Ideally the primary current should be zero under the open-circuit test. Could someone explain to me what are the physical effects corresponding to the two equivalent-circuit parameters that cause the primary current to be non-zero when the secondary winding is open-circuited? I was told that it isn't due to loss in the copper windings!!


Primary inductance

Primary current needs to remain finite to keep the core magnetised. It's the change in this magnetisation that generates the voltage in all the transformer windings, both primary and secondary. As the transformer becomes 'better', higher permeability core, smaller air-gaps in the magnetic circuit, this current becomes lower, and in the ideal transformer = infinite permeability/inductance case tends to zero.

This current only stores magnetic energy, so is reactive, and is represented by a shunt primary inductor in a transformer model.

While it does dissipate some power in the primary winding due to \$I^2R\$, as the magnetising current is usually an order of magnitude or two below rated current, this loss is usually negligible.

Core losses

Hysteresis loss in the core is due to the finite amount of energy required to take the core magnetisation round the full BH loop every cycle. Transformers tend to use 'soft' cores that have narrow BH loops to minimise this loss.

Eddy current loss in the core is due to closed conducting paths around areas of changing flux in the core acting as short-circuited parasitic secondaries. Cores are made out of thin insulated laminations to reduce the area and hence voltage round these loops, and high resistance materials to minimise the resulting currents flowing.

Both of the core losses dissipate energy, so look resistive in the transformer model.

  • \$\begingroup\$ Would you say that primary voltage is not zero (under short-circuit test) due to the secondary winding and resistance? \$\endgroup\$
    – William
    May 7 '19 at 10:23
  • \$\begingroup\$ If you s/c the secondary and feed the primary with a current, then the primary voltage will not be zero due mainly to winding resistance in both primary and secondary (in equal measure in a well designed single secondary transformer), and also due to leakage inductance (flux not shared between primary and secondary, there will always be some). \$\endgroup\$
    – Neil_UK
    May 7 '19 at 15:15
  • \$\begingroup\$ Thanks a lot. When you talked about the primary inductance, were you referring to the magnetiding core inductance or primary leakage reactance? \$\endgroup\$
    – William
    May 13 '19 at 3:04
  • \$\begingroup\$ @Khalid primary inductance => magnetising core inductance \$\endgroup\$
    – Neil_UK
    May 13 '19 at 5:15

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