Why do we need to take the conjugate in complex power calculations?

AI Thread Summary
Taking the conjugate of the current in the equation S=VI is essential for accurate complex power calculations, particularly to account for power losses in systems. The power loss in components like motors and transformers is proportional to the square of the current's magnitude, which includes both in-phase and out-of-phase components. Using the complex conjugate allows for the cancellation of cross terms, resulting in a correct calculation of real power (P) and reactive power (Q) without introducing errors. This approach simplifies the analysis of power systems by utilizing phasor notation to manage phase differences effectively. Overall, conjugating the current is crucial for ensuring accurate power delivery and minimizing losses in electrical systems.
anon6912
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Ive seen in some situations the equation S=VI is
being used with the conjugate of I

IN what situations do you have to take the conjugate and why?
 
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The complex power S = P + jQ, where P is the real power and Q is reactive power.

http://en.wikibooks.org/wiki/Circuit_Theory/Complex_Power"
 
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Thanks for the reply.
I understand what complex conjugate is.
But I do not know why you have to do it in S=VI
 
The principle reason regards loss in your power systems. The power loss in the windings of you motor / transformer / whatever varies to the square of the magnitude of the current. That is

Ploss = R * (Ii^2 + Iq^2)
Where R is your resistance,
Ii is the current that is in phase with your supply voltage
Iq is the current that is out of phase with your supply voltage

That wouldn't be so bad, except that the power being delivered to the shaft of your motor, to the load of your transformer, or to the whatsit of your whatever, is only going to be:
Pload = Ii x V
Where V is the line voltage.

So, Ii contributes nothing to your load, but does a disservice in your windings, wiring, and even out on the utility grids.

As for the complex number, that just gives a simple, phasor notation to keep track of this out of phase current. You can design in both Ii sources and Ii sinks. For example, given that an induction motor has an impedance that has a phase lag, you can drop in a parallel capacitor which will introduce a phase lead. Thus your building (and the utility) don't have to dissipate the power from Ii.

Mike
 
A short non-theoretical answer is that the amplitude of the power is the product of the magnitudes of the voltage and current phasors. If you conjugate one, the cross terms go away and you have a Pythagorean norm (sum of squares). If you don't conjugate, you get quadratic cross terms that give the wrong answer for power. Think of S=V*V / R instead if V*I. The right way to square a complex number to get it's magnitude is to multiply it by it's conjugate.

The situations: any time you have a time harmonic formulation with complex phasors or vector phasors. If you need a rigorous theoretical derivation I can give you references.
 
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