Question

A capacitor of capacitance, $C=25\mu \text{F}$ is fully charged to $300\text{V}$ . It is then connected across a $10\text{mH}$ inductor at $t=0$. The resistance in the circuit is negligible. Find the energy stored in the capacitor at $t=\frac{\pi }{3\omega }\text{s}$, where $\omega$ is the angular frequency.

• A. $0.9\text{J}$
• B. $0.7\text{J}$
• C. $0.5\text{J}$
• D. $0.3\text{J}$

Answer 1

The correct option is D $0.3\text{J}$

The charge $\left(q\right)$ varies with time $\left(t\right)$ as,

$q=q_0\cos \omega t=C{V}_0\cos \omega t$

Now, the charge at $t=\frac{\pi }{3\omega }\text{s}$ is,

$q=C{V}_0\cos (\omega \times \frac{\pi }{3\omega })=C{V}_0\cos \frac{\pi }{3}=\frac{C{V}_0}{2}$

$\Rightarrow q=\frac{25\times {10}^{-6}\times 300}{2}=3.75\times {10}^{-3}\text{C}$


So, energy stored in the capacitor, at $t=\frac{\pi }{3\omega }\text{s}$ will be,

$U_C=\frac{q^2}{2C}=\frac{(3.75\times {10}^{-3}{)}^2}{2\times 25\times {10}^{-6}}\approx 0.3\text{J}$

Hence, option $\left(D\right)$ is the correct answer.