CHAPTER 11 AC POWER ANALYSIS11.1 Instantaneous and Average Power11.2 Apparent Power and Power Factor11.3 Maximum Average Power Transfer11.4 Complex Power11.5 Conservation of AC Power 11.6 Power Factor Correction1IntroductionIn this chapter, our goals and objectives include Determining the instantaneous power delivered to an element Defining the average power supplied by a sinusoidal source Using complex power to identify average and reactive power Identifying the power factor of a given load, and learning means of improving it2Instantaneous and Average Power 1. Instantaneous power （瞬时功率 ）3i tOupUIcos- UIcos(2 t 1 )The average power is the average of the instantaneous power over one period.2. Average Power , Real POwer(平均功率、有功功率) 4Instantaneous power：瞬时功率Average power：平均功率/有功功率Power analysis to R:5Power analysis to L:Instantaneous power ： Average power ：Stored magnetic energy：Q is a measure of the energy exchange between the source and the reactive part of the load6Power analysis to C:Instantaneous power：Average power ： Stored electric energy：7inductor = 90capacitor = -90P = 0RX|Z|P+-R jX+-8Maximum Average Power TransferThe current through the load is ZLZth9Average power delivered to load is :Our objective is to adjust the load parameters RL and XL so that P is maximum. To do this we set P/ RL and P/ XL equal to zero. We obtain P/ RL=0 P/ XL=010For maximum average power transfer, the load impedance ZL must be equal to the complex conjugate of the Thevenin impedance Zth.Setting RL = Rth and XL = -Xth in Eq. (1) gives us the maximum average power as 11In a situation in which the load is purely real, the condition for maximum power transfer is obtained from Eq. (2) by setting XL = 0; that is,12Example： Given that v(t)=120cos(377t+450）V and i(t)=10cos(377t-100)AFind the instantaneous power and the average power absorbedBy the passive linear network of Fig.11.1.13Solution: The instantaneous power is given byp=vi=1200cos(377t+450)cos(377t-100) Applying the trigonometric identityOr p(t)=344.2+600cos(754t+350)WThe average power isp(t)=600cos(754t+350)+cos55014Example： For the circuit shown in Fig.,find the average power supplied by the source and the average power absorbed by the resistor.15Solution:The current through the resistor isThe average power supplied by the voltage source isAnd the voltage across it is16Which is the same as the average power supplied. Zero average power is absorbed by the capacitor.The average power absorbed by the resistor isExample:Determine the power generated by each source and the average power absorbed by each passive element in the circuit Fig11.3.17Solution:We apply mesh analysis as shown in Fig 11.3.for mesh1, For mesh 2,18For the voltage source ,the current flowing from it isI2=10.5879.10A. And the voltage across it is 60300V,so that the average power isV1=20I1+j10(I1-I2)=80+j10(4-4-j10.39) =183.9+j20=184.9846.210V19The average power supplied by the current source isP1+P2+P3+P4+P5=-735.6+320+0+0+415.6=0V3=-j5 I2=(5-900)(10.5879.10 )=52.9(79.10-900)2011.4 Effective or RMS ValueThe effective value of a periodic current is the dc current that delivers the same average power to a resistor as the periodic current.The effective value of a periodic signal is its root mean square (rms) value.21The power factor is the ratio of the average power to the apparent power, which is dimensionless.Apparent Power and Power Factor 1. Apparent power （视在功率 ）The apparent power (in VA) is the product of the rms values of voltage and current.1)pf is the cosine of the phase difference between voltage and current.2)pf is also the cosine of the angle of the load impedance.22Generally speaking, 0cos1X 0 , 0 , inductive, current lags voltageX 0iuC+-PC=UIcos =UIcos(-90)=0QC =UIsin =UIsin (-90)= -UI=U2/XC=I2XC0, we know ZL is inductive load. ZL=R+jXL(1)We setBecause ZL is inductive load,(2)(1)(2)Analyze the circuit of Fig. 11.46 to find the complex power absorbed by each of the five circuit elements.EXAMPLE11.6Solution：EXAMPLE11.8: The load in Fig. 11.44 draws 10 kVA at PF =0.8 lagging. If |IL |=40 A rms, what must be the value of C to cause the source to operate at PF =0.9 lagging?Solution:RL XL LGBEXAMPLE:Consider the circuit of Fig. 11.45. Specify the value of capacitance required to raise the PF of the total load connected to the source to 0.92 lagging if the capacitance is added (a) in series with the 100-mH inductor; (b) in parallel with the 100-mH inductor. Zeff = j100 + j300 | 200 = 237 54.25o. = PF = cos 54.25o = 0.5843 lagging. (a) Raise PF to 0.92 lagging with series capacitanceZnew = j100 + jXC + j300 | 200 = 138.5 + j(192.3 + XC) Solution:Solving, we find that XC = 133.3 = 1/wC, so that C = 7.501 mF (b) Raise PF to 0.92 lagging with parallel capacitanceZnew = j100 | jXC + j300 | 200 =