Question

In the electrical network at $t<0$, key was placed on $\left(1\right),$ until the capacitor becomes fully charged at $t=0$.
Position of key is changed to $\left(2\right)$ at $t=0$. The energy in the capacitor and the inductor will be same for the first time is

• A. $\frac{\pi }{4}\sqrt{LC}$
• B. $\frac{3\pi }{4}\sqrt{LC}$
• C. $\frac{\pi }{3}\sqrt{LC}$
• D. $\frac{2\pi }{3}\sqrt{LC}$

Answer 1

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

When key is positioned at 2, applying KCL in loop 2, we get
$\frac{q}{C}+L\frac{di}{dt}=0$

$L\frac{d^{2}q}{d{t}^2}+\frac{q}{C}=0$
$\frac{d^{2}q}{d{t}^2}=\frac{-q}{LC}\Rightarrow \frac{d^{2}q}{d{t}^2}+{\omega }^{2}q=0$
Hence,
$q=q_0\cos \omega t,\omega =\sqrt{\frac{1}{LC}}$
and $i=\frac{dq}{dt}=-q_0\omega \sin \omega t$

Let $U_C$ be energy stored in the capacitor
$U_C=\frac{q^2}{2C}=\frac{q_0^2{\cos }^2\omega t}{2C}$
Let $U_L$ be energy stored in the indecator
$U_L=\frac{1}{2}L{i}^2=\frac{1}{2}L{q}_0^2{\omega }^2{\sin }^2\omega t$
when $U_C=U_L$
$\frac{q_0^2{\cos }^2\omega t}{2C}=\frac{q_0^2{\omega }^{2}L{\sin }^2\omega t}{2}$
${\cot }^2\omega t={\omega }^{2}LC=1$
For minimum value of $t$,
$\omega t=\frac{\pi }{4}$
Hence, $t=\frac{\pi }{4}\sqrt{LC}$