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Finding the instantaneous power

  1. Feb 2, 2014 #1
    1. The problem statement, all variables and given/known data
    5WxmuXj.png

    2. Relevant equations
    P = IV

    3. The attempt at a solution
    The solution is there and they did it by multiplying everything in the time domain.

    However, I decided to try to multiply the phasor forms to get the power but that did not work...

    V*I = sqrt(2)cos(377t) * sqrt(2)cos(377t+pi)
    = (sqrt2<0)(sqrt2<180)
    = 2<180
    = 2cos(377t+pi)

    What is wrong with this?
     
  2. jcsd
  3. Feb 2, 2014 #2
    We want to simplify I first..
    ##√(2)cos(337t + \pi) \equiv -√(2)cos(337t) ##
    Therefore, ##I = -√(2)cos(337t) ##
    ## P = IV = (-√(2)cos(337t))(√(2)cos(337t)) = -2cos^2(337t) ##
    By the double-angle formulas:
    ##cos 2u \equiv 2cos^2u - 1##
    ##-cos 2u - 1 \equiv -2cos^2u##

    $$ -2cos^2(337t) \equiv -cos(754t) - 1$$
    $$ -cos(754t) - 1 \equiv cos(754t + \pi) -1 $$

    The key to this problem is recognizing that $$√(2)cos(337t + \pi) \equiv -√(2)cos(337t)$$
    Leaving your answer as ##-2cos^2(337t) ##would be fine since ##-2cos^2(337t) \equiv -1+1cos(754t + \pi). ##
     
    Last edited: Feb 2, 2014
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