(2)由换路定则 (O+)=i(0)=0 ll1(O4)=ul1(0)=6V lc2(04)=lc2(0)=6V (3)作0图,求初始值 ①l10)3 ic2(0(0 ) (0 ) 6V (0 ) (0 ) 6V (0 ) (0 ) 0 2 2 1 1 = = = = = = + − + − + − C C C C L L u u u u i i (3) 作0+图，求初始值 (2) 由换路定则 + - 3 + - 1 (0 ) uC1 + + - (0 ) uC2 + (0 ) uL + (0 ) C2 + (0 ) i C1 + i