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. Determine the first instant $t$, when energy in the inductor becomes one third of that in the capacitor.

• A. $3.31\times {10}^{-6}s$
• B. $8.05\times {10}^{-5}s$
• C. $7.05\times {10}^{-5}s$
• D. $6.05\times {10}^{-5}s$

Answer 1

The correct option is A $3.31\times {10}^{-6}s$

Since at $t=0$, charge is maximum ($=q_0$).
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 $q=\frac{\sqrt{3}}{2}{q}_0$
Since at $t=0$, charge in maximum or $q_0$ so we can write
$q=q_0\cos \omega t$ or $\frac{\sqrt{3}{q}_0}{2}=q_0\cos \omega t$

$\omega =\frac{1}{\sqrt{L{C}_1}}$
$=\frac{1}{\sqrt{0.2\times {10}^{-3}\times 2\times {10}^{-6}}}$
$=158110$ $rad/s$
or $\omega t=\frac{\pi }{6}$
or $t=\frac{\pi }{6\omega }=\frac{\pi }{6\times 158110}=3.31\times {10}^{-6}s$