JEE Advanced · 2006 · 09 Apr 2006 · Electromagnetic Induction

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The capacitor of capacitance C can be charged (with the help of a resistance R) by a voltage source V, by closing switch

S_{1}

while keeping switch

S_{2}

open. The capacitor can be connected in series with an inductor 'L' by closing switch

S_{2}

and opening

S_{1}

.

After the capacitor gets fully charged,

S_{1}

is opened and

S_{2}

is closed so that the inductor is connected in series with the capacitor. Then,

(A)at $t = 0$ , energy stored in the circuit is purely in the form of magnetic energy
(B)at any time $t > 0$ , current in the circuit is in the same direction
(C)at $t > 0$ , there is no exchange of energy between the inductor and capacitor
(D)at any time $t > 0$ , instantaneous current in the circuit may $V \frac{C}{L}$ ✓ Correct

Answer: (D)

Solution

When the fully charged capacitor is connected to the inductor, an LC oscillation begins.

At

t = 0

, the capacitor has charge

Q_{0} = C V

and the current is zero, so all energy is electrostatic, not magnetic (option A incorrect).

The charge oscillates as

q = Q_{0} cos (ω t)

where

ω = \frac{1}{L C}

. The current is

i = - \frac{d q}{d t} = Q_{0} ω sin (ω t)

, which changes sign (direction) periodically (option B incorrect).

Energy continuously exchanges between electric (capacitor) and magnetic (inductor) forms during oscillation (option C incorrect).

The maximum current is

i_{ma x} = Q_{0} ω = C V \cdot \frac{1}{L C} = V \frac{C}{L}

. At any time

t > 0

, the instantaneous current magnitude can equal this maximum value (option D correct).

JEE Advanced · 2006 · 09 Apr 2006 · Single correct option

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JEE Advanced · 2006 · 09 Apr 2006 · Electromagnetic Induction

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