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What I understand about working principles of ideal transformer:

We apply an alternating voltage at the primary side. This creates a flux inside the transformer core (\$\Phi_p\$). Some voltage is induced on the secondary side winding according to the Lenz's Rule (\$\mathcal{E}=-\frac{d\Phi_p(t)}{dt}\$). This voltage crates a current through the load (\$i_s(t)\$). And this current creates an opposing flux in the core (\$\Phi_s\$). The amount of this flux that is generated by the secondary winding is equal to the one generated by the primary winding (\$\Phi_s=\Phi_p\$) because the transformer drain that much of current from the primary side power supply according to he formula \$\frac{V_1}{V_2}=\frac{N_1}{N_2}\$. Therefore, there mustn't be any flux inside an ideal transformer.

But all these don't make any sense to me because of the following reasons.

Primary side inductance of an ideal transformer is so large, so no current flows through it if no load is connected at the secondary side. If we connect a load after powering up the primary side, there still shouldn't be any current at the secondary side, because no current if drawn from the primary side, thus there is no flux in the core. Even if we assume that there is some flux in the core, the opposing flux (\$\Phi_s\$) will cancel it out, and the net flux in the core will drop down to zero (\$\Phi_{net}=\Phi_p-\Phi_s=0\$). So the power transfer will stop.

Some flux must stay inside the core during normal operation, but all these facts(?) I told about claim that there mustn't be any flux in the core. Why does a contradiction like this occur? I understand that, in electronics, some circuit model won't function when built with ideal circuit elements (e.g.; flip flops won't take initial state, phase shift oscillator won't start, etc). We usually have some sort of realistic effect that starts the operation. Does the "magnetizing inductor \$L_m$" which we connect parallel to the primary side in the realistic transformer model have something to do with this?

How does flux exist/stay in the transformer core? Please explain it.


2 Answers 2


To help improve the understanding of why the flux in a transformer roughly stays constant under load consider these four pictures and look at each one in turn: -

enter image description here

Regards the infinite inductance part of the question, inductance is defined as total flux produced per amp i.e. a bigger inductor produces more flux per amp. However, when AC is used an inductor that is twice as big produces half the current but, (as per the above definition) it produces twice the flux per amp and because the current has halved, the flux remains the same. Keep taking inductance up and up and flux remains the same.

  • \$\begingroup\$ Thank you, very informative. Is this image from an online article? If so, can you please give the link? \$\endgroup\$ Commented May 5, 2014 at 10:35
  • \$\begingroup\$ @hkBattousai I drew the image last year sometime to answer another question. It could ideally do with a revamp!! \$\endgroup\$
    – Andy aka
    Commented May 5, 2014 at 10:44

The fluxes created by the primary and the secondary windings are not equal; your equotion \$\frac{V_1}{V_2}=\frac{N_1}{N_2}\$ is just an approximation.

If you connect a transformer to the mains, but not to the load, the current flowing through the primary winding will create some flux in the core. It's determined by mains voltage, frequency and primary winding inductance. If you connect a load to the transformer, the primary current will increase in a way to keep the core flux close to what it was without load.

An 'ideal' transformer has its primary inductance rising up to infinity, so the unloaded (magnetizing) primary current drops down to zero. But infinite inductance multiplied by infinitely small current results in some definite flux in the core.

  • \$\begingroup\$ Do you mean that there is always a small fixed amount of flux in the 'real' transformer core, and the transformer tries to keep the amount of that flux fixed? \$\endgroup\$ Commented May 5, 2014 at 9:25
  • \$\begingroup\$ Quite a large amount of flux actually - generated by the primary inductance - but not enough to start saturating the core. But yes, the transformer consumes additional primary current to "keep it fixed" when you draw current from the secondary. \$\endgroup\$
    – user16324
    Commented May 5, 2014 at 9:36
  • \$\begingroup\$ Does it mean that, when designing a transformer, I should make it so that inductance of the primary side should be the largest it can be which wouldn't saturate the core (with no load on the secondary side) when the maximum rated primary side voltage is applied? Because, even if we connect a load to the secondary side, this magnetizing flux will stay same around a certain fixed level, isn't it so? \$\endgroup\$ Commented May 5, 2014 at 9:49
  • \$\begingroup\$ @hkBattousai Yes, you are right. \$\endgroup\$ Commented May 5, 2014 at 10:09

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