 Sketch 15-30 i = e -051(6 cos 11.06t - 1.63 sin 11.06t) A I First we express the quantity in parantheses as a pure sine function. 6 cos 11.06t - 1.63 sin 11.06t. Expanding and equating coefficients, we obtain A =6.22 and 051 Thus, i = (6.22)esin (11.06t + 105.2°) A, for which The plot is shown in Fig. 15-31. £J = Let A sin (11.06t + £J O ) = 105.2° w = 11.06 rad/s, f= 1.76 Hz, and T = 2 s. I I I 1T 0.5 r 15.104 In the circuit of Fig. 15-24, C = 62.5 mF, Loo 2 H, and R = 12 n. the switch is closed at t = O. Solve for the transient current. I By KVL we have (after differentiating), 2 d i . di 2 dt2 -/- U -;It 1 . + 0.0625 1=0 Fig. 15-31 The capacitor is charged to 100 V and TRANSIENTS IN DC CIRCUITS 0 369 which has the characteristic roots PI = -4 and P2 = -2. The current is of the form i = A l e- 41 + A 2e- 21 A. i(O+) = 0 yields AI + A2 = O. Proceeding as in Probs. 15.90 and 15.91, we obtain t!!.(0+)=-50=-4A -2A 2 dt 1 Solving for AI and A2 yields 15.105 A2 = -25. and What is the current in the circuit of Prob. 15.104 at I At t = 0.5 s, i = 25(e15.106 AI = 25 4 e- (OS) - 2 (OS» t = 0.5 s? = -5.81 A. Determine the voltage across the capacitor of Prob. 15.104 at I At t Vc = 1 C J.dt I = t = 1 s. J(-41 -21) d e - e t 25 0.0625 = 25 (-21 -41) 0.25 2e - e = 25.23 V = I sec., we have vd1 s) = 100(2e- 2(1) 15.107 i = 25(e- 41 - e- 21 ) A. Thus, e- 4(1) - In the circuit of Fig. 15-24, let L = 4 H, R = 40 n, and C = 0.01 F, which is initially charged to 600 V. The switch is closed at t = O. What is the current 0.01 s after the switch is closed? Proceeding as in Prob. 15.106, the characteristic equation is 4p2 + 40p + 1/0.01 = 0, which has a pair of coincident roots, p = -5. The current has the form (see Prob. 15.86): i = Ale-sI + A 2te- sl. Since i(O+) = 0, AI = 0 and i = A 2te- sl. I di dt Thus, 15.108 i = -150te -SI A. At t = 0.01 s, + (0 ) = - Vo ""4 = -150 = A2 i = (-150)(0.01)e -S(OOI) = -1.43 A. Determine the steady-state current and the steady-state charge on the capacitor in the circuit of Prob. 15.107. I Since i = -150te- SI , i~O A as t~oo. Also, Q = CV= 0.01(0) = 0 C. Physically, sions imply that the capacitor is completely discharged through the resistor as t~ 00. 15.109 these conclu- In the circuit of Fig. 15-32 the parameters of coil 1 and coil 2 are, respectively, 1.5 Hand 8 nand 0.5 Hand 4 n. If C = is F and it is charged to 100 V, determine the current 0.2 s after the switch is closed. I The characteristic equation is 2p2 + 12p + 18 = 0, which has two coincident roots at p = -3. solution takes the form of i = A l e- 31 + A 2te- 31. Since i(O+) = 0, AI = 0 and i = A 2te- 31 . di + 100 . dt (0 ) = - 2 = -50 = A2 (as In Prob. 15.107) Thus, i = -50te- 31 A. At t = 0.2 s, C.d { i = -50(0.2)e(-3)(0.2) C = -5.49 A. e.:J 2- [:=J i __ 0 15.110 What is the voltage across coil 2 in the circuit of Prob. 15.109 at I Vcoil2 = L2 ~ + R2i = (0.5)(-50) ~ Fig. 15-32 t = 0.2 s? (te- 31 ) - (4)(50)te- 31 = -25(1 - 3t)e- 31 - 200te- 31 = -e- 31 (25 + 125t) V At t = 0.2 S, vcoi12 = _e(-3)(0.2)(25 + 125 x 0.2) = -27.44 V. The 370 15.111 0 CHAPTER 15 Sketch the current in the circuit of Prob. 15.109. I See Fig. 15-33. i,A t:s I 2. -2 -4 Fig. 15-33 15.112 Determine the time constant of the circuit shown in Fig. 15-24 for which R = 16 n, L = 2 H, C = 0.0122 F. Note: It will be incorrect to take LI R or RC as the time constant of this circuit. I We determine the time constant from the damping .;oefficient we have The corresponding characteristic roots are 15.87, and Cl' = 4 and T = ~ = 0.25 s. 15.113 p Cl' as follows. and By KVL, after differentiation, = -4 ± ;5. Thus, the form of solution is similar to that of Prob. Obtain an expression for the current in the circuit of Pr'Jb. 15.112, with Vc (0-) =480 V. I Since we have a pair of complex conjugate roots of the characteristic equation, the current is given by (as in Prob. 15.88) i = e -41(A 1 cos 5t + A2 sin 5t). i(O') ,= 0 requires AI = 0 and i = A 2e -41 sin 5t. di dt (0+) = -240 ~ A2[(I)(5)(I) or 15.114 What is the voltage across the inductor of the circuit of Prob. 15.112 at =L vL di d dt = 2 dt (-48e- 41 VL sin 5t) = -96e- 41 (-4 sin 5t + 5 cos 5t) = -96e(-4)(05)[_4 sin (5)(1).5) + 5 cos (5)(0.5)] = 84.86 V Determine the voltage across the capacitor of the circllit of Prob. 15.112 at I VII = or at Thus, Vc = (16)(-3.89) = -62.24 V = 16(-48e- 41 sin 5t) IJ L = 84.86 V (from Prob. 15.114) = -( -62.24 + 84.86) = -22.62 V. What is the period of oscillation of the current in I 16i th.~ circuit of Prob. 15.112? From Prob. 15.112, w Thus, t = 0.5 s. t = 0.5 s, VR 15.117 t = 0.5 s? t = 0.5 s, At 15.116 i = -48e- 41 sin 5t A and I 15.115 + 0] = 5 = 27Tf or T= 27T/5 = 1.25 s. Sketch the current in the circuit of Prob. 15.112. I See Fig. 15-34 where T = 0.25 sand T = 1.25 s. f= 5 27T = 1 T TRANSIENTS IN DC CIRCUITS D 371 Fig. 15·34 15.118 In the circuit of Fig. 15-24, R = 12 n, damped frequency of oscillation. I The characteristic equation is Cl' =6 and T= i p2 L = 1.0 H, + 12p + 1/0.01 = 0, from which the roots are Hence, or Thus 8 = 27T = 1.273 Hz f (0+) = -60 = A2(8) or i = -7.5e- 61 sin 8t A. What is the voltage across the capacitor of the circuit of Prob. 15.118 at I 1 Vc=c J .dt= -7.5 0.01 J(e I VL = L ~ = t = 0.2 s? 1' ]1 l~o2s=13.02V sm8t ) d t=- 750 100 [e- 6 (-6sm8t-8cos8t) -61' Verify that the result of Prob. 15.120 may also be obtained from I ~ (-7.5e- = 14.0 V at 61 = -(v L + V R) Vc at t = 0.2 s. sin 8t) = -7.5(-6e- 61 sin 8t + e- 61 8 cos 8t) t = 0.2 S v R = Ri= 12(-7.5e- 61 sin8t) = -27.12 Thus, v L +v R =14.04-27.12=-13.08V Prob. 15.120. 15.122 p = -6 ± j8. i = A 2 e- 61 sin 8t, Proceeding as in Prob. 15.113, ~: 15.121 Determine its time constant and the Obtain an expression for the current in the circuit of Prob. 15.118 if the capacitor is initially charged to 60 V. I 15.120 C = 0.01 F. =0.167s. w = 8 = 27Tt 15.119 and or at t=0.2s VC=-(VL +v R )=13.08V, which agrees with the result of Sketch the current for the circuit of Prob. 15.119. I See Fig. 15-35. ~,A Fig. 15·35 15.123 Given: R = 1 n, L = 2 H, and C = 1 F, which is charged to 100 V in the circuit of Fig. 15-24. mine the time constant and the frequency of damped oscillations. I The characteristic equation is T 2p2 1 = 4s = 0.25 + P + 1 = 0, which has the roots p = -0.25 ± jO.66. Thus, w = 27Tt=0.66 or f -- 66 °2· 7T -- 0.105 Hz Deter- 372 D 15.124 CHAPTER 15 What is the current in the circuit of Prob. 15.123 after one time constant? I Proceeding as in Prob. 15.119, di dt Thus, i = -75.7e~025t sin 0.66t A. At • (0 ) =, O.66A2 = -50 t =4s= or A2 = -75.7 T, i = _75.7e~(025)(4) sin (4)(0.66) = -13.24 A 15.125 Sketch the current for the circuit of Prob. 15.123. I See Fig. 15-36. Fig. 15-36 15.126 A circuit consisting of a voltage (or current) source has a source-free response and a forced response. The sum of these components is the circuit response. Using this procedure, find the current in the circuit of Fig. 15-37 if the switch is closed at t = O. I Forced response, . l,= 10 = V R Natural or source-free response (from Prob. 15.35). i"::: -loe~(RIL)t Complete response, (1) R v L Fig. 15·37 15.127 Obtain the solution given by Eq. (1) of Prob. 15.126 by solving the equation that governs the current in the circuit of Fig. 15-37. I The governing equation is di R' V 1'-;It + l= which may be solved by separating the variables as 11 ~ Ri di = By integration we obtain dt TRANSIENTS IN DC CIRCUITS ~ In (V R - Ri) = t + A 0 373 (1) Using the initial condition to evaluate the constant of integration, A =- L R InV (2) Substituting Eq. (2) into Eq. (1) and simplifying the resulting expression yields ._ R V (1 -e -RIlL) 1- 15.128 Sketch the current through and the voltage across the inductor of the circuit of Prob. 15.126. I . = 1. = R V (1 IL -e -RIIL)A v = L L ~ dt = Ve- RtIL V These are plotted in Fig. 15-38. VIR _ 15.129 --VL Fig. 15-38 Repeat Prob. 15.126 for vc, the voltage across the circuit of Fig. 15-39. I Forced response, Natural or source-free response, Vn =- Ve- tIRC (from Prob. 15.3) (1) Vc = vf + vn = V(l - e -tIRC) Complete response, R + V Vc c Fig. 15-39 15.130 Obtain Eq. (1) of Prob. 15.129 by using KVL. I For t > 0, KVL gives Ri which (since +~ f i dt = V i = dqldt) may also be written as dq 1 R-+-q=V dt C · b ecomes q = qn + qf = Qoe -tIRC + CV, where, from the initial condition, Tb e so IutlOn q = CV(l - e- IIRC ) and Vc = qlC = V(l - e- tIRC ). Qo = - CV. Thus 374 D CHAPTER 15 15.131 Sketch the current through and the voltage across ':he capacitor of Prob. 15.130. I ic = ~; = ~ [CII(l Vc e- IIRC VC-·- e- =, IIRC ~ e- = )] IIRC A V ) These are shown in Fig. 15-40. Vc -- -.,.,- -=----.- v _ Vc -j 15.132 In the circuit of Fig. 15-37, R = 0.0410, L = 0.17 H, to reach 2000 A if the switch is closed at (= O. I Solving for (yields e-024121) (= 0.1 s, i = 1.5(1 - 2000 = 6098(1 - or e-024121) L= 10 H, V = 12 V. and Find i at (= 0.1 s. e 'H" Ill) = 1.5(1 - e- 081 ) A = 0.115 A. e(-08)(01» From Prob. 15.128, V L Thus, =Lt!!.=Ve- RI1L 5= or d( 12e-81110 8( or - 10 v L 5 = In 12 ( = 1.094 s. What is the voltage across the inductor of the circuit of Prob. 15.133 at I O 81 =12e- . . At (=O.ls, v L (= 0.1 s? =12e(-08)lo·II=11.08V. A coil having L = 150 Hand R = 2000 is connected in series with a 100-0 resistor. connected to the circuit at (= O. Determine the voltage across the coil at (= 0.5 s. I At A 240-V dc source is Proceeding as in Prob. 15.126, i= (= 0.5 s, i = 0.8(1 - e(-2)(05» 1002~0200 (1 _. = 0.506 A. Ri = 100(0.506) = 50.6 V 15.137 Determine the time for the current Determine the time in the circuit of Prob. 15.133 when the voltage across the inductor is 5 V. I 15.136 V= 250 V. From Prob. 15.126, with the given values, At 15.135 A R = 8 0, i = ¥(1 - 15.134 and ( = 1.65 s. In the circuit of Fig. 15-37 we have I Fig. 15-40 r From Eq. (1) of Prob. 15.126, i = 6098(1 - 15.133 ~ e 300t1150) = 0.8(1 - e- 2t ) A Thus, li,o;1 = 240 - Ri = 240 - 50.6 = 189.4 V In the circuit of Fig. 15-41, the coil has a 10-0 resistance and a 6-H inductance. the switch is opened at (= 0, determine i. If R = 140, V = 24 V, and TRANSIENTS IN DC CIRCUITS ~-----~~ V~-------~ D 375 Fig. 15-41 I Forced response, . 24 = 10 + 14 = If 1A Natural or source-free response, i = 1 + Ae- 41 i=1+1.4e- 41 A. or i(0+)=~=2.4A 15.138 yields Thus, Determine the voltage across the coil of the circuit of Prob. 15.137 at (= 0.1 s. I From Prob. 15.137, i=1+1.4e- 41 A. At (=O.ls, i=1+1.4e(-4j(0.1)=1.94A. VR 15.139 A=2.4-1=1.4. = 14(1.94) = 27.16 V and Calculate the current i2 in the circuit of Fig. 15-42 at V L = 24 - v R = Thus, -3.16 V (= 0.3 s. I Notice that the 12-0 resistor has no effect on the current i 2 • Therefore, by inspection (or from Prob. 15.126), i2 = ¥(1 - e- 41/8 ) = 6(1 - e- 051 ) A. At (= 0.3 s, i2 == 6(1- e(-05)(03») == 0.836 A. 4,{1- ~ 8H Fig. 15-42 15.140 Determine the steady-state power supplied by the source of the circuit of Fig. 15-42. I 15.141 P = VI = 24(11 + 12 ) = 24( t1 + ¥) = 24(2 + 6) = 192 W In the circuit of Fig. 15-42, when the system reaches steady state, the switch is opened. through the inductor. I Under steady state, L T 15.142 15.143 = 8 R = 4 + 12 How much energy is stored in the coil of the circuit of Problem 15.141 at I Solve for the current At (= 0.3 s, i L = 6e(-2j(03) = 3.29 A = 0.5 s t = 0.3 s? WL = i L(i L)2 = -!(8)(3.29)2 = 43.3 J. and The circuit of Fig. 15-43 is under steady state with the switch at position 1. position 2. Find i. At I The current is of the form i = A + Be - RII L. Since i(oo) = M! i(O-) = i(O+) = Thus i = 0.25 + e -20001 A. = ~ 0.25 A = 1.25 A =0.25 B = i(O+) - A = 1 (= 0, the switch is moved to 376 0 CHAPTER 15 Fig. 15-43 15.144 The switch in the circuit of Fig. 15-44 is moved from l to 2 at I t = O. Vc<oo) = Find VC' -50V 1 RC =200 and Applying the above conditions to Thus, Vc Vc yields A = ve(oo) = -50 200t = -50 + 150eV. Fig. 15-44 15.145 t> 0 Determine the current in the circuit of Fig. 15-44 fO! . I 1= By KVL, v R +v e +50=0. v~ 7? =C from the results of Prob. 15.144. dv c dt Thus, v R = -150e- 200t V and i = - s'~e-200t = _0.03e- 200t A C dV e = (1O-6)(150)(-200)e-200t dt Otherwise, -0 03e- 200t A . = as expected. 15.146 Obtain the energy Wc and w R in the circuit of Prob. 15.144. w = It (V R)2 dt = 11.25(1 - e- 400t ) mJ I 15.147 In the circuit of Fig. 15-39, I R=600n, 15.148 t=O.ls, Find Vc at t=O.ls. ve=12(1-e-ollo24)=4.09Y. i I 15.149 V=12Y. V(l - e- tIRe ) = ]2(1 - e- tl (600)(400)(lO-6» V Determine the current in the circuit of Prob. 15.147 at At and From Eq. (1) of Prob. 15.129, Vc = At C=400/LF. R 0 R c t=O.ls, t=O.l s. = C dV e = ~~ e -tiRe = ~e-tI024 A dt R ic = 0.02e-Ol1024 = 0.0132 A = 13.2mA. For the circuit of Fig. 14-45 determine the current through and the voltage across the capacitor at at t=O-. t = 0+ and I At t=O+, vR+vc=24. 6 24/(2)(10 ) = 12 /LA. Since vR(O+)=Ri(CI") and vC<O+)=O, we obtain i(O+)=iC<O+)= TRANSIENTS IN DC CIRCUITS D 377 Fig. 15-45 15.150 Obtain the current in the circuit of Prob. 15.149 at I In general (see Prob. 15.148), ic = (VIR)e- 15.151 (= 70 s, (= 70 s. Since . RC = 2 X 6 10 x 35 X 10- 6 = 70, R=2MD V=24 or at tlRC ic = 12 x 1O-6e - l = 4.415 /-LA. In the circuit of Fig. 15-39, R = 1 MD, C = 20 /-LF, and -20 V, with the polarities as shown. Determine ic and Vc at V= 50 V. (= The capacitor is initially charged to and at (= 0+. 0- I since the switch is open. At (= 0+, idO+)R 15.152 + Vc = or V Calculate the current in the circuit of Fig. 15-39 at (= 0.5 s. I T = tI20 Thus, i=Ae- . At (=Os, i=ic(0+)=70/-LA ( = 0.5 s, i = 70e -05/20 = 68.27 /-LA. 15.153 that A=70xlO- 6 and i=70e- tI20 Q= CVc Thus, Q = 20 X 10- 6 X Vc = At 50Y 50 = 1 mC. In the circuit of Fig. 15-46 determine the current through the inductor and the voltage across the capacitor at (= 0- and at (= 0+. I ~: o.sF '~,;:. (~~ /8v ' - - -_ _L..-_ + 15.155 /-LA. How much charge is accumulated on the capacitor of the circuit of Fig. 15-39 when the circuit has reached steady state. I 15.154 so RC=20s Fig. 15-46 Find ic when the switch has been closed in the circuit of Fig. 15-46, at I In this case, we have a source-free RC circuit for which ic = Ae- ( = 0.5 s. tIRC = Ae-tll.S. or Thus A = -6 and ic = _6e- tl l.S. At (= 0.5 s, ic = _6e- o.sl l.S = -4.3 A. At (= 0+, 378 D 15.156 CHAPTER 15 Find iL when the switch has been closed in the circuit of Fig. 15-46, at t = 0.5 s. I In this case we have an RL circuit excited by aL l8-V source. Thus, . -' +.In -- '" 18 + A -61/5 A IL - If e 611S At t = 0, iL = 0 so that A = - ~ = -3 or i = 3 - 3eA. At t = 0.5 s, i 15.157 At t = 0 the switch in Fig. 15-47 is moving from position 1 to 2. = 3(1 - e-(6)(05)IS) With switch at position 1, the capacitor is charged to the battery voltage, 20 V. 10 3 )(500 x 10- 6 ) = 250 s, . 1.35 A. Solve for i. I le = dUe = (500 x 10 -6)[21)(_ 2s10)e-1/250j = -40e-1/250 = C Tt Since RC = (500 x II.A ,- 2 + 250 k!1 20 V Fig. 15-47 15.158 Determine the voltage across each resistor of the circuit of Fig. 15-47, at I 15.159 The two voltages are equal and lO-oe- 1250 = -9.96 V. VR = Ri = Ric = -·250 x 10'3 x 40 x 1O-6e-,/2so. At t = 1 s, In the circuit of Fig. 15-48, the switch is moved from 1 to 2 at I t = 1 s. t = O. VR =-250 X 10 3 x 40 x Determine i. . _.If +.In -- R V+! oe -R"L -__ 40+(20 40) -2110.5-20 2" 5-2 e - 16e 4 'A I - 0.5 H 20 V )WV Fig. 15-48 15.160 How much energy is dissipated in the 2-0 resistor of Prob. 15.159 in 0.25 s? I 15.161 WR = ' J.025 41 J.° (i2 R )dt=2 " (20+4e- )2dto:254Ws The switch in the RL circuit shown in Fig. 15-49 is moved from position 1 to position 2 at and U L with polarities as indicated. t = O. Obtain UR TRANSIENTS IN DC CIRCUITS D 379 2 2A t Fig. 15-49 I The constant-current source drives a current through the inductance in the same direction as that of the transient current i. Then, for t > 0, VR = Ri = 200e- 251 V 15.162 For the transient of Prob. 15.161 obtain PR and PLO I PR = vRi =400e- 501 W PL = vLi = -400e- 501 W Negative power for the inductance is consistent with the fact that energy is leaving the element, and, since this energy is being transferred to the resistance, PR is positive. 15.163 The switch in the circuit shown in Fig. 15-50 is closed at t = 0, at which moment the capacitor has charge Qo = 500 ,."C, with the polarity indicated. Obtain i and q, for t > 0, and sketch the graph of q. I The initial charge has a corresponding voltage Vo = QolC = 25 V, whence ve(O+) = -25 V. The sign is negative because the capacitor voltage, in agreement with the positive direction of the current, would be + on the top plate. Also vc<oo) = +50V and r = 0.02s. Thus, Vc = -75e- 50t + 50V from which 1 = CVe = -1500e- 501 + 1000,."C i = dq = 75e- 501 mA dt ~ 20V sooo + O.S J'F Fig. 15-51 Fig. 15-50 15.164 The switch in the circuit of Fig. 15-51 is closed on position 1 at constant, at t = r = 250 ,."s. Obtain the current for t> O. t = 0 and then moved to 2 after one time I It is simplest first to find the charge on the capacitor, since it is known to be continuous (at t = 0 and t = r), and then to differentiate it to obtain the current. For 0 $ t $ r, q must have the form q = Ae- 'iT + B. From the assumption q(O) = 0 and the condition at i( 0 + ) = dq I = 20 V = 40 mA dt 0+ 500n we find that A = -B = -10 ,."C, or q = 10(1- e-40001) ,."C From Eq. (1), q(r) = 10(1 - eO!) ,."C; determined for t ~ r as and we know that (0$ t $ r) q(oo) = (0.5 ,."F)(-40 V) = -20,."C. q = [q(r) - q(oo)]e-(I-T)iT + q(oo) = 71.55e-40001 - 20,."C Differentiating Eqs. (1) and (2), . dq 1= dt { = 40e - 40001 mA -286.2e-40001 mA (O<t<r) (t> r) (1) Hence, q is (2) 380 0 CHAPTER 15 son 2 100 100 V 200 -- n + 6A .!QH 6 0.1 H 5H Fig. 15·52 15.165 n Fig. 15·53 A series RL circuit has a constant voltage V appli,~d at (= O. At what time does vR = V L? I The current in an RL circuit is a continuous £t,nction, starting at zero in this case, and reaching the final value VIR. Thus, for ( > 0, ._ R V (1 1- where when T = LIR - e -liT) and is the time constant of the circuit. vR + vL Since = V, the two voltages will be equal ( - = In2 T that is, when 15.166 ( = 0.693T. A constant voltage is applied to a series RL circuit at 3.46ms and 5V at 25ms. Obtain R if L=2H. I (= O. The voltage across the inductance is 20 V at Using the result of Prob. 15.128, we have T 15.167 Note that this time is illdependent of V. = In (2 - (I -In v 2 VI _ - In Fig. 15-52 switch SI is closed at 25 - 3.46 _ , In 20 -In 5 - 15 ..04 ms (= L R= - = T Switch:\2 is opened at O. 2 15.54 x 10 (= 4 ms. 1 = 128.7 !1 Obtain i for (> O. I As there is always inductance in the circuit, the ,;urrent is a continuous function at all times. In the interval with the 100!1 shorted out and a time constant T = (0.1 H)/(50!1) = 2 ms, i starts at zero and builds toward 100 V/50!1 = 2 A, even though it n,~ver gets close to that value. Hence, o~ (~4 ms, i=2(1-e-'I~)A (0~(~4) (1) wherein ( is measured in ms. In particular, i( 4) .~ 2( 1 - e - 2) = 1.729 A. In the interval ( ?' 4 ms, i starts at 1.729 A and decays toward 1001150 = 0.667 A, with a time constant 0.11150 = ~ ms. Therefore, with ( again in ms, i = (1.729 - 0.667)e 15.168 -(1-4)1(213) + 0.667 = «( ?' 4) 428.4e- 3t!2 + 0.667 A In the circuit shown in Fig. 15-53, the switch is moved to position 2 at 34.7 ms. (= O. (2) Obtain the current i z at (= I After the switching, the three inductances have the equivalent Leq Then T = 5/200 = 25 ms, i=6e- 15.169 tl25 ~, !~ + 5(10) = 5 H 15 6 and so, with ( in ms, A i z (34.7) = 2e-347125 = 0.50 A and In the circuit of Fig. 15-54, with il and i2 as shown. obtain a differential equation for i l . I By KVL, . RIll • RIll Differentiating Eq. (1) with respect to (, di l • + LI -Ft + RIl2 = V • ~2 + (RI + R 2 )12 + L2 di = V (1) (2) TRANSIENTS IN DC CIRCUITS D 381 R, ~ + V-=- L, i2 Fig. 15-54 (3) and then eliminate i2 and di 21dt between Eqs. (1), (2), and (3). The result is a second-order equation for il: 2 d il RILl + R2LI + RIL2 di l R IR 2 . R 2V -+ -+--1 (4) dt 2 LIL2 dt LIL2 I LIL2 15.170 Obtain the characteristic equation for Eq. (4) of Prob. 15.169 and write the initial conditions. I The characteristic equation is The initial conditions are 15.171 In the circuit of Fig. 15-55, determine UC' I As far as the natural response of the circuit is concerned, the two resistors are in parallel; hence, r=R eq C=(511)(2ILF)= 10 ILS. By continuity, uC<O+)=uC<O-)=O. Furthermore, as t~oo, the capacitor becomes an open circuit, leaving 2011 in series with the 50 V. That is, i(oo) = ~ = 2.5 A uc<oo) Knowing the end conditions on Uc> we can write wherein t is measured in ILS. IOn = (2.5 A)(lO 11) = 25 V ue =[uC<0+)-uC<oo)]e- tl7 + uc<oo)=25(I-e- 'll o)V, + 2 p.F Fig. 15-55 15.172 Solve for i in the circuit of Prob. 15.171. I The current in the capacitor is given by i e = C dUe =5e- lllo A dt and the current in the parallel 10-11 resistor is 2.5(1 + e- 1120 ) A. 15.173 ilOo=ue/1011=2.5(1-e-lllo)A. The switch in the two-mesh circuit shown in Fig. 15-56 is closed at t>O. t = O. Hence, i=ie+ilOo= Obtain the currents i l and i 2 • for 382 0 CHAPTER 15 so 100 V 1).01 H Fig. 15·56 I (1) IOU 1 + i2) + 5i 2 = 100 From Eq. (2), i2 = (100-lOi 1 )115. Substituting In (2) Eq. (1), di d/ + KI3il = 3333 (3) The steady-state solution (particular solution) of Eq (3) is i 1 (00) = 3333/833 = 4.0 A; hence i 1 = Ae -8331 + 4.0A. The initial condition i1(0+)=0 now gives A=-4.0A, so that il=4.0(1-e-8331)A and i2 = 4.0 + 2.67e- 8331 A. 15.174 Set up a differential equation with initial conditions lor i in the circuit shown in Fig. 15-57, where the sources V and i are simultaneously turned on at t = O. At t = 0-, the capacitor is uncharged and the inductor is unenergized. I For t> 0, KCL at node a yields . 1- dUe C --- + i= 0 (1) .it and KVL around the left mesh yields . + U - V= 0 L d dt +iRI c (2) Now eliminate dUcldt between Eqs. (1) and the derivative of (2) to obtain d 2i L dt2 + R di i 1. ,it + c I = - C The capacitor voltage and inductor current must be continuous at be imposed on Eq. (3) read i(O) = 0 L ~I dt L R 0 (3) t = O. Therefore, the initial conditions to (4) =V a + v c + Vc 1 Fig. 15·57 15.175 In the circuit of Fig. 15-58, switch S represents a current-operated relay, the contacts of which close when and open when iL = 0.25 A. Determine the time period for one cycle of the relay operation. iL = 0.9 A I Let us measure time from a moment at which the switch opens. 0.25 A, with time constant L R= 1111 1 70::;: 30 = 11 00 s Then the inductor current builds up from TRANSIENTS IN DC CIRCUITS n 30 70 0 383 n + 100 V Fig. 15-58 toward a final value of 100/(70 + 30) = 1 A. That is, iL(t) = 1 - 0.75e-II00' A. This current reaches 0.9 A when l100t l = In 7.5, or tl = 1.83 ms. At this instant, S closes, producing a two-mesh circuit with mesh currents i 1 and i2 = i L' KVL gives or, eliminating i 1, (1) The solution of Eq. (1) that satisfies the initial condition i 2 (t1)=0.9A is i2(t) = iL(t) =0.125 + 0.775e- 800('-'ll. This current decays to 0.25 A when 800(t 2 -t l )=ln6.2 or t 2 -t l =2.28ms. Hencethe relay period is t2 = tl + 2.28 = 1.83 + 2.28 = 4.11 ms. 15.176 Assuming that Thevenin's and Norton's theorems are applicable to circuits containing inductances andlor capacitances, represent (a) a capacitance C, with an initial charge Qo' by a Thevenin equivalent circuit and (b) an inductance L, with an initial current 10 , by a Norton equivalent circuit. I (a) The open-circuit voltage across the capacitor of Fig. 15-59a is Vu = Qol C. Hence we obtain the Thevenin equivalent circuit of Fig. 15-59b, where the capacitor is uncharged. (b) The required circuit is shown in Fig. 15-60b, where the initial current through L is zero. L er L 10 L Vo - b (b) (b) (a) Fig. 15-60 Fig. 15-59 The switch of the circuit of Fig. 15-61a has been closed for a long time. voltage across the 200-D resistor. 1000 n b b b (a) 15.177 a a 10 a a It is opened at t = O. Determine the lOO p.F \,--....---:-1 f-_----. Vo BOO 0 sov 200 0 200 0 BOO 0 (b) (a) Fig. 15-61 384 D CHAPTE R 15 I Before S was opened, the network was under stead~{ state and the capacitance was charged to a voltage V;,. This voltage is the same as the voltage across the 800·n resistance; thus, by voltage division, V;, = lRH%(180) = 80 V. In view of Prob. 15.176a the circuit for t > (I may be represented as in Fig. 15-59b. We have I 800i + 200i + - - - - - -6 100 X 10The solution is 16e-- 10t V. IS.178 i= 8(1 -r/( It)()(I)( IOOx)O TIf<j(j e h) A. ft i dt = 80 () • and the voltage across the 200-n resIstor becomes The switch in the circuit of Fig. 15-62a is opened at Find the voltage across the 6-n resistor. 0; 1== U == 200i = prior to this the network was under steady state. ._ 36 (6/9) 1[(0 ) = 10;(31(-6-'-)-';'-(3-+-6-) == 2 A I Applying Prob. 15.176b, we obtain the circuit of Fig. l~i-52b, for which I! I 2 == -_- -t _-I} The solution to Eq. (1) IS S(IRe 00') = 12e - 0", v. U = 18e- 90 , V. ').1 ft U dt (1) 0 Hence the voltage across the 6-n resistance becomes Uw = IOD 3D nO I,!) 0.1 H 0.1 H (b) (a) Fig. IS-62 IS.179 In the circuit of Fig. 15-63, the switch is moved from position I to position 2 at I Thus IS.180 R i = 2e -25t A and UR = Ri == 200e- 2St = 100 n t = O. Find uR • L=4H V. Determine the initial energy stored in the induct or rf the circuit of Fig. 15-63. Next, calculate the energy stored at t == 0.1 s and. hence. find the energy dissipated in the resistor in 0.1 s. I At t=O.ls, i = 2e -2St == 2e- 25xO.l == 0.164 A WR w[(O.I)= !(4)(0.164)2=0.054J = R -- 0.054 = 7.946 J 2 /~ 100 n 2A 4 1-1 Fig. IS-63 TRANSIENTS IN DC CIRCUITS 15.181 0 385 Verify that the result of Prob. 15.180 is correct by solving the problem by direct integration. OI I WR=fl i 2Rdt=100 (2e- 251 )2dt L =~~ (e-501)~1 =7.946J := which agrees with the result of Prob. 15.180. 15.182 In the circuit of Prob. 15.164 determine (a) the charge at t = 0.1 ms; (b) the instantaneous energy stored in the capacitor; (c) the power dissipated in the resistor; (d) the power in the resistor; and (e) the voltage across the resistor at t = 0.1 ms. I (a) Since T = RC = (500)(0.5 x 10- 6 ) = 0.25 ms, q = 10 (I At Eq. (I) of Prob. 15.164 holds. _e- 40OOt ) iJ. Thus, C t = 0.1 ms q:= 10 (I_e-4000XOI x 10') = 3.3 iJ. C (b) O~t~T T~t O~t~ T (c) T~t (d) At t = 0.1 ms we use the first expression of part (c), since T = 0.25 ms. Thus, PR at becomes PR = 0.8e-ROOO(0.IXIO-3) = 0.36 W 3 v R (O.l) = (40e -4000xO I x 10- )(500) X 10- 3 = 13.4 V (e) 15.183 Sketch the current for the circuit of Prob. 15.164. I See Fig. 15-64. i. mA 40 14.7 0r---~~~------~~~ I =250JLs T __ I I I I I -105.3 Fig. 15-64 15.184 Determine the voltage across the capacitor of the circuit of Prob. 15.164 for I Since Vc = qlC, from Eq. (1) of Prob. 15.164 we have Vc = 20(1 - e-40001) V Similarly, from Eq. (2) of Prob. 15.164 we obtain, Vc = 143.1e-4000t - 40 V (T~t<OO) 0 < t < 00. t = 0.1 ms 386 15.185 0 CHAPTER 15 In the circuit of Prob. 15.163, determine the instant when the charge on the capacitor changes its polarity. Sketch q(t). I From Prob. 15.163, q(t) = 1000 -1500e- /Le For polarity reversal q(t) must go through zero; that is, t is given by 0= 1000-1500e- 50t or t=8.11 ms. The sketch q(t) is shown in Fig. 15-65. 50t q. J.LC 1000 - - - - - - - -- - - - - - - - Ol--~'------- I. ms -500 Fig. 15-65 15.186 In the RLe circuit of Fig. 15-66, V is a dc source. charge on the capacitor is zero. I Write a formal expression for the current The voltage equation is di L dt + RI 1 -- - e it. I dt = j if the initial V 0 or, differentiating with respect to time, d 2j di 1 L-+R-+-i=O dt 2 dt e Since Eq. (1) is homogeneous, if = 0 (1) and (2) where PI and P2 are the characteristic roots of Eq. (l): Pl=-2: P2 = - +V~~~r-Ic==-a+f3 2: -R21 r- LIe == - a - f3 (3) v Fig. 15-66 15.187 State the initial conditions to evaluate the constants of integration in Eq. (2) of Prob. 15.186. Define Wo == I/YrC, the resonance frequency, and express the characteristic roots of Eq. (3) of Prob. 15.186 in terms of a and WO" Comment on the significance of a. I The initial conditions are alld Since a = R /2L and Wo = 1v'IC, TRANSIENTS IN DC CIRCUITS ppp2=-a±Va2-w~ D 387 (1) Because a in Eq. (1) is a positive real number, the transient current decays in magnitude like an exponential function, and a is called the damping coefficient (see also Prob. 15.89). 