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

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Discussion Overview

The discussion centers on the necessity of using the complex conjugate in the equation S=VI for complex power calculations. Participants explore the theoretical underpinnings and practical implications of this approach in electrical engineering contexts, particularly in relation to power systems and phasor notation.

Discussion Character

  • Technical explanation
  • Conceptual clarification
  • Debate/contested

Main Points Raised

  • Some participants question the specific situations in which the conjugate of current must be used in the equation S=VI.
  • One participant explains that complex power S is defined as P + jQ, where P represents real power and Q represents reactive power.
  • Another participant discusses power loss in systems, noting that the power loss varies with the square of the current's magnitude and highlights the distinction between in-phase and out-of-phase currents.
  • A participant mentions that using the complex conjugate helps eliminate cross terms when calculating power, leading to a correct representation of power as a sum of squares.
  • It is suggested that the use of complex numbers allows for a clearer representation of phase relationships in electrical systems, particularly with inductive and capacitive components.

Areas of Agreement / Disagreement

Participants express varying levels of understanding regarding the necessity of the conjugate in complex power calculations, with some providing explanations while others seek clarification. No consensus is reached on a definitive answer to the original question.

Contextual Notes

The discussion includes assumptions about the reader's familiarity with complex power concepts and phasor notation. Some mathematical steps and definitions are not fully resolved, leaving room for further exploration.

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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