Given an AC circuit, the instantaneous power in the circuit can be determined by the formula $$P(t)=\frac {VmIm}2cos(\theta)+\frac {VmIm}2cos(\theta)cos(wt)-\frac {VmIm}2sin(\theta)sin(wt)$$ where \$ \theta \$ is the power factor angle, or the phase difference between voltage and current.

Complex power is defined as $$\vec S = P + jQ$$ where $$P=\frac {VmIm}2cos(\theta), Q =\frac {VmIm}2sin(\theta)$$ where \$ P\$ is real power, \$ Q \$ is reactive power, and \$ |\vec S|\$ is apparent power.

I completely understand the uses for complex power, and how it works nicely mathematically with voltage, current and impedance in AC.

What I don't understand is, why the "apparent" power is what we need to cater for when providing power and e.g. designing a power source for a circuit, as opposed to the maximum instantaneous power, which equals \$|\vec S| + P\$. After all, there will be a moment in time where the source does need to provide this amount of power, even if it isn't "useful".

What I also don't really understand is how these mathematical formula ultimately only end up involving \$ \vec S \$, and instantaneous power (i.e. \$ P(t) \$) is completely left out of the picture. For example, why is the real component of \$ \vec S \$ independent of the constant shift amount \$ \frac {VmIm}2cos\theta \$ for instantaneous power? Also, why is the \$ Q \$ term not negative in complex power, like it is in instantaneous power?

There are just a few gaps in my intuition which make it uncomfortable for me to properly understand complex power.

  • \$\begingroup\$ There's an answer touching this part here (see Alejandro's answer, too). \$\endgroup\$ Commented Jun 25, 2021 at 20:42

3 Answers 3


Answering your second question first: Your first equation is in the time domain, and your second is not. The use of complex numbers to represent electrical voltage and currents implies that the voltages and currents are sinusoidal and have a phase relationship with one another. Instantaneous values have no phase, so they have no meaning in the real + imaginary representation.

Now, why is apparent power so important? Consider a theoretical sinusoidal generator driving a purely reactive load (an ideal inductor or capacitor). In this case, the voltage and current of the load are 90 degrees out of phase, so there is no net power being consumed. At the instantaneous level, current is being provided to the load during part of the voltage waveform, and then the current is being returned to the voltage source during the other part. So the instantaneous power is sometimes positive and sometimes negative, but the net power is zero.

In the real world, that same inductor has a resistance, and the full current has to be provided from the generator and flows through long transmission lines, which are also resistive. To keep losses at a minimum, all of the conductors in the circuit have to handle this current, and their resistive component dissipates some of the power, proportional to the RMS current and independent of its phase relationship to the generator voltage. So in our example, we have to use large conductors and sustain resistive losses even though no net power is provided to our inductive load. So if we are running equipment with poor power factors, extra power must be generated and heavier conductors must be used. More average current is required to provide the same power.

I hope this makes sense. Good question!

  • 1
    \$\begingroup\$ Not a bad answer but one small correction. Any resistive element in the inductor would not contribute to the loss due to apparent power (that is all in the transmission lines and source). It would behave the same as a resistor would and effectively be tied into the real power part of the complex power. \$\endgroup\$ Commented Jun 25, 2021 at 1:59
  • \$\begingroup\$ Sorry if I confused you. I was attempting to compare it with the ideal inductor. In this case, the apparent power is the only "power" and the actual power is zero. There would be no losses associated with the reactive current, so there would be no need to concern ourselves with apparent power. If the inductor is not ideal (has a resistive component) the fact that there are large reactive currents means that there will be losses associated with this current, necessitating larger inductor wires to keep losses low. I probably should have said "reactive" instead of "inductive" load. \$\endgroup\$ Commented Jun 27, 2021 at 23:06

the "apparent" power is what we need to cater for when providing power

I'm not sure where that idea comes from. You typically "cater to"

  1. Peak Voltage
  2. Peak Current
  3. Storage capacity for different time constant to minimize voltage sag for load transients
  4. Thermal dissipation for whatever thermal time constant are relevant.
  5. Long term thermal dissipation

These are typically the relevant design specifications, ratings in data sheets and what kills parts (if not done properly).

For a very stiff power supply apparent power can be a stand-in for peak current but otherwise it's not particularly useful or relevant.


Aparrent power is what blows fuses, and melts transformers.

a cresty load, like a bridge rectifier circuit with oversized reservoir capacitors will have a higher peak power than aprarrent power, but the peaks are so brief that they don't have time to do damage, and looking tat the apparent power is all that's needed.

  • \$\begingroup\$ Nope. Current melts fuses. \$\endgroup\$
    – Hilmar
    Commented Jun 25, 2021 at 16:47

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