A condenser of capacity C is charged to a potential difference of...

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

Options

A\left(\frac{\mathrm{C} $\left(\mathrm{V}_1-\mathrm{V}_2\right)^2}{\mathrm{~L}}\right)^{\frac{1}{2}}$
B\frac{\mathrm{C} $\left(\mathrm{V}_1^2-\mathrm{V}_2^2\right)}{\mathrm{L}}$
C\frac{\mathrm{C} $\left(\mathrm{V}_1^2+\mathrm{V}_2^2\right)}{\mathrm{L}}$
D\left(\frac{\mathrm{C} $\left(\mathrm{V}_1^2-\mathrm{V}_2^2\right)}{\mathrm{L}}\right)^{\frac{1}{2}}$

D. \left(\frac{\mathrm{C} $\left(\mathrm{V}_1^2-\mathrm{V}_2^2\right)}{\mathrm{L}}\right)^{\frac{1}{2}}$

\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}

)