In an LCR circuit as shown below both switches are open initially. Now switch `S_(1)` kept open. (q is charge on the capacitor and `tau = RC` is Capacitive time constant). Which of the following statement is correct?

Options

A. Work done by the battery is half of the energy dissipated in the resistor
B. At `t = tau, q = CV//2`
C. At `t = 2 tau, q = CV (1 - e^(-2))`
D. At `t = (tau)/(2), q = CV (1 - e^(-1))`.

Answer

C

Explanation

In the figure, when switch `S_(1)` is closed and switch `S_(2)` is kept open, the condenser gets charged through resistance. At any time t, the charge acquired by the condenser is
`Q = Q_(0) (1 - e^(-t//RC))`, where `Q_(0) = CV` and `RC = tau`
At `t = 2 tau, Q = Q_(0) (1 - e^(-2tau//tau)) = CV (1 - e^(-2))`
Choice (c ) is correct.