Explanation

(b) The equivalent circuit of the given circuit is as shown
Resistances 6$\Omega$ and 2$\Omega$ are in parallel
$\therefore\ \ \ \ \ R^{'} = \frac{6 \times 2}{6 + 2} = \frac{3}{2}\Omega$
Resistances $\frac{3}{2}\Omega\ and\ 1.5\Omega\ are\ in\ series$
$\therefore\ \ \ \ R^{''} = \frac{3}{2} + 1.5 = 3\Omega\ $
Resistances 3$\Omega$ and 3$\Omega$ are in parallel
$\therefore\ \ \ \ \ R = \frac{3 \times 3}{3 + 3} = \frac{3}{2}$
The current, $I = \frac{V}{R}$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = \frac{9}{3/2} = 6A$