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In the equivalent circuit for a transformer, does only the no-load current I0 pass through core loss Rc and magnetizing reactance Xm? Or does load current also pass through Rc and Xm? enter image description here

The figure above suggests only the no-load current passes through Rc and Xm, but my textbook shows both the no-load current and the load current I2/a passing through the parallel path. Which depiction is correct?

enter image description here

Note: the textbook authors chose the direction of I2 to yield positive coefficients in the equations \begin{align} \mathrm V_1&=(r_1+j\omega L_{11})I_{1} + (j\omega L_{12})I_{2} \end{align} \begin{align} \mathrm V_2&=(j\omega L_{21})I_{1}+(r_2+j\omega L_{22})I_{2} \end{align}

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The first diagram shows I0 passing through Rc and Xm. The primary current Ip is the sum of I0 and the scaled output current that is not explicitly shown.

The second diagram shows the difference between I1 and I2/a passing through Gc and Bm, without calling it anything.

These two versions are equivalent, at least mathematically.

There are two specified currents at the node where Xp/X1, Rc/Gc, Xm/Bm and RsN2/x2a2 meet. The third is given by Kirchoff.

The current flowing through the Xm/Bm Rc/Gc components (let's call this the core current) is controlled essentially by the constant input voltage, as the components Rp/r1 and Xp/x1 tend to be very small. It is therefore more or less constant.

I find the first diagram easier to be intuitive about, as we can just say that the input current Ip is the sum of this core current, and the turns-scaled load current drawn by the ideal transformer.

My problem with the second one, though entirely consistent with the first, is that we have a small current drawn by the core components, which consists of the small difference between two large and variable currents I1 and I2/a. As such, I find it difficult to think other than that this core current could be very wild and variable. As it happens, I1 varies exactly as needed to be the sum of the more or less constant core current and I2/a, so thinking of the core current as a difference is unhelpful, at least to me, and it sounds like to you as well.

Neil_UK
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  • Seems like you're saying I2/a just "stops" at the junction without going anywhere, and that only the no-load current passes through Gc and Bm? How can I2/a just "stop"? – artist_and_not_EE_by_training Feb 22 '23 at 15:07
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    @artist_and_not_EE_by_training I1 is roughly -I2/a, so neither 'stop', they are just both practically the same current, their small difference (when the transformer is on load) being the GcBm current. See my update tot he answer. – Neil_UK Feb 22 '23 at 15:54
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does only the no-load current I0 pass through core loss Rc and magnetizing reactance Xm?

Yes, Approximately. The coupling magnetization current, represented by Io is not very sensitive to the load current unless you consider the drop in voltage across Zm from secondary loading.

Also, the current in Rc is much lower than Xm.

As @Neil answered, the two schematics are equivalent and do not contradict each other or this answer.

Hoagie
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does only the no-load current I0 pass through core loss Rc and magnetizing reactance Xm?

Correct.

does load current also pass through Rc and Xm?

No it doesn't.

my textbook shows both the no-load current and the load current I2/a passing through the parallel path

The load current does not pass through Rc/Xm aka Gc/Bm.

Andy aka
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  • "By writing Kirchoff's voltage equation around the path of each of the currents I1 and I2/a... the reader should find that" the equations above are satisfied exactly. It seems to me that the equivalent circuit is consistent with the equations only when we assume that I1 and I2/a share the parallel path. How else can V1 be a function of both I1 and I2? – artist_and_not_EE_by_training Feb 22 '23 at 14:54
  • Please be clearer about your point please @artist_and_not_EE_by_training – Andy aka Feb 22 '23 at 15:27
  • @artist_and_not_EE_by_training why should V1 depend on both I1 and I2? All the voltages and currents are set by solving the entire circuit including some kind of power source and load (or no load). They can't be calculated based on just the transformer by itself. – user253751 Feb 22 '23 at 17:44
  • @user253751 please see my edits above, where V1 is defined as a function of both I1 and I2 – artist_and_not_EE_by_training Feb 22 '23 at 18:38
  • @artist_and_not_EE_by_training L11, L12, L21 and L22 do not appear in the diagram. I suspect they are defined in a way that makes the equations correct. – user253751 Feb 22 '23 at 18:39
  • @Andyaka okay sounds good I will take a pass at re-articulating the question more clearly – artist_and_not_EE_by_training Feb 22 '23 at 18:52
  • It doesn't affect my answer because load current will not pass through the parallel components. It never does @artist_and_not_EE_by_training – Andy aka Feb 22 '23 at 20:04
  • @Andyaka is what you just said consistent with the quote/updated text I just provided? – artist_and_not_EE_by_training Feb 22 '23 at 20:14
  • Here's where you have stepped too far. You have altered the question beyond what makes my answer reasonable. Please roll back your question and please don't do this again without respecting the answers currently given. Nobody likes goal posts moving and quotes from the original question becoming redundant. This site is not a forum. **The load current does not pass through Rc/Xm aka Gc/Bm** no matter what convoluted math you apply. It's as simple as that. @artist_and_not_EE_by_training – Andy aka Feb 22 '23 at 21:08
  • Okay understood, thanks for the lesson in stack etiquette. Based on your response, I will restore the original question and move the updated question to a new post. – artist_and_not_EE_by_training Feb 22 '23 at 21:34
  • I cannot yet close this question because the answers seem to be inconsistent with my readings so if it's okay with you I'll close this in conjunction with the new question that I will post. Thank you both. – artist_and_not_EE_by_training Feb 22 '23 at 21:38
  • The original question is answered. The math at the bottom of the question may be ambiguous but, as I said, your basic question is answered by me and Neil as far as I can see and, that math is an irrelevant artefact. But, did you ever consider that the direction of I2/a is arbitrary (it can be a negative value AND point the wrong way and still be correct). Then rethink what this means to you now @artist_and_not_EE_by_training. Anyway, thanks for understanding about the SE etiquette. Closing the question is unnecessary if you grasp what I said about current direction appearing wrong. – Andy aka Feb 22 '23 at 22:06
  • Let us [continue this discussion in chat](https://chat.stackexchange.com/rooms/143114/discussion-between-artist-and-not-ee-by-training-and-andy-aka). – artist_and_not_EE_by_training Feb 23 '23 at 15:48