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Electric potential just after switch is opened.

  1. Oct 30, 2016 #1
    1. The problem statement, all variables and given/known data

    https://postimg.org/image/6dbwajk91/
    upload_2016-10-30_13-20-35.png
    (Image inserted by moderator to make it visible in-post)

    In the circuit the switch is closed for several seconds, then opened. Make a graph with the abscissa time in milliseconds, showing the potential of point A with respect to ground, just before and then for 10 milliseconds after the opening of switch. Show also the variation of the potential at point B in the same period of time.

    2. Relevant equations

    V=iR
    E=L*di/dt
    Kirchhoff's laws

    3. The attempt at a solution

    So is there any potential larger than ground(V=0) on point A just after the switch is opened? In one way I think there should be some potential left since discontinuous change is unnatural. But I cannot clearly prove this. Also, the problems asks me to get the variation of potential B as time passes, but since point B is grounded, I cannot help thinking that potential of B is just 0 regardless of time. Am I wrong?
     
  2. jcsd
  3. Oct 30, 2016 #2

    cnh1995

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    Homework Helper

    Your image is not attached properly. Try attaching it using "upload" button near the "post" button.
     
  4. Oct 30, 2016 #3

    cnh1995

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    Edit: Ok now it's visible. This reply got posted late due to some network connection problem.
     
  5. Oct 30, 2016 #4

    cnh1995

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    This would be correct.
    Potential at point A will be affected due to opening of the switch. How would you apply KVL to the circuit after the switch is opened?
     
  6. Oct 30, 2016 #5

    gneill

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    Staff: Mentor

    Well, it happens all the time when switches are opened (ignoring some non-ideal characteristics of real components).

    What do you know about how inductors behave when sudden changes occur in the surrounding circuit?
     
  7. Oct 30, 2016 #6
    After switch is opened, what I know is that I should set inductor's emf like L*di/dt then apply KVL. But I feel confused when I have to set initial potential at point A after switch is opened. After applying KVL with potential B=0, I can get a exact potential at point A. So my question will be like 'Is A=24V just after switch is opened? Why?'
     
  8. Oct 30, 2016 #7

    gneill

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    Staff: Mentor

    Point A will not be at 24 V immediately after the switch opens. It will be very different!

    An important thing to remember about inductors for these type of problems is that they do not permit their current to change instantaneously. They will generate whatever EMF is required to enforce this rule in the instant after a sudden change in the circuit.

    What this means in practice is that the instant after the switch opens the inductor will still be conducting the same current it did just before the switch opened. Your job in analyzing this moment in time is to find out where that current flows and what potential drops it causes.

    You should draw two versions of the circuit, one with the switch closed with the inductor current direction and magnitude indicated. The second one should be for the instant after the switch opens and showing the inductor current identical to the previous diagram.
     
  9. Oct 30, 2016 #8
    Inductor requires same current. Thanks! Now I think I can solve the equation.... So does it mean that potential at A changes discontinuously from 24V to some other voltage when switch is opened? Does this instantaneous voltage change occurs in other circuits with switch?
     
  10. Oct 30, 2016 #9

    gneill

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    Staff: Mentor

    Yes.

    Whenever you have reactive components (inductors, capacitors) then they will exhibit this behavior. Inductors can't change their current instantaneously and capacitors can't change their voltage instantaneously.
     
  11. Oct 30, 2016 #10
    Thanks a lot! :)
     
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