The circuit shown in the figure is in the steady state with switch $S_1$ closed. At $t=0,S_1$ is opened and switch $S_2$ is closed. Determine the first instant $t$, when the energy in the inductor becomes one-third of that in the capacitor. 

In a $LC$ circuit total energy remains always conserved.

$\therefore \text{Electrical energy stored in the capacitor }+\text{magnetic energy stored in the inductor }=\frac{1}{2}\frac{Q_0^2}{C_2}$

$U_E+U_B=\frac{Q_0^2}{2C_2}$

At the time $t=t_1,U_B=\frac{1}{3}{U}_E$

Hence, $U_E=\frac{3}{4}\left(\frac{1}{2}\frac{Q_0^2}{C_2}\right)$

And \$\frac{1}{2}\frac{Q^2}{C_2}=\frac{3}{4}\left(\frac{1}{2}\frac{Q_0^2}{C_2}\right)$

$Q=\frac{\sqrt{3}}{2}{Q}_0 \mathrm{or} Q_0\cos \omega t_1=\frac{\sqrt{3}}{2}{Q}_0$

$\omega t_1=\frac{\pi }{6}or{t}_1=\frac{\pi }{6\omega }=1.05\times {10}^{-5}=10.5\mu \text{s}$    Ans.