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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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  • \$\begingroup\$ An ideal source? Yes. An ideal independent source? No, it's an ideal dependent source; specifically, it's a current-dependent current source in one side. I've written an answer explaining this in detail. \$\endgroup\$ – Alejandro Nava Sep 25 '20 at 17:12
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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, because of this, it will take zero primary current when there is zero secondary current. 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\$ @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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    \$\begingroup\$ @MaryE an ideal transformer connected to an impedance on the secondary will look like that impedance on the primary except it will be multiplied in magnitude by the turns ratio squared. For a 1:1 transformer connected to a 10 ohm resistor, the primary will look like a 10 ohm resistor. Nothing to do with it being an ideal current source. \$\endgroup\$ – Andy aka Feb 2 '16 at 16:02
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    \$\begingroup\$ Ok I get it now. The current on the primary circuit depends on the impedance of the secondary circuit, therefore the current in the secondary circuit depends on its impedance (hiding in I1 in equation). So it is not an ideal current source, except when we have an ideal current source on primary, then an ideal transformer acts like an ideal current source as Eugene said. \$\endgroup\$ – Jack Feb 2 '16 at 16:05
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    \$\begingroup\$ @MaryE Yes, If you have an ideal current source feeding the ideal transformer primary then it looks like an ideal current source on the secondary. \$\endgroup\$ – Andy aka Feb 2 '16 at 16:13
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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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I think there's a confusion in some comments and maybe answers. It's important to distinguish between an ideal or non-ideal source, and, an independent or dependent source. An ideal source has no series impedance or shunt admittance, while a non-ideal source has. An independent current source provides the current given by its analytical expression regardless of the voltage across it, while a current-dependent current source provides a current proportional to another current.

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

An ideal transformer isn't an ideal independent current source. From one point of view, this doesn't make sense because a source has two terminals, while a single-phase two-winding transformer has four.

An ideal transformer is an ideal voltage-dependent voltage source in one side, and, an ideal current-dependent current source in the other side. This is shown in the following image, taken from Introduction to Electric Circuits (9th edition) by Richard Dorf and James Svoboda, where it's shown an ideal transformer with positive spatial orientation.

Ideal single-phase two-winding transformer and its equivalent circuit

The derivation of the equivalent circuit on the right is straightforward. You start with the equations \$N_1/N_2 = v_1(t)/v_2(t)\$ and \$N_1/N_2 = -i_2(t)/i_1(t)\$ (which are valid for the chosen reference polarity of \$v_1(t)\$ and \$v_2(t)\$, the chosen reference direction for \$i_1(t)\$ and \$i_2(t)\$, and the fact that the transformer has a positive spatial orientation.) Solve for \$v_2(t)\$ and \$i_1(t)\$ to get:

\$v_2(t) = \dfrac{N_2}{N_1} v_1(t) \tag 1\$

\$i_1(t) = -\dfrac{N_2}{N_1} i_2(t) \tag 2\$

Notice the first equation has units of volts, while the second of amperes. In other words, to model them as an equivalent circuit, we can assume we can get them by applying KVL and KCL. Furthermore notice in the right-hand side of both equations the voltage \$v_1(t)\$ and current \$i_2(t)\$ have a coefficient different from 1, which means they can't be modeled as an ideal independent voltage source and an ideal independent current source. But we can model them as ideal dependent sources! Since the coefficient for both is \$N_2/N_1\$, we can use dependent sources with a gain equal to that. Taking into account the signs, finally you get the equivalent circuit above.

To prove the equivalent circuit is correct, apply KCL at the upper node of the primary side. You get

\$ \text{(currents entering)} = \text{(currents exiting)} \implies i_1(t) + \dfrac{N_2}{N_1} i_2(t) = 0 \implies i_1(t) = -\dfrac{N_2}{N_1} i_2(t) \tag*{}\$

which is the same as equation (2). Apply KVL at the loop of the secondary side in clockwise direction to get

\$ \text{(voltage rises)} = \text{(voltage drops)} \implies \dfrac{N_2}{N_1} v_1(t) = v_2(t) \tag*{}\$

which is the same as equation (1). Thus the equivalent circuit is valid.

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  • \$\begingroup\$ Have you downvoted my answer. If so, maybe you could provide an explanation. All your answer appears to be doing is re-affirming that \$V_{OUT}\$ is \$V_{IN}\$ multiplied by the turns ratio. What has this got to do with the question? \$\endgroup\$ – Andy aka Sep 25 '20 at 17:29
  • \$\begingroup\$ Hi Andy. No, I didn't downvote your answer (nor any other.) Look here. The downvotes were already there. \$\endgroup\$ – Alejandro Nava Sep 25 '20 at 20:03
  • \$\begingroup\$ Yes, my answer affirms input voltage is proportional to output voltage, where the proportionality constant is the turns ratio (or the reciprocal of it.) But I indeed adress the question. The user said if an ideal tranaformer was an ideal current source. I said it was impossible to substitute the Tx with an ideal independent current source, and suggested an equivalent circuit using dependent sources. \$\endgroup\$ – Alejandro Nava Sep 25 '20 at 21:03

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