# AC Power: S^2 = P^2 + Q^2 ?

• posted
If waveform is purely sinusoidal, it was said the S^2 = P^2 + Q^2,
where S is apparent power, P is real power and Q is reactive power.
In this formula, I have two questions. Firstly, S, P, Q here are what
kind of power, average power? rms power (rms U * rms I)? Instantaneous
power (U *
I from moment to moment)? After clarify the definition, can
anyone show a prove?
The second question is that I think, for energy, apparent energy should
be a sum of real energy and reactive energy. I learned this from
Andrew's article:
I right on the concept? If so, do you see the energy relationship
between apparent/active/reactive is different with power?
Finally, for these AC concepts, is there a good text book in amazon.com?
• posted
------------------ Average power IS the product of rms U * rms I times power factor where power factor is the cosine of the phase angle between voltage and current P average= Urms* Irms cos (phase angle) Q average =- Urms* Irms sin(phase angle) (sign+ if I leads U ) S =P+jQ =Urms(conjugate of Irms) both expressed as complex numbers In terms of instantaneous power p(t) = Umax*cos(wt) and I(t)=Imax*cos(wt+a) where w is radian frequency and a is the phase shift of current with respect to voltage Integrate this over a cycle to get two terms - a constant term P and a double frequency term which does not contribute to average power but energy shuttled back and forth from the source and inductive/capacitive load-represented by represented by Q Since it is convenient to use rms voltage and current, it follows that we can express P and Q in terms of rms values and the phase angle of voltage with respect to current --------------------------
----------- Instantaneously, there is only power-the product of instantaneous voltage and current- averaging this over a cycle results in a constant P When we use rms U and I we are using a specific form of averaging. In terms or rms values, then it was recognized long ago that |V|*|I| magnitudes (S) don't represent real power ( i.e this represents the magnitude of apparent power) and that only the component of current that is in phase with voltage is what is real (non-zero average) power, The component of current 90 degrees out of phase with the voltage represents the effect of energy being shuttled back and forth.
There are many books available - look for electrical "circuit" theory. There are on-line references and cheap off line references include Schaum's Outlines Amazon has this and other references
This may help
• posted
It seems you know the things, thank you for the explaination. But, I am so fool and I have not get clear the point. Could you directly answer the question: In S^2 = P^2 + Q^2, S,P,Q are rms powers or instantaneous powers?
And, for the prove, I don't understand. Can I find it in what you mentioned Schaum's outline?
Do you agree what I mentioned: apparent energy should be a sum of real energy and reactive energy?
• posted
--------------- They are neither. They are found from a true average of instantaneous current and voltage over a period but can be expressed in terms of the product of rms voltage and current. The term "rms power" has been used but is wrong. It is better to use "true power" Rms power is a useful mode for considering heating of motors with varying loads but that is a different animal than what you are considering.----------
--------- it may not be but is in any general EE text. Calculus is used.
---------- As you have written "S^2 = P^2 + Q^2" This is a sum of squares as per Pythagorus' theorem for right triangles. true power and reactive are , in a phasor or power triangle, at right angles. True power P is the actual power involved with actual net energy transfer while reactive represents energy that is shuttled back and forth between source and load- average net energy is 0. Apparent power is simply what would appear if you measured the voltage and current and multiplied them together as if they were DC. Think of a right triangle with base =P , upright =Q and hypotenuse=S
check this reference