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MCQEasyJEE 2026LCR Circuits & Resonance

JEE Physics 2026 Question with Solution

A capacitor CC is first charged fully with potential difference of V0V_0 and disconnected from the battery. The charged capacitor is connected across an inductor having inductance LL. In tt s 25%25\% of the initial energy in the capacitor is transferred to the inductor. The value of tt is _____ s.

  • A

    πLC2\pi \sqrt{\frac{LC}{2}}

  • B

    πLC6\frac{\pi \sqrt{LC}}{6}

  • C

    πLC3\frac{\pi \sqrt{LC}}{3}

  • D

    πLC2\frac{\pi \sqrt{LC}}{2}

Answer

Correct answer:B

Step-by-step solution

Standard Method

Given: A capacitor CC is initially fully charged to potential difference V0V_0 and then connected across an inductor of inductance LL. At time tt, 25%25\% of the initial capacitor energy has been transferred to the inductor.

Find: The value of tt.

This is an LC oscillation problem. Initially, the entire energy is stored in the capacitor. During oscillation, energy shifts between the capacitor and the inductor.

The angular frequency is

ω=1LC\omega = \frac{1}{\sqrt{LC}}

The total initial energy is

U0=12CV02U_0 = \frac{1}{2} C V_0^2

The energy in the inductor at time tt is

UL(t)=U0sin⁡2(ωt)U_L(t) = U_0 \sin^2(\omega t)

Given that the inductor has received 25%25\% of the initial energy,

UL=14U0U_L = \frac{1}{4} U_0

So,

U0sin⁡2(ωt)=14U0U_0 \sin^2(\omega t) = \frac{1}{4} U_0 sin⁡2(ωt)=14\sin^2(\omega t) = \frac{1}{4} sin⁡(ωt)=12\sin(\omega t) = \frac{1}{2}

For the smallest positive time,

ωt=π6\omega t = \frac{\pi}{6}

Substituting ω=1LC\omega = \frac{1}{\sqrt{LC}},

tLC=π6\frac{t}{\sqrt{LC}} = \frac{\pi}{6} t=πLC6t = \frac{\pi \sqrt{LC}}{6}

Therefore, the value of tt is πLC6\frac{\pi \sqrt{LC}}{6}. The correct option is B.

Energy Partition Shortcut

Given: In an LC circuit, 25%25\% of the initial capacitor energy is transferred to the inductor.

Find: The earliest time tt.

Use the standard energy partition result:

UC=U0cos⁡2(ωt),UL=U0sin⁡2(ωt)U_C = U_0 \cos^2(\omega t), \qquad U_L = U_0 \sin^2(\omega t)

If UL=14U0U_L = \frac{1}{4}U_0, then

sin⁡2(ωt)=14\sin^2(\omega t) = \frac{1}{4}

So the first phase angle is

ωt=π6\omega t = \frac{\pi}{6}

Since

ω=1LC\omega = \frac{1}{\sqrt{LC}}

we get

t=πLC6t = \frac{\pi \sqrt{LC}}{6}

Therefore, the correct option is B.

Common mistakes

  • Using capacitor charge or current directly instead of the energy relation. The question asks about transferred energy, so use UL=U0sin⁡2(ωt)U_L = U_0\sin^2(\omega t), not only charge or current equations.

  • Taking sin⁡(ωt)=14\sin(\omega t) = \frac{1}{4} instead of sin⁡2(ωt)=14\sin^2(\omega t) = \frac{1}{4}. Since energy is proportional to the square of the oscillating quantity, the correct step is sin⁡(ωt)=12\sin(\omega t)=\frac{1}{2} for the first positive time.

  • Using the wrong angular frequency for the LC circuit. The correct value is ω=1LC\omega = \frac{1}{\sqrt{LC}}. Replacing it with LC\sqrt{LC} or any reciprocal error gives the wrong time dimensionally and numerically.

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