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Differential Equation Help

  1. Mar 13, 2008 #1
    1. The problem statement, all variables and given/known data
    Find the charge on the capacitor in an L-R-C circuit at time t = 0.001

    L = 0.05H, R = 2 ohms, C = 0.01F

    q(0) = 5
    i(0) = 0
    E(t) = 0


    2. Relevant equations



    3. The attempt at a solution

    [tex] L \frac {di(t)}{dt} + R \frac {dq(t)}{dt} + \frac {q}{C} = 0 [/tex]

    [tex] \frac {dq^2(t)}{dt^2} + \frac {R} {L} \frac {dq(t)}{dt} + \frac {q}{LC} = 0 [/tex]


    [tex] \frac {dq^2(t)}{dt^2} + 40 \frac {dq(t)}{dt} + 2000q = 0 [/tex]

    [tex] m^2 = 40m + 2000 = 0 [/tex]

    [tex] q(t) = e^{-20t} (c1 * cos(40t) + c2 * sin(40t)) [/tex]

    q(0) = 0 yields c1 = 5

    [tex] q'(t) = i(t) = -20e^{-20t} (c1 * cos(40t) + c2 * sin(40t) ) + e^{-20t}(-200 sin (40t) + 40*c2*cos(40t))[/tex]


    c1 = 5
    c2 = 5/2

    I get q(0.01) = 4.11 coulombs

    The book has q(0.01) = 4.568 coulombs

    Can you help resolve my error?

    Thanks
    -Sparky_
     
    Last edited: Mar 13, 2008
  2. jcsd
  3. Mar 13, 2008 #2

    tiny-tim

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    Science Advisor
    Homework Helper

    Hi Sparky! :smile:

    Isn't c1 = 0? :confused:
     
  4. Mar 13, 2008 #3
    sorry,

    my mistake
    q(0) = 5

    c1 = 5

    -Sparky
     
  5. Mar 13, 2008 #4

    Tom Mattson

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    Sparky:

    You got the wrong answer because your calculator was in degree mode. Put it in radian mode (like it should be), and you get the book's answer.
     
  6. Mar 14, 2008 #5
    Ahh!!

    I was using a calculator I wasn't familar with and didn't check the trig settings.

    Thanks so much!
     
  7. Mar 18, 2008 #6
    Do you agree with my q(t) =

    [tex] q(t) = e^{-20t} (5 cos(40t) + \frac{5} {2} sin(40t)) [/tex]

    ??


    There is a second part to this problem -

    Find the first time q is equal to 0.

    Thanks to Kreizhn I have

    [tex] 0 = e^{-20t} (5cos(40t) + \frac{5} {2}sin(40t)) [/tex]

    [tex] 0 = (5cos(40t) + \frac{5} {2}sin(40t)) [/tex]

    [tex] cos(40t) = -\frac{1} {2}sin(40t)) [/tex]

    [tex] 40t = -1.1.07 [/tex]

    [tex] t = -0.0276 [/tex]

    [tex] 40t = -1.1.07 + pi [/tex]

    [tex] 40t = 2.0345 [/tex]

    [tex] t = 0.0508 [/tex]

    The book gets t = 0.0669.

    Suggestions?

    Thanks
    -Sparky_
     
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