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 $L_2(L_1>L_2)$ of L then which of the following represents this graph correctly? (plots are schematic and not drawn to scale)

• A

  
• B

  
• Correct answer
  C

  
• D

  

Solution

Comparing to the damped pendulum. We write, $\mathit{\text{m}}\frac{\text{d}\mathit{\text{v}}}{\text{d}\mathit{\text{t}}}=-\mathit{\text{kx}}-\mathit{\text{bv}}$, $bv$ is resistive force,$\Rightarrow$ Amplitude, $\mathit{\text{A}}={\mathit{\text{A}}}_0{\text{e}}^{-\frac{\mathit{\text{bt}}}{2\mathit{\text{m}}}}$. Comparing results, we can write

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


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

$\Rightarrow \mathit{\text{A}}=\mathit{\text{q}}\text{; }\mathit{\text{R}}=\mathit{\text{b}}\text{, }\mathit{\text{m}}=\mathit{\text{L}}$
$\text{⇒}\mathit{\text{q}}={\mathit{\text{q}}}_0{\text{e}}^{-\frac{\mathit{\text{Rt}}}{2\mathit{\text{L}}}}$$\text{⇒}{\mathit{\text{q}}}^2={\mathit{\text{q}}}_0^2{\text{e}}^{-\frac{\mathit{\text{Rt}}}{\mathit{\text{L}}}}$$\text{⇒}{\mathit{\text{q}}}^2={\mathit{\text{q}}}_0^2{\text{e}}^{-\frac{\mathit{\text{Rt}}}{\mathit{\text{L}}}}$.

Exponentially decreasing function, $L_1$ graph has higher values than $L_2$.

Hence, the correct answer is option: 'C'.