15.188 Notice from Eq. (1) of Prob. 15.187 that pp P2 may be real and distinct, real and coincident, or a pair of complex conjugate roots. Write the forms of solutions for these three cases. I Case 1: a> WO' Here, f3 is real and positive, and f3 < a. The solution takes the form (1) In this case the circuit is said to be overdamped. i.e., the sum of two decaying exponentials. Case 2: a = wo' f3~0, It can be shown that as i Case 3: a < WO' Now f3 is a pure imaginary, or, equivalently, i = Eq. (1) goes over into al (AI + A 2 t)e- (2) f3 = jlf3l, and Eq. (2) becomes i = e-al(Aleill3ll + A2e-ill3ll) = Ae- al sin (1f3lt + "') (3) As given by Eq. (3) the response is a damped sine wave of frequency 1f31 (rad/s); the circuit is underdamped. 15.189 Sketch the forms of solutions given by Eqs. (1) through (3) of Prob. 15.188. I See Fig. 15-67. underdamped Fig. 15-67 15.190 In the circuit of Fig. 15-66, V=O, R=200n, L=O.lH, C=13.33ILF, and uC<0-)=200V. Determine the resonance frequency, damping coefficient, and the damped frequency of oscillation, f3, in rad/s. 1 I Thus, 1'0 J( = = 1O~~ f3 15.191 vTC = Wo = Va 137.85 Hz = 2 1 YO.1 x 13.33 x 10 - V 6 = 1 R = 2 200 a = 2i x 0.1 = 1000s w~ = V(10"? - (7.5 x 10 5 ) = 500S-1 For the circuit of Prob. 15.190, solve for i if the switch is closed at I Noting that a 2 > w~, we have, for 4 75 x 10 = 21Th) t>0 t = O. (with numerical values taken from Prob. 15.190), (1) With inductance in the circuit, i(O+) = i(O-) = 0, which is one condition for establishing the values of the constants Al and A 2 • The other condition is provided by the continuity of the capacitor charge, and hence of the capacitor voltage, at t = O. Thus, at t = 0+, KVL reads: 388 0 CHAPTER 15 di (O)R + L dt I + (l" Le I -di or (0) = 0 .= o' dt -2000 A/s Applying these two conditions to Eq. (1) and its time cerivative, 0 = A I + A 2 and - 2000 = - 500A I fromwhich A I =-2A, A 2 =2A, and i= __ 2e5001+2e-1500'A. 15.192 Repeat Prob. 15.191 for C= 10 ILF, I and the initial conditions are In this case Wo = a, dt These determine the unknown constants as Repeat Prob. 15.191 for C = 1 ILF, I As before, In this case a < WO. 1500A 2' other data remaining unchanged. d·1 ...!. 15.193 - A ,= - = 1000 S-I -2000 A; hence, a 2000 A Is 0' I = C, A2 = i = -2000te -1000, A. other data remaining unchanged. d·1 ...!. dt o· a = 1000s- 1 -2000A/s = but now I:l = v'10 6 - 10 7 = j3000 rad I s. Then i = e -I000jt 3 sin (3000t + <t». The constants are obtained from the initial conditions as <t> = 0, A3 = -0.667 A; hence, i = -0.667e -1000, sin 3000t A. 15.194 An inductance L, a capacitance C, and a resistance R are all connected in parallel. If the initial current through the inductance is 10 and the initial charge: on the capacitance is Qo, determine the natural behavior of the circuit in terms of the common voltage u(t) across the elements. I For t > 0, KCL gives u [I ('. ] R + L Jo v(uldu + 10 + C du dt = 0 or, taking the time derivative, d2u I du 1 C - 2 + .- - + - u = 0 dt R dt L (1) with initial conditions u(O) = Qo C , du C dt I 0 = - Equation (1) has the same form as Eq. (1) of Prob. 15.186. 15.186, u(t) = A,e P,1 + A 2e P2 ', where ( RC Qo + 10 ) (2) Thus, we have the solution from Eq. (2) of Prob. .rz---'; p,=-a+va a = 15.195 2~C =0 -w;; damping coefficient Wo = 1 vrc =0 resonant frequency Consider the natural behavior of a critically damped series RLC circuit (see Fig. 15-66) in which the initial current is zero. Determine the time at which the current reaches its maximum value in R = 5 nand L = IOmH. I For a critically damped circuit the natural currem IS given by Eq. (2) of Prob. 15.188, Since i = 0 at t = 0, we have AI = 0, and ,; ,= A 2te -al. For a maximum i, &t =A,e"( I-at ) =0 -d . from which 1 2L 2(10 x 10- 3 ) t= - = - = =4ms a R 5 i = (A 1 + A2t)e -al. TRANSIENTS IN DC CIRCUITS 15.196 If The circuit of Fig. 15-66 is critically damped. I R = 200 nand a = w~, 2 The condition for critical damping is R ( 2L 15.197 )2 At t=O, Then, i=O, and dildt=VIL = lO00A/s; udt) = -C 1 . = 10 ) F I.t t=0 (see Fig. i' 0 i(u) du = thus, i' 1 -6 IOxlO 0 a=.B....=1000s- J 2L with AJ=O, l000ue- lOoou A 2 =1000A/s, and i(t)=I000te- lOoo ,A. du = 100(1- e- IOOO'(1 + 1000t)] V The parameters of the circuit of Fig. 15-66 are such that the circuit is underdamped. With zero initial conditions, the switch is closed at t = O. Obtain an expression for the voltage across the capacitor. I Applying the initial conditions *10 = f = A(-a sin'" + 1131 cos "') i(O) = 0 = A sin '" which imply that sin'" = 0 and A cos'" = VILII3I. i(t) = a + I13rl2 = w~ "" lILC, 2 From this, using uc(t) = i L' i(u) du = V[ 1 Lrf31 Thus, Eq. (3) of Prob. 15.188 becomes e- a , sin 1f3lt we obtain -I~~' (a sin Il3lt + 1131 cos Il3lt)] = V[ 1 - s::a~ sin (1131t + cjJ)] in which the new phase angle cjJ is defined by cos cjJ == alwo' Because the circuit in Prob. 15.198 is underdamped, the voltage across C will undergo (damped) oscillations. Determine the time at which the first maximum of the capacitor voltage occurs. What is the value of the voltage at this instant? I t= 15.200 3 By Eq. (2) of Prob. 15.188, the current in the circuit is i(t)=(A J +A 2 t)e- a, 15.199 1 With the C determined in Prob. 15.196 and with zero initial conditions, the switch is closed at 15-66). If V = 100 V, find UC" I 15.198 determine the value of C. = LC = 4(100 x 10R2 (200)2 C 389 or = 4L Thus L = 100 mH, D From the expression for 7T/1131. We have i = C(ducldt) obtained in Prob. 15.198 the first maximum of uc(t) occurs at Sketch udt) determined in Prob. 15.198 and indicate the time when the first maximum of uC<t) occurs. maximum is known as the peak overshoot. I See Fig. 15-68. This 390 0 CHAPTER 15 vc Peak overshoot v ) 1T/1.81 15.201 t Fig. 15·68 Calculate the percent overshoot of the capacitor voltage in Prob. 15.198. I The function uc(t) is graphed in Fig. 15-68. It is ',een that the greatest deviation of the voltage from its steady-state value occurs at t = 7T/11:l1. Expressing this peak overshoot as a fraction, uc(TTI 1::1) - V Percent overshoot == -------- x 100% == 100e V 15.202 I = 200 n, and 5 or cot cjJ L = 100 mHo Calculate C such that the peak By Prob. 15.201, 100e- rr whence cos cjJ cot.p = In 20 = - - = 0.954 7T = a Iwo = 0.690. For this circuit, R 200 a=-= 2L 2(IOOxlO 3) = 1000 S-I from which 15.203 R V = 100 V, For the circuit of Prob. 15.198, overshoot is 5 percent. -rr cot.p = C ~~ = (~·gggr = 4.76 x 10- 7 = LC and 4.76 ><[0-7 100 ><10=3 = 4.76 JLF In an RC series circuit, C = 40 JLF, R = 40(' n, and the charge on the 625t q = 4000eJLe. Determine the energy dissipated in the resistor for 1 ms. capacitor varies as i = dq = ~ (4000e-6;'~t x 10- 6 ) = -0.25e- 625t A dt dt PR = i 2R = (0.25e- S25t )2400 = 25e- 125t W I 10 - 3 WR 15.204 = Io (25e- 125t dt) = 0.2(1- e- 125x 3 10- ) To what voltage was the capacitor of Prob. 15.203 initially charged? capacitance as well as from energy storage. = 23.5 (1) m] Obtain the result from the definition of I Since C = qlV, we have: 6 4000 X 10v:o = -Qo = ------= 100 V C 40 10- 6 >( At t = 00, the energy dissipated in the resistor is th,;: initial energy stored in the capacitor. (1) of Prob. 15.203, v: = I (0.2)(2) 6 = 100 V or 15.205 In the circuit of Fig. 15-66, initial conditions are zero. R = 3 kD, R 5-1 a=-=10s 2L The circuit is overdamped (a> wo). I SI = - a + I:l = -1. 70 s - 1 L = 10 H, 2 Wo = 1 LC Thus, from Eg. o 'J 40 X 10- C = 200 JLF, and -(10- 2 s =) and V = 50 V at t = O. Find i(t) if all TRANSIENTS IN DC CIRCUITS Since the circuit contains an inductance, KVL gives i(O+) = i(O-) = 0; di I =V O+O+Ldt 0+ l d'lt 0+ -d or q(O+) also, = q(O-) = O. Thus, at D 391 t = 0+, = 5 A/s Applying these initial conditions to the expression for i, 0= A,(1) from which 15.206 + A 2 (1) and A I =-A 2 =16.9mA 5 = -1.70A,(1) - 298.3A2(1) i=16.9(e-1.70t_ e -298.3t)mA. Sketch the transient current in the circuit of Prob. 15.205. I See Fig. 15-69. oL-----~----~~------~------~----~ __ t, ms 4() 10 Fig. 15·69 17.4 15.207 At what instant does the current in the circuit of Prob. 15.205 reach its maximum? I For the time of maximum current, dUdt = 0 = -2S.73e- L70t + 5041.3e-2983t. t = 17.4 ms. 15.208 A series RLC circuit, with R = 50 n, L = 0.1 H, and C = 50 }LF, has a constant voltage applied at t = O. Obtain the current transient, assuming zero initial charge on the capacitor. I a = ~ = 250s- 1 2 Wo 2L 1 = LC 5 = 2.0 x 10 s - 2 f3 = '-la 2 - W~ = This is an oscillatory case (a<wo)' and the general current expression is A2 sin 370.8t). The initial conditions, obtained as in Prob. 15.205, are l -d d'lt 0+ and these determine the values: 15.209 Repeat Prob. 15.20S for I Qo AI =0, = 2500 A2 = 2.70A. V = 100 V j370.S rad/s 25 i=e- 0t(A,cos370.St+ = 1000 A/s Then i = e- 250t (2.70 sin 370.St) A. }Le. In this case the second initial condition in Prob. 15.20S changes to 0+ L Thus 15.210 Solving by logarithms, See also Fig. 15-69. A2 = 1.35 and ~ I + dt 0+ Qo = V C or ~I dt 0+ = 100 - (2500/50) 0.1 = 500 A/s i = e -250'(1.35 sin 370.St) A. At t = 0, the switch is opened in the circuit of Fig. 15-70. All initial conditions are zero. Set up a differential equation for the node voltage u and state the initial conditions to evaluate the arbitrary constants in the corresponding solution. 392 D CHAPTER 15 I For t>O, we have (1) (2) 1 u = R.,i, + -c I' i z dt (3) du = R di d\ -1- + L dt 1 dt dt 2 (4) - - 0 and their first and second derivatives dZu 1 di d\ -= R - Z+ - -z dt Z Z dt C dt Equations (1) and (2) and the five Eqs. (4) compose a system from which i 1 and i z and their first and second derivatives may be eliminated, giving (5) which is the sought equation. To find the initial conditions on Eq. (5), note that at voltage is zero, and i z = I; hence, t = 0+, the capacitor (6) Moreover, Eq. (2) becomes R I = Ldi-1 I 2 dt 0+ and Eq. (1) and the fourth Eq. (4) give du I di 1 I dt 0+ = -- R2 dt 0+ I +C These two relations imply du I ( 1 dt 0+ 00(; - LR~) I -l + h r <> 1 5 t: () v Je ~L 15.211 In the circuit of Fig. 15-70, R 1 =60D, zero initial conditions find u(t). R2 R z =9CIH, L=I00mH, C=17.78JLF, I From Eq. (5) of Prob. 15.210 we have dZu 0.1 10 6 dv d1 + 150,(F + 17.78 The initial conditions are and The solution then becomes u = 30[2 + (l-75t)e- 750 '] V. 6 X 10 17.78 60 u= Fig. 15-70 and I=l.OA. For TRANSIENTS IN DC CIRCUITS 15.212 In the circuit of Fig. 15-66, I R = 1.0 n, L = 2 H, t = 0, 15.214 V = 50 V. J i dt = t = 0.5 s. 50 and the roots are ~ i = 0.5e -() 25t sin 50t A. At dt (0') = p VL = At t = 0.5 s, With i=0 = 25 A or A =0.5 2 i = 0.5e -025x05 sin (50)(0.5) = -0.058 A. t = 0.5 s, L t = = 0.5 s, 0.5 s. = 2 ~ (0.5e- OZ5t sin 50t) = e -OZ~t(50 cos 50t - 0.25 sin 50t) V vL = e -025 X05[50 cos (50)(0.5) - 0.25 sin (50)(0.5)] = 43.59 V t = 0.5 s? How much energy is stored in the capacitor of the circuit of Prob. 15.212 at t -0.25 ± j50. L * I At = .!": = 50 Determine the voltage across the inductor of the circuit of Prob. 15.212 at I Find i at the solution takes the form i = Ae - 025t sin 50t Thus, di 10" di + i + 200 2p2 + P + 5000 = 0, which has the characteristic equation 15.213 and 393 By KVL, 2 at C = 200 JLF, D from Probs. 15.212 and 15.213 we have Vn = 1i = 1(-0.058) = -0.058 V VL = 43.59 V By KVL, or 15.215 Vc = 50 - (43.59 - 0.058) = 6.47 V Figure 15-71 represents the primary circuit of a gasoline-engine ignition system. The contacts called "points" are closed and opened by a rotating cam. Assume the cam closes the points for a "dwell" period of 0.004 s, after which the contacts open. Determine the coil current 0.004 s after the contacts close. I When the contacts close, the capacitor is short-circuited and only an RL circuit remains for which 0.005di/ dt + i = 6. The solution becomes i = 6 + Ae -200t A. i = 0 at t = 0 requires A = -6. Thus, i = 6(1 - e- 200t ) A. At t = 0.004 s, i = 6(1 - e-200XOO(4) = 3.304 A. 0.2 p.F L - 0.005 H R -1.0 n Coil UPoinu" + Fig. 15-71 IV 15.216 Calculate the current in the circuit of Prob. 15.215 0.001 s after the contacts open. I When the contacts open, we have an RLC series circuit for which the initial conditions are i(O+) = i(O-) = 3.30 A By KVL we obtain the circuit integrodifferentia1 equation having the characteristic equation 106 /0.2 = O. The characteristic roots are p = -100 ± j3.2 x 10 4• Thus, i = e-100t(Al cos 32,000t + A2 sin 32,000t) A Since i = 3.304 A at t = 0, A I = 3.304. At t = 0, or KVL gives v R + V L = 6, since 3.304(1) + 0.005 di di (0+) = 6 0.005p2 + P + 394 D CHAPTER 15 di dt Thus, Thus, 15.217 (= 0.001 s, i = e -,.)Ot(3 .304 cos 32,000( + 0.027 sin 32,OOO() A and A2 = 0.027 which, at + (0 ) = 539.2 = 32,000A2 - 100A I i = 2.51 A. becomes Find the frequency of oscillation of the current wave in the circuit of Prob. 15.216. I From Prob. 15.216, w = 32,000 = 21Tf 15.218 32,000 f= - - = 5093Hz 21T .J}' State all the initial conditions for the circuit of Fig. 1:;-72, which is under steady state for is opened at ( = O. « 0, and the switch I L·60H R-301l Fig. 15-72 1F 15.219 15.21:~ Determine the current in the circuit of Prob. 0.4 s after the switch is opened. I The KVL equation is di + 30i + 60 dt 60p2 + 30p + 1 = O. from which the characteristic equation is i = if i(0+)=4 implies that Al +A2=4. 60 Thus, Solving for A, and A2 yields i = 3.89 A. 15.220 A, + in At = The characteristic roots are (0+) - 30i«n + vc<O+) = 120 di dt (0 ) = + = -0.38 and I) = vL = L or, at (= O.4s, VL i d' A2 = 4.38. Thus, = i = -0.38e-°.46t + 4.38e-o.04 ( = 0.4 s? _0.38e- 0 .46t + 4.38e- 004t = 60[(-0.38)(-O.46)e- 046t + (4.38)(-0.04)e- 004t ] = 60(0.145 - 0.172) = -1.62 V. Determine the total energy stored in the circuit of Fig. 15-72 at I Vc + v L + v R = 120. At W (= 0.4 s, = w L + Wc = !L(i)2 -0.46, -0.04. 0.46 Al + 0.04 A2 From Prob. 15.219, i = (=0, i d' p 0 + A,e- 046t + A2e-004t What is the induced voltage in the coil of the circuit of Fig. 15-72 at I 15.221 f i d( = 120 Vc = (= 0.4 s. 120 -f 1.62 - (3.89)(30) = 4.92 V. + !C(V(Y = H60)(3.89)2 + Hl)(4.92)2 = 466.07 J + A + TRANSIENTS IN DC CIRCUITS 15.222 395 A voltmeter is connected across the coil of the circuit of Fig. 15-72 to read the instantaneous voltage across the coil. What is the voltmeter reading at t = 0.4 s? I 15.223 D Vco;1 + Vc = 120. At t = 0.4 s, Vc = 4.92 V. Thus, vcoH = 120 - 4.92 = 115.0S V= voltmeter reading. An RL series circuit has R = 3 nand L = 6 H, and is excited by a decaying voltage source at t = O. Determine the current in the circuit at t = 5 s. V = 30e -0.51 V I The KVL equation is 6 ~ + 3i = dt 30e- osl (1) Thus, (2) which is the source-free response. Let if = A 2 te- 0 . sl A (3) i=if+i n (4) Then Combining Eqs. (1) through (4) yields i At 15.224 t=O, i=O implies Thus AI=O. = A le- o.sl + 5te- 0.sl i=5te- OsI A. At (5) t=5, i=5(5)e- osxs =2.05A. A turbine-driven dc generator driven at 30 V feeds power to an electromagnet whose inductance and resistance are 6.0 Hand 3.0 n, respectively. The resistance and inductance of the generator are negligible. Sketch the circuit and (a) determine the circuit current. (b) Tripping the turbine causes an exponential decay in turbine speed and voltage. Assuming the decay of generator voltage may be expressed by e gen = 30e- osl, determine the current 5 s after the turbine tripped. ~ = 10 A i=lOe-0.sl+5te-0.sl A. At I The problem is similar to that of Prob. 15.223 except for the initial condition, which is i(O+) = (see Fig. 15-73). Using Eq. (5) of Prob. 15.223, we have t = 5, i = 2.S7 A. lO=A t • Thus, Fig. 15·73 15.225 A series circuit consisting of a 250-}LF capacitor and a 1000-n resistor is connected to a generator whose voltage is represented by e gen = 60e -71 V. The capacitor is initially discharged, and the generator voltage is applied at t = O. Determine the current at t = 0.2 s. I The initial conditions are . + v(O+) 60 1(0 ) = -R- = 1000 = 0.06 A The KVL equation is 6 10 250 J i dt + lO00i = 60e -71 Ji.fdt= JA e- 7I dt=- ~I e- 71 +k=QfC l At t = 0, qf = 0, Thus k = 0, and 6 10 ( 250 - A) + lO00A T if = 0.14e- 71 A and i(O+) = 0.06 = A2 + 0.14 Finally, at t = 0.2 s, i = -1.42 mA. I = 60 or i or A2 = -O.OS = AI =0.14 A 2 e- 41 + 0.14e- 71 i = -0.OSe- 41 + 0.14e- 71 396 15.226 0 CHAPTER 15 In the circuit of Fig. 15-66, at t = 0.01 s. I R = 100 n, L = 5 H, C = 184 JLF, By KYL, d' 5~ dt The auxiliary equation is the form 6 + 100i 10 t -,-- p2 + 20p + 1086.96 = O. 184 f V = 470 Y. and Solve for i and evaluate i i dt = 470 The roots are p = -10 ± j31.42, and the solution takes i = e-loI(A I cos31.42t + A z sin 31.42t) Since i = 0 at t = 0, A 1= O. Since vc(O-)=vc(O+)=O, Thus, i = Aze- \(1 sin 31.42t A. ~ (0+) = ,~ dt ~ L = 470 5 = A 2e -101(31.42 cos 31.42t - 10 sin 31.42t) = 94 or di dt ( + 0 ) = 31.42 A 2 = 94 Thus, A 2 =2.99 and i=2.9ge- loI sin31.42tA. At (=O.Ols, i=2.9ge-loxOolsin(31.42)(0.01)=0.836A. 15.227 Sketch the current wave in the circuit of Prob. 12.226 and find the frequency of the wave. I See Fig. 15-74. w = 21Tf= 31.42 or, from Fig. 15-74, T=lIf=0.2s. (from Prob. 12.226) Thus, or . 31.42 f= - - = 5 Hz 21T /=5111. Fig. 15-74 15.228 What are the instantaneous powers in the R, L, and C of the circuit of Prob. 12.226? I Since i=2.9ge- lol sin31.42tA, PR = i 2 R = 100(2.9ge -101 sin 3 L42t)2 = 895e -201 sinz 31.42t W PL = L -i d' i = (v L)i = 5 [e - 101(94 cos 31..+2t - 29.9 sin 31.42t)](2. 9ge - 101 sin 31.42t) = 1405e -zol(sin 31.42t)(cos 31.42t) -- ·+47e -201 sin z 31.42t W Pc=(Vc)i=(-~ f idt)i= :~; (f 2.9ge-loISin31.42dt)(2.9ge-loISin31.42t) = (-149.5e -101 sin 31.42t - 469. 73e - 101 cos 31.42t)(2.9ge -101 sin 31.42t) = -447e -201 sin 2 31.42t - 1404.8e -=ol(sin 31.42t)(cos 31.42t) W 15.229 A transistor pulse amplifier is shown in Fig. 15-75a. If;J 0.5-Y pulse is applied to the transistor input, how long does it take for the base-to-emitter voltage to reach 90 percent of the pulse amplitude? I We first draw the equivalent circuit in Fig. 15-75b. for which The base-to-emitter voltage VBE is across the capacitor, VHl ,. = Vc = V(l - e'lIr() = 0.5(1 _ e- 115x 10-°) Solving for t for 0.9 V, we obtain t = 2.3r = 11.5 ns. TRANSIENTS IN DC CIRCUITS 0.5V D 397 r-l~+ ov--.J L:8~. ::o_J~ 1 Co> (a) o. 10 pF I , ~:BE (b) Fig. 15-75 15.230 The variable resistance and fixed-value capacitance given in Fig. 15-75 are used to determine the time during which voltage is applied to the electrodes of a welding machine. If this time is that taken for the capacitance to charge from 0 to 10 V, what is the time range covered by the extremes of the resistance values? I The voltage required to determine the end of the welding time is Vc = 10 V= 25(1 - e -tIRC) V. When R = 50 kD, RC = 250 ms. Thus t = 127.7 ms. When R = 1.5 MD, RC = 7.5 sand t = 3.83 s. Therefore, the welding arc time range may be varied from 0.125 s to 3.83 s by varying R. R = 50 kH to 1.5 MH v= 25 V Welding control circuit F=~ Electrodes Fig. 15-76 15.231 Figure 15-77a is the circuit of a relaxation oscillator used for generating a pulse waveform. As the capacitor C charges through Ri' the voltage ve rises. When ve becomes 6 V, the unijunction transistor internal resistance Re drops suddenly from a very high value to 400 D, discharging the capacitor. At this time an output voltage pulse Vo is produced. When V, drops to 2 V, Re again becomes very high, allowing the capacitor to recharge. The action automatically repeats itself, so that a series of output pulses occurs at regular time intervals. The resulting waveforms of ve and Vo are shown in Fig. 15-77b. What is the time t between consecutive pulses? I The complexity of Fig. 15-77a can be eliminated for this problem by considering the timing circuit equivalent in Fig. 15-77e. The switch simulates the transistor action. When the capacitor C is charging to 6 V, the switch is open; for discharging from 6 to 2 V, the switch is closed, connecting resistance R3 across the capacitor. R3 is the series equivalent of Re (400 D during capacitor discharge), and R2 = 560 D. While the capacitor C is charging between 2 and 6 V from the 20-V supply (switch open in Fig. 15-77e), the charging resistor is RI' The charging time constant is TC = RIC= 1 MD x 5OOpF= 500 j.LS The charging time is the difference between the times for charging to 6 and 2 V, V V tc = t6 - t2 = Tin V _ 6 V - Tin V - 2 V As subtracting logarithms corresponds to dividing their original numbers, V - 2V (20 - 2) V tc=Tin V-6V =5ooj.Lsxln (20-6)V = 125.7 j.LS While the capacitor is discharging hetween 6 and 2 V (switch closed in Fig. 15-77e), the equivalent circuit is that of Fig. 15-77 a. The series-parallel circuit in Fig. 15-77 d must be converted into a series circuit for direct evaluation of discharge time. The Thevenin principle used with the capacitor as the load can be applied to perform the conversion. Figure 15.77e shows the resulting Thevenin equivalent circuit. 398 D CHAPTER 15 t! = 20 V C = 500 pF l la} (c) (b) R' R, t-=..v= RJ = 1 MU t Vs 20 V = c= 960 U Vs C =500 pf 500 pF Fig. 15-77 , =V R3 vs RI + R R' = 3 = RI R 3 RI + R3 20" v 960n 00192V 1 Mn + 960 n = . = 1 Mn x 960 n 1 Mn + 960 n = 9 1n 95. Since the Thevenin voltage V ~ "" 0 V and the The venin resistance R' "" R 3 , for practical purposes it may be considered that C discharged directly into the 960 n of R 3 • The discharge extends from U e = 6 V to ue =2Y. T d "" Now, td R 3 C "" 960 n x SOO pF "" 480 ns "" 0.48 J-L s = Tin Uo Ue = 0.48 6V ItS x In 2 V = 0.5273 J.l.S The time between pulses is the sum of the charge and discharge times: 126.22 J.l.S. 15.232 t = te + td = 125.7 J-LS Apply Thevenin's theorem to the circuit of Fig. 15-78(/ 1:0 reduce it to a series RC circuit. t = lOms. I . _ VTh t = 10 ms, -I/T _ ~26 _ -117.674 rns A - 76.74 kfl e ie = 8.3 J-LA. ~ _ TIt - (~".>o )(100) ... "16.741ct).. loo = Then determine ie at The Thevenin circuit is shown in Fig. 15-78b, from which we have le - RTh e At + 0.5273 J.l.S +?I~o- IDt> /oo+b30 -:., .~2.'V Fig. 15-78 TRANSIENTS IN DC CIRCUITS 15.233 D 399 In the box of the circuit of Fig. 15-79a we have a square-wave generator. The switch is operated so that it stays in each position (1 or 2) for about 1 ms and rapidly switches between the positions. If the time constant of the circuit, T = RC, is about 0.1 ms so that the resistive and capacitive voltages stabilize between operations, sketch uC' u R ' and u;n' I See Fig. 15-79b. When the switch is at position 1, u;n = 0 V .md the RC network is short-circuited so that any charge on the capacitor will be dissipated in the resistor. When the switch is at position 2, u;n = 10 V is supplied to the RC network as a charging voltage. The resulting network supply voltage is shown as the square waveform at the bottom of Fig. 15-79b. 11 f'S"qTi"itre.wave-1 11 10V ij""m~t C~}_" 0-----1 . U I 1 ~:.:..V : : : Von 1 } +10V -v,. I I __________ JI L ~ ~ I = 10 V : f\ 0 _ 10 V = 0.1 ms or 1 ms "" 10 ~ Switch position 2 10V Von Switch position 1 rI I l ---1 LJ '-o 1 2 3 4 5 t (ms) (a) 15.234 Fig. 15-79 (b) The circuit of Fig. l5-80a performs differentiation. Thus, for the input voltage u;n of Fig. 15-80b we obtain the output voltage uout also shown in Fig. 15-80b. What is the reason for this shape of the waveform? I The rectangular wave u;n has unequal times for the pulse duration and the period between the pulses. Since the 23.5-jLs time constant of the differentiator circuit (47 kD x 500 pF) is short compared with the 200-jLs pulse duration and with the 800-jLs interval between pulses, the capacitor always has time to charge before the input pulse changes level. Thus the output voltage uout is always close to zero prior to an input level change. When the input level changes from 0 to 10 V, a positive swing, the output rises suddenly from 0 to + 10 V. When the input level drops from 10 to 0 V, a negative swing, the output rises from 0 to -10 V. The output voltage uout consists of spikes which rise almost instantaneously in either a positive or negative direction on an input level change, dropping virtually to zero before another input level change occurs. ~--------,- Vout 0 - 10 V (,200 J1.s + 10 V R = 47 kU f..Y--aoo J1.s ~ VII' o o 2 Time (ms) (a) 15.235 (b) Fig. 15-80 Figure 15-81a shows an integrator. Explain why the output voltage has the waveform shown in Fig. 15-81b for the given input voltage consisting of a train of pulses. 400 0 CHAPTER 15 +10 V vout 5V 4V 3V 2V 1V OV R = 82 kO o 2 3 5 4 t (ms) Fig. 15·81 (b) (a) I The 8.2-ms time constant of the integrator is long compared with the 800-p,s pulse duration and the 200 p,s between the pulses. During an 800-p,s period, the capacitor voltage increases by approximately 10 percent of the difference between its voltage at the start of the period and its maximum possible voltage 10 V. During a 200-p,s period, the capacitor loses about 2.4 percent 01 its voltage. Therefore the output voltage v ou • gradually builds up toward the maximum of 10 V. 15.236 How much energy is stored in the induct or of the I At t = 0+, Cl rcuit of Fig. 15-82 at i = 6/1 = 6 A = 10 , i(t) = lol·-llt = 6e- 21 A Thus, w L = !Li 2 = }( 1)(6e- 21 )2 At t = 0.1 s? t=O.ls, = 18e- 41 J wL=18e-4xol=12.07J. Fig. 15·82 15.237 Find the voltage Vo in the circuit of Fig. 15-83. i(O r", 0= ~2 = 3 A I KVL gives (for t > 0): ! f i dt + 2i + (smce the 2-F capacitor has i dt = 6 di dt 3. - + - 1=0 or since f i(O+) = 3, A =3, and 4 i:: 3e -0.751 A. Vo =1 or V2 F(O+) = 6) i=Ae- 0.751 Finally, J: 3e-°7~1 dt = 4(1- e- 0751 ) V Fig. 15.83 TRANSIENTS IN DC CIRCUITS 15.238 D 401 In the circuit of Fig. 15-84, the initial voltages on the capacitors are 12 V and 6 V with polarities as shown. Determine i(t). 10 6 12 - 6 I i(O + ) = 1X 10 6 = 6 r/lA = I0 T = RC = X 10- 6 X 2 X 10- 6 2 (2 + 1)10 6 = 3s Thus, t:.o + ('J. V 15.239 T:6V j:Lrf ',)1_ ~ t Fig. 15-84 Obtain the form of solution for vo(t) in the circuit of Fig. 15-85 when the switch is moved from position 1 to 2 at t = O. I By KCL: or which has the form of solution: ~~t~~_O~____~_~_T V b Fig. 15-85 15.240 In the circuit of Fig. 15-86, solve for i. I By KVL: Thus, i=3+Ae- 10t• 40 + 20 = 20i + 2 Since i(0+)=2, 10ft.. ~ A=-l or and di - + 10'1= 30 dt i=3-e- 10t A. , 017- . L :lH Fig. 15-86 402 0 15.241 CHAPTER 15 Find the voltage across the uncharged capacitor of the circuit of Fig. 15-87 if I v = 1 2e- V. or By KCL: Ae Vc = Solution is of the form: -31 + 2e- 1 A =-2 Hence I ~ 15.242 Jl- tl t __---'VVV ----r i o T ~.~ I ~_=r-~ ~ F ______________ __LL ____ Fig. 15·87 Determine the current through the 2-n resistor of the circuit of Prob. 15.241 and sketch the current. ' -3,) i= Vc = 2( _e ~.!..._ =e- 1 -e- 31 A I R 2 which is sketched in Fig. 15-87. i, A 0.2. ----- L-.---+_-+-:--4-_-+.-. 0.4 o.g 1.2. r., 15.243 , 2-0 • 2.4 ~ t, S. Fig. 15.88 Differential equations of the form: d; + p(t)y = Q(t) (1) are commonly encountered in circuit equations having one energy-storage element. of solution as: ye f P dl = rQef P Such equations have forms dl dt + A (2) J Apply this result to solve for the current in the circuit of Fig. 15-37, if V = 2e - I V, I and L = 2 H. By KVL we have: which is similar to (1) above. ie J2dl = 15.244 R = 4 n, f 2cj!..+4i=2e- 1 dt Thus, from (2) e-IeJ2dldt+A or -di + 2'l=e -I dt or or Solve for i in the circuit of Fig. 15-89. I By nodal analysis, at node 1: '(0 ) = '(0 ) = 6/ 3(3 + 3) I. 3+3+3 I =3A (1) TRANSIENTS IN DC CIRCUITS D 403 (;V + Fig. 15-89 6 - VI 3 VI VI - V - - = = - + - - -2 Thus, VI or 6 3 ==2.4+0.4v 2. At node 2, including i(O+), we have: (2) 6 - v2 VI - v 2 1 • - 3 - + - 3 - == 1(0+) + 2 f V2 (3) dt Substituting (1) and (2) into (3) and simplifying yields: 3 Since !f Vz dt = i f v 2 dt + 3.2v2 == - 1.2 (4) we have 3 f and v 2 dt == 6i V 2 di =2- dt Thus (4) becomes: 6.4 di . di + 61 - 1.2 Solution is of the form: i == - ~2 + Ae-(6/64)1 == _ 0.2 + t == 0, At i= 3 A = 3.2 requires that i:: - 0.2 + 3.2e -0.93751 A Hence, 15.245 What is the time constant of the circuit of Fig. 15-897 I From Prob. 15.244 we have: 1 - == 0.9375 or T 15.246 == 1.067 s t = 1 s7 From Prob. 15.244: WL t=l, which becomes, at == ! U 2 == !