Question

A capacitor of capacity $2\mu F$ is charged to a potential difference of $12V$. It is then connected across an inductor of inductance $0.6\text{mH}$. The current in the circuit at a time when the potential difference across the capacitor is $6.0V$, is $(n+0.6)A$. Then the value of $n$ is

Answer 1

Let us assume,
$q_0=$ Charge on the capacitor while charging
$q=$ Charge on the capacitor at potential difference of $6V$

We know, the angular frequency is given by
$\omega =\frac{1}{\sqrt{LC}}=\frac{{10}^5}{\sqrt{12}}$

The charge on the capacitor will vary according to following equation:
$q=q_0\cos \omega t$
$2\times {10}^{-6}\times 6=2\times {10}^{-6}\times 12\times \cos \omega t$
$\cos \omega t=\frac{1}{2}\Rightarrow \omega t=\frac{\pi }{3}$

$i=\frac{dq}{dt}=-q_0\omega \sin \omega t$
$i=-2\times {10}^{-6}\times 12\times \frac{{10}^5}{2\sqrt{3}}\times \frac{\sqrt{3}}{2}$
$i=0.6A$
$\Rightarrow n=0$