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The voltage across an inductor isn't -Ldi/dt

Discussion in 'Electronic Design' started by [email protected], Mar 4, 2007.

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  1. Guest

    I always thought the voltage across an inductor was -Ldi/dt. Yet it
    cannot be, because for a conductor with zero resistance and carrying a
    finite current, the electric field inside it must be 0 meaning there
    is no voltage across it's ends. When we talk about a voltage existing
    across an inductor, what we really mean is that the voltage across the
    terminals of the voltmeter is -LdI/dt.

    Take a look at this link for a full explanation:-

    http://ocw.mit.edu/NR/rdonlyres/Phy...DB-AB80-49E0-BFAB-BCE00159F00E/0/lecsup41.pdf
     
  2. Guest

    That's nice. Now could you take a few minutes out of your busy
    schedule to learn that "it's" means IT IS?
     
  3. Bob

    Bob Guest

    That *was* a dead giveaway, wasn't it?

    Next, he's going to explain how to use sheep's bladders to prevent
    earthquakes.

    :-}

    Bob
     
  4. Boris Mohar

    Boris Mohar Guest

    You can do that?
     
  5. Phil Allison

    Phil Allison Guest

    <

    ** The date on the top of that text is a worry ....




    ....... Phil
     
  6. John  Larkin

    John Larkin Guest

    Nope, it's E = + L di/dt using conventional notation.
    The electric field is nonzero because there's a changing magnetic
    field, which creates potential. The magnetic field is changing because
    of... dI/dT !
    Same thing, if you have a voltmeter. Same thing if you don't.
    You might do that.

    John
     
  7. Guest

    Yeah, 2002 was a bad year for me too. Lost my job and stubbed my big
    toe. Nail fell right off.
     
  8. Phil Allison

    Phil Allison Guest

    ** Google Groper half wit alert !!


    "Phil Allison"

    ** Yawn .....

    What a tedious moron.



    ...... Phil
     
  9. Guest

    There is no electric field inside a conductor of zero resistance
    carrying a finite current. If there was, the current would be
    infinite. Therefore there is no electric field inside an inductance of
    zero resistance carrying a finite current. The line integral of E
    across the terminals of the voltmeter - its measured voltage - equals
    the rate of change of magnetic flux through the circuit consisting of
    the inductor, the leads to the voltmeter and the terminals of the
    voltmeter.
    This is not obvious to me. With the voltmeter and leads connected to
    the inductor, only V across the voltmeter contributes to the total
    E.dl around the loop, since E is zero inside the leads and the
    inductor. So the voltage across the voltmeter equals the total E.dl.
    Take the leads away and E.dl across the voltmeter no longer equals the
    total E.dl around the loop because E is no longer zero where the leads
    were.
    What is your impression of the paper if you have read it?
    At least you have made an attempt to answer/correct my points. Thanks.
     
  10. Jim Thompson

    Jim Thompson Guest

    You need to read up on it. You are thoroughly confused. I hope you
    are not a student ;-)

    ...Jim Thompson
     
  11. John  Larkin

    John Larkin Guest

    In a volume of space that is subject to a changing magnetic field,
    there *is* a corresponding electric field. If there's a conductor in
    that space, the electric field is present inside the conductor, too.
    That's why you can have voltage across the ends of a transformer
    winding.

    No. Place a coil or loop in a time-varying magnetic field. Connect its
    ends to a twisted pair or a coaxial cable and run that cable a mile
    away, far from the magnetic field zone, and use the voltmeter there.
    Same voltage.

    Of course, if the voltmeter and its leads are in the field, you'll
    have to take the additional potentials into account. In the case of a
    voltmeter connected to the secondary of a transformer, the voltmeter
    isn't exposed to much field, but it still measures the winding
    voltage.
    As noted, you can move the voltmeter far away, out of the field, and
    accurately measure the non-zero coil or loop voltage. And a 100-turn
    coil pumps move voltage into the meter than a 5-turn coil, or a bigger
    loop more than a small one, so there must be voltage within the coil.

    There *is* an electric field inside a conductor which is immersed
    inside a changing magnetic field. "Ohm's law" is trashed by
    electrodynamics. The electric field is like a gravitational field...
    it permeates everything. If that were not true, every generator and
    transformer and motor on the planet would instantly stop working.

    In Hertz's seminal experiments, he detected em fields by using a
    single metal wire that *almost* made a full ring, leaving just a tiny
    gap. When exposed to a sufficiently strong em field, he observed
    sparks jumping across the small gap. No voltmeter needed.

