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Complex power calculations

  1. May 9, 2009 #1
    I'm confused about a point concerning complex power calculations. The formula my text gives is S=V(eff)I(eff)*. I'm confused about what the conjugate is for I(eff). The way I understand it, I(eff) = I(rms) and I(rms) = I(m)/√2. Since I(m) is a real number, and I(m)/√2 is real, how is there a conjugate.
    For example:
    i(t)= 20cos(ωt+165)
    I(m)=20
    I(eff)=20/√2
    ...so what is I(eff)*?

    I've read through the textbook book the only examples that I can find use the equation: S=(1/2)VI*.

    I would be grateful for any help you could provide to help me understand this.
     
  2. jcsd
  3. May 9, 2009 #2

    tiny-tim

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    Hi FrankJ777! :smile:

    Complex power is supposed to be complex.

    For a perfect resistor, the complex power is purely real,

    and for a perfect capacitor or inductor, it's purely imaginary …

    but for a mixture, it'll be "genuinely complex", but with a constant phase …

    (because we assume the current leads the voltage by a constant phase, and so 1/2 VI* subtracts the two phases, and keeps the total phase constant :wink:) …

    I think :redface:
     
  4. May 9, 2009 #3
    If
    V(t) = Vmax exp[jwt + phi] and
    I(t) = Imax exp[jwt + phi] then
    S = (1/2) V(t) I(t)* is the rms real power.
    But even if you use rms values, you still have both volt-amps and watts.
     
  5. May 10, 2009 #4
    Also...

    v(t) = Vmax/ exp[jwt]
    i(t) = Imax exp[j(wt + phi)]

    Veff (phasor) = v(t)/√2
    Ieff (phasor) = i(t)/√2

    I*eff = Imax/√2 exp[-j(wt + phi)] (note minus sign)

    so (product of phasors)
    Veff x I*eff = Vmax/√2 exp[jwt] x Imax/√2 exp[-j(wt + phi)]

    Seff = Vmax Imax/2 exp (-jphi)
    (plus exp of voltage time and minus exp of current conjugate time cancel, leaving only phi part, that's why conjugate is right for this)

    real power P = Real {S} = Veff Ieff cos (-phi)
    reactive power Q = Imag{S} = Veff Ieff sin (-phi)
    (so 'Reactive Power' is negative for inductive circuits where phi is positive)

    the phasors make a vector triangle:
    S^2 = P^2 + Q^2

    Sorry the formatting is not better. Still, I hope this helps.
     
  6. May 10, 2009 #5

    tiny-tim

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    Welcome to PF!

    Hi Dr.kW! Welcome to PF! :smile:

    (have a phi: φ and an omega: ω and try using the X2 and X2 tags just above the Reply box :wink:)

    Action replay (of Dr.kW :wink:):-
    v(t) = Vmax/ ejωt
    i(t) = Imax ej(ωt + phi)

    Veff (phasor) = v(t)/√2
    Ieff (phasor) = i(t)/√2

    I*eff = Imax/√2 e-j(wt + φ) (note minus sign)

    so (product of phasors)
    Veff x I*eff = Vmax/√2 ejωt x Imax/√2 e-j(ωt + φ)

    Seff = Vmax Imax/2 exp-jφ
    (plus exp of voltage time and minus exp of current conjugate time cancel, leaving only φ part, that's why conjugate is right for this)

    real power: P = Real {S} = Veff Ieff cos (-φ)
    reactive power: Q = Imag{S} = Veff Ieff sin (-φ)
    (so 'Reactive Power' is negative for inductive circuits where φ is positive)

    the phasors make a vector triangle:
    S2 = P2 + Q2 :wink:
     
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