Here. V = 100 ± 2 volt and I = 10 ± 0.5 ampere
Expressing limits of error as percentage error,
We have,
V = 100 volt ± \(\frac{2}{100}\) × 100% = 10 volt ± 2% and I = 10 ampere ± \(\frac{0.5}{10}\) × 100%
= 10 ampere ± 5%
∴ R = \(\frac{V}{I}\)
∴ %error in R = %error in V + %error in I
= 2% + 5% = 7%