Initially, the capacitor was uncharged. Now, switch $S_1$ is closed and $S_2$ is kept open. If time constant of this circuit is $\tau$, then

after time interval $2 \tau$, charge on the capacitor is $\mathrm{CV}\left(1-e^{-2}\right)$.

If the total charge stored in the LC circuit.is $\mathrm{Q}_0$, then for $t \geq 0$,

the charge on the capacitor is $$ \mathrm{Q}=-\mathrm{LC} \frac{d^2 \mathrm{Q}}{d t^2} . $$

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JEE Advanced Growth And Decay Of Current practice

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In Process 1, the energy stored in the capacitor E<sub>C</sub> and heat dissipated across resistance E<sub>D</sub> are related by

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