1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

    BvU

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    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

    NascentOxygen

    User Avatar

    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.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Voltage across an inductor
  1. Voltage across inductor (Replies: 12)

Loading...