I am happy to always not understand how does a transformer works ... One answer give me 10 questions ... (So I suggest you to not answer to this question, if you do not want to have more questions :D)

Here is the circuit :

enter image description here

So I put the "coupling" factor to 1, in order to have no leakage inductance and I set the value of the inductors very high in order to have a very low magnetizing current. So the primary current is approximately equal to the secondary current. My question is the following : How the primary current can reach an equilibrium to be equal to the secondary load current ? I will not accept answer which tells Pout is equal to ... So Pin is equal to Pout... There is something happenning into the transformer ...

Thank you very much and have a nice day !

  • 1
    \$\begingroup\$ Yes, it is happening in the transformer. You answered the question by yourself in a manner that you most like it. \$\endgroup\$ Commented Apr 20, 2020 at 7:44
  • \$\begingroup\$ Do you understand that the secondary voltage equals the primary voltage? Do you need proof of this? \$\endgroup\$
    – Andy aka
    Commented Apr 20, 2020 at 8:31
  • \$\begingroup\$ My mystake is probably because I influenced you by supposing an ideal transformer. But as "Jack is on the way" tell it in an ideal transformer the magnetic flux at t = 0 is equal to the one at t = 0+. This is true because the energy between the primary and the secondary is instantaneously transfered. But in a real transformer, the energy between the primary and the secondary cannot be transfered instaneously due to the magnetizing inductance. So at t = 0, how can the primary current could be equal to the secondary current without having any information about the secondary load ? \$\endgroup\$
    – Jess
    Commented Apr 20, 2020 at 9:15
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    \$\begingroup\$ Try to read this books.google.pl/… \$\endgroup\$
    – G36
    Commented Apr 20, 2020 at 13:30
  • 1
    \$\begingroup\$ The flux produced by a primary winding generates the back emf to counter the input voltage. But when secondary current start to flow it will "produce" a flux which exactly opposes the "direction" of a primary flux (right-hand rule), thus primary flux is reduced. And this resulting in a corresponding reduction of the back emf in the primary. The primary current then increases until the flux is again high enough to counter the applied primary voltage. \$\endgroup\$
    – G36
    Commented Apr 20, 2020 at 14:14

3 Answers 3


It seems to me that you have to be helped to recognize this: -

enter image description here

Picture from here.

In other words, if you apply 1 volt to the blue winding then 1 volt will appear on the red winding. This happens when the turns on both windings are in-phase and very closely coupled but I haven't given a proper reason why this happens yet.

Consider ONLY the blue winding that has 1 volt applied to it. The current into that winding is the magnetization current and, it ramps upwards governed by: -

$$\dfrac{di}{dt} = \dfrac{V}{L}$$

In that blue winding there is a back-emf that is equal to the applied voltage. This is an induced voltage and, if you introduced the red winding (the secondary) you would see the same induced voltage. It would be the same because the magnetic flux produced by the magnetization current in the blue winding is fully coupled to the red secondary winding.

Now what would happen if we attached a load resistor of 1 ohm to the secondary winding?

Current (1 amp) would flow into that 1 ohm resistor like so: -

enter image description here

That current (on the face of it) could create an extra magnetic flux in the transformer windings. The extra flux might (possibly) do two things. It might: -

  • cause the output voltage to rise (due to \$V = L\dfrac{di}{dt}\$)
  • cause the output voltage to fall (due to \$V = -L\dfrac{di}{dt}\$)

If it caused the output voltage to rise then we would have an unstable situation because, due to that voltage increase, there has to be more current flowing into the 1 ohm resistor and this means more flux and more voltage and this ends in disaster.

If it caused the output voltage to fall then the back emf in the primary would also fall and we would have a situation where the primary took an unholy amount of current.

But, the reality is that neither of the above happens. If either of the above happened we end with contradictions - if the secondary voltage rose driving more current through the 1 ohm load, the primary voltage also has to rise but it can't because, it is limited by the applied 1 volt hence, current has to simultaneously flow back into the 1 volt source on the blue winding whilst also flowing into the secondary load i.e. we have uncontrolled perpetual energy that self-destructs.

If the secondary voltage fell we have the contradiction of primary (blue) current rising uncontrollably which means flux rises uncontrollably which means secondary (red) voltage has to rise - a contradiction.

The sensible equilibrium is found when we examine the 3rd scenario: -

  • The rate of change of flux in the windings remains constant

This can only happen when the load current flowing from the secondary (red) is matched by a load current flowing into the primary (blue). These load currents produce equal and opposite magnetic fluxes and hence they cancel leaving the original magnetization flux and, that magnetization flux carries on doing what it did all along - setting the output voltage to be equal to the input voltage (1:1).

For a transformer with a non unity turns ratio, the secondary load ampere-turns are equal and opposite with the primary current ampere-turns that flow due to the load.

  • \$\begingroup\$ Thank you for your replies ! The response of how the equilibrium is reached ! :) \$\endgroup\$
    – Jess
    Commented Apr 20, 2020 at 13:47

You have set the inductances very high, which is what you'd expect in a near ideal transformer.

This means that any dI/dt in the total current through the transformer, where total = the signed sum of primary and secondary, ie the difference between primary and secondary currents, will be able to generate a very large voltage, as V=LdI/dt. This voltage is able to dominate the supply voltage and control the current through the primary, such that the difference in currents remains small.

If you reduce the inductances, you'll see that the primary current does not track the secondary so closely.

This is a qualitative argument that there exists a mechanism for balancing the currents, and that the strength of the mechanism is dependent on the inductance in the transformer. It would be interesting to develop it quantitatively.

You'll notice that the difference between the primary and secondary current is the magnetising current, so if we 'don't consider magnetising current' as you've asked, we're not going to get anywhere.

If you want the easiest quantitative answer, then going via energy balance is more straightforward. If no energy is lost in the transformer, then Pout = Pin. But of course this is non-mechanistic.


Please ignore the magnetizing current. Assuming that the secondary is open at time=0,Vin=Vinput,Vout=Vin/n(n=1 in your picture). Connect the load at time=0+.It will be a current "Is" in secondary.The magnetic core in the transformer always keep the magnetic flux unchanged(abrupt change of magnetic flux is impossible). But how?the transformer ask for current "Ip" from primary.Is=-n*Ip(n=1 in your picture),different direction.Then the magnetic flux at t = 0 and t = 0 + are finally the same. I hope above explanation can help you!


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