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Two resistance are measured in ohm and i...

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

A

0.03

B

`4.5%`

C

`0.67%`

D

`1.33%`

Text Solution

Verified by Experts

The correct Answer is:
D
`(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%`
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