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`

Two resistances `R_(1) = 100 +- 3 Omega and R_(2) = 200 +- 4 Omega` ar

1.	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`
Answer» Correct Answer - a Here `R_(1) = (100 +-3)Omega` `R_(2) = (200 +-4)Omega` The equaivalent resistent in parallel combination is `(1)/(R_(p))= (1)/(R_(1)) + (1)/(R_(2))` `(1)/(R_(p))=(1)/(100) +(1)/(100)= (1)/(100)` `R_(p) = (200)/(3) = 66.7 Omega` This error in equaivalent resistent is given by `(DeltaR_(p))/(R_(1)^(2)) + (DeltaR_(2))/(R_(2)^(2))` `DeltaR_(P) = DeltaR_(1) ((R_(p))/(R_(1)^(2)))^(2) + DeltaR_(2)((R_(p))/(R_(2)))^(2)` `=3((66.7)/(100))^(2) + 4 ((66.7)/(200))^(2) = 1.8 Omega` Hence, the equivalent resistance along with error parallel combination is `(66.7 +- 1.8) Omega`.

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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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