In the given circuit the capacitor (C) may be charged through resistance R by a battery V by closing switch `(S_1)`. Also when `(S_1)` is opend and `(S_2)` is closed the capacitor is connected in series with inductor (L).

At the start, the capicitor was uncharged. when switch `(S_1)` is closed and `(S_2)` is kept open, the time constant of this circuit is `tau`. which of the following is correct

Options

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

Answer

B

Explanation

Charge on capacitor at time `t` is
`q = q_(0) (1 - e^(-t//tau))`
Here, `q_(0) = CV` and `t = 2 tau`
`q = CV(1 - e^(-2tau//tau)) = CV (1 - e^(-2))`