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+-7ohm`
(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, `(DeltaR.)/(R.^(2))=(DeltaR_(1))/(R_(1)^(2))+(DeltaR_(2))/(R_(2)^(2))`
`DeltaR.=(R.^(2))=[(DeltaR_(1))/(R_(1))+(R.^(2))(DeltaR_(2))/(R_(2)^(2))]`
`=((66.7)/(100))^(2) 3+((66.7)/(200))^(2) 4=1.8`
Then, `R.=66.7+-1.8ohm` (Here, `DeltaR` is expressed as 1.8 instead of 2 to keep in conformity with the rules of significant figures).