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

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) = (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 pm 1.8 ohm` (Here, `Delta R` is expressed are 1.8 instead of 2 to keep in conformity with the rules of significant figures).