Calculate the amount of charge on capacitor of 4 µF in steady state. The internal resistance of battery is 1Ω.

The given circuit diagram is shown below

The battery connected in the circuit is a DC source and capacitor does not allow direct current to pass through it. Therefore, there will no current pass through upper branch of circuit.

The equivalent resistance of remaining circuit will be

             ${\mathrm{R}}_{\mathrm{eq}}={\mathrm{R}}_{\mathrm{in}}+4\mathrm{\Omega }$

Here, R_in is internal resistance of battery.

             ${\mathrm{R}}_{\mathrm{eq}}=1+5=5\mathrm{\Omega }$

From Ohm’s law, we know that

            $\mathrm{I}=\frac{\mathrm{V}}{{\mathrm{R}}_{\mathrm{eq}}}=\frac{5}{5}=1\mathrm{A}$

Now, voltage drop across branch AB is calculated as

         ${\mathrm{V}}_{\mathrm{AB}}=1\times 4=4\mathrm{V}$

Voltage across upper branch will also be 4V as the connection is parallel.

Now, charge flowing from upper branch is given as

         $\mathrm{Q}={\mathrm{C}}_{\mathrm{eq}}{\mathrm{V}}_{\mathrm{AB}} ...\left(\mathrm{i}\right)$

Calculating the equivalent capacitance,

        $\begin{array}{l}\frac{1}{{\mathrm{C}}_{\mathrm{eq}}}=\frac{1}{4}+\frac{1}{2+2}=\frac{1}{4}+\frac{1}{4}\\ {\mathrm{C}}_{\mathrm{eq}}=2\end{array}$

Substituting the value of equivalent capacitance and voltage drop in Eq. (i), we get

          $\mathrm{Q}=2\times 4\mathrm{V}=8\mathrm{\mu C}$

Therefore, the charge stored in 4µF will be 8µC.