Current Electricity

C and D are joined by a wire, so they are at a common potential $V$ (take $V_A=0$, $V_B=50$ V). Current balance at that node: $\frac{V}{1}+\frac{V}{3}=\frac{50-V}{2}+\frac{50-V}{4}\Rightarrow \frac{4V}{3}=\frac{3(50-V)}{4}\Rightarrow V=18$ V. Into C through $1\Omega :18A$; out of C through $2\Omega :16A$. The difference, 2.0 A, flows along CD.