The potential difference across the ends of a wire has been measured to be `(100 +- 5)` volt and the current in the wire as `(10 +- 0.2)` ampere. What is the percentage error in the computed resistance of the wire?

Explanation

Given V = `(100 pm 5)` volt and `I= (10 pm 0.2)` ampere
Now `R = (V)/(I)`
The maximum percentage error in R is
`((DeltaR)/(R))_("max")xx100=((DeltaV)/(V)xx100)+((DeltaI)/(I)xx100)`
`= ((5)/(100)xx100)+((0.2)/(10)xx100)`
`=5%+2%=7%`