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
    Commented Mar 2, 2015 at 20:17
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    $\begingroup$ 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. $\endgroup$ Commented Mar 3, 2015 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
    Commented Mar 3, 2015 at 4:47

4 Answers 4


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
    Commented Mar 2, 2015 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. $\endgroup$ Commented Mar 2, 2015 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
    Commented Mar 2, 2015 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 OT
    Commented Mar 3, 2015 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 OT
    Commented Mar 3, 2015 at 21:15

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
    Commented Mar 5, 2015 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
    Commented Mar 5, 2015 at 14:35
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    $\begingroup$ 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
    Commented Mar 5, 2015 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
    Commented Mar 5, 2015 at 21:35

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.


Another way to understand why kVA is different to kW, is the Power triangle.

enter image description here

Figure 1: power Triangle (source: Electrical technology)

The power triangle shows that the total (apparent) power is the vector sum of the active (or real) power and the reactive power.

  • Active Power $P$: this is the actual power that its used/is available (electrical engineers apologies for the poor choice of words here). This is the portion of the apparent power that is available to be converted to work/heat.
  • Reactive Power $Q$: this is portion of the power that is stored in the system. That power is stored in the capacitive or inductive elements of the systems (every system has those in some small degree). This is why the LED of a power supply takes some time to fade when you pull it off a plug.
  • Apparent power $S$: this is the quantity measured in kVA. It is what is calculated by measuring the voltage and the current and multiplying them together.

Why Apparent and Real power are not the same

When measuring an AC line, you can see, that the maximum values of voltage and current can be reached at the same time or not. See image below:

enter image description here

Figure 2: power Triangle (source: Electronics Hub)

When the current and the voltage reach the maximum and minimum values at the same time, then the Apparent power (S) and the Real Power (P) are equal. There is no reactive element.

If the current and the voltage do not reach the maximum and minimum values at the same time, then the Apparent power (S) is greater than the Real Power (P). In that case there are some times that the product of $V*I$ is positive and others that it is negative.

In the (ideal) case of sinusoidal waveforms of Voltage and Current, the delay between voltage and current when expressed in the angular quantity $\phi$, also determines the Power factor (which is $\cos(\phi)$).

  • $\begingroup$ Some corrections: 1) Active power is not "available power", it's the average power currently moving into the measured system (or out of it if negative), whether it's heating a resistor, lifting a load, charging a battery or whatever. 2) Reactive power is not power stored in the system. You can't store power because power is movement of energy and you can't store movement. Reactive power is actually power currently oscillating into and out of the system, on average contributing nothing to energy transfer. 3) LED taking time to shut down has absolutely nothing to do with reactive power. $\endgroup$ Commented Oct 7, 2022 at 6:35
  • $\begingroup$ regarding 1) as I mention in the post the available power is a poor choice of words, so definitely your description is more precise. Regarding 2) you are right, I should have said "temporarily stored, due to oscillation in the capacitative and inductive elements of the system". Being an mechanical engineer I draw the parallel from a mass damper spring oscillator, and energy is temporarily stored in a spring in a same way that a capacitor temporarily stores energy. So, I suppose our main difference is in temporarily. Regarding 3) if energy is temporarily stored then eventually it dissipates. $\endgroup$
    – NMech
    Commented Oct 7, 2022 at 6:55
  • $\begingroup$ About 2), yes: the key point is temporarily, as in within one period of oscillation. If it averages out to net positve or negative over a period of oscillation, there was active power there as well. About 3), if it's reactive then it will dissipate within 20 ms (@50 Hz voltage). Also, if it's lighting up the LED it's active. The delay happens due to stored energy, which must have been stored using active power, not reactive. The spring oscillator is a good example, but only the oscillating component of the power is reactive. The component that compressed the spring on average was active. $\endgroup$ Commented Oct 7, 2022 at 7:11
  • $\begingroup$ regarding 2) I would arque that it is not necessarily in one period of oscillation, it depends on the resistive/damping elements. In an LC circuit it will oscillate forever. In a RLC it will depend on R, compared to the other elements. Also, I would argue that spring energy is reactive, by direct comparison of the diff eqs for mass damper $m\cdot\ddot{x} + c\cdot\dot{x}+ k\cdot x = F(t)$, and the RLC $L\cdot\ddot{Q} + R\cdot\dot{Q}+ C\cdot x = E(t)$. The analogy between Spring and capacitor is evident. So if you argue that a capacitor stores reactive energy, the same goes for the spring. $\endgroup$
    – NMech
    Commented Oct 7, 2022 at 7:23
  • $\begingroup$ There is no reactive energy. If a capacitor accumulated energy over a period of oscillation, it received active power and stored it as electric potential energy. Reactive power can't contribute to accumulation of energy because it's, by definition, the component of power that did nothing. I only talk about single oscillation periods because active/reactive power is only defined for systems with constant-frequency constant-amplitude oscillations (or multiple of those). Otherwise, the model doesn't hold and you must deal with instantaneous powers and their integrals. $\endgroup$ Commented Oct 7, 2022 at 9:06

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