Two resistances are measured in ohm and are given as:
`R_(1)=3Omegapm1%`
`R_(2)=6Omegapm2%`
When they are connected in parallel, the percentage error in equivalent resistance is

Options

A. 0.03
B. `4.5%`
C. `0.67%`
D. `1.33%`

Answer

D

Explanation

`(1)/(R_(eq))=(1)/(R_(1))+(1)/(R_(2))rArr(dR_(eq))/(R_(eq)^(2))=(dR_(1))/(R_(1)^(2))+(dR_(2))/(R_(2)^(2))` ltbr. Given `dR_(1)//R_(1)=1//100,dR_(2)//R_(2)=2//100`
`(dR_(eq))/(R_(eq))=R_(eq){(dR_(1))/(R_(1)).(1)/(R_(1))+(dR_(2))/(R_(2)).(1)/(R_(2))}`
% error `=2{(1)/(3)+(2)/(6)}=4//3%`