Physics - Current Electricity Question with Solution | TestHub

From the given figure, current through lower branch of resistances which are joined in series is $i_1=\frac{10}{4+3}=\frac{10}{7}\mathrm{amp}$Again current through upper branch of resistances which are also connected in series, is $i_2=\frac{10}{8+6}=\frac{10}{14}\mathrm{amp}$ Now according to the Kirchhoff's voltage law $\begin{array}{rl}{V}_B& -V_A=8\times i_2-4\times i_1\\ & =8\times \frac{10}{14}-4\times \frac{10}{7}=0\end{array}$