A capacitor of 4 $\mathrm{\mu }$F is connected as shown in the circuit Figure. The internal resistance of the battery is 0.5 $\mathrm{\Omega }$. The amount of charge on the capacitor plates will be :

As capacitor offer infinite resistance for DC circuit. So current from cell will not flow across branch of 4$\mathrm{\mu }$F and 10$\mathrm{\Omega }$. So current will flow across 2 ohm branch.

So Potential Difference (PD) across 2$\mathrm{\Omega }$ resistance V = RI = 2 x 1 = 2 Volt. As battery, capacitor and 2 branches are in parallel. So PD will remain same across all three branches. As current does not flow through capacitor branch, so no potential drop will be across 10$\mathrm{\Omega }$. So PD across 4$\mathrm{\mu }$F capacitor = 2 Volt Q = C V = 2$\mathrm{\mu }$F x 2 V = 8$\mathrm{\mu }$C