# Complex power question?

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#### MikeMl

##### Well-Known Member
Suppose you have load R+jX that is fed with a Current Source.

What is the total power delivered by the current source?

What is the power in the R part of the load?

What is the power in X part of the load?

#### dknguyen

##### Well-Known Member
An AC current source?

Isn't it just

Apparent/Complex Power: S = (I^2)*Z = (I^2)(R + jX)
Real Power: P = Real(S) = S*cos(theta)
Reactive Power Q = Imaginary(S) = S * sin(theta)

Where "I" is a phasor and theta is the phase angle of the phasor. Can't seem to find a way to make the angle character here. But I suppose if "I" is just set to zero degrees and all other phase angles in the circuit are referenced off that, then you can just use treat it like a magnitude which gives you:

Real Power: P = Real(S) = magnitude(S)*cos(theta) = S*cos(0)= (I^2)*R
Reactive Power Q = Imaginary(S) = magnitude(S) * sin(theta) = S * sin(0) = (I^2)*X

EDIT: I'm a bit wary though since something like (I^2)*R is only valid for AC if "I" is RMS, but in here it's obviously the peak value/magnitude of the phasor. At the same time though it feels wrong to just throw in a 0.707 fudge factor...unless the magnitude of the phasor is already the RMS value... but that seems really weird to me since it does not seem physically representative to have a rotating RMS line. But maybe we really do just throw in a 0.707 fudge factor and the phasor magnitude represents the RMS just so we can use the conventional equations. I'm pretty damn sure there needs to be 0.707 in there, I just don't know how to work my way into introducing it in other than saying "it has to be there cuz RMS".

Okay, now I have the same question too lol.

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#### MikeMl

##### Well-Known Member
Yes, an AC current source at one frequency.
Assume we know the rms current of the source.

Off the top of my head without thinking about it too much, is the power in R just I^2R, while the power in X is zero?

#### dknguyen

##### Well-Known Member
Yes, an AC current source at one frequency.
Assume we know the rms current of the source.

Off the top of my head without thinking about it too much, is the power in R just I^2R, while the power in X is zero?

Yes, conceptually, the real power is all in R and will be I^2R where I is in RMS. The real power in X is zero. The reverse happens for reactive power. Just working up to it mathematically is...a bit less obvious.

#### Ian Rogers

##### User Extraordinaire
Forum Supporter
Off the top of my head without thinking about it too much, is the power in R just I^2R, while the power in X is zero?
Ideally... But X will always be a portion... The difference ( phase angle ) has to be added.. If you had 4 watts real power and 3 watts imaginary power you get a phase angle of 30 degrees so a apparent power of 5 watts..

But!! The leccy board can't bill you for the 1 watt unless they come down and check... This is why anything over 0.9 degree phase angle is not allowed...

#### MikeMl

##### Well-Known Member
Ideally... But X will always be a portion... The difference ( phase angle ) has to be added...

The power company delivers a constant voltage, and then bills me for "real" power provided that the power factor is not too far out...

My question was actually motivated by modeling an antenna that is fed from a constant current source. If the antenna radiates, but is reactive, it seems all of the power from the current source ends up being radiated (discounting wire losses) or coupled into the lossy earth under the antenna. None of the power is dissipated in the reactive part of the antenna's complex feedpoint impedance?

#### Ian Rogers

##### User Extraordinaire
Forum Supporter
Not into antenna's and I have no idea about their impedance.. Sorry I assumed from post 1 we were talking about power consumption on the whole... Are you modelling it for spice?

#### MrAl

##### Well-Known Member
Suppose you have load R+jX that is fed with a Current Source.

What is the total power delivered by the current source?

What is the power in the R part of the load?

What is the power in X part of the load?

Hi,

If you have a load that is R+jX with R the real part and X the imaginary part, then that is just called a complex impedance. The simplest example is R+jwL for a resistor in series with an inductor.

As you know, for a resistor the power is:
P=I^2*R

and if you have two resistors R1 and R2 in series the total resistance is R1+R2 and have current source I1 then the power in each is:
P1=I^2*R1
P2=I^2*R2

Now replace R2 with X and we get:
P1=I^2*R1 as before, and
P2=I^2*X

except now P2 is the reactive power and P1 is the real power.
The apparent power is the norm of the complex sum of the two P1+P2*j (note inclusion of the 'j' here) so it comes out to I^2*sqrt(R^2+X^2).

So in the end, R+jX acts like a series circuit made from a resistor and inductor, and in a series circuit the current is the same in all elements.

#### crutschow

##### Well-Known Member
None of the power is dissipated in the reactive part of the antenna's complex feedpoint impedance?
Reactive impedance never dissipates real power, so I don't understand your question.

#### MikeMl

##### Well-Known Member
Reactive impedance never dissipates real power, so I don't understand your question.

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