# AC current in a coil. Why is this statement true?

I can't understand the following statement:

• "A contactor coil with a small control voltage, e.g. 24V AC, consumes a higher coil current than a similar contactor coil for 230V AC."

Does this phenomenon have something to do with the fact that the self-induced voltage creates a current and that current generates a magnetic counter-flux which weakens the magnetic field in the input? Hence, I think that with the 230V, the counter-current will be higher and so the current in the circuit will be lower. Is this true?

A 230 V AC coil will have many more turns and therefore much more inductance than a 24 volt AC coil. Given that it is ampere_turns that generates the magnetic field inside a contactor, for the same pulling force on both, the current in the 230 V coil will be proportionately smaller due to it having many more turns.

Does this phenomenon have something to do with.....

No. It's all about making sure that the magnetic core inside the contactor does not saturate. The higher voltage coil has to have more turns in order to make inductance higher. Inductance is proportional to turns squared hence, doubling the number of turns means 4 times the inductance and one quarter of the current for the same applied AC voltage.

This in turn means that to get the same ampere_turns, the applied voltage will double for a coil with twice the turns. So, the turns have doubled and the current is half i.e. same ampere_turns.

All things being equal a 230 V coil will have approximately 9.6 x the number of turns of a 24 V coil.