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Magnetising current of inductor and transformer, onformation

  1. Mar 17, 2015 #1

    I've never thought of an inductor actually consuming any real power, however the other day I was concidering an open circuit secondary, transformer. This should be the same model as just an inductor should it not? (or rather this is how we can model an inductor too?)

    If this is so all the current that flows through it flows through the magnetising branch impedance? (To make the circuit, because no current flows into an ideal core) This would mean a tiny current is used to develop flux and a tiny power is dissipated, other than the winding resistance?
    Yet there is hardly any power being dissipated in the magnetising branch because the V on the branch is practically at 90deg to the I passing through it?

    Is this explanation correct? And does this mean that as the core saturates the magnetisation branch impedance changes? So then the calculated value from the open circuit test value is just an average?

    Last edited: Mar 17, 2015
  2. jcsd
  3. Mar 17, 2015 #2


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    I am thinking that if you do a bit of research on Saturable Core Reactors some of your questions will be answered. These devices are not generally given a lot of attention but are interesting.
  4. Mar 17, 2015 #3
    [/PLAIN] [Broken]
    "Magnetic Hysteresis results in the dissipation of wasted energy in the form of heat with the energy wasted being in proportion to the area of the magnetic hysteresis loop. Hysteresis losses will always be a problem in AC transformers where the current is constantly changing direction and thus the magnetic poles in the core will cause losses because they constantly reverse direction".
    Last edited by a moderator: May 7, 2017
  5. Mar 17, 2015 #4
    I will actually be doing some research into saturatable Reactors because I'm doing a project on Mag Amps, however in THIS case I'm only asking a binary yes or no's about if my explanations are correct:
    -Yes the model for OC is the model of a non-ideal inductor? (I'm used to seeing it model a TX not an inductor)
    -Yes all the current flows through the magnetising branch to make the circuit?
    -Yes there is not much power dissipated in the component because VI are almost 90deg to eachother?
    -Yes as the core saturates the magnetising impedance changes from that measured at OC?

    That material isn't new to me and it doesn't directly confirm or deny any of the aforementioned. I'm confident in my holistic understanding of the area but there are some aspects I wish to run past people to confirm as I'm doing here.
    Last edited: Mar 17, 2015
  6. Mar 18, 2015 #5

    jim hardy

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    Ahhh engineers always qualify their answers, don't they ?

    Yes. That's the mental model i use. What if you physically cut off the secondary windings? Then you'd have an inductor.
    The qualifier is we're ignoring capacitance to secondary winding.... and any thermal effects secondary has, like blocking heat flow out of primary.

    I agree, with caveat in first question


    Yes. Be aware that's a non linear phenomenon , current becomes peaky to push flux through higher reluctance of core.
    So it departs increasingly from a sine wave. Harmonic content grows....

    Magamps are really nonlinear. My 1930's Magnetic Ampifiers book says they defy precise qualitative analysis.

    Enjoy your foray into magamps. They cant be beat for rugged and reliable.
    They're enjoying a comeback in SMPS power supplies. Doubtless there are better computational methods available today.
  7. Mar 18, 2015 #6

    jim hardy

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    Re saturation

    well, it's the volt-second product that determines degree of saturation
    whether or not load current is flowing
    remember e=ndΦ/dt so Φ=n∫edt - there's no load current term in that !
    magnetizing current is a function of reluctance, and reluctance isn't really a constant it's a curve.
    Power transformers use the range where reluctance is fairly constant. Magamps push the iron well beyond that range.
    If you build a magamp for play you'll want a good core, old CT's should be a good source.
    Magnetics Inc of Pennsylvania website is a good information resource. Great folks there.
  8. Mar 19, 2015 #7
    Ah YES, I see what you're saying.
    I think of B(t) = -VPK*Cos(wt) / A.N.w
    which is I think along the lines of what you're alluding to.

    So when you do the OC Test, since you do it at rated excitation V meaning it's at rated reluctance, meaning OC Test impedance is rated impedance.

    COOL! Thanks for an excellent clarification, you should be a teacher.
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