Question

Initially key was placed on (1) till the capacitor got fully charged . Key is placed on (2) at t=0. The minimum time when the energy in both capacitor and inductor will be same:

• A. $t=\frac{\pi \sqrt{LC}}{4}$
• B. $t=\frac{\pi \sqrt{LC}}{2}$
• C. $t=\frac{\sqrt{LC}}{\pi }$
• D. $t=\frac{\sqrt{LC}}{2\pi }$

Answer 1

The correct option is A $t=\frac{\pi \sqrt{LC}}{4}$

Means energy of a capacitor will be half of its maximum energy when energy on both is same.

So,

$\frac{q^2}{2C}=\frac{Q^2}{2\times 2C}$

$\Rightarrow q=\frac{Q}{\sqrt{2}}$

$\Rightarrow Qcos\omega t=\frac{Q}{\sqrt{2}}$

$\Rightarrow \omega t=\frac{\pi }{4}$

$\Rightarrow t=\frac{\pi \sqrt{LC}}{4}$

So, the correct option will be $\left(A\right)$