Question

A capacitor $C$ is first charged fully with potential difference of $V_0$ and disconnected from the battery. The charged capacitor is connected across an inductor having inductance $L$. In $t$ s 25% of the initial energy in the capacitor is transferred to the inductor. The value of $t$ is _________ s.

• A. $\pi \sqrt{\frac{LC}{2}}$
• B. $\frac{\pi \sqrt{LC}}{6}$
• C. $\frac{\pi \sqrt{LC}}{3}$
• D. $\frac{\pi \sqrt{LC}}{2}$

Hint:

Remember the energy partitioning: $U_C = U_0 \cos^2(\omega t)$ and $U_L = U_0 \sin^2(\omega t)$. For 25% transfer, the phase angle is 30$^\circ$ ($\pi/6$).

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
This is an LC oscillation problem. Total energy oscillates between the capacitor and the inductor. Initially, all energy is in the capacitor. The energy in the inductor at time $t$ is given by $U_L(t) = U_{total} \sin^2(\omega t)$.

Step 2: Key Formula or Approach:
Angular frequency $\omega = \frac{1}{\sqrt{LC}}$.
Total initial energy $U_0 = \frac{1}{2} C V_0^2$.
Energy in inductor $U_L = U_0 \sin^2(\omega t)$.

Step 3: Detailed Explanation:
Given $U_L = 25% \text{ of } U_0 = 0.25 U_0 = \frac{1}{4} U_0$.
\[ U_0 \sin^2(\omega t) = \frac{1}{4} U_0 \]
\[ \sin^2(\omega t) = \frac{1}{4} \implies \sin(\omega t) = \frac{1}{2} \]
The smallest positive time $t$ satisfies:
\[ \omega t = \frac{\pi}{6} \]
Substituting $\omega = \frac{1}{\sqrt{LC}}$:
\[ \frac{t}{\sqrt{LC}} = \frac{\pi}{6} \]
\[ t = \frac{\pi \sqrt{LC}}{6} \]

Step 4: Final Answer:
The value of $t$ is $\frac{\pi \sqrt{LC}}{6}$.