Explanation

(d) The equivalent circuit is shown as
We can emit the resistance in the arm DF as balance condition is satisfied.
Therefore, the 3$\Omega$ resistances in arm CD and DE are in series.
$\therefore\ \ \ \ \ \ \ \ \ \ \ \ \ \ R^{'} = 3 + 3 = 6\Omega$
Similarly, for arms CF and FE, R’’=6$\Omega$
$R^{'\ }and\ R''$ are in parallel
$\therefore\ \ \frac{1}{R^{'''}} = \frac{1}{6} + \frac{1}{6} = \frac{2}{6} = \frac{1}{3}$
R’’’=3$\Omega$
Now, R’’’ and 3$\Omega$ resistances are in parallel
$\therefore\ \ \frac{1}{R} = \frac{1}{3} + \frac{1}{3}$
$\Longrightarrow R = 1.5\Omega$
Moreover, V across AB=3V and resistance in the arm=3$\Omega$
∴ Current through the arm will be
$= \frac{3V}{3\Omega} = 1A.$