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For two resistors R(1)andR(2), connected...

For two resistors `R_(1)andR_(2)`, connected in parallel, find the relative error in their equivalent resistance, if `R_(1)=(50pm2)OmegaandR_(2)=(100pm3)Omega`.

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`R_"eq"=(R_1R_2)/(R_1+R_2)=33.33 Omega` and `1/R_"eq"=1/R_1+1/R_2` we have `(DeltaR_"eq")/(R_"eq"^2)=(DeltaR_1)/(R_1^2)+(DeltaR_2)/(R_2^2)`
`rArr (DeltaR_"eq")/R_(eq)=((DeltaR_1)/(R_1^2)+(DeltaR_2)/(R_2^2))R_"eq"` or 0.036
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