Physics - Capacitor Question with Solution | TestHub

After a long time capacitor branch will behave like open circuit so circuit will be like,

Equivalent resistance is
$R=6+\left(4||\left(2+10\right)\right)\Omega =9\Omega$
and, the current through battery becomes $\frac{72}{9}=8\text{A}$
From the circuit shown, current through $10\Omega$ resistor becomes, $\frac{4}{(10+2+4)}\times 8=2\text{A}$

Potential available for capacitor is = potential drop across $10\Omega$
$\therefore$ Potential available for capacitor is: $10\times 2=20\mathrm{V}$
$\therefore$ Charge on capacitor $Q=\left(10\mu \text{F}\right)\left(20\mathrm{V}\right)=200\mu \mathrm{C}$