Question

The given RC circuit has two switches $S_1$ and $S_2$. The switch $S_2$ is closed till the capacitor $C$ attains its maximum possible charge $q_0$. Then, $S_2$ is opened and $S_1$ is closed simultaneously till the capacitor releases half of its total stored charge $q_0$ for a time interval $t_1$. Finally $S_1$ is opened and $S_2$ is closed till the capacitor attains a charge $\left(3/4\right)q_0$ for a time interval $t_2$. Find the ratio $\left(t_1/t_2\right)$

Answer 1

When the switch $S_2$ is closed, the capacitor is charged to a potential $V$. Now the switch $S_2$ is opened and the switch $S_1$ is closed.
The instantaneous charge on the capacitor is
$\Rightarrow q=q_0{e}^{\frac{t}{R_{1}C}};q_0=CV$
putting $t=t_1$
for $q=q_0/2$, we obtain $e^{\frac{t}{R_{1}C}}=\frac{1}{2}\Rightarrow t_1=R_{1}C\ln 2$
Again, the switch $S_1$ is opened and $S_2$ is closed. Therefore, the capacitor starts charging from charge $q_0/2$ to $\frac{3}{4}{q}_0$
Now instantaneous charge on the capacitor is
$q^{\prime }=q_0[1-e^{\frac{t}{{(R}_1+R_2)C}}]+\frac{q_0}{2}{e}^{\frac{t}{{(R}_1+R_2)C}}=q_0[1-\frac{1}{2}{e}^{\frac{t}{{(R}_1+R_2)C}}]$
At, $t=t_2,q^{\prime }=3q_0/4\Rightarrow \frac{3q_0}{4}=q_0(1-\frac{1}{2}{e}^{\frac{t}{{(R}_1+R_2)C}})$
$\Rightarrow e^{\frac{t}{{(R}_1+R_2)C}}=\frac{1}{2}\Rightarrow t_2=(R_1+R_2)C\ln 2$
The required ratio of the times $=\frac{t_1}{t_2}=\frac{R_{1}C\ln 2}{(R_1+R_2)C\ln 2}or\frac{t_1}{t_2}=\frac{R_1}{R_1+R_2}$