    Classic electromagnetics.

    John
     
  12. Jim Thompson

    Jim Thompson Guest

    Read it more carefully, you still are confused.

    (Physics 8.02 is MIT FRESHMAN physics, 2nd semester ;-)

    ...Jim Thompson
     
  13. Guest

    If the current is finite and resistance is 0, then the charge inside
    the conductor rearranges itself to create an electric field that
    cancels the applied electric field. The net electric field is
    therefore zero. Since the electric field is the measured force on a
    unit positive charge placed there and an infinitesimally small i.e. 0
    force is required to move it because the reistance is zero, then the
    electric field has to be zero.
    I think the answer is that Faraday's law is true for any passive
    region that modifies the electric field, and the electric field in
    Faraday's law is the total electric field in the passive region and
    not just the induced electric field. So that the electric field
    through the terminals of the voltmeter is that due to an
    infinitesimally small current that generates a voltage across it's
    internal resistance.
    For a finite I , E.dl through 0R leads is still zero, from definition
    of what an electric field is.
    Suppose a conductor requires 0.01V applied across it to push 1A
    through it. If it is now wound into an inductor and pushes 1A through
    1R, the voltage through the 1R will be 1V, and through the inductance
    still 0.01V. An electron in the inductance will still require the same
    force to move it for the same current, whether that force is from a
    static electric field or a changing magnetic field. If the force is
    the same, the electric field there is the same.
    The electric field inside a conductor depends on the resistance of the
    conductor. But for a conductor of 0 resistance carrying a finte
    current it is zero. Concentrate on the definition of an electric
    field.
    Yes. And the electric field in the gap was several orders of magnitude
    greater than that in the metal wire.

    Well, he makes the point that E.dl through an inductor is zero, yet
    you have difficulty understanding why. The paper was written by a
    distinguished professor at MIT.
    Once again, thanks for attempting to discuss my points.
     
  14. Jim Thompson

    Jim Thompson Guest

    [snip]

    You're mixing up DC and AC components with complete abandon.

    ...Jim Thompson
     
  15. John  Larkin

    John Larkin Guest

    It's not. Just quit being silly. If that were so, nothing would work.

    Since the electric field is the measured force on a
    OK, OK, I give up. Believe whatever you insist on believing.

    And don't send me a resume.

    John
     
  16. Robert Baer

    Robert Baer Guest

    I hate to tell you, but ohm's law is not trashed.
    Please consider the resistance of the coil and the current thru the
    coil, along with the (remote?) load.
     
  17. Phil Allison

    Phil Allison Guest

    "John Larkin"


    ** You did notice the date on the paper the OP quoted - didn't you ?




    ........ Phil
     
  18. You won't be able to maintain a steady-state DC voltage drop
    across a conductor with zero DC resistance, because that would
    imply infinite current. However, the transient or AC-voltage
    case the story is entirely different. Consider, for example,
    my 10kV 300kHz ferrite-core transformer, which has 22 turns on
    its secondary. The single-turn primary has 450 volts across
    its ends. Since it's a thick band of 1"-wide copper, with for
    all practical purposes, zero DC resistance, by your obviously-
    incorrect understanding this is not possible. BTW, the current
    in my transformer's single thick strip of copper is modest and
    due almost entirely to activities outside the copper strip.
     
  19. Fred Bloggs

    Fred Bloggs Guest

    That lecture note is poorly written, confused, and useless. A voltage
    drop, or more accurately potential drop, is understood to be a measure
    of energy expenditure per unit of charge injected through the component.
    There is no such energy expenditure in a pure inductance, it is all
    storage, and therefore there is no potential drop across it. Faraday's
    law introduces the concept of electromotive force due to induction, or
    e.m.f., and quantifies its interaction with the electric field in closed
    loops. It is a statement about the net potential drops around the loop
    external to the inductor and not the inductor itself. You cannot have a
    di/dt in the inductor without an external circuit loop, the loop will
    always be present, but because it can take many forms, it is simpler to
    consider that the algebraic sum of the external loop drops as appearing
    across the inductor in the sense that this is where one would place the
    measurement probes. Get real.
     
  20. Phil Allison

    Phil Allison Guest

    "Fred Bloggs"


    ** You did notice the date on the paper the OP quoted - didn't you ?

    Wake up.




    .......... Phil
     
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