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`.

Explanation

First, capacitor will discharge through `R_1`
`At t = R_1C`
`q_1 = CV(e^(-R_1C)/(R_1C)) = (CV)/(e )`
Now it will be charged through `R_1 +R_2`
`int_(q1)^q (dp)/(CE - Q) = int _(R_1C)^(2R_1C+R_2C) (dt)/((R_1+R_2)C)`
`q = CE(1-(1)/(e )) +(CV)/(e^2)`