In the shown electrical network at $t<0$, key was placed on $\left(1\right)$till the capacitor got fully charged. Key is placed on $\left(2\right)$ at $t=0$. Determine the time when the energy in both capacitor and inductor will be same for the first time.

When switch $2$ is closed, applying KCL in loop $2$ we get, 

$\frac{q}{C}+L\frac{di}{dt}=0$

or, $L\frac{d^{2}q}{d{t}^2}+\frac{q}{C}=0$

$\frac{d^{2}q}{d{t}^2}=\frac{-q}{CL}\Rightarrow \frac{d^{2}q}{d{t}^2}+{\omega }^{2}q=0$

Hence $q=q_0\cos \omega t,\omega =\sqrt{\frac{1}{LC}}$

and, $i=-q_0\omega \sin \omega t$

$U_C=\text{Energy stored in capacitor}=\frac{q^2}{2C}=\frac{q_0^2{\cos }^2\omega t}{2C}$

$U_L=\text{Energy stored in inductor}=\frac{1}{2}L{i}^2=\frac{1}{2}L{q}_0^2{\omega }^2{\sin }^2\omega t$

When $U_C=U_L$

$\frac{q_0^2{\cos }^2\omega t}{2C}=\frac{L{q}_0^2{\omega }^2{\sin }^2\omega t}{2}$

${\cos }^2\omega t={\omega }^{2}LC=1$

For minimum value of $t,\omega t=\frac{\pi }{4}$

Hence, $t=\frac{\pi }{4}\sqrt{LC}$ Ans.