JEE Main Alternating Current PYQ

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6. (JEE Main 2015 (Offline))

An $L C R$ circuit is equivalent to a damped pendulum. In an $L C R$ circuit the capacitor is charged to $Q_{0}$ and then connected to the $L$ and $R$ as shown below :

JEE Main 2015 (Offline) Physics - Alternating Current Question 91 English

If a student plots graphs of the square of maximum charge $(Q_{M a x}^{2})$ on the capacitor with time $(t)$ for two different values $L_{1}$ and $L_{2}$ $(L_{1} > L_{2})$ of $L$ then which of the following represents this graph correctly ?
$(p l o t s a r e s c h e m a t i c a n d n i o t d r a w n t o s c a l e)$

A.

JEE Main 2015 (Offline) Physics - Alternating Current Question 91 English Option 1

B.

JEE Main 2015 (Offline) Physics - Alternating Current Question 91 English Option 2

C.

JEE Main 2015 (Offline) Physics - Alternating Current Question 91 English Option 3

D.

JEE Main 2015 (Offline) Physics - Alternating Current Question 91 English Option 4

Correct Option is (C)

From $K V L$ at any time $t$

JEE Main 2015 (Offline) Physics - Alternating Current Question 91 English Explanation
$\frac{q}{c} - i R - L \frac{d i}{d t} = 0$

$i = - \frac{d q}{d t} \Rightarrow \frac{q}{c} + \frac{d q}{d t} R + \frac{L d^{2} q}{d t^{2}} = 0$

$\frac{d^{2} q}{d t^{2}} + \frac{R}{L} \frac{d q}{d t} + \frac{q}{L c} = 0$

From damped harmonic oscillator, the amplitude is

given by $A = A_{0} e - \frac{d t}{2 m}$

Double differential equation

$\frac{d^{2} x}{d t^{2}} + \frac{b}{m} \frac{d x}{d t} + \frac{k}{m} x = 0$

$Q_{max} = Q_{0} e - \frac{R t}{2 L} \Rightarrow Q_{max}^{2} = Q_{0}^{2} e - \frac{R t}{L}$

Hence damping will be faster for lesser self inductance.