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When watching a video by Veritasium about the SI units redefinition (5:29), a claim that the volt and unit of resistance (presumably the ohm) will change by about 1 part in 10 million caught my attention:

[...] I should point out that a volt will actually change by about 1 part in 10 million, and resistance will change by a little bit less than that. And that's because back in 1990, the electrical metrologists decided to stop updating their value of, effectively, plancks constant, and just keep the one they had in 1990. And there was a benefit to that: they didn't have to update their definitions or their instruments. [...] Well, now the electrical metrologists will have to change. But, that's a very tiny change for a very tiny number of people.

Apparently, the reason is that on 20 May, 2019, redefinitions of SI base units are scheduled to come into force. The kilogram will be redefined using the Planck constant, which, presumably, means that any change in value from the previous definition (the International Prototype of the Kilogram) would affect derived units depending on it, including the volt, ohm, farad, henry, siemens, tesla and (formerly) ampere.

  • Will the volt or ohm change, as Veritasium seemingly claims?
    • Are any other electrical units (listed above) affected?
    • If so, exactly how much will they have changed after the redefinition?
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    $\begingroup$ Click-baity headline! $\endgroup$
    – CJ Dennis
    Commented May 15, 2019 at 2:10

3 Answers 3

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Late last century electrical standards based on Josephson junctions became common. A Josephson junction together with an atomic clock can give an exquisitely precise voltage standard in terms of the Josephson constant. Unfortunately, the then-current definition of the volt relied on the definition of the SI kilogram, which introduced substantial uncertainty. So we could provide a very precise voltage standard, but because of the imprecise definition of the volt we were not sure how many volts it was.

Therefore, in 1990 the community came up with the conventional volt, denoted $V_{90}$, based on a fixed value of the Josephson constant, $K_{J-90}$. This conventional unit has served as a more accurate and reproducible standard for voltage since then, however its exact value in terms of SI $V$ was unknown due to the aforementioned lack of precision.

https://en.wikipedia.org/wiki/Conventional_electrical_unit

With the SI redefinition in a few days $K_J$ will now have an exact value, and that value is slightly different from the exact value assigned to $K_{J-90}$ by the 1990 convention. Therefore, the SI $V$ is also slightly different from the conventional $V_{90}$. Because both $K_J$ and $K_{J-90}$ are exact, the conversion between SI and conventional volts is also exact and therefore the conventional volt is abrogated. This means that electrical metrologists will need to stop using $V_{90}$ and use $V$ which has a slightly different value but the same precision. In other words, an accurate old 1 $V$ standard was much less precise than an old 1 $V_{90}$ standard, but an accurate new 1 $V$ standard will have the same precision as the abrogated 1 $V_{90}$ standard even though the value is slightly different.

So as Veritasium pointed out, it's a very tiny change for a very tiny number of people, although it is not that $V_{90}$ is changing, it is just being abrogated. And the value of $V$ is not changing, it is just gaining precision.

Here is a summary of the affected electrical units and the changes being made:

Unit Symbol Definition Related to SI SI value (CODATA 2014) SI value (2019)
conventional volt V90 see above $\frac{K_\text{J-90}}{K_J} \text{V}$ 1.000 000 0983(61) V 1.000 000 106 66... V
conventional ohm Ω90 see above $\frac{R_K}{R_\text{K-90}} \text{Ω}$ 1.000 000 017 65(23) Ω 1.000 000 017 79... Ω
conventional ampere A90 V90/Ω90 $\frac{K_\text{J-90}}{K_J} \cdot \frac{R_\text{K-90}}{R_K} \text{A}$ 1.000 000 0806(61) A 1.000 000 088 87... A
conventional coulomb C90 s⋅A90 = s⋅V90/Ω90 $\frac{K_\text{J-90}}{K_J} \cdot \frac{R_\text{K-90}}{R_K} \text{C}$ 1.000 000 0806(61) C 1.000 000 088 87... C
conventional watt W90 A90V90 = V902/Ω90 $\left(\frac{K_\text{J-90}}{KJ}\right)^2 \cdot \frac{R_\text{K-90}}{R_K} \text{W} $ 1.000 000 179(12) W 1.000 000 195 53... W
conventional farad F90 C90/V90 = s/Ω90 $\frac{R_\text{K-90}}{R_K} \text{F}$ 0.999 999 982 35(23) F 0.999 999 982 20... F
conventional henry H90 s⋅Ω90 $\frac{R_K}{R_\text{K-90}} \text{H}$ 1.000 000 017 65(23) H 1.000 000 017 79... H

From the exact value of $K_{J-90}$ in the link above and the exact value of $e$ and $h$ given here you can calculate that $\frac{K_{J-90}}{K_J} = \frac{ 71207857995393}{71207850400000}$ exactly. For the volt that works out to approximately $1 V_{90} = 1+1.06\times 10^{-7} \; V$, or ~100PPB.

