Physics - Current Electricity Question with Solution | TestHub

Balanced wheatstone bridge condition $$\frac{P}{Q}=\frac{R}{S}$$ No, current flows through galvanometer Now, $\mathrm{P}$ and $\mathrm{R}$ are in series, so Resistance ${\mathrm{R}}_1=\mathrm{P}+\mathrm{R}$ $$=10+15=25\Omega$$ Similarly, Q and S are in series, so Resistance ${\mathrm{R}}_2=\mathrm{R}+\mathrm{S}$ $=20+30=50\Omega$ Net resistance of the network as $R_1$ and $R_2$ are in parallel $$\begin{array}{l}\frac{1}{\mathrm{R}}=\frac{1}{{\mathrm{R}}_1}+\frac{1}{{\mathrm{R}}_2}\\ \therefore \mathrm{R}=\frac{25\times 50}{25+50}=\frac{50}{3}\Omega \end{array}$$ Hence, current, $\mathrm{I}=\frac{\mathrm{V}}{\mathrm{R}}=\frac{6}{50/3}=0.36\mathrm{A}$