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 $LC$ oscillator circuit
where $q=q_0\cos \omega t($ Since at $t=0,q=q_0=CV)$
$\omega =\frac{1}{\sqrt{LC}}$
$I=-\frac{dq}{dt}=q_0\omega \sin \omega t$
$I_{\max }=q_0\omega =\frac{CV}{\sqrt{LC}}=V\sqrt{\frac{\bar{C}}{L}}$