Question

At (= 0, switch S is closed. The charge on the capacitor is
varying with time as $Q=Q_0(1-e^{-at}).$ Obtain the value of $Q_0$
and $\alpha$ in the given circuit parameters.

Answer 1

$Q_0$ is the steady state charge stored in the capacitor.
$Q_0=C$ [PD across capacitor in steady state]
= C [steady state current through $R_2](R_2)$
$=C\left(\frac{V}{R_1+R_2}\right).R_2$
$\therefore Q_0=\frac{CV{R}_2}{R_1+R_2}$
$\alpha is1/{\tau }_{c}or\frac{1}{C{R}_{net}}$
Here, $R_{net}$ is equivalent resistance across capacitor after
short circuiting the battery. Thus,
$R_{net}=\frac{R_1{R}_2}{R_1+R_2}$ (As $R_{1}and{R}_2$ are in parallel)
$\alpha =\frac{1}{C\left(\frac{R_1{R}_2}{R_1+R_2}\right)}=\frac{R_1+R_2}{C{R}_1{R}_2}$