In Diar Need of HELP! PLEASE!!!!!(Current phase)

Discussion in 'Homework Help' started by lsecrease, Sep 4, 2008.

  1. lsecrease

    Thread Starter New Member

    Sep 4, 2008
    6
    0
    Ok, I realize this to be true but how do you prove that a circuit whose current is 90 degrees out of phase with the driving voltage consumes no power, averaged over an entire cycle???

    Thanks a MILLION!!!:):):)
     
  2. Ratch

    New Member

    Mar 20, 2007
    1,068
    3
    lsecrease,

    Why, it is childishly simple. Assuming sinusoidal waveforms, we get the average power by taking the average of the instantaneous power over a complete period. The instantaneous power of a "90 degrees out of phase" voltage*current is proportional to sin(x)*cos(x). The integral of sin(x)*cos(x) is (sin^2(x))/2. Pluging in the limits of the period 0 to 2*pi gives 0. Dividing by the period to get the average is still 0. Doing the same with and in-phase voltage*current means using either sin^2(x) or cos^2(x), which gives an answer of 1/2.

    Ratch
     
    Last edited: Sep 4, 2008
  3. Papabravo

    Expert

    Feb 24, 2006
    10,179
    1,800
    Can you do it without using calculus? I know it, but maybe the OP doesn't.
     
  4. Ratch

    New Member

    Mar 20, 2007
    1,068
    3
    Papabravo,

    Until the OP says otherwise, I will assume he does.

    Ratch
     
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