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I Discharging of an Inductor

  1. Nov 11, 2016 #1

    cnh1995

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    I was thinking about the discharging mechanism of an inductor and I have a question about it. It might sound stupid to the physics experts here, but I can't help thinking about it.
    Before I ask my question, I want to mention a few points I already know about an inductor..
    I know an inductor resists sudden changes in current through it. I am comfortable with the differential equations and their solutions for RL circuits with different types of excitation. So, I have enough understanding of the inductor as far as my EE domain is concerned.
    When an inductor is charging, the energy is stored in the magnetic field and that stored energy is used to drive the current while discharging of the inductor. I understand this approach.
    Now my question...
    Consider the following circuit.
    Screenshot_2016-11-11-22-08-44.png
    The switch was closed for a long time and the inductor was carrying a current of 1A. Now, when the switch is opened at t=0, current in the inductor is maintained momentarily at 1A and voltage across the 1 ohm resistor rises to 1V. So, at t=0+ (just after t=0), inductor current is 1A and induced emf=1V. But if the current is still 1A just after the switch is closed, how is emf induced in the inductor? There has to be some change in current for emf induction. My question is, how is this voltage of 1V developed across the 1 ohm resistor just after the switch is closed? If the resistor were 10 ohm, the induced emf at t=0 would be 10V. How does the inductor "know" that resistance value? I understand the behaviour mathematically, but I don't understand what maintains the current to 1A "just after" the switch is opened. I want to know the what happens at the electronic level just after opening the switch. I am having trouble understanding "what causes what" at t=0.
    Thanks in advance.
     

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  3. Nov 11, 2016 #2

    Dale

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    The current is 1 A, but it is not constant. It is changing.
     
  4. Nov 11, 2016 #3

    phinds

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    The inductor doesn't know, or care, anything about the resistor, it just knows that it's got 1 amp flowing and it by damn is going to KEEP that 1 amp flowing. If there is no resistive path, then it will develop infinite volts. That of course, is a non-real-world idealized case but explains why inductors will spark-gap if suddenly removed from a circuit --- it develops however much voltage is needed to create an ionized air path for the current. If there is a 1 ohm path for the 1 amp, then only 1 volt needs to be produced and if 10 ohms then 10 volts. The inductor doesn't care about the voltage.
     
  5. Nov 11, 2016 #4

    cnh1995

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    I think I am having trouble understanding the concept of di/dt at t=0.
    While charging, the initial value of di/dt is E/L, which is independent of the circuit resistance but initial value of di/dt depends on the resistance R. How does the inductor know the value of the connected resistance? For maintaining the current to 1A, emf induced should be 1V but for having an emf, the current has to change. How to visualize this situation?
     
  6. Nov 11, 2016 #5

    phinds

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    Which part of my explanation is not clear?
     
  7. Nov 11, 2016 #6

    cnh1995

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    How does the inductor know how many volts to generate at t=0? I mean if the resistance is 1 ohm, it will generate 1V and if it is 1k ohm, it will generate 1000V. How does it decide these values?
     
  8. Nov 11, 2016 #7

    phinds

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    AGAIN, it does NOT decide. It doesn't care. It just keeps on producing 1 amp. Ohm's Law does the rest.
     
  9. Nov 11, 2016 #8

    cnh1995

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    How does it "produce" this current? To "produce" a current, some voltage is needed. Now, if the resistance is 1 ohm, the inductor should generate 1V across the resistance first, then only 1A could flow through it. Am I missing something?
     
  10. Nov 11, 2016 #9

    Bystander

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    You have a magnetic field collapsing.
     
  11. Nov 11, 2016 #10

    cnh1995

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    Yes. But how does initial rate of collapsing of magnetic field depend on the external resistance? How does the field decide at what rate to collapse at t=0? Basically, what comes first at t=0? Induced emf or di/dt?
     
  12. Nov 11, 2016 #11

    Dale

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    No, for maintaining the current to 1 A the voltage must be 0. Any other voltage will lead to a changing current. The resistor prevents the voltage from being 0, which forces the current to change. The bigger the voltage given by the resistor the faster the current changes.

