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This extract is from a popular electronics book:

... a transformer of turns ratio \$n\$ increases the impedance by \$n^2\$. There is very little primary current if the secondary is unloaded.

I'm thinking about the emboldened part. Please see if the following argument is correct: If the secondary is unloaded, its current is zero, so the secondary can be neglected (as if it doesn't exist at all), and the transformer can be modeled as a single inductor with inductance of \$L\$. And according to the relations

$$X_L=2\pi\nu L$$

and

$$V=X_L I,$$

\$L\$ must be large if primary current is 'very little'.

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    \$\begingroup\$ Look up transformer equivalent model. The secondary current will dominate the primary so much you can offer omit the magnitizing current. At very light load however, it’s a significant portion of the primary current. \$\endgroup\$
    – winny
    Sep 11, 2018 at 15:41
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    \$\begingroup\$ Related question and answer \$\endgroup\$
    – Andy aka
    Sep 11, 2018 at 17:16

1 Answer 1

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Yes, you are correct. A transformer always has a "magnetizing current" (reactive) that is determined by its own inductance, whether or not it is loaded.

Loading the secondary imposes a second "resistive" (in-phase) current that varies with the load.

It sounds as if the book you quoted is referring to this in-phase component only, and ignoring the magnetizing current.

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    \$\begingroup\$ Magnetizing currents can be 10% with 10% load regulation on small transformers such that no load current is 10% reactive and Vout drops 10% to rated voltage, down to <1% on large grid XFMR’s \$\endgroup\$ Sep 11, 2018 at 16:27

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