Charge decay in an isolated series $LCR$ circuit

After the capacitor initially charged to $Q_0$ is connected to $L$ and $R$, the charge executes damped oscillations:

$L\frac{d^2Q}{dt^2}+R\frac{dQ}{dt}+\frac{Q}{C}=0$

For the oscillatory (underdamped) case,

$Q(t)=Q_0 e^{-\frac{R}{2L}t}\cos(\omega_d t+\phi)$

where

$\omega_d=\sqrt{\frac{1}{LC}-\frac{R^2}{4L^2}}$

Maximum charge envelope

The maximum value of charge at any instant is given by the envelope:

$Q_{\max}(t)=Q_0 e^{-\frac{R}{2L}t}$

Therefore,

$Q_{\max}^2(t)=Q_0^2 e^{-\frac{R}{L}t}$

Compare for two inductances $L_1$ and $L_2$ with $L_1>L_2$

For the two cases,

$Q_{\max,1}^2=Q_0^2 e^{-\frac{R}{L_1}t}, \qquad Q_{\max,2}^2=Q_0^2 e^{-\frac{R}{L_2}t}$

Since $L_1>L_2$,

$\frac{R}{L_1}<\frac{R}{L_2}$

so the decay for $L_1$ is slower.

Graph features

• At $t=0$, both curves start from the same value: $Q_{\max}^2(0)=Q_0^2$
• Both decrease exponentially with time.
• The curve for larger inductance $L_1$ stays above the curve for smaller inductance $L_2$ for $t>0$.

Correct option

Hence the correct graph is the one in which both start at the same initial value and decay exponentially, with the $L_1$ curve decaying more slowly than the $L_2$ curve.

That corresponds to Option C.