Resistances of 6 $\mathrm{\Omega }$ each are connected in the manner shown in fig. With the current 0.5 ampere as

shown in the figure, the potential difference Vp -V_Q is

Equivalent resistance R of the three resistances connected in parallel is given by
$\frac{1}{{\mathrm{R}}^{\mathrm{\prime }}}=\frac{1}{6}+\frac{1}{6}+\frac{1}{6}=\frac{3}{6}=\frac{1}{2}\text{ or }{\mathrm{R}}^{\mathrm{\prime }}=2\mathrm{\Omega }$
The equivalent resistance R' of the second circuit
$\frac{1}{{\mathrm{R}}^{\prime \prime }}=\frac{1}{6}+\frac{1}{12}=\frac{2+1}{12}=\frac{1}{4}\text{ or }{\mathrm{R}}^{\prime \prime }=4\mathrm{\Omega }$
The resistances .R' and R' are connected in series. So, the equivalent resistance of the circuit i'e,, between A and B
R= 2+ 4= 6 $\mathrm{\Omega }$
Potential difference between. A and B
$=\mathrm{R}\times \mathrm{i}=6\times 0\cdot 5=3\mathrm{V}$