I have a question about a standard approximation made about the magnetic flux in transformer analysis. Using the magnetic circuit model, the sum of the magnetomotive forces in a transformer are:

$$ \Phi_{core} R_{core} = N_1 I_1 - N_2I_2$$

The core reluctance \$ R_{core} \$ is very small for a transformer due to \$ \mu_r \$ being very high. Therefore, the approximation is made that the term \$ \Phi_{core} R_{core} \$ can be ignored. At first glance, I would accept this argument.

However, if you think of a KVL for an electrical circuit with two voltage sources separated by a very small resistor, the approximation doesn't make sense:

$$ IR = V_1 -V_2$$

Though the resistance is small, there is still a voltage drop between the two sources due to large current. Therefore, the term on the left is non-negligible.

So is the approximation that \$\Phi_{core} R_{core} = 0\$ wrong?

EDIT: There is a good answer to my initial question. I have a follow up question about what specific math\physics let you set the term to \$\Phi_{core} R_{core}\$ term to 0. For the case of two opposing voltage sources in series, if \$V_1\$ is more stiff, with a much higher decoupling capacitance than the capacitance of \$V_2\$, then after an initial transient, the voltage of \$V_2 \$ would be the same as \$V_1\$. Is there a similar train of thought that would enable \$N_1 I_1 = N_2 I_2 \$ after an initial transient?

  • \$\begingroup\$ It's not a generally accepted approximation - what makes you say that? Maybe some mickey mouse website? \$\endgroup\$
    – Andy aka
    Oct 1, 2021 at 7:21
  • \$\begingroup\$ For an ideal transformer, the current is related by the turns ratio via \$ I_2 = I_1 \frac{N_1}{N_2} \$. This is because the assumption is made that \$ \phi_{core} R_{core} = 0 \$ \$\endgroup\$
    – Brad
    Oct 1, 2021 at 18:35

1 Answer 1


It's a simple first order model, for a transformer with high or infinite permeability.

The important thing is that for a well designed high permeability core transformer, operating within its ratings, \$ \Phi_{core} R_{core} \$ is very small compared to either \$N_1 I_1\$ or \$N_2I_2\$. It's a small difference between two big numbers, so zero is a good approximation.

It's much the same sort of approximation you would make when you say an op-amp has infinite open loop gain, so the two inputs are at the same voltage. That works very well for understanding the basic circuit. If you want a more detailed model, then you let the gain be finite, and compute the finer details.

If you want to work with the magnetising current, then you need a finite permeability, and a non-zero left hand side.

It's not appropriate for air-core or low permeability transformers, or transformers being operated into saturation which collapses the permeability.

  • \$\begingroup\$ Thanks for your response; it makes perfect sense. My follow up question is what math/physics enables you to set the \$ \Phi R \$ term to 0? that For a voltage buffer amplifier, the error in input voltage is \$\frac{1}{A+1}\$, and as the open loop gain approaches infinity, the inputs become the same voltage. Is there some similar math argument that explains the approximation for a transformer? \$\endgroup\$
    – Brad
    Oct 2, 2021 at 14:25
  • \$\begingroup\$ @Brad Of course, permeability goes to infinity which means R goes to zero. Op amps get 'closer' to infinity with their gains of 10s of millions, whereas permeability only goes to thousands, but the important thing is that the difference between the NI terms is very small, a few percent, compared too their size. \$\endgroup\$
    – Neil_UK
    Oct 2, 2021 at 15:17

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