For a cicuit shown in fing 5.235 swhitch `S_1 is closed at t= 0,` then at `t = (2R_2+R_1)C, S_1` is opened and `S_2` is closed.

a. Find the charge on capactior at `t = (2R_2 + 2R_1)C.`
b. Find current through `R_2` (adjacent to the battery) at `t = (3R_1 +2R_2)C.`

Explanation

a. For t =0 to `t = (2R_(2)+R_(1))C,` capacitor gets charged from all
the three resistors. So
`q = CE[1-e^((t)/((2R_(2) +R_(1))C))]`.
Putting `t = (2R_(2) +R_(1))C`, we get
`q_(1) =CE[1 -e^(-1)] = (CE(e -1))/(e )`
Now battery is disconnected and the capacitor gets discharged
through `R_(1)` after `S_(1)` is opened and `S_(2)` is closed. So
`int_(q1)^(q2) (dq)/(q)-int_((2R_(2)+R_(1))C)^((2R_(2)+2R_(1))C)(dt)/(R_(1)C)` or `q_(2)=(q_(1))/(e )=(CE(e-1))/(e^(2))`
b. Since battery is disconnected at this time, there is no current
in `R_(2)`