# Why is apparent power not measured in watts?

I'm studying circuit analysis. We are currently learning about AC circuits, particularly how power is calculated.

I learned that there are four types of power: Complex, apparent, active and reactive, which makes sense, but I'm confused about the unit of measurement used for apparent power i.e. "Volt-ampere (VA)".

Isn't that just watts (W)? Power is defined as $$\P=UI\$$ so if V is multiplied by A, shouldn't that equal to watts? It's even more confusing when both apparent power and active power are measured in the same base SI units (kg⋅m^2⋅s^−3.)

I understand that they are two different quantities that measure two different things, but how can the units make sense?

• Have you searched for similar questions here? Do any of them already answer your question? Dec 8, 2022 at 19:21
• This question is essentially asking for tuition/discussion on the very basics of electronics. It's a Q&A site, rather than discussion forum, and can't be a personal tutorial service, which is effectively what you're asking for. You'll find mountains of information detailing the relationship between voltage, current and power freely available and already written in detail on the internet for you to research and learn from. That's surely your fastest and most fruitful route. Dec 8, 2022 at 20:10
• "Apparent Power" is a somewhat misleading term of art. The unit isn't Watts because it isn't power. At least, it's not power in the physicist's definition of the word. It's only sorta-kinda like power in the I'm-designing-an-electrical-switchyard sense of the word. One possible jumping-off point to learn more about it: en.wikipedia.org/wiki/AC_power Dec 8, 2022 at 20:23
• @TonyM well like I just said to Justme, I always do my own research before asking a question. This forum, and site in general, is a last resort with me; except for reading other questions and answers of course. If you're so sure that there are "mountains of information" regarding my question, then please refer me to one such source and I'll happily read it. Otherwise your comment is not at all helpful. Also, I don't at all think my question implies I want a discussion, that's your own opinion. I'm just asking for a clarification. Dec 8, 2022 at 20:34
• I don't want to go any further down this road because I'm not an E.E., but sometimes, unit analysis can give the same dimensions to quantities that mean very different things. When I charge my plug-in hybrid car, each hour on the charger increases the electric range by ten miles. I could say, "my car charges at ten miles per hour." But I don't, because when people hear "miles per hour," they think I'm talking about a speed, and it's not a speed except maybe in some weird sense. I'd guess that E.E.s distinguish "VA" from "Watts" for a similar reason: They don't want to be misunderstood. Dec 8, 2022 at 21:09

I'm confused about the unit of measurement used for apparent power i.e. "Volte-ampere (VA)". Isn't that just Watts (W)?

• Power is always the instantaneous product of voltage and current ($$\v\times i\$$) and, the unambiguous unit of that instantaneous product is the watt
• Additionally, If you average $$\v\times i\$$ over 1 cycle of an AC waveform, that is also measured in watts and, we call that quantity average power

For a voltage waveform driving a resistive load, the average power dissipated is mathematically the same as the RMS voltage × RMS current i.e. exactly the same number of watts as the average of $$\v\times i\$$. But, this is a special case for a purely resistive load.

So, if we have a load that is not purely resistive (influenced by capacitance or inductance), we can no longer say that the watts dissipated is RMS voltage × RMS current.

For instance, here is a pictorial example from this answer: -

• In red above are instantaneous volts × instantaneous amps (true power)
• I've also indicated the average power (red) in all the examples
• In all cases, RMS volts × RMS amps is the same value
• Top left shows a resistive load and produces true watts
• Top right shows a reactive load and clearly, the average power is smaller for the same applied RMS values of current and voltage
• Bottom left shows the case where the load is wholly reactive and the average power is zero despite the RMS values of voltage and current remaining the same.
• Bottom right is when the phase angle has moved so much, the "load" is now supplying power to the source.

The take-away thing here is that RMS voltage × RMS current is the same for all scenarios yet, only in the top left does RMS voltage × RMS current = true power.

Hence, we call the value of RMS voltage × RMS current "apparent power" and, we define its units as volt$$\\cdot\$$amps and not watts.

• My mistake was assuming that power is always defined as $$P = UI$$. That's not always the case. It is the case when concerning a resistor or just a purely resistive load, but in general, power is defined as work over time. Apparent power is indeed voltage times current, but it's not necessarily the actual work being done over time. So we are still multiplying the two units but we're not getting the actual work over time. So as @user71659 said in his answer, we call the unit VA as a convention. Dec 9, 2022 at 23:08
• No, there is a difference. Power is volts x amps and, if we don't know the load, we can't define the current hence, we can't say whether true watts are being consumed. Dec 10, 2022 at 0:34

I believe you are correct in saying from a physical units standpoint a Watt and a Volt-Amp are the same. I'd argue it's a convention to save extra words.

When you see a number specified in Volt-Amps, say a UPS, generator, transformer, or meter reading, you immediately know it's apparent power. If you see Watts, you can assume it's real power.

Note the other measurement: Volt-Amps Reactive (VAR). A "reactive" certainly isn't a unit, it's a description.

The alternative would be to always specify everything, e.g. Watts Real, Watts Reactive, Watts Apparent.

• in a way, it's similar to how not all newton-meters are joules. Dec 9, 2022 at 10:22
• Such as when a Newton-meter is a torque Dec 9, 2022 at 14:01

For an introduction to power factor and see the difference between apparent and average power, you can have a look at my latest tutorial on PFC. This first illustration shows the difference between average power which contributes the work in the physical sense, e.g. luminous flux, audio sound etc. and the apparent power actually produced by the utility company or the generator:

When you power a resistance, both current and voltage are in phase. Apparent and average power are the same and the utility "sees" 70 W to power. When you now power a nonlinear load such as a full-wave rectifier, absorbing the same 70-W average power as the resistance does, the utility will have to deliver more apparent power and a higher rms current will now circulate as determined by the formulas. Part of this extra current circulates and heats up the wires but does not participate to the work. You can find further explanations in this clear article.