(2)(- 0.2 + 3.2e -093751)2 J w L ==I.109J. Determine the voltage across the 6-D resistor of the circuit of Fig. 15-89 for I t > O. From Prob. 15.244: And Thus, 15.248 T How much energy is stored in the inductor of the circuit of Fig. 15-89 at I 15.247 Ae-0.93751 di V 2 d = 2 -dt = 2 -dt (- 0.2 + 3.2e -0.93751) VI = =2(-3.2 x 0.9375ro.9375'):: - 6e -0.93751 V 2.4 (i-rO.9375') V In the circuit of Fig. 15-39, V is a constant voltage Vo' Show that under steady-state, the energy supplied by the source is equally divided as that dissipated in the resistor and stored in the capacitor. 404 0 CHAPTER 15 I Since q = CVo(1- e- tIRC ), from Prob. 15.130, the energy stored in the capacitor, 1 q2 Wc ="2 1 C == :2 CV~(1- e-tlRcy J Total energy supplied by the source = qVo = CV~(1- WT As t~oo, Thus, 15.249 ~WT Wc = JCV~ = ~WT goes as dissipated energy in the resistor. Evaluate the dissipated energy in the resistor of the circuit of Fig. 15-39 over the period that the capacitor is fully charged. Thus, show that the energy dissipated is the same as the energy stored in the capacitor. I From Prob. 15.131: Vo l=j"ie r~ CtJ, Thus, since the capacitor is fully charged as wR = 15.250 e- tIRC ) J r i Rdt = R(; 2 rr e- Determine the initial conditions on i 1 , i 2 , and I t< Before the switch is closed, i3 21 RC / -tIRC energy dissipated in R dt = ~CV~ = from Prob. 15.248 Wc of the circuit of Fig. 15-90. ° i 4fl (0-)=i IH (0_)=2!4 =IA t> When the switch is closed, ° 2il = 6 - + V O.5 F (V05 PI 0 F (4)(6) = 4+2 = 4V or F + V\.O F) == 6 -4 =2 i)(O+) - i 3 (0+) = i) H(O+) = i l H(O_) (1) =1 A i 3 (0+) = il(O+) -1 = 1-1 = (lA Since V05F(0_)=V05F(0+) and V\.OF(O_)==v l )F(O+) . (0 ) _ . (0 ) = 11 + '/ + V O.5F ( 0 +) (2) 4 Because the capacitors are connected in seri{:s vo.s F V\.O F = _1_ = 2 0.5 or VU.s F = 2v\.0 F From (1) to (3) we finally obtain: 6v Fig. 15-90 (3) TRANSIENTS IN DC CIRCUITS t = 0. In the circuit of Fig. 15-91, the switch is moved from position 1 to 2 at 15.251 I By KVL: 0.1 di dt + (100 X 106 3 10 )i + 100 405 Determine Ji dt ° (1) = or Since D (2) 5 4 i(O+) = 10/10 = 10- A, Eq. (1) yields di dt 5 10 o:T 1(0+) = -lOOA/s • (0+) = - From Eq. (2), therefore, ~:~ (0+) = _[10 ~ (0+) + 10 i(0+)] = -[10 (-100) + 10 (105 6 6 5 4 ))= 8 10 A/s2 L /0 V 0.' H L Fig. 15·91 15.252 dvtldt(O+) In the circuit of Prob. 15.251, find and Ov. I Since i=(v t -v 2)IR dv 2Idt(0+), assuming vt(O+) = lOV or C di di v 2=L-=05dt . dt and 2 or 15.253 dV 2 d i 8 7 -d (0 +) = 0.5 - 2 (0+) = 0.5( -10 ) (from Prob. 15.251) = -5 x 10- Vis t dt In the circuit of Fig. 15·92 determine the voltage ve(O+) and its derivative dveldt(O+). I Since ve(O_)=vc(O+) and ve(O_)=O, ve(O+)=O. and Thus, dVe 6 -d (0+) = - = 12 Vis t 0.5 'l...n.. 1lV 15.254 i L (0 + ) = i L (0 - ) = E2 Now =6 A IH 0.5' l F Fig. 15.92 2 2 Determine d v Idt (0 +) in the circuit of Fig. 15·93. I dv v C-+-=/ or dt R 2 d v 1 dv 1 2 or dt2 (0+) = - RC dt (0+) = - RC2 Vis and v 2(0+) = 406 0 CHAPTER 15 v Fig. 15·93 15.255 Repeat Prob. 15.254 for the circuit of Fig. 15·94. * ±f I + or U dt == I or or Fig. 15·94 15.256 In the circuit of Fig. 15·95 determine the initial and final conditions on uL and UR' I Thus, and and UL(O+) =0 VR uR(oo) = R + R 1 Fig. 15·95 15.257 Find duLldt(O+) and duRldt(O+) in the circuit of Prob. 15.256. I Thus, 15.258 dUe ic dU dU L + -R -=-=dt C dt dt diL = UL dt L and and dU dt diL dt R r-= dU - R (0 ) =0 dt + In the circuit of Prob. 15.257 let the switch be closed for t < 0 and the circuit be under steady state. t = 0 the switch is opened. Solve for the initial and final conditions on U Land UR' Then at TRANSIENTS IN DC CIRCUITS 15.259 407 Determine ducldt(O+) for the circuit conditions stated in Prob. 15.258. I When the switch is open and C is charged dUe = __ dt 15.260 D -i L = ic = i. Then or C The switch in the circuit of Fig. 15-96 is opened at (at t=O+). t == 0. Determine the initial current i and its derivative I Since the capacitor is charged to a voltage V, we have, by KVL, ~ + (RI + R 2)i + V= L di V- (RI (it (0+)= + R 2 )[i(0+)) L bJ (1) i dt VR , =-R L A/s 2 't=o v Fig. 15-96 15.261 If V = RI = R2 = L = C == 1 in the circuit of Prob. 15.260, find I From Prob. 12.260 we have: 2 d i 1 ° = -2 (it (0+) - i(O+) di L dt 2 + (R I + R) 2 (it + 2 d i dt 2 (0+) 15.262 di C I. == d 2i/ dt 2 (0 +). d or 2 = -2(-1) - (1) = 1 A/s The switch in the circuit of Fig. 15-97 is moved from position 1 to 2 at on the three mesh currents. I Since i , H(O_) = i , H(O+) == Thus, i,(O+) == -11 3 ° =- 4 A d dti + 2 (iti + I. = t == 0. ° 2 Determine the initial conditions i,(O+) = i 2(0+) i 2 (0+)=-4A and Fig. 15-97 15.263 In the circuit of Fig. 15-98, the 4-D resistor is shorted out at t = 0. Obtain the initial conditions on the currents ic and i L • Write the general equations governing these currents for t > 0. 408 0 CHAPTER 15 Fig. 15-98 I v (0 ) == c + i~)(6) = 18 V 4+ 6 di L . 4 --- -+ 61 = 30 dt L 5i c 15.264 For the circuit of Prob. 15.263 determine J (1) + ic.it + vc(O+) = (2) 30 diLldt(O), di[/dt(oo), and dicldt(O+). I From Eq. (1) of Prob. 12.263 diL (0 ) = 30 - 6i L(0+) = ~~l...:::- 6 x 3 = 3 A/s dt + 4 4 di (00) = 0 dt --.f:. From Eq. (2) of Prob. 12.263 2.4 ic(O r) die - (0 ) = - - - dt + (5:1(1) 15.265 Write the equations governing voltages. I VI = -5 =- 0.48 Ajs and v 2 in the circuit of Fig: 15-99 and determine the initial conditions on these By KCL at nodes 1 and 2: 0= - J VI dt + 0.5 dV2 v2 + Jv dt Tt + 2' 2 Initial conditions: o.5F Fig. 15-99 15.266 What are the initial conditions on the first derivatives of the voltages VI and V2 in the circuit of Prob. 15.265? I 15.267 Repeat Prob. 15.265 if the lO-A current source in the circuit is replaced by a current source i = e- t A. I The voltage equations are: e - t = -VI 5 + 0= and J J VI V I dt - dt + 0.5 J V 2 dt dV2 v2 + JV dt Tt + 2' 2 ( 1) (2) TRANSIENTS IN DC CIRCUITS D 409 Repeat Prob. 15.266 for the data of Prob. 15.267. 15.268 I From Eq. (1) of Prob. 15.267: -I -e (0+) 1 dVI = 5" Tt (0+) + vJO+) - v 2 (0,) 1 dVI -1 = -5 -dt (0 + ) + 5 - or ° dV - dt2 (0 + ) =0 Thus, In the circuit of Fig. 15-100, find vL(O+), vL(oo), and vc(O+). 15.269 I Since vc(O_) = vdO+), Fig. 15-100 15.270 For the circuit of Fig. 15-101, write a set of integrodifferential equations to solve for v. I By KVL: 100 = 4 dj 1r dt + (2 + l)j + 2 Jo v= j+ j dt 1r 2 Jo j df with the initial conditions: 4H Fig. 15·101 15.271 The natural response of a circuit is given by: 2 -d j dt 2 for which the initial conditions are: I + 4 -di + 4'1= dt and i(0+)=2A ° dildt(0+)=4A/s. The auxiliary equation is p2 + 4p + 4 = (p + 2)2 = And the roots are p = -2. Solve fori. ° The solution is of the form: di= - 2A e -2t - i = Ae- 21 + Bte- 21 dt + B e -2t - 2B te -21 yields ~ dt (0 ) = 4 i and 15.272 4= -2A + B gives + = I For i max • Thus, dildt = °= 4(e- B =8 e- 2/(2 + 8t) A A critically damped circuit has the natural response 10t or i = 4te- IOt A. When does i reach its maximum value? lOte -10/) = 4e -IOt( 1 - lOt) - t = 0.1 s 410 0 15.273 CHAPTER 15 The current in a circuit is given by: d 2i dt 2 The initial conditions are: reaches its maximum value. and i(0+)=2A ,di tit . + :'- + 21 =0 dildt(O+)=lA/s. Determine the time when the current I The characteristic roots are: p =: -2, -1 -di = - 2 Ae -2t - Be t Thus, dt ~ dt Thus, A=-3 B=5 and i(O+) = 2 :rields 2=A+ B (0 ) yidds 1 = -2A - B + 2 A. or or t\ = Thus, 15.274 1 i=5e- t -3e- and di -dt (t \ ) = 0 For i max , = In q) = 0.182 s The current response of a damped RLC-series circuit is shown in Fig. 15-102, and is given by = Ae- at sin (wt + 8). Find: A, a, w, and 8 from the graph. i From Fig. 15-102, i = 0 at t = O. Thus, 8 =O. We construct an envelope to the curve for i, as shown by the dotted line from which A = 4, which is the amplitude of the envelope at t = O. I From the graph, T = 0.8 s 27T = -= ifS = 7.85 rad/s T 27T Thus, W Finally, from the envelope: 1 = 4e- O. Ho t. 4 ' 3 2 1 A .......... V I V o '" -... -- \ ~ I 1- 2 I-- t - 3 ,. 4 t~ -...E~velbpe I---- \. -1 / - }-- V - to-I- f',- 1--- -- - V - -. l'> I-/-' - - ~. t-. ... 't, s -- ~ I· 15.275 a = 1.733 or 1 sec ------~ Fig. 15-102 In the response shown in Fig. 15-103, the construction of the envelope does not yield an exact result. find (without any geometrical construction) A, a, w, and 8. i = Ae -at sin (wt + 8) A, I Since i = 0 at t = 0, 8 = O. i = Ae- at sin wt Thus, From Fig. 15-103, Thus, Finally, T=4ms. Thus, W=27T1T== 1570r/s, A = 0.97 i = 0.97 e and 133 ! a = 133 sin 1570t A If TRANSIENTS IN DC CIRCUITS D 411 Fig. 15-103 15.276 t = 0. The circuit of Fig. 15-104 is under steady state and the switch is opened at current i L • Also, determine its magnitude. di I di The roots are: (0+) = P 2 -diL + -10 dt 5 A/s 6 f i dt=O L = ±j316.22 iL = A cos 316.22t + B sin 316.22t Thus, w = 21Tf= 316.22 i L(0+)=2 ° Find the frequency of the requires A=2 and or f= 50.3 Hz The amplitude is B=O. i L =2A. Fig. 15-104 15.277 A 1-jLF capacitor is charged to 2 V. It is then discharged across an RL-parallel circuit having 6 L = 0.5 X 10 H. Determine the voltage across the RLC-parallel combination. I Let v be the required voltage. R =~ MU and Then: 1 x 10- 6 ~~ + (3 x 1O- 6 )v + (2 x 10- 6 ) f v dt = ° p = -2, -1. The characteristic roots are: Thus, dv (0 +) = 2 x 3 X 10- =6 =_ 2A + B I x 10-6 dt 8 and B =10 and v = lOe- t - 8e- 1t V 6 Thus, 15.278 A =- The response of a circuit excited by a dc source is given by: The initial conditions are i(O+) =0 I The characteristic roots are: and 2 d i di . -+2-+1=10 dt 2 dt di/dt(O+) = -5A/s. p = -1, -1. Solve for i. The solution becomes: i = 10 + Ae-' + Bte-' A=-lO Thus, 15.279 d' -.!=-Ae-'+Be-'-Bte-' dt B=A-5=-15 and di -(0 )=-5=-A+B dt + i=1O-lOe-'-15te-'A Figure 15-105 shows a basic circuit for producing a voltage of sawtooth waveform. When the switch is closed, the gas tube is nonconducting and the voltage across C rises. When it reaches V" the tube fires, allowing the capacitor to discharge to a voltage V2 • Thereafter the cycle repeats. Neglecting the discharge time (with respect to the charging time, find the voltage Vc and determine its period. 412 0 CHAPTER 15 Fig. 15·105 I Since Vc in an RC circuit is given by Vc"" V(l- e- 1IRC (see Prob. 15.131) ) for the given circuit we have (1) When Vc"" VI' let t = T. Then: (V2 -- V) T == RC In (VI -':::-V) 15.280 In the circuit of Fig. 15-105, I RC == 0.01 s, (V - VJ RC In (V _ VI) s = V = 2eO V, VI = 90 V, and V2 = 10 V. Sketch 1 cycle of VC' See Fig. 15-106. I I Q 1:. o 15.281 : /'1; +--I---+--t---II-1:"-'-~-+1-r'"~I''!''' : ~ 5 O.OGS' , Fig. 15·106 From (1) of Prob. 15.279, find dv cl dt(O +) and approximate the sawtooth curve of Fig. 15-106 by a straight line. Draw this line. I The straight line is drawn in Fig. 15-106. From (1) of Prob. 15.280 V- V2 dVe di- ('JJ = RC The equation to the straight line is 15.282 Compare the periods obtained in the exact 15.281. analy~i~ of Prob. 15.279 and the approximate analysis of Prob. I Let these periods be T, and Ta' From Prob. 15.279: Te == Rein V) V_ J (VI For the straight line, Vc = VI at t = Ta or 0 413 In the circuit of Fig. 15-107 we have: V= 240 V, R, = 48 n, R2 = 27 n, L, = 1.6 H, L2 = 0.9 H, All initial conditions are zero. The switch is closed at t = O. Solve for i , and i 2 • and TRANSIENTS IN DC CIRCUITS Fig. 15-107 T, Thus, = Ta In [(V- V2)/(V- VI») (VI- V2)/(V-V2) For the given numerical values: T, = In [(200 - 10)/(200 - 90») (90 - 10) /(200 - 10) 13 . = Ta 15.283 M = 0.72. I By mesh analysis: 1.6 di , Tt . di2 + 481 1 - 0.72 Tt = 240 -0.72 di , Tt . di2 + 2712 + 0.9 Tt = 0 Solving simultaneously for i , and i2 yields i , = 5 - 2.Se -751 15.284 - 2.Se -(75/4)1 A Sketch i , and i2 obtained in Prob. 15.283. I i2 = - 3.33e -751 + 3.33e -(75/4)1 A Also show the components of these currents. See Fig. 15-108. r i, , '" - - -.- . -.---:.,:-:;..;::.-:;:;;000---- o -3 Fig. 15-108 15.285 In the circuit of Prob. 15.283, all circuit parameters are unchanged, except M. Recall from Chap. 11 that the coefficient of coupling k is defined by k = M/V L I L 2 • If k is a variable, solve for the currents i , and i 2 • I Writing the KVL equations in terms of symbols, we have L, di , Tt . + RIll - M di2 Tt = V -M di • di2 Tt1 + Rh + L2 Tt Combining these equations for i I yields ( 1 _ M2) d\ + (RI + R2) di , + R,R2 i = R 2V LIL2 dt 2 L, L2 dt LIL2 I L2 Or, in terms of k and numerical values, Eq. (1) becomes =0 414 0 CHAPTER 15 (1 - e) d\ + hO. -d di + 900i dt t l -2 l = 30 V A similar equation may be obtained for i 2 • The solutions become . = 5 - 2 . 5e -[30/(I+k)]t - 25 i2 = 3.33[e- 30t /(I+k) I1 . e -[30/(I--k)]t .A 15.286 - e- 30t /(I-k)] A Discuss the effect upon the transient terms of variation in values of k in the circuit of Prob. 15.285. , As k ~ 1, one transient term predominates VIi hile the other decays to zero rapidly. time constants of the transients merge to form one t:xponential term. 15.287 , For For 15.288 r k= 0 i l =5-5e-·o A k= 1 =5 - il 2.5e- 15t i2 the k = 1. = 3.33e- 15t A At t = 0, SI is opened and S2 remains open. Write the loop equation for i I ' In the circuit of Fig. 15-109a, S2 has been open and SI closed. Draw an equivalent circuit to represent the initial conditions. (See Fig. 15-109b) V )R, R. L, L-a. ~ 1,\ C. "-) Fig. 15-109 (H In the circuit of Fig. 15-110, ignoring the effect of the rectifier, solve for ic and underdamped. , and k ~ 0, i 2 =OA A , 15.289 k =0 Solve for i l and i2 of the circuit of Prob. 15.285 for the specific values of As Vc> assuming that the circuit is From (3) of Prob. 15.188: . le = /3V (f3: • f3 + a 2) e -at sm t R where 2L 1 vc= C 1 and a=- Jicdt=V ['V 1-· a2 wo= vrc 2 at f3+f3 e- sin(f3t+8) ] 8 = tan- I (f3la) where I V ~~O -1 Lf ... tre ~ re Fig. 15-110 TRANSIENTS IN DC CIRCUITS 15.290 Sketch ic and Vc 0 415 obtained in Prob. 15.289. I See Fig. 15-111. Fig. 15·111 15.291 From Fig. 15-111 show when the rectifier comes into action. comes into play? I The rectifier comes into action when Vc What is the constant value of V c after the rectifier is going through its first maximum. At that time f3t = 1T and ic =0. 15.292 Show that the response of the circuit of Fig. 15-112 can never be oscillatory, regardless of the values of the circuit parameters. I The mesh equations may be written as Rli l + ~ I f i l dt - ~ I f i2 dt = V Eliminating i l from these equations, we obtain the characteristic equation Let its roots be p2+(_I_+_I_+_I_)p+ 1 =0 RICI R2C2 R I R 2CI C2 R 2CI p = H- A ± VB), where 1 1)2 B = ( RI Cl - R 2C2 2 (1 1) ( 1 )2 + R 2C I RI Cl + R 2C2 + R 2CI Because B cannot be negative, the roots of the auxiliary equation can never be complex. response of the circuit can never be oscillatory. v T C 'l. Fig. 15-112 Hence, the 416 15.293 0 CHAPTER 15 Repeat Prob. 15.292 if the capacitors Cl and C2 are replaced by inductors LI and L 2 , respectively. I In this case we have: The characteristic equation for i2 becomes RI RL RI) RIRL p2+ ( - + - - + - p+--=O L2 L2 LI LIL2 B = (RI _ R2)2 LI L2 which can never be negative. 15.294 +~~IR2L + 2(R I )2 + (RI)2 (L 2) LIL2 L2 Hence the response cannot be oscillatory. Determine the instantaneous power supplied by the current source in the circuit of Fig. 15-113 when the switch is opened at t = O. " RIR2) =5 (2X2) vC<0_)=vC<0+)=Is\j~I+R2 2+2 =SV I r=R I C=6s vC<t) = VC! + vc. = 10 + (5 _1O)eThus, Ps = Isv, == SO - 2Se- t16 tI6 = Vs W Fig. 15-113 15.295 Obtain the current i in the circuit of Prob. 15.294. Sketch the current. Verify that the total power supplied by the source, as obtained in Prob. 15.294, is the Saffii! as that absorbed by the resistor under steady state. I Since Vs = Vc 10 - ~;e -t16 -t16 Vc i = -R = ----;.~-- = 5 - 2. Se A I " The sketch for i is shown in Fig. 15-114. From Prob. 15.294, Ps = SO W. i(t) A 5 ----- --------;;;.=0:::.;=------- 4 2 L I I 15 20 25 I I. ...J.---.L.. I -L-L..L I. ...J._ _....... t(s) _ _-jf-L-L-L......L....1.....L..I....I-.l-1.....L.J.......J'--1-1 .J.-L-L..L o 5 10 Fig. 15-114 TRANSIENTS IN DC CIRCUITS 0 417 find iL(t) in the circuit of Fig. 15-115. 15.296 I The initial conditions are The characteristic equation for the circuit under transient condition is p2 + 4p + 2 = having the roots PP P2 = -3.414, -0.586. Thus, iL(O+)=O ° if- = A l e- 3 .414t requires that A I +A 2 =0. diL dt or Thus, Now, (0+) + A 2e --{l.586t A uL (0+)=-2i L (0+)-uC<0+)=-2V=Ldi L ldt(0+) uL(O+) -L- =- -3.414A I - 2 = - 0.5 = -4A/s 0.586A 2 = -4 Solving for A I and A 2 yields Al = 1.414 and A2 = -1.414 iL = 1.414(e-3.414t - e-0586t) A Finally, 20 Fig. 15-115 15.297 L = 1 Hand Repeat Prob. 15.296 for I C = 1 F. In this case the characteristic equation becomes p2 + 2p + 1 = which has the roots PP P2 = -1. ° The solution is iL = Ale- t + A 2 te- t Proceeding as in Prob. 15.296: iL(O+)=O requires that and AI=O iL = -2te- t A. 15.298 Repeat Prob. 15.296 for L = 5 Hand di L ldt(0+)=-2 =A 2 . Hence C = 1 F. I The characteristic equation is The roots are PP P2 = -~ ±n and dt The solution is of the form Applying the initial conditions yields Consequently, 15.299 diL (0+) 2 =-3 and A2 =-1 tI5 iL = _e- sin ~t A Sketch the currents obtained in Probs. 15.296 through 15.298. Verify that these represent, respectively, the overdamped, the critically damped, and the underdamped responses. I See Fig. 15-116. 418 0 CHAPTER 15 I~ (t) A 5 6 ___t---!----:~-"*'"-~--.J.-==!::==~I - - -... t(s) _--7 1.414e-3414, - 1.414e-O 5861 L =+ H. C= 1 F I~ (t) A -5 L -0.1 -0.3 6 .. t(s) -0.5 -0.6 -0.7 L" 1 H. C= 1 F -08 -0.9 -1 <PJ I~ (t) A 0.2 0.1 30 t(s) -0.1 -0.2 -0.3 -0.4 L==SH.C=lF -0.5 -0.6 (~) Fig. 15-116 TRANSIENTS IN DC CIRCUITS 15.300 Determine the current ix in the circuit of Fig. 15-117. solution. 0 419 First state the initial conditions and then obtain the I The initial conditions are iL(O_) = iL(O+) = - ~ A dUe (0 )= ie(O+) dt + C ( since ie = C dUe) Tt The characteristic equation is The roots are Thus, Applying the above-stated initial conditions yields A2 and _ 14 - Therefore, I' x and = -3- U - . " 1 +4'e-211'5 3 e - sl'n 4 tA 15 20 Fig. 15-117 15.301 Sketch ix(t) of Prob. 15.300 and verify that the circuit is underdamped. I Because the roots of the characteristic equation are a pair of complex conjugate numbers, the circuit is underdamped. The sketch shown in Fig. 15-118 also validates this fact. A -0.7 ---1--0.75 -0.8 -0.85 -0.9 .l.. 4 8-121/15) sin (~t) 15 - 1 -_-1-+--...1--..I..\--L---Jy <:..J.--""--3.L.----<. t(s) 5 -1.05 15.302 Fig. 15-118 Determine i(t) in the circuit of Fig. 15-119. I Thus, r=LlR=lOms i = 3- if = 2 + 1 =3 A e- 1001 A in = (2 - 3)e- '001 A 420 D CHAPTER 15 Fig. 15-119 15.303 Solve for vet) in the circuit of Fig. 15-120. I v(0+)=1O+5=15V v(oo) = 5 + HlO) = 7.5 V Hence, by inspection, vet) = (15 _7.5)e- 1 kO 1 kCl 100Ot + 7.5 = 7.5(1 + e -1000/) V -~ _.t... 1 pF 1=0 2 kO v(/) Fig. 15-120 15.304 In the circuit of Fig. 15-121a, the switch is initially closed at t = 0, opens and remains open for 0 < t < lO ms. At t = 10 ms it closes once again and remains closed for 10 ms < t < 20 ms. At t = 20 ms it opens again and closes at t = 30 ms. Sketch vet) for I) < t < 30 ms. I v(oo) = 0 r= RC=5ms vet) = lOe -20Cl O::s t::s 10 ms v(lO ms) = lOe- 2 = 1.35 V vet) = 10 + (1.35 - lO)e -(/-10)/5 rn. = '-8.65e -(/-10 rn.)/5 rn. + lO v(20 ms) = --S.65e -2 lO ::s t ::s 20 ms + 10 = 8.83 V vet) = 8.83e -(/-20 rn;) '5 rn. 20 ::s t ::s 30 ms v(30 ms) = 1.2 V Hence the sketch of Fig. 15-121b. 10 00 pF 500 +-------4------ (a..) (b) Fig. 15-121 15.305 In the circuit of Fig. 15-122, determine vet). I iL(O-)=O iL(oo)=~+2='¥A di i L(t)=-¥e- I / 2 +¥ vL(t) = vet) = L d: = ¥e- 2 I / V -~---- ~ t ,,.,s TRANSIENTS IN DC CIRCUITS D 421 40 + 40 2A 20 v(t) Fig. 15·122 15.306 Solve for i(t) in the circuit of Fig, 15·123, State all initial and final conditions, I i L(0 _) = i L(0 + ) = i(O + ) vc(oo) = 30 x 5 = 150 V =5 A and The roots of the characteristic equation are found to be (-1, -9), i(t) = iL(t) = Ale- t + A 2e- 9t + 5 A Thus, Applying the initial conditions gives and A2= ~ 3 Thus, 300 5A t 3H Fig. 15·123 15.307 Write a set of differential equations to solve for the node voltage v in the circuit of Fig, 15·124, I The required equations are v RI + C dv dt 12, , + I1 , = 1 c Fig. 15-124 15.308 In the circuit of Fig, 15·124, we have LI = L2 = 1 H, RI = 0,5 n, R2 i = 2 A, Obtain a differential equation to solve for the node voltage v, I = 2 n, C = 1 F, M = 0,5 H, and Substituting the numerical values yields dv , = 2 2 V+-+I dt I (1) di l di2 dt + 0,5 Tt = v (2) 422 D CHAPTER 15 di di2 dt dt 0.5 --1 -- - . + 21 2 = 0 (3) Multiplying Eq. (3) by 2 and differentiating with re,pect to t gives 2 d i1 dt 2 Substituting di2 d: =2v-2-1 dt dt - d\ di2 =0 dt + 2 --- + 4 d,,2 (4) and from Eq. (2) in Eq. (4) results in d\ di - 3 - 2 - 8 ---1 dt dt Substituting from Eq. (1) in Eq. (5) finally yields dv + 4 - + 8v = 0 and dt (5) /7 CHAPTER 16 LlStep, Ramp, and Impulse Functions 16.1 In Chap. 15, we considered the responses of circuits to suddenly applied sources by opening and/or closing of switches. Thus, a suddenly applied 1-V dc source may be represented by the graph of Fig. 16-1a. We term this a unit-step function, and denote it by u(t). Sketch the function v = u(t) - u(t - 4). I The steps are shown in Fig. 16-1b, which also shows the final result. Notice that v is a square wave of 4 s duration. l.( f':: (f) [ ~f, S let) 'IT, V t/ tv-, v I r _ u(f-4) "Ill -v.{t-+) ,. t/ I 1- "" S S 6 ( J,) 16.2 Fig. 16-1 From Prob. 16.1, we now have a technique to represent sources having finite durations. Sketch the following functions: (a) VI (t) = lOu(t) V; (b) vz(t) = (115V2 cos 377t)u(t) V; (c) v 3 (t) 20e-'u(t) V. = I See Fig. 16-2a to c. Notice that the functions are of the form v(t) = f(t)u(t). , 1 (t,I 10i------ (a) (b) 20 (c) Fig. 16-2 423 424 16.3 D CHAPTER 16 Sketch a unit-step function which begins at , t = T. This is a delayed unit step and is shown in Fig. :.6-3. ~t T Fig. 16-3 16.4 Make a table showing the sketches of the following Illlit-step functions: u(t), - u(t), u(t + t'), - u(t + t'), u(t- t'), -u(t- t'), u(-t), -u(-t), u(-t+ t') -u(-t+ t'), u(-t- t'), and -u(-t- t'). , The sketches are shown in Fig. 16-4. -u(/) U(/) o -u(1 U(I + I') + 1') -,' -I 0 -u(1 - 1') u(1 - 1') u(-/) -U(-/) ~I 0 u(-I + 1') ~,. ~-I -U(-I + 1') 0 u(-I - 1') WA-,' -] -1 -u(-I - 1') -,' ] 1$1 __ I 0 Fig. 16-4 sov Fig. 16-5 STEP, RAMP, AND IMPULSE FUNCTIONS 16.5 D 425 Evaluate v=2Su(-t-S)-50u(t)+SOu(t-1O) at (a) t=-6s, (b) -3s, (c) 3s, and (d) 11s. I See Fig. 16-S. First, each term is sketched separately, and then combined to obtain the following results: (a) 2SV, 16.6 (b) 0, (c) -SOY, (d) O. Obtain the current i, for all values of t, in the circuit of Fig. 16-6. I For lO-n t < 0, resistors: the voltage source is a short circuit and the current source shares 2 A equally between the two For t> 0, the current source is replaced by an open circuit and the SO-V source acts in the RL series circuit (R=20n). Consequently, as t~oo, i~-SO/20=-2.SA. Then, i(t) = [i(O +) - i(oo)]e -RilL + i(oo) = 3.Se -lOO, - 2.S A By means of unit-step functions, the two formulas may be combined into a single formula valid for all t: i(t) = 1u( -t) + (3.Se- IOO ' - 2.S)u(t) A (V) SOU(/) 100 lOO I i ~ 2u(-/) (A) 0.2 H Fig. 16-6 16.7 Determine vo(t) in the circuit of Fig. 16-7 when v(t) = 0.2Su(t). I By nodal analysis: Vo 1 dvo -6v I + 10 + 20 Tt = 0 v = 6i) and The constraints are VI = 4i l • Thus (1) becomes Vo 10 For t < 0, V = O. For t> 0, (1) 1 dvo + 20 Tt (2) =4v v = 0.2S, and (2) becomes dvo Tt + 2vo = 20 The solution for all t becomes 2n .-----~. + v(t) ' , A - .... '1 4n +1 +1 V 1 1 I -io F 6vI I vo(/} Ion Fig. 16-7 426 16.8 0 CHAPTER 16 Repeat Prob. 16.7 if I v(t) = O.25u(t - 1). t < 1. v = 0 for In this case Other conditions remain the same. Vo 16.9 = 10[1-- e -Z(f-I)]U(t - 1) Sketch. and compare, the responses obtained in I Hence the solution becomes Pro')~ .. 16.7 and 16.8. See Fig. 16-8. \ C (f ) 10 5 o o 0.55 1.05 1.55 2.0s 2.5s 2.0 2.5 fo} 10 5 o o 0.5 1.0 1.5 3.0 B Fig. 16-8 fh} 16.10 A series RC circuit, with R = 5 k!1 VI = and C = 20 25u(-t) V J-~F, has two voltage sources in series, V2 = 25u(t - t') V Obtain the complete expression for the voltage acro:;s the capacitor and make a sketch, if t' is positive. I The capacitor voltage is continuous. For t 5 0, VI results in a capacitor voltage of 25 V. both sources are zero, so that Vc decays exponentiall) from 25 V toward zero: Vc = 25e -fIRC = 25e -IOf V (05t5t') IOt In particular, vc(t') = 25e- ' V. For t?:. tf, Vc builds from VC(tf) toward the fimll value 25 V established by vz: Vc = [VC(tf) - vc(oo)]e-(f-f')IRC + v,,(ex) = 25[1- (e IOt ' -1)e- lOf ] V (t?:. t') Thus, for all t, Vc 16.11 = 25u(-t) + 25e- 10t [u(t) - u(t - t')] + 25[1- (ClOt' - 1)e- 10t ]u(t - t') Sketch the voltage obtained across the capacitor of Prob. 16.10. I See Fig. 16-9. V For 05 t 5 tf, STEP, RAMP, AND IMPULSE FUNCTIONS 0 427 25V o 16.12 " Fig. 16·9 The circuit of Fig. 16·7 is excited by the pulse shown in Fig. 16·10. I Determine vo(t). As determined in Prob. 16.7, for Vo we have Vo 1 dv o 1O+20Tt=v And v = 0.25[u(t) - u(t -1)] The net response may be considered to consist of two parts: V and Thus, 01 (t) = 2f 10(1- e- )u(t) V02 (t) = -10[1 - e -2(f-l)]U(t - 1) Vo = 10(1 - e -2f)U(t) - 10[1 - e -2(t-J )]u(t - 1) ",I' O.25V~--., o 1s Fig. 16·10 16.13 Sketch the response of the circuit of Fig. 16·7 to the signal of Fig. 16-10. I 16.14 The response is given by Vo of Prob. 16.12, which is sketched in Fig. 16-11. Represent the train of pulses shown in Fig. 16-12 in terms of Vo and unit steps. I In this case, each pulse may be expressed by a step and a delayed step. Hence, v(t) = Vou(t) - Vou(t - T) + Vou(t - 3 T) - Vou(t - 4 T) + Vou(t - 5 T) - Vou(t - 6 T) 16.15 If we integrate the unit-step function, we obtain a unit-ramp function. and sketch a unit ramp, r(t). I and r(t) = -d [r(t)] = u(t) = dt The unit ramp is sketched in Fig. 16-13. L' u(t) dt {O 1 -oo<t<O O::5t<oo Notice that r(t) = { ~ -oo<t<O O::5t<oo Express this statement mathematically 428 0 CHAPTER 16 10 5 ----~-----+----~------~------~I--------~.+~ 3 4 ' -5 -10 (0) V- 10 8.65 5 ----~-----+----~2--~~3~----~4----------t,~ Fig. 16.11 (b) v:fl Vo- T 2T 3T 4T 5T 6T Fig. 16·12 r(f) t /- + ~_ _ 1:, s Fig. 16·13 STEP, RAMP, AND IMPULSE FUNCTIONS 16.16 Sketch the following ramp functions: kr(t - 1) and D 429 kr(t + 1). I The sketches are shown in Fig. 16-14. Fig. 16-14 16.17 Represent the voltage waveform of Fig. 16-15 in terms of ramp functions. I We obtain the desired result by adding and subtracting ramps (see also Probs. 16.1 and 16.5). Thus, v(t) == r(t) - r(t - 1) - r(t - 1) == r(t) - 2r(t - 1) Fig. 16-15 16.18 A current of the waveform shown in Fig. 16-16 flows through a 2-mH inductor. voltage across the inductor. di v = LL dt In terms of unit ramps, i may be written as i(t) = r(t) - 2r(t -1) + r(t - 2) Thus, vL(t) = (2 x 10- 3 ) ~ [r(t) - 2r(t - 1) + r(t - 2)] == 2u(t) - 4u(t - 1) + 2u(t - 2) mY Fig. 16-16 16.19 Sketch the voltage across the inductor of Prob. 16.18. I See Fig. 16-17. 11." .... /-. v Fig. 16-17 Determine the corresponding 430 16.20 D CHAPTER 16 Figure 16-18 shows a voltage waveform. • Express i'l terms of unit steps and unit ramps. The given voltage may be expressed as udt) = u(t) - ~rU) + ~r(t - 4) + u(t - 4) Fig. 16-18 16.21 The voltage of Fig. 16-18 is applied across a 2-fLF capacitor. • ic =C du C dt =2[8(t) 16.22 d = C dt ~u(t) + (u(t) - ~r(t) ~u(t -- What is the current through the capacitor? + ~r(t - 4) + u(t - 4)] 4) + 8(t - 4)] (1) In (1) of Prob. 16.21, we have symbolically denoted the derivative d[u(t)]/dt by 8(t), which is known as the unit-impulse or unit-delta function. Sketch the two terms involving these functions in (1) of Prob. 16.21. • See Fig. 16-19. 