The values of two resistors are `R_(1)=(6+-0.3)kOmega` and `R_(2)=(10+-0.2)kOmega`. The percentage error in the equivalent resistance when they are connected in parallel is

Options

A. `5.125%`
B. `2%`
C. `10.125%`
D. `7%`

Answer

C

Explanation

`R=(R_1R_2)/((R_1+R_2))=(10xx6)/(10+6)=60/10kOmega=3.75 KOmega`
`implies (DeltaR)/R=(DeltaR_1)/R_2+(DeltaR_2)/R_2+(Delta(R_1+R_2))/((R_1+R_2))`
`=(0.3)/(6)+(0.2)/(10)+((0.3 +0.2))/(10+6) = 0.10125`
`implies (DeltaR)/R = 10.125%`