# Clarification on error and uncertainty propagation

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 $$V=(V\pm \Delta V)$$.

• Sum of normally distributed random variables. Commented Nov 17, 2019 at 22:52
• 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? Commented Nov 18, 2019 at 16:41

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.
• 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? Commented Nov 18, 2019 at 17:28