Question

A circuit containing capacitors $C_1$ and $C_2$ , shown in the figure is in the steady state with key $K_1$ closed and $K_2$ opened. At the instant $t=0,K_1$ is opened and $K_2$ is closed. If $C_t$, $\mu C$ is the charge on the plates of the capacitor at that instant and when energy in the inductor becomes one third of that in the capacitor, then find $x$ such that $x=\sqrt{3}\times C_t$:

Answer 1

Since at $t=0$, charge is maximum ($=q_0$).

$C_{eqv}=\frac{2\times 2}{2+2}=1$

$q_0=\frac{V}{C_{eqv}}=\frac{20}{1}=20$
Therefore current will be zero.
$\frac{1}{2}L{i}^2=\frac{1}{3}\left(\frac{1}{2}\frac{q^2}{C}\right)$
or $i=\frac{q}{\sqrt{3LC}}=\frac{q\omega }{\sqrt{3}}$
From the expression $i=\omega \sqrt{q_0^2-q^2}$
We have, $\frac{q\omega }{\sqrt{3}}=\omega \sqrt{q_0^2-q^2}$
or $q=\frac{\sqrt{3}}{2}{q}_0$
$q=\frac{\sqrt{3}}{2}{q}_0=\frac{\sqrt{3}}{2}\times 20=10\sqrt{3}\mu C$