Consider a two windings transformer (I'm not talking about flyback transformer):

• Magnetizing inductance is not equal to mutual inductance. Mutual inductance refers to the energy which is transfered from the primary to the secondary. The energy which is not transfered from the primary to the secondary refers to the leakage inductances and?
• Magnetizing inductance correspond to an inductance which is placed in parallel to the primary side of the transformer. So the energy stored in the magnetizing inductance (1/2LI²) is not transfered to the secondary. As some people says it "sets up" the transformer for then transfering energy to the secondary.
• Apparently in a forward transformer, if the magnetizing inductance was infinite, the flux into the transformer would be equal to 0 as the flux at the primary would be cancelled by the secondary.

What do you think about it?

Magnetizing inductance is not equal to mutual inductance.

It is for a 1:1 transformer. The mutual inductance (the turns that are 100% coupled between primary and secondary) represents the magnetization inductance. Those turns that don't couple are not mutual inductance but leakage inductance. Mutual inductance is $$\\sqrt{L_P\cdot L_S}\$$ hence, for a 1:1 transformer where primary and secondary inductance = L, M = $$\\sqrt{L^2}\$$ = L.

Mutual inductance refers to the energy which is transfered from the primary to the secondary.

The presence of mutual inductance allows energy to pass from primary to secondary.

The energy which is not transfered from the primary to the secondary refers to the leakage inductances

Leakage inductance is that inductance that doesn't couple primary and secondary

Magnetizing inductance correspond to an inductance which is placed in parallel to the primary side of the transformer.

Correct, for the model of a transformer where "the transformer" is internally an ideal power transfer component.

So the energy stored in the magnetizing inductance (1/2LI²) is not transfered to the secondary.

Correct (except for a flyback converter), but the magnetization inductance also facilitates primary to secondary power and energy transfer. It does two jobs.

Apparently in a forward transformer, if the magnetizing inductance was infinite, the flux into the transformer would be equal to 0

Not really a practical issue.

the flux at the primary would be cancelled by the secondary.

The flux due to load currents in the secondary are cancelled by flux in the primary due to those same load currents whatever the inductance.

• Thank you for your answer. There is still something which I do not understand. How mutual inductance can be equal to magnetizing inductance if the energy stored in the magnetizing inductance is not transfered to the secondary side ? It would mean that in a forward transformer no energy is transfered by the mutual inductance, BUT in a flyback transformer all the energy is transfered by the mutual inductance ... – Jess Jul 20 '20 at 11:26
• @Jess I never use mutual inductance in this respect; I use coupling factor, k and k is quite literally the percentage of the overall inductance measured on a winding that 100% couples to another winding. Mutual inductance is only needed (it seems) for doing exam questions raised by people who think they know best. I can't think of one example in real life where it is needed over and above k. I find the usage of mutual inductance as unnatural and superfluous. – Andy aka Jul 20 '20 at 12:04
• Check magnetic domains, also called Weiss domains in this Wikipedia article. In a transformer (where primary and secondary currents simultaneously circulate), the magnetizing current aligns the domains and energizes the core, coupling the primary and secondary. The domains alignment with magnetic field creates core losses and heat which are independent from the current in the sec side. You feel the mag. current at work with a classical unloaded ferromagnetic transformer left in the wall outlet: it's warm despite the absence of sec. current. – Verbal Kint Jul 20 '20 at 12:09
• What makes the core hot @Jess is not magnetization inductance but cores losses represented by a resistor in parallel with the magnetization inductance. This assumes that copper losses are not responsible for heating (i.e. it assumes a no-load situation). – Andy aka Aug 4 '20 at 8:55
• Ok ;) One day I will know how it works. I hope :D Thank you ! – Jess Aug 4 '20 at 9:28