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InterviewSolution
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Two resistances `R_(1) = 100 +- 3 Omega and R_(2) = 200 +- 4 Omega` are connected in series . Find the equivalent resistance of the series combination.A. `(66.7+-1.8)Omega`B. `(66.7+-4.0)Omega`C. `(66.7+-3.0)Omega`D. `(66.7+-7.0)Omega` |
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Answer» Correct Answer - A Here, `R_(1)=(100+-3)Omega,R_(2)=(200+-4)Omega` The equivalent resistance is given by `(1)/(R_(p))=(1)/(R_(1))+(1)/(R_(2)),(1)/(R_(p))=(1)/(100)+(1)/(200)=(3)/(200),R_(p)=(200)/(3)=66.7Omega` The error in equivalent resistance given by `(DeltaR_(p))/(R_(p)^(2))=(DeltaR_(1))/(R_(1)^(2))+(DeltaR_(2))/(R_(2)^(2)),DeltaR_(p)=DeltaR_(1)((R_(p))/(R_(1)))^(2)+DeltaR_(2)((R_(p))/(R_(2)))^(2)` `=3((66.7)/(100))^(2)+4((66.7)/(200))^(2)=1.8Omega` |
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