A voltmeter having a resistance of 1800 Ω is employed to measure the potential difference across a 200 Ω resistor which is connected to the terminals of a DC power supply having an emf of 50 V and an internal resistance of 20 Ω. What is the percentage decrease in the potential difference across the 200 Ω resistor as a result of connecting the voltmeter across it?

$$\begin{gathered} \mathrm{Current} \mathrm{I} =\frac{50}{( 200 + 20)}=\frac{5}{22}\mathrm{A} \\ :. \mathrm{Potential} \mathrm{drop} \mathrm{across} 200 \mathrm{\Omega } \mathrm{resistor} \\ \left(\mathrm{V}\right) =\frac{5}{22}\times 200 =\frac{500}{11}\mathrm{V} \end{gathered}$$

When a voltmeter of resistance 1800 Ω is connected across the 200 Ω resistor, the effective resistance R is given by

$$\begin{gathered} \frac{1}{\mathrm{R}}=\frac{1}{1800}+\frac{1}{200}\mathrm{which} \mathrm{gives} \mathrm{R} = 180 \mathrm{\Omega } \\ \mathrm{The} \mathrm{current} \mathrm{in} \mathrm{the} \mathrm{circuit} \mathrm{becomes} \\ \mathrm{I}\text{'} =\frac{50}{180 + 20}=\frac{5}{20}\mathrm{A} \\ \mathrm{The} \mathrm{potential} \mathrm{drop} \mathrm{become} \\ \mathrm{v}\text{'}=\frac{5}{20}\times 180 = 45 \mathrm{V} \\ \mathrm{Difference} \mathrm{V} - \mathrm{V}\text{'} =\frac{500}{11}-45=\frac{5}{11}\mathrm{V} \\ \mathrm{Percentage} \mathrm{decrease} =\frac{5}{11}\times \frac{11}{500}\times 100 = 1 \%. \end{gathered}$$