    As a fine point, in circuit theory you just have voltages at terminals. You don't speak of induced EMF since induced EMF is always contained within a circuit element which is treated as a black box with only terminal voltages and currents.
     
  13. Nov 11, 2016 #12
    I believe I can help. To explain, let's ignore the resistor for now. After you open the switch you have to think of the open circuit as a capacitor to ground (or depending on the design of the switch there may be some capacitance to the other side of the circuit too). This capacitance may be minuscule, but it is real, and it is critical for understanding what happens.

    At the moment you open the switch there is no charge on this capacitor. It takes essentially no foward emf to push the first charge into the end of the open circuit. The continuing flow through the inductor charges up the capacitor creating a back emf opposing the continued current. The rate of change of the current is the rate of change of the magnetic field in the inductor. The faster the capacitors back emf stops the current, the larger the forward emf of the inductor. These two emfs will balance.

    Q=CV so Vback = Q/C

    I is defined as dQ/dT so

    dVback/dt = I/C

    The foward voltage is

    Vforward = L dI/dt

    So equating the two

    L d^2 I / dt^2 = I/C

    So I(t) = I0 * exp(-t/sqrt(LC))

    and V(t) = I0 * sqrt(L/C) * exp(-t/sqrt(LC))

    The resistor complicates things and I'll leave it to you to put in the capacitance to ground and solve that one in a similar way.

    I hope that helps
     
  14. Nov 11, 2016 #13

    cnh1995

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    Thanks for the reply.
    Say the switch is opened at t=0. How is this event "detected" by the inductor? What happens first? Can I visualize this in terms of "things happening one after another"?
    Like just after the switch is opened, does the inductor send some EM signal around the circuit and detect a high resistance path, which makes the initial voltage across the terminals high? How does this interaction occur at t=0? I am asking this because I can visualize how current flows in a normal resistive network using the surface charge feedback mechanism but here, I am unable to visualize how the resistance decides initial di/dt. Is there any such initial mechanism or it's complete nonsense and I have to think differently?
     
  15. Nov 11, 2016 #14
    Or to put it more succinctly, the current will fall at a rate that develops a forward emf in the inductor equal to the back emf of the open circuit.
     
  16. Nov 11, 2016 #15

    Dale

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    No, not in circuit theory. In circuit theory each circuit element is a black box which enforces the current-voltage relationship at its terminals at all times. You cannot say that one causes the other because they all happen at the same time.

    You would indeed need to think differently. Specifically, you would need to use the full Maxwell's equations if you wish to consider time scales that are so short that you could distinguish which order things happen. One of the explicit assumptions of circuit theory is that you are only interested in longer time scales.
     
  17. Nov 11, 2016 #16

    Dale

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    By the way, if you were to use a full EM solution then you would find that the EM wave propagates first from the switch, not the resistor or inductor. The order of the resistor and inductor changes would depend on the distances and conductor geometry. It all gets absurdly complicated.
     
  18. Nov 11, 2016 #17

    cnh1995

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    So can I say that using "some sort of EM feedback", the inductor "knows" the resistance of the discharging path and then generates a voltage accordingly at t=0 such that initial di/dt is equal to the voltage across the resistor?
     
  19. Nov 11, 2016 #18
    This is the best response to a muddled thread !!!
    Suppose that the current falls at a rate to produce an induced emf greater than the back emf
    Suppose that the current falls at a rate to produce an induced emf smaller than the back emf....
    There is only one logical answer... I think the logical technique is called 'reduction absurdium'...something like that
    The circuit 'knows' exactly what to do.
     
  20. Nov 11, 2016 #19

    Dale

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    I wouldn't say that, I don't like anthropomorphizing if it can be avoided.

    If I were using circuit theory then I would only say that the inductor enforces its current-voltage relationship at all times. That is all. Similarly for the resistor. Neither needs to receive any other information from the rest of the circuit in order to do its job. I would explicitly avoid any discussion about information in EM waves in the context of circuit theory.

    If I were using Maxwell's equations then I would only say that the fields obey Maxwell's equations at each location and the materials similarly obey their constitutive equations at each location. No knowledge of the rest of the circuit is needed by any part.
     
  21. Nov 11, 2016 #20
    Thanks, lychette, even if apparently you are the only one who thinks so!
     
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