Turning an inductor into transformer and leaving the secondary open

Question

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 1

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

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.

Answer 2

It doesn't sound like you want the electric/electronic explanation, but instead more of a physical/physics explanation. So I'll focus on that.

Think of your secondary as a conductor of some given length. Don't think about the primary, yet. Don't think about the varying magnetic field nearby it. Just think of the conductor. It's a veritable sea of electrons floating around in the "conduction band" due to thermal agitation. That's what a conductor IS.

If it helps any (and it's not important for thinking about this), the number of electrons is in an equilibrium state which merely means that an equal number of electrons are entering the conduction band (due to thermal agitation) as leaving it (due to recombination.) So it's not a static "sea" but it is a "stable" one.

These electrons repel each other. That's a given. So what takes place is that they get as far from each other as possible within the volume of the wire. So this means they are uniformly distributed, except at the very surface, where a few more can accumulate to create a very slight static charge (not many -- we are taking about ones and tens, not millions or billions by any stretch.) So the basic idea here is that the density is essentially uniform throughout for all practical purposes.

Now add to this an impressed external magnetic field that is changing. This magnetic field generates a non-Coulomb electric force that will act on the electrons in your secondary wire.

But this is like putting a weight onto the floor, pressing down on that floor due to gravity. The floor doesn't really move much. It might flex a little for a moment. But quite quickly counter-acting forces act to counter the added force and no more work is done. No matter how long the weight sits on the floor, no work is performed or required. It just sits there.

Similarly, these non-Coulomb forces could accelerate the conduction band electrons into motion.... if there was somewhere for them to go. But there isn't. You haven't connected anything up to it. So all that happens with these varying non-Coulomb forces is to press this way, then that way, then another way, on these conduction band electrons. They might move around just a little bit. But only a little. This is because as soon as one tries to move, it immediately gets repelled still more by some neighboring conduction band electron (itself trying to move, perhaps.) And because electric repulsion forces are so strong (about 40 orders of magnitude greater than gravity), in almost no time at all that "sea" has "flexed slightly" in response and ceases to move more. After all, the ends of the wires can only accumulate a few extra electrons (at most) before the repulsion is enough to completely stop any additional motions in the wire.

So we are talking about a few (one or two or three) electrons accumulating on one end of the coil's end, then the same few on the other end, then back again. There might be some slight changes at the wire surface in between. But the gist here is "nothing much changes or happens." It's just a slight jostling around.

The work done by these tiny motions is similarly tiny and probably not easily measurable. (I don't know of any standard COTS instrumentation capable of it, though I'm sure that physicists have from time to time played around, trying to.)

Now, if you provide a load of some kind, then this can allow those non-Coulomb forces to operate. And they can be easily powerful enough to induce sizable currents. (It only takes one or two electrons at a 90-degree corner in a large copper wire to cause an amp of current to make such a turn with ease -- such is the incredible power of electric forces.)

Answer 3

For a 100% efficient transformer power in does equal power out and, the residual current you see on the primary does not represent any power therefore there is nothing contentious happening. That current is called the magnetisation current and will be 90 degrees lagging the primary voltage hence, power in is zero.

After all, a perfect inductor takes current when a voltage is applied but does that perfect inductor get hot i.e. does it waste power? It will only do so if it has a winding resistance or there are core losses, hysteresis losses or eddy current losses.

Answer 4

I would like to answer the following aspect of your question: "I have learned that the product of voltage and current across the primary and secondary windings must equal to have conservation of energy".

This is a partial truth applicable only to the load current. When they say that phrase, they are ignoring the branch of magnetization of the core (magnetization current).

Answer 5

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!

Other answers focus on the difference between ideal and real transformers but that's just smoke screen. We can perfectly well stick with an ideal transformer for explaining this.

An inductor viewed as an electric device does not do "conservation of energy" since it periodically converts electric energy into a magnetic field and then withdraws it again. Ideally, all stored energy is eventually retrieved again, so the conservation of energy holds in the long run though not at any given point of time: if you time-integrate the product of voltage and current entering an inductor, the long-term integral will end up zero.

You can perfectly well use just the primary coil as an inductor. The secondary coil can be left open and its voltage monitored as an indicator of the flux changes in the core due to the action of the primary coil.

The only problem is when you interrupt the current when it is non-zero: in that case the collapse of the magnetic flux will cause very high voltages across both of primary and secondary coils and may kill either.

This is the "failure mode" of an inductor that needs to be catered for in some manner, independent of whether there is a second coil coupled with the inductor's magnetic flux.