Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Double check apparent power definition

  1. Oct 27, 2015 #1
    Hey just to make sure,
    can we say apparent power S = V^2 / Z

    for a complex impedance Z?
    And do we have to worry about conjugates at all?

  2. jcsd
  3. Oct 27, 2015 #2


    User Avatar
    Science Advisor
    Gold Member

    The total power S = VI. Where S, V and I are all complex. I prefer that to $$\frac{V^2}{Z}$$ because VI works for any circuit.

    Is apparant power the same as total power? I'm not exactly sure what you mean by apparent power.

    Edit: no conjugations were needed.
  4. Oct 27, 2015 #3


    User Avatar
    Gold Member

    I always thought apparent power is the voltage across the complex impedance multiplied by the current through the complex impedance. So, V * I, or V^2 / Z, or I^2 * Z. I don't think I have heard of a definition of total power. But that just might be semantics.
  5. Oct 27, 2015 #4
    First, you have the complex power:
    \mathbf{S} = \mathbf{V}\overline{\mathbf{I}}
    where ##\mathbf{V},\mathbf{I}## are RMS phasors. The overline means to take the complex conjugate.

    The apparent power is defined as:
    |\mathbf{S}| = |\mathbf{V}\overline{\mathbf{I}}| = |\mathbf{V}||\overline{\mathbf{I}}| = |\mathbf{V}||\mathbf{I}| = \frac{|\mathbf{V}|^2}{|\mathbf{Z}|}
    which is consumed by the impedance ##\mathbf{Z}## with the voltage ##\mathbf{V}## across it.
  6. Oct 27, 2015 #5
    Yeah Averagesupernova, I think that's right, I always think of it as being the hypotenuse of the reactive and real power.
  7. Oct 27, 2015 #6
    Hi Miles, thanks.
    Ah ok, it is for magnitudes. But what about if I still wanted it all to be in terms of comples voltages, currents and powers?

    Specifically because I'm trying to work out a newton raphson here:


    Using the aforementioned power equation. I'm trying to work out the complex voltages for 4 busses in Matlab. Where I have worked out the matricies for the complex power for each bus and the admittance matrix.

    Thank you
  8. Oct 28, 2015 #7
    ##\mathbf{V}## and ##\mathbf{Z}## are complex.

    When you need the apparent power, then you could just, for instance, use the 'abs' function in MATLAB to find ##|\mathbf{V}|,|\mathbf{Z}|##.
  9. Oct 28, 2015 #8
    Hey Miles,
    Yeah I actually tried that yesterday, but now I'm thinking I can't use apparent power. So is it incorrect to say that:

    Complex power = complex voltage * complex admittance?
    like P + Q = V^2 / (R + X)

    I think one thing that might be wrong is my jacobian.
    From what I remember NR uses the form y = f(x)
    where y is a constant (in this case the powers) and x is the voltages ('volts' matrix), so my function is

    so the jabobian is:

    ? Or have I not differentiated that properly?
    Last edited: Oct 28, 2015
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Similar Discussions: Double check apparent power definition
  1. Double Power Supplies (Replies: 4)