Physics - Current Electricity Question with Solution | TestHub

Let the resistance of the voltmeter be $R\Omega$. The voltmeter is in parallel with the $5\Omega$. So the total resistance in the circuit is

$\left(\frac{5R}{5+R}+2\right)\Omega$

Applying Ohm's law in the whole circuit, $V=IR$,

$I=\left(\frac{3}{2+\frac{5R}{5+R}}\right)\text{A}$

It is given that the potential difference across $5\Omega$ is $2\mathrm{V}$.

Thus, by Ohm's law,

$$\begin{gathered} \left(\frac{3}{2+\frac{5R}{5+R}}\right)\times \left(\frac{5R}{R+5}\right)=2 \\ \Rightarrow \frac{15+3R}{10+7R}\times \frac{5R}{5+R}=2 \\ \Rightarrow 15R=20+14R \\ \Rightarrow R=20\Omega \end{gathered}$$