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RL Circuit Confusion

  1. Sep 18, 2016 #1
    1. The problem statement, all variables and given/known data
    Screen Shot 2016-09-18 at 5.06.26 PM.png

    2. Relevant equations
    V=IR


    3. The attempt at a solution
    This problem is really confusing me. For VL(0-) and IL(0-), the voltage and current when t<0, wouldn't both those values just be 0? the 0- means before the switch is closed right? So before it is closed the right branch does nothing right? So It only has a current and voltage when you close the switch right? if so I found that the voltage VL would be the same as the voltage across the 10 ohm to the left of it, which is just a voltage divider giving me an value of 60V. If that is correct how can I find the current? Can I just say i=60V/10ohms and that would give me my current?
     
  2. jcsd
  3. Sep 18, 2016 #2

    phyzguy

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    No. Remember that the voltage across the inductor is given by V = L dI/dt. This means that the inductor "resists changes in current". Since the current in the inductor is zero before the switch is closed, it must also be zero immediately after the switch is closed. In order for the current through the inductor to change from zero to any non-zero value instantaneously would require an infinite voltage, since dI would be non-zero and dt would be zero.
     
  4. Sep 18, 2016 #3
    so when the switch closes and we look at the circuit at t=infinity how can we find the voltage and current across the inductor? If we have V = L (di/dt), we know L, but how can we find (di/dt). How can we find iL?
     
  5. Sep 18, 2016 #4
    at t=infinty doesn't an inductor just act like a wire? so we can say that VL at infinity would be 0. And to find the current we can just use the 60V from the voltage divider I mentioned earlier and the 10 ohm resistor on the right to find the current through there right? I = V/r = 60/10 = 6A
     
  6. Sep 18, 2016 #5

    phyzguy

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    At t = ∞, everything has settled down and nothing is changing anymore. So dI/dt = 0. Then what are the currents? At times between 0 and ∞, you need to write a differential equation that relates the currents in the resistors to the currents in the inductor and solve it.
     
  7. Sep 18, 2016 #6
    How can I do that? Is there some formula I ahve to memorize or is there some way to derive it from the circuit. I'm not sure how I would write my diff eq.
     
  8. Sep 18, 2016 #7
    V = Ri + L(di/dt)

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

    If di/dt is 0 at infinity then the first equation would give us V=iR for the inductor. but we don't know i or r.

    Using the second equation and plugging in infinity for time we get that i=(V/R)

    What we do know is that the voltage across the inductor and the 10 ohm resistor would be 60V. So can we use that information to help us find the voltage and current at t=infinity?
     
  9. Sep 18, 2016 #8

    phyzguy

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    This isn't right. It's correct to say that at t = ∞ the inductor is just a wire and has no voltage drop across it. So the circuit is just reduced to the 120 volt source and the three resistors. So what is the current through the three resistors in that case?
     
  10. Sep 18, 2016 #9
    the 2 on the right would add in parallel to 5ohms and then that would add in series to the 10 on the top left to give you and equivalent resistance of 15ohms. Then I=V/R gives you 8A. So that means that 8A are flowing through the inductor at t=infinity. Right? So for voltage could we also use V=IR and the current we just found? V=8A(10ohms)?
     
  11. Sep 18, 2016 #10

    gneill

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    Staff: Mentor

    If I may suggest, since the question only asks for information concerning the inductor (currents, voltages), start by replacing the voltage supply and resistor network with its Thevenin equivalent.
     
  12. Sep 18, 2016 #11

    phyzguy

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    No. 8A is flowing out of the voltage source, and 8 amps flows through the left-most 10 Ohm resistor, but then the current divides. So how much flows through the right-most 10 Ohm resistor and then into the inductor?
     
  13. Sep 18, 2016 #12
    Hmm. Im not too sure how I could find that. Since 8A is coming from the 120 source, when we go over the 10ohm resistor we will drop to 40V. the voltage across the leftmost 10ohm resistor would be V=ir = 8A(10) = 80V. So 120-80 = 40V going into the branches and splitting. 40 V down to ground and then 40V into the rightmost 10ohm and the inductor. So that would mean that V=IR --> I=40/10 = 4A is flowing through that branch. So that means that 4A has to be flowing through the inductor. Right?
     
  14. Sep 18, 2016 #13

    cnh1995

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    Right.
     
  15. Sep 19, 2016 #14
    hmm ok. How can we find the voltage now. VL at infinity. If it acts like a wire, would it just be 0V at t=infinity?
     
  16. Sep 19, 2016 #15

    cnh1995

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    Yes.
     
  17. Sep 19, 2016 #16
    Ok cool. Now for part b, would I just use the equations for charging a inductor? So I = Io (1-e^t/T)? Or would I have to set up a differential equation somehow. I am not sure how to setup a differential equation for RL/RC circuits
     
  18. Sep 19, 2016 #17

    gneill

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    Again I suggest that you replace the source and resistor network with its Thevenin equivalent. The circuit will become a trivial RL series circuit that you can "solve" by inspection.
     
  19. Sep 19, 2016 #18

    cnh1995

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    Follow gneill's advice in #17. Are you familiar with the Thevenin's theorem?
     
  20. Sep 19, 2016 #19
    Yes I am, but I am not sure how to apply it here. How can I find a Thevenin equivalent with an inductor in the circuit? Do I find the Thevenin assuming t = infinity? So the inductor just acts like a wire and then I find the thevenin equivalent? If so then wouldn't my Vth just be 120V and my Reg be 15ohms like I said earlier? Would that me my Thevenin equivalent circuit? How to I recreate the circuit once I find Vth and Req even?
     
  21. Sep 19, 2016 #20

    gneill

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    The inductor is the load. Remove the load (inductor) and find the Thevenin equivalent looking into the now open terminals. Or, if you're worried about maintaining the switch action, break the circuit just before the switch:

    upload_2016-9-19_10-58-34.png
     
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