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 interval `tau` charge on the capcitor is `(CV)/(2)`
B. after interval `2 tau` charge on the capcitor is `(CV) (1-e^(-2))`
C. the work done by the voltage 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

Explanation

(b) for charging of R-C circuit , `Q=Q_(0)(1-e^(-t//tau)]` when the charging is complete, the potential difference between the capacitor plate will be V. The charge stored in this case will be maximum.
Therefore, `(Q_0)=CV`
when `t=2(tau), Q=CV[1=e^((-2 tau)/(tau))]`
`CV[1_e^(-2)]`
.