Question

A capacitor of capacity $2\mu F$ is charged to a potential difference of $12\text{V}$. It is then connected across an inductor of inductance $0.6\text{mH}$. The current in the circuit at a time when the P.D. across the capacitor is $6\text{V}$ is (in ampere).

Answer 1

$\omega =\frac{1}{\sqrt{LC}}=\frac{{10}^5}{\sqrt{12}}$
$q=q_0\cos \omega t$
$2\times {10}^{-6}\times 6=2\times {10}^{-6}\times 12\cos \omega t$
$\cos \omega t=\frac{1}{2}\Rightarrow \omega t=\frac{\pi }{3}$
$i=\frac{dq}{dt}=-q_0\omega \sin \omega t$
$i=0.6\text{A}$