6.如图所示电路，利用网孔法求I1，I3、V。3.5V+0.65210252102I352(10 +10+5)I -101, -101, = -9 +3.5I, = 0.6v(10+5+5)I -10I -5I, =0V= 51,V=IVI, =0.1A, I, =0.2A
0.6v 5 5 10 10 5 9V 3.5V I1 I2 I3v 1 2 3 2 3 1 2 3 (10 10 5) 10 10 9 3.5 0.6 (10 5 5) 10 5 0 5 I I I I v I I I v I + + − − = − + = + + − − = = 1 3 I I = = 0.1A, 0.2A v =1V
7.如图所示含理想运放的电路，（1）利用结点分析法求Vo/vs：（2）求由电压源vs看进去的输入电阻。RRRV-RR2kRRRRRRRRV3V, =0RRV.V4=V3VV2RVR2V3VsR, =R+(R/ /R)=1.5RCommon mistake运放输出结点不能列电流方程
+ - + - vs vo R R R R R R R + - + - Vs Vo v1 v2 v3 v4 R R R R R R R 1 2 2 1 3 4 o 2 4 3 1 1 1 1 1 ( ) - =0 1 1 1 1 ( ) - =0 1 1 1 ( ) =0 0 + + − + − + − = = s v v v R R R R R v v v R R R R v v R R R v v v o 2 = - 3 s v v ( / / ) 1.5 R R R R R in = + = Common mistake 运放输出结点不能列电流方程
Chl&Ch2 ExercisesThe absorbing power of A element is 2W, find theabsorbing power of the voltage source and current source3A1229ASolution:IA =1A, Iv =1A, Pv=2WVsA = 5V, PA = -15W
2V 2 3A A 1 A 2V 2V I I P = = = 1A, 1A 2W ， 3A 3A V P = = − 5V, 15W Ch1&Ch2 Exercises The absorbing power of A element is 2W, find the absorbing power of the voltage source and current source . Solution:
When the current source increases 3A, the I. increases 1AUse superposition method to find R and VV12V.1x32R5A3R=6ohmx3=1NSolution:R+336VV = 5x(3//6)-12x3+6
5A 3 12V R Vx x I 3 3 1 3 x I R = = + x 3 = 5 (3// 6) -12 = 6V 3+ 6 V When the current source increases 3A，the I x increases 1A. Use superposition method to find R and Vx。 Solution: R=6ohm
Apply equivalent transformation to simplify the circuit12a12622A02V1Ab oSolution:2V10aoao10hCboaoac20or20bobo
2V 6 a b 1 1 2A 1A 1A 2V a b 1 1 1A2V 4V a b 2 1A 2 a b 1A a b 2 2V Apply equivalent transformation to simplify the circuit Solution: or