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Inductor response to applied voltage

  1. Jun 16, 2015 #1
    I'm having a bit of trouble wrapping my head around why inductors behave the way they do in certain circuits. Every physical explanation for why ##V = -L\frac{dI}{dt}## that I've seen explains how the voltage across the inductor is developed as the current changes: the changing current means a changing magnetic field, which induces an electric field along the wire. This electric field provides an EMF which, if the inductor has zero resistance, equals the potential difference that develops across the ends of the inductor.

    However, in many—most, even—circuits of interest it is a (possibly time dependent) voltage that's applied across an inductor and current thus builds up according to the integral of the above equation. While I can see well enough how this works mathematically—the equation is valid whether ##I## or ##V## is the independent variable—I have a hard time turning the above argument around in order to understand physically what's happening. In terms of Maxwell's equation (and perhaps the Lorentz force law, if necessary) how can you derive the relation ##\frac{dI}{dt} = -V/L## when ##V## is applied across the inductor? To take the simplest example, consider an ideal inductor in series with a battery: the current through the inductor simply increases linearly as ##I(t) = \frac{V}{L}t## in the opposite direction as the applied voltage. How should I understand this behaviour from first principles?
     
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  3. Jun 16, 2015 #2

    NascentOxygen

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    You have a confounding negative sign that doesn't belong.

    With an inductor, the voltage across it is related to its current as v = L. di/dt

    You apply a voltage to it and the current slowly builds (providing you're starting from zero).. Compare this with a resistor where the current immediately jumps to its final value. These resistor and inductor currents are in the same direction.
     
  4. Jun 16, 2015 #3
    OK, well, my confusion was not over the direction of the current, though in any case I think we're just using different sign conventions for what directions are positive for the EMF and the current. You're using the passive sign convention, I was just using the same positive direction for both like EM books (including the one in front of me, Griffiths) often do.

    That doesn't really have anything to do with my question though.
     
    Last edited: Jun 16, 2015
  5. Jun 17, 2015 #4

    Svein

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    The [itex]V_{L}=-L\frac{dI}{dt} [/itex] equation does not have anything to do with the applied voltage, but the EMF generated across the inductor as a result of a change in the current through the inductor. Thus, the voltage that drives the current through the inductor is [itex]V_{b}+V_{L}=V_{b}-L\frac{dI}{dt} [/itex].
     
  6. Jun 17, 2015 #5
    ##V_L## absolutely has something to do with applied voltage—in the example in my OP, it's equal to the applied voltage. That must be true for Kirchoff's voltage law to be satisfied. Connecting an ideal inductor to a battery that supplies voltage ##V_b## causes a current whose growth satisfies [itex]\frac{dI}{dt} = -V_b/L [/itex] to develop through the inductor. To put my question another way: connecting an inductor to a voltage source causes current to flow through the inductor at just the right rate so as to produce an EMF that exactly equals the applied voltage. Appealing to the underlying facts of electromagnetism, why is that?
     
  7. Jun 17, 2015 #6
    Also: your expression is not correct. The total voltage across the inductor is simply ##V_b##. Again, this is necessarily true by Kirchoff's law.
     
  8. Jun 17, 2015 #7

    Svein

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    I did not say that. I said "Thus, the voltage that drives the current through the inductor is..". Which means:
    • If the current tries to rise too fast, VL increases and reduces the driving voltage, reducing the current.
    • If the inductor is suddenly disconnected, there is no Vb to drive the current. This means that [itex]\frac{dI}{dt} [/itex] is suddenly large negative, so [itex] V_{L}=-L\frac{dI}{dt}[/itex] is suddenly large positive - a phenomena we have to deal with whenever we try to disconnect a relay or a solenoid.
     
