(b) 2mH1|+ 302-0)(s0A 解:换路以后电路为 2m1 3(202(O5A
(b) 解：换路以后电路为 5 (t)A + _ 2 2mH L 1 3 ( ) 1 u t ( ) 1 u t i (t) L + _ 2 2mH L 1 3 ( ) 5A 1 u t ( ) 1 u t i (t) L
2mH|+ 32140)95 求等效电阻 3 21 加压求流法 i=3l1+ R Bu A!2 l 2 7(ms) R2/7
求等效电阻 加压求流法： 7( ) 2 / 7 2 ms R L = = = + _ 2 2mH L 1 3 ( ) 5A 1 u t ( ) 1 u t i (t) L + _ 2 1 3u1 u1 + _ 2 1 + - u i 3u1 u1 1 1 1 2 7 2 3 u u i = u + = = = = 7 1 2 i u i u Req
已知uc(0)=1V,求t>0的i(和mc(t) K t=U 4 解:1.由换路定则Lc(0)=lc(0)=1V 4 稳态时:l(c(0)= 11×1=2V
三、已知uC(0- )=1V, 求 t>0 的i1 (t)和uC(t) 解： uC (0+ ) = uC (0− ) =1V K + - t = 0 4V + - 1 1 1 u (t) C ( ) 1 i t F 5 4 1. 由换路定则 1 2V 1 1 4 ( ) = + 稳态时： uC =
2.作0+图 网孔方程2 4 +2 7 则i1(0) A 3 b A 3
1 2 1 2 4 i i i i i i a a b a b = − + = − 网孔方程 − = 则 A 3 5 A 3 7 (0 ) i 1 + = i a = i b = + - 4V + - 1V 1 1 1 i1 2. 作0+图 ia ib
3作图 则(∞)= =2A 1+1 R=1+1/1=1.592 z=RC=1.5×=1.2(s) 5
则 2A 1 1 4 ( ) 1 = + i = 3. 作 ∞ 图 +-4 V 1 1 1 i 1 1.2(s) 54 = RC = 1.5 = Req = 1 + 1// 1 = 1.5