Question

The capacitor shown in figure has been charged to a potential difference of $Vvolt$, 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 $t=2R_{1}C+R_{2}C$.

Answer 1

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

$q=q_i+(q_f-q_i)(1-e^{-t{\tau }_C})$

After total time $2R_{1}C+R_{2}C$ or $t=R_{1}C+R_{2}C$, one time constant in above equation, charge will remain

$q=\frac{CV}{e}+(CE-\frac{CV}{e})(1-\frac{1}{e})$

$=EC(1-\frac{1}{e})+\frac{VC}{e^2}$