There are other cases where the units are the same but they aren't physically the same. For example, Volts from a source like a battery and volts from a voltage drop. Then there's the forward EMF and back EMF. All are Volts, measure the same, but not the same when you interpret it physically. Another is the force applied and the reaction force. Both Newtons, both kinda the same, but not really. Or a more non-physics example: income vs debt. Both dollars, but not the same. Then of course there's Watts. Watts supplied vs Watts consumed. Similar, but not the same.

The math isn't the end all. There's the interpretation of the math.

And in this way, Volt-Amps is not Watts and shares less between counterparts than any of the preceeding examples because Volt-Amps is calculated by taking the product of the peak amplitude of the voltage and current. The peak amplitudes don't necessarily coincide in time so they cannot possibly be actual Watts. It's not even the complimentary.

In actuality, Volt-Amps is non-physical as it is just a bookkeeping trick because it is easy for humans to work with sine waves on paper napkins. It's not real power. It's a book-keeping metric.

Actual Watts is the product of a voltage and current that coincide in time.

• But apparent power/VA does have a important physical meaning: If my load is rated for x VA, then on average, x Joules/second are moving on its power cord. That's why we care about apparent power, because we don't want to burn up the power cord, or its equivalent, transmission lines. Dec 9, 2022 at 10:31
• @user71659 You can get that just from looking at current on its own. Or reactive power vs real power which are both flows of real power and measurable. Apparent power isn't required at all and is more of a paper napkin thing with sinusoids. Dec 9, 2022 at 14:26
• We rate transformers in VA, not A because VA is obvious whereas there's 2 different currents, and then you have to track which side you're talking about, and with two-way power flows it becomes confusing. In bigger power systems, where you need to move x VA from city 1 to city 2, amps also becomes difficult to track because different power lines run on different voltages. Dec 9, 2022 at 18:44
• @user71659 That still doesn't make it less of an abstract book keeping metric than something like current or real power. Dec 9, 2022 at 18:48

They're training you very old school, presenting loads as if they are basically sinusoidal, big clunky old motors and ballasts. I find the "watts vs VA" question to be better served by modern electronic examples.

You're building a holiday display for the village. The mayor's wife bought 400 LED Christmas light strings at deep discount. They are wired as a single series string of ninety LEDs and a diode to guard against reverse current. They're only on for half the power cycle (they shimmer) but this is very typical of modern LED Christmas lights. Meaning they are on for half the AC cycle. The plugs are polarized and modifying them is off limits.

For math simplicity, power is British 240V AC single-phase. Each string has a 6 watt draw. (this is our WATTS unit). The 400 strings draw 2400W together or total 10 amps. Remember, their power draw looks like this because the diodes are all facing one direction.

Now, you are sizing the generator to drive this load (and no others). It's an ideal world, so a 2400 VA rated generator can carry a 2400 VA load. *

Will a 2400 "watt"/VA generator do the trick? Hint: Will a 10A fuse do the trick? **

NO! It won't! Because the LEDs are drawing 2400W on average but only drawing power half the time, that means they are drawing 4800W or 20A when they draw power.

The fuse would blow because it would run hot. * As would the generator windings, which is why we can't use the generator's kinetic energy as energy storage.

Without any means of time-shifting, the generating capacity at the bottom half of the sinewave is wasted and useless.

The load may only use half the sinewave, but we must generate the whole sinewave. Thus, our generator must generate the full 4800 VA.

And that is how I think of the VA unit. Watts is the part of the sinewave you actually use. VA is the whole sinewave you must generate.

* Hint, hint.

** By the way, a sidebar on thermal matters. Work it through Ohm's Law and you see that for a given resistance and ampacity, thermal rise is proportional to the square of current. Thus, twice the current half the time == 4x the heat half the time == 2x the heat in net for the same watts. Really. Crunch the numbers, you'll see.

• hmm, but would the fuse run hot just because the actual power draw only happens on one half-cycle? The average power heating the fuse over multiple cycles (or just one full cycle) here is still 2400 W (corresponding to 10 A) and the fuse isn't likely fast enough to blow over just one cycle. After a few cycles, the energy dissipated in the fuse is the same regardless of how the total energy is split over time. It would be different if the split was half an hour on, half an hour off, because then there'd be plenty of time for the fuse to heat during the active period. Dec 9, 2022 at 9:27
• @ilkkachu no, see the last paragraph of the answer. Assume your fuse has 1 Ohm resistance. At 10A this will mean 10V drop, so 100W that heat your fuse. However, at 20A this means 20V drop, so 400W. Of course you'll only use the 400W half the time, but that still means double the heat generation over time. The wattage of the whole installation is irrelevant for the fuse; the only things that matter is current (Amps) through the fuse and resistance of the fuse. The rest of your 2400 watts goes to the LEDs, the fuse doesn't care. Dec 9, 2022 at 11:57
• @GuntramBlohm, good point, the power on the fuse isn't the same as the power on the load (d'oh). Of course it's still not just the amperage, the on-off period also matters, but not as much as I stated before. That 400 W half-time, or 200 W on average would correspond to about 14 A full-time wrt. heat generation, not the whole 20 A. Dec 10, 2022 at 12:15