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Transformer Core Loss Lissajous

  1. Nov 23, 2016 #1
    1. The problem statement, all variables and given/known data

    A single phase transformer circuit is assembled with an RC load. A voltmeter is connected to across the capacitor and an ammeter in series with the primary. The output of the ammeter is connected to the X-channel of an oscilloscope and the output of the voltmeter is connected to the Y-channel.

    The oscilloscope is set to X-Y mode and the resulting display is a Lissajous figure that closely resembles the B-H hysteresis loop.

    Explain why this is the case...

    2. Relevant equations

    N/A

    3. The attempt at a solution

    The current in the primary is the current that establishes the flux in the core i.e. provides the magnetomotive force which effectively represents the behavior of the H field.

    Since the e.m.f. produced by the transformer windings is proportional to the rate of change of flux the voltage lags the flux by 90deg.

    Lissajous figures represent phase differences between two signals. In this case the frequency of the two signals is the same but anti-phase. If the current was perfectly sinusoidal the figure would be an ellipse however due to saturation effects the current is not perfectly sinusoidal which is observed by the resulting Lissajous figure.

    I think this answer is on the right lines however I'm confused as to why the need for an RC load? Also, the magnetising current should be more prominent under no load however in this there is a 150kohm and 2microF capacitor which represents some load impedance on the transformer.
     
  2. jcsd
  3. Nov 26, 2016 #2

    NascentOxygen

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    Have you calculated the phase of the sinusoid across the capacitor, relative to the voltage applied to the RC combo for your powerline frequency?

    How would you describe in words the shape of the BH 'loop' for an ideal power transformer core?
    You seem to be saying a phase difference of "90 deg" is being in "antiphase"?
     
  4. Nov 26, 2016 #3

    NascentOxygen

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    That seems an unusual usage of the terms ammeter and voltmeter but I understand what they intend.
     
  5. Nov 26, 2016 #4
    Hi thanks for your feedback.

    I haven't calculated the phase of the volt-drop across the capacitor since in this arrangement I am only supplied with the magnitude of the volt-drop across the capacitor. Also, I wouldn't be sure how to analyse this since there will be some impedance in the transformer windings which will influence the power factor. I suppose just by observation the capacitor would have the effect of correcting the power factor slightly but I'm confused as to why the capacitor is needed for the B-H curve analysis.

    Also, to answer your second question; in an ideal transformer the B-H curve would be linear for all values of H in other words the magnetic permeability is a true constant. Also since the hysteresis contributes to losses in the core an ideal transformer core would not exhibit coercivity.

    Also, thank you for spotting that mistake.
     
  6. Nov 26, 2016 #5

    NascentOxygen

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    But you know R and C, so you could?
    It might be best to initially focus on what is needed to set up this x-y display so with an ideal transformer your setup would be displaying a recognizable linear B-H relationship. (B=μH)
     
  7. Nov 26, 2016 #6

    jim hardy

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    Reference https://www.physicsforums.com/threads/transformer-core-loss-lissajous.894511/

    it has nothing to do with the transformer.

    Nascent is giving you a hint by asking about phase. At risk of meddling, i'll strengthen the hint a little bit.....

    For the moment ignore the transformer and go to the basics of your RC circuit.
    You said R is 150 K ohms. What is Xc of your 2μf capacitor at line frequency ? I wager it's small compared to 150 KΩ .
    Which means the capacitor has little effect on current through the RC circuit.
    Which means current through the capacitor is very nearly in direct proportion to voltage across the RC circuit;
    Now, voltage across a capacitor being the integral of current through it, means what ?
    It means your RC circuit makes an integrator where voltage across capacitor very nearly represents the integral of voltage applied to the RC circuit. Certainly as close to a perfect integral as you can see by eye on a 'scope screen that has probably 3% accuracy.

    So your RC circuit is an integrator. That's the key to this homework exercise.
    I can tell you from real world practical experience that an integrator is a handy tool to have when studying inductance . I've done exactly this experiment myself for studying iron cores.

    So you think real practical now,
    why might whoever set up this homework exercise want you to think about observing the integral of voltage?

    @NascentOxygen: it's counterintuitive to beginners so i thought he needed a nudge in right direction... no offense meant....

    Try it with non-sine magnetizing current, like a triangle wave...
     
  8. Nov 26, 2016 #7

    NascentOxygen

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