内容电路理论chapter

1、5-5 Impedance, admittance and phasor diagram1. Impedance阻抗In sinusoidal circuit：XR|Z|ZreactanceresistanceZ +无源线性+u-i欧姆定律的相量形式当无源网络内为单个元件时有：R+-C+-L+-Z可以是实数，也可以是虚数R-C串联的分压Z = R+jX = R+j(XL - XC)XL = XC= 0u 、i in phaseZ is resistiveXLXC0u leads iZ is inductiveXLXC 0，the two-terminals circuit is inducti

2、ve;while B0, irreversible part，dissipated by RInstantaneous power/瞬时功率p(t) p(t)=ui=2UIcos t cos(t ) =UIcos +UI cos(2t) constantTwice p i tOuUIcosUIcos(2 t )RLCP1. Average power/real power P（平均功率,有功功率）The average power is the average of the instantaneous power over one period.P is the only useful pow

3、er. It is the actual power dissipated by the load.Average power is absorbed by all the resistors.If the circuit contains only one element，theniOne element +uUnit: WRURIIU2R2RRR=00The reactive power Q is a measure of the energy exchange between the source and the reactive part of the load.2. Reactive

4、 power Q（无功功率 ）Q is the maximum instantaneous power transferred back and forth between source and L-C。Q=UIsin Unit：Var /乏RLCQ0ULIL- UCICThe apparent power (in VA) is the product of the rms values of voltage and current.3. Apparent power S（视在功率 ）4. Power factor （功率因数）Power factor：cos Unit：VA ( 伏安)S=U

5、IPower companies require that the customers power factor be high.Power factor depends on the load network itself.SQPPower triangle功率三角形RImpedance triangle阻抗三角形有功功率: P=UIcos 单位：W无功功率: Q=UIsin 单位：var视在功率: S=UI 单位：VAThe wattmeter is the instrument for measuring the average power.A current coil is conne

6、cted in series with the load, the voltage coil is connected in parallel with the load.i1i2R电流线圈电压线圈测量时，P、U、I 均不能超量程。5. Complex Power （复数功率 ）Complex power (in VA) is the product of the rms voltage phasor and the complex conjugate of the rms current phasor, its real part is real power P and its imagin

7、ary part is reactive power Q.Conservation of complex power：The complex, real, and reactive powers of the sources equal the respective sums of the complex, real and reactive powers of the individual loads.Example1: The voltage across a load is u(t)= 60cos(377t-10 )V and the current through the elemen

8、t in the direction of the voltage drop is i(t)=1.5cos(377t+50)A. Find: (1) the complex and apparent powers, (2) the real and reactive powers, and (3) the power factor and load impedance.Fundamentals of electric circuits, page 452S = 45 -60” , S =45VA, P22.5W, Q=-38.97Var, pf=cos(-10-50)=0.5, Z=40-60

9、”Example 2: Us =100 V, R1=1 W, R2=Xc=2 W， (1)find the complex power, the average power and reactive power of the circuit. (2)Determine the wattmeter reading.Solution:wattmeter reading : W =or : W =例三表法测线圈参数。已知f50Hz，且测得U=50V，I=1A，P=30W。求线圈等效参数。解RL+_ZVAW*方法一方法二 又6. 功率因数提高设备容量 S (额定)向负载送多少有功要由负载的阻抗角决定。

10、P=UIcos=ScosS75kVA负载cos =1, P=S=75kWcos =0.7, P=0.7S=52.5kW一般用户： 异步电机 空载cos =0.20.3 满载cos =0.70.85日光灯 cosj =0.450.6 (1) 电源不能充分利用，电流到了额定值，但功率容量还有； (2) 当输出相同的有功功率时，线路上电流大 I=P/(Ucos)，线路损耗大。功率因数低带来的问题：解决办法：并联电容，提高功率因数 (改进自身设备)。分析12LRC+_并联电容后，原负载的电压和电流不变、有功功率和无功功率不变，即：负载的工作状态不变。但是电源发出的无功功率减小了，电路的功率因数提高了。

11、特点：Capacitor can be determined by power triangle：12PQCQLQ从功率这个角度来看 :并联电容后，电源向负载输送的有功UIL cos1=UI cos2不变，但是电源向负载输送的无功功率UIsin2UILsin1减少了，减少的这部分无功是由电容来补偿的。所以，提高了功率因数。5-6 Maximum average power transferMaximum power transfer when ZL is restricted + ZSZZS=RS+jXS Z=RL+jXLRL=RSXL=SFor maximum average power t

12、ransfer, ZL must equal the conjugate of the Thevenin impedance. In this case we say that the load is matched to the source.Example: Determine the load impedance ZL that maximizes the average power drawn from the circuit. What is the maximum average power? 5-7 Sinusoidal Steady-State AnalysisSeries a

13、nd parallel reductions of impedances,voltage and current division, the direct method, source transformationThevenin and Norton equivalent, Superposition Theorem node-voltage and mesh-current analysis are all available for use in solving the phasor circuit in exactly the same fashion as they were use

14、d to solve resistive circuits.The only difference is the need to use complex algebra. i =0 u=0 u=RiExample 1: ,find using Thevenin theorem.2j62jj2Solution：(1) find + 2j62-j2j + Zo(3) Find in Thevenin equivalent (2)find ZoExample 2 :us=10 2 cos1000t, find i1，i2 and the complex power given by us.3 + u

15、s4mH + 500Fi1i22i13 + j4 + j2 Solution 1：loop analysis：3 + j4 + j2 0Example 3 :u=100 2 cos314t V, I=IC=IL, the average power that the circuit dissipated is 866W. Find R，L and C. + RUL=50VUR=86.6VI=IC=IL=10AR=8.66 ,L =15.9mHC=318.5FLCExample 4 : Find the conditions that is independent of the frequenc

16、y. R1C1= R2C2TDS2014数字示波器，100MHz，四通道，1GS/S采样率 The circuit for measuring R and L of a coil is shown below. Us = 100V, f =50Hz，r1=6.5, the reading of V reaches the smallest value of 30V when r3=5 and r4=15. Find r2 and L.r2 =4.2 , L=0.04H如图所示电路中，已知 R1=30, U1= 105V, 电压表的读数也为105V。 试求R2、XL和电路的无功功率。 + -jX

17、3Example 5: The reading of each meter is：V1 = V，V =220V， A2 =30A， A3 =20A，W =1000W，find R、X1 、X 2 、X 3WVA2A3*X1jX 2RV1answer：R=10，X1 =10 ，X 2=3.2 ,X 3 =4.79 Example 6、下图中已知：I1=10A、UAB =100V，求： A 、UO 的读数两种解题方法：1.利用复数相量运算2.利用相量图AAB C25UOC1解法1:利用复数相量运算则：即：设：为参考相量，ABU.UO读数为141伏读数为 10安A解法2:利用相量图求解设：45由已知