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1 Kulshreshtha, D. C. @ McGraw-Hill Education Basic Electrical Engineering McGraw-Hill Education © 2010 BASIC ELECTRICAL ENGINEERING D. C. KULSHRESHTHA, PowerPoint Slides PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this PowerPoint slide may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this PowerPoint slide, you are using it without permission. Next 2 Kulshreshtha, D. C. @ McGraw-Hill Education McGraw-Hill Education © 2010 Basic Electrical Engineering Chapter 3 Network Analysis–Part I D.C. Kulshreshtha Next Thought of The DAY Whatever THE MIND OF MAN can CONCIEVE and BELIEVE, it can ACHIEVE. Friday, May 12, 2017 Ch. 3 Network Analysis- Part I Next 3 Topics to be Discussed Electric Circuit, The Resistance Parameter, The Capacitance Parameter, The Inductance Parameter, Energy Sources-Classification Ideal Voltage Source. Ideal Current Source. Friday, May 12, 2017 Series and Parallel Combinations. Practical Voltage Source. Practical Current Source. Source Transformation. Kirchhoff’s Laws. Ch. 3 Network Analysis- Part I Next 4 Electric Circuit It is a closed path, composed of active and passive elements. Active Element : It supplies energy to the circuit. Passive Element : It receives energy and then 1) either converts it to heat, as in a Resistance (R). 2) or stores it in (a) Electric Field, as in a Capacitor (C). (b) Magnetic Field, as in an Inductor (L). Friday, May 12, 2017 Ch. 3 Network Analysis- Part I Next 5 Energy Sources Classification Independent Source Voltage Source DC Source Ideal Source Friday, May 12, 2017 Or Or Or Or Dependent Source Current Source AC Source Practical Source Ch. 3 Network Analysis- Part I Next 6 Independent Ideal Voltage Source Source Load Note that the source determines the voltage, but the current is determined by the load. The source has zero internal resistance. Friday, May 12, 2017 Ch. 3 Network Analysis- Part I Next 7 The voltage source is said to be idle if the output terminals are open such that i = 0. When turned off (killed or made inactive), so that v = 0, it is equivalent to a short circuit. Reference Marks : One terminal is marked plus and the other minus. (Oversimplification; one mark can be omitted.) When actual polarity is opposite to the reference marks, the voltage is a negative number. Friday, May 12, 2017 Ch. 3 Network Analysis- Part I Next 8 Independent Ideal Current Source Source Load Note that the source determines the current, but the voltage is determined by the load. The Source has infinite internal resistance (Ri). Friday, May 12, 2017 Ch. 3 Network Analysis- Part I Next 9 The current source is said to be idle if the output terminals are shorted together, such that v = 0. When turned off (killed or made inactive), so that i = 0, it is equivalent to an open circuit. Reference Marks : An arrow is put. When actual direction of current is opposite to the reference (arrow) direction, the current is a negative number. Friday, May 12, 2017 Ch. 3 Network Analysis- Part I Next 10 Do you observe duality ? The roles for the current and voltage are interchanged in the two sources. Friday, May 12, 2017 Ch. 3 Network Analysis- Part I Next 11 Practical Voltage Source It is represented by an ideal voltage source in series with an internal resistance (RSV). Friday, May 12, 2017 Ch. 3 Network Analysis- Part I Next 12 Practical Current Source It is modelled as an ideal current source in parallel with an internal resistance (RSI). Friday, May 12, 2017 Ch. 3 Network Analysis- Part I Next 13 Source Transformation A practical current source can be converted into its equivalent practical voltage source, and vice versa. This