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.

Question

Accepted Answer

B. 1. When a charged capacitor is connected to an inductor, energy oscillates between the capacitor's electric field and the inductor's magnetic field, with charge on the capacitor varying as
$$
q(t) = Q_0\cos(\omega t), \qquad \omega = \frac{1}{\sqrt{LC}}
$$
2. Initially, all the energy resides in the capacitor.
[IMAGE: LC circuit diagram showing charge oscillating between capacitor and inductor over time]
$$
U_0 = \frac{Q_0^2}{2C}
$$
3. The **energy stored in the capacitor** at time $t$ follows from the charge expression.
$$
U_C(t) = \frac{q(t)^2}{2C} = U_0\cos^2(\omega t)
$$
4. By **conservation of energy**, whatever leaves the capacitor goes into the inductor as magnetic energy.
$$
U_L(t) = U_0-U_C(t) = U_0\sin^2(\omega t)
$$
5. Given that 25% of the initial energy is transferred to the inductor, set up the equation.
$$
U_0\sin^2(\omega t) = \frac{1}{4}U_0 \;\Rightarrow\; \sin^2(\omega t) = \frac{1}{4} \;\Rightarrow\; \sin(\omega t) = \frac{1}{2}
$$
6. Solve for $\omega t$ using the smallest positive angle satisfying this.
$$
\omega t = \frac{\pi}{6}
$$
7. Substitute $\omega=\dfrac{1}{\sqrt{LC}}$ and solve for $t$.
$$
\frac{t}{\sqrt{LC}} = \frac{\pi}{6} \;\Rightarrow\; t = \frac{\pi\sqrt{LC}}{6}
$$
Hence, the value of $t$ is $\dfrac{\pi\sqrt{LC}}{6}$ s, so the answer is **option B**.