电工部分习题解答(1,2,3)

t iC?iC(?)??iC(0?)?iC(?)?e =

??

2?1.666667teA 33-12 图3-27所示电路原已处于稳定状态，试用三要素法求S闭合后的uC 及iC。

S3ΩS6Ωi t = 0i t = 0iL50 CμFu+c+U+iC-9 V1 H-S18 VRISR212Ω-8Ω10.5 A

图3-27 习题3-12图

解：uC(0?)?uC(0?)??ISR2??6V

t?018?6?时的电路, iC(0?)?R?3A 1uC(?)=18V iC(?)?0A

??R0C?R1?50?10?6?4?10-4s

uC=uC(?)+[uC(0?)-uC(?)]e?t? ?18?24e?2500tV

?t iC?iC(?)??iC(0?)?iC(?)?e?

=3e?2500tA

3-13 图3-28所示电路原已处于稳定状态，试用三要素法求S 闭合后的iL，3ΩS6Ωi2i t = 01iL+9 V1 H+IR9 V12Ω--

图3-28 习题3-13图

i1，i2。

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+9 V-S t = 0C50 μFR8Ωi-uc+0.5 A

解：iL?(0?)?iL(0?)?96?1.5A 换路后t?0?时刻的电路中，由结点电压法得

9?1.5?9UAB(0?)?3611?6V 3?6i9?61(0?)?3?1A

i)?9?62(0?6?0.5A

i(?)?913?3A

i(?)?926?1.5A

iL(?)?i1(?)+i2(?)?4.5A

??LR?1?0.5s 06//3 i?i?tL?iL(?)?L(0?)?iL(?)?e?=4.5-3e?2tA i1?i1(?)??i1(0?)?i1(?)?e?t?=3-2e?2tA

i2?i2(?)??i2(0?)?i2(?)?e?t?=1.5-e?2tA

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