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)`.

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

Options

A. after time interval `tau,` Charge on the capacitor is CV/2
B. after time interval `2 tau`, charge on the capacitor is CV`(1 - e^(-2))`
C. the work done by the voltae source will be half of the heat dissipated when the capacitor is fully charged.
D. after time interval `2 tau`, charge on the capacitor is CV`(1 - e^(-1))`

Answer

B