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Question about transformer reflected source

  1. Jun 14, 2015 #1

    kelvin490

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    The following show the idea how a Thevenin equivalent source is used to replace the primary circuit of transformer.
    https://www.dropbox.com/s/0kaazx1d1srgirz/reflectedsource.jpg?dl=0
    https://www.dropbox.com/s/0kaazx1d1srgirz/reflectedsource.jpg?dl=0

    In textbook it is commonly stated that the load impedance sees an equivalent source and the open-circuit voltage is given by Voc=NV1=NVs, usually it is said that there is no voltage drop across the source impedance therefore V1=Vs in this equation. However, the primary circuit itself is a closed circuit, there should be some current and voltage drop across impedance of the supply, how V1 can be equals to Vs?

    The second question is, the short circuit current is found by NIsc=Vs/Zs. It implies that when the secondary circuit is shorted, the primary current will be Vs/Zs. It seems that in this case the impedance of the primary coil is ignored. Why?
     
  2. jcsd
  3. Jun 15, 2015 #2
    I would say it's because that particular impedance is usually very small compared to the load impedance and can, therefore, be neglected since it has no huge contribution to the final answer.

    As for your second question, the short circuited current is Vs/Zs. I don't see where you're having trouble: The primary circuit impedance is in the denominator.
     
  4. Jun 15, 2015 #3

    kelvin490

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    I don't know why the short circuited current is Vs/Zs, the primary coil wiring around the iron coil should have its impedance (and it induces flux), it should be added to the Zs to get the total impedance of the primary circuit, is that right?
     
  5. Jun 15, 2015 #4
    That's right, but here it seems you're using a simplified idea of a transformer rather than the model that's used to take the things you're talking about into account. The model here indicates Isc = Vs/Zs since you short circuit across the load and the current will no longer flow through the load impedance. If you wanted to be more precise you could say that Isc = Vs/(Zpri+Zsec+Zs), but this is outside the scope of the picture you've provided and generally has little purpose in the teaching of conceptual ideas. To take into account coil impedance you'd use the non-ideal transformer model that includes components for the resistive and inductive coil losses, along with the magnetization and hysteresis losses of the core.
     
  6. Jun 16, 2015 #5

    jim hardy

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    Because whoever drew that image drew the primary as only a coil, it has no Z term like he shows for secondary.

    He probably figured but didn't say out loud that primary Z is lumped in with that Zs block.

    Scrutinize your source document and see if maybe there's a statement to that effect hidden in the fine print ?
     
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