Question

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 1

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`.