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 for that matter what the magnetizing inductance does.

  • \$\begingroup\$ Perhaps what you're misunderstanding is that the flux Φ itself is the "magnetic current", not dΦ/dt. \$\endgroup\$ – Hearth Oct 23 '18 at 22:04
  • \$\begingroup\$ I understand that's how it was defined but the voltage induce would be equal to the rate of change of magnetic flux not the magnitude of the flux. This can be compared to amperes law where the magnetic field is proportional the rate of change of electric charge not the magnitude of the electric field itself. I don't see why a magnetic flux needs to be set up to begin with just like a build-up of electric charge does not create a magnetic field. \$\endgroup\$ – Colin Hicks Oct 23 '18 at 22:14
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    \$\begingroup\$ I'm not sure I understand what you're not understanding, honestly. The flux needs to be "set up" because it has to get there somehow, it's not just magically there. \$\endgroup\$ – Hearth Oct 23 '18 at 22:31
  • \$\begingroup\$ Since an answer would entail the behavior of the core itself, could you please refer to specific core types. Iron,soft ferrite, high density ferrite, Permalloy (nickle-iron), etc \$\endgroup\$ – Sparky256 Oct 23 '18 at 22:31
  • \$\begingroup\$ I don't understand why a given magnitude of B field needs to be maintained in the core. The magnetizing inductance will generate a given level of B field as far as I understand, but why does there need to be a given magnitude of B field. If the B field at one instant was 0 but increasing at a given rate of change that would induce a voltage. It seems to me that "setting up" is unnecessary as the rate of change is whats important. Intuitively the "set up" flux is like the "DC" value of the changing flux that doesn't matter. \$\endgroup\$ – Colin Hicks Oct 23 '18 at 22:35

Transformers are feedback systems, with summation of input and feedback FLUX occurring within the core.

When the secondary load current changes, the flux summation changes, the voltage induced in the PRIMARY changes, and the voltage changes across the impedance between the external power source and the primary. This change of voltage across the impedance causes a change in current drawn from the external power source.


simulate this circuit – Schematic created using CircuitLab


Let we look what the rate of change is. A magnetic flux of a given transformer is: $$\Phi=\Phi_{max}sin(\omega t)$$

Rate of change is given as differential over time:

$$\dfrac{d\Phi}{dt}=\Phi_{max}\ \omega\ cos(\omega t)$$

And the induced voltage is: $$u_i=-N\dfrac{d\Phi}{dt}$$

You can clearly see that rate of change as you are refering at is not magnitude invariant. It is clearly seen (second equation) that an induced voltage will vary with magnitude of flux (\$\Phi_{max} \$) but also with frequency (\$\omega\$) and finally with number of turns (N).


This is a great question Colin, one which I puzzled over long and hard myself before coming to an intuitive understanding.

Magnetizing inductance and magnetizing current are two aspects of the same departure of a real transformer from the ideal transformer model; in particular, as you mention, the fact that a real transformer core has finite permeability.

The permeability of an ideal transformer core approaches infinity. Interestingly, this means the magnetizing inductance of an ideal transformer is infinite, not zero. It does not appear in the ideal circuit because it models as an open circuit, not because it's not present. As a consequence, the flux needed in the ideal core to induce a voltage in the secondary is produced by a vanishingly small current (read zero). So the non-ideality of a real transformer is that it's inductance is less than infinite and therefore requires a current to flow to produce flux.

There are three other important factors to note about this:

  1. The change in flux produces a voltage. It does not by itself produce a current. Essentially, the \$\frac{d\phi}{dt}\$ produces a voltage and the voltage produces a current if it can, based on the impedance of the secondary circuit.

  2. For a given permeability and frequency, \$\frac{d\phi}{dt}\$ is proportional to the peak flux. If the peak (or peak-to-peak) flux is lowered, the voltage on the secondary drops. This means that the magnetizing current is the minimum current that can flow without dropping the output voltage. (Here I'm thinking of magnetizing current as sinusoidal AC. The same principles apply in switching transformers fed with square-wave input voltages but the complicating factors that introduces are perhaps better left until later.)

  3. Nothing says the magnetizing current has to be in-phase with the input voltage, in fact, in a well-designed transformer, in general, the magnetizing current will be near 90-degrees out of phase. The real portion of the complex magnetic current gives a measure of the resistive and core losses. The imaginary part is pulsing back and forth, producing flux but not consuming power.

Now, when a load is placed on the secondary, a secondary current flows. This current produces a magnetic field (flux) that opposes that produced by the primary. This reduces the effective inductance of the primary which allows more current to flow through it, which produces more flux. This feedback cycle finds equilibrium when the magnetizing flux is restored and the output voltage is maintained. So the magnetizing flux is there at the same level at all times during the transformer's normal (i.e. non overload) operation. There might be a boat-load of other flux happening in both directions as power is transferred in a loaded state, but the net flux is always at the constant level produced by the magnetizing current.

From this you can see that the primary must always produce more flux than is cancelled out by current flow in the secondary and this surplus is provided by the magnetizing current. An open-circuit secondary takes the secondary current out of the picture and allows ready measurement of this magnetizing current.

This might make some sense of the expression "setting up the flux", although I think that's one of those less-than-instructive expressions that only makes sense when you already understand and don't need it any more :)


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