A capacitor of capacity 'C' is charged to a p | vedclass

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(C) The total energy in an $LC$ circuit is conserved.Initially,the energy stored in the capacitor is $U_i = \frac{1}{2}CV_1^2$.When the potential diff

Vedclass · Accepted Answer

$\sqrt{\frac{C}{L}}(V_1^2 - V_2^2)^{1/2}$

Vedclass · Answer

(C) The total energy in an $LC$ circuit is conserved.Initially,the energy stored in the capacitor is $U_i = \frac{1}{2}CV_1^2$.When the potential difference across the capacitor becomes $V_2$,the energy stored in the capacitor is $U_c = \frac{1}{2}CV_2^2$.The energy stored in the inductor at this instant is $U_L = \frac{1}{2}LI^2$.By the law of conservation of energy: $U_i = U_c + U_L$.$\frac{1}{2}CV_1^2 = \frac{1}{2}CV_2^2 + \frac{1}{2}LI^2$.$LI^2 = C(V_1^2 - V_2^2)$.$I^2 = \frac{C}{L}(V_1^2 - V_2^2)$.$I = \sqrt{\frac{C}{L}(V_1^2 - V_2^2)}$.Thus,the correct option is $C$.

$A$ capacitor of capacity '$C$' is charged to a potential difference of '$V_1$'. The plates of the capacitor are then connected to an ideal inductor of inductance '$L$'. The current through the inductor when the potential difference across the capacitor reduces to '$V_2$' is:

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