Does flux exist inside an ideal transformer with a shorted secondary?

Discussion in 'Physics' started by nickw1881, Dec 25, 2009.

  1. nickw1881

    Thread Starter Member

    Dec 25, 2009
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    I am learning about transformers, but one question that is bothering me, which no book seems to clearly answer, is whether or not there is flux in the core of an ideal transformer with a shorted secondary winding.

    If current in a loop flows to oppose changes in flux, then in an ideal transformer with a shorted secondary, no flux should ever exist in the core, because as soon as flux increases from zero, opposing flux from the secondary winding would cancel it out.

    Non-ideal transformers have at least some flux in their cores, because they have measurable reluctance, or even an air gap, which is why equivalent circuits have a parallel inductor and resistor. Assuming the transformer is well built, this amount of flux should fairly small as long as the transformer is driven at the proper frequency. So here is the question that has been bugging me for years:

    Why does every textbook and core datasheet make it seem like the B-H curve is the be-all end-all of transformer core selection?! Unless the transformer is only lightly loaded there IS NO FLUX to worry about. What am I missing here?
     
  2. t_n_k

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    Mar 6, 2009
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    If there is no flux there can be no voltage - primary or secondary emf is related to the rate of change of flux in the magnetic circuit. In an ideal transformer the flux just never changes from no load to "full load" - irrespective of the magnitude of the winding currents.
     
  3. t_n_k

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    Perhaps I should have said the magnitude of the flux never changes for the ideal case - if the flux doesn't change then there's no voltage.
     
  4. nickw1881

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    Dec 25, 2009
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    How does the mmf generated by the secondary winding (in response to incoming flux) link back to the primary winding if it doesn't generate opposing flux in the core?
     
  5. t_n_k

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    Of course the other thing I didn't mention was the short circuited secondary case - which was the main matter you raised.

    In principle one can't consider an ideal s/c transformer driven by an ideal voltage source - there would be an infinite current drawn which is a nonsense or at least a troublesome conundrum.

    So it would probably be more sensible to consider the (ideal) transformer driven by a current source. Suppose the primary is driven by a constant sinusoidal current ip(t).

    The shorted secondary current must satisfy the condition

    M d(ip(t))/dt=Ls d(is(t))/dt

    where is(t) is the secondary current, M is the mutual inductance (primary to secondary) and Ls is the secondary inductance. In an ideal case M=√(LpLs), where Lp is the primary inductance.

    In words, the mutually induced voltage is just canceled by the secondary inductive voltage drop.
    For the ideal transformer under the aforementioned (constant current driven) conditions, the mutually induced voltage is the same whether the secondary is shorted or not. If the mutually induced voltage is the same then the rate of change of flux and the flux itself must be the same - given both are interdependent sinusoids (albeit with a phase difference).
     
  6. nickw1881

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    Dec 25, 2009
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    I believe the equations are correct. What I lack is a real physical understanding of what is happening in the core. Without having an experience of what gravity means or knowing what the field looks like, I can read F=Gm1m2/R^2 and plug in numbers but I wouldn't be able to comprehend what it really means. Alas, I cannot "see" inside a transformer core.

    Let me re-state my question a different way. (x' means first derivative of x-function with respect to time.)

    Consider this thought experiment: An ideal 1:1 transformer with a shorted secondary is driven by an ideal sine current source. At t=0+, the current begins to increase in the primary winding, which causes flux to flow through the core and the secondary winding. We will say flux (B field) and flux' are now >=0. With flux' >0, current must flow in the secondary winding to generate an opposing magneto-motive force (H-field) to the changing flux.

    If the flux density is constant whether the transformer secondary is shorted or opened, then something else must be linking the energy induced in the secondary back to the primary.

    What is the physical mechanism, the fundamental force that links two coils? If it's not the B-field (the magnetic flux itself), then is it the H-field (the force driving the flux)? I'm pretty sure it's not a telegram from a little man in the secondary to a little man in the primary?

    If flux does link the coils, then it is impossible for flux to exist in an ideal transformer core, because the flux created by the secondary would cancel out any flux created by the primary. I have myself PERSONALLY observed a magnet floating 2mm above a superconductor, the downward flux of the magnet perfectly matched by upward flux from current induced in the superconductor. There was no net flux between them... or was it the mmf from the current in the superconductor that opposed the increasing flux?
     
    Last edited: Dec 27, 2009
  7. nickw1881

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    Dec 25, 2009
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    Maybe you will see what I am not understanding if I draw MY INTERPRETATION of what a B-H curve for a transformer would look like under two conditions: open and short circuit. I assume the under short circuit, maximum energy will be transferred, H will be very high, but B will be the same as it is at no load. Am I right?[​IMG]
     
    Last edited: Dec 27, 2009
  8. t_n_k

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    Mar 6, 2009
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    I understand your points in the thought experiment and the dilemma concerning opposing mmfs.

    As to your point on the physical means of energy transfer - I would say this is a result of magnetic induction arising from time varying magnetic fields.

    It's not an easy matter giving a physical explanation. Perhaps it's a bit like Zeno's paradox - sorry I don't intend to make light of the issue. Presumably one could apply your arguments with equal validity to the case of bringing the poles of two identical bar magnets near each other and considering what resultant flux exists along the longitudinal axis in the gap between the poles. Each magnet has its own flux but for the like poles case, the resultant flux would presumably be zero at some point between the poles. There is of course no mechanical energy transfer occurring if the magnets are physically constrained.

    Hopefully other far wiser / smarter members could add their comments to the discussion.

    Keep in mind also when I referred to a 'constant' flux I actually had in mind a sinusoidally varying flux of constant amplitude - not a static unchanging flux.

    I would make another point about the notion of an ideal transformer. What does that really mean? I think it means a device which has neither losses nor stored energy - I guess in principle one might argue that in the absence of stored energy there can be no magnetic field - and no flux. Is this what you have in mind? Although, the zero flux notion would apply for any secondary loading condition from open to short circuit.

    On a "lighter note" you may have already seen this - but the link below is to an interesting [if somewhat longish] video that was mentioned on the forums some time back ...

    http://videolectures.net/mit802s02_lewin_lec16/

    Good luck with your thinking this matter through.
     
  9. nickw1881

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    Dec 25, 2009
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    I was thinking about my observation of the magnet above the superconductor. If mmf (h-field) has a real force upon a substance exerting a magnetic field (b-field), then that would explain why the magnet floats. Since there is air between the two (low permeability), the h-field might be exerting a strong force to prevent the magnet from falling, but only minimal opposing flux. The guy in the video says the superconductor exerts a strong b-field. In a transformer core, an h-field will produce a b-field based on the permeability of the core material.

    Or perhaps, the h-fields have an equal effect on the b-field such that they cancel each other, and the b-field remains constant as it would if the secondary coil did not exist. That way the only change observed would be the apparent change in impedance of the transformer.

    Is my drawing correct in my previous post?
     
    Last edited: Dec 27, 2009
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