Explanation

(d) In the given circuit 4$\Omega$ resistors are connected in parallel, this combination is connected in series with 1$\Omega$ resistance.
$\therefore\ \frac{1}{R^{'}} = \frac{1}{4} + \frac{1}{4} = \frac{2}{4} = \frac{1}{2}$
$\Longrightarrow \ \ R^{'} = 2\Omega$
Also, R’’=2 $\Omega$ +1$\Omega$ =3$\Omega$
From Ohm’s law, $V = iR$
$\therefore\ \ \ i = \frac{V}{R} = \frac{6}{3} = 2A$