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 series with an inductor `L` by closing switch `S_(2)` and opening `S_(1)`.
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

Answer

B

Explanation

`Q=Q_(0)(1-e^(-t//tau))`
`Q=CV(1-e^(-t//tau))` after time interval `2tau`.