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Inductance and RL circuits

  1. Oct 30, 2009 #1
    pic : http://s982.photobucket.com/albums/ae310/christianerik/?action=view&current=untitled.jpg

    If the switch is closed at t = 0, find the
    current in the inductor at t = 0.224 s. Answer
    in units of A.
    &
    Find the current in the switch at that same
    time.

    I know that i must use i = V/R(1 - e^-t/L/R) but cant seem to get the right answer.

    Any help is appreciated


    ................5 ohms
    -------------^^^^^^^------I
    I........................................I
    I.......10 ohms............1.4 H...I
    I------^^^^^^^^---&&&&--I
    I........................................I
    I.......S.........30 V.....5 ohms.I
    I-------/..---I I----^^^^^--I


    3. The attempt at a solution

    i = V/R(1 - e^-t/L/R + V/5 = (30 - V)/5

    (V/10)( 1 - e^-(0.224*10/1.4) + V/5 = -V/5 + 6

    0.1V(0.798)+0.4V = 6

    (0.4+0.0798)V = 6

    V = 6/0.4798 = 12.5 v

    The inductor current at t = 0.224 s

    i = V/R(1 - e^-t/L/R)

    i = (12.5/10)[ 1 - e^-(0.224(10/1.4) ]

    i = 0.9976 A

    The switch current:

    Is = i + V/5 = 0.9976 + 12.5/5 = 3.497 A


    What am I doing wrong?
     
    Last edited: Oct 31, 2009
  2. jcsd
  3. Oct 30, 2009 #2

    rl.bhat

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    Homework Helper

    At t = 0, you have to find out the voltage across LR circuit.
    When the switch is open, what is the total resistance of the circuit? Note that when there is no current or a steady current, L has no reactance.
    At t = 0, when the switch is closed, find the current drawn from the source.
    Then find the voltage across LR branch.
    Then using relevant equation find the current t given time t.
     
  4. Oct 30, 2009 #3
    i = V/R(1 - e^-t/L/R + V/5 = (30 - V)/5
    i thought that was what i was doing
     
  5. Oct 30, 2009 #4

    rl.bhat

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    12.5 volt across LR branch is not the voltage at t = 0.
     
  6. Oct 30, 2009 #5
    hmmm..... I calculated 16.52?
     
  7. Oct 30, 2009 #6

    rl.bhat

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    Show your calculation.
     
  8. Oct 30, 2009 #7
    v=30-30(e^(-.244(1.4/5)
     
  9. Oct 30, 2009 #8

    rl.bhat

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    No. It is not correct. Work out according to my post#2
     
  10. Oct 31, 2009 #9
    I got one answ... I(0.224) = 1.2*(1 - e^-2) = 1.038 A
    ::::
    5 + 5*10/(5 + 10) = 8.33
    30/8.33 = 3.6 A
    5*10/(5 + 10)=3.33
    3.33*3.6 = 12 V
    12/10 = 1.2 A
    so I have that now how would I find the current in the switch at that same
    time?
     
  11. Oct 31, 2009 #10

    rl.bhat

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    Use law of conservation of current.
     
  12. Oct 31, 2009 #11

    rl.bhat

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    I(0.224) = 1.2*(1 - e^-2) = 1.038 A
    How did you write this step?
     
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