1 '1. ~ S; [+) -Lt, o 1. o ' 0:-4-) ..,. t,' Fig. 16·19 v( I} , I'll p.25V~--------~ 1.0V - o o 0.025 5 (a) v( t) 0.5 V t - - - -.. o o o O.ol25 s o (b) 0.00625 5 (c) Fig. 16-20 STEP, RAMP, AND IMPULSE FUNCTIONS 16.23 See Fig. 16-20. From Fig. 16-20c it follows that the pulse tends to the impulse. Determine the current through a 2-mH inductor across which the voltage has the waveform shown in Fig. 16-21. I v = 20u(t) - 30u(t - 1) + 10u(t - 2) Thus, i = L1 It _~ v dt = 210 3 It -00 [20u(t) - 30u(t - 1) + lOu(t - 2)] dt lO3 = 2 = [lOr(t) -lSr(t -1) + Sr(t - 2)] x lO3 [20r(t) - 30r(t - 1) + 10r(t - 2)] A lo 10 Fig. 16·21 16.25 431 Sketch the following pulses of volt ages: v(t) = 0.2Su(t) - 0.2Su(t - 0.025); v(t) = O.SOu(t) - O.SOu(t - 0.012S); v(t) = 1.00u(t) - l.OOu(t - 0.00625). Observe that as the period gets shorter, the pulse tends to the impulse function. I 16.24 0 Sketch the current i of Prob. 16.24. • See Fig. 16-22, which shows all the steps of adding (and subtracting) the ramps. l()r[t) l' [- ,.;- -r-(t-r ')J - - forlt)- /.::;r-(t- I ) 10 10 5 '5 t- t 1- 1.67 t- -s _10 I') 10 10 rH)- 10; r{i- I) , + 10r(l"-2-) :;./.; 5" '5 ~ 2. Fig. 16·22 t '2.. 3 4-t CHAPTER 17 ~ Duals and Analogs~ 17.1 Write the loop equation for the circuit of Fig. 17-1a, a'ld the node equation for the circuit of Fig. 17-1b. Verify that the two equations are identical, and thus their solutions have the same form. The networks are duals of each other. • For the circuit of Fig. 17-1a: (1) (2) L, R, (a.) 17.2 From (1) and (2) of Prob. 17.1, it follows that the roles of voltage and current have been interchanged. a table to show the pairs of dual quantities. • Resistance, R Inductance, L i dt Loop Short circuit Conductance, G Capacitance, C J v dt Node Open circuit The following procedure is followed, in general, to obtain the dual of a network. 1. Place a dot within each loop of the given network. These dots correspond to independent nodes. Number these dots (and the respective nodes). Place a node external to the network. This is the datum node. 2. Connect all internal dots in adjacent loops, traversing only one element at a time. For each element traversed in the original network, connect the dual element on the dual network. Continue this process until the number of possible paths through single elements is exhausted. 3. Join all internal dots to the external dot crossing all external branches. Using this procedure, obtain the dual of the network shown in Fig. 17-2a. • The procedure is shown in Fig. 17-2a, and the dual in Fig. 17-2b. 1. ,'_ - --., R, iV\t--,---{!-;;:-----. , " ' . . . ___--""'0 - --~- (4) 432 Make TABLE 17.1 Pairs of duals J 17.3 Fig. 17-1 (b) o (h) Fig. 17-2 DUALS AND ANALOGS 17.4 D 433 Repeat Prob. 17.3 for the circuit of Fig. 17-3a . • See Fig. 17-3b. -- -" ,- ! (' t I- - " o \ 17.5 ......,. '" \,. / - - - - - -yo- - - - - - _----, " ... I ,/ (6) / Fig. 17.3 Draw the dual of the network shown in Fig. 17-4a. • The dual is shown in Fig. 17-4b. /.. ,I ,i ' ' .. - - -, , \ ..... , ,1 - ti,l I I \.. -[2.. - - _. - ... (D F ~ 6 +'7-0 I \ I "j '~-- --.. ---.1:=-= =:= __ _J/ o ) o Fig. 17-4 17.6 Using the dual shown in Fig. 17-4b, solve for the current in the 1/l4-ilresistorofthecircuitofFig.17-4a. '",' +' =~+A -(14+1D)r/6",1+A -4r If In 10 + 14 e e I i(O,) == ¥o = 2.4 A. Thus, A == 2.4 - 1", 1.4. i=1+1.4e- 4r Finally, 17.7 A Verify that the result of Prob. 17.6 can be obtained by using the original circuit of Fig. 17-4a. • In the dual the current corresponds to the voltage. or V. Thus, by nodal analysis: Thus, the problem reduces to the determination of dv 6 dt + 24v '" 24 With the initial condition /' v(O +) = v(O _) == 2.4 V " '\ the solution becomes L v = 1 + 1.4e -4r V. 1 / / I 1/ \ " r ~ \ \ ,\,,-, ----,- ....... 0 (a.) -" I. -R, I I 1 C. .3 " (6) Fig. 17-5 V 6F 434 17.8 0 CHAPTER 17 Find the dual of the network shown in Fig. 17-5a. • 17.9 See Fig. 17-5b. Repeat Prob. 17.8 for the network of Fig. 17-6a. • See Fig. 17-6b. o-" (a.) 17.10 .... - o (h) Fig. 17·6 Repeat Prob. 17.8 for the network of Fig. 17-7a • See Fig. 17-7b. -.....1<., ", L I r I I \ I \ "- f _ I r~' I -"'2 {J.", \ \ c '2.. L 0 \, '- ....... - -?-- 0---- - - - - - - (h) (a.) 17.11 Fig. 17·7 Figure 17-8 shows a purely mechanical system. As:;uming that all forces are linear, apply Newton's law to write the equation of motion of the system, where F is an externally applied force. • By Newton's law F = Fma,,~ + Ffriction + Fspring d"x F = M --dt 2 or +b dx dt + kx (1) Fig. 17·8 17.12 Compare (1) of Prob. 17.11 with (1) of Prob. 17.1 and make a table of force-voltage analogy. • In terms of electric charge q and dropping the subscripts, we may rewrite (1) of Prob. 17.1 as 2 d q dq L --- R dt 2 dt Hence, we obtain the Table 17.2. 1 +- q= v C (1) DUALS AND ANALOGS D 435 TABLE 17.2 Force.voltage analogy Force, F Velocity, x Damping, b Mass, M Spring constant, k 17.13 Voltage, v Current, i Resistance, R Inductance, L Elastance = reciprocal of capacitance, 11 C Compare (1) of Prob. 17.11 with (2) of Prob. 17.1 and make a table of force-current analogy. I On the basis of duality, if force is taken to be analogous to current, the Table 17.3 is obtained. TABLE 17.3 Force·current analogy Force, F Velocity, x Damping, b Mass, M Spring constant, k 17.14 Current, i Voltage, v Conductance, G Capacitance, C Reciprocal of inductance, 11 L Draw the circuits for the analogs obtained in Probs. 17.12 and 17.13. I These circuits are, respectively, shown in Figs. 17-9a and b. lA) c Fig. 17-9 17.15 A construction procedure for drawing electrical analogs for mechanical systems is as follows: 1. 2. 3. 4. Denote all masses by capacitors to ground. Node volt ages on capacitors correspond to velocities. Draw current sources to denote forces. Connect all other components between the nodes in an arrangement which parallels the connections between the masses in the mechanical system. Using the above procedure, draw an electrical analog for the system shown in Fig. 17-lOa. I By inspection, using the above procedure, we obtain the circuit of Fig. 17-lOb. 436 D CHAPTER 17 Fig. 17·10 17.16 Obtain a circuit on the basis of force-voltage analogy for the system of Fig. 17-lOa. I The circuit of Fig. 17-11 a shows the force-current analogy (just obtained in Prob. 17.15). We draw the dual network in Fig. 17-11b to obtain the force-voltage analog. I ( I I ( I I I I \ \ \ \.- '- - - - - - - -~ - - - -- I - 0 Fig. 17·11 17.17 Draw an electrical analog for the mechanical system shown in Fig. 17-12a. I See Fig. 17-12b. Ca,) (b) Fig. 17·12 17.18 Repeat Prob. 17.17 for the system shown in Fig. 17·13a. I See Fig. 17.19 Repeat Prob. 17.17 for the system shown in Fig. 17-14a. I See Fig. 17.20 17-13b. 17-14b. A wheel mounted on bearings has a lO-kg· m 2 inertia and a bearing friction of 0.05 N . mlrad/s. At t = 0, a lO-N . m torque is applied to the wheel, starting from rest. Write its equation of motion and solve for its speed. I The equation of motion is dUI 10 Thus, -;k + 0.05w = 10 w = 200(1- e-0005T) rad/s DUALS AND ANALOGS Fixed frame (a.) Rn R24 LI2 (b) Fig. 17.13 (a) (b) 1++ M g • c .... M. L ++ Fig. 17·14 k. R ++ b D 437 438 17.21 0 CHAPTER 17 Draw an electrical analog for the wheel of Prob. 17.:W. I See Fig. 17-15. Fig. 17-15 17.22 How much energy is stored in the wheel of Prob. 17.20 under steady state? I 17.23 From Fig. 17-15: After the wheel of Prob. 17.20 has a steady speed, the torque is removed. wheel make before it comes to rest? I The speed corresponds to the discharge of the st'Jred energy and is given by w=200e- O radls (}('SI or rad or 17.24 8'leady 'Iale = (4 x 10') rad = 6360 revolutions Draw a torque-current analog for the wheel of Prob. 17.20. nodal equation. I How many revolutions will the Label all parameters numerically, and write the See Fig. 17-16. dv v 10 --'- - = 10 dt 20 where v corresponds to w. loA Fig. 17-16 17.25 The equation of motion of a mechanical system is (Fig. 17-17a) 2 d x2 M dt2= k(x J Obtain an electrical analog for the system if M =. 1 kg, on the spring is moved to the right at a speed 2 ml >. I Let x2 ) - - b dX 2 Tt k = 4.01 N Im, or Thus, (1) becomes di2 + 0.2i2 + 4.01 Tt which is represented by the circuit of Fig. 17-17b. f (i2 - 2) dt = 0 (1) b = 0.2 N Im Is. At t = 0, point 1 DUALS AND ANALOGS i 0 439 ~/b 27:?77T (a) Fig. 17-17 (b) 17.26 Draw a voltage analog for the system of Prob. 17.25. I See Fig. 17-18, obtained from duality. Fig. 17·18 17.27 Determine the instantaneous compression of the spring of the system of Prob. 17.25. I The instantaneous compression is x 2 • obtain Substituting the numerical values and rewriting (1) of Prob. 17.25 we The solution is x 2 = 2t - 0.09975e- 01t sin (2t + 174.3°) m which is the required compression. 17.28 After the system of Prob. 17.25 has reached steady state, the spring is detached at point 2. the equation of subsequent motion. I At The equation is t = 0, i z = 2 m/so Thus, i2 =2e- O. 2t 17.29 How far does the mass in Prob. 17.28 travel before it comes to rest? I 17.30 m/s X 2 = f i2 dt = 10(1 - e- 02 ,) or m (X 2 )steadY state = 10 m The equation of motion of a mechanical system is 10 ~~ + 120x + 360 f x dt = 98.1 Represent the system by an electrical analog. I See Fig. 17-19. Fig. 17-19 Obtain and solve 440 17.31 0 CHAPTER 17 The dynamics of an electromechanical system is given by Obtain a corresponding set of analog equations involving only electrical quantities. I Let {38 1 = v2 • Then we have where 17.32 Draw an electrical analog for the system of Prob. 17.31. I See Fig. 17-20. + + Fig. 17-20 17.33 The electromechanical dynamics of a system is given by Express these equations in terms of purely electric<ll quantities. I 17.34 Let XI = i2 and 41 = i l. Thus: Represent the system of Prob. 17.33 by an electrical analog. I See Fig. 17-21, where Co = l/S o and Cm~' I If'. b Fig. 17·21 17.35 Under what conditions is the circuit of Fig. 17-21 physically realizable. I Since the capacitors cannot assume negative values, we must have and ~ransients in AC Circuits CHAPTER 18 18.1 A 60-Hz 2400-V (rms) generator is connected in series with a switch and a load whose resistance and inductance are 0.24 nand 0.030 H, respectively. Assume the impedance of the generator is negligible. If the switch is closed at the instant the voltage wave passes through 0 V rising in the positive direction, determine the current. , By KVL: 0.03 - V * + 0.24i = 2400v'2 sin 377t Z= If = 2400 L!r 0.24 + j(377)(0.03) = 212.14/-88.8° Thus, i = Ae -Sf + 300 sin (377t - 88.8°). A = 299.93. Thus, 8f 18.2 A n The initial condition requires that i = 299.93e- [Note: = Ae-(0.24/0.03)f = Ae- Sf i if = 212.14v'2 sin (377t - 88.8°) A + 300 sin (377t - 88.8°) 0 = A + 300 sin (-88.8°), A In problems of this chapter, phasors are indicated by overbars rather than by boldface type.] Determine the current 0.1 s after the switch is closed in the circuit of Prob. 18.1. steady-state current. , From Prob. 18.1, at i(oo) ~ = A 212.14 From Prob. 18.1 i = 0.24 + j11.31 = 11.313/88.8° n - 2400 L2!r If = 11.313/88.8° = 212.4/1.22° i = Ae -Sf + 300 sin (377 t + 1.22°) Thus, A in = Ae- 8f A A i = 0 at t = 0 yields A = -6.39 8f i = -6.3ge- + 300 sin (377t + 1.22°) A or Repeat Prob. 18.2 for the data of Prob. 18.3. , From Prob. 18.3 at t = 0.1 s i(O.l s) = -6.3ge -(8)(0 I) + [sin (377 x 0.1 + i(oo) = 212.14 A 18.5 = A Repeat Prob. 18.1, assuming that the switch is closed at the instant the voltage wave is at its positive maximum. , 18.4 Also calculate the t = 0.1 s i(O.l s) = 299.93e-(8)(Ol) + 300 sin (377 x 0.1 - 88/8) = -165.17 5 3 18.3 or ~;~~) ]300 = 3.78 A (same as in Prob. 18.2) Set up the differential equation and the initial conditions for the circuit of Fig. 18-1a excited by the voltage wave of Fig. 18-1b. , Thus From Fig. 18-1b, 2 T=2XO.05=0.ls, * j=1IT=1I0.1=lOHz, ege. = 500 sin (62.83t + 150°) + 180i = 500 sin (62.83t + 150°) and w=21T(1O)=62.83rad/s. V with 441 442 D CHAPTER 18 L-2H R - 180 fl (a) Fig. 18.1 (b) 18.6 Solve for i in the circuit of Fig. 18-1 a. - I If V = l~ = 1.610 sin (62.83t i(O+)=O, A=-2.28sin115°=-2.07. i = -2.07e- 18.7 901 = Ae- 901 Hence, + 2.28 sin (62.83t + 115°) . = - 2.07 e -90xO.O! + 2.28 SIn • (' 62.83 x om + t = om s, A + 2.28 sin (62.83t + 115°) How much energy is stored in the coil of the circuit of Fig. 18-1a at I At A + 115") = 2.28 sin (62.83t + 115°) i or Since (500/V2) /150° ° lH(] + j2(62.83) = 1.61 ~ Z= I A t = 0.01 s? l1Y) 57.3° = 0.26 A WcOi ! = !L(i)2,~ !(2)(0.26)2 =67.6mJ 18.8 A 450-V 60-Hz generator supplies a series circuit consisting of a switch, a 40-.n capacitive reactance, and a coil whose inductive reactance and resistance are 50 nand 2.0.n, respectively. The switch is closed 20° after the voltage wave passes through zero rising in the positive direction. (a) Sketch the circuit; (b) sketch the voltage wave and indicate time-zero; (c) determine the steady-state current. I (a) See Fig, 18-2a, (c) i = 2 + j(50 - 40) = 10.2 /78.7°.n (b) See Fig. 18-2b. y' =450 /20° - V If = V Z= 450 /20° 10.2 m2' = 44.12 / -58.7° A (a..,) (b) Fig. 18·2 TRANSIENTS IN AC CIRCUITS 18.9 0 443 Determine the instantaneous current in the circuit of Prob. 18.8. 50 L = 377 = 0.133 H I 1 C = 377 x 40 = 66.31 JLF The current is governed by 0.133 The solution is i = if + in' Since ~ + 2i + 6~~;1 J i dt = 450v'2 sin (377t + 20°) If = 44.12/- 58. 7° A if = 44.12v'2 sin (377t - 58. 7 °) (1) A The characteristic equation of the circuit is 6 2 10 0.133p + 2p + 66.31 = 0 and the roots are PI' P2 = -7.5 ± j336.8. Thus, (2) I Combining (1) and (2) to obtain i, and applying the initial condition i(O _) = i(O+) = 0, yields Al = 53.3. TodetermineA 2 weuse vc(O_)=vc(O+)=O and di(O+)ldt= 450y2sin 20 % .133:0 1636.55. After considerable manipulation we obtain A2 = -30.25. i = 62.4 sin (377t - 58.7°) Hence, + e- 751 (53.3 cos 336.8t - 30.25 sin 336.8t) IS.10 A (3) How much average power is absorbed by the circuit of Prob. 18.9? I To determine the average power, we consider only the steady-state terms in the current and voltage at the terminals of the given circuit. Thus, p == WIIII cos cp == (450)(44.12) cos [20 - p = [2R = (44.12)\2) == 3893.1 W Otherwise, 18.11 How much energy is stored in the inductor of the circuit of Prob. 18.9 at I At t == 0.01 s from (3) of Prob. 18.9, we obtain t = 0.01 s? i = -17.85 A. WL = ~L(i)2 = HO.133)( -17.85)2 = 21.19 J Thus, 18.12 (-58.7)] == 3890.3 W Determine the value of the capacitance that should be connected in parallel with the original capacitor, or in series with the original capacitor, so that the circuit of Prob. 18.9 is under resonance. 1 I fr 1 == 21TVLC =60= 21Tv'0.133C or C= 52.9 JLF Therefore, a series-connected capacitor is required to bring the equivalent capacitance down to the required value. For a series connection we have or 18.13 Calculate the power dissipated in the circuit of Prob. 18.9 under resonance. I At resonance, Thus, 18.14 i. =R. V R 450 2 /=-=-=225A The voltage across a series RL circuit is v = 150 sin (500t + 0.785) V. Given: R = 50.n and L = 0.2 H. Find the voltage across the resistor at t = 0.002 s. Use the method of undetermined coefficients. I The circuit equation for t > 0 is di dt + 250i = 750 sin (500t + 0.785) (1) 444 0 CHAPTER 18 The solution is in two parts, the complementary function (iJ and the particular solution (ip)' so that i = ic + ip • The complementary function is the general solution of (1) when the right-hand side is replaced by zero: ic = ke- 250r• The method of undetermined coefficients for obtaining ip consists in assuming that ip = A cos SOOt + B sin SOOt since the right-hand side of (1) can also be expressed as a linear combination of these two functions. di d; = -500A sin SOOt + 500B cos SOOt Then Substituting these expressions for ip and dV dt into (l) and expanding the right-hand side, -5OOA sin SOOt + 500B cos SOOt + 250A cos SOOt + 250B sin SOOt = 530.3 cos SOOt + 530.3 sin SOOt Now equating the coefficients of like terms, -500A + 250B = 530.3 Solving these simultaneous equations, 500B + 250A and A = -0.4243 A., = 530.3 B == 1.273 A. ip = -0.4243 cos SOOt + 1.273 sin 500t = 1.342 sin (SOOt - 0.322) i == ic + ip = ke- and At t = 0, i = O. t = 0.002 s, 250r + 1.:142 sin (SOOt - 0.322) = 0.425 A, A A and, finally, + 1.342 sin (SOOt - 0.322) A i = 1.1 A. Thus, 18.15 k Applying this condition, i = 0.425e- At 250r VR In the circuit of Fig. 18-3, i(O + ) =(50)(1.1) =55V = i(O _) = 0.5 A. Determine i for t > O. I The voltage equation is i = 100 + jlOO = loov'2/4SO n . = If. + In. I i(O+) = 0.5 implies that 0.5 == -0.5 + A. - = (l00/V2) !.Sf.. = ~ /-45° A If (loov'2) /45° 2 1 Sill . ( 10'. t - 450) + A e -IOOOr = v'2 Thus., A =1 A and '1. (00 ~/o ~ t V 18.16 Sketch the current obtained in Prob. 18.15. I See Fig. 18-4. J 0.' H Fig. 18-3 TRANSIENTS IN AC CIRCUITS D 445 Complete solution I t,s /"Steady state Fig. 18-4 18.17 It may be shown that the current i in the circuit of Fig. 18-5a is given by: i= ,/ 2 Vm yR +(l/wC) 2 . ) [VO + Vm cos (8 + f3) sin 13] -IIRC sm (wt + 8 + 13 R e (1) where 13 = tan- t (l/wCR) and Vo is the initial voltage on the capacitor. For the following numerical values, sketch the current: Vm =lOV, Vo =-5V3V, R=5!l, C=30.63j.tF, 8=60°, and w=377rad/s. t 103 ) 13 = tan- ( 377 x 30.63 x 5 = 60° , The current is sketched in Fig. 18-5b. R i, A Fig. 18·5 18.18 In the circuit of Prob. 18.17, it is feasible to close the switch at an instant so that no transient occurs. Determine the condition for which there will be no transient in the circuit. 446 0 CHAPTER 18 , For no transient to occur, the coefficient of the transient term in (1) of Prob. 18.17 must be zero. the required condition is Hence, ) o + f3 - "21T = sm• -t( Vm ~} sin f3 18.19 Repeat Prob. 18.18 by determining the voltage acroi;s the capacitor of the circuit of Fig. 18-5a. I Considering only the transient term in (1): Vet< = At (= 0, 1 C I . f3 ) sm . ( w ( + f3 + () (r d ( = (Vm sm Vo = (Vm sin (3) sm (() + f3 - -"21T) + k e -·IIRC I) + k The value of k will be zero if ~) = (Vm sin In sin (() + f3 - I) which is the same condition as derived in Prob. 18.1.8. 18.20 Sketch the voltages v and transients. I Vc of the circuit of Fig. V~-5a. From the sketch, find the condition on Vo to avoid The voltages are sketched in Fig. 18-6 from which it may be seen that the transient can be avoided only if Wol < Vm sin f3. Fig. 18·6 18.21 The circuit of Fig. 18-3 is under steady state with the switch in position 2. The switch is thrown to position 1 at the instant when the current is going through a positive maximum. Sketch the current for « 0 and for (>0. I See Fig. 18-7. Fig. 18·7 18.22 Refer to Prob. 18.21. If the voltage of the source continues independent of the switch position, are there any subsequent instants of time when the switch could bl! thrown back to position 2 without causing any transients? TRANSIENTS IN AC CIRCUITS 0 447 , Yes. The switch could be thrown back to position 1 without causing a transient at any instant when the actual current has the same value as the continuation of the original function. The first such moment [determined graphically (see Fig. IS-7) or by trial-and-error] is t = 1.295 ms. R = 10 n, In the circuit of Fig. IS-5a, 18.23 , ~) C == 10 JA-F, = 0, lOe~1 v= and sin t V. Solve for i. The voltage equation is lOi + 10 5 i(O+)=O, with vC<O_)=vc(O+)=O. I i dt == lOe~1 sin t ( 1) Let the forced solution be if == e-I(C I sin t + C2 cos t) (2) Substituting (2) in (1) yields 4 10 C I - 4 Cl + 10 C 2 == 1 C2 = -1 Cl = - C2 = - 10-4 Thus and i = 1O- 4 e-l(cos t - sin t) + Ae- I041 Thus i( 0 + ) = 0 requires that A = -10 - 4. . Hence, 1=10 18.24 . -1041 A ] t = 0.1 ms. From Prob. IS.23 i(O.1 ms) = 63.2 JA-A 18.26 -t [e (cost-smt)-e Determine the voltage acrOss the resistor of the circuit of Prob. IS.23 at , 18.25 -4 , or v R = Ri = (10)(63.2 x 10- 6 ) = 0.632 mV What is the energy stored in the capacitor of the circuit of Prob. IS.23 at Vc = v - vR == 0.9999- 0.632= 0.3679mV t = 0.1 ms. w = ~ C( V C)2 = 0.677 X 10- or 12 J The circuit of Fig. IS-S is under steady state with v = 100 sin 377t V. The switch is closed at t = 0, and the circuit is allowed to come to steady state again. What is the initial condition on the current and what is the steady-state current? , For i t < 0, = 10 3 + j(377 - 5300) = 5023/-7SS n 100 L!r Im = 5023/-7SS == 19.9/7SS mA = 3.97 + j19.5 The reactive component of the current cannot change instantaneously. For t > 0, R = O. mA Thus, Thus, (Im)steady <tate = _ 100 j4923 == 20.3/90° mA or i" = 20.3 cos 377t Fig. 18-8 mA 448 18.27 D CHAPTER.18 , t > O. Solve for i for the circuit of Prob. 18.26 for where if is the same as iss of Prob. 18.26. The characteristic roots are PI' P: ,= ±j103v'2 Thus, and i= in = Cl cos lO:'\l2t + C2 sin 10 3v'2t 3 20.3 cos 377t + Cl cos 10 3v'2t + C sin 10 v'2t 2 From Prob. 18.26, Vcm = 19.9 X 10- 3 x 5300 LQ8.~ Thus, vc(O_) = vc(O+) = -21 V=' 90°) = 105.5 I-uS = 103.5 - j21 -~ (0+) i(O+) = 19.5 mA These initial conditions yield Cl = -0.8 X 10- 3 and C2 = 14.85 X 10- 3 Finally, therefore, i = 20.3 cos 377t - 0.8 cos 18.28 w3v'2t + 14.85 sin 10 3v'2t mA Obtain a differential equation to solve for the current through the 1-0 resistor in the circuit of Fig. 18-9. , For the two mesh currents i l and i 2 , we have di + 1, 2i 1 + -~ dt3 From (2): I (1) (i 1- 2 i ) dt = vs I (i2 - i I) dt + i2 = 0 (2) 3 I (3) (.'1 - i 2) dt = i2 . . di2 31 1 == 31 2 + Tt and di I dt =, (4) 2 di2 1 d i2 dt + 3 df (5) Substituting (3) to (5) in (1) yields d~2 ili". . + 5 -d" + 91 2 = 6t + 9 sm 4t dt t -2 20 1H 10 Fig. 18-9 18.29 , List the initial conditions required to solve for ix in the circuit of Fig. 18-10. Fig. 18·10 TRANSIENTS IN AC CIRCUITS 18.30 Solve for ix in the circuit of Fig. 18-10. I By steady-state circuit analysis if = 0.36 cos (31 - 146.9°) A The characteristic equation is The roots are p2 + ~p + 1= 5 . PI' P2 = - . Thus, In = 0 v'23 6 ± J -6- v'23 1 + C' v'23 ) e -5/61( Cl cos -62 sm -6- 1 The initial conditions on ix are ix(O+) Consequently, Hence, = iL(O+) - 5 = 0 A Cl =0.3 and C2 = -0.43 v'23 1 - 0.43 sm . -6v'23 1) ix = 0.36 cos (31 - 146.9°) + e -5/6 ( 0.3 cos -6I A 0 449 CHAPTER 19 \\ Circuits with Multifrequency Inputs ~ 19.1 Linear circuits, excited by multi frequency sources, may be solved by superposition. On this basis, solve for the current in the circuit of Fig. 19-1a under stt:ady state. Given: e 10- = 30 sin 21TlOt V and e 20 _ = 5Q sin 21T20t. Obtain an expression for the instantaneous current. - ~J- IT L -_ _ _- " " " " - R-41l L - 28 mH (a) "20- R-Hl R-Hl L - 28 mH L - 28 mH (b) (c) Fig. 19·1 , First, we decompose the circuit of Fig. 19-1a into two components, each with its own single-frequency input, as shown in Fig. 19-1b and c. We now have: For Fig. 19-1b ZIO_ = R + jX L . 10 _ == 4 + j21T(1O)(0.028) ==4+j1.76 = 4.37 /23.75° n E 1o _ = 30 L!r v'2 == 21.21 V 21.21L!r 110 _ = 4.37 /23.75° == 4.85/-23.75° A For Fig. 19·1c Z20_ = R + jX L . 20 - = 4 + j21T(20)(0.028) =4 + j3.52 = 5.33/41.34° n E 20 _ = 50L!r v'2 120 _ = 5.33/41.34° = 35.36 V 35.36 L!r = 6.63/-41.34° A The phasor components (1 10 _,1 20 _) of the circuit ,;urrent rotate at different angular velocities. addition cannot be used to obtain the resultant rms current. Hence phasor The resultant instantaneous value can be deternined by expressing the component currents in the time domain, and adding. Thus, i lO _ = 4.85v'2 sin (21TlOt - 23.75°) From the superposition theorem, iT = i lO _ + i 20 _ = 6.86 sin (21TlCIt - 23.75°) + 9.38 sin (21T20t - 41.34°) 450 (1) CIRCUITS WITH MULTIFREQUENCY INPUTS 19.2 0 451 Plot (1) of Prob. 19.1 and verify that a nonsinusoidal is produced by sinusoidal inputs of different frequencies. I See Fig. 19-2. Fig. 19-2 19.3 , Determine the instantaneous power drawn by the circuit of Fig. 19-1a. (1) where all the instantaneous voltages and currents may be substituted from Prob. 19.1. 19.4 Sketch the four components of instantaneous power obtained in (1) of Prob. 19.3. , 19.5 See Fig. 19-3. From Fig. 19-3a through d infer if the corresponding average power will be zero or nonzero. , In Fig. 19-3c and d, the voltage and current waves have different frequencies, and the average power calculated over one or more periods of the lower-frequency wave is zero. In Fig. 19-3a and b, the current and voltage waves have the same frequency, and the average power calculated over one or more periods is not zero. A comparison of Fig. 19-3a, b, c, and d indicates that only current and voltage waves of the same frequency provide a nonzero value for the average power. 19.6 Calculate the average power associated with each instantaneous power wave of Fig. 19-3, using the relationship P = El cos cp. Hence determine the total average power supplied to the load. = (21.21)( 4.85) cos 23.75 0 = (35.36)( 6.63) cos 41.340 =94W = 176W The total power supplied to the load is PT = P 10 - + P20 - = 94 + 176 = 270 W 19.7 19.8 , Repeat Prob. 19.6 by using the 12R relationship. PT = P 10 - + P20 - = 270 W Multifrequency (MF) currents flow through a resistor R. Obtain an expression for the rms value of an equivalent single-frequency current producing the same loss in R. , The power dissipated in R due to the MF currents is (1) 452 D CHAPTER 19 196W t,s (II) -57W Ib) 261 W Cc) 320W Id) Fig. 19·3 CIRCUITS WITH MULTIFREQUENCY INPUTS D 453 Define Ieq as the rms value of an equivalent single-frequency current that would cause the same power dissipation in R as do the multifrequency components: (2) Equating (1) and (2) and solving for Ieq yields 1eq = ,V/12[I + 12[2 + ... + t [g + ... + 12[n (3) Since Ieq is the square root of the sum of the squares of the component currents, it is defined as the root-sum-square current or rss current. Thus (2) and (3) may be written as I", = VI~I + I~2 +: .. + I~g + ... + I~" where Ifl' If2' etc., are rms values. 19.9 Following the procedure of Prob. 19.8, obtain a general expression for the root-sum-square voltage E rss ' , The power drawn by a resistor R may be calculated from the E2/ R relationship, provided that the multifrequency voltage drops across the resistance component are known. Thus, E2 E2 E2 E2 =::.t.!+::13.+ ... +~+ ... +:::..l!!. P R MF R R (1) R Defining Eeq as the rms value of an equivalent single-frequency voltage drop across the resistance component that would cause the same power dissipation in R as does the multi frequency components, then E~q P MF =7? (2) Equating (1) and (2) and solving for Eeq, E2 E2 E2 R R R ~=::.t.!+::13.+ E2 ' E2 ... +~+ ... +:::..l!!. R R (3) Eeq ='V/ E2[1 + E2[2 + ... + E2Ig + ... + E2In = E rss 19.10 Generalize the result (3) of Prob. 19.9 for rss voltage drops across L, C, and Z in a circuit excited by MF sources. , E Tss,L = VE2 L1 + E2L2 + ... + E2Lg + ... + E2Ln E rss,C = VE2 Cl + E2C2 + ... + E2Cg + ... + E2en where 19.11 , 19.12 19.13 19.14 e = (6 sin 188.50t + 4 sin 18,850t) V. Determine the frequency The voltage impressed across a 10-0 resistor is of each voltage component. , , , 21Tfl = 188.5 or f 21Tf2 = 18850 or f2 = 188.5 21T = 3 kHz 1 = 30 Hz Calculate E rss for the circuit of Prob. 19.11. Efl 6 = v'2 = 4.24 V 4 E[2 = v'2 = 2.83 V Ens = V (4.24)2 + (2.83)2 = 5.1 V Determine I rss for the circuit of Prob. 19.11. I [I = ~ = R 4.24 = 0 424 A 10 . 1[2 2.83 = 10 = 0.283 A I rss = V(0.424)2 + (0.283)2 = 0.51 A How much average power is dissipated in the resistor of Prob. 19.11? 454 19.15 D CHAPTER 19 Two sinusoidal generators, an rms-reading ammeter, and a coil are connected in series. The inductance and resistance of the coil are 0.010H and 6.0n. respectively. The generator voltages are expressed by e l = 10 sin 377t and e 2 = 25 sin 754t, respectively. Sketch the circuit and determine the ammeter reading. I The circuit is shown in Fig. 19-4. 12 = 120 Hz II = 60Hz Z60_ = 6 + j3.77 Z120_ = 6 + j7.54 =7.09/32.1°n El = = 9.64/51Sn 10 v'2 = 7.07 V E2 = 7.07 I1 = 7.09 = 0.997 1 = 17.68 = 1 83 A 2 9.64 . I", = ammeter readin~; ,= 6f1... 19.16 V(1)2 + (1.83)2 = 2.09 A o .of H Fig. 19-4 What is the average power dissipated in the resistor of Prob. 19.15? I 19.17 25 v'2 = 17.68 V P = (l"YR = (2.09)26 = 26.21 W A 100 is!..-V 120-Hz generator and an 80 is!..