I'm trying to get my head wrapped around transformer operation and in the process regretting the times I snoozed in my Electromagnetics class as a EE student back when I was a lad :)

I'm looking for an intuitive understanding, but not just an analogous one. I'd like it to be grounded in the actual physics of what's happening. I've found several excellent sources on the web, but they all seem to skirt this question.

I've come across a few interesting hints and am now tantalizingly close, I think, but still yearning :)

Fact 1: Although varying sinusoidally, the "peak-to-peak" flux, so to speak, in a transformer's core is essentially constant (for a given voltage applied to the primary), regardless of the load.

My intuitive hypothesis was that variation in the "strength" of the flux was what transferred the energy from the primary to the secondary, but this fact would seem to contradict that theory. I had thought that the primary makes a bunch of flux based on the current flowing through it and the secondary sucks it up to make current of its own. No dice, it seems.

Then of course there's the fact that the formula for flux involves only voltage, time (frequency), and turns :)

Fact 2: The current in the primary is (approximately) 90 degrees out of phase with the voltage at no load, and approximately aligned in phase at full load.

This fact seems very promising and also curiously satisfying. It would imply that the Volt-Amps (VA) of the primary is constant and only the power factor changes as the current load on the secondary increases.

But I still don't get how the energy is actually being transferred. It seems vaguely like the flux is just there as an energy conductor or something and some other phenomenon is actually doing the energy transfer bit.

Can someone see what I'm missing and explain what's actually happening in there?

  • \$\begingroup\$ Transformers require AC, which results in a changing flux. That in turn induces a current in the secondary. \$\endgroup\$ Apr 14, 2015 at 19:44
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    \$\begingroup\$ the peak value of the flux is constant regardless of load. But the actual value is continually changing and reversing with the AC waveform. \$\endgroup\$ Apr 14, 2015 at 20:25

4 Answers 4


The answer has been totally rewritten to fulfill the request of a mathematical, format treatment, as asked for by StainlessSteelRat. The bold-face part in the center of this answer is the actual answer to your question. The remaining part is to show how I get to that formula and to connect your two facts (which I consider true) to the answer.

Take a look at the following formulas are important to understand the transformer:

  • Faraday's law of induction (for a coil with \$n\$ turns): \$U=-n\frac{\mathrm d}{\mathrm dt}\Phi\$.
  • Hopkin's law (aka magnetic Ohms law): \$\Phi = \frac{nI}{R_m}\$

Faradays law describes how the rate of change of the magnetic flux \$\Phi\$ through a single turn of a coil is related to the voltage \$U\$ induced in that coil. Hopkin's law describes the connection between the current \$I\$ applied to a long coil with \$n\$ turns and the flux \$\Phi\$ caused by that current. The value \$R_m\$ (magnetic resistance) depends on the geometry of the core, so it is a constant value for a given transformer.

If a sinusodial voltage \$U(t)=U_0\sin(\omega t)\$ is applied to a coil, induction of faradays law yields

\$\Phi(t) = -\frac{U_0}{\omega n}\cos(\omega t)\$ (The integration constant has been chosen in a way that the average flux vanishes)

In a transformer, there are two coils on a shared core. The resulting flux in the core \$\Phi_c\$ is the sum of the flux generated by both coils on the core:

\$\Phi_c = \Phi_p + \Phi_s\$

The law of induction relates the derivative of total flux through a coil (whether it is caused by the current of that coil or is caused by external fields) to the induced voltage. So you can write faradays law for both the primary and the secondary winding as

\$U_p = -n_p\frac{\mathrm d}{\mathrm dt}\Phi_c\$; \$U_s = -n_s\frac{\mathrm d}{\mathrm dt}\Phi_c\$.

Note that while the voltages on the primary and secondary side (as well as the turn counts) are different, the shared total flux \$\Phi_c\$ appears in both equations. It can be eliminated yielding the well-known transformer equation

\$U_s = -\frac{n_s}{n_p}U_p\$

More step-by-step: By integration the \$U_p\$ equation (as already shown for sinusodial voltage above), the flux \$\Phi_c\$ caused by a voltage source on the primary side can be calculated. By differentiating \$\Phi_c\$ again in the \$U_s\$ equation, the secondary voltage caused by the flux can be calculated.

