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enter image description here

In the above diagram, the parallel branch containing Rc and jXm represents the excitation branch.

My question: I just can't seem to understand why the core loss current and the magnetization current had to be modeled using a parallel branch?

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    \$\begingroup\$ It's rather trivial: the magnetizing current is not related to the load attached on secondary, it is always constant, therefore we can draw in parallel branch. Think logical, is there any other possible way to draw it? I guess no. \$\endgroup\$ – Marko Buršič Mar 25 '16 at 8:39
  • \$\begingroup\$ Because it's pure loss. \$\endgroup\$ – Brian Drummond Mar 25 '16 at 10:57
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If you didn't connect anything to the secondary you would just have, in effect, an inductor formed by the primary. If you measured the inductance (for instance) you would get a few henries for a standard AC power transformer.

That measurement implies the inductance is in parallel with the primary connections. It's a big value inductance of course because you don't want massive reactive currents being taken by electronic appliances or the electicity companies would be up in arms about it. So that's why a regular AC transformer has up to a 1000 primary turns.

So, if that big inductance were in series with the "ideal" transformer in the transformer model, you would, at best get a few tens or maybe a hundred milli amps into the primary on full load on the secondary and that doesn't make a very good transformer.

To double check this, think again about it being in series with the ideal transformer in the model. Now, with the secondary unloaded there would be absolutely zero reactive current into the primary and this is just not the case.

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  • \$\begingroup\$ @Eliza If this answers your question, please consider formally accepting it or maybe leave a comment about what might be confusing you. \$\endgroup\$ – Andy aka Mar 1 '17 at 9:36

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