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, $$\begin{array}{rl}\frac{1}{R_{1v}}& =\frac{1}{R_1}+\frac{1}{R_v}\\ & =\frac{1}{1000}+\frac{1}{4000}\\ & =\frac{5}{4000}\\ \therefore R_{1v}& =800\mathrm{\Omega }\end{array}$$ As $R_{1v}$ and $R_2$ are in series, the total resistance is, $$R=R_{1v}+R_2=2800\mathrm{\Omega }$$ Now,
Total voltage supply, V = 100 V.
So, $$\begin{array}{rl}V& =IR\\ 100& =I\times 2800\\ I& =\frac{1}{28}A\end{array}$$ Potential difference across 1000 $\mathrm{\Omega }$ resistor is, $$\begin{array}{rl}& =I\times R_{1v}\\ & =\frac{1}{28}\times 800\\ & =28.57V\end{array}$$ Hence, the reading shown by the voltmeter is 28.57 V.