-V 60-Hz generator are connected in series with a 60-V battery and a coil. The resistance and inductance of the coil are 3.0 nand 2.65 mH, respectively. Sketch the circuit and determine the impedance at each frequency. I The circuit is sketched in Fig. 19-5. Z120_ = 3 + j21T(120 )(0.00265) = 3.6/33.66° n Z60_ = 3 + j21T(60)(0.00265) '" 3.16/18.43° n ZO_ = 3 + jO = 3 is!.. n Fig. 19·5 19.18 Calculate Irss for the circuit of Prob. 19.17. I 1120 _ = 100 3.6 = 27.78 A I", = 19.19 160 _ =, 80 3.16 = 25.32 A 10 _ = V(27. 78)2 + (~5.32)2 + (20.0)2 = 42.58 A Determine the voltmeter reading in the circuit of Fig. J9-6. I The voltmeter reads Em which is given by Em = V(120~-:j::-~: 40)2 + (300)2 = 326 V 60 "3 = 20.0 A CIRCUITS WITH MULTIFREQUENCY INPUTS 120 V 60 Hz D 455 300 V 20Hz 40V Fig. 19-6 19.20 Find the ammeter reading in the circuit of Fig. 19-6. I Z60_ = 10 + j[21T60(0.05)] = 10 + j18.85 = 21.34 /62.05° n Z20_ = 10 + j[21T(20)(0.05)] = 10 + j6.28 = 11.81 /32.14° n Zo_ = 10 + j[21T(0)(0.05)] = 10 + jO = 10 ~ n 120~ 160 _ = 21.34 /62.05° = 5.62 /-62.05° A 300~ 120 _ = 11.81 /32.14° = 25.40 (-32.J40 A 40~ 10_= ~ =4.0~A irss = 19.21 V(5.62)2 + (25.40)2 + (4)2 = 26.32 A = ammeter reading How much total power is drawn from the sources of the circuit of Fig. 19-6? I 19.22 The current and voltage to a load are expressed by eT = 20 + 30 sin (377t) Determine voltage. I + 50 sin (1130t + 20°) iT = 15 sin (377t) + 14 sin (1130t - 36°) (a) the frequency of each component of the driving voltage; (b) the rss current; (c) the rss (a) The three component frequencies are 1. 0 Hz (dc) 2. 21Tf = 377 3. 21Tf = 1130 19.23 f= 60Hz f = 180 Hz 15)2 (14)2 VI( v'2 + v'2 = 14.5 A (b) i", = (c) E", = '1(20) + I 2 (30 ) 2 v'2 + (50 v'2 ) 2 = 45.8 V We define the apparent power SMF and the power factor pfMF relating to a multi frequency circuit by and Using this definition, find the apparent power drawn by the circuit of Prob. 19.22. I SMF = E"J", = (45.8)(14.5) = 665 V· A 456 19.24 D CHAPTER 19 Determine the total active power and the power factor of the circuit of Prob. 19.22. I P6[)P 180- = V I 180- ~/ "1;0cos -'<" J, + 196 = 421 W 50 14 =.v"'2 •V"'2L. cos 56° = 196 W ~ and pfMF = PMF SMF 421 = 665 = 0.63 Obtain the impedances corresponding to the different frequencies in the circuit of Prob. 19.22. I 19.26 180- H,O-- P MF = 0 + 225 19.25 ° 30 15 ')(/ \i o( - = V60- I 60- cos-'\." )I,0 - =.v"'2~ •v"'2~ cos 0 = 225 W Z 0- = Vo_ = 20 1 _ 0 _ 0 = 00 n. .. ZOO_ r V60 _ 30/v 2: = -1= 15t/2 =2n 60- V1BO _ Z180_ = -11BO- = 50/V2 41' '2 1 y ~ = 3.57 n Three sinusoidal generators and a battery are connected in series with a coil whose resistance and inductance are 8.0 nand 26.53 mH, respectively. The frequency and rms voltage of the respective generators are 20 Hz, 15 Y; 60 Hz, 30 Y; and 80 Hz, 50 V. The battery voltage i~. 6 V. Sketch the circuit and determine the rss voltage. I The circuit is shown in Fig. 19-7. Em = Y(15)2 + (30)2 + (50)2 + (6)2 = 60.51 Y 'V BA 19.27 Z20_ = 8 + j21T(20)(0.02653) = 8.67 /22.62° n ZBO_ = = 8 + j21T(60)(0.02653) = 12.81 /51.34° n 8 + j21T(80)(0.02653) = 15.55 /59.04° n ZO_ = 8 + jO = 8 ~ n Z60_ Calculate the I", in the circuit of Fig. 19-7. 30 160 _ = 12.81 = 2.34 A 15 120_ = 8.67 = 1.73 A I 50 I Bo _ = 15.55 =3.22.~ 10_= I rss = Y(1.73)2 + (2.34)1 Thus, 19.29 Fig. 19·7 Determine the complex impedances at the different frequencies of the circuit of Fig. 19-7. I 19.28 :H_S~_H 6 '8 =0.75A + (3.22)2 + (0.75)2 = 4.40 A Find the average power, apparent power, and the power factor of the circuit of Fig. 19-7. I P MF = (JrsYR = (4.4)28 = 154.88 W SMF = ErsJrss = (60,51)(4.4) = 266.2 y. A pfMF = 19.30 P MF S MF 154,88 = -"-66--::' = 0.58 ~ .~~ In the circuit of Fig. 19-8 eT = 141.42 sin 21T30t + 141.42 sin 21T90t ir = 9.80 sin (21T30t - 30°) + 5.66 sin (21T90t - 60°) Determine I rss ' I 9.80 130 _ = y'2 = 6.93 A 5.66 190 _ = y'2' = 4.00 A less = Y(6.93)2 + (4.00)2 = 8.00 A CIRCUITS WITH MULTIFREQUENCY INPUTS D 457 } - - - - - - { ' V )---..., Impedance coil 1- - - - - - I I I Il _L_ _ _ _R _ .JI '-----{ V )-_....J Fig. 19·8 E,. 19.31 Find Em in the circuit of Fig. 19-8. I 19.32 _141.42_ E 30- -v'2 - - 1 OOV Determine 1: 30 _ and Z90_ v'2 = 100 V for the circuit of Fig. 19-8. z 30- = I 141.42 E 90 _ = E 30 _ 1 _ = 100!.J!. /"1110 . 6.93/_300=14.44LlQ.=(12.5+J7.22)n 30 E 90 _ Z90_ = -I- 100 !.J!. = 4/-600 = 25 /"'-11 0 • L2Q. = (12.5 + J21.65) n 90- 19.33 Determine the resistance and the inductance of the coil of the circuit shown in Fig. 19-8. I From Prob. 19.32: = (12.5 + j7.22) n 21T(30)L = 7.22 L = 0.0383 H Z30_ Thus, 19.34 R = 12.5 n X L30 _ = 7.22 n or 38.3 mH How much active power is drawn by the coil of the circuit of Fig. 19-8? I = (100)(6.93) cos 300 = (100)(4.00) cos 600 =600W =200W P MF = 600 + 200 = 800 W 19.35 Verify that the result of Prob. 19.34 will be obtained by the 12R method. I P30 - = I;o_R = (6.9W(12.5) =600W P MF = 600 + 200 = 800 W 19.36 = (4.00)2(12.5) =200W Repeat Prob. 19.34 using the definition of I"s' I From Prob. 19.30, Thus, 19.37 P90 - = I~o_R I"s = 8 A. PMF = (I",.R)2R = (8)2(12.5) = 800 W Repeat Prob. 19.34 using the relationship P MF =(E",.R)2IR, multifrequency voltage drop across the resistance component. I E3D-.R = 130_R = 6.93(12.5) = 86.63 V E"s.R = V(86.63)2 + (50.0W = 100 V where E",.R is the root-sum-square of the E 90 _. R = 190_R = 4.00(12.5) = 50.00 V thus (100)2 P MF = 12.5 = 800W 458 19.38 D CHAPTER 19 Obtain the results of Probs. 19.34 through 19.37 from I 19.39 P MF = (E",.R)(l",.R)' P MF = (00)(8) = 800 W Determine the apparent power and the power factor of the circuit of Fig. 19-8. I SMF = E,j", = (HI.42)(8.00) = 1131 y. A P M1 S-- pfMF= 800 == 1131 =0.71 IVII 19.40 A lO-n resistor is connected in series with the follc-wing series-connected generators: e I = 50 sin (377t + 40°) = 40 sin (377t + 20°) e] e 3 = 80 sin 150t Determine the frequency of the individual genera ton,. I 19.41 t I 377 = 60Hz f 2 = ._27T 377 =-=60Hz 27T cin~lLit Obtain the rss voltage across the resistor of the 150 t, = -2 = 23.87 Hz . 7T of Prob. 19.40. I Since voltages eland e2 have the same frequency, it is necessary to calculate the resultant voltage at that frequency before calculating the rss value. El + E2 = 50 /40° 40 /20° v'2 + v'2 = (27.08 + j22.73) + (26.58 + j9.67) = (53.66 + j32.40) = 62.7 /31.12° Y E3 The rms value of the 23.87-Hz wave is E",.R = V E!o_ 19.42 = 80 Iv"2 = 56.6 V. + E~:\X7 _ = V (62.7)2 + (56.6)2 = 84.5 Y What is the rss current through the resistor of the ::ircuit of Prob. 19.40? I 19.43 Thus. V60 _ = 62.68 = 6 268 A I60- = Z60_ 10' 56.57 " 12387 - = '-j(,' = 5.657 A I", = V (6.268) 2 + (5.657) 2 = 8.44 A Determine the power delivered to the resistor of the circuit of Prob. 19.40. I 19.44 A coil whose resistance and inductance are 50 n 1IId 88 mH, respectively, is connected in series with three sinusoidal generators. The generator volt ages are e l = 400 sin 377t, e 2 = 100 sin 754t, e 3 = 50 sin 1131t. Determine the frequency of each generator. I 19.45 27Ttl = 377 27Tt2 = 754 27Tt3 = 1131 tl =60Hz t2 = 120 Hz t, = 180 Hz Determine the rss voltage across the coil of the circuit of Prob. 19.44. I El = 400 v'2 = 282.84 Y 100 E.= .. M = 70.71 Y . '/2 E2 = 50 v'2 = 35.36 Y E", = V(282.84)2 + (71).71)2 + (35.36)2 = 293.68 Y 19.46 Determine the rss current in the circuit of Prob. 19.44. I Z60_ = 50 + j377(0.088) = 60 /33.57" (1 Z120- = 50 + j754(0.088) = 83/53.0° n Z180_ = 50 + j1131 ((1,()88) = 111.38/63.33° n 11 = 282 .84 = 4.71 A 60 12 = I", = V(4.71)2 .~~~ ,~3 = 0.85 A + (0.85)2 + (0.32)2 I = 35.36 3 111.38 = 4.8 A = 0.32 A CIRCUITS WITH MULTIFREQUENCY INPUTS E", = Y(35)2 + (10)2 + (8)2 = 37.3 V What is the rss current in the circuit of Prob. 19.47? 19.48 I At 60Hz: Z60 160 = 3 + j(1.67 - 8.99) = 7.91 /-67.72° n 35 m: = 7.91/-67.72° = 4.42 /67.72° = 4.42 A 1 At 180Hz: Xc = 21T(180)(295 x 10 6) = 3 n Z180 1 180 At 300Hz: = 3 + j(5 - 3) = (3 + j2) = 3.61 /33.69° n = R ° 10 m: = 3.61 ~ =2.77 /-33.69 =2.77 A XL = 21T(3OO)( 4.42 x 10Z300 3 ) = 8.33 n Xc 1 = 21T(3OO)(295 x 10- 6 ) = 1.80 n n + j(XL - XC> = 3 + j(8.33 -1.80) = 7.19 /65.33° 1300 = 8m: 7.19 ~ = 1.11 /-65.33° = 1.11 A I rss = Y(4.42)2 + (2.77)2 + (1.11)2 = 5.33 A Obtain the reading of an rms voltmeter connected across the coil of Prob. 19.44. I V eoil 60 Similarly, = 160Zeoil60 = 4.42 /67.72°(3 + j1.67) = (4.42 /67.72°)(3.43 /29.10°) = 15.16 /96.82° V Veoil180 Veoil300 = (2.77 /-33.69°)(3 + j5) = (2.77 / -33.69°)(5.83 /59.04°) = (1.11 / -65.33°)(3 + j8.33) v,.•• 19.50 coil = Y(15.16)2 = = 16.15 /25.35" V (1.11 / -65.33°)(8.85 /70.19°) = 9.82 /4.86° V + (16.15)2 + (9.82)2 = 24.2 V = voltmeter reading Determine the rss voltage across the capacitor of the circuit of Prob. 19.44. I VC60 = IC60Xc60 = 4.42(8.99) = 39.74 V VC180 = IC180XC180 = 2.77(3) = 8.31 V VC300 = IC3ooXC3oo = (1.11)(1.80) = 2.00 V Vrss •C = Y(39.74)2 + (8.31)2 + (2.00)2 19.51 = 40.65 V Obtain the active power, apparent power, and the power factor for the circuit of Prob. 19.44. I P MF = RI;ss,R = 3(5.3W = 85.23 W pfMF 19.52 459 A series circuit containing a 295-JLF capacitor and a coil whose resistance and inductance are 3 nand 4.42 mH, respectively, are supplied by the following series-connected generators: 35 Vat 60 Hz, 10 V at 180 Hz, and 8 V at 300 Hz. Determine the rss driving voltage. 19.47 19.49 D = SMF = ErsJrss = (37.27)(5.33) = 198.65 V' A PMF 85.23 SMF = 198.65 = 0.43 Determine the instantaneous current in the circuit of Prob. 19.44. I i=i60+iI80+i300 = 4.42V2 sin (21T60t = 6.25 sin (377t + 67.72°) + 2.77V2 sin (21T180t - 33.69°) + 1.11V2 sin (21T3OOt - 65.33°) + 67.72°) + 3.92 sin (1131t - 33.69°) + 1.57 sin (1885t - 65.33°) 460 19.53 D CHAPTER 19 A 30-n resistor is connected in series with a 500- I AF capacitor, a 120-V battery, an ammeter, and three sinusoidal generators. The generator voltages are 100 sin 30t, 50 sin 80t, and 70 sin lOOt V. Sketch the circuit and include a voltmeter across the capacitor-resistor combination. Determine the voltmeter reading. I See Fig. 19-9. El 100 E 2 -- = v'2 = 70.71 V ~ v'2 -"5 - - ..•c· 6 V E3 == 70 v'2 E4 = 120 V = 49.50 V Em = y(70.71)2 + (35.36)2 + (49.5(,), + (120)2 == 152 V== voltmeter reading Fig. 19.9 19.54 Determine the ammeter reading in the circuit of Fig 19-9. I II = 70.71 73.11 == 0.97 A I", 19.55 = 12 = Z4 = Z3 == 36.06 n 00 49.50 13 = 36.06 == 1.37 A 35.36 39.05 = (.91 A 14 =OA Y(0.97)2 + (0.91)2 + (l.3n 2 + 0 == 1.91 A == ammeter reading Calculate the power dissipated in the resistor of the c:,rcuit of Fig. 19-9. I From Prob. 19.54, I", = 1.91 A. Thus, 19.56 Z2 = 39.0S !1 ZI == 73.11 n PMF = (1.91)\30) = 109.44 W A load consisting of an 8.842-J.LF capacitor and a ceoil whose inductance and resistance are 88.4 mH and 5 n, respectively, is connected in series with three sinll~oidal generators. The generator voltages are 100 sin 377t, 100 sin (377t + 50 and 100 sin 1131t V. Sketch th( circuit and determine the frequency of each generator. 0 ), I The circuit is shown in Fig. 19-10. f = I 377 = 60Hz 27T 1131 377 f 2 = --= 60Hz 27T f 3 = - 27T = 180Hz ~~~~OG~---1r-----~ 8. fJ42.jf 19.57 Fig. 19.10 Determine the rss voltage across the load of the cireuit of Fig. 19-10. I First, we combine the 60-Hz volt ages to obtain E 60 _ m .. = 100 ~ + 100 /50 0 = 181.26/25" V E 60 _ = 181.26 v'2 = 128.17 V E rss = Y(128.17f + (70.71)2 == 146.38 V E 180 _ = 100 v'2 = 70.71 V CIRCUITS WITH MULTIFREQUENCY INPUTS 19.58 461 Calculate the rss current in the circuit of Fig. 19-10. I Z60_ = 266.71 n Z180_ =5n 160 _ = I", = V(0.48)2 19.59 D 128.17 70.71 1180 - = -5- = 14.14 A 266.71 = 0.48 A + (14.14)2 == 14.15 A How much active power is delivered by all the sources of the circuit of Fig. 19-1O? I 19.60 A series RLC circuit excited by several multi frequency sources is shown in Fig. 19-11. expressions for I"., E",.R' PMF> SMF' and pfMF • VM E .. , I to IPt Vlt Et I I I I Fig. 19·11 Eoi' 1=- I g Zg ERg = IgR I rss = V12 + 12 + 12 + ... + 12 0 1 2 g E" •. L' E",.e' E" •. z are similar to the preceding formula. P MF = Po + PI + P 2 + ... + Pk 2 E2 ,,,.R = E",.RI",.R == RI",.R = ~ Write the general CHAPTER 20 \\ Circuits with Nonsinusoidal Sources~ 20.1 Nonsinusoidal periodic waves may be expressed as (when certain conditions are satisfied): f(t) = ! A 0 + A I cos wt + A 2 cos 2wt + A 3 cos .~wt + ... + B I sin wt + B2 sin 2wt + B3 sin 3wt + . . . This series is known as Fourier series. The coeffil:ients As and Bs in (1) are given by w (h-Iw An = :;;: 2 (T Jo f(t) cos n,:ut dt Bn = ; Jo 27Tnt = T Jo f(t) cos ---y;- dt (h-IW and (1) 2 ( f(t) sin null dt = T Jo 2 f(t) sin ~n (2) t (3) dt On the basis of (1) to (3) obtain the Fourier series for the waveform shown in Fig. 20-!' , The waveform is periodic, of period 27T Iw in t or 27T in wt. It is continuous for 0 < wt < 27T and given therein by f(t) = (10/27T)wt, with discontinuities at wt = n27T where n = 0,1,2,. . .. The· Dirichlet conditions are satisfied. The average value of the function is 5, by inspection, and thus !a o = 5. For n > 0, (2) gives A n = -7T1 f2" (-2 10) wt cos nwt d( wt) = 0 7T Thus the series contains no cosine terms. 10 [ ~-t ~iJl nwt + 21 cos nwt 27T 11 11 --2 Us~g 1f2"(1O) - d(wt) wt sin nwt 27T. B = n 7T - 0 ]2" = -7T102n (cos n27T - cos 0) 2 2 0 = 0 (3). we obtain = 10 ~--2 .!7T [ t 1 ]2" = - ~ cos nwt + - sin nwt n n2 0 -10 7Tn Using these sine-term cg.efficients and the average term, the series is f (t ) =/ 5 - 10. 10 . 2 10. 5 10 ~ sin nw t - sm wt - - sm u)t - -- sm 3wt - ... = - - L.J 7T 27T 37T 7T n~l n (4) fH) I~ IL-_-_-_-_-_-L.-_lw, 47T 20.2 Fig. 20-1 Based on the concept of Prob. 20.1, and on the contents of Chap. 19, draw an equivalent circuit for the circuit of Fig. 20-2a, excited by the generator voltage of the waveform of Fig. 20-2b. , The equivalent circuit is shown in Fig. 20-2e, which has sinusoidal sources whose algebraic sum produces the same nonsinusoidal output voltage as shown in Fig :l0-2b, where e gen = e fl + e'2 + en + ... + e fn Once the sinusoidal components that make up the nonsinusoidal wave are known, the respective current and power components may be determined using the techniques developed in Chap. 19. 20.3 For the waveform shown in Fig. 20-3, find An' A 3 , and B 4 • , 462 The constant term can be easily determined from the net area of the wave for one cycle. Thus, CIRCUITS WITH NONSINUSOIDAL SOURCES R.L D 463 C ~ Lold ----0R,L C ~ Load (e) Fig. 20·2 4V t,4 ,...-- > t" 0 Of 4" 2" 1 6" 1 ( 2" Ao = 21T (net area)o = 21T 4 x 1 Fig. 20·3 41T) = 2.67 3"" t,,/3 4cos(3a) da =;4 (Sin3a)4,,/3 4 . -3= 31T (sm41T) =0 A3 =; Jo 0 1 (4,,/3 . 4 ( cos 4a )4"/3 1 (161T ) 1 B4=;Jo 4sm(4a)da=; - - 4 - 0 = - ; cos---cosO =-;(-0.5-1)=0.48 3 20.4 What is the frequency of the fourth harmonic of the waveform of Fig. 20-3, if the period of the wave is 0.04 s? I The frequency of the nonsinusoidal wave is 1 1 f= r= 0.04 =25Hz The fourth harmonic is four times the frequency of the nonsinusoidal wave. Hence, f4 = 4(25) = 100 Hz 20.5 Determine the Fourier-series coefficients Ao and B2 for the periodic voltage wave in Fig. 20-4. I 1 ( net area)2rr 1 (area of large triangle - area of small triangle) Ao = 21T 0 = 21T A o =..!. [! (71T 21T 2 4 x 140) - !2 (~4 x 20)] = 60 464 D CHAPTER 20 160 i > 140 120 100 80 60 ____+-~----~------~~~----L------~~~-----------Pa :1" " Fig. 20-4 If the dc term is to be calculated by the use of olculus, the equation for the straight-line function between a = 0 and a = 27T must be determined. The equation for the straight line is 80 v = ma + b = --- 20 oX - from Fig. 20-4 7T 1 12~ (80 1 (80 A = -- a - 20 ) da = -- -a - 20a )2~ = 60 o 27T 0 7T 27T 7T 2 0 2 Hence: 1 f2~ 1 '2", (80 ) B 2 =f(a)sin2ada=-J -a-20 sin2ada 7TO 7TO 7T {4 [( 22I.SIn 2a )2" (I2 20 = -:;;;: 20.6 0 - oX cos 2a )2,,] (12 cos 2a )2"} = -25.46 0 - - 0 Determine the Fourier coefficients A o ' A k , and Bk for the periodic voltage wave shown in Fig. 20-5. I 1 where f(a)=2 a = 27T. 1 ( 2" = 27T (net area)o = 27T 27T - 3 Ao for a=O Ak = -1 1T = f" 0 to a=7T; 47[) = 0.63 f(a)='O f for 7 2 cos (ka) da + -1 1T "n 1r a=7[ 1T to Jo 1T f(a) cos (ka) da a=i7T; ocos (ka) da + -1 f2" 771'/4 f(a)=-3 for (-3) cos (ka) da 2 [2sin ka I" + 0 + (-3) ~i~!:! 1" ] = ~ (sin 7k7T) 7T k 0 k 7,,/4 k7T 4 f2" f(a) sin (ka) da = -1 f" 2sin (ka) 1T 0 = -1 [2(-COS ka) 7T k I" + 0 0 .:tit + -1 1T 0 2 3 -cos ka 1" ] k 7..-/4 1 ~,-k7T f7"/4 Osin (ka) da + -1 1T 7T f2" 7'tr/4 - 3sin (ka) da [ -2(cos k7T - 1) + 3(I7- k 7-T ) ] cos 4 ~2 "0 > i" L... .--.-1--------'-----.----1--.... er 2" " -3 a=i7T ! Similarly, Bk = -1 1 (2" = ;: Ak 3" 4" S" L... Fig. 20-5 to CIRCUITS WITH NONSINUSOIDAL SOURCES 20.7 D 465 Evaluate the first three harmonics in the waveform of Prob. 20.6. I 1 ( -2 cos 2TT + 2 + 3 - 3 cos 4 14TT) = 0.48 B2 = 2TT 1 ( -2 cos TT + 2 + 3 - 3 cos 4 7 TT) = 1.55 B. = -:;;: 1 ( -2 cos 3TT + 2 + 3 - 3 cos 4 21TT) = 0.97 B3 = 3TT Thus the dc term and the first three harmonics of the driving voltage in Fig. 20-5 are v = 0.63 - 0.68 cos a + 1.55 sin a - 0.48 cos 2a + 0.48 sin 2a - 0.23 cos 3a + 0.97 sin 3a 20.8 One cycle of a nonsinusoidal periodic voltage wave has a value of 1 V from 0 to TT/2, 0 V from TT/2 to TT, and -0.5 V from TT to 2 TT. Sketch the wave and determine (a) the dc term; (b) the coefficients of the third harmonic. I The waveform is shown in Fig. 20-6. (a) 1 [net areal o2rr = 2TT 1 [(TT) Ao = 2TT "2 (1) - TT(0.5) ] = 0 (b) A3 = - 1 TT 1 B3 = TT - 20.9 (I.~ - l"'/2 (1) cos 3a da + -1 J.2rr (-0.5) cos 3a da = -0.106 TT '" 0 1"'/3 (l)sin3ada+-1 f2rr (-0.5)sin3ada=0.212 0 TT Fig. 20-6 - Determine the rms value of the third harmonic of the voltage wave from the results of Prob. 20.8. I e3 = -0.106 cos 3a + 0.212 sin 3a = -0.106 sin (3a + 90°) + 0.212 sin 3a or E 3 • ma • = -0.106/90° + 0.212!..sr. = 0.237 /-26.6° or 20.10 rr E= 0.237 v2 =0.1676V For the voltage wave shown in Fig. 20-7, determine the constant term and the coefficients of the fifth harmonic. v f- r--- 3~ rr0 -1 r- "2 " 2,. Q -2 r- -3 r- -4 f-5 r- Fig. 20-7 466 D CHAPTER 20 I Ao = 1 As=- 1T Bs = -1 f f rr !2 Let a = wt. rr 57T] = -1.75 I f2rr 3cos5ada+-5cos5a da =0.191 1T !2 1r I f2rr 3 sin 5a da + -7T rr -5 sin 5a da = 0.828 Express the fifth harmonic component of the voltage in Fig. 20-7 in the time domain. I e s = 0.191 sin (5a -I- 90°) + 0.828 sin 5a E s . max = 0.191/90° + 0.828 JJr. = jO.191 or -+ 0.828 or Determine the amplitude of the voltage wave shown in Fig. 20-8. 1 frr/)O A)=- 7T 0 1 6cosada+- 7T 1 lrr!lo. 1 7T 7T B) = - 6sm a da + - 0 2/Srr 4 2 0 -2 2 "!! i" 2 .! 10 f2/Srrrr !2 6. rr!lO 6 . rr/2 -6cmlrda=-[smalo --[smal 2l5rr =0.497 jrr/2. 6- -6sm a 7T do~ 7T 6 rr!lO 6 rr!2 ,= - [-cos al o - - [-cos al 2/srr = -0.497 7T 7T ~" 2• --6 20.13 E S _max = 0.849 /7r V e s = 0.849 sin (Swt + 7r) V Hence, 20.12 7T 0 7T 0 20.11 21 [3( ~:) - Fig. 20-8 The following symmetry checks determine the absence of even-numbered terms, cosine terms, or sine terms in the Fourier series, and thus reduce the number of cakulations. Half-Wave Symmetry If the bottom half of a nonsinusoidal wave is the min'or image of the top half but displaced from it by 7T rad as shown in Fig. 20.9a, the wave is said to have half-wave symmetry. A wave possessing half-wave symmetry will have no de term and no even harmonics. That is, if f(a) = -f(a + 7T), Half-wave symmetry is unaffected by the location of the vertical axis. Even Symmetry If the symmetry of the nonsinusoidal wave about the 'V(~rtical axis is similar to that shown in Fig. 20.9b, the wave is said to have even symmetry. A wave possessing even symmetry will have no sine terms. That is, if f(a) = f(-a), Bp B 2 , B 3 , · •• , Bn = O. Odd Symmetry If the symmetry of the nonsinusoidal wave about the vertical axis is similar to that shown in Fig. 20-9c, the wave is said to have odd symmetry. A wave possessing odd symmetry will have no cosine terms. That is, if f(a) = -f(-a), AI' A 2 , A 3 , · . · , An = O. CIRCUITS WITH NONSINUSOIDAL SOURCES D 467 (Q) (h) ~~----4------4------~------~----------------_Q I ",-0 Fig. 20-9 (d On the basis of the above, determine the type of symmetry present in the voltage waveform shown in Fig. 20-10. State which coefficients are zero in the Fourier series. I The waveform has an even symmetry. Hence the coefficients B) = B2 = .. = Bn = 0, and only cosine terms will be present in the Fourier series. v 2f- - 1 f- ro!! ~" 2 0 " 2" '" -If- -2 20.14 Fig. 20-10 Shifting the vertical axis of the graph of a periodic wave can reduce the number of terms in Fourier series. In Fig. 20-11a the wave has half-wave symmetry. Thus, only odd harmonics are present. Shift the vertical axis so that sine terms may also be absent from the Fourier series. I The result of shifting the vertical axis is shown in Fig. 20-11h. For this choice of the vertical axis, only odd cosine terms are present in the Fourier series. 468 D CHAPTER 20 I I I I a-O (a) I I I a-O Fig. 20-11 (h) 20.15 Shift the vertical axis of the wave of Fig. 20-11a so that only odd sine terms are present in the Fourier serie I See Fig. 20-12. In this case we have no cosine terms, no even harmonics, and no constant term. I I I I Fig. 20-12 a-O 20.16 Determine the coefficients A o' A 3 , and 8 3 , for the 3OO-Hz voltage wave shown in Fig. 20-10. A =..!. [net area]2rr = .~ [-2(!!) + 1( 'IT) - 2(!!)] = -0.5 o 2'IT 0 2'IT 2 2 I Henc~" The periodic wave exhibits even symmetry. only cosine terms are present. B3 =0 A, . 20.17 1 = - 1T l"/2 -2cos3a da + -1 J 3 2 / " 1T 0 1T,'2 1cos3a da + -1 1T 12". 3/21T Therefore, -2cos3a da =0.637 Write the third harmonic voltage of the wave of Prob. 20.16 in the time domain. I But e3 = 1).637 cos 3a a = wt = 2'IT300t. Thus, e 3 = 0.637 cos 3(2'IT3OO)t = 0.637 cos 5654.87t = 0.637 sin (5654.87t + 90°) 20.18 What is the rms value of the third harmonic voltage of the waveform shown in Fig. 20-10? I From Prob. 20.17 E 3 • max = 0.637 20.19 or E= For the wave shown in Fig. 20-13, obtain A o ' A:;, and 8 5 , 0~7 = 0.45 V V CIRCUITS WITH NONSINUSOIDAL SOURCES D 469 1 ~ "2 !3 2 0 -1r 1 1 ~" 2" " a ---- i Fig. 20·13 I Because of even symmetry, Bs = O. 1 Ao = 2 7T [net area]"" 1 As=- 1T 20.20 1 2 7T [6( 7T) - 3( 7T)] = 1.5 = 1 1"'/2 6cos5adx+-1 J,3/2'" -3cos5ada+-1 2" 1T 0 7r/2 1T 6cos5ada=1.15 3/21T Determine the frequency of the fifth harmonic of the voltage wave shown in Fig. 20-13. I From Fig. 20-13 T= 0.01 s 20.21 1 f= - = 100 Hz T or f, = 5(100) = 500 Hz Express the fifth harmonic of the voltage wave of Fig. 20-13 in the time domain. I From Prob. 20.19 ES •max = 1.15V From Prob. 20.20 w = 27T(500) = 3140rad/s e s = 1.15 cos 3140t Thus, 20.22 V The amplitude and phase angle of any harmonic may be obtained by combining the respective sine and cosine components for that frequency. For example, the addition of the two components of the k harmonic is Ak cos kwt + Bk sin kwt = C k sin (kwt + 4>n (1) where Ck == amplitude of harmonic k and 4>k = phase angle of harmonic k. Express Ck and 4>k in terms of Ak and B k. I Converting the cosine term to an equivalent sine term, Ak sin (kwt + 90°) + Bk sin kwt = C k sin (kwt + 4>~) Expressing Eq. (1) in phasor form, A k /90° + B k ~ = Ck 14>~ (2) Hence, (3) 20.23 Write the Fourier-series equation represented by the following Fourier coefficients of a certain 20-Hz nonsinusoidal periodic voltage wave. Ao = 8.00 A) = -6.63 A2 = 0.26 A3 = -0.97 A4 B) B2 = -4.99 B3 = 3.33 B4 = -2.49 = 10.00 = 0.26 I Using (2) and (3) of Prob. 20.22, the calculated values for Ck and 4>k are As = -0.52 B5 = 1.99 A6 = 0.26 B6= -1.65 470 D CHAPTER 20 k I,Hz Ck,V t/lk' degrees 0 1 2 3 4 5 6 0 20 40 60 80 100 120 8.00 lc2.00 5.00 3.47 2.50 2.06 1.67 -33.54 +177.02 -16.24 +174.04 -14.64 + 171.05 Using the calculated data, the Fourier-series representation of the nonsinusoidal voltage wave is e = 8 + 12 sin (21T20t - 33.54°) + 5.0 sin (21T40t + 177.0n + 3.47 sin (21T60t - 16.24°) + 2.5 sin (21T80t + 174.04°) + 2,(16 ~iIl (21T100t - 14.64°) + 1.67 sin (21T120t + 171.05°) 20.24 The frequency spectrum of a nonsinusoidal periocli<: wave is a plot of harmonic amplitude versus harmonic frequency. The plot is in the form of a bar graph or line graph as shown in Fig. 20-14. No frequencies exist between the plotted lines. Plot the frequency spectrum of the wave expressed by the coefficients given in Prob. 20.23. =~~~-:-:-:-~~~~~ ~ ~+------\ - .- --~ 3 5 Humoni. Fig. 20-14 I The frequency spectrum is plotted in Fig. 20-IS. 10 I;> -i iE" « 5 - '---A....-'------L--L-J__ o 20.25 > - I_ 20 40 60 80 100120 Fig. 20-15 Determine the rss voltage (see Chap. 19) of the ""ave given in Prob. 20.23. I E = ~(8)2 + (E..)2 + (~)2 + (~±Z)2 + (2.5)2 + (2.06)2 + (1.67)2 = 12 6 V2 V2 ,t;! V2 V2 V2 . 9V <SS CIRCUITS WITH NONSINUSOIDAL SOURCES 20.26 D 471 The following Fourier coefficients represent the components of a certain 20-Hz periodic current wave in a 6-.0 resistor: Ao = -8.70 A. = -4.64 A2 = 4.14 A3 = -0.59 B. = -6.06 B2 = 0.47 B3 = -0.24 As = 0.22 B4 = -0.38 Bs = 0.26 Plot the frequency spectrum. I The Fourier coefficients are Ao = -8.70 Ck = Bk + jAk C. = (-6.06 - j4.64) = 7.63 / -142.56° C2 = (0.47 + j4.14) = 4.17 /83.52° C3 = (-0.24 - jO.59) = 0.64 / -112. W C4 = (-0.38 + jO.57) = 0.69 /123.69° Cs = (0.26 + jO.22) = 0.34 /40.24° The frequency spectrum is plotted in Fig. 20-16. i , 7 !,- ... ., z 1- 0 20.27 "'to I I &0 to 100 Determine the rss current in the resistor of Prob. 20.26. I 20.28 2.0 I <SS =~(-87)2 +v1 (7.63)2 (4.17)2 (0.64)2 (0.69)2 (0.34)2=1068A +v1 +v1 +v1 +v1 . • How much average power is drawn from the resistor of Prob. 20.27? I 20.29 Determine the de term and the coefficients of the first and second harmonics of the periodic wave shown in Fig. 20-17. 1 I 1 [1(6)] (27T) = 3 Ao == 27T [net area] = 27T The equation of the ramp is determined by y=mx+b 1 A. == -:;;: v=(2~)a+0=(~)a loe"- (-:3; : a ) cos a da = 3 [1 7T2 12 cos a + 11 (3~ )sin a da = -32 [12 B = • 7T ,,- 0 7T 7T 12 I sin a ]2,,- = 0.304 0 sin a - ~ cos a ]2,,- = -1.910 1 0 c. = (-1.910 + jO.304) = 1.93 /170.96° 1 A2 = -:;;: lor h ( -:3; : a ) cos 2a da = 3 [1 7T2 22 cos 2a + ~ sin 2a ] 02,,- = 0 1 (h ( 3 ) 3 [1 ] 2,,B2 = -:;;: )0 4 a sin 2a da = 7T2 22 sin 2a - ~ cos 2a 0 = -0.955 C2 = B2 + jA2 = (-0.955 + jO) = 0.955 /180° 472 D CHAPTER 20 Fig. 20-17 20.30 Write the time-domain equation represented by the coefficients of Prob. 20.29. I The frequency of the nonsinusoidal wave is 1 1 f= 1; = (1.02 = 50 Hz v = 3 + 1.93 sin (2'IT50t + 170.96°) + 0.955 sin (2'ITIOOt + 180°) = 3 + 1.93 sin (314.16t 20.31 + 170.96°) + 0.955 sin (628.32t + 180°) Determine the dc term and the coefficient of the fifth harmonic for the voltage wave of Fig. 20-18. I [112:\(6)2'IT) - 3( 2'IT)] = 0 1 [net area]o2rr =. 21T 1 Ao = 2'IT The equation representing the ramp function is (--~.\~ + 0 = v= 1 As=. 'IT 1rr/2 0 Bs = -1 'IT f (12 ) 1 - a cos5adc/+'IT 'IT rr 2 / 0 'lT/2J 12rr (12) - a sin 5Q~ dx + -1 'IT 'IT Cs = Bs + jAs = 12 a 'IT 3/2rr -3cos5ada=0.142 f2rr 3/2rr - 3 sin 5a da = 0.255 0.255 + jO.142 = 0.292 /29.11° --+---'---- er f+---O.OOll----j Fig. 20-18 20.32 Write the time-domain equation represented by the coefficients in Prob. 20.31. 1 I v 20.33 1 f= T = il:ool , = 1000 Hz = 0 + 0.292 sin [2'IT1000(5t) + 29.11°] = 0.292 sin (31,415. 95t + 29.11 0) Show that the Fourier series containing sine and cosine terms may be written as f(t) = ... + where I A_2e-i2wl + A .. le -iwl and + Ao + Ale iwt + A2ei2wl + ... (1) (2) Expressing each of the sine and cosine term:; in the trigonometric Fourier series by its exponential equivalent, the result is a series of exponential term~; CIRCUITS WITH NONSINUSOIDAL SOURCES D 473 Rearranging, fi( t)== ... +(A 2 _ B2)e-j2WI +(AJ _ BI)e-iW(+ Aa +(AJ + BI) iW(+(A2 + B2) i 2w (+ ... 2 2j 2 2j 2 2 2j e 2 2j e (3) Now defining An and A_n as given by (2), (3) may be rewritten as (1). 20.34 Evaluate the coefficients An given by (2) of Prob. 20.33. I To obtain the evaluation integral for the An coefficients, we multiply (1) of Prob. 20.33 on both sides by e -inw( and integrate over the full period: frr f(t)e- inw ( d(wt) = ... frr A_ 2e- i2W ( e- jnw ( d(wt) + frr A_Je- jW ( e- jnw ( d(wt) 2rr jnw f2rr jW inw + f A e- ( d(wt) + A e ( e- ( d(wt) + ... 