Two resistors of resistances `R_(1)=100 pm 3` ohm and `R_(2)=200 pm 4` ohm are connected (a) in series, (b) in parallel. Find the equivalent resistance of the (a) series combination, (b) parallel combination. Use for (a) the relation `R=R_(1)+R_(2)` and for (b) `1/(R')=1/R_(1)+1/R_(2)` and `(Delta R')/R'^(2)=(Delta R_(1))/R_(1)^(2)+(Delta R_(2))/R_(2)^(2)`

Explanation

(a) The equivalent resistance of series combination
`R = R_(1) + R_(2) = (100 +- 3)ohm + (200 +- 4)ohm = (300 +- 7)ohm`.
(b) The equivalent resistance of parallel combination
`R' = (R_(1)R_(2))/(R_(1) + R_(2)) = (200)/(3) = 66.7ohm`
Then, from `(1)/(R') = (1)/(R_(1)) + (1)/(R_(2))`
we get, `(Delta R')/(R'^(2)) = (Delta R_(1))/(R_(1)^(2)) + (Delta R_(2))/(R_(2)^(2))`
`Delta R' = (R'^(2)) (Delta R_(1))/(R_(1)^(2)) + (R'^(2)) (Delta R_(2))/(R_(2)^(2))`
`= ((66.7)/(100))^(2) 3 + ((66.7)/(200))^(2) 4 = 1.8`
Then, `R' = (66.7 +- 1.8) ohm`