Explanation

(d) The given circuit having parallel and series combination of resistance 3$\Omega$, we can calculate as
$R_{1} = 3 + 3 = 6\Omega$
$R_{2} = 3 + 3 = 6\Omega$
$\frac{1}{R_{3}} = \frac{1}{R_{1}} + \frac{1}{R_{2}}$
$= \frac{1}{6} + \frac{1}{6} = 3\Omega$
$R_{4} = 3\Omega$
$\frac{1}{R} = \frac{1}{R_{3}} + \frac{1}{R_{4}}$
$= \frac{1}{3} + \frac{1}{3} = \frac{2}{3}$
$R = 1.5\Omega$