conversion is valid only for the external load connected across the terminals of the source. Friday, May 12, 2017 Ch. 3 Network Analysis- Part I Next 14 Friday, May 12, 2017 Ch. 3 Network Analysis- Part I Next 15 Equivalence between Voltage Source and Current Source Two sources would be equivalent if they produce identical values of VL and IL, when they are connected across the same load. Friday, May 12, 2017 Ch. 3 Network Analysis- Part I Next 16 Series and Parallel Combinations What would be the net emf of the combination if two ideal voltage sources of 2 V and 4 V are connected in series so as to aid each other? Click Ans. 6 V What would be the net emf of the combination if two ideal voltage sources of 4 V and 4 V are connected in parallel ? 4 V or 8 V ? Click Ans. Obviously, it should be 4 V Friday, May 12, 2017 Ch. 3 Network Analysis- Part I Next 17 What would be the net emf of the combination if two ideal voltage sources of 2 V and 4 V are connected in parallel ? 2 V or 4 V or 3 V ? Click The question seems to be quite tricky! Click Ans. The question is wrong. The question contradicts itself. Ideal Voltage Sources in parallel are permissible only when each has the same terminal voltage at every instant of time. What is its dual ? Friday, May 12, 2017 Ch. 3 Network Analysis- Part I Next 18 Ideal Voltage Sources Connected in Series Friday, May 12, 2017 Ch. 3 Network Analysis- Part I Next 19 Ideal Current Sources Connected in Parallel Friday, May 12, 2017 Ch. 3 Network Analysis- Part I Next 20 Practical Current Sources Connected in Series Friday, May 12, 2017 Ch. 3 Network Analysis- Part I Next 21 Practical Voltage Sources Connected in Parallel Friday, May 12, 2017 Ch. 3 Network Analysis- Part I Next 22 Example 1 : Reduce the network shown in figure to its simplest possible form by using source transformation. Friday, May 12, 2017 Ch. 3 Network Analysis- Part I Next 23 Solution Friday, May 12, 2017 Ch. 3 Network Analysis- Part I Next 24 Friday, May 12, 2017 Ch. 3 Network Analysis- Part I Next 25 Example 2 In the given figure, (a) If RL = 80 Ω, find current iL. (b) Transform the practical current source into a practical voltage source and find iL if RL = 80 Ω again. (c) Find the power drawn from the ideal source in each case. Friday, May 12, 2017 Ch. 3 Network Analysis- Part I Next 26 Solution : 20 (a) iL (100) 20 mA 20 80 (b) iL 2 20 mA 80 20 Click (c) vL iL RL (20 mA )(80 ) 1.6 V Click P vL I (1.6 V)(100 mA ) 160 mW Click In the second case, P iL V (20 mA )( 2 V) 40 mW Friday, May 12, 2017 Ch. 3 Network Analysis- Part I Next 27 Benchmark Example 3 Take the benchmark example of the circuit given in figure. Using source transformation, determine the voltage v across 3-Ω resistor. Friday, May 12, 2017 Ch. 3 Network Analysis- Part I Next 28 Solution : Transforming the 4-A current source into a voltage source, Friday, May 12, 2017 Ch. 3 Network Analysis- Part I Next 29 Combining the two voltage sources, Friday, May 12, 2017 Ch. 3 Network Analysis- Part I Next 30 Again transforming the voltage source into current source, Friday, May 12, 2017 Ch. 3 Network Analysis- Part I Next 31 Combining the two current sources we get Fig. (e). Transforming this current source into voltage source (Fig. f ) Friday, May 12, 2017 Ch. 3 Network Analysis- Part I Next 32 Combining the two resistances, we get Fig. (g). Finally, using voltage divider, we get 3 v 5 2.5 V 33 Friday, May 12, 2017 Ch. 3 Network Analysis- Part I Next 33 Kirchhoff’s Laws (1) KCL : Algebraic sum of currents meeting at a junction of conductors in a circuit is zero. It is simply a restatement of the principle of conservation of charge. b I j 1 Friday, May 12, 2017 j 0 Ch. 3 Network Analysis- Part I Next 34 (2) KVL :The algebraic sum of voltages around a closed circuit or a loop is zero. It is simply a restatement of the principle of conservation of energy. k v j 1 Friday, May 12, 2017 j 0 Ch. 3 Network Analysis- Part I Next 35 Polarity of Voltages Note that polarity of the voltage (emf) across a battery does not depend upon the assumed direction of current. Friday, May 12, 2017 Ch. 3 Network Analysis- Part I Next 36 Applying KVL 1. 