a
(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 \({R_4} = \frac{3}{{150}} \times \frac{{60}}{{(30 + 60)}}\)\( = \frac{3}{{150}} \times \frac{{60}}{{90}} = \frac{2}{{150}}\,A\)
\({V_4} = {i_4} \times {R_4} = \frac{2}{{150}} \times 30 = \frac{2}{5}V = 0.4\,V\)