Saturday 18 March 2017

Turning an inductor into transformer and leaving the secondary open



Let's say I have an inductor of whatever type (ferrite core, air core). This inductor is a part of a circuit, an oscillator for example. Now I wind some magnet wire around it and I leave the ends unconnected so that this new winding is an open circuit.


Now the inductor is essentially turned into a transformer with the primary coil acting as an inductor in the bigger circuit and with the secondary open. I have learned that the product of voltage and current across the primary and secondary windings must equal to have conservation of energy. But as the "secondary" is now open, it has zero current, so the primary should have zero current too! If the "transformer" was previously an inductor in another circuit, does it mean that the inductor (and therefore the circuit) stops working (stops conducting, except for maybe some small magnetizing current)? This does not sound logical, I would assume that the new winding makes no real difference to the circuit, but how do we get around the fact that the inductor is now essentially a transformer?



Answer



Don't confuse ideal transformers with real ones. Below is the model for a real transformer.


enter image description here


The stuff up to \$I_o\$ branch is your inductor. The rest is the transformer part. Note the first resister in that path is basically infinite when there is no load \$R_S = \infty\$ and so that whole part of the circuit can be ignored.


NOTE: Image borrowed from this cross-post.


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