Question

$$\begin{matrix} & R_{AB} = (1\Omega/3\Omega)\text{~}\text{in series with}\text{~}(2\Omega/4\Omega) \\ & \ = \frac{3 \times 1}{3 + 1} + \frac{2 \times 4}{2 + 4} \\ & \ = \frac{3}{4} + \frac{8}{6} = \frac{9 + 16}{12} = \frac{25}{12}\Omega \end{matrix}$$

Now total current through cell

$$\begin{matrix} & I = \frac{50}{25/12} = 24\text{ }A \\ & I_{1\Omega} = \frac{3}{4} \times 24 = 18\text{ }A,I_{3\Omega} = \frac{1}{4} \times 24 = 6\text{ }A \\ & I_{2\Omega} = \frac{4}{6} \times 24 = 16\text{ }A,I_{4\Omega} = \frac{2}{6} \times 24 = 8\text{ }A \\ & \text{~}\text{Using junction rule at}\text{~}C,I_{CD} = 18 - 16 = 2\text{ }A\text{~}\text{(From}\text{~}C\text{~}\text{to}\text{~}D\text{~}\text{)}\text{~} \end{matrix}$$