①0)0图 (0)-i(0 duc(o) ic(0t) 22i()(0)+ dt 10v42①2(0)=5V ic(0)=i(0)+i2(0)-6(0)=5-25=25A di1(0+)_2(0) =0A/s dt14 + uC (0+ ) - t=0+图 + 10V - + uL (0+ ) - iC (0+ ) iL (0+ ) 2 4 i1 (0+ ) i2 (0+ ) C (0 ) = L (0 ) + 2 (0 ) − 1 (0 ) = 5− 2.5 = 2.5A + + + + i i i i 0A/s (0 ) d d (0 ) L = = + + L u t i L 5V/s (0 ) d d (0 ) C = = + + C i t u C