2rr A einw ( e - inw( d(wt) + ... + fo + () 0 0 1 n (1) The definite integrals on the right side of (1) are all zero except J~rr An d(wt), which has the value 21TAn' Then An = 211T 20.35 frr f(t)e- inw ( d(wt) or An = ~ r f(t)e- i2rrn ('T dt (2) Express the real Fourier coefficients A nand Bn in terms of complex coefficients An' defined by (2) of Prob. 20.33. I Since f(t) is real, A_n = A~ and only positive n need be considered. Hence from (2) of Prob. 20.33, we obtain: and 20.36 An =2Re(An) (1) Bn = -2 Im (An) (2) Equation (1) of Prob. 20.33 is known as the exponential Fourier series. the waveform shown in Fig. 20-1. Find the exponential Fourier series for I In the interval 0<wt<21T the function is given by f(t) = (10/21T)wt. By inspection, the average value of the function is _ An - Ao = 5. Substituting f(t) in (2) of Prob. 20.34, we obtain the coefficients An' jnw [e- ( . J2rr _. (_jn)2(-lnwt-1) 0 -121Tn 2rr -jnw( _ 10 1 r ( 10) Jo 21T wte d(wt)-(21T)2 21T 10 Inserting the coefficients An in (2) of Prob. 20.33, the exponential form of the Fourier series for the given waveform is -,"2w( . 10 -jW( . 10 jw( . 10 j2w( f(t) = ... - 1 - e - 1- e +5+1- e +1- e + ... (1) 41T 21T 21T 41T . 10 20.37 Verify that (1) of Prob. 20.36 is consistent with (1) of Prob. 20.1. I Ao = 0 by inspection. From (1) of Prob. 20.35: 10 ( .10) = - -1Tn B = -2 Im 1 n 21Tn Hence, 20.38 JF,() t . 2 wt - -10. == 5 - -10. SIn wt - -10 SIn SIn 3 wt - ... 1T 21T 31T Obtain the frequency spectrum of the waveform of Fig. 20-1. I The series has only sine terms, as obtained in spectrum is shown in Fig. 20-19. (1) of Probs. 20.1 and 20.37. Hence the desired frequency 474 D CHAPTER 20 5 ~-L_ _L-~__~_~I o 20.39 234 __~I~___ 5 6 n Fig. 20-19 Plot the first four terms of the series: JF,() t = 5 - -10. 10. SIn wt·- ;;-SIn 2 wt - -10. SIn 3 wt - ... 7T 37T LT.' Verify how closely the resultant of these four terms approximates a sawtooth. I See Fig. 20-20. 10 5 ~----------~~--------- Fig. 20-20 20.40 The voltage v and the current i in a circuit are givt:n by v = Vo + L Vn sin (nwt + 4>n) and Obtain expressions for instantaneous and average powers. I Instantaneous power is given by p = vi = [Vo + L Vn sin (nwt + 4>n)][ 10 + L In sin (nwt + !/In)] Since v and i both have period T, their product must have an integral number of its periods in T. (Recall that for a single sine wave of applied voltage, the product vi has a period half that of the voltage wave.) The average may therefore be calculated over one period of the voltage wave: p = ~ r [Vo + L Vn sin (nw! +- 4>n)][ 10 + L In sin (nwt + !/In)] dt Examination of the possible terms in the product of the two infinite series shows them to be of the following types: the product of two constants, the product (of a constant and a sine function, the product of two sine functions of different frequencies, and sine functIOns squared. After integration, the product of the two constants is still Volo and the sine functions squared with the limits applied appear as (VJ)2) cos (4)n - !/In); all other products upon integration over the period T are zero. Then the average power is (1) where On = 4>n -!/In is the angle on the equivalent impedance of the network at the angular frequency nw, and Vn and In are the maximum values of the respl~ctive sine functions. 20.41 Obtain the expression for the average power in a circuit excited by a single-frequency source, from (1) of Prob. 20.40. I In the special case of a single-frequency sinusoidal voltage, Vo = V2 = V) = ... = 0, and (1) of Prob. 20.40 reduces to the familiar CIRCUITS WITH NONSINUSOIDAL SOURCES 20.42 A series RL circuit in which R = 5 0 and L = 20 mH has an applied voltage 25 sin 3wt V, with w = 500 rads/s. Find the instantaneous current. D 475 v = 100 + 50 sin wt + I Compute the equivalent impedance of the circuit at each frequency found in the voltage function. Then obtain the respective currents. At w = 0, Zo = R = 5 0 and I = Vo = 100 = 20 A o At w = 500 rad/s, it = At 3w = 1500 rad/s, . I) = R 5 Zt = 5 + /(500)(20 x 10-)) = 5 + jlO = 11.15/63.4° 0 V~~ax sin(wt-(Jt)= 1:.~5 sin(wt-63.4°) = 4.48 sin (wt-63.4°) Z) = 5 + j30 = 30.4/80.54° 0 V),max. ~ SIn (3 and A and (J ) 25 . ( 5 . ( wt - ) = 30.4 SIn 3wt - 80. 4°) = 0.823 SIn 3wt - 80.54°) A The sum of the harmonic currents is the required total response: i = 20 + 4.48 sin (wt - 63.4°) + 0.823 sin (3wt - 80.54°) 20.43 A Determine the power dissipated in the resistor of the circuit of Prob. 20.42. I This current has the effective value leff 2 2 2 = '120 + (4.48 /2) + (0.823 /2) = V41O.6 = 20.25 A which results in a power in the 5-0 resistor of P = l~ffR = (410.6)5 = 2053 W 20.44 Calculate the power contributed by each harmonic in the circuit of Prob. 20.42. powers is the same as the power obtained in Prob. 20.43. I At At Po = Vo/o = 100(20) = 2000 W w =;,0: Pt = w = 500 rad/s: At 3w = 1500 rad/s: WJt cos (Jt = H50)(4.48)cos63.4°=50.1 W p) = ~ V) I) cos (J) = ~ (25)(0.823) cos 80.54° = 1.69 W P= 2000 + 50.1 + 1.69 = 2052 W Then 20.45 Verify that the sum of these Determine the instantaneous and effective voltages across the resistor of Prob. 20.42. dissipated in the resistor. Hence obtain the power I The Fourier-series expression for the voltage across the resistor is and v R = Ri = 100 + 22.4 sin (wt - 63.4°) + 4.11 sin [3wt - 80.540) V 2 V • = '1100 + H22.4)2 + H4.11)2 = ylO,259 = 101.3 V R eff Then the power dissipated in R is 20.46 P = V~,eff/ R = (10,259)/5 = 2052 W. A voltage of the waveform shown in Fig. 20-21 is applied to a capacitor C. Find the current. .. wt ----.----r-----=-JI--~--..__---.l Fig. 20-21 476 D CHAPTER 20 I In the interval -7T < wt < 0 the voltage function is v = Vrnax + (2Vrna .l7T)wt; and for 0 < wt < 7T, v = Vrnax - (2Vrna .l7T)wt. Then the coefficients of the exponential series are determined by the evaluation integral from which Vn = 4Vrna.l7T2n2 for odd n, and Vn = 0 The phasor current produced by Vn (n odd) is for even n. Vn 4Vrn.,17T2n2 . 4Vrnax wC In = Zn = -li,nw C = ] 7T 2n with an implicit time factor e jnwt The resultant current is therefore +00 4V w C +00 jnwt i(t) = Ine,nwt =: j rnax2 _e'TT n • L L -00 -00 where the summation is over odd n only. 20.47 Obtain the Fourier series for the square wave shown in Fig. 20-22. I In the interval 0 < wt < 7T, f(t) = V; and for 7T < wt < 27T, f(t) = - V. The average value of the wave is zero; hence A o/2 = O. The cosine coefficients are obtained by writing the evaluation integral with the functions inserted as follows. An = ; = 0 {r {r V cos nwt d(wt) + f" (- V sin nwt d(wt) + V = -7Tn ~ sin nwtJ: - [~ sin nwtI"} for all n Thus the series contains no cosine terms. Bn = ; V) cos nwt d(wt)} = .; {[ (-cos n7T Then Bn = 4VI7Tn wave is for Proceeding with the evaluation integral for the sine terms, f" (- V) sin n,,,t d(wt)} = .; {[ - ~ cos nwtJ: + [~ cos nwtI"} 2V (1 - cos n7T) 7Tn b n = 0 for n = 2, 4, 6, ... , The series for the square + cos 0 + cos n27T - cos r':7T) = n = 1,3,5, ... , and 4V sm . 3 wt + -4V. 5 t = 4V. - Sill wt + -Sill wt + ... f() 7T 31T 57T A V 0 -V 20.48 fT .) •• fT wt 3fT f----- Fig. 20-22 Plot the frequency spectrum for the waveform shown in Fig. 20-22. Fig. 20-23 CIRCUITS WITH NONSINUSOIDAL SOURCES 0 477 I The line spectrum for this series is shown in Fig. 20-23. The series contains only odd-harmonic sine terms, as could have been anticipated by examination of the waveform for symmetry. Since the wave in Fig. 20-22 is odd, its series contains only sine terms; and since it also has half-wave symmetry, only odd harmonics are present. 20.49 Repeat Prob. 20.47 for the waveform shown in Fig. 20-24. I The wave is an even function since f(t) = f(-t), and if its average value VI2 is subtracted, it also has half-wave symmetry, i.e., f(t) "" - f(t + 7T). For -7T < wt < 0, f(t) = V + (VI7T)wt; and for f(t) = V ~ (VI7T)wt. Since even waveforms have only cosine terms, all bn = O. For n ~ 1, An = -1 JO 7T -'" = -V {J'" 7T [V + (VI7T)wt] cos nwt d(wt) + -1 f'" [V - 7T - '" 7T, (VI7T)wt] cos mwt d(wt) 0 cos nwt d(wt) + JO ~t cos nwt d(wt) - -'" 0 < wt < f'" 7T 0 ~t cos nwt d(wt) } 7T {[ 21 cos nw t + -wsm t . nw ]O t - [ 21 cos nw t + -wt. sm nw t ]"'} n 7T -'" n n 0 V 2V = ~ {cos 0 - cos (-n7T) - cos n7T + cos O} = ~ (1- cos n7T) 7Tn 7Tn As predicted from half-wave symmetry, the series contains only odd terms, since 2,4,6, ... , For n"" 1,3,5, ... , An = 4V/7T 2 n 2• Then the required Fourier series is "" 2V 7T An = 0 for n "" V 4V 4V 4V f( t) = - + - 2 cos wt + - - cos 3wt + - - cos 5wt + ... 2 7T (37T)2 (57T)2 Fig. 20-24 20.50 Obtain the frequency spectrum of the waveform of Fig. 20-24. I The coefficients decrease as 11 n 2, and thus the series converges more rapidly than that of Prob. 20.47. This fact is evident from the line spectrum shown in Fig. 20-25. eft ~" I () 20.51 2 3 4 5 6 7 Fig. 20-25 8 Repeat Prob. 20.47 for the waveform of Fig. 20-26. I By inspection, the waveform is odd (and therefore has average value zero). Consequently the series will contain only sine terms. A single expression, f(t) = (VI7T)wt, describes the wave over the period from to + 7T, and we will use these limits on our evaluation integral for b n' 1. J'" ~ = - 2V (cos n7T) n n -'" n7T As cos n7T is + 1 for even nand -1 for odd n, the signs of the coefficients alternate. The required series is Bn"" 7T (VI7T)wt sin nwt d(wt) = -'" [.; sin nwt - wt cos nwtJ'" -7T 7T ,,() 2V {sm . wt t "" -:;; J' t :1 • 2 sm wt + :31. sm 3wt - t 4 . sm 4wt + ... } 478 D CHAPTER 20 -4 ~ __ 1Tf----:;V"jt:-'-+-. Fig. 20-26 20.52 Obtain the frequency spectrum of the waveform of Fig. 20-26. , The coefficients decrease as 1I n, and thus the !;eries converges slowly, as shown by the spectrum in Fig. 20-27. Except for the shift in the origin and the average term, this waveform is the same as in Fig. 20-1; compare the two spectra. CA 2V/1T 234 () 20.53 5 6 7 R Fig. 20-27 9 Find the Fourier series for the waveform of Fig. 20-28. , In the interval 0 < wt < 7T, f(t) = (VI7T)wt; and for 7T < wt < 27T, f(t) = O. By inspection, the average value of the wave is V14. Since the wave is neither even nor odd, the series will contain both sine and cosine terms. For n > 0, we have An = -1 f'" (VI7T)wt cos nwt d(wt) = -2: V i,- -'2 1 cos nwt + -wt. V (cos n7T sm nwt J'" ="""'22 7TO When n is even, 7T-n cos n7T -1 =0 and an = O. n When n is odd, an = 07Tn 2 2 -2VI(7T n ). The Bn coefficients are 1f'" (VI7T)wtsmnwt . d( wt)="2 V [1. wt ]'" =--(cosn7T)=(-1) V -:;smnwt--cosnwt 7T n°' n 7Tn B =- 7T 0 Then the required Fourier series is n 1) 0 V 2V 2V 2V f(t) = - - - 2 cos wt - - - cos 3wt - - - - cos Swt - ... 4 7T (37T)2 (S7T)2 V . + - sm wt 7T - V 27T . V n+l -V 7Tn . sm 2wt + - sm 3wt - ... 37T v () 20.54 1T Fig. 20-28 Plot the frequency spectrum for the waveform shown in Fig. 20-28. , The even-harmonic amplitudes are given directly by IBnI since there are no even-harmonic cosine terms. However, the odd-harmonic amplitudes must be computed using Cn = YA~ + B~. Thus Cl = Y(2V/7T2)2 + (VI7T)2 = V(O.:·77) The line spectrum is shown in Fig. 20-29. CJ = V(0.109) C, = V(0.064) CIRCUITS WITH NONSINUSOIDAL SOURCES D 479 c", o 20.55 23456789 Fig. 20-29 Repeat Prob. 20.53 for the waveform shown in Fig. 20-30. I The wave shows no symmetry, and we therefore expect the series to contain both sine and cosine terms. Since the average value is not obtainable by inspection, we evaluate a o for use in the term a o/2. 1'" V sm. wt d() wt = -V [ -cos wt ]'" = -2V Ao = -1 0 7TO 7T 7T Next we determine an: A = -1 n 7T 1'" V sin wt cos nwt d(wt) 0 V [-n sin wt sin nwt - cos nwt cos wt]'" 2 = 7T - n +1 0 = - With n even, An =2V/7T(I- n2); and with n odd, An =0. n = 1, and therefore we must integrate separately for A I • AI Now we evaluate Bn: B = -1 n = -1 7TO f"'v'sm sm. wt 7TO ( cos n7T + 1) ~ sin 2wtd(wt) = 0 nwt d() wt = -V [nsinwtcosnwt-sinnwtcoswt]'" 2 =0 -n+l 7T Here again the expression is indeterminate for BI 2 n ) However, this expression is indeterminate for 1'" V sin wtcos wtd(wt) = -V 1'" 7TO V 7T(1 - =;:1 Jo('" Vsm. 2 n = 1, wtd(wt) 0 and b I is evaluated separately. = ;:V [wt 2" - sin2wt]'" --4- 0 ="2V Then the required Fourier series is f(t) = ~7T {I + ~2 sin wt - 3~ cos 2wt - 2 15 cos 4wt - 235 cos 6wt - v - f - - - - - l o -_ _--''-_ _---I__. . . wt o 20.56 Fig. 20-30 Repeat Prob. 20.54 for the waveform shown in Fig. 20-30. I See Fig. 20-31. VI," 0 L, 2 3 4 i 5 , 6 I 7 Fig. 20-31 ... } (1) 480 20.57 D CHAPTER 20 Repeat Prob. 20.55 for the waveform shown in Fig. 20-32. I The function is described in the interval -7T < UJl < 0 by f(t) = - V sin wt. same as that in Prob. 20.55, i.e., ~Ao = VI7T. For the coefficients An' we have = -1 An IO 7T For n even, separa tel y . An (-V sin wt) cos nt!lt d(wt) ~'" = 2VI7T(1- n2); and for n odd, I An = V 2 (1 7T(I-n) = 0, The average value is the + cos n7T) except that n =1 must be examined O AI = 1 7T (-Vslnwt)coswtd(wt)=O -'" For the coefficients B n' we obtain I O Bn = except for 1.7T (- V sin wt) sin nwt d(wt) = 0 -'" n = 1. Thus the series is V{ 1 - f(t) = - 7T 2 2 cos 4wt - -2 cos 6wt - ... } -7T. SIn wt - - cos 2wt - 2 3 15 35 ~~~ o 20.58 1T 21T (1) Fig. 20-32 Compare (1) of Prob. 20.57 with (1) of Prob. 20.55. Verify that the graph of Fig. 20-30 can be obtained from that of Fig. 20-32. Thus, the two waveforms have ':he same frequency spectrum. I If V sin wt is subtracted from (1) of Prob. 20.55, or from Fig. 20-30, (1) of Prob. 20.57, or Fig. 20-32, is obtained. Hence, the two waveforms have the same frequency spectrum. 20.59 Obtain the Fourier series for the waveform shown in Fig. 20-33. I With the vertical axis positioned as shown, the wave is even and the series will contain only cosine terms and a constant term. In the period from -7T to + 7T used for the evaluation integrals, the function is zero except from -7T/6 to +7T/6. 1 Ao=- 7T I""6 1 An='- V Vd(wt)=-3 I""6 7T ~",/6 Since sinn7T/6=1/2,V3/2,1,V3/2,1I2,0,-1I2, ... series is for V f(t) = "6 + ~",/6 7T 2V Vcosnwtd(wt)=-sinn n7T 6 n=I,2,3,4,5,6,7, ... , 2V { 2 1 cos wt + T V3 ( iI ) cos 2wt + 1( '13 ) cos 3wt + T V3 (1) + -:;; '4 cos 4wt ~ G) cos 5wt - ~ ( ~ ) cos '1l!lt - V f(t) = 6 or ... } 2V ~~ 1 . n7T + -- L'.i 7T Fl n SIn - 6 cos nwt v .~°r-~ -1T/6 1T/6 ________~______~-,~~WI 1T Fig. 20-33 respectively, the CIRCUITS WITH NONSINUSOIDAL SOURCES 20.60 D 481 Plot the frequency spectrum of the waveform shown in Fig. 20-33. I The line spectrum, shown in Fig. 20-34, decreases very slowly for this wave, since the series converges very slowly to the function. Of particular interest is the fact that the 8th, 9th, and 10th harmonic amplitudes exceed the 7th. With the simple waves considered previously, the higher-harmonic amplitudes were progressively lower. VI," lv o 20.61 2 3 4 5 6 7 8 9 10 Fig. 20-34 11 Find the exponential Fourier series of the waveform shown in Fig. 20-22. I In the interval -7T < wt < 0, f(t) = - V; and for 0 < wt < 7T, f(t) = V. The wave is odd; therefore, A o = 0 and the An will be pure imaginaries. A= J... {Jo (_V)e- jnwf d(wt) + (" Ve- jnwf d(wt)} = ~ {_[_1_. e-jnwf]O + [_1_. e- jnwf ]"} 27T -" )0 27T (-]n) -" (-]n) ° n .V (-eO + e jn " + e- jn " _ eO) = j ~ (e jn " -1) (-]27Tn) n7T jn jn For n even, e " = + 1 and An = 0; for n odd, e " = -1 and The required Fourier series is An = - j(2Vln7T) (half-wave symmetry). . 2V -j3wf . 2V -jwf . 2V jwf . 2V j3wf +]-e -]-e -]-e - ... f( t ) ='''+]-e 37T 7T 7T 37T 20.62 Sketch the frequency spectrum from the results of Prob. 20.61. I The graph in Fig. 20-35 shows amplitudes for both positive and negative frequencies. Combining the values at + nand - n yields the same line spectrum as plotted in Fig. 20-23. /A../ 2V -+~-+-.-+-+-+~-+~-~n -5 -4 -3 -2 -\ 20.63 0 2 3 4 5 Fig. 20-35 Obtain the trigoriometric Fourier series coefficients from those of the exponential series of Prob. 20.61, and compare the results with those of Prob. 20.47. I The trigonometric-series cosine coefficients are and 4V Bn = -2ImA = n n7T for odd n only These coefficients agree with those obtained in Prob. 20.47. 20.64 Find the exponential Fourier series for the waveform shown in Fig. 20-24. I In the interval -7T < wt < 0, f(t) = V + (VI7T)wt; and for 0 < wt < 7T, f(t) = V - (VI7T)wt. The wave is even and therefore the An coefficients will be pure real. By inspection the average value is V12. 482 D CHAPTER 20 An = ~ {fO [V + (VhT)wt]e~jnwt d(wt) + (" [V - ~" 27T = V 2 {fO 27T -" For even n, (" Jo (-wt)e~jnwtd(wt)+f" 7Te~jnwtd(wt)} ~" ]0 2V '-jwt V jwt j3wt + - 2V - - e ~j3wt + ----c; e + - + -2V -2 e + - 2V -2 e + ... (-37T)2 (-7T)' 2 (7T) (37T) Find the amplitudes of harmonics in the waveform of Fig. 20-24. n = 2, 4, 6, .. . n=1,3,5, .. . V Co = 2 I 20.66 wte~jnwtd(wt)+ V {[ ~jnwt [~jnwt ]"} V = - 2 ~ (-jnwt-1) - ~ (-jnwt-1) = ~ [1- e jn ,,] 27T (-]n) -" (-]n) 0 7T n 2 2 jnrr e = + 1 and An = 0; for odd n, An = 2VI7T n . Thus the series is f(t) = ... 20.65 (VhT)wt]e~jnwt d(wt)} Jo Obtain the exponential Fourier series for the waveform shown in Fig. 20-30. I 0 < wt < 7T, and from 7T to 27T, f(t) = O. Then jnwt 1 1" . V [ e·]" V( -jmr + 1) A = -2 Vsinwte~}nwtd(wt)= -2 ---2-(-jnsinwt-coswt) = e( 2) n 7T 0 7T (1-n) 0 27T1-n In the interval f(t) = V sin wt; For even n, An = VI7T(1- n2); for odd n, An = O. However, for n = 1, the expression for An becomes indeterminate. L'Hopital's rule may be applied; i.e , the numerator and denominator are separately differentiated with respect to n, after which n is allowed to approach 1, with the result that Al = -j(VI4). The average value is Ao 1 = -27T Jo(" V sin w.' d(wt) = 2 7T [-cos wt]~ = ~ 7T V Then the exponential Fourier series is V ~j4wt V ~j2wt . V ~Jwt V . V jwt V j2wt V j4wt --e +]-e +--]-e - - e ---e - ... 157T 37T 4 7T 4 37T 157T f( t ) = " ' - - - e 20.67 Determine the amplitudes of harmonics in the waveform of Fig. 20-30. I The harmonic amplitudes are (n = 2, 4, 6, ... ) (n = 1) (n=3,5,7, ... ) V Co =A 0 = 7T - 20.68 A voltage waveform is given by O<t< TI4 TI4<t<3TI4 3TI4<t< T Determine its Fourier coefficients. I By inspection Ao = O. Also, the waveform ha> an even symmetry. Finally, with the period T= 27Tlw 2V 1T'4 A I = -T Since w T = 2'7T, o 211 f3T'4 cos wt dt - -1:- = 2V [Sin wT -(sin wT 4 we obtain T/4 cos wt dt 3~~~-sin 4 Thus, Bn = O. 2V IT + -T 3T/4 cos wt dt wT)+(sinwT-sin 3wT)] 4 4 CIRCUITS WITH NONSINUSOIDAL SOURCES Al V 4V = - (1 + 2 + 1) = TT 4V and nTT TT , n = 1,5,9, ... An = -4V ( nTT n=3,7,1l, ... o 20.69 D 483 n = even Obtain the frequency spectrum for the waveform of Prob. 20.68. I See Fig. 20-36, and compare with Fig. 20-27. c'" Fig. 20-36 20.70 Find the average power in a re$istance 5 sin 3wt + 2 sin 5wt A. R = 10 .0, if the current in Fourier-series form is I The current has an effective value leff = power is P = l~ffR = (64.5)10 = 645 W. 20.71 V!(1O)2 + !(5)2 + !(2)2 = \1'64.5 = 8.03 A. i = 10 sin wt + Then the average Obtain the result of Prob. 20.70 by adding the harmonic powers. I The total power is the sum of the harmonic powers, which are given by! VrnaJrnax cos O. But the voltage across the resistor and the current are in phase for all harmonics, and On = O. Then v R = Ri = 100 sin wt + 50 sin 2wt + 20 sin 5wt and 20.72 P = ~ (100)(10) + !(50)(5) + !(20)(2) = 645 W. Find the average power supplied to a network if the applied voltage and resulting current are v = 50 + 50 sin 5 x 10 3t + 30 sin 10 4t + 20 sin 2 x 10 4t i = 11.2 sin (5 3 V 4 x 10 t + 63.4°) + 10.6 sin (10 t + 45°) + 8.97 sin (2 x 10 4t + 26.6°) A I The total average power is the sum of the harmonic powers: P = (50)(0) + !(50)(11.2) cos 63.4° + !(30)(1O.6) cos 45° + H20)(8.97) cos 26.6° = 317.7 W 20.73 Obtain the constants of the two-element series circuit with the applied voltage and resultant current given in Prob. 20.72. I The voltage series contains a constant term 50, but there is no corresponding term in the current series, thus indicating that one of the elements is a capacitor. must be a resistor. leff = Y!(11.2)2 Since power is delivered to the circuit, the other element + !(1O.6)2 + !{8.97)2 = 12.6A The average power is P = l~ffR, from which R = P/ l~ff = 317.7/159.2 = 2.0. At w = 10 4 rad/s, the current leads the voltage by 45°. Hence, 1 1 = tan 45° = w CR or 1 C = (10 4 )(2) = 50 jLF Therefore the two-element series circuit consists of a resistor of 2.0 and a capacitor of 50 jLF. 484 20.74 D CHAPTER 20 The voltage wave shown in Fig. 20-37 is applied to a series circuit of trigonometric Fourier series to obtain the voltage across the resistor. R = 2 kll L = 10 H. and Use the v. V Fig. 20-37 I The applied voltage has an average value Vrna /7T. The wave function is even and hence the series contains only cosine terms, with coefficients obtained by the following evaluation integral: 600 n7T 1 J"'/2 An=300coswtcosnwtd(wt)= (1 2)COS2 7T 7T - n V ~",/2 Here, cos n7T/2 has the value -1 for n = 2, 6,10 .... , and +1 for n = 4, 8,12, ... , For n odd, 2 = O. However, for n = 1, the expression is indeterminate and must be evaluated separately. cos n7T/ TABLE 20.1 n nw, rad/s R,kfl nwL,kfl Zn,kfl 8n 0 1 2 4 6 0 377 754 1508 2262 2 2 2 2 2 0 3.77 7.54 15.08 22.62 2 4.26 7.78 15.2 22.6 0° 620 75.1° 82.45° 84.92° 1 J"'/2 2 • 300 [wt· sin 2wt]"'/2 300 A = 300cos wtd(wtl ,= - + --=) 7T ~",/2 . 7T 2 4 ~",/2 2 V v = -300 {7T 1 + - cos wt + -2 cos 2wt - -2 cos 4wt + -2 cos 6wt - ... } V 7T 2 3 15 35 In Table 20.1, the total impedance of the serie~; circuit is computed for each harmonic in the voltage expression. The Fourier coefficients of the current ~.eries are the voltage series coefficients divided by the Zn; the current terms lag the voltage terms by the pha~;c angles On' Thus, I = 300/7T o 2 mA, . _ 300/2 0 4.26 cos (wt - 62 ) I) - mA, . 12 = 600/37T ~ cos ° (2wt -75.1 ) mA, ... Then the current series is ._300 1- 300 0 600 ° 27T + (2)(4.26) cos (wt - 62) + ~3;(7.78) cos (2wt -75.1 ) 600 0 600 ° 157T{15.2) cos (4wt - 82.45 ) +- 357T(22.6) cos (6wt - 84.92) - ... mA and the voltage across the resistor is vR = Ri = 95.5 + 70.4 cos (wt - 62°:, -I- 16.4 cos (2wt -75.1°) - 1.67 cos (4wt - 82.45°) -I- 0.483 cos (6wt - 84.92°) - ... 20.75 V Sketch the spectra of the applied voltage and uR to show the effect of the inductance on the harmonics. Assume w = 377 rad/s. I Figure 20-38 shows clearly how the harmonic amplitudes of the applied voltage have been reduced by the lO-H series inductance. CIRCUITS WITH NONSINUSOIDAL SOURCES D 485 ~",v 3OO/fr o 2 (a) 20.76 3 4 5 6 o 7 234567 Spectrum of u (b) A current waveform is shown in Fig. 20-39a. Spectrum of Fig. 20-38 UJt Sketch its derivative. I The derivative is shown in Fig. 20-39b. ____,-____~~----+-----~----_,r_--~~----_.--~wl ~A'~ d(wl)' 20tfr -fr () I fr 2fr 10 wl -20/fr Fig. 20-39 20.77 The current in a lO-mH inductance has the waveform shown in Fig. 20-39a. the voltage across the inductance, given that w = 500 rad/s. Obtain the trigonometric series for I The derivative of the waveform of Fig. 20-39a is graphed in Fig. 20-39b. This is just Fig. 20-22 with V= -20hr. Hence, from Prob. 20.47 di 80. . . d(wt) = - 7T 2 [sm wt + ism 3wt + } sm 5wt + ... ] and so 20.78 VL A di 400[. \. \. ] = Lw d(wt) =- 7 sm wt + '3 sm 3wt + "5 sm 5wt + ... v In Prob. 20.68, let w = 1. Obtain the trigonometric Fourier series for the waveform given in Prob. 20.68. Sketch the given waveform. Assume V = 7T / 4. I The sketch is shown in Fig. 20-40. Using the Fourier coefficients obtained in Prob. 20.68, with w = 1, the required Fourier series is v(t) = cost-icos3t+}cos5t-··· (1) 486 D CHAPTER 20 ""/4 ~% 1112. 0 t -rrr T -7T"/+ 20.79 Fig. 2()'40 If the voltage wave of Prob. 20.78 is applied to a series RL circuit, having R = In and determine the phasor currents for the first five harmc.nics of the input voltage wave. I From (1) of Prob. 20.78, the phasor voltages are VV = le w J - l 1 :!oe ~jl80° The admittances for the various harmonics are . 1 Y(Jn) = l+jn Y(j5) = ~ e~j788° a.nd Hence, A A 20.80 Obtain the current in the circuit of Prob. 20.79 in I + 0.105 cos (3t - 251.6°) + 0.039 cos (5t -78.8°) +... Ps or = vsis = (~ cos 5t)[0.039 cos (51 --78.8°)] = 0.0078 cos 5t cos (5t -78.8°) Ps = ! (0.0078) cos 78.8° = 0.7575 mW Sketch the frequency spectrum of the waveform given by 1 f ()t = 4 -;:1 (.sm t + :21.sm 2t + :31.sm 3t + ... ) I See Fig. 20-41. 0." tJ./ Fig. 20·41 20.83 A What is the average power associated with the fifth harmonic current in the circuit of Prob. 20.79? I 20.82 time domain. The required current is given by i(t) = 0.707 cos (t - 45°) 20.81 th,~ A Repeat Prob. 20.82 for f I See Fig. 20-42. ()t = -1(1 + -2 cos t TT TT :2 +~ .3 2cos 4t + ... ) cos 2t - - 15 L = 1 H, CIRCUITS WITH NONSINUSOIDAL SOURCES 0 487 c,., 6 .40.) 0.( Fig. 20-42 20.84 Find the amplitude of the third harmonic in the waveform shown in Fig. 20-43. I Because the function is an even function, only cosine terms are nonzero. 4lT/2 2t A 3 = -T o T - Thus, 21T 2 4 (3t) dt = - 2 (cos 31T - 1) = - - 2 T 91T 91T cos - fff) J7vv T 20.85 2T ~t . Fig. 20-43 Determine the amplitude of the fifth harmonic of the waveform shown in Fig. 20-44. I Because of even symmetry, A5 4 [ e Bn = O. Thus, (2 21T = 4 Jo t cos "4 (St) dt + J1 cos 21T ] 4 ( 51T) "4 (5t) dt = 251T 2 cos "2 - 1 = - 4 251T 2 L t 20.86 Fig. 20-44 Obtain the amplitude of the fundamental of the waveform shown in Fig. 20-45. I r 2 [(bT at . 21T r a(t - T) 21T ] B =1' Jo bTslllTdt+JbT -(l_b)TcosTtdt A _2[(bT at 21T a(t-T) 21T ] a(cos21Tb-l) 1' Jo bTcosTtdt+JbT - (l_b)Tcosrtdt = 21T2b(1-b) 1 - 1 Thus, C =' I A2 1 a sin 21Tb = 21T 2b(1-b) + B2 = aYl - cos 21Tb VII Y21T 2b(1- b) Fig. 20-45 (1) 488 20.87 0 CHAPTER 20 From the result of Prob. 20.86, obtain the amplitude of the waveform of Fig. 20-43. I a == 1 and Figure 20-43 is a special case of Fig. 20-45, when Prob. 20.86 yields yT- C']S 211(0.5) CI = , V211-(O.S)(l - 0.5) 20.88 b == 0.5. Substituting these values in (1) of 4 11 ==±2 Verify the result of Prob. 20.87 by a direct evaluatJOn of the Fourier coefficients for the fundamental of the waveform shown in Fig. 20-43. 4lT/2 2t 211 2 4 A == - cos -- .. £It == - (-1 + cos 11) == - I ToT T 112 112 I which is the same solution as in Prob. 20.87. 20.89 In a special case, a == 1 and b == 0.75 in the waveform of Fig. 20-45. If this waveform represents the voltage across a 1-n resistor, determine power com~sponding to the fundamental component of power. I From (1) of Prob. 20.86, 1yT- cos 211(0.75) Cl == V2(112)(075)(1 _ 0.75) == 0.382 or VI == 0.382 sin wt 0 . 2 .146slll wt W PI == !«(1.146) == 73 mW and 20.90 (V I )2 PI==~== V Compare the power obtained in Prob. 20.89 to the case when the applied voltage has the waveform shown in Fig. 20-43. I From Prob. 20.87, Thus, VI == 4 2 11 . Sill (4)2 . 2 PI == (112)2 Sill wt v wt 1 (' 4 -;- PI == -2 , ?T W )2 ==82.13mW which is (obviously) greater than the power (73.0 mW) calculated in Prob. 20.89. 20.91 Determine the Fourier series for the waveform shown in Fig. 20-46. I Thus, f(t) == 10 cos 1ft 1-;; 2 lTn t 20 A == 10 cos '!!- dt == o To T 11 4lT/2 11t 211nt A == lOcos -- cos - - dt n ToT ' T Hence, 40 (1 1 211t f(t) == --; 2 + '3 cos -f -- and 1 411t 1 611t 15 cos T + 35 cos T fco 10 --t-----'----4-~'L T 'l-r ~T Fig. 20-46 -.,. ) CIRCUITS WITH NONSINUSOIDAL SOURCES 20.92 0 489 Determine if the functions shown in Fig. 20-47 are odd or even. I (a) Even, (b) odd. _~_+---1-_~t _T 'i Fig. 20·47 20.93 Repeat Prob. 20.92 for the functions shown in Fig. 20-48. I (a) Odd, (b) odd. 1fl+> d V/ _T/-,.. -l 7"h Lbj 20.94 Fig. 20-48 Repeat Prob. 20.92 for the functions of Fig. 20-49. I (a) Even, (b) odd. fit) --~~--~----~f Fig. 20·49 (6) 20.95 Determine the coefficients of the exponential Fourier series for the waveform shown in Fig. 20-3. I 20.96 Evaluate C 2 and C 3 for the waveform of Prob. 20.95. I 20.97 (1) From (1) of Prob. 20.95: Find the coefficients of the exponential Fourier series for the waveform shown in Fig. 20-43. I ! C0-2 - 2lT/2 -2 ( (-T) e -j2""IT d(_- - -2C 2 2 n ToT 4 1T n n odd 490 20.98 0 CHAPTER 20 Repeat Prob. 20.97 for the waveform of Fig. 20-44. Co = ~ I e =~ n 20.99 4 [eJo te-jhnr/4 dt + Jl(3 e- j2"nt/4 dt + J4 - (t - 4)e- j2"n r/4 dt] = --h (cos 1Tn - 1) 1T n 2 J Repeat Prob. 20.97 for the waveform of Fig. 20-45. e I = n ~ [(bT .!!!... e-j2"nIlT dt + ( T Jo I or 20.100 -a(t - T) -jhnrlT dt] JbT (l-b)T e bT en 1=~{V2(1-cos21Tnb) 47.