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Q.
If voltage $V=(100 \pm 5) V$ and current $I=(10 \pm 0.2) A ,$ the percentage error in resistance $R$ is:
Physical World, Units and Measurements
Solution:
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 $R$
$=\frac{\Delta V }{ V } \times 100+\frac{\Delta I }{ I } \times 100$
$$
=\frac{5}{100} \times 100+\frac{0.2}{10} \times 100=7 \%
$$