Note that up to now, the current did not really take place in the discussion of the transformer. The equation (and thus the integrated \$\Phi_c\$) is valid for all load conditions of an ideal transformer driven by an ideal voltage source. (This is your "Fact 1")

To add the effects of the load, consider the two parts that make up the resulting flux:

\$\Phi_c = \frac{n_p I_p}{R_m} + \frac{n_s I_s}{R_m}\$

As we assume the primary side is an ideal voltage source, there is nothing known about the the current \$I_p\$, as the source would deliver every current needed so that the primary voltage is as required. If we assume a simple ohmic load, the load current is known, though, it is \$I_l = U_s/R_s\$, with \$R_s\$ being the resistance of the load. At this point, signs get very important. If I call one terminal of the secondary "ground" and the other one "live", the voltage between the "live" and the "ground" terminal has a clearly defined sign. On the other hand, the current has to flow from "ground" to "live" inside the transformer as it flows from "live" to "ground" in the load and the other way around. So if secondary voltage and load voltage are defined as the voltage from the "live" to the "ground" terminal, they are obviously equal-signed, whereas the currents, measured as "current flowing into the respective live terminal" are oppositely signed. This means \$I_s = -I_l\$, so the formula for the flux can be written as

\$R_m\Phi_c = n_p I_p - n_s I_l = n_p I_p - n_s \frac{U_s}{R_s} = n_p I_p - \frac{n_s^2}{n_p}\frac{U_p}{R_s}\$

The first term of the right hand side shows the flux "created" by the primary winding, and the second term the flux "consumed" by the secondary side. The part of the flux consumed by the secondary side corresponds to the energy transmitted from the primary to the secondary. Solve that equation for \$I_p\$ to obtain:

\$I_p = \frac{R_m\Phi_c}{n_p} + \frac{n_s^2}{n_p^2}\frac{U_p}{R_s}\$

substituting a sinusodial voltage for \$U_p\$ and using the integrated term for \$\Phi_c\$, one obtains:

\$I_p(t) = -\frac{R_m}{n_p^2 \omega}U_0\cos(\omega t) + \frac{n_s^2}{n_p^2}\frac{U_0}{R_s}\sin(\omega t)\$

The first term on the right-hand side is out-of-phase with \$U(t)=U_0\sin(\omega t)\$, so it describes reactive power and is load-independent, while the second term is in-phase, so it describes active power and is load dependent. If the transformer is unloaded, the second term vanishes and the first reactive term remains, causing the 90° phase shift, while on a highly loaded transformer, the second term dominates so that the phase shift gets very close to zero. This is your "Fact 2".

  • \$\begingroup\$ Ah HA! So my making and eating flux intuition wasn't completely mistaken then, is that right? \$\endgroup\$
    – scanny
    Apr 14, 2015 at 22:11
  • \$\begingroup\$ No, I think that intuition is quite good. The formula you quote not containing current is just about the flux in the core, which is the flux made but not eaten. \$\endgroup\$ Apr 14, 2015 at 22:30
  • \$\begingroup\$ I expanded my answer to contain a paragraph about the part about the "unused" flux that remains in the core. \$\endgroup\$ Apr 14, 2015 at 22:36
  • \$\begingroup\$ I re-read the second paragraph, and I will stick to it, possibly its unclear what I meant. Yes. If the current in the primary increases, the flux generated by the primary increases. But that only happens when the secondary current increases, too, which increases the counter-flux. The sum of flux and counter-flux stays the same, though, and that is what I call the "flux in the core". Also, you are right that the turn determines the voltage ratio. This is because for each turn, U=-d/dt(phi) is valid, with phi being the "flux in the core". \$\endgroup\$ Apr 15, 2015 at 5:54
  • \$\begingroup\$ But if the flux does not take place in the energy transfer, how does energy get transferred? The flux amplitude has to do with current not voltage. How can "flux amplitude is proportional to the primary voltage" apply if the voltage does not change and flux does? The flux in the core will be in phase with the current since it produces it. If the load is lagging it will lag, etc. Give me a formula. As an engineer, I can never accept the magic happens. \$\endgroup\$ Apr 15, 2015 at 11:44

From what I know, the delta in primary coil current (as per an AC system) is what builds up energy inside the metal core of the transformer, which in turn generates a magnetic field in the opposite direction of the current flow (so it is in effect resisting the current flow). But, at the same time, in the secondary coil this magnetic field causes induction, which generates a current flow in the secondary circuit, which is what gives the current there. The fact that two different principles are at play generates the phase difference.
But I'll let any expert poke holes in my explanation for my own learning and understanding :)


