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I asked this question in electronics stack exchange as well but I thought it would also be applicable here as my question revolves largely around Maxwell's equations.

I am learning about transformers and in the equivalent circuit the magnetizing inductance of the excitation branch is supposed to represent the current needed to "set up" the flux in the core due to finite permeability of iron.

However, this does not explain to me why flux needs to be "set up" to begin with. The voltage induced is not a function of the magnitude of the flux but rather the rate of change of flux. The flux could at one instant be zero and changing quickly and the same voltage would be induced in a winding as if the magnitude were larger with the same rate of change.

This seems to be one of the biggest problems when analyzing "magnetic circuits" to me as there is a clear discrepancy in the duality with respect to Maxwell's equations. The rate of change of magnetic flux would seem to be the "magnetic current" especially for energy conservation and duality for Faraday's law with Ampere's law.

Can someone explain why there needs to be flux "set up" in the core and how finite permeability would change the relations for an inductor.

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For a physical transformer, the self-inductance of the primary and secondary are both finite since the magnetic core has finite permeability.

Assuming perfect coupling and zero winding resistance, the coupled inductor equations are (in the phasor domain)

$$V_1 = j\omega L_1 I_1 + j\omega MI_2$$

$$V_2 = j\omega MI_1 + j\omega L_2 I_2$$

where

$$M = k\sqrt{L_1L_2}$$

and $k=1$ for perfect coupling.

enter image description here

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When the secondary is left open (no load attached to the secondary), the (phasor) current $I_2 = 0$ thus

$$I_1 = \frac{V_1}{j\omega L_1}$$

For an ideal transformer, $I_1$ should be zero when $I_2 = 0$ and so this is a departure from ideal transformer behavior. Clearly, placing an inductor with inductance $L_1$ in parallel with the primary of an ideal transformer models this departure and this parallel inductance is known as the magnetization inductance.

This magnetization inductance is due to the finite permeability of the core. If the core permeability were 'infinite', the primary (and secondary) inductance would be 'infinite', and then there would be no need for a parallel magnetization inductance.

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    $\begingroup$ Wow that answers a lot. It seemed like at first with infinite inductances there would be infinite voltages on both sides but the mutual inductance on either side would bring the voltage EXACTLY to the level of the ideal transformer parameters. Since i2 would be negative in your diagram the self and mutual inductance would oppose each other to provide the finite voltages of the transformer. My next question is then why is the magnetizing inductance only considered on one side? $\endgroup$ – Colin Hicks Oct 24 '18 at 0:02
  • $\begingroup$ The explanation given for "setting up flux" however seems to be very misleading though $\endgroup$ – Colin Hicks Oct 24 '18 at 0:03
  • $\begingroup$ you have a typo, in general $M=k\sqrt{L_1L_2}$ $\endgroup$ – hyportnex Oct 24 '18 at 13:10
  • $\begingroup$ @hyportnex, the second sentence starts with "Assuming perfect coupling..." so it's not a typo. If you think it is misleading, I will edit my answer to include $k$ and the additional verbage "for perfect coupling, $k = 1$". $\endgroup$ – Alfred Centauri Oct 24 '18 at 16:48
  • $\begingroup$ To me "perfect coupling" for a transformer means that the coupling is lossless and without radiation but you maybe right in that it might mean something else to the question's poster. When I went to school the $k=\pm 1$ case was called tight coupling. $\endgroup$ – hyportnex Oct 24 '18 at 22:12
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I'll use a mechanical analogy.

Let's say you dig a hole with a shovel. In order to transfer power from your hands to the blade, you need to stress or "energize" the handle. If the shovel is stuck, you are not performing any useful work, but a little work would still have to be preformed to energize the handle by slightly bending it. The stiffer the handle, the less work is required to energize it, given the same applied force and the same stress.

The role of the flux in a transformer is similar to the role of the mechanical stress in a shovel: it is a mechanism for transferring electrical power. In order to transfer power from the primary to the secondary, a transformer has to be energized or magnetized and some work has to be performed, even in the absence of a load. The greater the inductance (the higher permeability of the core, the number of turns, etc.), the less work is required to energize it, given the same applied voltage and the same flux.

So, in order to transfer power, a transformer has to be magnetized (energized as any other power transfer device) and, since it does not have infinite inductance, the magnetizing will require some extra work.

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  • $\begingroup$ Does this imply it is there to bring the B field to certain level for operation? Kind of like transistors? $\endgroup$ – Colin Hicks Oct 24 '18 at 0:40
  • $\begingroup$ @ColinHicks It (magnetization) is there to bring the flux to a certain level. A transistor, even a power transistor is not a good analogy, since it does not transfer power from input to output - the power comes from the power supply. Think about devices that actually transfer power, like a gear box or a transmission line, to understand the concept of energizing. BTW, this is not a textbook concept - I just use it for better understanding of power transfer and you can use it, if it helps you. $\endgroup$ – V.F. Oct 24 '18 at 0:52

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