\(\frac{1}{\text{R}}\) = \(\frac{1}{2}\) + \(\frac{1}{3}\) = \(\frac{5}{6}\) R = \(\frac{6}{5}\) = 1.2\(\Omega\); So p.d through the 3\(\Omega\) and 2\(\Omega\) resistors in parallel = 1.5 x 1.2 =1.8V So the current through the 3\(\Omega\) resistor(I) = \(\frac{\text{V}}{\text{R}}\) = \(\frac{1.8}{3}\) = 0.6A (recall V = IR)