Let current through $4\mathit{\Omega }$ resistance is $\mathit{i}$ then current through $6\mathit{\Omega }$ resistance is $\left(0.5-\mathit{i}\right)$
$\therefore \mathit{i}\times 4=\left(0.5-\mathit{i}\right)\times 6$
$4\mathit{i}=3-6\mathit{i}$
$10\mathit{i}=3$
$\mathit{i}=0.3\mathrm{A}$
$\therefore$ Current through $4\mathit{\Omega }$ resistance
$=0.3\mathrm{A}$
andcurrent through $6\mathit{\Omega }$ resistance
$=0.5-0.3=0.2\mathrm{A}$