Question

The capacitor shown in the figure has been charged to a potential difference of $V$ volts, 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. The charge on the capacitor at $t=2R_{1}C+R_{2}C$ is $CE(1-\frac{1}{e^n})+\frac{CV}{e^{(}n+1)}$. Then the value of n is

Answer 1

At $t=0$, the circuit is given by

The capacitor discharges from time $t=0$ to $t=R_{1}C$
Thus, $q=CV{e}^{\frac{-t}{R_{1}C}}$

At $t=R_{1}C$,

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

When the switch $S_1$ is opened and $S_2$ is closed at $t=R_{1}C$, the circuit becomes


The capacitor charges from $t=R_{1}C$ to $t=R_{1}C+(2R_1+R_2)C$

$q=\frac{CV}{e}+(CE-\frac{CV}{e})(1-e^{\frac{-(t-t_0)}{(R_1+R_2)C}})$

where $t_0=R_{1}C$

At $t=(2R_1+R_2)C$, the charge is given by

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

$=CE(1-\frac{1}{e})+\frac{CV}{e^2}$