It is said that, though only active power can actually do work, we have to consider apparent power to dimension the source.

But what would seem important to me for that would be to know the maximum absolute value of the instantaneous power, which would be the apparent power plus the active power (S+P, not P+Q (to be extra clear, there is nothing hidden in that notation, it's a pure boring arithmetic sum of two real numbers)).

Is that correct?


As Charles Cowie answered, it is not the peak of the instaneous power \$p(t)=u(t)i(t)\$ which is of any use most of the times. So, why is S important when dimensioning power source and input? What physical value exactly is important and which S represents more or less?


I said that \$\max(p(t))=S+P\$ which is maybe a bit surprising and unjustified. Let's say that \$u(t)=\sqrt{2}\,U\cos t\$ and that \$i(t)=\sqrt{2}\,I\cos(t+\phi)\$ (frequency or pulsation isn't important for my calculation, so WLOG, let's admit \$\omega=1\$).

Then \$p(t)=u(t)i(t)=2UI\cos(t)\cos(t+\phi)=UI(cos(2t+\phi)+\cos(\phi))\$. You can find that with trig identities. \$p(t)\$ takes its maximum value when \$2t+\phi=0\$, so you have \$\max(p(t))=UI(1+\cos(\phi))=UI+\cos(\phi)UI=S+P.\$

  • 3
    \$\begingroup\$ You need to know the actual current. That is why you need to know the apparent power (or the real power and the power factor). \$\endgroup\$
    – user57037
    Commented Jan 22, 2020 at 7:03
  • \$\begingroup\$ Aparrent power is what melts your cables when you only sized them for real power. \$\endgroup\$ Commented Jan 22, 2020 at 8:55
  • 1
    \$\begingroup\$ @JonasDaverio Well I see you've edited your question after my previous comment, conveniently removing the word "sum" ; ) But what exactly is your question? The maximum value of the instantaneous power is the peak value of the active power. That is, if by "instantaneous power", you mean the "instantaneous product" of voltage and current... \$\endgroup\$
    – ManRow
    Commented Jan 22, 2020 at 11:01
  • 1
    \$\begingroup\$ Your question is actually not very clear. It is always important to know how many amps are required. Switches are typically rated for amps. Fuses are rated for amps. Circuit breakers are rated for amps. Generators may need to be de-rated if the power factor is not 1.0. So it is important to know the power factor and/or the actual number of amps. Short term overload (due to motor startup or capacitive inrush) is a whole other topic. You need to clarify your question or it may end up being closed. \$\endgroup\$
    – user57037
    Commented Jan 22, 2020 at 17:36
  • 1
    \$\begingroup\$ If the voltage and apparent power are known, and if the current and voltage are assumed to be sinusoidal, then the peak and RMS current can be easily calculated. \$\endgroup\$
    – user57037
    Commented Jan 23, 2020 at 0:44

4 Answers 4


The physical value that is important is the current that something will need to conduct, and the maximum power it might be asked to provide.

Let's briefly talk about what all these abstract mathematical concepts actually represent physically:

Real power (\$ P \$) is the rate at which energy is consumed to do work of some sort (often simply in the form of waste heat, or mechanical motion, or sound, or light, so on and so fourth).

Reactive power (\$ Q \$) is the rate at which energy is stored by a circuit. Energy that is stored in the electric field of a capacitor or the magnetic field of an inductor still has to come from somewhere. This energy storage does take energy from the circuit and it does so at a certain rate (yielding power). Unlike real power, reactive power is not simply consumed but instead returns back into the circuit in a phase-dependent manner. Put simply: energy stored during the one sign of the AC wave will also be returned to the circuit during the opposite sign of the wave.

Complex power (\$ S \$) is the combined real and reactive power. Combined meaning vector sum.

Apparent power (\$ \left| S \right| \$) is the magnitude of \$ S \$, or the length of the hypotenuse/composite vector.

So complex power already represents the total possible power that we need to worry about.

One cannot simply add them together because they are orthogonal to each other. This is ultimately because reactive impedance, which is measured in ohms just like real resistance, causes a voltage drop. The only difference is the drop is going towards storing energy rather than dissipating it.

This drop results in voltage and current no longer being in phase. For example, a capacitor will cause voltage shift behind current because some of that current is going to charging that capacitor, and there is a time-dependent voltage drop as a result. Then 90 degrees later in the AC cycle, the capacitor is going to discharge and keep the voltage higher than one would expect. The net result is voltage lags behind current. A phase shift.

Power is, of course, \$ V x I \$. And with a purely resistive load (no reactive power, only real power), voltage and current are always in phase, and thus the apparent power equals real power, and is simply the RMS voltage multiplied by the RMS current.

