Electronic – Why doesn’t an open-secondary transformer violate Ampere’s law

Look at the following scenarios and then it should be clearer why maximum transformer flux happens when off-load: -

The above are perfect idealizations of inductors and transformers. And if you are having problems with any of the steps then maybe someone else can do this question some justice.

Why is flux greatest when off-load - if there were no-losses in the windings (R or leakage inductance) then off-load or on-load the flux in the core remains exactly the same (scenario 4). But, because the secondary current causes small volt-drops in the primary winding, the voltage that is able to magnetize the core is slightly reduced thus the flux in the core is slightly reduced and the core is less able to saturate.

The permeability of the material (when not close to saturating) is determined by the physical properties of the material and not the opposing fluxes in primary and secondary due to load current.

The vast majority of flux does not leak into space around the core - it is contained within the core and cancels out to zero for load-currents leaving what was originally there - the magnetizing flux.

When a turns ratio is applied ampere-turns from primary might be 1 x 10 whereas on the secondary (for 10:1 turns ratio) they will be 10 x 1 - ampere-turns drive flux and these cancel out just the same as a 1:1 transformer.

Here is the link to the question posed by Jim related to this one.

Lenz's Law Lenz's law says

"An induced current creates a magnetic field that opposes the changing flux that initially gave rise to the induced current".

Bar magnets This is not the same in bar magnets held N-N and S-S. In Bar magnets (held steady wrt each other) there is no induction and therefore only a force exists and I think this is to do with Newtons 3rd Law?

Twisted pair cable Also think about twisted pair cable - there is no net flux emanating beyond a small distance. of course there are very local fluxes around each wire but these cancel at a very short distance.

What you may be getting confused about is the "ideal transformer" in the equivalent circuit. You should not regard it as having any magnetic qualities at all. Try and see it like this: -

Whatever voltage you have on the input to the ideal transformer, \$V_P\$ is converted to \$V_S\$ on the output of this "theoretical" and perfect device. It converts power in to power out without loss or degradation such that: -

\$V_P\cdot I_P = V_S\cdot I_S\$

The ratio of \$V_P\$ to \$V_S\$ happens to be also called the turns ratio and almost quite literally it is on an unloaded transformer because there will be no volt-drop across R3 and L3 and only the tiniest of volt-drops on R1 and L1.

This means you now have a relatively easy way of constructing scenarios of load effects and recognizing the volt drops across the leakage components that are present.

The equivalent circuit of the transformer is really quite good once you accept that the ideal power converter is "untouchable" and should just be regarded as a black box. For instance you can measure both R1 and R3 and, by shorting the secondary you can get a pretty good idea what L3 and L1 are. With open circuit secondary you can measure the current into the primary and get a pretty good idea what \$L_M\$ is too.