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Average power.

  1. Oct 27, 2009 #1
    A resistor draws a current i=8sinwt at a voltage of v=200sinwt. Calculate the average power dissipated in the resistor.

    What i did is p=ui = 1600sin^2 (wt) and i got stuck:P i dont think it's the right equation.. the answer should be 800W .. and that's nothing like it.
    Could someone help me ?
     
  2. jcsd
  3. Oct 27, 2009 #2

    tiny-tim

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    Hi dortec! :smile:

    (have an omega: ω and try using the X2 tag just above the Reply box :wink:)

    Hint: for a resistor, the power factor, cosφ, is 1, and so Paverage = VrmsIrms :wink:
     
  4. Oct 27, 2009 #3
    Can you rephrase what you just said please. i actually didnt understand. wat's rms? and what do u mean by have an omega.. can u re-explain please:S
     
  5. Oct 27, 2009 #4

    tiny-tim

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    Hi dortec! :smile:

    rms means "root-mean-square" … do you know what that is?

    (if you don't, you can get the same result by finding the average value of sin2ωt, which you get by integrating it from 0 to 2π)

    (and where you typed w, I assumed you would have preferred ω. :wink:)
     
  6. Oct 27, 2009 #5
    nope i dont know what root-mean-square:S and my bad for the omega thing:P.. can u please show me how to solve it?
     
  7. Oct 27, 2009 #6

    tiny-tim

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    ok, never mind if you haven't done root-mean-square …

    just find the average value of your original P(t) = 1600sin2ωt, which you get by integrating it from 0 to 2π/ω, and dividing by … ? :smile:

    (alternatively, just write 1600sin2ωt in terms of cos2ωt and/or sin2ωt, and then the average is obvious :wink:)
     
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