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1、Chapter 11 AC Power Analysis,要求深刻理解与熟练掌握的重点内容有： 正弦稳态电路的有功功率、无功功率、视在功率和功率因数的概念及计算，复功率的概 念及最大功率传输。 难点：提高功率因数、功率分析,11-1 Introduction,11-2 Instantaneous and Average Power,11-3 Maximum Average Power Transfer,11-4 Effective or RMS Value,11-5 Apparent Power and Power Transfer,11-6 Complex Power,11-8 Power
2、Factor Correction,11-10 Summary and Review,11-7 Conservation of AC Power,11.1 Introduction,In 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
3、and reactive power Identifying the power factor of a given load, and learning means of improving it,integration 积分法,integrand 被积函数,the load parameter 负载参数,the complex conjugate 共轭复数,trigonometric identity 三角恒等式,Domestic load 家用电器,quadrant 象限,pressing iron 压力熨斗,inertia 惯性,11-2 Instantaneous and Avera
4、ge Power,1. Instantaneous power is p(t)=v(t)i(t),The average power is the average of the instantaneous power over one period.,Substituting instantaneous power p(t) gives,2. Average Power.,Thus, average power is,3. Average Power absorbed by R,L,C.,In Summary: A resistive load (R) absorbs power at all
5、 times, while a reactive load (L or C ) absorbs zero average power.,电抗性负载,11.3 Maximum Average Power Transfer,The current through the load is,Average 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
6、 zero. We obtain, P/ RL=0, P/ XL=0,For 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,In a situation in which the load is purely real, the cond
7、ition for maximum power transfer is obtained from Eq. (2) by setting XL = 0; that is,Example11.1： Given that v(t)=120cos(377t+450）V and i(t)=10cos(377t-100)A Find the instantaneous power and the average power absorbed By the passive linear network of Fig.11.1.,Solution: The instantaneous power is gi
8、ven by p=vi=1200cos(377t+450)cos(377t-100) Applying the trigonometric identity,gives,Or p(t)=344.2+600cos(754t+350)W,The average power is,p(t)=600cos(754t+350)+cos550,Example11.2： For the circuit shown in Fig.11.2,find the average power supplied by the source and the average power absorbed by the re
9、sistor.,Solution:,The current through the resistor is,The average power supplied by the voltage source is,And the voltage across it is,Which is the same as the average power supplied. Zero average power is absorbed by the capacitor.,The average power absorbed by the resistor is,Example11.3:Determine
10、 the power generated by each source and the average power absorbed by each passive element in the circuit Fig11.3.,Solution: We apply mesh analysis as shown in Fig 11.3. for mesh1, For mesh 2,For the voltage source ,the current flowing from it is I2=10.5879.10A. And the voltage across it is 60300V,
11、so that the average power is,V1=20I1+j10(I1-I2)=80+j10(4-4-j10.39) =183.9+j20=184.9846.210V,The average power supplied by the current source is,P1+P2+P3+P4+P5=-735.6+320+0+0+415.6=0,V3=-j5 I2=(5-900)(10.5879.10 )=52.9(79.10-900),Example 11.4 Given the time-domain voltage,Find both the average power
12、and an expression for the instantanous power that result when the corresponding phasor voltage is applied across an impedance Z=2600.,Solution: The phasor current is V/Z=2-600A,and average power is:,The time-domain voltage,time-domain current,And instantaneous power,11.4 Effective or RMS Value,The e
13、ffective 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.,11.5 Apparent Power and Power Factor,1.Reactive Power,2.Apparent Power,3.Power Factor,The p
14、ower factor is the cosine of the phase different between voltage and current . It is also the cosine of the angle of the load impedance.,11.6 Complex Power,1.Define,So that,2.Power triangle,Complex power (in VA) is the product of the rms voltage phasor and the complex conjugate of the rms current ph
15、asor. As a complex quantity, its real part is average power P and its imaginary part is reactive power Q.,11.7 Conservation of AC Power,Why ?,11.8 Power Factor Correction,功率因数的提高,1,0,例、f=50HZ，U=380V，cos1=0.6，P=20KW，欲使功率因数提高到0.9，求C=?,解：,1,0,可推导出一个公式：,表示并联一个多大的电容可提高多少功率因数。,If the pf=0.8, then,where 1
16、is the phase difference between voltage and current. We obtain the apparent power from the real power and the pf as,Example11-5: When connected to a 120V(rms), 60Hz power line, a load absorbs 4kW at a lagging power factor of 0.8. Find the value of capacitance necessary to raise the pf to 0.95.,Solution:,The reactive power is,When the pf is raised to 0.95,The real power P has not changed. But the apparent power has changed; its new value
