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. $\bigg(\frac{C(V_1-V_2)^2}{L}\bigg)^\frac{1}{2}$
• B. $\frac{C(V_1^2-V_2^2)}{L}$
• C. $\frac{C(V_1^2+V_2^2)}{L}$
• D. $\bigg(\frac{C(V_1^2-V_2^2)}{L}\bigg)^\frac{1}{2}$

Answer 1

Answer: (D)

$\bigg(\frac{C(V_1^2-V_2^2)}{L}\bigg)^\frac{1}{2}$

Answer 2

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, $\omega=\frac{1}{\sqrt{LC}} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ...(i)$
$\therefore \ \ cos\ \omega t=\frac{CV_2}{CV_1}=\frac{V_2}{V_1} (\because q=CV)$
Current through the inductor
$\ \ \ \ \ I=\frac{dq}{dt}=\frac{d}{dt}(q_0 cos \omega t)=-q_0 \omega sin\omega t$
$\ \ \ \ \ |I|=CV_1\frac{1}{\sqrt{LC}}[1-cos^2 \omega t]^{1/2}$
$\ \ \ \\\ \ \ =V_1\sqrt\frac{C}{L}\bigg[1-\big(\frac{V_2}{V_1}\big)\bigg]^{1/2}=\bigg[\frac{C(V_1^2-V_2^2)}{L}\bigg]^\frac{1}{2}$
(using (i))