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Max Voltage across an inductor

  1. Mar 3, 2014 #1
    1. The problem statement, all variables and given/known data

    The current in a 28mH inductor is known to be −10A for t≤0and (−10cos(400t)−5sin(400t))e^(−200t) A for t≥0. Assume the passive sign convention. At what instant of time is the voltage across the inductor maximum? What is the maximum voltage?

    2. Relevant equations

    v(t) = L*di/dt

    3. The attempt at a solution

    v(t) is at a max when di/dt is max since L is constant.

    Taking the derivative of the equation for t≥0, I have:

    di/dt = 5000e^(-200t)*sin(400t)

    Finding the maximum using my calculator, I found t = 0.0028 ms.

    Plugging this back into di/dt, di/dt = 2569.8

    v(0.0028) = 0.028 * 2569.8 = 72.3 V.

    Now both of these answers are correct, but my question is there a better way to find t than just plugging it into my graphing calculator and using the built in function to find the maximum? Is there a way this can be solved on a basic/scientific calculator? I know I could take the 2nd derivative of i(t) and set it equal to 0 but there is an infinite number of roots for t<0 and a good number of them for t>0. My professor never solved a problem similar to this in class so I'm not sure what his method is.
  2. jcsd
  3. Mar 3, 2014 #2


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    Science Advisor
    Homework Helper
    2017 Award

    Your ##{dI\over dt}## looks a little strange. Differentiating ## I = \left (A \cos(\omega t) + B \sin (\omega t) \right )e^{-\alpha t}## is differentiating a product. How come you end up with only one sin term ?

    Never mind, I can reproduce, sorry. Filling in numbers is useful sometimes....

    So now differentiate again and rewrite as ##A^\prime \sin(\omega t + \phi)\ \exp(-200t)##.

    From the ##\exp(-200t)## that descends monotonically it is obvious the first maximum is the maximum.
    Last edited: Mar 3, 2014
  4. Mar 3, 2014 #3


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    Staff: Mentor

    There is no easy out, that's what you'll have to do. As BvU pointed out, you are looking for the
    local maximum that lies 0 <t< T/2 because as time goes on the oscillations get smaller (it's a decaying exponential you have there).

    I checked your answer; and I agree with it.
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