I am trying to calculate transformer ratio of an unknown transformer by measuring primary and secondary voltage. I have 2 standard voltmeter with small error. My job is to take the 4 readings and find the transformer ratio. enter image description here

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My manual says that for each individual reading, I should find the ratio and final ratio is average of those values.
My question is that my should I take the average value of the transformer ratios. Wouldn't taking $( average secondary voltage)/(average primary voltage)$ give more accurate value? If there are many reading (say 100) then which method would give more accurate value? [I said many readings as I think that for a few readings, either may give correct values]

  • $\begingroup$ If you know the distribution of the error of the voltmeter you can calculate an interval with some probability. Read a wikipedia page about error estimation before you should look it up. It also showed how to estimate errors in derivatives. Don't remember which page it was unfortunately. $\endgroup$
    – Emil
    Commented Feb 7, 2017 at 5:18
  • $\begingroup$ I am not quite following. I don't have a lot of knowledge in this. Any good reference book for this? $\endgroup$ Commented Feb 7, 2017 at 5:44
  • $\begingroup$ Nope I was just saying stuff from my memory. But en.m.wikipedia.org/wiki/… and links in there look like they are relevant (theory for the stuff in JamalS answer). $\endgroup$
    – Emil
    Commented Feb 7, 2017 at 20:41

2 Answers 2


Your manual is generally right, especially if the position of the tab of the left transformer varies widely and the voltmeter(s) are used at different scales depending on the position of the tab of the left transformer (including what happens with autoranging digital voltmeters, and voltmeters with manual range used correctly). In these cases, averaging the ratios of Vin/Vout readings will tend to give a more accurate result than computing the ratio of the average Vin to the average Vout.

A general rule when averaging measurements is: only average quantities that should be equal per your model. And here, assuming that you vary the position of the tab of the left transformer, both Vin and Vout will vary widely, so it makes little sense to average Vin (or Vout) measurements. On the other hand, the ratio Vin/Vout should remain about constant, so it makes sense to average the ratios of measurements.

Another way to look at this is that measurements made at high voltage will tend to dominate those at low voltage when you average the Vin, average the Vout, and then divide; thus the error-cancelling effect of making different measurements will tend to be less effective than when averaging the ratios.

This can be illustrated by an example. Assume the Vin/Vout ratio is always exactly 25; that a measurement is made at Vin=50V with +5% error on Vin and -5% error on Vout (52.50V and 1.90V); and another measurement is made at Vin=10V with -5% error on Vin and +5% on Vout (9.50V and 0.420V). The average of ratios yields 25.13, while the ratio of averages yields 26.72, which is considerably farther from 25.

On the other hand, if measurements made at low voltage have far more relative error (for example, because the same range of the voltmeter is used for all Vin measurements, and most of the error comes from an offset of the voltmeter), then de-emphasizing the measurements made at low voltage will lead to a better result, and that's a (then desirable) side effect of the ratio-of-averages method.

Thus, things can only be made rigorous with some model of the error of the voltmeters.


You are measuring $V_{\mathrm{in}}$ and $V_{\mathrm{out}}$ with errors $\alpha_{\mathrm{in}}$ and $\alpha_{\mathrm{out}}$ respectively. Now, if we form the ratio, then the error in $V_{\mathrm{in}}/V_{\mathrm{out}}$ is given by,

$$\alpha_{\mathrm{in/out}} = \sqrt{\left(\frac{V_{\mathrm{in}} + \alpha_\mathrm{in}}{V_{\mathrm{out}}}-\frac{V_{\mathrm{in}}}{V_{\mathrm{out}}} \right)^2 + \left(\frac{V_{\mathrm{in}}}{V_{\mathrm{out}}+ \alpha_\mathrm{out}} -\frac{V_{\mathrm{in}}}{V_{\mathrm{out}}}\right)^2}$$

using the functional approach to error propagation. You propose two choices; either averaging the ratio or the respective voltages and then taking a ratio.

If averaging the ratios, then the error in the average $\langle V_\mathrm{in}/V_\mathrm{out}\rangle$ is given by the standard error,

$$\sqrt{\frac{1}{N(N-1)} \sum_i \left( \frac{{V_\mathrm{in}}_i}{{V_\mathrm{out}}_i} - \langle V_\mathrm{in}/V_\mathrm{out}\rangle\right)^2}.$$

If instead you average the voltages first, then the error is the functional approach above, but using the standard error for $\alpha_\mathrm{in}$ and $\alpha_\mathrm{out}$ and the voltages are the means instead of the individual values.

Comparing these two approaches, which do you think gives a lower final error for the ratio?


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