Explanation

(d) In the given circuit the resistors of 2$\Omega$ and 3$\Omega$ are connected in parallel hence, equivalent resistance is
$\frac{1}{R^{'}} = \frac{1}{2} + \frac{1}{3} = \frac{5}{6}$
$\therefore R^{'} = \frac{6}{5}\Omega$
Also in steady state, the circuit is shown as. Resistor’s of
$\frac{6}{5}\Omega$ and 2.8$\Omega$ are connected in series.
Hence, $R’’ = \frac{6}{5}\Omega + 2.8\Omega$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = 1.2\Omega + 2.8\Omega = 4.0\Omega$
From Ohm’s law,
$Current\ i = \frac{V}{R} = \frac{6}{4.0} = 1.5A$