But when reactive loads cause voltage and current to no longer be in phase, this doesn't change the apparent power, but it does impact real power by reducing it. If voltage is lagging current, then when originally you might have had peak voltage, say, 10V, coincide with peak current, say 1A, resulting in 10W peak power, you might now only have 9V when you hit peak current. And peak voltage is no longer reached when current is at its peak. This causes an unavoidable reduction in real power.

So they are orthogonal vectors, one is only able to steal power from the other, both fighting over the apparent power which isn't going to change for a given number of ohms, no matter how those ohms are split between real and reactive impedance.

The angle is, of course, the phase shift between voltage and current. At 90 degrees, voltage peaks when current is zero, and current peaks when voltage is zero. All power is reactive, but no work is being done.

So why does this even matter? Why does one need to size a power supply based on apparent power?

Because electrons are still moving even if they aren't doing anything. An amp is an amp and it doesn't stop being an amp just because it isn't actually doing anything.

100A, even if it is simply stored in a capacitor only to be re-released next cycle without doing anything, is still 100A of current conducting through cables. Its still 100A that a transformer must withstand flowing through its windings. It is still 100A the core must withstand without overheating.

Ultimately, what actually matters for this sort of thing isn't how much power is actually being used to do work, it is how much power must be carried. And this is always the apparent power.

This is why power factor is a big deal. Power factor is the ratio of real power to apparent power. And as far as things like an electrical grid is concerned (remembering that the grid itself has to worry about what it needs to carry), if you're using 1W of power but have some huge capacitor across the mains resulting in 1kW of apparent power... they're going to charge you for that because that 1kW still has to flow through the grid and still results in resistive losses through the transmission lines and even worse, it is doing so for no reason.

This is something that has no end of thorough theoretical examinations that delve heavily into the math, and that really is the correct way to ultimately understand all this, but I think it can help a lot to have a good conceptual idea of why the math is what it is, and why all this stuff matters. I hope this helped!

  • \$\begingroup\$ What an excellent answer - nothing to add! \$\endgroup\$
    – N.G. near
    Commented Mar 25, 2021 at 6:49
  • \$\begingroup\$ My reaction will maybe be a bit disappointing, but these are things I already know. I think it is a great answer to explain what the apparent, real, and reactive power means. What I want to know is why, in your explanation, at the end, you would find the apparent power more useful to know how much energy you waste in your transmission, than let's say the current. There's a detail I'm missing. \$\endgroup\$ Commented Mar 25, 2021 at 9:20
  • \$\begingroup\$ When you say " it is how much power must be carried. And this is always the apparent power.", I don't understand what it means mathematically and why it would be true. \$\endgroup\$ Commented Mar 25, 2021 at 9:21
  • \$\begingroup\$ @JonasDaverio Because the "source\load" and the wires need to be able to withstand this "100A" of current without melting down. Also, you need to build a source that will be able to "produce" this amount of current (thick wires to reduce I^2 R_wire losses). \$\endgroup\$
    – G36
    Commented Mar 25, 2021 at 11:30
  • 1
    \$\begingroup\$ @JonasDaverio: Apparent Power is often more useful to know than current because it is voltage-agnostic. It can go through transformers and still have the same value. \$\endgroup\$
    – Theodore
    Commented Mar 25, 2021 at 20:01

I feel that the other answers are a little long and don't address the point.

Usually large instantaneous power doesn't destroy components by itself. It's temperature. Mains power is around 60Hz. That time is usually shorter than the thermal time constant of stuff in the power supply. So its temperature will not change much within a cycle. It takes many cycles for the devices to heat up and break. And since voltage is cyclic, current draw is roughly cyclic, and apparent power is averaged over a cycle, apparent power is good enough when we need to know the average power in a time long enough to heat things up.

PS: It is possible for a short pulse to break stuff in the power supply (like transistors). But the energy that pulse will need to carry is usually much larger than the power supply can generate in a single cycle. Examples include lightning and energy dumped by an inductor when its switch is opened. But those are separate concerns from choosing wires that are thick enough.

  • \$\begingroup\$ Same comment as for other: everything here is a qualitative description, which I get. But what I want is where S fits in the picture, e.g. how you would use S to quantitatively describe the physical quantity you are actually concerned about (here it would be heat losses again, I think). \$\endgroup\$ Commented Apr 1, 2021 at 9:14
  • \$\begingroup\$ @JonasDaverio Current heats wires. Apparent power is current times voltage. Voltage is usually fixed. \$\endgroup\$
    – 30tah8uu
    Commented Apr 15, 2021 at 10:58


Apparent power S is the AC power X-Y vector of P (real) and Q (stored or reactive).

