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So I have to find the maximum possible error $dR$ in calculating equivalent resistance for three resistors, $\displaystyle\frac{1}{R}=\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}$ if the maximum error in each is 0.5%; $R_1=25\ \Omega$, $R_2=45\ \Omega$, $R_3=50\ \Omega$.

Now, originally I did $dR_1=dR_2=dR_3=0.005$, and then did $\displaystyle\frac{dR}{R^2}=\frac{dR_1}{R_1^2}+\frac{dR_2}{R_2^2}+\frac{dR_3}{R_3^2}$ and solved for $dR$... now I realize now that that doesn't make any sense.

I recall when doing an example problem we took the derivative like normal except when doing the chain rule, replacing it with the partial; for example, when $A=\ell w$, $dA = \frac{\partial A}{\partial \ell} d\ell+\frac{\partial A}{\partial w} dw$ (finding error in area of rectangle) and plugged in what I know. How would that work here? Was I close in my original attempt? I feel like I'm not sure where to put all the partials now that there's a bunch of reciprocals everywhere.

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up vote 3 down vote accepted

The dRi denotes the absolute error, but that 0.5% is a relative error. Therefore, you should use

\begin{align} dR_1 &= 0.5\% \times R_1 = 0.125\, \Omega \\ dR_2 &= 0.5\% \times R_2 = 0.225\, \Omega \\ dR_2 &= 0.5\% \times R_3 = 0.25\, \Omega \end{align}

The rest is correct for obtaining a conservative value of dR.

(However, if you can ensure the quantities are distributed normally and uncorrelated, the errors are added together quadratically: $$ \Delta(A+B) = \sqrt{\Delta A^2 + \Delta B^2}, $$ i.e. $$ \left(\frac{\Delta R}{R^2}\right)^2 = \left(\frac{\Delta R_1}{R_1^2}\right)^2 + \left(\frac{\Delta R_2}{R_2^2}\right)^2 + \left(\frac{\Delta R_3}{R_3^2}\right)^2 $$

This may be too advanced for your course.)

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This is easy, since $R$ is monotonic as a function of $R_1,R_2,R_3$ (which is obvious from physical considerations: increasing one of the resistances can only increase the total resistance). Thus, to compute error bounds for $R$, it suffices to consider the minimum and maximum possible values for your three resistors: the minimum corrected value of $R$ is $f(0.995R_1, 0.995R_2, 0.995R_3)$, and the maximum corrected value is $f(1.005R_1, 1.005R_2, 1.005R_3)$, where $f(x,y,z) = (x^{-1} + y^{-1} + z^{-1})^{-1}$.

But a further simplification is that $f$ is homogeneous of degree 1, i. e., $f(cx, cy, cz) = cf(x,y,z)$ (again, this is obvious intuitively: choosing different units for your resistances multiplies both the individual and total resistances by the same proportionality constant). Thus, $f(0.995R_1, 0.995R_2, 0.995R_3) = 0.995R$, and $f(1.005R_1, 1.005R_2, 1.005R_3) = 1.005R$. Thus, the maximum relative error in the total resistance is also 0.5%.

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I got this result by just setting $dR_n=\frac{R_n}{200}$ but when I first read it it sounded way too complex. Thanks though! – Eugene Bulkin Oct 5 '10 at 21:55
@Eugene: Beware: your method won't generally give you an error bound; it only provides an "estimate" (which can generally be arbitrarily bad). You got lucky in this case because of the degree-1 homogeneity of function you're estimating. Sometimes an estimate is actually what you want (as opposed to a hard error bound), but since you asked for the maximum possible error, it's important to know that your method is incorrect. – Darsh Ranjan Nov 5 '10 at 20:25

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