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Inductance in an RL circuit

  1. Nov 13, 2009 #1
    This is not a homework problem, but practice. I have the answer. I need to know why.

    1. The problem statement, all variables and given/known data

    Capture.JPG

    2. Relevant equations

    [tex]i=\frac{E}{R}(1-e^{\frac{-t}{\tau}})[/tex] "charging" an Inductor

    3. The attempt at a solution

    I have solved part a) correctly.

    [tex]i=\frac{E}{R}(1-e^{\frac{-t}{\tau}})[/tex]
    [tex].8\frac{E}{R}=\frac{E}{R}(1-e^{\frac{-t}{\tau}})[/tex]
    [tex].2 = e^{\frac{-t}{\tau}}[/tex]
    [tex]ln .2 = \frac{-t}{\tau}[/tex]
    [tex]t= 8.45\mu s[/tex]

    Part B is stumping me. I have a walkthrough of the problem but I don't understand why they do what they do. Here's the walkthrough.
    Maybe I'm just missing something stupid, but how do they end up with [itex]-t/\tau[/itex]= -1? If [itex]\tau[/itex] is the time constant of the inductance (L/R), why do they "assume" or how do they calculate that t=1 and [itex]\tau[/itex]=1?
     
  2. jcsd
  3. Nov 13, 2009 #2

    Delphi51

    User Avatar
    Homework Helper

    In the problem and the example you have "at time t = 1.0*τ" so that is why
    e^(t/τ) = e^(1.0*τ/τ) = e^1
     
  4. Nov 13, 2009 #3
    Oh. I see. I think I was just misunderstood what they were saying. Thanks.
     
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