Explanation

(c) The given circuit may be redrawn as shown in adjacent figure. Resistance of parallel combination of $2\Omega$ and $6\Omega$,
$R_{1} = \frac{2 \times 6}{2 + 6} = 1.5\Omega$
Now resistance of $ABC$ arm$= 1.5 + 1.5 = 3\Omega$
and total network resistance $R = \frac{3 \times 3}{3 + 3} = 1.5\Omega$
$\therefore$ Total current supplied by the battery $i = \frac{6V}{1.5\Omega} = 4A$