Power (energy) transfer in iron core transformer has never been given a thorough treatment outside of the math realm. An interesting fact is illustrated by te Poynting Theorem. In the case of the transformer on load, the core flux is almost constant, however the "LEAKAGE" flux is load dependent anc can reach staggering levels, that is flux intensity. That is flux outside of the core around the Primary winding while on load. that flux, is not dependant on core permeability, as even if the core had infinite permeability, the load flux (Called Leakage) would still exist. Many sources claim this so called "leakage" flux is essential for power flow thru the transformer. I agree, but I think it is fundamentally wrong to call it a "LEAKAGE" flux. The core GUIDES energy to the secondary. The core flux is 90 degrees lagging the primary voltage so the only energy associated with the core is purely inductive storage that gets cycled to and from the supply every quarter cycle and hysteresis and heating losses due to eddy currents in the core. It would seem that the core flux induces EMF in the secondary coil and the "LEAKAGE" flux surrounding the loaded primary and in phase with the voltage satisfies the poynting theorem for energy flow. As an illustrative example, consider the old constant current street lighting transformers of yesterday, where the secondary winding was free to move up and down away from the primary to keep the series current stable; when on low load the electromagnetic forces developed from those (LEAKAGE) fluxes were weak and the coils were physically close, when on heavy load, the secondary winding would be repelled far away from the primary which would reduce the energy flow thru the transformer. Textbooks explain by stating the leakage flux around each winding, now far apart choke current flow, if close together they would tend to cancel each other. I am saying, maybe the primary "LEAKAGE" flux cannot not be close to the secondary, thereby reducing the power flow. Just my thoughts, but of intense interest to me........ Tom Cosby

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    \$\begingroup\$ Welcome to EE.SE, Tom. Your answer could do with some paragraph breaks (2 x enter) and fixing some random capitalisation. It's a rather difficult read as is. Signatures are discouraged as they automatically appear beneath your post. \$\endgroup\$
    – Transistor
    Jul 11, 2017 at 12:35

Let us first repeat the question

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 when we connect a load to the transformer, which results in power transfer from primary to secondary? In short, which magnetic parameter is responsible for power transfer between primary and 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.

Now let me answer your question

In the case the secondary coil is open, the primary coil is like a self-inductance. It only produce the reactive power. The power is sends out to the secondary coil. But the secondary coil do not receive it. Hence the power returns. There are no power is consumed for the transformer.

In the case the secondary coil is connected with a load. The secondary coil has current. These current will create a electromagnetic field to the primary coil. This is the reaction of the secondary coil to the primary coil. The primary coil has to overcome this reaction. This need to do some active power to the secondary coil. This active power will be received by the secondary coil. This way the power is transferred from the primary coil to the secondary coil.

For transformer the primary coil and secondary coil satisfy the energy conservation law (or mutual energy theorem):

$$ -\iiint_{V_{1}}\boldsymbol{E}_{2}^{*}(\omega)\cdot\boldsymbol{J}_{1}(\omega)dV=\iiint_{V_{2}}\boldsymbol{E}_{1}(\omega)\cdot\boldsymbol{J}_{2}^{*}(\omega)dV $$

The energy from the primary coil to the secondary coil is the mutual energy flow. There is the mutual energy flow theorem $$ -\iiint_{V_{1}}\boldsymbol{E}_{2}^{*}(\omega)\cdot\boldsymbol{J}_{1}(\omega)dV=(\xi_{2},\xi_{1})=\iiint_{V_{2}}\boldsymbol{E}_{1}(\omega)\cdot\boldsymbol{J}_{2}^{*}(\omega)dV $$ where $$ (\xi_{2},\xi_{1})=\iint_{\Gamma}(\boldsymbol{E}_{1}\times\boldsymbol{H}_{2}^{*}+\boldsymbol{E}_{2}^{*}\times\boldsymbol{H}_{1})\cdot\hat{n}d\Gamma $$ The energy flow from the primary coil to the secondary coil is not the energy flow of Poynting vector. The energy flow of Poynting vector belongs to the self-energy flow. The self-energy flow do not transfer energy. The energy flow is only transferred by the mutual energy flow. If the load of the secondary coil is a resistance the mutual energy flow is active power. If the load of secondary coil is capacity, the mutual energy flow is reactive.

The above energy conservation law and mutual energy flow theory is not only suitable to the transformer, it is also suitable to a antenna system with a transmitting antenna and a receiving antenna. Actually if we move the secondary coil away from the primary coil, the primary coil becomes the transmitting antenna, and the secondary coil becomes the receiving antenna.

In case the of antenna system, the transmitting antenna send the retarded potential. The receiving antenna sends advanced potential.

The above electromagnetic theory is not the classical electromagnetic field theory. Classical electromagnetic field theory do not accept the advanced wave, because it Violation of causality. But there are lot of scientists believe the advanced waves are real objective existence. The above electromagnetic field theory is the mutual energy theory developed by shuang-ren Zhao

For advanced wave see the transactional interpretation of QM of John Cramer or absorber theory of Wheeler and Feynman.