- 2 n 2b( 1 - b) Repeat Prob. 20.97 for the waveform of Fig. 20-46. TI2 20.101 ~-I 1Tt -j2"nrIT dt =10 e n = T-l f-T/2 10 cos-e T 2T I Z [j("IT)(1-2n)r + e -j("IT)(1+2n)r dt] e f = -1'2 20 cos 1Tn2 1T(1-4n) A series RC circuit, having R = 1 nand C = 1 F. is excited by a voltage source of the waveform of Fig. 20-40. Assuming w = 1 rad/s, determine the phasor current for the nth harmonic. I In terms of exponential Fourier series we have Consequently, -j(,,/2)(n -I) I == e _ _ __ n n- j Hence, 20.102 Repeat Prob. 20.101 if the input voltage has the waveform of Fig. 20-1. I In this case: 1 Yn= l-(j/n) 20.103 n Determine the average power for a network having v = 2 cos (t + 45°) + cos (2t + 45°) + cos (3t - 60°) V i = 10 cos t + 5 cos (2t - 45°) and I 20.104 'lO 1T(n - j) I = --,Jc-_ P = H2 x lO) cos 45' + HI x 5) cos 90° = 7.07 W In a three-phase network we have ia = 15 sin (wt - 30°) A A Determine it' I Since i, 20.105 In a network I Thus, v = -ia - ib = 15 sin (wt + 150°) + 0.5 sin (3wt + 135°) = 2 sin t + sin 3t and i = sin 2t + 0.5 sin 4t. What is the average power? p = vi == (2 sin t + sin 3t)(sin 2t + 0.5 sin 4t) P=O. A /7 CHAPTER 21 i / Laplace Transform Method 21.1 The Laplace transform method is a powerful technique for solving circuit problems. We define a Laplace transform as follows: Let f(t) be a time function which is zero for t=:;O and which is (subject to some mild conditions) arbitrarily defined for t > 0. Then the direct Laplace transform of f(t) , denoted .P[f(t)], is defined by .P[f(t)] = F(s) = r f(t)e- SI dt (1) Thus, the operation .P[ ] transforms f(t), which is in the time domain, into F(s) , which is in the complex frequency domain, or simply the s domain, where s is the complex variable a + jw. Using this definition find the Laplace transform of (a) the unit-step function and (b) the exponential decay function Ae -al. I .P[u(t)] = (a) (P[A e -al] = (b) 21.2 oL 1 00 (P[' oL Sill ] wt = (. o ) Sill wt e 1 (l)e- SI dt = - S [e-SI]~ 1 =S 100 A e -ale -si dt= -- A- [-(a+S)I]OO Ae =a+s o 0 s+a -sld [-S(Sinwt)e-SI-e-slwcoswtJoo w t= 2 2 = -2--2 s +w 0 s +w Find the Laplace transform of a derivative df(t)ldt . .p[ df(t)] = (00 df(t) e-sl dt )0 dt dt I Integrating by parts, 1.5 21.4 oo o Find the Laplace transform of sin wt. I 21.3 l 'p[d~~)]=[e-S1(t)]~+ r - f(t)(-se-SI)dt=-f(O+)+s r f~t)e-Sldt=-f(O+)+sF(s) Prepare a table of Laplace transforms of some commonly encountered functions. I See Table 21.1. The initial-value theorem states that f(O+) = 21.5 !i..ll! sF(s) (1) Apply this theorem to find the current i(O+) in a circuit, if the transform of the current is J(s) _ 2s + 10 - s(s + 1) I 21.6 i(O+) = lim sJ(s) = lim s[ 2(S + 11°)] =2 A s-+oo s-+oo ss + The voltage across a circuit is given by v = 4e- 1 V. What is the initial voltage? I 21. 7 Express v of Prob. 21.6 in the transform domain. Prob. 2l.6. I From Table 21.1: Then apply the initial-value theorem to verify the result of .P[4e- l ] = _4_ = V(s) s+1 v(O+) = s_oo lim sV(s) = s_oo lim s(~I) =4V S+ 491 492 0 CHAPTER 21 TABLE 21.1 Table of Transform Pairs F(s) j(t) 1 df(t)ldt 2 sF(s) -- d 2f(t) dt (0+) s2F(s) - 3 dnf(t) dt n snF(s)_sn-Idf(o )_Sn-2d2f_ ... _dr-I(0) dt + dt 2 dt n- I + 4 g(t) 11 12 13 14 15 16 17 18 19 = r 0 ~10. + g(O+) f(t) dt s the unit-step function u(t) 8(t) the unit impulse function t t" -1/(n - I)! n is an integer E -at lE -at n-I s 1/s 1 1 Is" 1 Is" 1/(5 + a) 1/(.\ + af (n - l)!/(s + aY' -at t E sin wt cos wt sin (wt + 0) cos (wt + 0) -at • E Sill wt -at E cos wt -at . tE Sill wt tE -at cos wt + w 2) + w 2) [s sin 0 + w cos 0]/(s2 + w 2) [s cos 0 - w sin 0]/(s2 + w 2) 2 WI [(s + a)2 + w ] (s + a)/[(s + a)2 + w 2] 2CJ1(S + a)/[(s + a)2 + W 2]2 [k + a)2 - w 2]/[(s + a)2 + W 2]2 WI(S2 S/lS2 t ~ 0 and considered as being multiplied by u(t). It is assumed that all f(t) exist for The transform of the voltage across a circuit is I f(t) = (] t < O. for qs:, ,= 1/s(s + 1). Determine the initial voltage. S_'OC The circuit of Fig. 21-1a is labeled in the time domain. Draw a corresponding circuit in the transform domain. R R 0 L ~qo+) L - L ((0+ ) ( h.J tAJ I Each of the functions from 7 to 19 can be v(O+) = lim sV(s) = lim [ s+( 1)] =OV sss_OC 21.9 sf(O+) - dT 5 6 7 8 9 10 21.8 f(O+) Fig. 21-1 Denoting the initial current in the inductor i(O _) by a voltage source Li(O + ), we obtain the circuit of Fig. 21-1b. 21.10 The current in an RL series circuit is given by '" di "-. t 4'1= 0 dt Express the current in the s domain. I or 2[ 2 ~ + 4i] = 2[81(s) - i(O+)] 2i(O+) l(s) =, 2s + 4 + 41(s) = 0 (1) LAPLACE TRANSFORM METHOD 21.11 493 If the initial current in the inductor of the circuit of Prob. 21.10 is 5 A, determine i(t). I From (1) of Prob. 21.10 we have: 10 J(s) 5 = 2s + 4 = s + 2 Thus 21.12 0 A Determine the inverse Laplace transform of F(s) _ 3s + 15 - (s + 1)2 + (3)2 I In order to use Table 21.1 of Laplace transform pairs, we rewrite F(s) as 3(s+I)+12 F(s) 3(s+l) 12 = (s + 1)2 + (3)2 = (s + 1)2 + (3)2 + (s + 1)2 + (3)2 = FI(s) + F2(S) Now using results 9 and 10 of Table 21.1, we obtain 5£-l[F(s)] 21.13 = f(t) = 5£-l[FI(S)] + 5£-I[F2(s)] = 3e- cos 3t + 4e- 1 sin 3t (1) Combine the trigonometric functions in (1) of Prob. 21.12 to express the result in terms of e - I and a sine function. I It may be shown that A cos x + B sinx = v' A2 + B2 sin [x + tan-I (AIB)] = v' A2 + B2 cos [x - From (1) of Prob. 21.12, we let A=3 and f(t) = e-'(3 cos 3t + 4 sin 3t) = 5e-' sin (3t + 36.9°). 21.14 1 B=4. tan-I (BIA)] tan-I (AIB)=tan- 1 (3/4)=36.9° Hence, and Express the voltage of the circuit of Fig. 21-2 in the s domain, if v(O+) = 0. -t"lr (of ) Fig. 21-2 I By KCL we have 1 dv 2v + - - - u(t) = 2 dt Taking the Laplace transform of both sides 2V(s) + 1 2[sV(s) or 21.15 V(s) = ° - v(o+)]- 1 S= ° 2 s(s + 4) (1) Solve for v(t) in the circuit of Fig. 21-2, using the result of Prob. 21.14. I In order to use Table 21.1, we rewrite (1) as V(s) = __ 2_ = kl + ~ = (k l + k2)S + 4kl s(s+4) s s+4 s(s+4) Equating the coefficients in the numerator yields kl Thus, 0.5 and = ° k2 = and 4kl = 2. -0.5 V( ) _ 0.5 -0.5 s-s+s+4 and Finally, = kl + k2 V () t = oL(0-1(0.5 S - s0.5) + 4 = 0.5(1 - e- 41 ) = 0.5(1 - e- 41 )u(t) v t2:0 v 494 21.16 0 CHAPTER 21 According to the final-value theorem we have = lim sF(s) f(CYJ) $--+0 Apply the final-value theorem to (1) of Prob. 21.14 to obtain the steady-state value of v(t). I Vsteady ,tate = lim sV(s) s-o -= lim s [ 2 4)] ( = 0.5 V s-o ss + which agrees with the result of Prob. 21.15. 21.17 The current (in the s domain) through a circuit is gh'en by 6 /(s)- - 5(:;+2)(s+3) What is i(oo)? I By the final-value theorem we have i(oo) = lim s/(s) = lirn ( 2~~s + 3) = 1 A S~O ,~o s s + 21.18 The splitting of V(s) into two functions in Prob. 21.l5 is done on the basis of partial-fraction expansion. general, the F(s) whose inverse transform is required can be expressed as the ratio of two polynomials, In A(s) a I,m sm-\+"'+as+a F(s) = __ = ~__ ta m -\ \ 0 B(s) s" + bn_\sn \ + .. , + h\s + ho Furthermore, the denominator of F(s) can be factored as B(s) = (s + a\)(s + n ° 2 )'" (s + an) = TI (s + a\) ;=1 where each value s = -a\ is a root of B(s). To develop the general method, first consider F(s) for which A(s) is of lower degree than B(s); that is, m < n. In addition, it is assumed that B(s) has di~tinct roots (each of the a; is different). For this situation, F(s) can be expanded as A(s) K\ K2 Kn ~ K; F(s) = = - - + - - - + ... + - - = L . J - B(s) s+a\ 5--a2 s+a n ;_\s+a; A method to evaluate K\ is to multiply both sides of Eq. (1) by (s + a\) which gives (s + a\)A(s) B(s) -'-----:...:,---'--'- = By setting s = -al' K + (s "- a ) [ -K2- + ... + -Kn- ] \ \ S + a2 s + an (1) (2) the right-hand side of Eq. (21 is zero except for K\ so that K = (s t a\)A(s) I \ B(s) s- -"'1 Repetition of the process for each root yields the value of K; for each term in Eq. (1). then be expressed as K = (s -+ aJA(s) , B(s) I (3) S=-"', Apply this procedure to obtain the partial-fraction expansion of 8(s + 2) (s -+ l)(s + 3)(s + 5) K\ K2 K3 F(s) = s -+~- + s + 3 + s + 5 F(s) = I From Eq. (3) K =(S+l)F(S)1 \ s=-\ =~~~~I (s+3)(s+5) s--\ K =(S+3)F(S)1 = ~i~~~1 2 s=-3 (s+1)(s+5) s=-3 The general result can = 8(-1+2) =1 (-1+3)(-1+5) = 8(-3+2) =2 (-3+1)(-3+5) LAPLACE TRANSFORM METHOD K3 =(S+5)F(S)1 = s--5 0 495 8(s+2) I = 8(-5+2) =-3 (s+1)(s+3) s--5 (-5+1)(-5+3) These values give 123 F(s) = s + 1 + s + 3 - s + 5 21.19 Obtain f(t) for the F(s) of Prob. 21.18. I 21.20 f(t) = 5£-I[F(s)] = 5£-1(_1_ + _2_ _ _ 3_) = e- I + 2e- 31 _ 3e-51 s+l s+3 s+5 Obtain the partial-fraction expansion of 4 (s + I) F(s) = S(S2 + 2s + 2) I The roots of the denominator of F(s) are s = 0, -1 ± j1. The partial-fraction expansion of F(s) is Kl K2 K3 F(s) = - + + ----"-s s + 1 - jl s + 1 + jl The values of Kl' K 2 , and K3 are found by use of Eg. (3) of Prob. 21.18: K 1 = sF(s) Is-o = 4(s + 1) I S2 + 2s + 2 s=o K2 =(S+1-jl)F2(S)! K 3 =(S+1+jl)F(S)1 s=-I+jl = =2 4(S+1).! = __2_=v'2€-jI35" s(s+l+Jl) s=-I+jl -l+jl I . = 4(s+1~ =_2_._=v'2€+j135" s=-I-}1 s(s + 1- Jl) s--I-jl -1 +}1 The resulting F(s) becomes 2 v'2€ -jI35" v'2€ +jI35" F(s) = - + 1 . + . s s + - Jl s + 1 + Jl 21.21 Determine f(t) for the function F(s) of Prob. 21.20. I From Table 21.1 f(t) = 2 + v'2€ -jI35" . €(-1 +jl)1 + v'2€ +jI35" . €(-I- j l)1 Rearrangement of the terms gives f(t) = 2 + v'2€-/[€ +j(/-135") + €-j(/-135")] The use of Euler's identity, cos x == !(€jX + €-jX), permits f(t) to be expressed as f(t) = 2 + 2v'2€ - I cos (t - 135°) 21.22 In Prob. 21.18, we assumed that A(s) is of lower degree than B(s); that is, m < n. For the case where A(s) is of the same degree as or higher degree than B(s), that is, rewritten as m '2: n, F(s) can be (1) where (1) is obtained by long division. remainder Al (s) / B(s). Hence obtain Then the rules developed in Prob. 21.18 may be applied to the 14] f(t) = 5£_I[S2 + 7s + S2 + 3s + 2 = 5£-I[F(s)] I Performing the required long division and factoring the denominator yields 4s+12 Al(S) F(s) = 1+ (s+1)(s+2) =1+ B(s) The partial-fraction expansion of Al (s) / B(s) is 496 0 CHAPTER 21 Al(S) 4s + 12 K2 - = -- - - = -Kl- +-B(s) (s+I)'s+2) s+1 s+2 Evaluation of Kl and K2 gives K = 4s + 121 = 8 1 S +2 s=-1 K = 4s + 121 = -4 2 s+1 s=-2 and which result in 8 4 F(s) :, 1 + - - - - s+1 s+2 The inverse transform is obtained f(t) = 5t'-I[F(s)] = 8(t) + 8€ 4€ -21 -I - Tht term 8(t) is the unit-impulse function. 21.23 Consider F(s) with repeated roots: A(s) F(s) = (s + a)3 = Kl K2 K3 (1) (s-:j. a)3 + (s + a)2 + (s + a) Multiplication of both sides of Eg. (1) by (s + a)3 gives in which the degree of A(s) is no greater than 2. (s + a)3 F(s) = Kl + K2{s + a) + K3(S + a)2 Setting (2) s = -a in Eg. (2) yields (3) K2 is obtained by differentiating both sides of Eg. (2) with respect to s and setting s = -a. d and for " The result is ds. [(s + a) F(s)] = K2 + 2K3(S + a) (4) d [(.I + a) 3 F(s)] 1s=-a K2 = ds (5) s =-a s ,: -a determines the value of K3 as Differentiating both sides of Eg. (4) and setting 2 d [(s + a)3 F(s)] 1s=-a K3 = 2:1 ds2 (6) For the general case of repeated roots F(s) is expanded as F(s) = A(s) (s+a)k = _!i~_ + (s+at The constants Kl to Kk are evaluated in a fashion term has the form ~;imilar K2 (s+a)k Kk 1 +".+-s+a to the development in Egs. (1) to (6). 1 ,i"- 1 k 1 Kv = (v _ 1)1 ~{;-;=T [(s + a) F(s)] '=-a (7) The general (8) Apply (8) to obtain the partial-fraction expansion of s+2 F(s) = ss+1 ( )3 (9) I The expansion for F(s) is By use of Eq. (2) of Prob. 21.18. Ko = I sF(s) ",0 =~I (s + 1) s=o =2 The coefficients Ku K2, and K3 are determined flom Egs. (3), (5), and (6), respectively, as 0 497 LAPLACE TRANSFORM METHOD KI = (s s + 21 , - - I + 1)3F(s) I ,=-1 = -s- K2 = -d [(s + 1) 3 F(s)] 1 ds ,--I K3=~2 d [S+2]1 = -ds -S = --21 2 == -2 ,--I S ,--I 2 d [(S+1)3 F (S)]1 ds 2 ,=-1 These values of the K terms yield =~2 2 F(s) = 21.24 =-1 d:[S+2]1 ds S"--I 1 s- (s + 1)3 - =~[~]I =-2 2 S ,=-1 2 2 (s + 1)2 - (s + 1) (10) Find the inverse transform of (9) of Prob. 21.23. I To obtain the inverse transform we use the partial expansion of (9) as given by (10). Hence, f(t) = 2 - ~2 € - t 21.25 - 2t€ - t - 2€ - t = 2 - € - f ( ~2 + 2t + 2) Certain useful properties of Laplace transforms are summarized in Table 21.2. delayed unit step function to illustrate entry 1 in Table 21.2. Apply Laplace transform to a TABLE 21.2 Properties of the Laplace Transform Property Time domain Frequency domain 1 Time delay f(t - T)u(t - T) € f(t) F(s)/[I-€T'] 2 Periodic function f(t) = f(t + uT) O:St:5t -'TF(s) F(s) = 3 Time scaling f(at) 4 Frequency differentiation (multiplication by t) tf(t) r where f(t)€ -si dt 1 - F(s/a) a _ dF(s) ds f(t) 5 Frequency integration (division by t) 6 Frequency shifting f(t)€ -af F(s + a) 7 Initial-value theorem lim f(t) = f(O+) }i.ll! sF(s) 8 Final-value theorem lim where limit exists t~O lim sF(s) ,~O I Evaluation at the indicated limits gives 5£[ u(t - T)] = G)€ -sT == 5£[ u(t)]€ -sT The effect of the delay in the application of the step function by T seconds is reflected in the € -,T term which multiplies the Laplace transform of the unit-step function. In the general case, delay in the time domain is indicated by the € -,T factor in the frequency domain. 21.26 Determine 5£[ te -f]. I where f(t) = € -t. By use of entry 4 in Table 21.2 5£[tf(t)] = _ dF(s) ds Performing the requisite differentiation, and 498 0 CHAPTER 21 dF(sl 1 --ds-=(s+1)2 W[ so that -II tE oL 1 = (s + 1)2 which is entry 10 in Table 21.2. 21.27 Apply the frequency-shifting property to evaluate .:t·te-']. 5£[t€ -I] = .5"[. -J(t)] = F(s + 1) I and where f(t) = t. By use of entry 7 in Table 21.2 5£[ t] 1 =, "2 s = F(s) which causes 21.28 The function 5 I(s) , = s+2 is the frequency-domain response of an RL circuit. Determine f(O+) and lim f(t). (_00 I From the initial-value theorem i(O+) = lim sl(.,) = lim s_~ s_~ S 5S +2 = 5A 5S = 0 and from the final-value theorem lim i(t) = lim d(s) = lim 21.29 s-O 5-+0 I_X, S +2 Find the time-domain current i(t) if its Laplace transform is J(s) =, s -10 --4--2 S I +s Factoring the denominator, ,= J(s) s - 10 /(s - j)(s + j) we see that the poles of J(s) are s = 0 (double pole) and The partial-fraction expansion of J(s) is therefore J(s) = ! - 10 ~ s s s = ±j (simple poles). (O.S -- j5) _1_. - (0.5 _ j5) _1_. S-] s+] and term-by-term inversion using Table 21.2 gives i(t) 21.30 = 1 -lOt - (0.5 + j5)e il - (0.5 -- j5)e- il = 1 -lOt - (cos t - 10 sin t) Obtain the transformed equivalent circuits for R, L, and C, including initial conditions. I The transformed network equivalents for the three elements R, L, and C are based on the Laplace transforms of their respective volt-ampere characteri~,tics. These relations are 5£[v(t) = Ri(t)]~ 5£ [ v(t) = L .'£ [ v(t) = -1 C since V(5) = RJ(s) di(t)] Tt ~ V{!') = I' -x . sLJ(s) - Ll(O +) ] v(O+) v(t) dt -. V(s) = - 1 J(s) + Cs s fo (1) (2) (3) i(t) dt = q(O_) and q(O_)/C is v(O_). By continuity of capacitanc,~ voltage, v(O_) = v(O+), which is the term appearing in Eq. (3). Equivalent circuit representations for the relations in Eqs. (1) to (3) are shown in Fig. 21-3. LAPLACE TRANSFORM METHOD ; (t) 1 R 2 O---M/'....._-o + v (f) I (s) R 1 + - V !!1!.. ~~-----4rrOlnfOl!Oi',\..-~~ (tJ - !ii~ V (s) of • 2 0 ; (t) -. ~ J i(t)dt C 1 + v (t) I(.)=C. V(.)-C.,(O+) • ., (0+ )1. + _·-V (s)-- - 0 (I) ~ ~Y~ 2 I( 0 +_V(d_(e) I C 1(.)+ .. - (d) .,(t)= C.,(O+ ) - I (s) db 1 0 + V (5) 2 0 - (h) (8) 21.31 - V(.)= L.,(.)- Li(O+) Li(O+ ) (C) V(.)= (~) 0 (b) .,(t)= L di dt v 2 N\f -0 (0) + 499 ., (.) = RI (s) .,(t)=Ri(t) - 0 Fig. 21·3 Find the transformed network representations of independent voltage and current sources. I The transformed network representations of current and voltage sources are simply the Laplace transforms of the time functions which define the source current and voltage as illustrated in Fig. 21-4. + ., (t) V (s) (b) (0) i (t) I (s) (c) (d) Fig. 21·4 500 21.32 0 CHAPTER 21 Draw circuits showing dependent sources in the transformed domain. I See Fig. 21-5. +0 +0 ., (t) A.,(t) V (s) -0 AV(s) -0 (0) (b) +0 +0 ., (t) g.,(t) V (s) -0 gV(s) -0 (d) (c) ; (t) 1 r; (t) I (s) J rl (s) (e) (I) Ai (t) I (s) 1 AI (s) (g) 21.33 In an RL series circuit find i(t) for t > O. R Fig. 21·5 (h) = 5 n, L = 2.5 mH, i(O +) and = 2 A. If a source of 50 V is applied at t = 0, I In the s domain we have sL/(s) + RI(s) - Li(O +) = V(s) [0 or I(s) = .()t = w-l(lOs Thus, 21.34 1 oL - - 's (-8) + s + 2000 g ---- s+2000 )-10 - 8 e -20001 A In the circuit shown in Fig. 21-6a, an initial current il is established while the switch is in position 1. At t -:= 0 it is moved to position 2, introducing both a capacitor with initial charge Qo and a constant-voltage source V2 . Obtain the s-domain circuit. I The s-domain circuit is shown in Fig. 21.6b. RI(s) + sL/(s)·in which Vo=QoIC and i(O+)=i\=V\IR. T'le s-domain equation is ., [.11,0+) + I(s) Vo se + sC = V2 S LAPLACE TRANSFORM METHOD e R D 501 R sL Vo le se ~----------~+-.-----------~ (b) (a) Fig. 21·6 21.35 Find the Laplace transform of e- ar cos wt, where a is a constant. I Applying the defining equationX[f(t)] = W[ - a l ] e cos wt = oL 21.36 If X[f(t)] = F(s) I By definition, J; f(t)e- SI dt to the given function, we obtain x -(.<+a)1 d [ -( s +) a cos wte -(s+a)1 + e -(s+a)l· w sm wt s+a cos wte t= 2 2 = 2 2 o (s + a) + w 0 (s + a) + w ]X l show that X[e-a'f(t)] = F(s + a). X[f(t)] = J(~ f(t)e- Sl dt = F(s). X[ e-a1f(t)] = Apply this result to Prob. 21.35. Then r [e-a1f(t)]e- st dt = r f(t)e-(s+a)1 dt = F(s + a) (1) Applying (1) to line 17 of Table 21.1 gives W[ - al ] e cos wt = ( oL S S +a )2 2 +a +w as determined in Prob. 21.35. 21.37 Find the Laplace transform of f(t) = 1 - e- al, where a is a constant. r (1 - e-al)e- SI dt = re-SI dt - X[l- e- al ] = I = [_ ! es 21.38 SI + _1_ e-(s+a)I]X = ! s +a .0 s r e-(s+a)1 dt __1_ = s+a __ a_ s(s + a) Determine I Using the method of partial fractions, and the coefficients are 1- 2 A=-2- I s - a s=O X-1[ Hence 1 =-2 C = s(s a 1 S(S2 - a 2 ] = X- 1[ -lIa s ) 2 2 ] + X- 1[ 1/2a s +a ] 1 + a) Is=a = 2a1 + X- 1[ 1I2a 2 2 ] s- a The corresponding time functions are W -1 [ oL 1 S(S2 - ] a 2 ) al l I I (e + e -al) = - -12 + - 1 2 e -at + - 1 2 ea = - - 2 + -2 = -12 (cosh at - 1) a 2a 2a a a 2 a 502 21.39 0 CHAPTER 21 Find I Using the method of partial fractions, we have s+1 s(s + 2)2 Then Hence X-' [ S +-11 B =s - 1 s +1 1 A = (s + 2)2 S~O = 4 2 S S(S2 +1 S + 4s + 4) B, B2 -j---+--S +2 (s + 2)2 A =-I ] X-" = and 2 s=-2 -. ] X-' [ I ] [s'] + X-' [' s +2 + (s + 2)2 2: The corresponding time functions are found in Table 21.1: w-,[ J oL S(S2 21.40 s+l = -1 - -1 e -21 + -1 te -21 + 4s + 4) 4 4 2 In the series RC circuit of Fig. 21-7, the capacitor has initial charge 2.5 me. At t = 0, the switch is closed and a constant-voltage source V == 100 V is applied. Use the Laplace transform method to find the current. lOO Y + Fig. 21·7 I The time-domain equation for the given circuit after the switch is closed is Ri(t) + ~ [Q" + J: i(r) dr] = V 1 6 [(-.::.5 x 10- 3 ) + (' i(r) dr] == V 50 x 10 Jo Qo is opposite in polarity to the charge which the source will deposit on the capacitor. transform of the terms in (1), we obtain the s-domain equation or lOi(t) + 1 / o (s) - 2.5 >< 10 -3 J(s) 50 )<:tc~:~ + 50 x 10 J(s) = ~ or 6 (1) Taking the Laplace 100 s S 15 (2) + (2 X 10 3 ) The time function is now obtained by taking the i:JVerse Laplace transform of i(t) == 21.41 x-,[--.]~--] = 3 s + (2)< 10 15e- 2x ,03 , A ) In the RL circuit shown in Fig. 21-8, the switch is III position 1 long enough to establish steady-state conditions, and at t = 0 it is switched to position 2. Find the resulting current. ~ SOY J(XI Y 250 0.01 H Fig. 21·8 LAPLACE TRANSFORM METHOD D 503 I Assume that the direction of the current is as shown in the diagram. The initial current is therefore io = -50/25 = -2 A. The time-domain equation is 25i + 0.01 ~ = 100 (1) Taking the Laplace transform of (1), 25/(s) + O.Ols/(s) - O.Oli(O+) = 100ls Substituting for i(O+), 25/(s) + O.Ols/(s) + 0.01(2) 100 I(s) = s(O.Ols + 25) and = 100ls 10 4 s(s + 2500) 0.02 O.Ols + 25 2 s + 2500 Applying the method of partial fractions, 10 4 s(s + 2500) -;-----:::::-::-:~ A B = -s + ----=-=:-:-::s + 2500 4 with 4 Then I(s) 4 10 s + 2500 A = = S- B= -10 s =4 1 s-O 4 s + 2500 2 s + 2500 Taking the inverse Laplace transform of (2), we obtain 21.42 i= 1 s ... -2500 =-4 4 s 6 s + 2500 4 - 6e- 2500' A. (2) According to the Heaviside expansion formula, if all poles of R(s) are simple, the partial-fraction expansion and termwise inversion can be accomplished in a single step: (1) where a p a 2, . .. , an are the poles and Q'(a k ) is dQ(s)lds evaluated at Apply (1) to determine X I Here pes) = 250 -I[ (s + 100)(s 250 ] + 50) Q(s) = s = ak • = f(t) l + l50s + 5000 Q'(s) = 2s + 150 and f,( ) 250 - lOO, 250 -50, t =e +- e -50 50 Hence, 21.43 l' _ 5 e 100, + 5e -50, Apply (1) of Prob. 21.42 to obtain f,() ,t = X-I [ 2 4.5s 1.25 ] (l+4X106)(s+103) - s+103 I In this case we have: pes) = 4.5l, Q(s) = S3 + 103s2 + 4 X 106s + 4 X 109, 10 6 , a I = - j2 X 10 3, a2 = j2 x 10 3, and a3 = _10 3• Then (1) Q 'Cs) = 3s 2 + 2 X 103s + 4 X f,()- P(-j2x103) -j2XIO', P(j2X103) j2XI03, P(_103) _10', 25 _10 3, t e + e + e - 1 e Q'(-j2 x 10 3) Q'(j2 X 10 3) Q'(_103) . J' = (1.8 - jO.9)e-j2XI03, + (1.8 + jO.9)ej2XI03, - 0.35e- 103 , = -1.8 sin 2000t + 3.6 cos 2000t - 0.35e _103, = 4.02 sin (2000t + 116.6°) - 0.35e- lO " 21.44 In the series RL circuit of Fig. 21-9 an exponential voltage t = O. Find the resulting current. (2) u = 50e -1001 V is applied by closing the switch at 504 0 CHAPTER 21 100 v 0.2 H Fig. 21·9 i~ I The time-domain equation for the given circuit Ri + l. di dt = (1) U In the s domain, (1) has the form R/(s) + sL/(s) - Li(O+) V(s) = Substituting the circuit constants and the transform of the source, 50 or lO/(s) + s(0.2)/(s) = s + HiD (2) V(s) /(s) = 50/(s + 100) in (2), 250 (s + 100)(s + 50) = Now, using the result of Prob. 21.42, . = oL cp-l[/()] S I 21.45 250 =:50 e -50, = - 5e -lOO, + 5 e -50, =- e -lOO, + ---50 50 A The series RC circuit of Fig. 21-10 has a sinusoidal voltage source u = 180 sin (2000t + cP) V and an initial charge on the capacitor Qo = 1.25 me with polarity as shown. Determine the current if the switch is closed at a time corresponding to cP = 90°. 400 v + 25,.F Fig. 21·10 I The time-domain equation of the circuit is 40i(t) + 25 x \0 6 (1.25 x 11) 3 + f: i( r) dr) = 180 cos 2000t (1) The Laplace transform of (1) gives the s-domain equation 40/(s) + or 1.25 x 10- 3 25 x 10 6S /( 4 X 10 4 + - - - /(s) = S 180s -;:---"""7 S2 4.Ss! s) = (S2 +4 x which is the same as (1) of Prob. 21.43. 106K~ + 103 ) - S + 4 X 10 6 1.25 + 10 3 Hence, frolr. (2) of Prob. 21.43 we have i(t) = 4.02 sin (2000t + Il6.6°) - 0.35e -10 21.46 (2) 3 , Find the initial current in the circuit of Prob. 21.45. I Applying the initial-value theorem to (2) of Prob. 21.45, we obtain i(O+) = lim [s/(s)] S~OO = 4.5 -1.25 = 3.25 A A (3) LAPLACE TRANSFORM METHOD 21.47 D 505 Verify the result of Prob. 21.46 from circuit considerations and from (3) of Prob. 21.45. I At t = 0 the current is given by the instantaneous voltage, consisting of the source voltage and the charged capacitor voltage, divided by the resistance. Thus io = (180 sin 90° The same result is obtained if we set 21.48 I~~5XXI~0:3) /40 = 3.25 A t = 0 in (3) of Prob. 21.45. In the series RL circuit of Fig. 21-11, the source is u = 100 sin (500t + cP) V. if the switch is closed at a time corresponding to cP = o. Determine the resulting current v Fig. 21·11 I The s·domain equation of a series RL circuit is R/(s) + sUes) - Li(O+) cP The transform of the source with = 0 V(s) = (1) is V( ) = (100)(500) S Since there is no initial current in the inductance 5 X 5/(s) + O.OIs/(s) = S2 + 25 10 X S2 + (500)2 Li(O+) = O. 4 104 or Substituting the circuit constants into (1), 5 X 10 6 /(s) - ~------c;--- - (S2 + 25 X 104)(s + 500) (2) Expanding (2) by partial fractions, -l+j) (-I-j) /(s) = 5 ( s + j500 + 5 s - j500 10 (3) + s + 500 The inverse Laplace transform of (3) is i = 10 sin 500t - 10 cos 500t + lOe - 5001 = lOe - 5001 + 14.14 sin (500t - 45°) 21.49 Rework Prob. 21.48 by writing the voltage function as u = lOOej5001 Now V(s) = lOO/cs - j500), A (1) V and the s-domain equation is 5/(s) + O.OIs/(s) = 100 '500 s- J or /(s) 10 4 = -:----===-.,....--=c:-::.,.(s - j5(0)(s + 5(0) Using partial fractions, /(s) = 10 - jlO + -10 + jlO s - j500 s + 500 and, inverting, i = (10 - jIO)e i500 • + (-lO + jlO)e- 50o • = 14.l4e i(5oo'-,,/4) + (-lO + jlO)e- 5001 A (2) The actual voltage is the imaginary part of (1); hence the actual current is the imaginary part of (2). i = 14.14 sin (500t - 1T/4) + lOe -500. 21.50 A In the series RLC circuit shown in Fig. 21-12, there is no initial charge on the capacitor. at t = 0, determine the resulting current. If the switch is closed 506 0 CHAPTER 21 2H IH sov 0.5 F Fig. 21·12 , The time-domain equation of the given circuit is Ri + L Because i(O +) = 0, ~d~t' + .~I': Jo(' i(r) dr == V (1) the Laplace transform of (I) is . 1 ) V R/(s) + sLIIs) . + -c s /(s == -s (2) 1 50 2/(s) + Is/(s' + 0.5s /(s) == (3) s or 50 /(s) == Hence S2 50 (4) + 2s + i = (s + 1 + j)(s + 1 - j) Expanding (4) by partial fractions, i2S /(s) == (s -t:t+ j) j25 - (s + 1 - j) (5) and the inverse Laplace transform of (5) gives i == j25{ e(- 1- ilt - e(-l ri)l} = 50e -I sin t 21.51 A In the two-mesh network of Fig. 21-13, the two loop currents are selected as shown. equations in matrix form. 50 Write the s-domain 2H -I ;2 2F v 50 Fig. 21.13 , Writing the set of equations in the time domain, 5i l + i [Qo+ L il(r)dr] +5i2 == v and (1) Taking the Laplace transform of (1) to obtain the corresponding s-domain equations, (2) When this set of s-domain equations is written in matrix form, 21.52 Obtain a circuit in the s domain corresponding to the circuit of Fig. 21-13. , See Fig. 21-14. LAPLACE TRANSFORM METHOD D 507 50 V(I) Fig. 21·14 21.53 In the two-mesh network of Fig. 21-15, find the currents which result when the switch is closed. 50 100 V Fig. 21·15 I The time-domain equations for the network are . lOll di l di2 + 0.02 dt - 0.02 dt = 100 0.02 di . dt2 + 51 2 - 0.02 di dtl =0 (1) Taking the Laplace transform of set (1), (10 + 0.02s)I/(s) - 0.02sI2(S) = l00ls (5 + 0.02s)I2(S) - 0.02sI1 (s) = 0 (2) From the second equation in set (2) we find (3) I2(S) = I 1 (S)C /250) which when substituted into the first equation gives s + 250 ] 10 3.33 II(s) = 6.67 [ s(s + 166.7) = -; - s + 166.7 i l = 10 - 3.33e-16671 Inverting (4), (4) A Finally, substitute (4) into (3) and obtain I2(S) = 6.67( s + :66.7) 21.54 i2 = 6.67e-166.71 whence Apply the initial- and final-value theorems in Prob. 21.53. I The initial value of i l is given by . . . [ (S++1667 250 )] =6.67A . 11(0+)=hm[sIl(s)]=hm 6.67 s_x s_t% S and the final value is The initial value of i2 is given by i2(0+) = 5lim [SI1(S)] = lim [6.67( S +. :667)] = 6.67 A _00'" . $_X> and the final value is i2(OO) = 5-0 lim [SI2(S)] = 5_0 lim [6.67( S + :667)] =0 . A 508 21.55 D CHAPTER 21 Verify the results of Prob. 21.54 from Fig. 21-15. I Examination of Fig. 21-15 verifies each of the above initial and final values. At the instant of closing, the inductance presents an infinite impedance and the currents are i{ = i2 = 100/(10 + 5) = 6.67 A. steady state, the inductance appears as a short circuit; hence i{ = 10 A, i2 = O. 21.56 Then, in the Solve for i{ in Prob. 21.53 by determining an equivalent circuit in the s domain. I In the s domain the 0.02-H inductor has impedance Z(s) = 0.02s. Therefore, the equivalent impedance of the network as seen from the source is Z(s) = 10 + (0.02~1(5) = 15(s + 166.7) 0.02s + 5 s + 250 and the s-domain equivalent circuit is as shown in Fig, 21-16. The current is then V(s) _ 100 [ s -+ 250 ] _ [ s + 250 ] I{(s) = Z(s) - -s- 15{s.t=-166.7) - 6.67 s(s + 166.7) This expression is identical with (4) of Prob. 21.53, and so the same time function i{ is obtained. 