JEE Main · 2011 · 10 Apr 2011 · Offline · Electromagnetic Induction

A fully charged capacitor C with initial charge

q_{0}

is connected to a coil of self inductance L at

t = 0

. The time at which the energy is stored equally between the electric and the magnetic field is:

(A) $\frac{π}{4} L C$ ✓ Correct
(B) $2 π L C$
(C) $L C$
(D) $π L C$

Answer: (A)

Solution

In an

L C

circuit the charge oscillates as

q = q_{0} cos ω t

with

ω = \frac{1}{L C}

. The capacitor stores

\frac{q ^{2}}{2 C}

and the inductor holds the rest of the constant total energy. The two shares are equal when the capacitor keeps half the total, that is when

q = \frac{q _{0}}{2}

, so

cos ω t = \frac{1}{2}

and

ω t = \frac{π}{4}

. The time is therefore

t = \frac{π}{4 ω} = \frac{π}{4} L C

JEE Main · 2011 · 10 Apr 2011 · Offline · Single correct option

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