Physics - Current Electricity Question with Solution | TestHub

At steady state the capacitor will be fully charged and thus there will be no current in the $1\Omega$ resistance. So the effective circuit becomes
Net current from the $6\mathrm{V}$ battery, $I=\frac{6}{\left(\frac{2\times 3}{2+3}\right)+\frac{2.8}{1}}=\frac{6}{1.2+2.8}=\frac{3}{2}=1.5\mathrm{A}$ Between $\mathrm{A}$ and $\mathrm{B}$, voltage is same in both resistances, $2{\mathrm{I}}_1=3{\mathrm{I}}_2$ where ${\mathrm{I}}_1+{\mathrm{I}}_2=\mathrm{I}=1.5$ $\Rightarrow 2{\mathrm{I}}_1=3\left(1.5-{\mathrm{I}}_1\right)\Rightarrow {\mathrm{I}}_1=0.9\mathrm{A}$