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

To solve this problem, we need to understand the LC circuit dynamics involving a capacitor and an inductor. This is an example of simple harmonic oscillation in an LC circuit.

The energy stored in the capacitor when it is fully charged is given by:

E = \frac{1}{2}C V_0^2

This energy is transferred back and forth between the capacitor and the inductor. Initially, all the energy is in the capacitor, and it starts transferring to the inductor.

After time t, it is given that 25% of the initial energy remains in the capacitor, meaning 75% of the energy is now in the inductor. Let us calculate this time.

1. The energy in the capacitor at time t is:

E_C = 0.25 \cdot \frac{1}{2}C V_0^2

2. The energy in the inductor at time t is:

E_L = \frac{1}{2}L I^2 = 0.75 \cdot \frac{1}{2}C V_0^2

Since energy oscillates between the inductor and capacitor as a sinusoidal function, the time taken to go from maximum energy in the capacitor (100% to 25%), is a quarter of the period of the oscillation (i.e., moving from 0 degrees to 90 degrees in sinusoidal terms).

The angular frequency \omega of an LC circuit is given by:

\omega = \frac{1}{\sqrt{LC}}

The period of oscillation T is given by:

T = \frac{2\pi}{\omega} = 2\pi\sqrt{LC}

Thus, a quarter of the period (25% transfer is reached) is:

\frac{T}{4} = \frac{2\pi\sqrt{LC}}{4} = \frac{\pi\sqrt{LC}}{2}

However, the 75% transfer scenario corresponds not to a quarter period but corresponds to the remaining quarter period of one sinusoidal oscillation, because 25% energy in capacitor implies that more than one-fourth period has occurred, signifying the circuit has oscillated to where:

\theta = \sin^{-1}(\sqrt{0.25})

This occurs at \frac{\pi}{6} rad, hence the corresponding time at which this occurs is:

t = \frac{T}{12} = \frac{\pi\sqrt{LC}}{6}

So, the correct answer is \frac{\pi \sqrt{LC}}{6} seconds. This aligns with the correct option.

Thus, the value of t when 25% of the energy is transferred is \frac{\pi \sqrt{LC}}{6}.