Question

The capacitor shown in figure has been charged to a potential difference of `V` volt, so that it carries a charge `CV` with both the switches `S_1 and S_2` remaining open. Switch `S_1` is closed at `t = 0`. At `t= R_1C` switch `S_1` is opened and `S_2` is closed. Find the charge on the capacitor at `t= 2R-1C + R_2C`.

Answer 1

Correct Answer - B::C
When `S_1` is closed and `S_2` open, capacitor will discharge. At time `t=R_1C` , one time constant, charge will remain `q_1=(1/e)` times of `CV` or `q_1=(CV)/e`
When `S_1` is open and `S_2` closed charge will increase (or may decrease also) from `(CV)/e` to `CE` exponentially. Time constant for this would be `(R_1C+R_2C)`. Charge as function of time would be
`q=q_i+(q_f-q_i)(1-e^(-t//tau_C))`
`q=(CV)/e+(CE-(CV)/e)(1-e^(-t//tau_C))`
After total time `2R_1C+R_2C` or `t=R_1C+R_2C` one time constant in above equation, charge will remain
`q=(CV)/e+(CE-(CV)/e)(1-1/e)`
`=EC(1-1/e)+(VC)/e^2`