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Current in RL circuit

  1. Mar 3, 2014 #1
    1. The problem statement, all variables and given/known data

    The circuit is in steady state before the switch is closed at t=0 I need to find how the current changes with time after the switch is closed.

    3. The attempt at a solution

    I read that at steady state inductor behaves like a short circuit, so does this mean when the switch closes all the current flows through the inductor? If so what happens after a long time has passed since the inductor opposes change in current flow? I am really confused...
     

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  3. Mar 4, 2014 #2

    rude man

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    When the switch closes you no longer have steady-state. So no, the inductor does then no longer look like a short circuit.

    But if the switch is left on long enough, you once again have a steady-state comndition.
     
  4. Mar 4, 2014 #3
    Why wouldn't it be steady state?
     
  5. Mar 4, 2014 #4

    rude man

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    Because you just flipped the switch, applying +10V to both components. Before that there was no voltage on either one.
     
  6. Mar 4, 2014 #5
    So at t=0,

    Current in the circuit = V/R = 10v/2kohms

    and when t approaches infinity,

    Current in the circuit = infinity? since the current won't flow through the resistor

    Also, if i sum the current flow at the node (assuming all the currents are flowing out of the node)

    -i(t) - V(t)/R - 1/L * Vdt from 0 to t

    Does it look right?
     
  7. Mar 4, 2014 #6

    rude man

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    Right!
    That's not an equation, and it's not the total outflow of current at the node either. (I sneaked an ∫ sign in for you).

    Correct would be i(t) = V/R + (V/L)∫i dt from 0 to t.
     
  8. Mar 4, 2014 #7

    Thnx but where did u get I from?
     
  9. Mar 4, 2014 #8

    rude man

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    Same place you did. i(t) is the current shown in your diagram.

    My equation says "current i(t) into the node = current out of the node."
     
  10. Mar 4, 2014 #9
    Sorry i meant the i in the integral part on the right hand side of the equation.
     
  11. Mar 4, 2014 #10

    rude man

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    i is shorthand for i(t).
     
  12. Mar 4, 2014 #11
    Isn't the equation for voltage across inductor,

    V = L di/dt

    Then when you integrate to find the current,

    i = 1/L [itex]\int[/itex]Vdt = V/L [itex]\int[/itex]dt

    In your equation the current through the inductor is

    i = V/L [itex]\int[/itex]i dt, i don't get how there is an i there...
     
  13. Mar 4, 2014 #12

    rude man

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    The reason you don't get it is because I screwed up! :redface: You have the right expression.

    So now you see how the current builds up to ∞ with time?
     
  14. Mar 4, 2014 #13
    lol. Well, the equation becomes

    i(t) = v(t)/R + (v(t)/L)*t

    When t is infinity i(t) is infinity, and when t=0, i(t)=V(t)/R

    So plugging in everything i get,

    i(t) = 0.005 + 1000t

    Thanks for everything :)
     
  15. Mar 4, 2014 #14

    rude man

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    Congrats for getting the right answer despite my stumbling!
     
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