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Power to work of an inductor

  1. Feb 1, 2013 #1
    4UVXHx6.png

    How can you get from the first expression to the second. I've done calculus, just need a fresher. I don't know how to deal with the dt/dt

    thanks
     
  2. jcsd
  3. Feb 1, 2013 #2

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    What you have written is not clear. Is "Li" a single function or is this the constant, L, times the function i?

    If the latter this is easy. [itex]L\int_{-\infty}^t i(\tau) (di/d\tau)d\tau= L\int_{-\infty}^t i di = L\left[(1/2)i^2(\tau)\right]_{\infty}^t[/itex] which, if [itex]\lim_{\tau\to\infty} i(\tau)= 0[/itex] is [itex](1/2)i^2(\tau)[/itex]

    If Li is a single function, then it is NOT always true.
     
  4. Feb 1, 2013 #3

    berkeman

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    Staff: Mentor

    Yeah, L is the constant value of the inductance. i(t) is the current as a function of time.
     
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