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Chapter 4 AC Networks 4.1 Introduction Alternating Waveforms o o o Sinusoidal AC Voltage Square Wave Triangular Waveform Function Generators are devices that can create all of these waveforms. These devices will be used in lab. Sketch the waveforms and circuit symbols for the waveforms above. 4.2 Sinusoidal AC Waveform AC Generators (a.k.a. alternators) are fueled by: o o o o o o water power oil gas nuclear fusion wind solar power Power to a shaft turns a rotor inside of a stator. Rotor - rotating magnetic Stator - stationary coil of wire Simple Harmonic Motion (SHM) Position versus time for simple harmonic motion is shown as sine or a cosine Uniform Circular Motion Video 90°, p/2 radians y 180°, p radians q x 270°, 3p/2 radians 0°, 0 radians p radians deg rees 180 180 deg rees radians p p Example : 45 4 p Example : 60 3 Wave Description Amplitude (Vpeak) - maximum displacement from the zero line Period (T) - the time required for one cycle Frequency (f) - the number of cycles that appear in a time span 1 cycle per second = 1 Hertz T=1/f Example Problems Sketch the following waveform: v=Vpeaksinq Example 4.2: o o Determine the period and frequency of the waveform in Figure 4.7. (5 cycles in 4ms for v versus t) Example 4.3: o How long will it take a sine wave with a frequency of 0.2 kHz to complete 10 cycles? Example Problems Example 4.4: o For v=20sinq, at what angle will the voltage rise to 5 Volts? Angular Velocity (w) - rotation rate of a vector (or generator shaft) w is measure in radians per second. New Waveform Equation: v=Vpsinwt 2p w 2pf T Sinusoidal Voltage Relationships v = Vpeaksin q v = Vpeaksin wt v = Vpeaksin 2pt/T v = Vpeaksin 2pft Which form you use depends on what you know Example Problems Example 4.5: o Given i=2010-3 sin 400t, find the time at which i rises to 10 mA. Phase Relationship f = phase angle For initial intersections before 0 v=Vpsin(wt+f) For initial intersections after 0 v=Vpsin(wt-f) Resistive Circuit Only for a resistive circuit can current and voltage be in phase. If the current and voltage go through zero at different times, they are said to be out of phase Current and Voltage out of Phase View as two vectors rotating around the circle o Must mean they have the same w V f I w Sinusoidal Plot Example Problems Example 4.6: o o o o (a) Write the sinusoidal expression for each waveform appearing in Fig. 4.12. (Current Amplitude = 5mA, f i = +60°, w = 400 rad/sec) (Voltage Amplitude = 100 V, f v = -30°, w = 400 rad/sec) (b) What is the phase relationship between the two waveforms? Example Problems Example 4.7: o Determine the phase relationship between the following two waveforms: v=8.6 sin(300t+80°) i=0.12 sin(300t+10°) Graphs of Ex. 4.7 4.4 Effective (RMS) Value These mean the same thing for AC circuits: o o o “equivalent DC voltage” “effective voltage” “rms voltage” RMS = root mean square Draw the “relating effective values” plots. 1 Veff Vp 2 I eff 1 Ip 2 4.5 Average Value For a sine wave, the average value is zero. For other waves…. area (under waveform) G(average value) T ( period ) Example Problems Example 4.9: o Write the sinusoidal expression for a voltage having an rms value of 40 mV, a frequency of 500 Hz and and initial phase shift of +40. o Vp = 1.414(Vrms) = 1.414(40 mV) = 56.56 mV w = 2pf = (2p)(0.5 kHz) = 3.142 X 103 rad/s v = 56.56 mV sin(3142t + 40°) o Example Problems Example 4.10: Calculate the average value of the waveform below over one full cycle. 