If voltage `V=(100pm5)` V and current `I=(10pm0.2)` A, the percentage error in resistance `R` is

Options

A. `5.2`%
B. 0.25
C. 0.07
D. 0.1

Answer

C

Explanation

Given, voltage, `V=(100pm5)`V
Current, `I=(10pm0.2)`A
`therefore` Resistance, `R=(V)/(I)`
Maximum percentage error in resistance
`((DeltaR)/(R)xx100)=((DeltaV)/(V)xx100)+((DeltaI)/(I)xx100)`
=`((5)/(100)xx100)+((0.2)/(10)xx100)` = 5+2=7%