Question

Find v(t) for t> 0 in the given circuit. Assume / = 4u(t) A. t=0 1H mo of + 4.5 A ( 102 4F 522 DI - The value of v(t) = [A + (B)ect + (D)e-4.949tju(t) V where A = B = CE and D =

Answer 1

For t=0 (t.o) Capacitor acts as open circuit, inductor acts as I=0 ( Ult)=0 4+20) Switch is closed. Short Circuit and Current - 4.5A 5.2Ś Rulo > 10) = 45(10+5) division & 100 v6) illo)= BA So vlo) = 5x115) = 15v → Fort-ot, Capacitor is Shošted with Vó), inductor is opefted with 16) across it. Switch is opened ) I=4A 3A 5.2 w ( DY V10) fida V.18) 16V 50 QOV 1) ЧА 154 It is a series Circuit So Same current flows in each element ilo) = -3A-C dvo) at dveo) -3 at LP /sec. sy for to the circuit can be drawn in frequency domain. 345 Ie(s) 07 + Apply source Transform Technique Vis) 512 1 US 3v I.1 KUL com (s)/5+ 5 + + 4) 15 S 120-3+5+1 69(6+ 5 t's

us+205 + 1 2 SO I (5) - -( 5+38) us S Ic(s) = -4 15+35) (us+ 205+1) v(s) = 5 + ts Iels) 1.4 (5+35) us 45² 205+1 15 ( 5+35) $ s(ust 205+1) les en 205 +97 us²+205+1 1 10 S 205 +97 us +205+1 55+974 10 S 5²7 55+Y4 10 58 +24.25 S (5+4.949) (5+0.0505) 4.899 0.101 10 3 (5+0.0505) (5+4.949) Apply Inverse Laplace Transforms. V+) -110 - 4.899 e -0.0505t -4.9494 -0.101e Juct) In orden .4.quat vt) = 100.101e ct -4.9494 Given [A + Be toe Just) A = 10 V B = -4.89qv с = -0. 0505 0 - -0.101