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Wattage delivered using triple integral help

  1. Oct 16, 2011 #1
    Useful equation.

    Avg. Power
    p(t)=(V^2(t))/R

    My attempt at instantaneous power was

    p(t,V,R)= ∫(0->1 for t ∫0->5 for V and ∫0->.1 for R V^2(t)/RdvdRdt

    Integrating I go the triple integral of V^3t^2/6R^2

    Substituting my values in gave a wattage of 1,250 watts/m^2 at t=1 second, v=5 volts, and R=1 ohms

    Using Ohms Law V=IR or I=V/R I calculate now as I write this a current of
    I=5/.1=50 amps

    Is this correct?
     
  2. jcsd
  3. Oct 16, 2011 #2
    I do not know where you came up with the triple integral thing...you need to review this concept.

    If anything, it should be the integral of V/R.

    Is your R really varying? If not, it would simply be the integral of V

    By the way, p(t) = (v(t))^2 / R is not average power, it is instantaneous power

    The average power would by the integral of the instantaneous power over a period of time divided by that amount of time.
     
  4. Oct 16, 2011 #3
    If R was varying would I be right for instantaneous power? The triple integral follows the rules for setting a triple integral in relation to the variable. The values given are set for R being constant though they should be the same in that case.
     
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