b
$\mathrm{R}_{\mathrm{N}}=\frac{25}{12} \,\Omega, \mathrm{I}_{\mathrm{N}}=24 \mathrm{\,A}$
$\mathrm{I}_{1}=\frac{3}{4} \times 24=18 \mathrm{\,A}, \quad \mathrm{I}_{2}=\frac{1}{4} \times 24=6 \mathrm{\,A}$
$\mathrm{I}_{3}=\frac{4}{6} \times 24=16 \mathrm{\,A}, \quad \mathrm{I}_{4}=\frac{2}{6} \times 24=8 \mathrm{\,A}$
Current through conducting wire
$=\mathrm{I}_{1}-\mathrm{I}_{3}=2 \mathrm{\,A}$
