The following show the idea how a Thevenin equivalent source is used to replace the primary circuit of transformer.

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In textbook it is commonly stated that the load impedance sees an equivalent source and the open-circuit voltage is given by \$V_{oc}=N\cdot V_1=N\cdot V_s\$, usually it is said that there is no voltage drop across the source impedance therefore \$V_1=V_s\$ 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 \$N\cdot I_{sc}=V_s/Z_s\$. It implies that when the secondary circuit is shorted, the primary current will be \$V_s/Z_s\$. It seems that in this case the impedance of the primary coil is ignored. Why?

  • \$\begingroup\$ Please double check the formula's I edited. \$\endgroup\$ – jippie Jun 14 '15 at 12:01
  • \$\begingroup\$ @jippie The formula's are correct, thank you for editing. \$\endgroup\$ – Kelvin S Jun 14 '15 at 12:05

That's an excellent question - it really gets to the heart of how transformers work.

If the load is infinite (open circuit), and the transformer is ideal, no current will flow in the primary. (If you try it with a real transformer, you'll find a small current flows, but way less than you would expect from just considering the primary as an inductor).

If the load is 0 (short circuit), and the transformer is ideal, the primary will act like a short.

Here's a good explanation of why this is true (hint: it's all because of back EMFs): http://www.physics.usyd.edu.au/~khachan/PTF/Transformer%20explanation.pdf


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