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I'm reviewing the derivation for the ideal transformer equations and I can't seem to understand something.

The textbook I'm reading states,

A transformer is said to be ideal if it has the following properties:

  1. The coils have very large reactances (L1, L2, M -> ∞)
  2. The Coupling coefficient is equal to unity (k=1)
  3. The Primary and Secondary coils are lossless (R1 = R2 = 0)

It then goes on to list the ideal transformer equations.

V2/V1 = N2/N1 and I2/I1 = N1/N2

What has me confused is the requirement that L1 and L2 and the mutual inductance should approach infinity. If the inductances are infinite, doesn't that mean the impedance is infinite and the transformer could never conduct current?

Or if it conducted any amount of current through the primary, wouldn't there be an infinite voltage across the primary?

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What has me confused is the requirement that L1 and L2 and the mutual inductance should approach infinity.

I don't think we can expect people to understand component-value properties without first stating what the top-level ideal requirements are. For instance, when you list this: -

The coils have very large reactances (L1, L2, M -> ∞)

These are implications from the following top-level requirements: -

  • An un-loaded transformer doesn't draw current from a voltage source.
  • A 1:1 transformer (for example) with 1 volt on the primary produces exactly 1 volt on the unloaded secondary.

If we have a top-level requirement, do the statements in your question seem unreasonable?

The next one is this: -

The Coupling coefficient is equal to unity (k=1)

This is redundant and covered in the first point where M approaches infinity.

The third is this: -

The Primary and Secondary coils are lossless (R1 = R2 = 0)

This isn't difficult to understand from the point of view that an ideal transformer must transfer power 100% efficiently. It also modifies the 2nd bullet point I made above like this: -

  • A 1:1 transformer (for example) with 1 volt on the primary produces exactly 1 volt on the loaded or unloaded secondary.

So, do you see that the "ideal properties" listed in your question are either redundant or, are implications that satisfy the top-level requirements?

If the inductances are infinite, doesn't that mean the impedance is infinite and the transformer could never conduct current?

You recognize that an infinite inductance draws zero current from a finite voltage source however, that doesn't mean it produces zero magnetic flux; flux is proportional to amps multiplied by turns and, if you have infinite inductance, there will be infinite turns and, a finite flux.

Because there is both flux and a coupling between primary and secondary, there has to be a secondary induced voltage that can source current into a load.

Then, if the secondary produces a voltage that can source a load current, by implication of the power coupling efficiency being 100%, the same power must be drawn into the primary winding.

I'll also add a note about the "infinite inductance" statement. I think it's confusing to talk about "infinite inductance". Wouldn't it be better to say that the primary inductance is large enough so that the unloaded transformer takes negligible current from a voltage source.

However, saying it has "infinite inductance" is a more efficient use of words.

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  • \$\begingroup\$ I'm still confused, If 1V is applied across the primary of a 1:1 ideal transformer and a 1Ω load is placed across the secondary would we expect 1A to flow through both the primary and secondary? I don't see how 1A could flow through a very large inductance. \$\endgroup\$ May 28, 2023 at 15:33
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    \$\begingroup\$ Well, I think you are nearly on the money for understanding a regular transformer on that basis; the two coils (pri and sec) work in flux opposition when it comes to current (as it would be with a differential signal on the CM choke). \$\endgroup\$
    – Andy aka
    May 28, 2023 at 18:48
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    \$\begingroup\$ That's basically it but with one modification that you should work through the math on; the primary "inductance" reflects the impedance of the secondary load multiplied by the turns ratio squared. In other words, what we call a voltage transformer is, in fact, a sub-set of the generalized thing called an impedance transformer @ElliottWinsted \$\endgroup\$
    – Andy aka
    May 28, 2023 at 19:02
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    \$\begingroup\$ Okay one more question. Does magnetizing inductance of a real transformer then represent the inability for a transformer to cancel out flux between primary and secondary? \$\endgroup\$ May 28, 2023 at 19:06
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    \$\begingroup\$ @ElliottWinsted it does indeed. It's the inductance of the primary (measured when the secondary is unloaded). That inductance always remains the same under-load and under no-load. The reflected load impedance from the secondary becomes in parallel with the magnetization inductance. \$\endgroup\$
    – Andy aka
    May 28, 2023 at 19:09
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With an unloaded transformer, the primary acts simply as an inductance. It's true that if you were to pass a current through an infinite inductance, then you would have an infinite voltage.

There are two observations that solve this conundrum though. The first is that it's the voltage that is fixed when we use a transformer, not the current. What an infinite inductance would mean is that the primary current drawn, which is the magnetising current, is zero. This is not a problem, even in ideal theory. We do not have to 'let the inductance tend to infinity' as a limit, zero magnetising current is just fine for ideal theory.

The second is that a infinite inductance is of course not possible. In a good transformer, we try to make the magnetising current as low as possible, by making the inductance as high as possible. Any magnetising current is a 'defect' that we have to allow for, when modelling a real transformer as an ideal one.

When we load a transformer, the voltage across the secondary causes a current to flow through the load. This load current also flows through the transformer secondary.

Let's assume we have an ideal infinite inductance transformer. In a sense, the infinite voltage that this secondary current would generate, if that was all that happened, is a good intuitive pointer to the mechanism by which the secondary current causes an opposing primary current to flow. When the primary current and the secondary current are added, taking note of their signs, the net current through the core is still zero, and as a result the primary and secondary voltages are still finite.

