# How to calculate Apparent Power

Discussion in 'Homework Help' started by JDR04, Jul 23, 2012.

1. ### JDR04 Thread Starter Active Member

May 5, 2011
339
4

I'm given an ac circuit that has 15V and 2A. I need to calculate Apparent Power.

I'm not so confident in what I have tried so would appreciate anybodies help on this one.

Pa = Apparent power expressed as volt-amperes
E = Voltage expressed as volts
I = current expressed as amperes.

So..... Pa = E x I therefore Pa = 15 x 2 = 30. Pa = 30 volt-amperes.

Could somebody be kind enough to let me know if I am correct. If I am wrong I would appreciate it if you could point out to me where I went wrong.

Thanks guys - JDR04

2. ### #12 Expert

Nov 30, 2010
18,078
9,616
I believe you are right because the thing with the phase angle figured in is called, "True Power".

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3. ### tracker Member

Jun 1, 2011
41
1
it's right.
the source apparent power is the maximum power the source can deliver so it's the product E.I

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4. ### WBahn Moderator

Mar 31, 2012
22,861
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This is assuming that your 15V and 2A are RMS values. They probably are, but read the question carefully just to be sure that they aren't throwing you a curve ball by giving you peak or peak-peak values.

5. ### JDR04 Thread Starter Active Member

May 5, 2011
339
4
Good point Wbhan, never thought of that. I did however check after you pointed it out and it is RMS values. Easy to overlook.......thanks a lot =JDR04

6. ### hexram New Member

Feb 29, 2012
8
0
Apparent power is a complex quantity. It should be expressed as a complex number, i.e., real part and imaginary part if in rectangular mode or module and angle if in polar mode. If we assume that the given 15V and 2A are the modules of the voltage and the current there where I need to calculate apparent power, then the best I can say is that the module of the apparent power is 30VA, that is, assuming RMS values as should be the case. What can not be answered is the angle of the apparent power, which should be equal to the difference between the voltage angle and the current angle; this is because $S = V I^*$, where $I^*$ is the conjugate of the current, that is, the complex number with the same module and the negative of the angle of the current.

7. ### The Electrician AAC Fanatic!

Oct 9, 2007
2,561
459
Apparent power is not a complex quantity; complex power is a complex quantity. See:

http://en.wikipedia.org/wiki/Complex_power

8. ### mlog Member

Feb 11, 2012
276
36
So in other words, apparent power is the magnitude of complex power.

9. ### hexram New Member

Feb 29, 2012
8
0
May I dissent? AC circuits have no other kind of power than apparent power, denoted as S and expressed as the product of the complex V to the conjugate of the complex I. Real power, as its name implies, is the real part of the apparent power (mostly known as P and expressed as S times the cosine of the difference between the angle of the voltage and the angle of the current); reactive power is the imaginary part of the apparent power (mostly known as Q and expressed as S times the sine of the aforementioned difference between the angles).
When solving AC circuits we use a mathematical model based on complex numbers instead of using trigonometric equations; power, as we measure it physically, is represented by the real and imaginary parts of this model of power that we call "apparent power". What we are charged for by the electric company is real power. Reactive power is present as part of the operation of inductors and capacitors in AC circuits; they are accounted for by the electric company through the use of the "power factor", which is just the cosine of the difference of the voltage angle and the current angle.
I am not very fond of Wikipedia because I cannot trust the authority of the people who publish in there. Any book of electrical circuits (my personal favorite is Hayt and Kemmerly) may give a better explanation of these topics.

10. ### WBahn Moderator

Mar 31, 2012
22,861
6,827
Then, by all means, don't use it. Or, follow the links to the source material that most techy Wikipedia articles are pretty good about providing.

Perhaps these will be more to your liking:

Gee, I haven't opened this book for nearly a quarter of a century, but that last sentence is almost verbatim the way that I explain apparent power (and which you can find at least a couple examples of in other threads).

A couple of others from my shelf:

Some texts skirt the issue (probably not intentionally, just as a consequence as the organization of topics they have chosen) by introducing apparent power before delving into complex representations. So they define apparent power as the product of V and I and since they have only introduced V and I as being magnitudes, apparent power is a real number that is the product of the magnitudes. They are explicit about stating that the quantities involved are effective rms values. But you don't really know what their true intention about distinquishing between complex power because they don't always get to complex power at all (or at least not that is easy to track down via the index). This appears to be the case with:

Vincent Del Toro, Electrical Engineering Fundamentals, 2nd Ed, 1986, Prentice Hall, p264.

Allan R. Hambley, Electrical Engineering: Principles and Applications, 3rd Ed, 2005. Printice Hall, p217.