12 10 8 6 4 2 0 -2 0 -4 -6 2 4 6 8 10 12 What are the following for a wall outlet? f Veff = Vrms Vpeak Vave Ieff = Irms Ipeak Iave f = 60 Hz Veff = Vrms = 120 V Vpeak = 2 Vrms = 170 V Vave = 0 V Ieff = Irms = limited by breaker Ipeak = 2 Irms Iave = 0 A Effective Values Average values for I or V are zero If we use IV to measure power with a dc meter, we will get zero o o Yet the resistor gets hot – power is dissipated I and V are not always zero – you can get shocked from ac just like dc We need a new definition for power Use Effective (rms) Values PR = VRIR = IR2R = VR2/R Look just like the dc forms of power o But we use rms values instead 4.7 The R, L, and C Elements Ohm’s Law for Peak Values Resistors: Vpeak=IpeakR Capacitors: Vpeak=IpeakXC Inductors: Vpeak=IpeakXL Inductive Reactance Recall the form of inductance i v L t Since I for ac is constantly changing, there is a constant opposition to the flow of current o o o Inductive Reactance XL = wL = 2pfL Measured in ohms Energy is stored in the coil, not dissipated like in the resistor. Example An inductor of 400 mH is connected across a 120 V, 60 Hz ac source. Find XL and the current through the coil a. b. XL = 2pfL = 2p(60 Hz)(0.4 H) XL = 151 W I = V/XL = 120 v/151 W I = 0.8 A XL vs. f graph for ideal inductor XL = 2pfL o o Should give a straight line graph of slope 2pL Higher the L the greater the slope XL f V and I for an Inductor From when i v L t Current we see that voltage is greatest the change of current is greatest. Highest positive v Zero here Most negative here Voltage and Current (Coil) Current Voltage Voltage leads current by 90° Example Problems Example 4.13: o The current iL through a 10-mH inductor is 5 sin(200t +30). Find the voltage vL across the inductor. Capacitive Reactance The effect of a capacitor is to prevent changes in voltage o Keep the potential difference from building up in the circuit Since the build up of potential is minimized, the flow of current is reduced. o Capacitors act like a resistance in an ac circuit. Capacitive Reactance V is changing in ac circuits o o But Q = VC, so the charge changes also Since I Q , current is greatest when change in t voltage is greatest. Current Voltage I leads V by 90° CIVIL CIVIL is a memory aid. For a capacitor C, current I leads the voltage V by 90 degrees. For an inductor L, voltage V leads the current I by 90 degrees. Capacitive Reactance 1 Xc 2p f C Xc also measured in ohms What is the capacitive reactance of a 50 mf capacitor when an alternating current of 60 Hz is applied? 1 1 Xc 2pfC 2p (60 Hz )(50 X 10 6 f ) X c 53 W Form of Reactance (Fig. 4.28) 300 250 XC (W ) 200 150 100 C=1mF C=2mF 50 0 0 1000 2000 f (Hz) 3000 Example Problems Example 4.14: o The voltage across a 2-mF capacitor is 4mV (rms) at a phase angle of -60. If the applied frequency is 100kHz, find the sinusoidal expression for the current iC. 3.6 Phasors and Complex Numbers A vector is a quantity has both magnitude and direction. A scalar quantity has only a magnitude. Examples: o o Scalar: 50mph Vector: 50mph North A phasor is a complex number used to represent a sine wave’s amplitude and phase. A complex number can be written as C = A + Bj where C is a complex number, A and B are real numbers and j 1 Imaginary C=A+Bj q Real What is the magnitude of C? C A B 2 2 What is the direction of C? B q tan A 1 Phasor Diagrams We make the real axis the resistance The imaginary axis is reactance o o Inductive reactance is plotted on the +y axis Capacitive reactance is plotted on the –y axis The vector addition of R and X is Z XL XC X = XL - XC R Phasors Form Reactance First X = XL – Xc Form Impedance Z R 2 X 2 Calculate phase angle f Tan 1( X XL XC X R R X R Z R ) Example Series R-L Circuit A coil has a resistance of 2.4 W and a inductance of 5.8 mH. If it is connected to a 120-V, 60 Hz ac source, find the reactance and the current Solution XL = 2pfL = 2p(60 Hz)(5.8 X 10-3 H) XL XL = 2.19 W R Z R 2 X 2 ( 2.4W )2 ( 2.19W )2 Z 3.25W f Tan 1 ( X R ) Tan 1 ( 2.19W 2.4W ) f 42.4 X Z R f Solution (cont’d) I = V/Z I = 120 V/3.25 W = 36.9 A ================================ Notice that if f = 120 Hz, then XL = 2p(120 Hz)(0.0058 H) = 4.37 W Z = 4.99 W f = 61.2° I = 24 A Example Series R-C Circuit What is the current flow through a circuit with 120 V, 60 Hz source, an 80 mf capacitor, and a 24 W resistor? Solution 1 1 XC 2p f C 2p ( 60 Hz )( 80 X 10 6 f ) X C 33W XC Z R 2 X 2 ( 24W )2 ( 33W )2 Z 40.9W tanf 33W R R 24 1.38 f 54.1 V 120V I 2.93 A Z 40.9W f X Z Solution (cont’d) What