Defining Average Power

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I interpret the bold text as trying to say that the complex portions will contribute to the real part of the complex product. The complex product is another way of getting the same number as the dot product of the two vectors involved.

If you base your zero phase point on the incoming voltage waveform, then the voltage waveform will have only a real component.
Once you have established that, then the current will be at some phase angle relative to the voltage, and only the real part of the complex number that represents the current will contribute to the power.

If you set your zero phase reference point so that both voltage and current are imaginary, then indeed the imaginary part will have to be taken into consideration when calculating real power.

but you need BOTH to perform a dot product
the result is the same
but with only the real components a phase angle can't be determined
hence niether can P

what you are saying
v= 1
i = 1 + 1j
S = 1 - 1j
of course in this case you only need the real
but not in the general case
 
v = 1 + 3 j
z = 2 + 4j
i = ?
does i have a reactive component?
acting on X = 4j ohm
and real or active on R = 2 ohm???

if the load z is 0 + 10j and v = 100 + 0j
is the current reactive or imaginary? Not 'real' or active for certain
will an amp clamp 'see' it?
How about you look back at what was said? Seems to me the OP was talking about a V1&I1 product that produced P (from the real components) and a V2&I2 product that produced Q (from the reactive components).

The 1s and 2s can both be sinusoidal. It did not seem to me that the OP meant separating the complex V&I into its real and imaginary parts. Maybe he did mean that but it I did not read it that way.

Maybe you are mixing the sinusoidal and complex representations? Not sure. Not sure I'm that interested. Just trying to get you & Bes on the same page. I think I'll let you two play.
 
but you need BOTH to perform a dot product
the result is the same
but with only the real components a phase angle can't be determined
hence niether can P

what you are saying
v= 1
i = 1 + 1j
S = 1 - 1j
of course in this case you only need the real
but not in the general case

It is not the general case for engineering problems in textbooks. Nor is it the case for local power calculations of elements inside networks containing reactances. It would be nice if everybody knew how to do the calculations you showed, and any EE must be able to do them.

But if you are making a real world measurement and calculation involving a single phase supplyou will not have an external reference for phase. You will take the voltage as your phase reference and only the current will have a non-zero phase angle, or equivalently a reactive component.
I do not know offhand of any power quality meters, for example, that do not take one of the phase voltages as the time reference.

If you are doing three phase measurements and calculations, then I agree 100% that at least two voltage amplitudes will have imaginary components.

It all comes down to what you are trying to do.
 
but P NOT EQUAL TO 7 x 1 lol
Exactly. But I don't think that was what the OP meant. I read it more like "resistive current and reactive current" like is often used when speaking loosely even though there are not really two currents in the wire.
 
Exactly. But I don't think that was what the OP meant. I read it more like "resistive current and reactive current" like is often used when speaking loosely even though there are not really two currents in the wire.
Resistive current and reactive current, unlike the real and imaginary parts of the current, have meaning only when taking the phase of the voltage as zero. Or, rather, resistive current is defined by being in phase with the applied voltage, while reactive current is all at either 90 or 270 degrees.

The post to which Ingenieur was responding made a general statement about the "real part", not the resistive part.
 
That doesn't make the average power zero though.
Correct. Power is typically an offset sinusoidal waveform. Average power is equal to the amount of offset:

  • [(V∙I)MAX+(V∙I)MIN]/2
...where V and I are instantaneous values through one cycle with steady-state sinusoidal voltage and current.
 
Resistive current and reactive current, unlike the real and imaginary parts of the current, have meaning only when taking the phase of the voltage as zero. Or, rather, resistive current is defined by being in phase with the applied voltage, while reactive current is all at either 90 or 270 degrees.

The post to which Ingenieur was responding made a general statement about the "real part", not the resistive part.
Perspective. Only the OP knows what he meant. Which voltage? Which current? The source voltage on a series circuit or the element voltages? The source current on a parallel circuit or the element currents?

Obviously us seasoned guys know circuit theory backwards and forwards and "us" includes Besoeker and Ingenieur. I just saw no point in them arguing two different points. At least get on the same page and if they have a difference there (don't think they would) then debate that.
 
FWIW.......something I did earlier

1V1A1W_zps788ce72e.jpg
 
FWIW.......something I did earlier
All real, and as sinusoidal waveforms they have both real and imaginary rectangular coordinate parts over time. No "reactive" parts. That is what I was trying to point out to Ingenieur.
 
:thumbsdown:

Would you please discontinue making snide remarks.
That was snide???
It's simply the truth.

Some members on here seem to try to look clever.
My own perspective is to try to keep as simple as possible and stick to the relevant and necessary facts.
 
Telling the truth (or should I say your opinion of the truth) is fine in many cases. I believe the exceptions are where telling the truth is unnecessary and discourteous.
I'm not going to debate this matter - either with you or anyone else.
Can we get back on topic, please?

Power can be averaged over any period but doing so over less than one cycle or a non-integral number of cycles can give misleading results.
It is real so any quadrature components have no bearing on its value.
 
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