In the circuit shown, the current in 3Ω resistance is

1 A $\frac{1}{7} A$ $\frac{5}{7} A$ $\frac{15}{7} A$

$\frac{5}{7} A$

Resistance, $\mathrm{R}_{\text {net }}=2+\frac{6 \times 2}{6+2}=\frac{7}{2} \Omega$ ∴ $\frac{10}{(7 / 2)}=\frac{20}{7} \mathrm{A}$ Current through $3\Omega$ resistance is $\mathrm{i}=\left(\frac{2}{2+6}\right) \mathrm{i}=\frac{1}{4} . \mathrm{i}=\frac{5}{7} \mathrm{A}$