②由换路定理,画出≠=0的电路图, (04) +2(t) L r3 ic(t) R r1(,)/cli/ R U R Qu(O i2(04)=5A, (0)=10 ic(0+)=2.5A, 1(0+)=2A,2(04)=0V, 进一步可求各阶导数的初始值 2.5=5AF 0.5②由换路定理,画出t=0+的电路图, R1 R3 (0 ) uC (0 ) L i US uL (0) (0 ) C i (0 ) R1 i (0 ) 5A, L i uC (0 ) 10V (0 ) 2.5A, iR1 (0 ) 0V, uL (0 ) 2.5A, C i 进一步可求各阶导数的初始值 L C uL (t) US R1 R2 R3 ( ) 1 i t R i (t) C 2.5 5A/F 0.5 1 (0 ) 1 0 C c i dt C du