The capacitor of capacitance C can be charged (with the help of a resistance R) by a voltage source V, by closing switch $\mathrm{S}_{1}$ while keeping switch $\mathrm{S}_{2}$ open. The capacitor can be connected in series with an inductor 'L' by closing switch $\mathrm{S}_{2}$ and opening $\mathrm{S}_{1}$. If the total charge stored in the LC circuit is $\mathrm{Q}_{0}$, then for $\mathrm{t} \geq 0$

(C) the charge on the capacitor is $\mathrm{Q}=-\mathrm{LC} \frac{\mathrm{d}^{2} \mathrm{Q}}{\mathrm{dt}^{2}}$

(A) the charge on the capacitor is $\mathrm{Q}=\mathrm{Q}_{0} \cos \left(\frac{\pi}{2}+\frac{\mathrm{t}}{\sqrt{\mathrm{LC}}}\right)$

(B) the charge on the capacitor is $\mathrm{Q}=\mathrm{Q}_{0} \cos \left(\frac{\pi}{2}-\frac{\mathrm{t}}{\sqrt{\mathrm{LC}}}\right)$

(D) the charge on the capacitor is $\mathrm{Q}=-\frac{1}{\sqrt{\mathrm{LC}}} \frac{\mathrm{d}^{2} \mathrm{Q}}{\mathrm{dt}^{2}}$

JEE Advanced · 2006 · 09 Apr 2006 · Electromagnetic Induction

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}

.

If the total charge stored in the LC circuit is

Q_{0}

, then for

t \geq 0

Answer: (C)

For an LC circuit with initial charge

Q_{0}

and zero initial current, the charge oscillates as:

Q = Q_{0} cos (ω t)

where

ω = \frac{1}{L C}

Q = Q_{0} cos (\frac{t}{L C})

, which doesn't match options A or B (they have phase shifts of

π /2

).

From Kirchhoff's voltage law:

\frac{Q}{C} + L \frac{d i}{d t} = 0

Since

i = \frac{d Q}{d t}

, we have:

\frac{Q}{C} + L \frac{d ^{2} Q}{d t ^{2}} = 0

Rearranging:

Q = - L C \frac{d ^{2} Q}{d t ^{2}}

(option C correct).

Option D has incorrect dimensions and is not the correct differential equation.

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