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I thought my electric car charging unit uses 6.6 kW of power. However, I found the label and it actually says 6.6 kVA. When I saw this I thought something along the lines of...

Well, $ P=VI $, therefore kVA must be the same thing as kW... strange, I wonder why it's not labelled in kW.

So a quick Google search later, and I found this page, which has a converter that tells me 6.6 kVA is actually just 5.28 kW. The Wikipedia page for watts confirmed what I thought, that a watt is a volt times an ampere.

So what part of all this am I missing, that explains why kVA and kW are not the same?

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    $\begingroup$ Note that for most countries with stable power networks the regulations require good enough power factor for such big loads that kVA ~= kW; the mentioned site just blindly applied a power factor of 0.8 which imho is highly unrelaistic for an electric car charging unit. $\endgroup$ – PlasmaHH Mar 2 '15 at 20:17
  • $\begingroup$ In physics, both would be the same... in engineering, kW counts the net power transferred to the car, while kVA counts the power transferred along the wire in both directions. $\endgroup$ – immibis Mar 3 '15 at 3:31
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    $\begingroup$ I think the answers are pretty good, but I just wanted to point out, from a linguistics perspective, that the best reason I've seen for kVA is that Engineers wanted to make it very clear they were not kW, which was too useful of a unit to double up on. Keeping the Volts and the Amps separate was a convenient notation to denote that they should be treated differently, even if both of them are units of power. $\endgroup$ – Cort Ammon Mar 3 '15 at 4:47
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The problem is that the formula $P=I\ V$ is correct when dealing with DC circuits or with AC circuits where there is no lag between the current and the voltage. When dealing with realistic AC circuits, the power is given by $$ P=I\ V\ \cos(\phi), $$ where $\phi$ is the phase difference between the current and the voltage. The unit kVA is a unit of what is called 'apparent power' whereas W is a unit of 'real power'. Apparent power is the maximum possible power attainable when the current and voltage are in phase and real power is the actual amount of work which can be done with a given circuit.

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    $\begingroup$ Note: the cos($\phi$) part ONLY applies when both voltage and current are sinewaves. It does not apply when current is spiky (through a "dumb" rectifier) or when either is distorted by any means. See my answer for more details. $\endgroup$ – AaronD Mar 2 '15 at 21:29
  • $\begingroup$ @AaronD You are correct that the situation is slightly more complicated when the signals are not sin waves, but the $cos(\phi)$ term still applies. Its just that $\phi$ is now a function of frequency in the Fourier domain and the power you are most likely interested in is the integral over all frequencies. In practice it might be easier just to measure the power directly as you mention in your answer. $\endgroup$ – Chris Mueller Mar 2 '15 at 22:17
  • $\begingroup$ Okay, technically you're correct - you're converting the problem into a bunch of sinewaves so that the cos($\phi$) term can work again - but I really doubt that most people would understand what that means and do it right. The difference between 50Hz and 60Hz labels might even be a stretch beyond, "It's incompatible." $\endgroup$ – AaronD Mar 2 '15 at 22:50
  • $\begingroup$ What I think is awesome, as a mathematician, that the 'rest of the power' (ie, that power not given in the above answer as 'real power'), rocks off in the imaginary direction. You actually get power moving in the imaginary direction. How cool is that? $\endgroup$ – Sam T Mar 3 '15 at 21:14
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    $\begingroup$ I'm not 100% about this bit (hence the separate comment), and as such if it is wrong (which I don't think it is), please just shout and I'll ditch it, but the power is then given by the formula $$P = I V (\cos(\phi) + i \sin(\phi)) = I V e^{i \phi}$$ and so we see that if we take the modulus/length of this, then we get $\vert P \vert = I V$. $\endgroup$ – Sam T Mar 3 '15 at 21:15
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Both watts and volt-amps come from the same equation, $P=IV$, but the difference is how they're measured.

To get volt-amps, you multiply root mean square (RMS) voltage ($V$) with RMS current ($I$) with no regard for the timing/phasing between them. This is what the wiring and pretty much all electrical/electronic components have to deal with.

To get watts, you multiply instantaneous voltage ($V$) with instantaneous current ($I$) for every sample, then average those results. This is the energy that is actually transferred.


Now to compare the two measurements:

If voltage and current are both sinewaves, then $\text{watts} = \text{volt-amps} \times \cos(\phi)$, where $\phi$ is the phase angle between voltage and current. It's pretty easy to see from this that if they're both sine waves and if they're in phase ($\phi = 0$), then $\text{watts} = \text{volt-amps}$.

However, if you're NOT dealing with sine waves, the $\cos(\phi)$ relationship no longer applies! So you have to go the long way around and actually do the measurements as described here.

How might that happen? Easy. DC power supplies. They're everywhere, including battery chargers, and the vast majority of them only draw current at the peak of the AC voltage waveform because that's the only time that their filter capacitors are otherwise less than the input voltage. So they draw a big spike of current to recharge the caps, starting just before the voltage peak and ending right at the voltage peak, and then they draw nothing until the next peak.

And of course there's an exception to this rule also, and that is Power Factor Correction (PFC). DC power supplies with PFC are specialized switching power supplies that end up producing more DC voltage than the highest AC peak, and they do it in such a way that their input current follows the input voltage almost exactly. Of course, this is only an approximation, but the goal is to get a close enough match that the $\cos(\phi)$ shortcut becomes acceptably close to accurate, with $\phi \approx 0$. Then, given this high voltage DC, a secondary switching supply produces what is actually required by the circuit being powered.

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  • $\begingroup$ After you multiply the instantaneous voltage by the instantaneous current to get instantaneous power, do you really need to take the RMS of the power at each instant, or can you take the simple average? $\endgroup$ – David Cary Mar 5 '15 at 2:11
  • $\begingroup$ @DavidCary: I think you might be right. For the case that they're pure sinewaves and $\phi = 90deg$, half of the samples will be positive power and half negative, and the answer should be zero. I'll edit my answer. $\endgroup$ – AaronD Mar 5 '15 at 14:35
  • $\begingroup$ It's simple average. RMS is derived from this averaging and presuming, that u = Ri and that U = RI, where u/i are actual values and U/I are RMS. $\endgroup$ – Crowley Mar 5 '15 at 21:29
  • $\begingroup$ @AaronD: If we suppose that the power factor $\cos\phi_r$ consists of phase angle $\phi$ and form factor $\phi_f$ we can still use the formula $P=UI\cos\phi_r$ but evaluation of this form factor and the way how to combine it with phase angle aren't simple. $\endgroup$ – Crowley Mar 5 '15 at 21:35
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When an AC line is driving an inductive or capacitive load, then the load will spend some of its time taking power from the source, but will also spend some of its time feeding power back to the source. In some contexts, a device which draws a total of 7.5 joules each second and returns a total of 2.5 joules may be regarded as though it was drawing 5 watts (especially if whenever the device is returning power some other load is ready to consume it immediately). Something like a transformer, however, will suffer conversion losses not only during the part of the cycle when the load is drawing power, but will also suffer losses during the part of the cycle when the load is feeding it back. While a transformer would probably dissipate less heat driving the above load than one which drew 10 joules/second and returned zero, it would dissipate more than when driving a load which drew 7.5 joules/second and returned zero.

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protected by GlenH7 Jan 20 '16 at 22:40

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