For circuit shown in figure switch $S_1$ is closed at t = 0, then at $t=\left(2R_2+R_1\right)C,S_1$ opens & $S_2$ closed.

The charge on capacitor at $t=\left(2R_2+2R_1\right)C$ is  $Q_1$

and the  current through (adjacent to battery) at $t=\left(3R_1+2R_2\right)C$ is $i_1$.
 

 upto $t=\left(2R_2+R_1\right)C$ the charging of capacitor takes place through the resistors $R_1,R_2\text{ and }{R}_2$

At time t charge on the capacitor is given by , Q = EC (1 - $e^{\frac{t}{R_1+2R_2}})$

So at t = ($R_1+2R_2)C,$ charge on the capacitor $Q_1$ = EC($1-e^{-1})$ 

 After $t=\left(2R_2+R_1\right)C$ the discharging of capacitor takes place through $R_1$ . So current through the battery is zero .