JEE Main · 2021 · 27 Aug 2021 · Shift 1 · Capacitance

In the steady state no current flows through the branch that contains the capacitors, so there is no potential drop across the $6\Omega$ resistor, the internal $1\Omega$, or the $4\Omega$; the full source emf of $5$ V appears across the capacitor network. That network is a $4\mu \mathrm{F}$ capacitor in series with the parallel pair $2\mu \mathrm{F}$ and $4\mu \mathrm{F}$, whose combined capacitance is $6\mu \mathrm{F}$. The series capacitance is $\frac{4\times 6}{4+6}=2.4\mu \mathrm{F}$, so the charge driven round the series path is $Q=2.4\times 5=12\mu \mathrm{C}$. This charge sits on the $6\mu \mathrm{F}$ parallel section, across which the voltage is $\frac{12}{6}=2$ V. The charge on the $4\mu \mathrm{F}$ capacitor of this parallel pair is $q=4\times 2=8\mu \mathrm{C}$.