100/s Z(s) Fig. 21·16 21.57 In the two-mesh network shown in Fig. 21-17 there is no initial charge on the capacitor. i{ and i2 which result when the switch is closed at t = O. Find the loop currents 100 --;,) 50V -~ 0.2 F 400 Fig. 21·17 I The time-domain equations for the circuit are lOi{ + 0~2 l' i{ dr + 10(, ,= 50 The corresponding s-domain equations are 50 1 1O/t (s) + -02 I\(s) + 10/2(s)'~ -. S Solving, which invert to 21.58 S 5 I{ (s) = s + 0.625 A 1 1 Iz(s) = ~ - s + 0.625 /2 = 1- e-O,6251 A Referring to Prob. 21.57, obtain the equivalent impedance of the s-domain network and determine the total current and the branch currents using the current-division rule. I The s-domain impedance as seen by the voltage SO'Jl'ce is Z(s) = 10 + 40(1I0.2s) =, 80s + 50 = lO(s + 5/8) 40+1I0.2s 8s+1 s+1/8 The equivalent circuit is shown in Fig. 21-18; the resulting current is LAPLACE TRANSFORM METHOD VIs) '" 50/5 D 509 Z(5) Fig. 21·18 1 s = V(S) = 5 s + 118 () Z(s) s(s+5/8) Expanding I(s) in partial fractions, 1 4 i=1+4e- sII8 +-from which A s s+5/8 Now the branch currents II(S) and 12(S) can be obtained by the current-division rule. we have I(s) = - 40 ) 5 I\(s) = I(s) ( 40 + 1/0.2s = s + 5/8 1/0.2s 1 ) S- 12(S) == I(s) ( 40 + 1I0.2s == - and 1 s + 5/8 Referring to Fig. 21-19, A and A lll(') J;) 1(5) ;: l' 1/0.2. n 40 Fig. 21·19 21.59 Find i in the circuit of Fig. 21-20, if the initial conditions are all zero and the switch is closed at 0.5 F sov 5 n 5n Fig. 21·20 I The network has an equivalent impedance in the s domain 2 Z(s) == 10 + (5 + lIs)(5 + 1I0.5s) = 125s + 45s + 2 10 + lis + 1I0.5s s(lOs + 3) Hence the current is I( ) _ V(s) _ 50 s(lOs + 3) s - Z(s) - S (125s 2 + 45s + 2) 4(s + 0.3) (s + 0.308)(s + 0.052) Expanding I(s) in partial fractions, 1/8 3118 I(s) == s + 0.308 + s + 0.052 21.60 and . 1 -0.30S< +-e 31 -0.052< z=-e 8 8 Apply the initial- and final-value theorems to the s-domain current of Prob. 21.59. I A t = O. 510 21.61 0 CHAPTER 21 Verify the results of Prob. 21.60 from the circuit of Fig. 21-20. I Examination of Fig. 21-20 shows that initially the total circuit resistance is R = 10 + 5(5) I 10 = 12.5 nand thus i(O +) = 50112.5 = 4 A. Then, in the steady statl~, both capacitors are charged to 50 V and the current is zero. 2 21.62 Find the Laplace transform of t e 3 ' by applying the shifting property, item 6 in Table 21.2. I 2:(t 2) = 2/s3. Then 2:(t 2 e3') = 2/(s - 3)3. 21.63 Find 2:(e- 2' sin 4t). I 2:(sin4t)=4/(l + 16). Then, by the shifting property, 2:( 21.64 Find e 2:[F(t)] -2, 4) 4 t = (;+:;~2 + 16 = . Sin 4 S2 + 4s + 20 t > 21T13 t<21T13 F(t) = {~os (t - 21T13) if I 2:(cos:) = s -2-- +1 S Then using line 1 of Table 21.2, 2:[F(t)] 21.65 2 If = L = e -2"s/3 2:[F(t)] 2:[f(t)] I 21.67 Given "/3 o e-S'(O) dt + f~ e- H cos [t - (21T13)] dt 21T13 x L o 2:[F(at)] = 10r -:'"s/3 se , e -su cos u du = s· +- 1 10r e-S'F(at)dt= 2:(sin tit) = tan -1(11 s). 10r e-SU,aF(U)du=~F(~) a a e-S(ula)F(ll)d(ula)=! a Apply the result of Prob. 21.66 to find 2:(sin atlt). (f)(sin at) -_ -1 eL(f)(sin at') -_ --1 tan --at at. Cl eL -1( - 1) s Ia 1 tan = - a _I(a) s (f)(sin at) _- tan - I (a) --t s Thus, eL If F(t) has period T > 0 then show that 2:{F(t)} = I We have = L~ e- s(u+2,,/3) cos u du 0 = F(s), show that 2:[F(at)] = (lIa)F(sla,. I 21.68 se- 2 'TTs/3 -s'2-+-1- Verify the result of Prob. 21.64 by applying (1) of Prob. 21.1. I 21.66 =, 2:{F(t)} r e-SlF(t) dtj(1- e- ST ) (1) = (00 e-S'F(t) dt = (T e-stF(t) clt + f2T e-S'F(t) dt + f3T e-S'F(t) dt + ... In the second integral let 10 t 10 T = u + T, in the third integral let 2T t = u + 2 T, etc. Then 2:{F(t)} = JOT e-SUF(u) du + LT e-S(U+T)P(lt + T) du + LT e- S(U+2T)F(u + 2T) du + ... = LT e-SUF(u) du + e- sT LT e-SUF(u) du + e- 2sT LT e-SUF(u) du + .. , = (1 + e- sT + e- 2sT + ... ) LT e'SUp(u) du = LT e-SUF(u) du/(1- e- ST ) where we have used the periodicity to write 1+ , + F(u + T) = F(u) , F(u + 2T) = F(u), . .. , and the fact that ,2 + ,3 + .. 1 1-, 1,1 < 1 LAPLACE TRANSFORM METHOD 21.69 D 511 Graph the function (a) F(t) = {~in t O<t<1T 1T < t< 21T extended periodically with period 21T. (b) Find .:e{F(t)}. I (a) The graph appears in Fig. 21-21. F(t) I (b) From (1) of Prob. 21.68 since = 2"s 1 1- e 2"s {e-SI(-Ssint-cost)}!" S2 + 1 0 1 1- e 2"s {1+e-"S} 1 S2 + 1 = (1- e "S)(S2 + 1) 1- e1 1 = 21.70 The current, in the s domain, in a circuit is e -sI F(t) dt = 1 - e1-2"s " o Fig. 21-21 we have T = 21T, 2 !t'{F(t)} ,... t J(s) = 3/(s + 2). 1" 0 • e -sI Sin t dt Determine the initial and final currents. I By the initial-value theorem: By the final-value theorem: i(oo) = lim sJ(s) = s_o 21.71 i(t) = .:e-1[J(s)] = Thus, Q .:e- 1 { We have 2 = 2, Q 3 21 A s+2 i(oo)=OA 2S2-4 (s + 1)(s - 2)(s - 3) } p(s)=2i-4, Q(s)=(s+1)(s-2)(s-3)=s3-4i+s+6, Q'(s)=3i-8s+1, Then the require" inverse is, by the Heaviside expansion theorem, = 3. P(-1) -I P(2) 21 Q'(-I) e + Q'(2) e Find .:e- 1 (_3_) = 3e- i(0+)=3A Find , 21.73 = 0A Verify the results of Prob. 21.70 by expressing the current in the time domain. I 21.72 ls+2 P(3) 31 _ -2 + Q'(3) e - 12 e .:e- 1 { -I + -±. 21 -3 e 3s+1 } (s _1)(S2 + 1) + 14 31 4 e = _ .!. 6e -I _ ~ 21 3e +~ 31 2e Q1=-1, 512 D CHAPTER 21 2 Q/(s) = 3s - 2s + 1, a l = 1, a 2 = j, Then, by the Heaviside expansion formula, the required inverse is I We have P(s) = 3s + 1, Q(s) = (s - 1)(s2 + 1) ,= S3 a3 = - j since S2 + 1 = (s - J)(s + J). + s -1, / P(l) t P(j) it P(-j) _it _ 4 i 3j + 1 it -3j + 1 -it Q/(l)e+ Q '(j)e +Q/(_j)e -i e +=2_ 2j e +_2+ 2j e = 2e t + (-1 - ~ j)(cos t + j sin t) + (-1 + !i)(cos t - j sin t) = 2e t - cos t + ~ sin t - cos t + ~ sin t = 2e t - 2 cos t + sin t 21.74 The voltage across a circuit is given by d2~ -,+v=t dr The initial conditions are v(O + ) = 1 and dv Idt( (I + ) = -2. Find v(t). I Taking the Laplace transform of both sides: S2V(S) - sv(O+) - + V(s) = l/s 2 V(s)(/ + l) - s + 2 = 1// or 1 V(s) = /(S2 + 1) + Thus, s 2 1 ;Z-=-i ,,' ~2 + 1 =?- s + +1 - S2 3 S2 +1 v(t) = 2- I V(s) = t + cos t - 3 sin t Hence, 21.75 ti(O+) The current in a circuit is given by 2 d i di . 2t - 2 -' 3 -' + 21 = 4e dt dt i(O +) = - 3 A and dil dt(O +) Find i(t) if the initial conditions are (1) = 5 AI s. I From (1) and the initial conditions: /(S)(S2 - 3s + 2) + 3s - 14 = 4/(s - 2) -3s 2 + 20s -- 2~· -7 4 4 /(s)= (s-1)(s-2)2- = s-l + s-2 + (S-2)2 or Thus, 21.76 i(t) Solve for i(t) if i(O+) = '-7e t + 4e 2t + 4te 2t = 0, dildt(O+) = 1 and 2 d i di . -t. - 2 + 2 - -- 51 = e sm t dt dt Taking Laplace transforms and substituting initial cc'nditions yields 1 2 2 s + 2s + / s _ / + 2s + 3 _._ 1 [ 1 ]+2 [ 1 ] ( ) - (S2 + 2s + 2)(S2 + 2s + 5) '-"3 (s + 1)2 + 1 "3 (s + 1)2 + 4 2 s /(s) - 1 + 2s/(s) + 5/(s) = or Hence, 21.77 i(t) The current in a circuit is periodic. governed by = ~e-!(sil1 t + sin 2t) A Two values of the current are i(O+) = 1, i(1T12) I In the transform domain: 2 s /(s) or di • SI(O+) /(S)(S2 + 9) - s - A '-"lit (0+) + 9/(s) = = '-2!.-- s -' 4 Solving for /(s) and breaking into partial fractions yields where S2 s +4 di A = - (0 ) dt + = -1. Find i(t) if it is LAPLACE TRANSFORM METHOD 4( S) A s I(s) = 5" S2 +9 + S2 +9 + 5(S2 +4) . I(t) or = A. 4 5" cos 3t + "3 Sin 3t + i(i)=-1=-4-~ A=I~ or i(t) = ~ (cos 3t + sin 3t) + ~ cos 2t Hence, 21.78 1 5" cos 2t The currents in a coupled circuit are given by di{ Tt = The initial conditions are . . 21{ -31 2 and i{(0+)=8A and Determine i{(t). i 2(0+)=3A. I In the transform domain, the equations become and Solving for I{(s) yields 8s-17 5 3 I{(s) = (s+I)(s-4) = s+1 + s-4 21. 79 or A The charge q in an electric circuit is given by d2 d dt; + 8 d; + 25q = 150 = E If all initial conditions are zero, determine the current through the circuit. I In the s domain we have or 150 Q(s) = S(S2 + 8s + 25) and i(t) = Repeat Prob. 21.79 if s 6(s + 4) (S+4)2+9 24 (s + 4)2 + 9 Q = 6 - 6e- 4 ' cos 3t - 8e- 4 ' sin 3t or 21.80 6 ~; = 50e -4, sin 3t A E = 50 sin 3t. I Proceeding as in Prob. 21.79, we have 150 Q(S)(S2 + 8s + 25) = Z--9 s + or 150 Q(s) = (S2 + 9)(l + 8s + 25) 75 1 75 s 75 1 75 s+4 =-------++---c;--26 S2 + 9 52 S2 + 9 26 (s + 4)2 + 9 52 (s + 4)2 + 9 Thus, and 21.81 q(t) = ~(2sin3t - 3 cos3t) + ~e-4'(3 cos3t + 2 sin 3t) . ) dq I(t = dt = 75 ( 2 cos 3t + 3 Sin . 3t ) - 52 25 e -4'(. 52 17 Sin 3t + 6 cos 3t ) The mesh currents in an electric circuit are given by . di{ di2 . - 51 {-d- t + 2 -dt + 101 2 = 0 D 513 514 D CHAPTER 21 I Transforming and applying the initial conditions yields 55 /j(s)(s + 20) + 15/2 (s) = S /j(S)(S + 5) -/2(s)(2s + 10) = 0 Solving for /2 (S): 55 1 2 /2(S) = s(2s + 55) = ; - 2s -:;~~~ 21.82 or A The current in a circuit is given by d2 • de! + 8i = 4 cos wt = v(t) 2 i(O +) Determine i(t) if the initial conditions are I In = dildt(O +) and 10 A 2 = - 24s --2 S or . l(t) Repeat Prob. 21.82 for w 2 = 4, = w = 2 2 10 cos 2t + -( - " 1 - (cos 2t - cos wt) 41) -- 4) = l+4 2s + (s2+4f v(t) = u(t - a). the transformed domain: 2[S2/(S) -lOs) + 8/(s) e- as =-s 10s e - as /(s) = - - + - or S2 +4 Repeat Prob. 21.82 if 8 (1- - -s) S S2 +4 t> a t< a i(t) = {1O cos 2t + A[1 - cos 2(t - a)) 10 cos 2t Thus, I In i(t) = 10 cos 2t + !t sin 2t and is the resonance frequency. Repeat Prob. 21.82 if I In v(t) = 8(t). this case we have 2[l/(s) -lOs) + 8i(s) = 1 or /(s) i(t) = 10 cos 2t + Thus, 21.86 s +w and explain its significance. lOs 21.85 s +4 From (1) of Prob. 21.82 we have /(s) 21.84 (1) +w +- (-Z-S- -~) 10s 2 + s +4 w -4 I(s) = Thus, I o. the transform domain, with the given initial conditions, we have 2[s /(S) - lOs _. 0) + 8/(s) 21.83 = l = lOs 1 l + 4 + 2(S2 + 4) *sin 2t Find the Laplace transform of the rectangular pulse :;hown in Fig. 21-22a. {Lt) /( (i) -------- -+--~--+-t ~t- T \ _ «.(f_T) L..-----(J) Fig. 21-22 D LAPLACE TRANSFORM METHOD I The given pulse may be synthesized as [(t) = u(t) - u(t - T), as shown in Fig. 21-22b. 515 Then 1 1 _ T, 1 - e - Ts F(s)=2[u(t)-u(t-T)]=s-se = s 21.87 Solve Prob. 21.86 by applying differentiation and shifting theorems. I Differentiating the pulse with respect to transform is 1 - e - Ts. t gives the two impulses shown in Fig. 21-23, whose Laplace Hence, 2[~ [(t)] = sF(s) - [(0+) = 1- e- Ts or F(s) = 1- es Ts ~[f(t) L -__ ~ ________ ~t T Fig. 21-23 21.88 Find and compare the Laplace transforms of the following two functions: gl(t) = e-a(t-Tlu(t - T) g2(t) = e-atu(t - T) where T is positive. The two signals are depicted in Fig. 21-24. It is important to note that although gl(t) is the exponential function e- at delayed at T, g2(t) does not involve any shifting; it is simply the signal e- at multiplied by the delayed unit step u(t - T). Thus g2(t) = 0 for t < T. The Laplace transform of gl (t) is obtained by at direct application of the shifting theorem. Since the Laplace transform of e- is 1/(s + a), the Laplace at transform of the shifted e- is 1 _T, G (s ) = - - e I s+a The signal g2(t) may be formed by first shifting e-atu(t) to the left by T, multiplying it by u(t), and then shifting it back by T. This process is illustrated in Fig. 21-24c to e. The Laplace transform of e-a(t+Tlu(t) is 2[e-a(t+Tlu(t)] = e-aT2[e-atu(t)] = e- aT _I_ s+a Now shifting the function e-a(t+Tlu(t) to the right by T gives (1) o·~--~-------===~ (0) 1 o • I Fig. 21-24 516 21.89 D CHAPTER 21 Verify (1) of Prob. 21.88 by using (1) of Prob. 21.1. I The result can be verified by substituting g2(t) directly in the Laplace transform integral. f (00 [1 ] -aT G2(S)=2[g2(t)1=Jo~e-aIU(t-T)e-Sldf= r_e-ale-Sldt= - S+ae-(s+a)1 r= :+a e- TS 21.90 00 00 Determine the Laplace transform of the triangular pulse shown in Fig. 21-25a. 1~ oTT T _~(t_L)u(t--2T) • t T 2 (b) Fig. 21·25 I The triangular pulse is considered to be the sum of three ramp functions as shown in Fig. 21-25b. The expression for the pulse signal is given by 2(t 2:T) u (t -"2T) + T 2 tu(t) - T 4 (t g(t) = T T)u(t - T) Taking the Laplace transform on both sides of the ast equation, we get G(s) = 21.91 ~ TS2 - -±e- Ts12 TS2 -+- -~.1"!,z e- Ts = ~ TS2 (1- 2e- Ts12 + e- TS ) (1) Obtain the Laplace transform of the pulse of Fig. 2l-25a by differentiation. I We take the derivative of the original pulse sigm,l with respect to t successively until impulses occur. The first and the second derivatives of g(t) are shown in Fig. 21-26b and c. The Laplace transform of d 2g(t)/dt 2, as obtained by inspection, is 2[ d~~t) ] = ~ (1- 2e- Ts12 + e- Ts ) Using the derivative theorem, we write 2[ d!~t) ] j (1- 2e- Ts12 + e- Ts ) = s2G(S) - sg(O .. )·- g(I)(O_) = Solving for G(s) from this equation gives the same result as in (1) of Prob. 21.90, since g(O_) and g(J)(O_) are zero in this case. c~"" .. t OTT 2" (4) dq(t) 2 dI T 0 -~ T 2" d 2q(tl T 2" (b) ~B(I-T) ---;;;z -}8 (I) 0 .. t T T 4-8(1-{) • t Cc) Fig. 21·26 LAPLACE TRANSFORM METHOD 21.92 D 517 Find the Laplace transform of the periodic signal shown in Fig. 21-27. I~ oTT 2T 3T .. Z Fig. 21-27 I We notice that the waveform of this periodic signal during one period is described by the triangular pulse in Fig. 21-25, whose Laplace transform is given by (1) of Prob. 21.90. Prob. 21.25 we have Therefore from line 2 of Table 21.2 of TSI2 1 [2 -Ts12 -Ts] 2 1- eGT(s) = 1 _ e Ts TS2 ( 1 - 2e + e ) = TS2 1 + e Tsl2 21.93 Determine the Laplace transform of the unit impulse train shown in Fig. 21-28. 9T'/ ) 1 t t t t L _8(1) OL----~T:----;:2~T:--~3T;---4f;T;----5T I The Laplace transform of g(t) for ° t< T ~ is 2[5(t)] = 1. GT(s) = 1 21.94 I Fig. 21-28 Thus from Table 21.2 we obtain 1 -e Ts Find the Laplace transform of the train of pulses shown in Fig. 21-29. Fig. 21-29 I Proceeding as in the last two problems, G(s) = 1- e-' s 21.95 Given: x+3i+2x=3, i(O+)=O, and s and x(O+)=-l. Solveforx(t). I In the transformed domain: S2X(S) - sx(O+) - i(O+) + 3sX(s) - 3x(0+) + 2X(s) = 3/s X(t) = (~ - 5e- t + ~e-2t)u(t) or 21.96 For the network shown in Fig. 21-30, write the general mesh equations (in terms of symbols) including all initial conditions. R.=30hms v (I) Fig. 21-30 518 D CHAPTER 21 I 21.97 The required equations are Write the transformed equations corresponding to tile general equations of Prob. 21.96. I Taking the Laplace transform of both sides of the loop equations, we get V(s) = (R1 + R Z )I1(S) + Ls/1(S) - Li1(O.) - R2/2(S) - LsI2(S) + Li 2(0+) 1 vc(O+) 0= - R2/1 (s) - Ls/1 (s) + Li 1 (0 +) + R2/2(S) + Ls/2(S) - Li 2(0+) + -C 12(S) + - s s Rearranging terms, these equations become 21.98 Substitute the numerical values from the circuit of Fig. 21-30 in the transformed equations of Prob. 21.97. Solve for i1(t) in terms of v = Vu(t) and the unknown initial conditions. I Substituting the values of the network parameters in the transformed equations, we get V(s) + iL(O+) = (s + 6)/1(S) - (s + 3)/2(S) (1) S 12(S) . vc(O+) -IL(O+) - - s - = ·_(s + 3)/1(S) + s + 3 + where i 1 (0+) - i 2 (0+) has been replaced by Solving for 11(S) from (1) and (2) yields iL(O~), (1) (2) which is the initial current in the inductor L. [V(s) + i L(O+ )](s + 3+ lis) + (s + 3)[ - i L(O+) - vc(O +) Is] 1 (s) - "---..:.--'------"--'--'...:..:..:.....1 (s+6)(.>t3+lIs)-(s+3)2 [V(s) + i L(0+)](S2 +:ls + 1) + (s + 3)[ -iL(O+)s - vc(O+)] .h 2 + lOs + 6 Taking the inverse transform of 11 (s) finally gives i1(t) = ~ u(t) + [0.178V + 0.19i L(0+) - 0,42vc(0+)]e-0788'u(t) + [-0.012V - 0.19i L (0 ,.) + 0.087vc(0 + )e- 254 'ju(t) 21.99 Find the voltage across the capacitor of the circuit of Fig. 21-30. I From (1) and (2) of Prob. 21.98 we have 1 ( s) = -,-(s_+_6~)[=-----"iL..o.(0-,+),-s_- vJO.)] - 2 + s (s + 3)[ V(s) + iL(O+)] 3s:: + \Os + 6 Taking the inverse transform yields i2(t) = [0,42V + 0,45i L(0 +) - 0.99vc(0 + )e- 0788 ']u(t) + [-0.087V - 1.451 L (0 +) + 0.657v c (0 + )e- 254 'ju(t) The voltage across the capacitor for (1) t ~ 0 can be obtained from (2) Substituting i2(t) from (1) in (2): vc(t) = 0.5Vu(t) + [-0.5341/ - O.57i L(0+) + 1.25vc(0 + )]e -0 788'U(t) + [0.034V + 0.57i L(0 + ) - 0.25v c (0+ )]e -254'U(t) LAPLACE TRANSFORM METHOD 21.100 Write the loop equations for the transformer shown in Fig. 21-31a. corresponding equivalent circuit. D 519 Obtain the transformed equations and the + V2(S) (hJ I Fig. 21-31 For the transformer circuit shown, the loop equations in the time domain are Taking the Laplace transforms of this set of equations, we get V1(s) = L1sI1(s) ± MsI2(s) - Ljij(O+) =+= Mi 2(O+) V2(s) = ±Mslj(s) + L 2s1 2(s) - L 2i 2(O+) =+= Mij(O+) where plus and minus signs are used in front of M to indicate variations in winding orientations of the transformer. The two transform equations are represented by the circuit of Fig. 21-31b. 21.101 Let r == 1/ L. Write the node equations for th~ transformer of Fig. 21-31 a, in the time domain and in the s domain. Obtain an equivalent circuit in the s dpmain. I The node equations are I il(t)=r j (' Jo+ vl(r)dr+rM (' Jo+ v 2 (r)dr+i\(O+) Taking the Laplace transforms on both sides of these equations, we get The corresponding transform network is shown in Fig. 21-32. + + Fig. 21-32 520 21.102 D CHAPTER 21 Find the transform network in loop form for the network shown in Fig. 21-33a. I See Fig. 21-33b. RI ::0 't. L~C 1 T R2 v (I) vc(l) (0) + V\s) (h) 21.103 Fig. 21-33 Repeat Problem 21.102 to obtain the transform netwJl'k in node form. I See Fig. 21-34. VIs) il(o+) "2(0+) s-- -s- Fig. 21-34 21.104 In the network of Fig. 21-35a, S is switched from a neutral position to position a at t = O. At moved from a to b. Obtain transform networks corr,~~ponding to these switching operations. I See Fig. 21.105 t = T, S is 21-35b and c. Find iL(t) in the circuit of Fig. 21-35a. I From Fig. 21-35b: v LsIL(s) = - - RJl (s) + LiL(O+) s Solving for IL(s), and since iL(O+) is zero, we get The inverse Laplace transform of IL(s) is iL(t) (1- e- = : I R1IIL ) 005t<T (1) LAPLACE TRANSFORM METHOD D 521 1 v-=- I.. v s -- Fig. 21·35 At t = T+, the switch is moved from position a to position b. Substituting t == T+ in (1) yields (1- e- RtT +/ L ) iL(T+) == : t which is the initial inductor current at t == T+. Now the transform equivalent network for t> T is shown in Fig. 21-35c. Notice that the equivalent-voltage source has a voltage strength of Li L(T+), which corresponds to an impulse in the time domain. The state-transition equation of the network for t> T is written from Fig. 21-35c. Solving for IL(s), The inverse Laplace transform of IL(s) is (l_e- R,T+IL)e-(Rt+ R2)(t-T)/L iL(t) = : t>T I 21.106 Sketch the complete waveform for i L (t) for I t > O. See Fig. 21-36. i,(/) .l. Rlr--r-----------~-~-~-~-~~-------~ (1- e-R1T+IL) e-(R1+R2 )(t- niL RI ~----~-------=~======--I T Fig. 21·36 522 21.107 D CHAPTER 21 In the network of Fig. 21-37a, SI is closed at ( = 0, with vc(O +) = O. After SI is closed for T seconds, S2 is closed. Corresponding to this switching, obtain the transformed equivalent circuit. I See Fig. 21-37b and c. R StiT~O '\IV'v 11+ 'c e v-=- -1 t-:..T 52 111 (0) R v+l s '\IV'v -1 1 ]e,c e,yc lo +I • O (b) R 1 '\IV'v r-1 T+ eJ Vc (5) I 1 Fig. 21-37 (c) 21.108 Find vc(t) in the circuit of Fig. 21-37a. I From Fig. 21-37b we have Vc(s) Rsv - ~ + CvdO+) CsVds) = Solving for Vds) from this expression, and since vc(O+) is zero, we have V Vc(s) = s(1 + RCs) The inverse transform of Vds) is, simply, vc(t) = V(1- e- tIRC ) 0«< T (1) Thus the capacitor voltage rises exponentially toward IV' volts with a time constant of RC. At ( = T seconds, the switch S2 is closed, and SI remains closed. The transform network for ( > T is shown in Fig. 21-37c. The initial voltage on the capacitor at ( = T+ is obtained by substituting ( = T+ in (1); then we have vd T +) == V(I- e- T +/ RC ) For (> T, (2) from Fig. 21-37c, we obtain CsV (s) = C ~ Rs - Vc~2 R - Vc(s) + Cv (T+) R C Solving for Vds) from this expression, we get (3) The inverse transform of Vds) is LAPLACE TRANSFORM METHOD ~ (1 - vc(t) = [ e -2(/- T)IRC) + vc(T+)e -2(/-T)IRC ] U(t - D 523 (4) T) where vdT+) is defined in (2). When t approaches infinity, vdt) approaches V/2. This result may be obtained by letting t approach infinity in (4) or by applying the final-value theorem to (3). Substituting vdT+) from (2) in (4) and simplifying, we have vd t ) = 21.109 Sketchvdt)for e- TIRC >! [~ + V( ~ andfor e- TIRC - e-TIRC)e -2(/-T)/RC ]u(t - T) <!. I See Fig. 21-38. V~---------------------------- [f +V(t - e- TI RC)e- 2 ( 1- T)/RC]U(/_T) O~~--------------------------~I (0) Vr---------~~-------------- ~~~~~L-~~==~____----- [~+ V( ~ - e- TIRc) e -2(1- T) RC]uV- T) (b) 21.110 Fig. 21·38 For the network shown in Fig. 21-39a, obtain the transform network. I See Fig. 21-39b. L -- ~ i"l'luttl~ iL (I) fR. -+______________ L -_ _ _ _ _ _ fR' ~---- __ et + Vc (I) ~ (0) L'L (0+) + /(5)= I S Cs (b) 21.111 Vc Fig. 21-39 For a step input I, determine the voltage across the capacitor of the circuit of Fig. 21-39b. I The circuit is in the s domain. Thus, (1) 524 D CHAPTER 21 Next, the current in the capacitor is equated with lhe difference between through R z . Thus IL(s) + CvC<0+) and the current Solving for Ve(s) in (1) and (2), we get Ve For a step-function input, _ RI/(s) + (RI + Ls)CvC<0+) + LiL(O+) 1+(R I +Ls)(Cs+IIR z) ( s) I(s) = lis; VC<s) = 21.112 (3) (3) becomes IRI + s(RI + Ls)CvC<0+) + LsiL(O+) s[LCs z + (RIC + LlRz)s + 1 + RIIR z] Obtain the steady-state voltage across the capacitor of the circuit of Prob. 21.111, and show that the result is independent of initial conditions. I By the final-value theorem: IRI + s(RI + Ls)Cve(O+) + sLiL(O+) 2 LCs + (RIC + LlRz)s + 1 + RIIR z lim ve(t) = lim sVC<s) = lim Hoo 5-0 5-0 which is independent of ve(O+) and iL(O+). 21.113 Verify that the result of Prob. 21.112 can be directly obtained from the circuit of Fig. 21-39a. I In the steady state, the inductor acts as a short circuit and the capacitor acts as an open circuit; thus the current i(t) is divided between RI and R z. We easily observe that . hm lI (t) '_00 e 21.114 RIR2 1+ R z I = R Obtain the s-domain circuit corresponding to the circuit of Fig. 21-40a. I See Fig. 21-40b. 2 pF lv'1x 1O- 4 c052 x 107 t] oft) ~--~------~------+~--~r-~-----+~~ I t I id ) I 5000 vI (t) 50pF ~. x 10" ---5- ~------ t V2 X 10-4 5 52 +4 x 10'~ 4000 0 __----~~~--~r---~----~--~ + ~r, (s) + 2X10'IO - - 5--0 100/, (~) VI (.) 4000 \ Fig. 21·40 LAPLACE TRANSFORM METHOD 21.115 D 525 Determine V2 (s) in the circuit of Fig. 21-40b. I From Fig. 21-40b, with zero initial conditions, 4 V\(s) -\2 -\2 v'2 X 1O- s +50xlO sVI (s)+2xlO S[VI(S)-V2(S)]- 2 \4 =0 500 s +4xlO \2 V2 (s) 2 x 10- S[V2(S) - VI (s)] + 400 + 100/\(s) = 0 The constraint equation for the dependent source is Combination of the three equations yields VI (s)(2 x 10- 3 + 52 x lO- 12s) - V2(s)(2 x 1O- 12s) = ~ x 1O-4~4 s +4xlO - VI (s)(2 x 1O- 12s - 0.2) + V2(s)(2.5 x 10- 3 + 2 x 1O- 12s) = 0 The third step is to obtain the simultaneous solution for V2 (s), which, after rearrangement of terms, is V (s) 2 21.116 - (/ +4X 2v'2 x 10 6 (s _lOll) 10 14 )(s2 + 53.4 X 10 8s + 5 X 10\6) Obtain the partial fraction of V2 (s) in (1) of Prob. 21.115. (1) Hence determine v 2 (t). I From (1) of Prob. 21.115 we have V _ 2(S) - (s - j2 X 2v'2 X 10 6s(s _lOll) 10 )(s + j2 X 10 7 )(s + 0.94 X 107 )(s + 53.3 7 K\ --..!---: S - j2 X 10 7 + Kr s + j2 X 107 + K2 s + 0.94 X 10 7 X 10 8 ) K3 + -_""":""_-." s + 53.3 X 10 8 Solving for K I , K 2, and K3 yields, K2 = 1.022 K3 = -0.0105 or 21.117 Obtain the transformed network for the circuit shown in Fig. 21-41a. I See Fig. 21-41b. 6n + 8n 110V 2H B lei\~ 10 A + + 110 s - ~ 8 ~ 6 2s B Fig. 21-41 526 21.118 D CHAPTER 21 Determine the current i\(t) in the circuit of Fig. 21-41a. I i \ (0 +) = 110 (3 + 8) = 10 A, the mesh equations are With the initial condition 110 - -s- + (3 + s)/\(s) -10+ 8[/I(s) - (s)] =0 -8[I\(s) - 12(S)] + (6+ 2s)/2(S) =0 Rearrangement of terms gives 1\ (s)(l1 110 s + s) - 8/2(S) = -- + 10 + 12(s)(14 + 2s) = 0 -8/\ (s) After simultaneous solution and factoring, lO(s + l1)(s + 7) I\(s)= s(s +3)(s + 15) 80 1 16 s+3 80 e -3t + -16 e -15t . () t = -154 -- -I 9 9 9 or 21.119 154 1 ==9 ~ -9 t = 0.1 s 1 s+15 A I Determine i2(t) at +9 in the circuit of Fig. 21-41a. I From the mesh equations of Prob. 21.118 we have 40(s + 11) 88 1 12(S) = s(s + 3)(s + 15) == 9 ~ 80 -9 1 8 1 s + 3 - <) s + 15 Consequently, or at 21.120 i 2 (0.1) = 2.994 A t=O.ls Replace the circuit to the left of AB (Fig. 21-41) h} its Thevenin equivalent and solve for i2(t). I VTh(S) and ZTh(S) are found from Fig. 21-42a and b, respectively. of the given circuit, as shown in Fig. 21-42c. 10 + 110 -S ~ 3 A + Thus we obtain the Thevenin equivalent A .\ 8 ~r 8 ~ 8 (a) I.,{.S) (b) Z = 8(s+3) s + 11 A --, + 80 V =- 6 2$ 8 Cc 1 Fig. 21-42 LAPLACE TRANSFORM METHOD I From Fig. 21-42a VTh(S) = I(s) x 8 D 527 -(110/ s) + (3 + s)/(s) - 10 + 8/(s) = 0 and _ (1lO/s) + 10 I( s) l1+s from which v and, after rearrangement, Th (s) = 80(s + 11) = 80 s(s+l1) s The Thevenin impedance is computed from the circuit of Fig. 21-42b as the parallel combination of the 8-0 resistance and the (3 + s)-O impedance. Z () = 8(3 + s) = 8(3 + s) S Th 8+3+s l1+s Finally, from Fig. 21-42c: 40(s + 11) S(S2 + 18s + 45) 80/s 12(s) == [8(3 + s)/(l1 + s)] + (6 + 2s) Upon factoring, 12(s) == s(s 40(s + 11) + 3)(s + 15) which is the same result as obtained in Prob. 21.119. 21.121 The current in a transformed circuit is given by Is _ S2 + 6s + 5 ( ) - S(S2 + 4s + 5 ) Find i(t). S2 + 6s + 5 1 -j I(s) = s(s+2+jl)(s+2-jl) =s+ s+2-jl I t == 0 A! -F capacitor initially charged to 1 V is discharged at Determine i(t) for t > O. I j + s+2+jl i(t) = 1 + 2e -2t sin t or 21.122 Thus, d'l -dt + 2i + 2 By KVL: _ro R == 2 nand i dt = 0 2 [I(s) + q(O+)] = 0 s + 2/(s) + - [sl(s) - i(O+)] or It across a coil having Since the capacitor was initially charged to 1 V, 2q(0+) s 1 s ---==-1 Thus, 21.123 I(s) = S2 1 + 2s + 2 (s + If + 1 or i(t) = e -t sin t u(t) -101[ (s) + (s + 20)/2(s) =0 For a transformed network we have (s Determine i2(t) at + 20)/[ (s) 100 -10/2(s) == - S t = 0.1 s. I Solving for lis) yields I (s) == 2 S(S2 or At 21.124 t = 0.1 s = 3.33 _ _ 5_ + 1.67 s s + 10 s + 30 i2(t)=3.33-5e-lOt+1.67e-30t A 1000 + 40s + 300) == 1000 s(s + lO)(s + 30) i 2 (0.1) = 1.574 A The capacitor in the circuit of Fig. 21-43a is charged to 10 V. I See Fig. 21-43b. Obtain the transformed network, L = 1 H. 528 D CHAPTER 21 ~ 'l.{l. IF ;I- " '2. !> (~) l b) 21.125 Determine the current i(t) in the circuit of Fig. 21·43a. I From Fig. 21-43b: 10 10 ----2(s + 1) /(s) == s[2 + (2/s)] 21.126 Fig. 21-43 i(t) == 5e- 1 or In the circuit of Fig. 21-44a, the switch is moved from a to b at I t == o. A Obtain the transformed network. See Fig. 21-44b. 6 Fig. 21-44 21.127 Determine i(t) in the circuit of Fig. 21-44a. I 21.128 From Fig. 21-44b: 30 /(s) == 3s 10 or + 6 == S+2 i(t) = lOe -21 Obtain the transformed circuit for the circuit of Fig. 21-45a. I See Fig. 21-45b. ~1* 0 (') 21.129 l2< Fig. 21-45 Find i(t) in the circuit of Fig. 21-45a. I 21.130 A From Fig. 21-45b: lO/s /(s) == 2s + (1 /2s) 10 = 2(S2~- 1) or i(t) == 10 sin 1/2 Find the transformed circuit for the circuit of Fig. 21·46a, where S is moved from a to b at I See Fig. 21-46b. (h) Fig. 21-46 t == o. LAPLACE TRANSFORM METHOD 21.131 529 Solve for i(t) in the circuit of Fig. 21-46a. lOs I From Fig. J(s) = 2(S2 21-46b: Thus, 21.132 D + Ss Ss n == ?+ ! = S2 + ~ i(t) == S cos (tI2) A Determine the transformed circuit corresponding to that shown in Fig. 21-47a. I See Fig. 21-47b. Fig. 21-47 21.133 Solve for i(t) in the circuit of Fig. 21-47a in which the switch is moved from a to b at I From Fig. 21-47b: + (1 16s) + (1 13s)] 10 + 1 == s + ~ A What is the voltage across the 3-F capacitor in the circuit of Prob. 21.133 at the s domain. v I At 21.135 W = 6s+3 = 2s i(t) == We -112 or 21.134 ~ W J(s) == s[1 V 3 F(S) == (s) J(s) 20 20 ( I =""3s"" = 3s(2s + 1) =="3 S - t = 1, v(l) = 2s 2) +1 t = 1 s. or V 20 -t/2 () t == - (1- e ) 3 2.62 V 21-48b. . .h Fig. 21-48 Solve for i(t) in the circuit of Prob. 21.135. I From Fig. 21-48b W-W /(s) = s[2 + (1I3s) 21.137 Formulate the problem in Obtain the transformed circuit for that shown in Fig. 21-48a in which the 3-F capacitor is initially charged to 20 V, the 6-F capacitor to W V, and the switch is closed at t == O.