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) = 66.7 ` ohm
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)`
`=((66.7)/(100))^2 3 + ((66.7)/(200))^2 4 = 1.8`
Then , `R. = 66.7 `pm `1.8 ohm (Here , ` DeltaR` is expressed as 1.8 instead of 2 of keep in conformity with the rules of significant figures.