Question

Section 5.2 5.12 In Fig. 5.20, the capacitor is initially charged to a voltage V At time I = 0 the switch is closed. The charge stored on the capacitor then flows out through the inductor. At some later time the charge on the capacitor has fallen to zero. Find the current through the inductor at that time. FIGURE 5.20

Answer 1

Given circuit XY The above circcutis called Tank circuit, un SL) According to KVL ... Ldlift + 4 fic co. Appling Laplace transform: SLIG) ++ [2C9 - Com] -o : Here negaline sign before cus is used because Current ist) deaves the positive terminal of c and enters its negative terminal. I(E)* [se+b] = you lose the IQ) = entren]

T = No to start was a 16) - You are here to 10% ke wo Kite VO I() = WOL L stwo by Applying inverne Laplace to the above Expression ict) = : No sinuot = sin wot Vo Toril = Vojų {in wat sin wat V = L dilt) = Ld (o d Şinno = nosak T 1 Lvo , 9, Сод to E (DOL V = Vocos wat Sin Gebot et V. Hot-V. = 1 Vo COL = Ye un coconut): -vo = Voca-copus H) - Vo Vc = -Vococoot.

for Halt cycle et radians. The current will be conducted Wota to the the toLC (sec). - Im vorul Wot toot L : - wat Energy flow se» / cvoro llevoll (Lyo/LMO...) for a tank circuit , the following points are applicable. 1) Nature of Current is sinusoidal. 2) peat value of current is voch 3) conduction line of the circuit is for the see w At the end of Conduction i voltage across Capacitor gets - 0 reversed.