Paragraph :
The capacitor of capacitance \(C\) can be charged (with the help of a resistance \(R\) ) by a voltage source \(V\), by closing switch \(S_1\) while keeping switch \(S_2\) open. The capacitor can be connected in series with an inductor \(L\) by closing switch \(S_2\) and opening \(S_1\).

Question :
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

1. A after time interval \(\tau\), charge on the capacitor is \(C V / 2\)
2. B after time interval \(2 \tau\), charge on the capacitor is \(C V /\left(1-e^{-2}\right)\)
3. C the work done by the voltage source will be half on the heat dissipated when the capacitor is fully charged
4. D after time interval \(2 \tau\), charge on the capacitor is \(C V\left(1-e^{-1}\right)\)