About the history of the energy conservation law. Around 1900 Lorentz reciprocity theorem was appeared. $$ \iiint_{V_{1}}\boldsymbol{E}_{2}(\omega)\cdot\boldsymbol{J}_{1}(\omega)dV=\iiint_{V_{2}}\boldsymbol{E}_{1}(\omega)\cdot\boldsymbol{J}_{2}(\omega)dV $$

Welch's reciprocity theorem was appeared in 1960: $$ -\iiint_{V_{1}}\boldsymbol{E}_{2}(t)\cdot\boldsymbol{J}_{1}(t)dV=\iiint_{V_{2}}\boldsymbol{E}_{1}(t)\cdot\boldsymbol{J}_{2}(t)dV $$

Rumsey's reciprocity reciprocity theorem appeared in 1963 which has the same form as the mutual energy theorem.

Mutual energy theorem is appeared in 1987. The important thing is Shuang-ren Zhao thought this theorem is not only a reciprocity theorem but a energy theorem.

In the end of 1987 de Hoop introduced the cross-correlated reciprocity theorem $$ -\int_{t=-\infty}^{\infty}dt\iiint_{V_{1}}\boldsymbol{E}_{2}(t)\cdot\boldsymbol{J}_{1}(t+\tau)dV=\int_{t=-\infty}^{\infty}dt\iiint_{V_{2}}\boldsymbol{E}_{1}(t+\tau)\cdot\boldsymbol{J}_{2}(t)dV $$ The mutual energy theorem is Fourier transform of the cross-correlated reciprocity theorem. Welch's reciprocity theorem is a special case of the cross-correlated reciprocity theorem. Rumsey's theorem is same as the mutual energy theorem. Hence, these 4 theorem are same they are reciprocity theorem and also energy theorem. Lorentz reciprocity theorem can be obtained from the mutual energy theorem by conjugate transform.

The mutual energy flow theorem was introduced by shuang-ren Zhao in 2017. According to this, the mutual energy theorem is not only a energy theorem but a localized energy conservation law.

The mutual energy flow is a updated version of Poynting vector energy flow. It is more suitable in case two objects are considered rather than only one object. A transformer is clear has two objects the primary coil and the secondary coil. There are action and reaction between the two objects, hence the mutual energy flow theorem is needed to describe them.

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    \$\begingroup\$ "The energy flow from the primary coil to the secondary coil is not the energy flow of Poynting vector. " Could you cite sources for this claim? It seems to conflict with, for example researchgate.net/publication/… \$\endgroup\$ Apr 29 at 13:15
  • \$\begingroup\$ Mutual energy theory is introduced not for transform but more for antenna system include a transmitting antenna and a receiving antenna and the wave-particle duality problems. But the theory is general and also should be applied to transformers. You can find the citation as following: 1. Shuang ren Zhao. A new interpretation of quantum physics: Mutual energy flow interpretation. American Journal of Modern Physics and Application, 4(3):12_23, 2017. 2. Shuang-ren Zhao, Photon Can Be Described as the Normalized Mutual Energy Flow. Journal of Modern Physics Vol.11 No.5, May 202. \$\endgroup\$ Apr 29 at 13:52
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    \$\begingroup\$ Wikipedia describes SCIRP, the publisher of Journal of Modern Physics as a predatory publisher of questionable quality. en.wikipedia.org/wiki/Scientific_Research_Publishing Downvoting this answer. \$\endgroup\$ Apr 29 at 14:05
  • \$\begingroup\$ The problem of energy flow for a transformer is still open. There are only a very few papers try to solve this problem with Poynting vector. But all these a few papers usually cannot be good understand. The theory for mutual energy flow is the combination of 3 fields one is in physics that is the absorber theory of Wheeler and Feynman and transactional interpretation of Cramer. The second is the reciprocity theorem of Lorentz, Welch, Rumsey, Zhao, de Hoop. The third is the theory of energy flow of Poynting. Any way the problem is still Open. There are many possibilities. \$\endgroup\$ Apr 29 at 14:37
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    \$\begingroup\$ @Mutualenergy - Hi, You seem to be promoting something (in this case, a specific theory). Please note that this site rule applies. If you have any affiliation with what you describe in your answer, then you must edit the answer to disclose & explain that affiliation - otherwise further action, possibly including the deletion of the answer, might follow. Thanks. \$\endgroup\$
    – SamGibson
    Apr 29 at 16:01

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