北京交通大学：《电路》课程教学资源（讲稿）中期测验详解——中英文练习题 Ch1Ch2

中期测验讲解1．已知图中元件3放出功率3W，求（1）电压V3；（2）元件2、5的吸收功率P、P4A53+1V2V422 +12AP, =(1+2+2)×2=10W口1AP, = 4 ×(2+2) = 16WCommon mistake关联/非关联

5 4 • • • • 3 2 2V 1 6 4A 1A 2A 3 3 = 1V 2 1 V = + 2 P = + +  = (1 2 2) 2 10W 5 P =  = 4 (2+2) 16W 中期测验讲解 Common mistake 关联/非关联

2．如图所示电路，已知Vab=9V，试求：（1）v的值；（2）若Vs增加1V，Vab的变化量10-VabVab =2V(1)444Q4040av=-1V402404Q6（2）此时电路等效如图IV40404040Common mistake42减少2Vh△v = 0.5V, 2△v = 1A, △vab =-2×1=-2V

4Ω 4Ω a b 10V vs 2v v 4Ω 4Ω 4Ω 4Ω 10 2 4 4 ab ab v v v − − = v = −1V （1） , （2）此时电路等效如图 4Ω 4Ω a b 1V 2v v 4Ω 4Ω 4Ω 4Ω 0.5V, 2 1A, 2 1 2V ab  =  =  = −  = − v v v Common mistake 减少2V

3.利用等效变换方法化简图示二端电路2Aao36039Common mistake302无效伴随323232boCommon mistake方向2Aao?69220302Obo

3. 利用等效变换方法化简图示二端电路。 3Ω 3Ω 3Ω 3Ω 3Ω 3Ω 3V3Ω 6Ω 2A a b 3V3Ω 6Ω 2A a b 1A 3Ω 2A 6Ω 1A 2Ω Common mistake 无效伴随 Common mistake 方向

4如图所示电路，求（1）ab端口左侧的戴维南等效电路：（2）RL可获得的最大功率6i4Q6Qa020VRL65bCommon求开路电压Voc=20×0.5=10Vmistake求短路电流，此时I=i；6 × 101 - 6i + 4i = 206Isc=i=2.5AR。=4Q2外接电阻等于4Q2时获得最大功率ve= 6.25WP.三ARL

20V 6i 6Ω 4Ω a b 6Ω RL i 求短路电流，此时Isc =i； 求开路电压 I sc=i=2.5A R0 =4Ω 外接电阻等于4Ω时获得最大功率 Common mistake

5．如图所示电路，利用结点法求V。，V，、I。Common mistake假设为电流源V.=30.5211Vb22IVV=1-10.50.5C(=+4)V22I-V.-V.2V.-V,=IV. =2VV,=1VI = 1A

1A 3V 0.5 2  4 1 Va Vb Vc I I 3 1 1 1 0.5 0.5 1 1 ( 4) 2 2 2 c a c b c c b a b V V V I V V I V V I V V I  =   − = −     + − =    − =    − =       = = = 1A 1V 2V I V V b a Common mistake 假设为电流源

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