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(D) According to the principle of conservation of energy in an $LC$ circuit,the total energy remains constant.Initial energy stored in the capacitor i

Vedclass · Accepted Answer

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

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(D) According to the principle of conservation of energy in an $LC$ circuit,the total energy remains constant.Initial energy stored in the capacitor is $U_i = \frac{1}{2} C V_1^2$.When the potential difference across the capacitor is $V_2$,the energy stored in the capacitor is $U_c = \frac{1}{2} C V_2^2$.The remaining energy is stored in the inductor as magnetic energy,$U_L = \frac{1}{2} L I^2$.By conservation of energy: $\frac{1}{2} C V_1^2 = \frac{1}{2} C V_2^2 + \frac{1}{2} L I^2$.Rearranging for $I^2$: $L I^2 = C(V_1^2 - V_2^2)$.Thus,$I^2 = \frac{C(V_1^2 - V_2^2)}{L}$.Therefore,the current is $I = [\frac{C(V_1^2 - V_2^2)}{L}]^{1/2}$.

$A$ 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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