Question

A voltmeter with resistance 500 $\Omega$ is used to measure the emf of a cell of internal resistance 4 $\Omega$. The percentage error in the reading of the voltmeter will be

• A. 0.2%
• B. 0.8%
• C. 1.4%
• D. 2.2%

Answer 1

The correct option is B 0.8%

$i=\frac{\epsilon }{504};V=\epsilon -4i=\epsilon -\frac{\epsilon }{504}\times 4$

$\Rightarrow V=\epsilon (1-\frac{1}{126})=\epsilon \frac{125}{126}\Rightarrow$ % error

$=\frac{1}{126}\times 100=0.8$%