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

Answer

C

Explanation

When switch `S_(1)` is closed and switch `S_(2)` is kept open. The capacitor is charging through a resistor `R`. Charge on a capacitor at any time `t` is given by

`q=q_(0)(1-e^(-t//tau))=CV(1-e^(-t//tau)) [As q_(0)=CV]`
At `t=(tau)/2, q=CV(1-e^(-tau//2tau))=CV(1-e^(1//2))`.
At `t=tau, q=CV(1-e^(-t//tau))=CV(1-e^(-1))`
At `t=2tau, q=CV(1-e^(-2tau//tau))=CV(1-e^(-2))`