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Instantaneous power as a function of time

  1. Aug 21, 2015 #1
    1. The problem statement, all variables and given/known data
    If v(t) = t - 4 and i(t) = 3t, find the instantaneous power p(t) as a function of time.

    2. Relevant equations


    3. The attempt at a solution
    p(t) = ∫ v(t)*i(t) dt
    p(t) = ∫ (t-4)*(3t) dt

    Is it correct to do this? Or am I supposed to take the derivative of the functions v(t) and i(t) first and then multiply them and take the integral? Thanks!
     
  2. jcsd
  3. Aug 21, 2015 #2

    Zondrina

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    Usually power is defined as ##p(t) = v(t) i(t)##. In a more general sense ##p(t) = \frac{dW(t)}{dt}## where ##W(t)## is the work function.
     
  4. Aug 21, 2015 #3

    phinds

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    Why do you want to bring calculus into it? Power is voltage times current. P = V * I, which in this case is P(t) = V(t) * I(t)

    EDIT: I see zondrina beat me to it.
     
  5. Aug 21, 2015 #4
    So I don't need to integrate? I was just using a formula that I found in my notes.
     
  6. Aug 21, 2015 #5

    Zondrina

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    No integration is required. Perhaps what you are referring to is the change in energy:

    $$\Delta W = \int_{t_1}^{t_2} p(t) \space dt = \int_{t_1}^{t_2} v(t) i(t) \space dt$$
     
  7. Aug 21, 2015 #6

    phinds

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    Using forumlae without understanding them is a terrible idea. Forget the forumlae. Focus on the concepts.

    In specific answer to your question, re-read post #3

    EDIT: HA ... again zondrina beat me to it.
     
  8. Aug 21, 2015 #7
    Yeah thats the one!
     
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