Explanation

(c) As 3$\Omega$ and 6$\Omega$ resistances are
In parallel their equivalent resistance will be 2$\Omega$. Here 2$\Omega$ and 4$\Omega$ are in series, their equivalent resistance will be 6$\Omega$. From current distribution law
$i_{1} = \frac{12 \times 18}{18} = 12A\ $
$i_{2} = \frac{6 \times 18}{18} = 6A$
Now, 12A current is entering in parallel combination of 3$\Omega$ and $6\Omega$ again from current distribution law
$i_{1}^{''} = \frac{6 \times 12}{9} = 8A\ $
$i_{2}^{''} = \frac{3 \times 12}{9} = 4A.$
∴ Potential difference across 3$\Omega$ resistance
=8×3=24V