Physics - Current Electricity Question with Solution | TestHub

The given circuit diagram can be drawn as shown below
The equivalent resistance of circuit is given by $$\begin{array}{rl}& \frac{\mathrm{l}}{{\mathrm{R}}_{\mathrm{eq}}}=\frac{\mathrm{l}}{{\mathrm{R}}_{\mathrm{AE}}}+\frac{\mathrm{l}}{{\mathrm{R}}_{\mathrm{DF}}}\\ & =\frac{1}{(20+30)}+\frac{1}{(30+20)}\\ & =\frac{1}{50}+\frac{1}{50}=\frac{2}{50}\\ & \Rightarrow {\mathrm{R}}_{\mathrm{eq}}=25\Omega \\ & \end{array}$$ As the resistance of two branches is same i.e. $50\Omega$. So, the current ${\mathrm{I}}_1={\mathrm{I}}_2$ $$\Rightarrow \begin{array}{rl}\mathrm{I}& ={\mathrm{I}}_1+{\mathrm{I}}_2\\ \frac{2}{25}& =2{\mathrm{I}}_1\Rightarrow {\mathrm{I}}_2={\mathrm{I}}_1=\frac{1}{25}\mathrm{A}\end{array}$$ $\therefore$ The voltage across $\mathrm{AB}$ $${\mathrm{V}}_1={\mathrm{I}}_1{\mathrm{R}}_1=\frac{1}{25}\times 20$$ and voltage across $CD$ $$\begin{array}{rl}{\mathrm{V}}_2=& {\mathrm{I}}_2{\mathrm{R}}_2=\frac{1}{25}\times 30\\ \therefore \text{ Voltmeter reading }& ={\mathrm{V}}_2-{\mathrm{V}}_1\\ & =\frac{30}{25}-\frac{20}{25}\\ & =\frac{10}{25}=0.4\mathrm{V}\end{array}$$