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.
Best Answer
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.)