0-图 5A 由0-图 (2)由换路定则 n1(0_)=1A (0)=in1(0)=1A 22(01)=2A i2(0+)=i2(0)=2A l(0)=1V lc(04)=lc(0)=1V(0 ) 1V (0 ) 2A (0 ) 1A 2 1 = = = − − − C L L u i i 0-图 1 + - + - 1 1 1 L1 uL1 uL2 L2 C i 5A - uC + (2) 由换路定则 (0 ) (0 ) 1V (0 ) (0 ) 2A (0 ) (0 ) 1A 2 2 1 1 = = = = = = + − + − + − C C L L L L u u i i i i 由0-图