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?
EDIT:
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?
EDIT2:
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.\$
Best Answer
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!