Calculating uncertainty when combining parallel resistors A recent question asks students to calculate the total resistance of parallel resistors R1 and R2 with uncertainty. R1 = 120 Ω (+/-10%). R2 = 500 Ω (+/-10%).
We would normally use the equation R = ( R1-1 + R2-1 )-1
The exam board's answer is 97 Ω (+/-10%). I understand the value for total resistance but can someone please give the simplest rule to explain the 10% uncertainty with some mathematical reasoning? 
 A: The largest value of $R$ will be when $R_1$ and $R_2$ have their largest values. So putting $R_1= 132\ \Omega$ and $R_2= 550\ \Omega$, we get 106.5 $\Omega$, whereas if we put in the 'central' values, $R_1= 120\ \Omega$ and $R_2= 500\ \Omega$, we get 97 $\Omega$. The discrepancy as a percentage of 97 $\Omega$, is 10%.
The moral: elementary methods are sometimes the best! You can do this calculation by compounding individual uncertainties:
The percentage uncertainties in $\frac{1}{R_1}$ and $\frac{1}{R_2}$ are the same as in $R_1$ and $R_1$ because taking the reciprocal is a division. So, as Michael Seifert has calculated,$$G_1=\frac{1}{R_1}=(8.33±0.83)\times 10^{-3} \Omega^{-1}$$
$$G_2=\frac{1}{R_2}=(2.00±0.20)\times 10^{-3} \Omega^{-1}$$
We now have to add these conductances, so (in this naïve treatment) we add the absolute uncertainties. [We'd also add absolute uncertainties if we were subtracting quantities.]
$$\text{So}\ \ \ \ G=G_1+G_2=(10.33±1.03)\times 10^{-3} \Omega^{-1}$$
The percentage uncertainty in $G$ is therefore 10%. Taking the final reciprocal (a division operation) leaves the percentage uncertainty as 10%.
$$\text{So}\ \ \ \ R=\frac{1}{G}=(97±10) \Omega.$$
For divisions and multiplications you add percentage uncertainties, so if there is only one variable, as in $\frac{1}{R_1},$ the percentage uncertainty stays as it is. For quantities that are added or subtracted you add absolute uncertainties. For multiplication and division the addition of percentage uncertainties is, strictly, a first order approximation, but the whole treatment is oversimplified anyway (see M Seifert's answer).
A: There is no mathematical reason to give a 10% percent error on the final resistance; there is, however, and engineering one.
First: If the percentage number encoded in the resistors' stripes were a 1-sigma Gaussian error, then @Michael Seifert's answer would be correct: add the resistors harmonically and the errors are reduced in quadrature.
But: Those number are not errors, they are tolerances. So $R=100\pm10\,\Omega$ is supposed to mean:
$$ R({\rm Ohm}) \in [90, 110] $$
In which case, @Philip Wood's answer is a better answer: to get the range, use the range.
But there is a problem: there are 5% and 1% resistors too. And they aren't included in the 10% stock. Hence, your 100 Ohm resistors exclude any ones that satisfies:
$$ R({\rm Ohm}) \in [95, 105] $$
which makes for bimodal input distributions (this is simply a terrible exam question, with a wrong answer, btw).
At this point, you have to Monte Carlo it, with the result that:
$$ R = 96.6\pm 6.2\,\Omega  $$
and
$$ 87 < R < 106.5 $$
See figure.

(note: a probability density per Ohm is in fact measured in Siemens ;-)
A: You can find the uncertainty if you carefully calculate things step by step.  The conductance $G = 1/R$ of each resistor is
$$
G_1 = \frac{1}{R_1} = (8.33 ± 0.83) \times 10^{-3} \, \mho 
$$ $$
G_2 = \frac{1}{R_2} = (2.00 ± 0.20) \times 10^{-3} \, \mho
$$
Note that the fractional uncertainties in the conductances are the same as those in the resistances (i.e., 10%).  From here, you can calculate the uncertainty in the total conductance $G_{eq} = G_1 + G_2$, and then calculate the uncertainty in the equivalent resistance.
As an aside:  if the uncertainties in the resistances are independent and uncorrelated, the uncertainty in $R_{eq}$ will actually be smaller than 10%, since the absolute uncertainties in $G_{eq}$ add in quadrature rather than directly.  Your students may or may not be expected to know this;  but if the given answer is 10%, I expect they're not expected to know it.
