Explanation

(d) Here, 2$\Omega$ and 2$\Omega$ are in parallel
$\therefore\ \ \ \frac{1}{R} = \frac{1}{2} + \frac{1}{2}\ \ $
$R = \frac{2 \times 2}{2 + 2} = 1\Omega$
Now, internal resistance (1$\Omega$), 2$\Omega$, 4$\Omega$ and resistance R in series.
$\therefore R_{net} = 1\Omega + 2\Omega + 4\Omega + 1\Omega = 8\Omega$
Hence, current
$I = \frac{V}{R} = \frac{4}{8} = 0.5A$