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Do we get 2x the frequency when doing P=VI for AC Circuits?

  1. Jan 24, 2016 #1
    I know that when you get a current from a voltage (like calculating current through resistor), the current and voltage equations have the same frequency. Does this still hold for power P = VI ?

    So assume V in Euler form, V(t) = Vcos(wt+theta) + jVsin(wt+theta) = V*ej(wt+theta) and in Phasor, V<theta = Vcos(wt+theta)

    And assume I in Euler form, I(t) = Icos(wt+phi) + jIsin(wt+phi) = I*ej(wt+phi) and in Phasor, I<phi = Icos(wt+phi)

    Then P = V(t)*I(t) = V*ej(wt+theta) * I*ej(wt+phi)

    To get P = VI*ejwt+jwt+jphi+jtheta, call jphi + jtheta = jpsi
    Then P = VI *e2jwt+jpsi
    Then P = VI *ej(2wt+psi)
    Then P = VIcos(2wt+psi)+jVIsin(wt+psi)

    Real component is what we want, so P = VIcos(2wt+psi)

    2wt (power frequency) isn't the same as wt (the frequency of current and voltage). Did I get the right result?
     
  2. jcsd
  3. Jan 24, 2016 #2

    meBigGuy

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    Think of the power as peaking on the positive voltage peak and the negative voltage peak, which is twice the frequency.
     
  4. Jan 24, 2016 #3

    jim hardy

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    meBigGuy answered it

    here's a picture to help

    age=http%3A%2F%2Felectrical4u.com%2Felectrical%2Fwp-content%2Fuploads%2F2013%2F03%2Factive-power.png

    any time volts and amps have same sign their product is positive
    and that happens twice per line cycle

    see - you learned that in first year algebra. Much of learning is really just discovering what we already know.
     
  5. Jan 24, 2016 #4

    meBigGuy

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    That's the picture I was looking for!
     
  6. Jan 24, 2016 #5

    Averagesupernova

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    Not to hijack the thread but notice that there is now also a DC component. This is like AM modulation in that 2 signals are multiplied together to get sum and difference frequencies. In this case it so happens that the two input signals are the SAME frequencies.
     
  7. Jan 25, 2016 #6

    meBigGuy

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    The trig identity for squaring a sinewave (since power is E^2/R) shows the DC and the 2x frequency
    sin2(x) = ½[1 – cos(2x)]
     
  8. Jan 28, 2016 #7

    jim hardy

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    sin^2x.jpg

    I envy you who are fluent in math.
     
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