电工与电子技术（毕淑娥）第二章课后习题参考答案

6ΩI2+15V_4Ω+5V_2ΩIS10A6ΩaI2'4Ω2ΩIS10A+15V_b

解：当15V电压源和10A电流源共同作用时，

15?106Uab??30V 11?6430I2'??7.5A

4当5V电压源作用时

6ΩaI2''4Ω+5V_b

I2''??5??0.5A 6?4I?I2'?I2''?7.5?0.5?7A

2-10在题图2-10中，已知Us1?12V, US2?2V, IS?10A,R1?R2?2?,R3?R4?3?。用叠加定理求电压Uab。

IsR1+US1_R3Uab_R4+R2R1+U_S2+US1_R3Uab+ISR2R1R2+Uab''aaR4+U_S2R3aR4__b

b

b

解：两个电压源作用时，

Uab'??US1?US2?(R1?R3)?US1

R1?R2?R3?R4Uab'??12?2(2?3)?12??5?12?7V

2?2?3?3Is单独作用时，

Uab''?R3?R4R1?R2?IS?R1??IS?R4?0V

R1?R2?R3?R4R1?R2?R3?R4共同作用时

Uab?Uab'?Uab''?7V

2-11 在题图2-11 中，已知IS?1A,US?10V,R1?6?,R2?3?,R3?4?。用戴维南定理求I2.

ISI2+US_+ISR1_UOCR1_R3++US_+R1R2R3R1ReqR3Uoc_I2R2Req

解: UOC?US?ISR110?1?6?R1?ISR1??6?1?6?8.4V

R1?R36?4Req?R1//R3?6//4?2.4?

I2?UOC8.4??1.56A

Req?R22.4?32-12已知电路如题图2-12所示，用戴维南定理求I和电流源的功率Pi。 解：

3Ω2Ω3Ω_9V++U_3A+UOC_2Ω_9V++UOC_I3Ω3AI3Ω4Ω4ΩReq

UOC?2?3?4?3?9V Req?2?4?6?

I?UOC9??1A

Req?36?3U?3?3?1?3?12V

P??UI??3?12??36W

R1?4?,R2?1?,R3?3?。2-13在题图2-13电路中，已知US?3V, IS1?3A,IS2?1A，

用戴维南定理求I3

R2IS1R1IS2I3+US_R3+IS1R1_R1R2++UOC_I3+US_R3UOCIS2_Req

解：UOC?IS1R1?IS2R2?IS2R1?3?4?1?(4?1)?7V

Req?R1?R2?1?4?5?

I3?UOC?US7?3??0.5A

Req?R35?32-14在题图2-14中，已知IS?3A, US1?US2?6V，R1?R2?R3?R4?3?,R5?7?,用戴维南定理求I5

R1R3I5R5+U_S2ISUS1R2R3ISR2+US1_R4R2R4+U_S2

R2+R3+

+_R3UOCI5R5US1?ISR2R2R4+U_S2(US1?IS)?R2R2+_R4+U_S2UOC_Req

解： UOCUS1?IS)?R2?US2R2(2?3)?3?6??R4?US2??3?6?9V

R2?R3?R43?3?3(Req?(R2?R3)//R4?6//3?2?

I5?UOC9??1A

Req?R52?72-15已知电路如题图2-15所示，(1)用戴维南定理求I（2）电压源的功率Pu

I1I2+10V3ΩI2ΩI2+10VI13Ω+UOC_1A+3ΩReq_-1A6Ω2Ω-6Ω2Ω6Ω

解：UOC?10?1?2?8V

Req?2?

I?UOC8??2A

Req?22?2再利用戴维南定理求解I1

UOC'?10?1?6?16V Req?6?

I1?UOC1616??A

Req?36?39I2?I1?I?1634?2?A 9934P??UI2??10???37.78W

92-16在题2-16中，已知IS?2A, US?12V，R1?6?,R2?3?,R3?1?。当IS如图示方向时，电流I=0，当IS反方向时，I=-1A，求含源一端口网络的戴维南等效电路。 解：

R1ISR3+UOC_+US_IReqR2

设含源一端口网络的开路电压为上正下负UOC，等效电阻为Req

当IS如图示方向时，电流I=0，此时UOC?IS?R2?US?2?3?12?18V

当IS反方向时，列右侧网孔的KVL方程可得：(I?IS)?R2?I?(R3?Req)?UOC?US 所以Req?8?