JEE Advanced · 2006 · 09 Apr 2006 · Current Electricity

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}

.

Initially, the capacitor was uncharged. Now, switch

S_{1}

is closed and

S_{2}

is kept open. If time constant of this circuit is

τ

, then

(A)after time interval $τ$ , charge on the capacitor is CV/2
(B)after time interval $2 τ$ , charge on the capacitor is $CV (1 - e^{- 2})$ ✓ Correct
(C)the work done by the voltage source will be half of the heat dissipated when the capacitor is fully charged.
(D)after time interval $2 τ$ , charge on the capacitor is $CV (1 - e^{- 1})$

Answer: (B)

Solution

For an RC circuit charging from zero, the charge at time t is given by:

Q (t) = Q_{0} (1 - e^{- t / τ}) = C V (1 - e^{- t / τ})

where

τ = R C

is the time constant.

At

t = τ

Q = C V (1 - e^{- 1}) \approx 0.632 C V

, not CV/2 (option A incorrect).

At

t = 2 τ

Q = C V (1 - e^{- 2}) \approx 0.865 C V

(option B correct, D incorrect).

When fully charged, energy stored in capacitor is

\frac{1}{2} C V^{2}

. Work done by source is

C V^{2}

. Heat dissipated is

C V^{2} - \frac{1}{2} C V^{2} = \frac{1}{2} C V^{2}

, which equals the energy stored, not half the work done (option C incorrect).

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