Explanation

(a) Equivalent resistance of the given network
$R_{eq} = 75\Omega$
∴ Total current through battery,
$i = \frac{3}{75}$
$i_{1} = i_{2} = \frac{3}{75 \times 2} = \frac{3}{150}$
Current through resistance
$R_{4} = \frac{3}{150} \times \frac{60}{(30 + 60)}$
$\ \ \ \ \ = \frac{3}{150} \times \frac{60}{90}$
$\ \ \ \ \ = \frac{2}{150}A$
$V_{4} = i_{4} \times R_{4}$
$\ \ \ \ \ \ = \frac{2}{150} \times 30$
$\ \ \ \ \ \ = \frac{2}{5} = 0.4\ volt$