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I define every circuit element of a primary coil of a transformer that is connected to a voltage source and calculate \$I_1\$ (current of the primary loop). I then connect it to a secondary coil with a known number (\$N\$) turns.

Regardless of the values of the secondary coil, I always can calculate \$I_2\$ (current in secondary coil) by using $$I_2=\frac{N_2}{N_1}I_1$$ Does it mean that an ideal transformer is an ideal current source?

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Does it mean that an ideal transformer is an ideal current source?

An ideal transformer will have infinite magnetization inductance and will therefore take zero current. This means zero current on the output also. This doesn't make it an ideal current source.

Look at the equivalent circuit: -

enter image description here

An ideal transformer doesn't have any series elements so basically they become shorts. It doesn't have any parallel losses so Gc becomes open circuit and the only contentious thing left is the primary magnetization inductance shown with a blue square surrounding it. For an ideal transformer this is infinite in value and no current can flow into the primary if there is no load on the secondary. If there is load on the secondary then the primary impedance looking in will have an impedance of: -

\$P_Z = (\dfrac{N_1}{N_2})^2 \times S_Z\$

You could argue that a CT (current transformer) might be an ideal current source but it isn't. With a high primary current of (say) 100 amps it still has a primary magnetization inductance that will naturally limit the open circuit secondary output voltage despite rumours of thousands of volts being talked about on some websites.

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  • \$\begingroup\$ But mathematically I2=N2/N1.I1, and every example I solve there a value for current (Sinusoidal with amplitude and phase) \$\endgroup\$ – Jack Feb 2 '16 at 15:49
  • \$\begingroup\$ We are talking about AC, right? \$\endgroup\$ – Eugene Sh. Feb 2 '16 at 15:51
  • \$\begingroup\$ @MaryE That DOES NOT make it an ideal current source. You cannot have current in the primary of an ideal transformer if the secondary is open circuit and, any current that might flow into the primary of a transformer with the secondary open circuit flows ONLY into the magnetization inductance and DOES NOT couple to the secondary. If it has magnetization inductance it's not an ideal transformer by definition. \$\endgroup\$ – Andy aka Feb 2 '16 at 15:52
  • \$\begingroup\$ @Andyaka Well, you can't have ideal current source in open circuit as well. \$\endgroup\$ – Eugene Sh. Feb 2 '16 at 15:53
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    \$\begingroup\$ So in this sense the transformer with forced primary current will be equivalent to a current source \$\endgroup\$ – Eugene Sh. Feb 2 '16 at 15:55
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An ideal current source will produce the specified current independent of load. See this Wikipedia page for more details

A transformer secondary has an effective impedance (due to the turns ratio and the impedance on the primary) so the current will change dependent on the load and the primary circuit.

Therefore, the secondary can't be modelled as an ideal current source unless the primary is connected to an ideal current source.

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  • \$\begingroup\$ The question is about an ideal transformer. \$\endgroup\$ – Andy aka Feb 2 '16 at 15:39
  • \$\begingroup\$ Hopefully clarified my answer now. \$\endgroup\$ – rolinger Feb 2 '16 at 15:46
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    \$\begingroup\$ Secondary can be modelled as an ideal current source, if the primary is connected to ideal current source. \$\endgroup\$ – Eugene Sh. Feb 2 '16 at 15:49

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