Question

The capacitor of capacitance $C$ can be charged (with the help of a resistance R) by a voltage source $V$, by closing switch $S_{1}$ while keeping switch $S_{2}$ open. The capacitor can be connected in series with an inductor $L$ by closing switch $S_{2}$ and opening $S_{1}$.
After the capacitor gets fully charged, $S_{1}$ is opened and $S_{2}$ is closed so that the inductor is connected in series with the capacitor. Then,

• A. at $t=0$, the energy stored in the circuit is purely in the form of magnetic energy.
• B. at any time $t > 0$, the current in the circuit is in the same direction.
• C. at $t > 0$, there is no exchange of energy between the inductor and the capacitor.
• D. at any time $ t > 0$, the instantaneous current in the circuit may $\vee \sqrt{\frac{C}{L}}$

Answer 1

Answer: (D)

On closing $S_{2}$ and opening $S_{1}$,
it becomes an $L C$ oscillator circuit
where $q=q_{0} \cos \omega t\left(\right.$ Since at $\left.t=0, q=q_{0}=C V\right)$
$\omega=\frac{1}{\sqrt{L C}}$
$I=-\frac{d q}{d t}=q_{0} \omega \sin \omega t $
$I_{\max }=q_{0} \omega=\frac{C V}{\sqrt{L C}}=V \sqrt{\frac{\bar{C}}{L}}$