(2)白换路定则,i1(0)=i(0)=0.2A 0)=(0)=4V,作t=0等效图 +u1(0+)-i2(0+) R R1+u1(0 i(0+)20) (0)Ri(0 t=0图 (3)求初始值(0)=(0)=02A9 R1 R 2 R3 + uC (0+ ) - t=0+图 + uS - iL (0+ ) +uL (0+ )- +u1 (0+ )- i2 (0+ i ) C (0+ ) i3 (0+ ) (2)由换路定则， ,作t =0+等效图 i L ( 0 ) = i L ( 0 ) = 0.2A + − C (0 ) = (0 ) = 4 V + − u uC (3)求初始值 1 (0 ) = (0 ) = 0.2A + + L i i