11 例2:1(0)=2A,求l(1)t≥0 R=3 分析: l1 L=4H+ 1先求等效电阻Re 0.5U R=1 I+0.5u 由KV得:U=3*+[0.5u+*1 0.5U=4I→R=U/8 2求 l 4 0.5 r 8 3.求)()=(0,27=(0)7=2 4.求u(t)(t)=L -16e11 例2： i(0− ) = 2A , 求 u(t) t  0 R= i(t) 1 L=4H R=3 0.5U U(t) I 分析： 1 1.先求等效电阻Req: I1=I+0.5u 由KVL得：U=3*I+[0.5u+I] *1 →0.5U=4I →Req=U/I=8 2.求  0.5 8 4 = = = R L  3.求i(t): t t t i t i e i e e 2 ( ) (0 ) (0 ) 2 − − − − = + = =   4.求u(t) t e dt di t u t L 2 16 ( ) ( ) − = = −