This is a purely fictional example: suppose I want to find the voltage drop across a resistor $R$ where a current $I$ flows. My resistance supplier established a tolerance of 5% for $R$, so I know that the true value of my resistance is somewhere between $[R-0.05R;R+0.05R]$ (let's call $\Delta R=0.05R$). Also the measurement device for the current only gives me values until the second decimal place, that is, the value of current that I measure is somewhere between $[I-0.01;I+0.01]$ ($\Delta I=0.01$). That being said, what is the error/uncertainty in $V=IR$?

One could argue that: $$ \Delta V=\bigg|\frac{\partial V}{\partial I}\bigg|\Delta I+\bigg|\frac{\partial V}{\partial R}\bigg|\Delta R $$

which makes sense, because it gives me a rectangle of possible values for my new variable $V$. But I have also seen (and used extensively) the uncertainty propagation formula:

$$ \Delta V=\sqrt{\bigg(\frac{\partial V}{\partial I}\Delta I\bigg)^2+\bigg(\frac{\partial V}{\partial R}\Delta R\bigg)^2} $$

This also makes some sense, in a way that it gives me an ellipse of possible values for $V$, but which one is more appropriate and in which cases? What is the difference between these two formulas?

The goal is to present the result as $V=(V\pm \Delta V)$.

  • $\begingroup$ Sum of normally distributed random variables. $\endgroup$
    – Farcher
    Nov 17, 2019 at 22:52
  • $\begingroup$ Yes, that helps but only understanding the second expression. In the sense that it gives me standard deviation of my variable V, that is, I will get $V\pm \Delta V$ 68% of the times. But what about the first one. How does it differ? Does the first assume a uniform distribution? $\endgroup$
    – Bidon
    Nov 18, 2019 at 16:41

1 Answer 1


Your choice of an example considering resistance complicates matters which I will mention later.

If the error is the maximum error then using the first equation you are finding an error based on the assumption that both values have a maximum error simultaneously which is not a very likely event.
This means that the error found this way is an over estimate.

For resistor the complication is that a supplier quotes a tolerance as a percentage and usually produces resistors with a series of tolerances, eg $1\%,\,2\%, \,5\%,\,10\%,$ etc.

So if you have some $5\%$ $100\Omega$ resistors it is possible/probable that your resistors are all have values in the range $95\Omega$ to $98\Omega$ and $102\Omega$ to $105\Omega$ because all the resistors with a smaller tolerance have been removed from the batch for sale at a higher price.
It is also possible that your batch of $5\%$ $100\Omega$ resistors contain only resistors in the range $95\Omega$ to $98\Omega$ because all resistor above the nominal value have been "trimmed" during processing to a value closer to the nominal value.

  • $\begingroup$ could we say that the second equation assumes a normal distribution of the values of $I$ and $R$ and so it gives me standard deviation of my variable V, that is, I will get V±ΔV 68% of the times, whereas the first one assumes a uniform distribution for both variables with some maximum deviation? $\endgroup$
    – Bidon
    Nov 18, 2019 at 17:28

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