In the given circuit (figure), 

Resistors 2 $\mathrm{\Omega }$ and 2 $\mathrm{\Omega }$ are in series, and 2 $\mathrm{\Omega }$ and 4 $\mathrm{\Omega }$ are in series, then their resultant are in parallel. Produce net resistance .R = 2.4 $\mathrm{\Omega }$.

$\begin{array}{l}\mathrm{I}=\frac{20}{\mathrm{r}+\mathrm{R}}=\frac{20}{1.6+2.4}=5\mathrm{A}\\ {\mathrm{I}}_1+{\mathrm{I}}_2=5\mathrm{A} .........\left(\mathrm{i}\right)\\ {\mathrm{V}}_{\mathrm{AB}}=4{\mathrm{I}}_1=6{\mathrm{I}}_2\text{ or }2{\mathrm{I}}_1=3{\mathrm{I}}_2 ...........\left(\mathrm{ii}\right)\end{array}$

From Eqs. (i) and (ii), ${\mathrm{I}}_1=3\mathrm{A},{\mathrm{I}}_2=2\mathrm{A}$

${\mathrm{V}}_{\mathrm{A}}-{\mathrm{V}}_{\mathrm{P}}=6\mathrm{V},{\mathrm{V}}_{\mathrm{A}}-{\mathrm{V}}_{\mathrm{Q}}=2{\mathrm{I}}_2=4\mathrm{V}$

Thus, ${\mathrm{V}}_{\mathrm{Q}}-{\mathrm{V}}_{\mathrm{P}}=2\mathrm{V}$