K. Heumann, Basic Principles of Power Electronics, 1986, Springer-Verlag, p116.

In all, I reviewed twelve texts (about half of which don't use apparent power at all for one reason or another -- usually because the focus is more on electronics and what little A.C. power is covered is very superficial). None of them is consistent with the claim that apparent power is the same as complex power. Everyone that discusses apparent power is clearly using it as a fundamentally real quantity and several are explicit about the distinctioh between complex power and apparent power.

11. ### hexram New Member

Feb 29, 2012
8
0
Yes, you are right.

Power, when dealing with AC circuits, is a tad less than a mess; you begin talking about instantaneous power (this is the product of voltage and current on any element, both taken as functions of time - $p(t) = v(t)i(t)$), then you continue with average power (this is the average of the instantaneous power over an interval - $P = \frac{1}{t_2-t_1}\int_{t_1}^{t_2} p(t)dt$), then you take a side step to talk about effective values ($I_{eff} = \sqrt{\frac{1}{T}\int_{0}^{T} i^{2}dt$, for instance), and finally you say that $P = V_{eff}I_{eff}$ is not a power as such, but rather an apparent power. Now, just for the sake of it, try replacing $v(t) = V_{m}cos(\omega t + \theta)$ and $i(t) = I_{m}cos(\omega t + \phi)$ in the expressions above and you may get a glimpse of why electrical engineering was not a very popular option back in the 1900's.

But there is light at the end of the tunnel. If we create a model of our AC circuit in which every argument is represented by complex numbers, then we may spare all the derivatives and integrals and instead use normal complex arithmetic operations to solve it. For the power we "invent" a complex number that we call complex power, whose absolute value is the apparent power, and we define it as $\bar S = \bar V \bar I^{*}$, where the asterisk represents the conjugate of the complex number that is the current in our model; the only caution remaining is that we have to employ effective values instead of maximum values, i.e., $\bar V = V_{eff}e^{j\theta}$ and $\bar I = I_{eff}e^{j\phi}$, but even this is a minor issue when dealing with sinusoidal voltage and current, because $V_{eff} = \frac{V_{m}}{\sqrt{2}}$ and $I_{eff} = \frac{I_{m}}{\sqrt{2}}$ in this case.

Last edited: Aug 12, 2012
12. ### WBahn Moderator

Mar 31, 2012
22,861
6,827
I don't follow what you mean by electrical engineering not being a very popular option back in the 1900's. Seems like it was pretty popular to me. What supposedly happened in the 2000 time frame to suddenly make it popular in the 2000's when it, supposedly, wasn't in the 1900's?

The use of complext numbers (phasors) to represent AC quantities falls directly from the application of Fourier analysis to the differential equations describing circuits. As you say, these transform the derivatives and integrals into algebraic operations, but Fourier analysis has been around for over two centuries.

And we certainly didn't invent the notion of a complex conjugate just so that we could express complex power.

The notion of complex power, however, is an artificial construct and does not come out of Fourier analysis. Fourier analysis is only applicable to linear systems and power is non-linear. Thus is should not be surprising that we can't just take the Fourier transform of the voltage signal and multiply it by the Fourier transform of the current signal to get the Fourier transform of the power. In fact, the complex power, S, is not the Fourier transform of anything. It is was constructed in order to permit some easy book keeping for some specific non-linear properties of sinusoidal systems in steady state and can't be pushed much beyond that very specific purpose.

13. ### hexram New Member

Feb 29, 2012
8
0
Sir, I am not interested in starting a flame war with you. If your personal opinion of the popularity of electrical engineering in the 1900's differs from mine, then so be it; maybe if you do some research on this topic you will find some facts that will certainly enlighten your actual criteria, but then it is nothing of my business to do something about it. If your opinion on Fourier analysis is so great, then maybe a brief read of "Theory and calculation of alternating current phenomena" by Charles Proteus Steinmetz and Ernst J. Berg, 1897 can also enlighten your actual knowledge, but then it will not change the fact that I have been working as an electrical engineer for the last 30 years. Maybe you failed to read the first statement of my previous post; here it is: "Yes, you are right."

This is my last post in this forum. Thank you for reading.

14. ### WBahn Moderator

Mar 31, 2012
22,861
6,827
What flame war? I'm not the one that all of a sudden is taking a general discussion as a personal affront. I simply stated that my experience does not support the claim that electrical engineering was not a popular option in the 1900's and that it somehow has become popular only in the last dozen years or so. Apparently I should never have asked for an explanation of what brought about the sudden and recent change since you have been working as an electrical engineer for 30 years (and, presumably, one of the very few since it was so unpopular). I had no idea when I received a full scholarship in electrical engineering 29 years ago that it was so unpopular; I guess I was fooled by how many colleges and universities offered EE degrees and how full the classes were.