Two resistors of resistance 1000 Ω and 2000 Ω are joined in series with a 100 V supply. A voltmeter of internal resistance 4000 Ω is connected to measure the potential difference across 1000 Ω resistor. Calculate the reading shown by the voltmeter.

Two resistors of resistance 1000 Ω and 2000 Ω are joined in series with a 100 V supply. A voltmeter of internal resistance 4000 Ω is connected to measure the potential difference across 1000 Ω resistor. Calculate the reading shown by the voltmeter.
Solution:

The equivalent reistance of \(R_1\) and \(R_v\) in parallel is, \[\begin{align*} \frac{1}{R_{1v}}&=\frac{1}{R_1}+\frac{1}{R_v}\\ &=\frac{1}{1000}+\frac{1}{4000}\\ &=\frac{5}{4000}\\ \therefore R_{1v}&=800 \hspace{0.1cm}\Omega\\ \end{align*}\] As \(R_{1v}\) and \(R_2\) are in series, the total resistance is, \[R=R_{1v}+R_2=2800 \hspace{0.1cm} \Omega\] Now,
Total voltage supply, V = 100 V.
So, \[\begin{align*} V&=IR\\ 100&=I\times 2800\\ I&=\frac{1}{28} \hspace{0.1cm} A\\ \end{align*}\] Potential difference across 1000 \(\Omega\) resistor is, \[\begin{align*} &=I\times R_{1v}\\ &=\frac{1}{28}\times 800\\ &=28.57\hspace{0.1cm} V\\ \end{align*}\] Hence, the reading shown by the voltmeter is 28.57 V.