In Ohm's law experiment, potential drop across a resistance was measured as `v=5.0` volt and current was measured as `i=2.00amp`. Find the maximum permissible error in resistance.

Options

A. `1.5%`
B. `2.5%`
C. `1%`
D. `5%`

Answer

B

Explanation

`R=(v)/(i)=vxxiimplies((dR)/(R ))_(max)=(Deltav)/(v)+(Deltai)/(i)`
`v=5.0 "volt to" Deltav=0.1 "volt"`
`i=2.00amp "to" Deltai=0.01 "amp"`
`%((dR)/(R ))_(max)=((0.1)/(5.0)+(0.01)/(2.00))xx100%=2.5%`
value of `R` from the observation
`R=(v)/(i)=(5.0)/(2.00)=2.5Omega`
So we can write `R=(2.5+-2.5%)Omega`