方法一:解:将ab两端断开,利用加压求流法: 20g2 ⅠS=5∠0A 4U12 b U=2l2=2(-U1)+4U1=2(s+3U/1)…(1) 又:Uoe=UL+U/1=-J 20(s-U/1)+U/1 20s+(1+j20)U1 (2)解:将ab两端断开,利用加压求流法: 2 2 ( ) 4 2( 3 1) ... (1) 2 1 1 U oc I I S U U I S U 又: 20 (1 20) ... (2) 20( ) 1 1 1 1 j I j U U U U j I U U S oc L S 2 1 4U 2 I + - 5 0A IS - + j20 UL 1 U 1 a b 方法一: