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Physics going on in a transformer

Former Community Member over 4 years ago

I’m trying to get a better understanding of the physics going on in a UK site transformer, those that reduce the UK mains voltage to 110V.

I’m fully aware of the basic operating principles of a transformer and the basic Power in equals V*I . Power out is the same as power in and induced voltage is given by Faradays law dependent upon the number of turns, and changing magnetic flux with time. With these we can work back from a load to determine the current draw from the secondary side of a transformer.

What I would like clarified is, depending upon the varying load on the secondary side, what at different loads is happening to the magnetic field strength in the core and the power in the primary side.

For example when I have a 1800W tool connected to my 3.3Kva 110V transformer and the power is on in the primary side but the load tool is not switched on. Obviously there is current flowing in the primary coil since there are no switches on these transformers apart from the socket switch. This will create a magnetic field which will induce an emf on the secondary windings.

If there is no load on the secondary side, is the magnetic field strength in the core and the power in the primary at its maximum even with no load on the secondary?

Or is primary input current much lower in comparison to what it could be if the transformer secondary is loaded to its maximum current output? Does putting a load on the secondary cause a feedback through the magnetic field resulting in the current increasing in the primary, hence increasing the magnetic field?

Thanks

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0 David Stone over 4 years ago

There are two other effects too, which I will add to Mikes's description. First the unloaded transformer inductance produces a back EMF as the mains cycle varies in voltage, which opposes the input voltage so the current is small. I=L dv/dt is the equation. There are some resistive and magnetic losses so there is a small current flow, in phase with dv/dt, that is at about 90 degrees to the voltage changing. Load on the secondary decreases the effective inductance (a shorted turn type effect) so the current in the primary increases in direct relation to the load Power (kVA). The losses increase with load, both resistive and magnetic, but these are small compared to the load. Typical large transformer efficiencies are in the high 90%, depending on the size, the quality of the magnetic material, and the disposition of windings on the core.
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0 David Stone over 4 years ago

There are two other effects too, which I will add to Mikes's description. First the unloaded transformer inductance produces a back EMF as the mains cycle varies in voltage, which opposes the input voltage so the current is small. I=L dv/dt is the equation. There are some resistive and magnetic losses so there is a small current flow, in phase with dv/dt, that is at about 90 degrees to the voltage changing. Load on the secondary decreases the effective inductance (a shorted turn type effect) so the current in the primary increases in direct relation to the load Power (kVA). The losses increase with load, both resistive and magnetic, but these are small compared to the load. Typical large transformer efficiencies are in the high 90%, depending on the size, the quality of the magnetic material, and the disposition of windings on the core.
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Physics going on in a transformer

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