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

Homework Help: Apparent power calculation

  1. Nov 26, 2012 #1
    A circuit has a power factor 0.72 lagging and the power dissipated is 375w.

    (A) Apparent power

    Relevant equations:
    ohms law V = I x R
    apparent power (S) = supply voltage (Vs) x Current (I)

    Current =

    = 12Amps

    ∴ Apparent power (S) = Supply voltage (Vs) x Current (I)
    120V x 12 Amps

    = 1440 VA

    I have seen another equation where it states that true power is the power dissipated at 375W therefore my second attempt which I am very unsure of as I cannot locate the stated equation within my book:

    altenative attempt
    S (Apparent power) =
    P (True power)
    PF (power factor)


    = 520.83333

    please help?
    1. The problem statement, all variables and given/known data

    2. Relevant equations

    3. The attempt at a solution

    Attached Files:

  2. jcsd
  3. Nov 26, 2012 #2
    The current is V/(Z+R) where Z is an unknown impedance.

    This is right. I take it the question is asking for the unknown impedance Z?

    I think maybe a little review of what S is will help. Suppose a sinusoidal voltage V is applied to an impedance Z. Then a current I will flow that is also sinusoidal but at some phase offset. In the time domain the voltage is Vcos(wt) and the current is Icos(wt-θ) where θ could lead or lag depending on the load.

    The instantaneous power is
    P(t) = VI cos(wt)cos(wt-θ) = VI [cos2(wt)cos(θ) + sin(wt)cos(wt)sin(θ)]

    The average power over a period is
    Pav = (VI/T) ∫[cos2(wt)cos(θ) + sin(wt)cos(wt)sin(θ)]dt
    = (VI)(w/(2∏)) ∫(1+cos(2wt))cos(θ)/2 dt
    = (VI)(w/2∏)cos(θ)(1/2)(2∏/w)
    = (VI/2)cos(θ)
    = VrmsIrmscos(θ)

    The complex power S is defined as
    S = VI*
    where * is complex conjugate and both V and I are rms voltages.

    The complex conjugate on I is taken so that the angle of S will be the difference in phase between V and I, which is what is important in the average power calculation above. If you sketch S on the complex plane, its magnitude will be |VrmsIrms| and its angle will be θ, the angle between the V and I phasors. If you take the real part of S,

    Re(S) = |S|cos(θ) = VrmsIrmscos(θ)

    This is the average power consumed by the load. So S projected on the real axis in the complex plane is the real power. S projected on the imaginary axis is indicative of the power being stored and released in the reactive components (this can be seen from the instant power equation above); no average power is consumed by reactive components. Over a period, energy is consumed by the reactive components and then the same amount is released; this extra current must be absorbed and supplied by the source through the cycle.

    In your problem you are given the real power of 375W. This is S projected on the real axis. So the magnitude of S is 375/cos(θ) = 520.8VA as you found.

    Since you now know |S|=|VrmsIrms| and you know Vrms, you also know |Irms|. Given V, |I| and information on the angle between V and I, you should be able to determine the total impedance seen by the source.
    Last edited: Nov 26, 2012
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook