# The three powers of AC Reactance

Discussion in 'General Electronics Chat' started by sjgallagher2, Apr 22, 2013.

1. ### sjgallagher2 Thread Starter Active Member

Feb 6, 2013
128
12
My mind is getting brutalized by all this new reactance knowledge I'm getting! Let me start with a circuit:
All there is, is a 100V AC source, 10A (effective). In the circuit there is one 10Ω resistor R, and one 10Ω inductor XL.
With this circuit I know that I can vectorally (is that a word?) add the two resistances to find the impedance: Z=14.14Ω. To find the phase difference between voltage and current:
θ=sin-1 (XL/Z) = 45°.
Finding power is where things get difficult to comprehend for me. First, I can use the simple P=VI,
P=100*10
P=1000 VARs
Okay that's good, but using the impedance as the resistance in another formula:
P=I^2 R
P=100*14.14
P=1414 VARs.

Well great, now I have two types of power. But that's only the apparent power(s), there's also true power!
True power = VI cos θ
=1000*0.7071
=707.1 watts

So help me! I have two apparent power values, and one true power value, what is what and how the hell do I use these to help design circuits or anything? What's the value of these values? Where does each come in to play?

TL;DR I can find 2 apparent powers and 1 true power in an RL circuit, which is what and how do I use them.

2. ### WBahn Moderator

Mar 31, 2012
23,578
7,214
You didn't indicate whether, in your circuit, the resistor and the inductor were in series or parallel.

You do not have "two apparent powers". You have apparent power and you have reactive power. Apparent power, with units of volt-amperes (VA) is the simple product of RMS voltage and RMS current. Reactive power is NOT apparent power, it is the power being shuttled in and out of the reactive components.

It would have been more illustrative if your resistor and your inductor didn't have the same impedance magnitude, since that makes it easier to get them mixed up.

If your components are in series, then your total impedance has a magnitude of 14.14Ω, as you indicated, and your RMS current (assuming your 100V is RMS) would be 100V/14.14Ω=7.07A. Where are you getting that it is 10A?

Your problem is not consistent. Think about it. If your RMS current is 10A, then your RMS voltage across a 10Ω resistor is 100V. Since that is also your source voltage, you have nothing left to support a voltage across a reactive component.

3. ### sjgallagher2 Thread Starter Active Member

Feb 6, 2013
128
12
Very true, I shouldn't have rushed to put together an example huh? You're input is exactly what I needed to set me straight! They were in series by the way haha I guess you assumed that. Neither will I confuse apparent power and reactive power again, nor will I confuse my current and voltage! I feel kinda dumb now haha

4. ### WBahn Moderator

Mar 31, 2012
23,578
7,214
No need to feel dumb (and we all deserve to from time to time, that's not gonna change for any of us). It is a bit difficult to get your mind wrapped around AC concepts because you probably have just started becoming reasonably comfortable with DC concepts and now the whole apple cart is getting jiggled (sinusoidally?).

Last edited: Apr 22, 2013
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5. ### crutschow Expert

Mar 14, 2008
20,866
5,928
You actually have three types of power -- apparent power, reactive power, and real power. Real power (Watts) is the part that does the work, where the volts and amps are in-phase. Reactive power (Volt-Ampere-Reactive or VAR) is the part that does no work (the part where the current and voltage are 90° out of phase). Apparent power (VA) is the volt-amp product, which ignores any phase shift, and includes both reactive and real power.

Power companies are interested in keeping the VAR value low since it does no work and they do not charge directly for it, but they still have to supply the reactive current which adds to losses in the power lines and transformers.

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