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

Now total current through cell

$$\begin{aligned} \& I=\frac{50}{25 / 12}=24 \mathrm{~A} \\ \& I_{1 \Omega}=\frac{3}{4} \times 24=18 \mathrm{~A}, I_{3 \Omega}=\frac{1}{4} \times 24=6 \mathrm{~A} \\ \& I_{2 \Omega}=\frac{4}{6} \times 24=16 \mathrm{~A}, I_{4 \Omega}=\frac{2}{6} \times 24=8 \mathrm{~A} \end{aligned}$$

Using junction rule at $C, I_{C D}=18-16=2 \mathrm{~A}$ (From $C$ to $D$)