1.

For two resistors `R_1` and `R_2`, connected in parallel, find the relative error in their equivalent resistance. if `R_1=(50 pm2)Omega ` and `R_2=(100 pm 3) Omega`.

Answer» `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


Discussion

No Comment Found

Related InterviewSolutions