Question

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

• A. $\left(\frac{C\left(V_{1}-V_{2}\right)^{2}}{L}\right)^{\frac{1}{2}}$
• B. $\frac{C\left(V_{1}^{2}-V_{2}^{2}\right)}{L}$
• C. $\frac{C\left(V_{1}^{2}+V_{2}^{2}\right)}{L}$
• D. $\left(\frac{C\left(V_{1}^{2}-V_{2}^{2}\right)}{L}\right)^{\frac{1}{2}}$

Answer 1

Answer: (D)

In case of oscillatory discharge of a capacitor through an inductor, charge at instant $t$ is given by
$q=q_{0}\,cos\omega t$
where, $\omega=\frac{1}{\sqrt{LC}} \ldots\left(i\right)$
$\therefore cos\,\omega t=\frac{q}{q_{0}}=\frac{CV_{2}}{CV_{1}}=\frac{V_{2}}{V_{1}} \left(\because q=CV\right)$
Current through the inductor
$I=\frac{dq}{dt}=\frac{d}{dt}\left(q_{0}\,cos\,\omega t\right)=-q_{0} \omega sin \omega t$
$\left|I\right|=CV_{1} \frac{1}{\sqrt{LC}}\left[1-cos^{2}\,\omega t\right]^{1/ 2}$
$=V_{1}\sqrt{\frac{C}{L}}\left[1-\left(\frac{V_{2}}{V_{1}}\right)^{2}\right]^{1 /2}$
$=\left[\frac{C\left(V_{1}^{2}-V_{2}^{2}\right)}{L}\right]^{1 /2}$ (using (i))