例2 0.1 1002i 0.4H 100① 十 50 100μF宁u 104/S= i(0-)=0.25Auc0-)=25v I(S)= 9+0.1-3 A 1(S)=S+125-j96.8+ A +125+j96.8 100+0.4s+104/S 0.25S+62.5 0.25S+62.5 11(S)-s2+250s+25000 A= S+125+j96.8 S=-125+j96.8 0.25S+62.5 =0.204∠-52.2° (S+125)2+9375ֻ2 100PF 100: + – 50v iL + – uc k 0.4H 100: iL (0-)=0.25A uC(0-)=25v 100: + – + 0.4S S 50 – S 25 IL (S) – + 0.1 104 /S 50 25 S IL (S)= +0.1– S 100+0.4s+104 /S IL (S)= 0.25S+62.5 S 2+250S+25000 0.25S+62.5 (S+125)2+9375 = A= |S= –125+j96.8 0.25S+62.5 S+125+j96.8 =0.204‘ – 52.2º IL (S) = + A S+125–j96.8 A S+125+j96.8 *