Suppose we have applied a voltage to a transformer with an open secondary, we will get a certain flux in the core and a certain voltage at the secondary depending upon the turns ratio.

Now, suppose we connect resistance to the secondary and therefore current will start to flow through it and a reflected current will flow in the transformer's primary winding. But the magnitude of flux remains the same with or without the load.

So my question is: What exactly changes inside the magnetic core when we connect a load to the transformer, which results in power transfer from primary to secondary?

As far I understand, change in flux induces a voltage but does not carry any power, and change in flux remains the same with or without the load.

I asked the exact same question on electronic stack exchange and someone suggested to ask it on this forum for a more complete answer from a physics point of view.


1 Answer 1


To analyse a transformer completely is very difficult so I will assume an ideal transformer with no losses.

With the secondary coil not connected the voltage and current in the primary coil are $90^\circ$ out of phase just as you would have for a self inductor.
There is a flux change in the secondary coil and this does induce an emf but as there is no complete conducting circuit there is no induced current and so no electrical energy is dissipated.
So when an ideal transformer has nothing connected to the secondary coil the net energy supplied by the voltage source connected to the primary over a period is zero.

Connecting a resistor across the secondary results in the voltage and the current in the primary are in phase and so energy is supplied to the transform which is dissipated a heat in the resistor connected across the secondary.
The transformer transfers energy from the voltage source to the resistor with $100\%$ efficiency without itself dissipating any energy.
As far as the voltage source is concerned connecting it to a resistor with resistance $R$ via a transformer with turns ration $n = N_{\rm primary}/N_{\rm secondary}$ is equivalent to connecting the voltage source directly to a resistor of resistance $R\,n^2$.

  • $\begingroup$ I understand the equations, what I want to understand is something more fundamental. I want to understand which physical entity carries the power from primary which is dissipated in the secondary. The only thing as far as I understand connecting primary with secondary is the flux, but the magnitude and frequency of flux is identical with or without the load, so the only quantity connecting primary and secondary didn't change, how come the power transfer changed? $\endgroup$ Jul 3, 2021 at 7:44
  • $\begingroup$ The flux depends on the resistance $R$, here is your mistake. Consider $R\to 0$. The current in the secondary side becomes so strong that prevents the primary side from any flux time-variations, I guess (an infinite induction case). $\endgroup$ Jul 3, 2021 at 7:51
  • $\begingroup$ Generally, it is (retarded) interaction benween charges of both sides that carries energy. $\endgroup$ Jul 3, 2021 at 7:53
  • $\begingroup$ I am sorry but can you elaborate a little, I thought voltage is the time derivative of flux and flux's magnitude has nothing to do with current. $\endgroup$ Jul 3, 2021 at 7:55
  • $\begingroup$ The thing which changes is the phase between the voltage and the current. $\endgroup$
    – Farcher
    Jul 3, 2021 at 8:00

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