A current of 2 A flows in the system of conductors as shown in the figure. The potential difference V_P - V_R will be:

-1 V

Resistance between Q and S,

$R'=\frac{5 \times 10}{5+10}=\frac{10}{3} \Omega$

Potential difference across Q and S,

$V_{Q}-V_{S}=\frac{2 \times 10}{3}=\frac{20}{3} V$

Current through arm QPS, $I_1=\frac{20}{3 \times 5}=\frac{4}{3} A$

$V_{Q}-V_{P}=\frac{4}{3} \times 2=\frac{8}{3} V$

Current through arm QRS, $I_2=\frac{20 / 3}{10}=\frac{2}{3} A$

$V_{Q}-V_{R}=\frac{2}{3} \times 3=2 V$

From equations (i) and (ii),

$V_P-V_R=\left(V_Q-V_R\right)-\left(V_Q-V_P\right)$

$=2-\frac{8}{3}=\frac{-2}{3} \approx-1 V$