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    $\begingroup$ @tellus, if your "dead accurate" voltage reference is 100 ppb less accurate than the national standards labs' Josephson junction standards, then it likely won't need to change. If your "dead accurate" voltage reference is more accurate than the standards labs' standards, then you should have shared it with them so they could replace their standards with yours. $\endgroup$
    – The Photon
    Commented May 13, 2019 at 16:41
  • $\begingroup$ Damn, you beat me to the punch. Nice answer. $\endgroup$
    – Emilio Pisanty
    Commented May 13, 2019 at 20:49
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    $\begingroup$ @Tellus You don't need to ask permission to improve questions or answers. Once posted, everything is freely editable so that people can make improvements when warranted. An important aspect of our site is community moderation and collective ownership. $\endgroup$
    – tpg2114
    Commented May 13, 2019 at 23:25
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    $\begingroup$ +1 because it is a really good answer. I do have one complaint though: in this sentence "electrical metrologists will need to stop using $V_{90}$ and use $V$ which has a slightly different value but the same accuracy" it should be precision rather than accuracy. You are using the technical terminology in the following sentence but it got mixed up in this one (if they had the same accuracy they would be equally close to the actual value but the whole point of the change is that $V_{90}$ is not accurate enough) $\endgroup$
    – user137661
    Commented May 15, 2019 at 0:27
  • $\begingroup$ Will 0 V still be set at the standard reduction potential of the standard hydrogen electrode? $\endgroup$
    – Aidan
    Commented May 15, 2019 at 1:02
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Yes, the volt really will change.

If you have a Really Good Voltmeter that’s capable of one part in 10-million accuracy, you’re already used to the idea that you have to get it periodically calibrated against a chain of standards leading back to some national standards body.

The next time you go in for that calibration on your RGV after May 20, the calibration will change.

Few people actually have such devices. People who make such precise measurements generally know about these issues and have been making plans to deal with the changes.

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    $\begingroup$ You don't have to recalibrate, you just need to multiply by the correction factor. The more involved task is probably to recalculate the uncertainty values. $\endgroup$
    – user71659
    Commented May 13, 2019 at 19:22
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    $\begingroup$ @user71659 How do you know what correction factor to use if you haven't calibrated your meter recently, and don't know how inaccurate it has become? $\endgroup$
    – Igby Largeman
    Commented May 14, 2019 at 2:47
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    $\begingroup$ @IgbyLargeman The correction factor in this case is fixed, it's like changing inches to cm. Separately, your calibrated measurement has an uncertainty, the $\pm$ value, and is known through the instrument specs and calibration paperwork. For a typical instrument with traceable calibrations, you'd have some uncertainty that's guaranteed for a year as long as the instrument is within some temperature range and has been running for an hour, passes self-tests, etc. $\endgroup$
    – user71659
    Commented May 14, 2019 at 2:53
  • $\begingroup$ @user71659 I see, thank you. $\endgroup$
    – Igby Largeman
    Commented May 14, 2019 at 22:47
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The redefinition redefines units with exact values for $h,\,e,\,k_B,\,N_A$. The old definitions of metres and seconds are retained, so specifying $h$ redefines the kilogram. The ampere (1 coulomb per second) will change because specifying $e$ redefines the coulomb. (The old definition takes the ampere as fundamental, specifying $\mu_0$ as $4\pi\times 10^{-7}$ in SI units.) The volt changes because $eV$ is an $e$-dependent force, which when multiplied by a metre-second gives the units of $h$. Since the volt change depends on $h$ s well as $e$, you needn't count powers of $e$ to realise the ohm will slightly change (but since $1\Omega=1\text{Js}/\text{C}^2$, $e$ matters here too). By the same logic, the farad (second per ohm) changes, as do the Henry (ohm second), siemens (1/ohm) and tesla ($\text{Vs}/\text{m}^2$).

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    $\begingroup$ @Tellus I don't have numbers for 1 to hand; the Wikipedia article might. Note that if $X\propto Y^n$ the fractional change in $X$ is approximately $|n|$ times that in $Y$ provided it's small, which it will be in this case by design. $\endgroup$
    – J.G.
    Commented May 13, 2019 at 11:38