A current of 3A flows in a circuit the potential difference between the junctions of 6$\Omega$ & 4$\Omega$ and 4$\Omega$ & 2$\Omega$.

It is a Wheatstone bridge ;

$\mathrm{Then} {\mathrm{R}}_1 = 6 + 4 =10\mathrm{\Omega }$

${\mathrm{R}}_2=3+2=5\mathrm{\Omega }$

${\mathrm{i}}_1:{\mathrm{i}}_2=\frac{1}{{\mathrm{R}}_1}:\frac{1}{{\mathrm{R}}_2}=1:2$

$\Rightarrow {\mathrm{i}}_1=\frac{3}{3}=1 \mathrm{A}$

∴ potential difference across 4Ω =  potential difference between A and D= i₁(4) = 4 V