Chapter 11 AC Circuit Power Analysis PDF

Chapter 11 AC Circuit Power Analysis Copyright © 2013 The McGraw-Hill Companies, Inc. Permission required for 1 reproduction or display. Objectives At the end of this chapter, you should be able to: 1. Compute real power, reactive power, apparent power, and/or power...

Chapter 11 AC Circuit Power Analysis Copyright © 2013 The McGraw-Hill Companies, Inc. Permission required for 1 reproduction or display. Objectives At the end of this chapter, you should be able to: 1. Compute real power, reactive power, apparent power, and/or power factor for a circuit or element. Copyright © 2013 The McGraw-Hill Companies, Inc. Permission required for rep 2 Introduction  This chapter examines different representations of power in an AC circuit.  These power representations include:  Instantaneous power  Average power  Apparent power  Complex power  Reactive power Copyright © 2013 The McGraw-Hill Companies, Inc. Permission required for rep 3 Introduction  In addition to these power representations, other related measures are defined.  These include:  Effective (RMS) current and voltage  Power factor Copyright © 2013 The McGraw-Hill Companies, Inc. Permission required for rep 4 Instantaneous Power The instantaneous power is p(t)=v(t)i(t). At all times t, power supplied = power absorbed Copyright © 2013 The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 5  Power from Sinusoidal Source If in the same RL circuit, the source is Vmcos(ωt), then and so the power will be Double Constant Frequency Term Term Copyright © 2013 The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 6 Average Power  The average power over an arbitrary interval from t1 to t2 is  When p(t) is periodic with period T, the average power is calculated over any one period: Copyright © 2013 The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 7 Average Power: Sinusoidal Steady State If v(t)=Vmcos(ωt+θ) and i(t)=Imcos(ωt+ϕ), then 1 P  Vm I m cos(   ) 2  Copyright © 2013 The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 8 Average Power for Elements  The average power absorbed by a resistor R is 2 1 1 2 V PR  Vm I m  I m R  m 2 2 2R  The average power absorbed by a purely reactive element(s) is zero, since the current and voltage are 90 degrees out of phase: PX 0 Copyright © 2013 The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 9 Example: Average Power Find the average power being delivered to an impedance ZL= 8 − j11 Ω by a current I= 5ej20° A. Only the 8-Ω resistance enters the average-power calculation, since the j11-Ω component will not absorb any average power. Thus, P = (1/2)(52)8 = 100 W Copyright © 2013 The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 10 Example: Average Power Find the average power absorbed by each element. Answer: PL=0 W PC=0 W, PR=25 W Pleft=-50 W Pright=25 W Copyright © 2013 The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 11 Maximum Power Transfer  An independent voltage source in series with an impedance Zth delivers a maximum average power to a load impedance ZL which is the conjugate of Zth: ZL = Zth*  That is, if Z L RL  jX L & Z th Rth  jX th  Then, RL Rth X L  X th & Copyright © 2013 The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 12 Maximum Power Transfer Derivation First, solve for the load power: 2 1 2| Vth | RL  2 2 (Rth  RL )  (X th  X L ) Clearly, P is largest when XL+Xth=0 Solving dP/dRL=0 will show that RL=Rth Copyright © 2013 The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 13 Effective Values of Current and Voltage  Effective value of a periodic current (voltage) is equal to the value of the direct current (voltage) required to deliver the same average power to a resistor R.  The same power is delivered to the resistor in the circuits shown. periodic, period T Copyright © 2013 The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 14 Effective Values of Current and Voltage  It can be shown the effective value of a current (voltage) is obtained by taking the square-root of the mean or average value of the square of the instantaneous current (voltage).  Effective value often called root-mean-square or rms value Copyright © 2013 The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 15 Effective (RMS) for Sine Wave  For sine waves: V  1 eff Vm 0.707Vm 2 1 I eff  I m 0.707 I m 2  Average power can calculated using effective values 2 V P Veff I eff cos    P I R 2 eff P eff R Copyright © 2013 The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 16  Apparent Power & Power Factor  Ifv(t)=Vmcos(ωt+θ) and i(t)=Imcos(ωt+ϕ), then 1 P  Vm I m cos(   ) Veff I eff cos(   ) 2  The apparent power is defined as VeffIeff and is given the units volt-ampere VA.  Apparent power is the power that would  apparently be absorbed if a load impedance behaved exactly like a resistor. Copyright © 2013 The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 17  Apparent Power & Power Factor Power factor is defined as average power P PF   apparent power Veff I eff  fora resistive load, PF=1  fora purely reactive load, PF=0  generally, 0 ≤ PF ≤ 1  Copyright © 2013 The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 18 Power Factor: Lagging & Leading  Since the power factor for sine waves is PF cos(   )  The information as to whether current leads or lags voltage is lost, so we add the adjective to the power factor term.  An inductive load has a lagging PF (I lags V).  A capacitive load has a leading PF (I leads V). Copyright © 2013 The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 19 Example: Power Factor Find the average power delivered to each of the two loads, the apparent power supplied by the source, and the power factor of the combined loads. Answer: 288 W, 144 W, 720 VA, PF=0.6 (lagging) Copyright © 2013 The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 20