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, $\omega =\frac{1}{\sqrt{LC}}\dots \left(i\right)$
$\therefore cos\omega t=\frac{q}{q_0}=\frac{C{V}_2}{C{V}_1}=\frac{V_2}{V_1}\left(\because q=CV\right)$
Current through the inductor
$I=\frac{dq}{dt}=\frac{d}{dt}\left(q_{0}cos\omega t\right)=-q_0\omega sin\omega t$
$\left|I\right|=C{V}_1\frac{1}{\sqrt{LC}}{\left[1-co{s}^2\omega t\right]}^{1/2}$
$=V_1\sqrt{\frac{C}{L}}{\left[1-{\left(\frac{V_2}{V_1}\right)}^2\right]}^{1/2}$
$={\left[\frac{C\left(V_1^2-V_2^2\right)}{L}\right]}^{1/2}$ (using (i))