Two resistors of resistances `R_(1)=100pm3` ohm and `R_(2)=200pm4` ohm are connected (a) 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 `(DeltaR')/(R'^(2)) = (DeltaR_(1))/(R_(1)^(2))+(DeltaR_(2))/(R_(2))`

Explanation

(a)The equivalent resistance of series combination
`R = R_(1)+R_(2) = (100 pm 3)` ohm + `(200 pm4)` ohm
` = 300pm7` 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)^(2))4`
1.8
Then, R. = `66.7pm1.8` ohm
(Here, `DeltaR` is expressed as 1.8 instead of 2 to keep in confirmity with the rules of significant figures.)
Error in case of a measured quantity raised to a power
Suppose Z = `A^(2)`,
Then,
`DeltaZ//Z=(DeltaA//A)+(DeltaA//A)=2(DeltaA//A)`.
Hence, the relative error in `A^(2)` is two times the error in A.
In general , if `Z=A^(p)B^(q)//C^(r)`
Then,
`DeltaZ//Z = p(DeltaA//A)+q(DeltaB//B)+r(DeltaC//C)`.