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)`
`DeltaR.= (R.^2) (DeltaR_1)/( R_1^2) + (R.^2) (DeltaR_2)/( R_2^2) `
`= ((66.7)/( 100))^(2) 3+ ((66.7)/( 200) ) 4= 1.8`
Then , `R. = 66.7 pm 1.8 "ohm"` (Here, `Delta R` is expressed as 1.8 instead of 2 to keep in conformity with the rules of significant figures) .