  9. Jun 17, 2015 #8
    This "driving voltage" you're talking about does not sound like a very coherent concept, and is not present in any discussion about induction I've read before. There is the voltage across the inductor that you can actually measure with a voltmeter. This is ##V_b##. There is the emf that opposes the increase of current, which is also ##V_b## by Kirchoff's law. ##V_b##, the voltage across the inductor, is by definition the voltage that drives current through the inductor, just as it would be the voltage driving current through a resistor or any other circuit component. The quantity [itex]V_{b}-L\frac{dI}{dt} = 0[/itex], and as the current is steadily rising interpreting this as the "voltage that drives current through the inductor" does not make any sense. How could a current increase with time if it's being driven by a quantity equal to zero?
     
  10. Jun 17, 2015 #9

    Svein

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    Sorry, this is not so.
     
  11. Jun 17, 2015 #10
    It's either zero or ##2V_b## depending on the sign convention you're intending (which you've been inconsistent about) since, as I've said several times, the magnitudes of ##V_L## and ##V_b## are equal by Kirchoff's law. With the sign implied by your comment that, "If the current tries to rise too fast, ##V_L## increases," it is zero.

    Thank you for your time, but I would prefer to have someone else try to answer. This talk of "driving voltage" is fundamentally incorrect.
     
  12. Jun 17, 2015 #11

    Svein

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    Kirchoff's law says something about the sum of currents. If you are singlehandedly trying to rewrite the laws of reactances, feel free to do so, but do not expect anybody to agree with you.
     
  13. Jun 17, 2015 #12
    Kirchoff's current law says that. Kirchoff's voltage law says that the sum of voltages over components in a loop is zero, and hence requires the magnitude of ##V_L## to equal the magnitude ##V_b## in this scenario. I am not rewriting anything; you are apparently unaware of, and mistaken about, some very basics ideas of circuit theory.
     
    Last edited: Jun 17, 2015
  14. Jun 18, 2015 #13

    Physically what is happening is exactly the same as what happens when a force is applied to a mass to accelerate it. F = ma and VL = L di/dt

    Electrical inductance is the same as physical or inertial mass and electrical current is the same as velocity, di/dt is acceleration.

    If you understand one, you understand the other.
     
  15. Jun 19, 2015 #14

    tech99

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    May I try another approach. When you connect a battery to any circuit, there is initially a fast impulse as current starts to flow, and this contains high frequencies. Each electron which starts to move has to build a magnetic field, and this stops it from moving instantly - it exhibits mechanical inertia - so it cannot fully respond to a fast impulse. If there is iron near it, for instance, it has to build an even greater magnetic field. The energy it is storing away in the field is a consequence of its inductance, and anything which increases the magnetic field increases the inductance and also the inertia effect. As the initial impulse dies away, a steady current can flow and the magnetic field is by then fully built. As others have said, it is an exact parallel to accelerating a mass.
     
  16. Jun 19, 2015 #15

    NascentOxygen

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    Edit: In an RL circuit there is no current impulse when a battery is connected. Consequently, there is no current impulse to "die away".
     
  17. Jun 19, 2015 #16
    There is a voltage spike across the inductor that does die away as current flows, so in that sense there is an impulse or transient response.
     
  18. Jun 19, 2015 #17

    NascentOxygen

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    I wouldn't call that a spike, but it certainly is a voltage that dies away. The post I responded to I read as referring to a current impulse on switch-on, and there isn't one.
     
  19. Jun 19, 2015 #18

    sophiecentaur

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    This was in the OP and it is the crux of the problem / dilemma. When you stick zeros and infinities into the values in what are perceived as practical problems, you can get daft answers. Include a finite series resistance and you will get a no-nonsense answer.
     
  20. Jun 19, 2015 #19

    tech99

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    Yes, I agree, but there is a voltage impulse in the form of a step function.
     
  21. Jun 19, 2015 #20

    NascentOxygen

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    An impulse is a short-duration pulse, a practical substitution for a delta function. A step is not an impulse; though if you have two steps, one up quickly followed by one down, you can create an impulse.
     
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