Question

If voltage $V=(100\pm 5)V$ and current $I=(10\pm 0.2)A,$ the percentage error in resistance $R$ is:

• A. $5.2\%$
• B. $25\%$
• C. $7\%$
• D. $10\%$

Answer 1

Answer: (C)

Given : Voltage $V=(100\pm 5)V$
Current $I=(10\pm 0.2)A$
According to ohm's law, $V=IR$ or $R=V/I$ Taking $\log$ on both sides, $\log R=\log V-\log I$
Differentiating, we get;
\frac{\Delta R}{R}=\frac{\Delta V}{V}-\frac{\Delta I}{I}
For maximum error, $\frac{\Delta R}{R}=\frac{\Delta V}{V}+\frac{\Delta I}{I}$
Multiplying both sides by 100 for taking percentage, We get, $\frac{\Delta R}{R}\times 100=\frac{\Delta V}{V}\times 100+\frac{\Delta I}{I}\times 100$
Percentage error in resistance $R$
$=\frac{\Delta V}{V}\times 100+\frac{\Delta I}{I}\times 100$

=\frac{5}{100} \times 100+\frac{0.2}{10} \times 100=7 \%