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I know that real power is the power consumed by the circuit. But what about average power? Are they the same?

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In a steady-state circuit, if you average the "instantaneous power" over a cycle, you get "real power"... so yes, they are the same.

enter image description here

Furthermore, this is how power meters work, to show what the real power consumed. They average the instantaneous voltage and current (vectors) over a few seconds.

If you know what power factor is, see this answer for more details: https://electronics.stackexchange.com/a/50034/16096

edit:

unless you are taking about some appliance like an air conditioner.. whose power consumption changes as it turns on and off.. then the "average power" would be averaged over the day

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  • \$\begingroup\$ Its not the only way! Microwave power measurements can be taken by measureing the heating effect en.wikipedia.org/wiki/Bolometer thus measuring true delivered power \$\endgroup\$
    – N.G. near
    Jul 31, 2016 at 17:31
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A few examples of V x I and the power waveforms they produce: -

enter image description here

Average power x time (i.e. watt seconds or kWh) is what you are billed on.

You could call real power average power if you want or you could use it to refer to the waveshape of v*i.

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So real power or complex power?

I believe it is a matter of representation of physical quantities, not of physical quantities themselves.

E.g. take voltage or current: they are scalar real quantities.
Then it has been found that using Steinmetz's transform to turn them into complex numbers makes calculations on steady-state sinusoidal quantities easier.
But a complex number is a mathematical abstraction, just even thinking of an imaginary voltage somewhere in a real circuit is totally nonsense.

Though at the end of complex calculation we can return back from representation domain to real physical quantities, get our scope and check them on real circuit.

So real scalar quantities, their complex representation which more generally can be just about anything as long as it is reversible not loosing any of the initial information and back to real life.
This is the round trip which is by the way quite common in engineering (e.g. Laplace, Fourier...)

The same may be applied to power:

In the real world (whatever it means) we have instantaneous power which is a real scalar.
Then using the same complex representation above it can be shown that multiplying voltage by conjugate current we get a complex representation S of power such as its real part P is mean power (real one) while Q, the imaginary one is reactive. $$ S=VI^*\quad P=\text{Re}\{S\}\quad Q=\text{Im}\{S\} $$

Again we have gone from physical quantities, complex representation and back to real.

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  • \$\begingroup\$ Please don't have discussions in an answer. Keep it to the comments. \$\endgroup\$
    – pipe
    Jul 31, 2016 at 21:10
  • \$\begingroup\$ Sorry I tried but was too long to fit in comments. And anyway I believe it is relevant to the question, just in case I could remove recall to N.G.near \$\endgroup\$
    – carloc
    Jul 31, 2016 at 21:15
  • \$\begingroup\$ Yes, I'm sure this answer is fine - I just don't want to see back-and-forth discussions between users in the answer field. It's problematic in a couple of ways. \$\endgroup\$
    – pipe
    Jul 31, 2016 at 21:17

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