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

Vedclass · Accepted Answer

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

Vedclass · Answer

(D) According to the law of conservation of energy in an $LC$ circuit,the total energy remains constant.Initially,the energy stored in the capacitor is $U_i = \frac{1}{2} C V_1^2$.When the potential across the capacitor becomes $V_2$,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 energy conservation: $\frac{1}{2} C V_1^2 = \frac{1}{2} C V_2^2 + \frac{1}{2} L I^2$.Rearranging the terms: $\frac{1}{2} L I^2 = \frac{1}{2} C (V_1^2 - V_2^2)$.Solving for current $I$: $I^2 = \frac{C(V_1^2 - V_2^2)}{L}$.Therefore,$I = \sqrt{\frac{C(V_1^2 - V_2^2)}{L}}$.

$A$ capacitor of capacitance $C$ is charged to a potential $V_1$. It is then connected to an ideal inductor of inductance $L$. When the potential across the capacitor drops to $V_2$,what is the current flowing through the inductor?

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