2. 3. 4. Select a closed loop. Mark the voltage polarity (+ and -) across each element in the closed loop. Go round the selected loop, and add up all the voltages with + or – signs. Any one of the following two rules can be followed : Friday, May 12, 2017 Ch. 3 Network Analysis- Part I Next 37 (i) Rule 1 : While travelling, if you meet a voltage rise, write the voltage with positive sign ; if you meet a voltage drop, write the voltage with negative sign. (ii) Rule 2 : While travelling, write the voltage with positive sign if + is encountered first; write the voltage with negative sign if – is encountered first. We shall be following Rule 1, as it has a strong analogy with the physical height (altitude) of a place. Friday, May 12, 2017 Ch. 3 Network Analysis- Part I Next 38 Example 5 : Use KVL to find vR2 and vx. Friday, May 12, 2017 Ch. 3 Network Analysis- Part I Next 39 For finding vR2, we write KVL eqn. going around loop abgha clockwise : Click 36 vR 2 4 0 vR 2 32 V If you choose to go around the loop anticlockwise, you get Click 4 36 vR 2 0 vR 2 32 V Giving the same result. Friday, May 12, 2017 Ch. 3 Network Analysis- Part I Next 40 There are two ways to determine vx 1) We can consider this voltage as the voltage across the gap from d to f. Writing KVL (habcdfgh) : Click 4 36 12 14 vx 0 vx 6 V 2) Knowing vR2 , apply a short-cut (bcdfgb) : Click 12 14 vx 32 0 vx 6 V Friday, May 12, 2017 Ch. 3 Network Analysis- Part I Next 41 Important Note about KVL The assumed direction of current through a resistor and the polarity of voltage across it are always in conformity. The end into which the current enters is marked positive. Passive-element sign convention. Friday, May 12, 2017 Ch. 3 Network Analysis- Part I Next 42 Example 6 : Find the current supplied by the 60-V source in the network. Friday, May 12, 2017 Ch. 3 Network Analysis- Part I Next 43 Solution : We need not find the currents I1, I2 and I3. Instead, we reduce the network. Friday, May 12, 2017 Ch. 3 Network Analysis- Part I Next 44 Example 7 : Determine the value of current I. 2–3–I–4=0 Friday, May 12, 2017 or I = -5 A Ch. 3 Network Analysis- Part I Next 45 Example 8 Using KCL and KVL, determine the currents ix and iy in the network shown. Friday, May 12, 2017 Ch. 3 Network Analysis- Part I Next 46 Solution : Using KCL, the currents in other branches are marked as shown. Writing KVL equations for the loops 1, 2 and 3, 100 10 I1 5I x 0 5I x 0 I y 10 I1 100 5I x 50 2( I1 I x ) 2 I y 0 7 I x 2 I y 2 I1 50 2 I y 50 3( I1 I x I y ) 0 Friday, May 12, 2017 3I x 5I y 3I1 50 Ch. 3 Network Analysis- Part I Next 47 Writing the above equations in matrix form, 5 7 3 0 2 5 10 2 3 I x 100 I y 50 ; I 50 1 Using Calculator, we solve for Ix and Iy, I x 3.87 A; Friday, May 12, 2017 and Click Click I y 0.51 A Ch. 3 Network Analysis- Part I Next 48 Review Electric Circuit, The Resistance Parameter, The Capacitance Parameter, The Inductance Parameter, Energy Sources-Classification Ideal Voltage Source. Ideal Current Source. Friday, May 12, 2017 Series and Parallel Combinations. Practical Voltage Source. Practical Current Source. Source Transformation. Dependent Sources. Kirchhoff’s Laws. Ch. 3 Network Analysis- Part I Next 49