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

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Aicharya · Accepted Answer

$\begin{aligned} & \frac{\mathrm{CV}^2}{2}+\frac{1}{2} \mathrm{Li}^2=\text { constant } \ & \text { Initially } \mathrm{i}=0, \mathrm{~V}=\mathrm{V}_1 \ & \text {Total energy }=\frac{1}{2} \mathrm{CV}_1^2 \ & \mathrm{~When } \mathrm{V} \rightarrow \mathrm{V}_2 \ & \frac{1}{2} \mathrm{C}\left(\mathrm{V}_2\right)^2+\frac{1}{2} \mathrm{Li}^2=\frac{1}{2} \mathrm{CV}_1^2 \ & \mathrm{Li}^2=\mathrm{C}\left(\mathrm{V}_1^2-\mathrm{V}_2^2\right)^2 \ & \mathrm{i}^2=\frac{\mathrm{C}}{\mathrm{L}}\left(\mathrm{V}_1^2-\mathrm{V}_2^2\right) \ & \mathrm{i}=\sqrt{\frac{\mathrm{C}}{\mathrm{L}}\left(\mathrm{V}_1^2-\mathrm{V}_2^2\right)}\end{aligned}$ ^

A condenser of capacity $C$ is charged to a potential difference of $V_1$. The plates of the condenser are…

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