Question

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

• A. $\frac{\pi \sqrt{LC}}{4}$
• B. $\frac{\pi \sqrt{LC}}{2}$
• C. $\frac{5\pi \sqrt{LC}}{4}$
• D. $\frac{5\pi \sqrt{LC}}{2}$

Answer 1

Answer: (A), (C)

The correct options are
A $\frac{\pi \sqrt{LC}}{4}$
C $\frac{5\pi \sqrt{LC}}{4}$

For given situation,$\frac{q}{C}+L\frac{di}{dt}=0$
$\Rightarrow \frac{d^{2}q}{d{t}^2}+\frac{q}{LC}=0$

$\Rightarrow \frac{d^{2}q}{d{t}^2}+{\omega }^{2}q=0$
$\because \omega =\frac{1}{\sqrt{LC}}$

We know that,
$\Rightarrow q=q_{0}cos\left(\omega t\right)$
Differenciating w.r.t time
$i=-q_0\omega sin\left(\omega t\right)$
According to given condition, the energy in capacitor and inductor are same,
$\therefore \frac{q^2}{2C}=\frac{1}{2}L{i}^2$
$\Rightarrow \frac{q_0^{2}co{s}^2\omega t}{2C}=\frac{1}{2}L{q}_0^2{\omega }^{2}si{n}^2\omega t$

$\Rightarrow co{t}^2\omega t=1$

$\Rightarrow \omega t=\frac{\pi }{4},\frac{3\pi }{4},\frac{5\pi }{4},\frac{7\pi }{4}.........$
$\Rightarrow t=\frac{\pi \sqrt{LC}}{4},\frac{3\pi \sqrt{LC}}{4},\frac{5\pi \sqrt{LC}}{4},\frac{7\pi \sqrt{LC}}{4}....$