Two conductors of resistance `R_(1) (50 +- 2)` ohm and `R_(2) = (100 +- 4)` ohm are connected in (a) series and (b) parallel. Find the equivalent resistance.

Explanation

(a) `R_(eq) = R_(1) + R_(2) = (50 + 2) + (100 +- 4)`
`= (150 +- 6) ohm`
(b) `R_(eq) = (R_(1) R_(2))/(R_(1) + R_(2)) = (50 xx 100)/(50 + 100) = (100)/(3) ohm = 33.3 ohm`
`(1)/(R_(eq)) = (1)/(R_(1)) + (1)/(R_(2))`
`(Delta R_(eq))/(R_(eq)) = (Delta R_(1))/(R_(1)^(2)) + (Delta R_(2))/(R_(2))`
`Delta R_(eq) = (R_(eq)^(2)) ((Delta R_(1))/(R_(1)^(2))) + (R_(eq)^(2)) ((Delta R_(2))/(R_(2)^(2)))`
`= ((100)/(3))^(2) (2)/((50)^(2)) + ((100)/(3))^(2) (4)/((100)^(2))`
`= (8)/(3) + (4)/(9) = (24 + 4)/(9) = (28)/(9) = 3.1`
`R_(eq) = (33.1 +- 3.1) ohm`