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. $0.4 \%$

Answer 1

Answer: (B)

$I=\frac{E}{500+4}=\frac{E}{504}$
Reading of voltmeter $=500 I=\frac{500 E}{504}$

$\%$ error $=\frac{E-\frac{500 E}{504}}{E} \times 100$
$=\frac{400}{504} \simeq 0.8 \%$