  • S = P+jQ (not S+P )
  • i.e. [VA] = [W] + [VAR] for units

This is why all transformers are rated in apparent power or VA rather than Watts as the reactive load degrades the rating of the transformer. This is true for electronic diode bridge caps and magnetic motor loads. Both store energy and the VAR load subtracts from VA rating for Watt power loads. (although margins are used in industry to stay well below max rating) It is also why the new design standard for PSU's > = 100W must be active PFC to 99% and passive PFC is used on grids to reduce the average VAR as much as possible with switched grid caps in PU 19" racks or oil-filled caps.

The peak is the magnitude of the apparent vector is the one you are concerned about conduction losses in distribution transformers or grids. is more significant when the load is opened for arc suppression. Although for conduction as @30tah8uu has rightly mentioned it is the thermal rms energy temp rise that matters most as well as saturation effects in transformers ( from stored Remanence and reclosure phase difference)

For the consumer, they are only charged for P as all power meters ought to be measuring P_rms only and the supplier is responsible for losses in Q.

In Industry there may be both P and Q meters with different charge rates or P and p.f. so p.f.c. is considered to offset costs of Q power.

In many countries where power quality is good ( like in North America with 10%) there is never a need for "Power Conditioners" which are as common as routers in India and Pakistan where power quality in older regions is poor. These happen to make the grid even more unstable.

  • When the grid is near max capacity and too many Power Conditioners sense low voltage, a relay change taps to get more voltage which draws more current and loads the distribution down further causing a runaway condition until the grid faults out! Such a great invention that creates havoc for everyone some of the time (?) weekly daily? . here never !

So the solution is a major cost infrastructure upgrade to increase the power capacity with redundancy so provide 99.9% service for RMS power or kW-hours so that Peak power is almost irrelevant for consumer service and Apparent power is only important for Industry costs where arc welding, metallurgy or heavy AC machines are used.

  • \$\begingroup\$ feel free to add or subtract in edits or comments \$\endgroup\$ Commented Mar 29, 2021 at 14:53
  • \$\begingroup\$ I don't really understand how it answers "What is the important of apparent power". You talked about conduction losses, but only in a qualitative way. Could you show by a simple example how you would use S to quantitatively measure heat losses? \$\endgroup\$ Commented Apr 1, 2021 at 9:10
  • \$\begingroup\$ Like I said, it is causes losses in Distribution$$Pd=I^2R$$ . If losses increase with the square of I, Apparent vs. Real, why do you need a math example? Why are qualitative references not enough? \$\endgroup\$ Commented Apr 1, 2021 at 10:19
  • \$\begingroup\$ Sorry was it a mistake to assume you knew how current and power loss are related? \$\endgroup\$ Commented Apr 1, 2021 at 10:25
  • \$\begingroup\$ I don't really get why you are making fun of my question. Like you say, losses increase with I^2, but how does that relate to apparent power which is I*U, and why would cos(phi) be related to that? \$\endgroup\$ Commented Apr 1, 2021 at 19:13

When the load contains both resistance and reactance, this product represents neither real power nor reactive power, since it appears to represent power, it is called apparent power.

S=V times I

It can also be written as

S=I^2 times Z


S = VI/1000 (kva)

The relationship between P,Q,S may be expressed as a complex number

S = P+jQL

If the circuit is capacitive instead of inductive, then

S = P-jQc

The power relationships may be written in generalized form as

S = P + Q


S = V times I*

Where I* is the conjugate of current I.

These relationships hold true for all networks regardless of what they contain or how they are configured.

The time varying power is referred as instantaneous power. Thus,

P = V times I

Positive p represents power to the load and negative p represents power returned from the load.

Power to a resistive load

P = Vm times Im/2 times (1-cos2omegat)

Power to an inductive load

PL = V * I * sin2omegat

Power to a capacitive load

Pc = -VIsin2omegat

Importance of Apparent power:

both resistive and inductive loads are utilized by home and business. As you add more inductive equipment, the ratio between these two types of loads becomes important as you add more inductive equipment. Working power and reactive power make up Apparent Power, which is called kVA, kilovolt-amperes. We determine apparent power using the formula, kVA2 = kV*A.

Going one step further, Power Factor (PF) is the ratio of working power to apparent power, or the formula PF = kW / kVA. A high PF benefits both the customer and utility, while a low PF indicates poor utilization of electrical power.

  • \$\begingroup\$ Hi, Maverick. The question is, "what is the importance of apparent power?" You haven't answered that. You can edit your question at any time to improve it and avoid downvotes. \$\endgroup\$
    – Transistor
    Commented Mar 31, 2021 at 6:30

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