An $LCR$ circuit is equivalent to a damped pendulum. In an $LCR$ circuit the capacitor is charged to $Q_0$ and then connected to the $L$ and $R$ as shown below.
If a student plots graphs of the square of maximum charge $( Q_{Max} ^2 )$ 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? (plots are schematic and not drawn to scale)
JEE MAIN 2015, Diffcult
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From $KVL$ at any time :
$\frac{q}{c}-i R-I \cdot \frac{d i}{d t}=0$
$\mathrm{i}=-\frac{\mathrm{d} q}{\mathrm{dt}} \Rightarrow \frac{\mathrm{q}}{\mathrm{c}}+\frac{\mathrm{dq}}{\mathrm{dt}} \mathrm{R}+\frac{\mathrm{Ld}^{2} \mathrm{q}}{\mathrm{dt}^{2}}=0$
$\frac{d^{2} q}{d t^{2}}+\frac{R}{I} \frac{d q}{d t}+\frac{q}{I c}=0$
From damped harmonic oscillator, the amplitude is given by $\mathrm{A}=\mathrm{A}_{0} \mathrm{e}-\frac{\mathrm{dt}}{2 \mathrm{m}}$
Double differential equation $\frac{\mathrm{d}^{2} \mathrm{x}}{\mathrm{dt}^{2}}+\frac{\mathrm{b}}{\mathrm{m}} \frac{\mathrm{dx}}{\mathrm{dt}}+\frac{\mathrm{k}}{\mathrm{m}} \mathrm{x}=0$
$\mathrm{Q}_{\max }=\mathrm{Q}_{\mathrm{o}} \mathrm{e}-\frac{\mathrm{Rt}}{2 \mathrm{L}} \Rightarrow \mathrm{Q}_{\max }^{2}=\mathrm{Q}_{\mathrm{o}}^{2} \mathrm{e}-\frac{\mathrm{Rt}}{\mathrm{L}}$
Hence damping will be faster for lesser self inductance
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