happens is the frequency doubles? XC = 16.6 W Z = 29.2 W f = 34.6° I = 4.11 A Example - Series RLC Circuit XL XC R X R X = XL - XC X Z R Z2 = R2 + X2 A 240 V, 60 Hz line is connected to a circuit containing a resistor of 15 W, an inductor of 0.08 H, and a capacitor of 80 mf. Compute the circuit impedance and current. Solution X L 2p f L 2p ( 60 Hz )( 0.08 H ) 30.2W 1 1 XC 33.2W 6 2p f C 2p ( 60 Hz )( 80 X 10 f ) X X L X C 30.2W 33.2W 3.0W Z R 2 X 2 ( 15W )2 ( 3W )2 15.3W I V Z 240V 15.3W 15.7 A Power Factor In dc circuits we had P = IV and this still holds in purely resistive circuits with ac In general, I and V are out of phase in ac circuits. o o At times V and I are negative and power is drawn from the circuit. Thus the true power must be less than IV Power Factor i I p sin wt v V p sin( wt f ) f is the phase between i and v p iv I pV p sin wt sin( wt f ) true power(P) Power Factor IV P IV ( power factor ) Sum of p over one cycle P T I pV p P cos f I rmsVrms cos f 2 Power Factor Power Factor = cos f X Z f X R R cos f Z R R2 X 2 For purely inductive or purely capacitive circuits, f = 90° and cos f = 0 no power is dissipated Control of the Power Factor Most household appliances are highly resistive and require high PF. Industry tends to be rather inductive. Some substations have capacitor banks timed to break in at certain times when inductive load changes, adjusting the PF to the desired value Example An industrial plant service line has an impressed power of 100 KV-A on a 60 Hz line at 440 V. An installed motor offers a total ohmic resistance of 350 W and an inductance of 1 H. Find… a) b) c) d) e) f) The inductive reactance The impedance The current The power factor The power used by the motor The size capacitor required to change PF to 95%. Solution a) X L 2pfL 2p (60 Hz)(1 H) 377 W b ) Z R 2 X L2 ( 350W )2 ( 377 W )2 514.4W c) I V Z d ) PF R 440V Z 514.4W 350W 0.855 A 514.4W 0.68 e ) P ( 100 KV A )( PF ) 68 KW Solution (cont’d) PF 0.95 R Z 368.42W 0.95 Z 2 R 2 ( X L X C )2 ZR Z 2 R 2 ( X L X C )2 X L Z 2 R 2 X C 377 W ( 368.42W )2 ( 350W )2 262W 1 XC 2p f C 1 1 C 10.1mF 2p f X C 2p ( 60 Hz )( 262W ) Lightning Arrestors A high voltage line is likely to transmit a high voltage, high frequency electric discharge if struck by lightning. o The high frequencies may travel to a substation, causing arcing in generators. Solution Choke Coil is an inductor of very few turns (low L). It offers low reactance to 60 Hz. but stops high frequencies Alternate path to ground offered (spark gap and series resistor). This path has high R to low frequencies and low R to high frequencies. Series Resonance Z R 2 ( X L X C )2 If XL = XC, then Z = R and the circuit is purely resistive. This will occur at a certain frequency 1 2p f L 2p f C 4p 2 f 2 LC 1 1 f res 2p LC By altering L or C, the circuit can be made to resonate at any frequency. This condition makes X = 0 V and I will be in phase minimum Z and maximum I Variable Capacitors (varicaps) One set of plates moves with respect to the other. Since C depends on plate area, C varies as plates are shifted. Used in tuning a radio. All frequencies are excited on the primary (left) coil and induced onto the secondary coil. We tune the capacitor to reach resonance, which maximizes the current flow. Power Transmission In the early days power was transmitting as dc. o o Applications used low voltage (10 – 110 v) Power losses (I2R) meant power couldn’t be transmitted far. Solution is to lower the current and raise the voltage (P = IV) o o We can get the same power and not suffer the I2R losses. Problem: applications cannot use the high voltages Transformers Used to change the voltage either down (step-down) or up (step-up) Principle of Operation Primary is connected to the ac source. o o Alternating current causes an alternating magnetic field Voltage is induced on the secondary The same number of lines of magnetic force pass through each coil o Induced voltage in each turn of wire should be the same Law of Transformers