The capacitor shown in the 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 $t=R_{1}C$ switch $S_1$ is opened and $S_2$ is closed, Find the charge on the capacitor at $t=2R_{1}C+R_{2}C$.

When $S_1$ is closed and $S_2$ open, capacitor will discharge. At time $t=R_{1}C$, one time constant, charge will remain $q_1=\frac{1}{e}$ times of $CV$ or 

$q_1=\frac{CV}{e}$

When $S_1$ is open and $S_2$ closed, charge will increase (or may decrease also) from $\frac{CV}{e}$ to $CE$ exponentially. Time constant for this would be $(R_{1}C+R_{2}C)$. Charge as function of time would be 

$$\begin{gathered} q=q_i+\left(q_f-q_i\right)\left(1-e^{-t/\tau }C\right) \\ q=\frac{CV}{e}+\left(CE-\frac{CV}{e}\right)\left(1-e^{-t/{\tau }_C}\right) \end{gathered}$$

After total time $2R_{1}C+R_{2}C \mathrm{or} t=R_1{C}_{+}{R}_2{C}_1$ the time constant in above equation, charge will remain 

$q=\frac{CV}{e}+\left(CE-\frac{CV}{e}\right)\left(1-\frac{1}{e}\right)=EC\left(1-\frac{1}{e}\right)+\frac{VC}{e^2}$