文档介绍：(A)(C)(C)(B)(B)(B)(B)(B)(B)(B)(B)(B)(A)(B)(B)(B)(B)(1)略(2)元件1和2为电源,元件3,4和5为负载(3)(-560-540+600+320+180)*w=/(2110/8+R)=8/110,所以R≈?,WR=(8/110)2×≈=U/I=6/50*310?=120?,应选者(a):220/(R1+315)=,得R1≈314?.220/(R2+315)=,得R2≈0?.(1)并联R2前,I1=E/(0R+2Re+1R)=220/(++10)≈,I2=E/(0R+2Re+1R∥2R)≈50A.(2)并联R2前,U2=R1*I1=212V,U1=(2Re+1R)*I1=,U2=(1R∥2R)*I1=200V,U1=2Re+1R∥2R=210V.(3)并联R2前,P=212*=,P=200*50==I1+I2=,I4=I5-I3=-=,I6=I2+I4=,所以不管S断开还是闭合abR=5R∥(1R+3R)∥(2R+4R)=200?.:aU=1U=16V,bU=<[(45+5)≈]+45>×16/<[(45+5)∥]∥+45>≈=(45+5)∥×bU/总R≈bU/10=,同理dR≈cU/10=:当滑动端位于上端时,2U=(R1+RP)1U/(R1+RP+R2)≈,2U=R2*1U/(R1+RP+R2)≈-:等效电路支路电流方程:IL=I1+I2E2-RO2*I2+RO1*I1-E1=0RL*IL+RO2*I2-E2=0带入数据得I1=I2=20A,IL=:先利用叠加定理计算R1上的电流分成两个分电路1U1单独作用:解A5212111R)//R(RRUI43211'1???????2IS单独作用:*1·I)//(RRRRIS32144''1???????所以A56III''1'11???,A53I*??:根据KCL得则1A1-2I-II123???40V2*1020IRUU20V,1*20IRU2212311????????1A电流源吸收的功率:20W1*20IUP111???2A电流源吸收的功率:-80W2*-40I-UP222???R1电阻吸收功率:20W1*20IRP2231R1???R2电阻吸收功率:40W2*10IRP2222R???:将电流源转换成电压源,如下图则(1//1)1121I1????,53I3?:将两个电压源转换成电流源再