Question

In the circuit, initially switch $S$ has been open for a long time. It is then closed for a long time. The charge on the capacitor (in $\mu C$) changes by an amount of
Here, $C=30\mu F,E_1=1\text{V},E_2=3\text{V},R_1=0.2\Omega ,R_2=0.4\Omega$.

Answer 1

When switch $S$ is open



Charge on capacitor is $q_1=3C$

When switch $S$ is closed,


Current in the circuit at steady state is
$I=\frac{E_2-E_1}{R_1+R_2}=\frac{3-1}{0.4+0.2}=\frac{2}{0.6}=\frac{20}{6}=\frac{10}{3}$
$\frac{q_2}{C}=E_2-I{R}_2=3-\frac{10}{3}\times 0.4$
$q_2=\frac{5C}{3}$
Change in charge of the capacitor is
$\Delta q=q_1-q_2$
$\Delta q=3C-\frac{5}{3}C=\frac{4C}{3}=\frac{4\times 30}{3}=40\mu C$