K t=0 原图 ①10v 13 C2 L Qa(0+)b3 l2(0+) Cl C2 0+图
+ - uC1(0+) 3 uC2 (0+) +uL - iC1 iC2 1 + - 0 +图 + 原图 - K t=0 10V 2 3 C1 C2 L +uL - iC1 iC2 1
C2(O+)=i(O+)=0 C1(0)=-llc1(0)/3=-2A l1.(O+)=ll2(O+)-uc1(O+)=0V L ua(0+)[3 ①ua2(0) CI C2 0+图
( 0 ) ( 0 ) ( 0 ) 0 V ( 0 ) ( 0 ) / 3 2 A ( 0 ) ( 0 ) 0 L C 2 C 1 C1 C1 C2 L = − = = − = − = = + + + + + + + u u u i u i i +-u C 1(0 + ) 3 uC2 (0 + ) + u L - iC1 iC2 1+- 0 + 图
(b) t=0 L2 5A IL1及L,L12 ①sV 解:换路前处于稳态,作0-图
+ - K t=0 5V 1 uL1 iC 5A 1 C + - uL2 + - 1 1 1 L1 L2 (b) 解： 换路前处于稳态，作0-图
0-图 5A LI L2 由0-图,得 1(0)=1A in2(0)=2A llC(0)=-1V
( 0 ) 1 V ( 0 ) 2 A ( 0 ) 1 A 21 = − == −−− CLLuii uL1 iC 5 A 1 +- u L 2+- 1 1 1 0 - 图 由 0 -图，得
由换路定则,得 i1(O)=i1(O)=1A i12(O)=i12(0)=2A uc(o=ucO)=-IV 作0图
(0 ) (0 ) 1V (0 ) (0 ) 2A (0 ) (0 ) 1A L2 L2 L1 L1 = = − = = = = + − + − + − uC uC i i i i 作0 +图 由换路定则，得