Find $V_P-V_Q$ in the circuit shown in figure.

In the given circuit, net resistance between P and Q points is $\mathrm{R}_{\mathrm{PQ}}$ and net resistance between Q and R points is $\mathrm{R}_{\mathrm{QR}}$, then $\mathrm{R}_{\mathrm{PQ}}=\frac{10 \times 8}{10+8}=\frac{80}{18}=\frac{40}{9} \Omega$ $\mathrm{R}_{\mathrm{QR}}=\frac{10 \times 18 \times 25}{10 \times 18+18 \times 25+25 \times 10}=\frac{4500}{880}=\frac{225}{44} \Omega$ So, circuit will be In the series circuit, voltages will be divide in the ratio of their resistances. $\begin{aligned} & \therefore \quad \frac{\mathrm{R}_{\mathrm{PQ}}}{\mathrm{R}_{\mathrm{QR}}}=\frac{\mathrm{V}_{\mathrm{PQ}}}{\mathrm{V}_{\mathrm{QR}}} \\ & \therefore \quad \frac{\left(\frac{40}{9}\right)}{\left(\frac{225}{44}\right)}=\frac{\mathrm{V}_{\mathrm{PQ}}}{\mathrm{V}_{\mathrm{QR}}} \text { or } \frac{1760}{2025}=\frac{\mathrm{V}_{\mathrm{PQ}}}{\mathrm{V}_{\mathrm{QR}}} \\ & \text { or } \quad \frac{1760}{3785}=\frac{V_{P Q}}{10} \Rightarrow V_{P Q}=4.65 \mathrm{~V} \\ & \end{aligned}$