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 maxi

Question

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 $(Q_{M a x}^{2})$ 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? (plots are schematic and not drawn to scale)

TestHub · Accepted Answer

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

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

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

$\Rightarrow A = q; R = b, m = L$

$\Rightarrow q = q_{0} e^{- \frac{Rt}{2 L}}$ $\Rightarrow q^{2} = q_{0}^{2} e^{- \frac{Rt}{L}}$ $\Rightarrow q^{2} = q_{0}^{2} e^{- \frac{Rt}{L}}$ .

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

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