Two resistors of resistances `R_1 = 100 pm 3` ohm and `R_2 = 200 pm 4 ` ohm are connected (a) in series , (b) in parrallel. 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 (DeltaR')/(R^(,2)) = (DeltaR_1)/(R_1^2) + (DeltaR_2)/(R_2^2)`

Explanation

a) The equivalent resistance of series combination
`R=R_(1)+R_(2)=(100 pm 3)`
ohm + (200 `pm` 4) ohm = 300 `pm` 7 ohm.
b) The equivalent resistance of parallel combination
`R.=(R_(1)R_(2))/(R_(1)+R_(2))=200/3=6.67ohm`
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)^(2))+(R.^(2))(DeltaR_(2))/(R_(2)^(2))`
`" "=((6.67)/100)^(2)3+((66.7)/200)^(2)4=1.8`
`" "`Then, `R.=66.7 pm 1.8 ohm`
(Here, `DeltaR` is expresse as 1.8 instead of 2 to keep in conformity with the rules of significant figures.)