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RL circuits and unit step functions

  1. Mar 19, 2012 #1
    1. The problem statement, all variables and given/known data

    http://img828.imageshack.us/img828/1149/70833278.jpg [Broken]

    Now this is a part of the solution that was given I was having difficulty understanding.

    3. The attempt at a solution

    i(0) = 0 - understood

    u(t -1) for 0 < t < 1 = 0

    I can understand why the voltage from this source would be zero for that time period.

    For any time t > 0 -> u(t) = 1, so V = 1

    1) By i(∞) do they mean for the 0 < t < 1 period? So current just before t = 1? If so,

    a) How is i(∞) = [itex]\frac{1}{6}[/itex]? There's also a 3Ω resistor in there, why is that being ignored? Is it because for i(∞), the inductor is considered to be acting like a short-circuit thereby implying the current has to flow through 6Ω resistor, but ignores the 3Ω due to the path of least resistance?
    Last edited by a moderator: May 5, 2017
  2. jcsd
  3. Mar 19, 2012 #2


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    Staff: Mentor

    i(∞) should be i after a long time, when currents and voltages have stabilized at steady state values.

    I can't see why they say i(∞) = [itex]\frac{1}{6}[/itex].

    Can't see a reason.

    If the source on the right was of opposite polarity, then i would settle out at [itex]\frac{1}{6}[/itex].
    Last edited: Mar 19, 2012
  4. Mar 19, 2012 #3
    Perhaps I should have posted the entire solution, although I was struggling to get the concept behind the first part, here it is in its entirety:

    http://img828.imageshack.us/img828/1149/70833278.jpg [Broken]
    http://img832.imageshack.us/img832/6140/33110861.jpg [Broken]
    Last edited by a moderator: May 5, 2017
  5. Mar 19, 2012 #4


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    Staff: Mentor

    So they are saying if you were to sketch the current, i(t), then at time 0 current is 0 and while t<1 the current is an exponential that, left alone, would level off at 1/6 A. However, that behaviour continues only until t=1, at which case the other unit step is applied and causes i(t) to now head on an exponential decay towards a final value of ½ A.
  6. Mar 19, 2012 #5
    Hmm yea that makes sense, I think I just got confused as to what they meant about i(∞), but it's clear now.

    So does this mean the current through the inductor is [itex]\frac{1}{6}[/itex] because that 3Ω resistor is not considered because that inductor is now acting like a short-circuit at i(∞)?
  7. Mar 19, 2012 #6


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    Staff: Mentor

    During 0<t<1 the 3Ω is not part of the calculation of the final current because the inductor is a short-circuit at t(∞) so shorts out the 3Ω so that none of the current from the 6Ω will pass through the 3Ω. Then at t=1, u(t-1) kicks in and changes the final value that i is headed towards. The inductor still acts like a short circuit at t=∞ but now there are two components of current passing through it.
  8. Mar 19, 2012 #7
    Awesome, just wanted to make sure my reasoning was correct, thanks for that!
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