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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 isA. 0.03B. `4.5%`C. `0.67%`D. `1.33%` |
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Answer» Correct Answer - 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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