Apparent Power and Complex Conjugate Confusion 1

[patsweet]

Hey all,

I'm trying to make sense of why apparent power is calculated using the complex conjugate of I, and not just regular I. The equations in my textbook go something like this...

P = VI*cos(theta)
Q = VI*sin(theta)

... where theta is the angle of the impedance element, and thus the lead/lag difference between the voltage and current of this system. The text goes on to say...

S = P + jQ
= VI(cos(theta) + jsin(theta))

... then, using Euler's identity...

S = VI*e^(j*theta)
= VI@ angle theta
= VI* {here the * indicates complex conjugate}

It's that last step that is confusing me. How can the angle's polarity just be reversed at the last minute like that?

Thanks for all your help!

 

[electricpete]

In the first case, theta could represent angle of impedance into which the power is flowing as you say:
P = VI*cos(theta)
Q = VI*sin(theta)
(although it should be pointed out P and Q are used in many networks where there is not a single impedance that can be easily associated with P and q).

Since Z = V / I, we know the angle of V minus angle of I is the angle of Z.
i.e. that’s what we called theta.

The quantity S = (V I*) has an angle which is the sum of the angle of V and the angle I*.
The angle of I* is the negative of the angle of I.
So the angle of S is the angle of V minus the angle of I. i.e. it is what we called theta.


=====================================
(2B)+(2B)' ?

 

[patsweet]

Hi electricpete,

Awesome. This makes sense. Ultimately, apparent power has the same theta at the impedance (even if, as you mentioned, there's no ONE specific impedance element).

Thanks a ton!
Pat

 

[ijl]

According to Grainger (for example) the conjugate I* must be used because of the _agreement_ that the reactive power Q of an inductive circuit (lagging current) is positive. (i.e. it has been _agreed_ that Q has the same angle as the impedance, as patsweet puts it )

 

[odlanor]

P = VIcos(theta) in watt
V= V1 + jV2 and I = I1 +jI2; then average power is algebric sum of the product of the real parts and the product of imaginary parts.
Thus, P = V1I1 + V2I2 in watt
This P may be obtained from the conjugate method of calculating
power. That is,
S = P + jQ = VI* = (V1+jV2)(I1-jI2)
= V1I1 + V2I2 + j (VI2-V1I1) ; real part is P.

 

[jghrist]

In a nutshell, inductive current lags the voltage, but to get inductive power to be positive, you have to use the complex conjugate. It's all a matter of convention.