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It is easier to visualize initially when there is resistance in the windings. Imagine a transformer with 1 turn primary and 1 turn secondary, and 1H of in each winding inductance and 1 ohm of primary resistance. We apply 10V to the primary and a 4 ohm load to the secondary. With no load, flux initially must rise at a rate of 10wb/second to induce 10V across the primary, current will increase at a rate of 1A/second due to the inductance of the primary. Voltage will be induced in the secondary resulting in secondary current. The moment secondary current flows it will create a flux opposing the flux set up by the primary thereby reducing the induced voltage in the primary. This results in a difference between the applied voltage and the induced voltage, causing current to flow. This current is limited by the primary resistance. In this case, the instant voltage is applied to the primary, flux must be at 0. Induced voltage in the primary and secondary are equal. Induced voltage in the primary must be 10V - 1ohm * primary current. Secondary voltage = primary voltage. Secondary current = primary voltage / 4 ohms. Solving this gives 8V induced in each winding resulting in 2A in the 4ohm load. This results in 10V - 8V = 2V drop across the primary resistance of 1 ohm. This results in 2 amps flowing in the primary. This system is in balance.

If we reduce the primary resistance, the same principle will apply, the secondary current will oppose the flux set up be the primary, tending to reduce induced primary voltage the therefore increase primary current until balance is achieved. This applies all the way to zero resistance, where no drop in induced voltage is possible and the contribution of the primary and secondary to total flux must exactly balance initially. Following the initial application of voltage to the primary, the flux must increase steadily at the rate determined by applied voltage / Primary turns, but as the current required to produce a given flux is inversely proportional to inductance, this current falls towards zero as inductance approaches infinity.

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If there is nothing connected to the secondary, you are correct. It is just an infinite inductor and no current can pass through it.

If you allow a compensating current to flow through the secondary, however, no internal magnetization is induced in the core and the effective inductance, looking into the primary becomes zero.

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No, you seem to think the transformer as one coil only. The point of being an ideal transformer is that all energy coming in to primary is fully coupled to the secondary.

The inductance and impedance being infinite means there are no non-idealities. If there is no load on secondary, it means power out is zero so power in is zero so no current flows in the primary, because impedance and inductance are infinite so the transformer itself is not a load.

And no it does not mean voltage is infinite if current flows. If you have a transformer with supply and load, the primary has supply voltage and current is determined by the power taken by the load from secondary.

You seem to confuse that with back-EMF of an inductance, where you do get a theoretically infinite spike of voltage if you abruptly try to disconnect a current going through the coil. That is a different scenario.

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  • \$\begingroup\$ "The point of being an ideal transformer is that all energy coming in to primary is fully coupled to a magnetic field": no. An ideal transformer does not maintain any energy in a magnetic field: the magnetic field is zero at any point of time, and forcing it to be zero at all time is what forces primary and secondary currents to be coupled. \$\endgroup\$
    – user107063
    May 28, 2023 at 10:44
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The primary and secondary voltage ratio = the ratio of the numbers of the turns and the inverse for the currents are limit values when the common non-ideality factors are reduced towards zero. It's the math limit of the behaviour of a non-ideal transformer which is in a circuit and works stationarily after the initial transient.

Your problem is you skip the math limit concept and start to think like you had an ideal transformer with infinite inductances and zero stray quantities. And then you meet the impossible current anomaly. There even cannot be any compensating secondary current without the primary current. You have built an impossible situation by yourself.

Understanding the math limit was a difficult thing. The concept got the logical integrity in 1800's (see note 1). Before it the best mathematicians like Euler made groundbreaking creations without handling the limit concept properly. They operated with infinite quantities and their inverses successfully just like we can use succesfully the ideal transformer - as long as we do not let us get stucked in logical contradictions like your zero current.

Note 1: First logically sound formulation was done by Bolzano in 1817. A little later many others started to use the same idea. At least Cauchy seems to have found it without knowing Bolzano's writing.

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doesn't that mean the impedance is infinite and the transformer could never conduct current?

This is correct with a slight clarification: the transformer can never conduct net current (with "net" being modified by the winding ratio between primary and secondary). That is, primary and secondary currents are forced to be in a fixed ratio. It is similar for the voltages.

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You make this too difficult. The point of the ideal model is to make it simpler. In the ideal case, the ratio of the two currents on the primary and secondary windings is the ratio of the number of turns. You already stated this in formula form. There is no "loss" of current due to magnetization. Current in equals current out, when adjusted for the turns ratio.

There is no dilemma of some current through an infinite impedance, because, seen from the primary side, you have TWO currents to be added together: magnetization current for the transformer, plus the output current to the load on the secondary winding, divided *) by the turns ratio. The ideal transformer has such a great core material that it needs close to zero magnetization (H) to get all the needed magnetic flux (B), so that the magnetization current is invisibly low, and also such great windings without any leakage etc, so that the equations reduce to those that you quoted.

*) or multiplied, depending how you define the turns ratio

Formulas are great, but you should also be thinking in diagrams. Here is the equivalent diagram of the ideal transformer, where V2 is adjusted for the winding turns ratio.

equivalent diagram of transformer with ideal coupling

Now you see that V1 can have a current even if L is infinite, because the current goes to V2. Because this is an ideal transformer with infinite L, you might even leave L away, so ALL primary current goes to the secondary (adjusted for turns ratio).

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