Vp Vs Np Ns Strictly true only when no current flows, since counter currents are set up (remember inductors?) For Np < Ns we have Vp < Vs a step-up transformer For Np > Ns we have Vp > Vs a step-down transformer Transformers (cont’d) If no losses occur (real transformers are 90 – 95% efficient), then o VpIp = VsIs thus, Ip Vs N s I s Vp N p A step-down transformer gives Is > Ip A step-up transformer means a decrease in current Example Generator furnishes 10 A at 550 V o Assume the transmission line has 20 W loss o o Ploss = I2R = (10A)2(20) W = 2000 W 2000/5500 = 36% power loss Let’s step-up the voltage to 5500 V o o o P = VI = 5500 W Then Is = (VpIp)/Vs = (550V)(10A)/5500V = 1A Now the losses are Ploss = I2R = (1A)2(20) W = 20 W Or 0.36% loss When the power gets to the user, we put in a step-down transformer Parallel RC Circuit General Scheme o Form impedance for each branch o From Kirchhoff’s Current Law iT = i1 + i2 i1 V sin( w t f1 ) Z1 i2 V sin( w t f2 ) Z2 The possibility of different phase angles means parallel impedances do not add as simply as for dc circuits Total Current and Definitions 1 1 iT V sin( w t f1 ) sin( w t f2 ) Z2 Z1 Z 1 R12 X 12 Define GR Z 2 R22 X 22 Z 2 and BX G is called the conductance B is called the susceptance Z2 Rule for Adding Parallel Impedances 1 ( G1 G2 )2 ( B1 B2 )2 ZT tanqT B1 B2 G1 G2 Note: by convention, XC always carries a negative sign in computing the impedances. Y = 1/ZT is called the admittance Example Compute Branch Impedances 1 Z1 R 2 2 2 4p f C1 1 X1 2p f C R1 X1 G1 2 B1 2 Z1 Z1 Z 2 R2 X2 0 2 1 G2 1 Z 3 R 2p f L3 2p f C3 2 3 2 1 R2 B2 0 X 3 2p f L3 1 G3 R3 Z 32 B3 2p X3 Z 32 f C3 Resistive Elements If parallel impedance contains only resistors, we have o G = 1/R B=0 1 Y G2 B2 ZT 1 1 2 R R And the equation of parallel resistances results. Example Z1 R 5W X1 0 1 G1 R Z2 X X C G2 0 B2 2 C B1 0 XC 1 2p f C 2 Z2 XC B2 0.0377 S Y ( G1 G2 )2 ( B1 B2 )2 ZT 4.91W 1 ( 5W ) 2 ( 0.0377 )2 0.204 S Parallel Resonance Add L and C in parallel Z2 X X L G2 0 B2 1 1 X L 2p f L Z 3 X C2 X C G3 0 B3 XC 1 2p f C 2 Z3 XC 2 L Parallel Resonance 1 ZT G2 G3 2 B2 B3 2 B2 B3 2 B2 B3 1 2p f C 2p f L 1 1 - 4p 2 f 2 LC ZT 2p f L ZT 2p f L 1 - 4p 2 f 2 LC Resonance occurs when the denominator becomes zero Parallel Resonance f res 1 2p LC Same relationship as for series resonance o o In series resonance the impedance is a minimum (Z=R) In parallel resonance the impedance is a maximum (Y=0) Resonance Comparisons Series Parallel Z Min Max Y=1/Z Max Min I Max Min V=IZ Min Max Tuning a Resonant Circuit Both series and parallel circuits have a resonance at f res 1 or, wres 2p LC 1 LC Resistance in the circuit is significant in determining how much current passes at frequencies different from resonance. Z R w L 1 2 Factor out R, Then wL Z R 1 Z R 1 2 wC 1 w L - 1w C 2 R wL 2 1 1 R 2 w 2 LC Quality (Q) Factor At resonance, wo, define Q0 The current in the circuit is… I V V Z R w0 L R 1 w L 2 1 1 1 2 R w LC 2 At resonance (series) Z = R, so we can write the current at resonance as IM = V/R Q-Factor Using the definition and a little manipulation, we can derive… I IM 1 w w0 1 Q w0 w 2 0 2 Q-Factor High Pass (RC) Filter 1 Z R 2 X C2 R 2 2p f C I Vi Z Vi 1 R 2p f C 2 2 2 High Pass Filter I is the same through both R and C Vo RI Vo Vi RVi 1 R 2 2 p f C 1 1 1 2p f RC 2 When f is small, denominator is large and the voltage ratio is small. 2 When f is large, denominator is almost one Typical Results R = 0.1 MW C = 0.16 mF Why is it called a high pass filter? Half Power Point Note the point at which o o Vo Vi 1 2 This will make (Vo/Vi )2 = ½ Recall that one form of power is V2/R, so the output power is half the input power at this frequency. Low Pass (RC) Filter 1 2 2 2 Z R XC R w C I Vi Z 2 I Vo IZ C IX C wC Low Pass Filter I Vo wC Vo Vi 1 w C R w C Vi 2 2 Rw C 2 1 Vo 1 2 Vi 1 w RC Low frequency means denominator is small and Vo / Vi 1 High frequency means denominator is large and Vo / Vi is small Low Pass Results Note the half power frequency again.