Question 143 of 190

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 $\left({Q}_{Max}^{2}\right)$ on the capacitor with time (t) for two different values ${L}_{1}$ and ${L}_{2}({L}_{1}\gt {L}_{2})$ of L then which of the following represents this graph correctly? (plots are schematic and not drawn to scale)

ABCD

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Answer: (C)

Comparing to the damped pendulum. We write, $\text{m}\dfrac{\text{d}\text{v}}{\text{d}\text{t}}=-\text{kx}-\text{bv}$, $bv$ is resistive force,$\Rightarrow$ Amplitude, $\text{A}={\text{A}}_{0}{\text{e}}^{-\dfrac{\text{bt}}{2\text{m}}}$. Comparing results, we can write

$\dfrac{\text{q}}{\text{c}}=+\text{L}\dfrac{\text{d}\text{i}}{\text{d}\text{t}}+\text{iR}$, as charge decreasing,


$\dfrac{\text{q}}{\text{c}}=\text{L}\dfrac{{\text{d}}^{2}\text{q}}{\text{d}{\text{t}}^{2}}-\dfrac{\text{d}\text{q}}{\text{d}\text{t}}\text{R}$

$\Rightarrow \text{A}=\text{q}\text{;}\text{R}=\text{b}\text{,}\text{m}=\text{L}$
$\text{⇒}\text{q}={\text{q}}_{0}{\text{e}}^{-\dfrac{\text{Rt}}{2\text{L}}}$$\text{⇒}{\text{q}}^{2}={\text{q}}_{0}^{2}{\text{e}}^{-\dfrac{\text{Rt}}{\text{L}}}$$\text{⇒}{\text{q}}^{2}={\text{q}}_{0}^{2}{\text{e}}^{-\dfrac{\text{Rt}}{\text{L}}}$.

Exponentially decreasing function, ${L}_{1}$ graph has higher values than ${L}_{2}$.