JEE Main 2026 — Jan 21, Shift 2 — Physics solution

JEE Main2026Jan 21, Shift 2Physics

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
$\frac{π L C}{3}$
B
$\frac{π L C}{6}$
C
$\frac{π L C}{2}$
D
$π \frac{L C}{2}$

Solution

In an LC oscillator, the charge on the capacitor varies as $Q (t) = Q_{0} cos (ω t)$ , with $ω = \frac{1}{L C}$ and period $T = 2 π L C$ .

Capacitor energy: $U_{C} (t) = \frac{Q ^{2}}{2 C} = U_{0} cos^{2} (ω t)$ . If 25% of the initial energy has moved to the inductor, then $U_{C} = 0.75 U_{0}$ , so $cos^{2} (ω t) = \frac{3}{4}$ and $cos (ω t) = \frac{3}{2}$ .

This gives $ω t = \frac{π}{6}$ , i.e. $t = \frac{T}{12} = \frac{2 π L C}{12} = \frac{π L C}{6}$ .

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