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

Options

A. 0.052
B. 0.25
C. 0.07
D. 0.1

Answer

C

Explanation

Given voltage `V = (100 pm 5) V`
Current `I = (100 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`
Differentiatig, we get ,
`(DeltaR)/(R) = (Delta V)/(V) - (Delta I)/(I)`
For maximum error, `(Delta R)/(R) = (Delta V)/(V) + (Delta I)/(I)`
Multiplying both sides by 100 for taking percentage ,
We get, `(Delta R)/(R) xx 1000 = (Delta V)/(V) xx 100 + (Delta V)/V xx 100 + (Delta I)/(I) xx 100`
Percentage error in resistance R
` =(Delta V)/(V) xx 100 + (Delta I)/(I) xx 100`
`= 5/100 xx 100 + (0.2)/(10) xx 100 = 7%`