# Taking the inductor off

say we have Direct current source with a resistance and an inductor. what happens

1. when we open the circuit?

2. when we break the inductor off from the circuit via dielectric stick(something not conductor I mean)

I suspect the answer of the first is, we probably observe a spark since Voltage would be extremely high due to extreme value of the time rate of change of the current. Yet, I couldn't come up with a idea what might be happen in the second case?

• Same thing. Probably a spark. Aug 2, 2021 at 14:47

It looks like your "dielectric stick" is trying to get towards the idea of interrupting the circuit with a perfect insulator. If you have an ideal insulator and an ideal inductor, you have one of those situations with an unstoppable force and an immovable object. The ideal inductor will keep the current the same while the ideal insulator will try to force the current to 0. The laws of electromagnetism simply break down at this point.

In practice, nothing is perfect. Those ideals do not exist. You can arc through air (and even a vacuum, if necessary). There are magnetic couplings between the inductor and other objects. There is a resistance to the material in the inductor. All sorts of things will keep it from ideal.

That being said, the effect you are looking at has a name: flyback. It's a real issue which arises when trying to switch highly inductive loads like motors. The result is almost always an arc across the insulator (typically air), although if you open the switch fast enough, you can just destroy your inductive motor instead! The arc is a problem because it does physical damage to the contacts of the switch. Thus, we take great care to construct circuits that don't arc like that.

Sometimes we do have switches that we let arc, like this glorious 500kV switch opening. But usually we take care not to.

One extreme example of this is in HVDC, which is using very high DC voltages to carry power across countries. These systems, of course, have to interact with our normal AC power grid, so there's conversion stations where they turn AC into DC by simply switching the polarity every time the AC goes from positive to negative. They do this with giant Thyristors banks which are a sight to behold, simply because of what we had to do in order to prevent arcing.

The human on the bottom gives you a great sense of scale. It also gives you a great hint that these banks are currently de-energized. A lot of the time I have to put disclaimers about not risking one's life with electricity... but this is one of those scenes that really provides its own disclaimer!

Now these converters are designed to switch the polarity right as the voltage crosses 0, to avoid flyback like issues. However, I link them to point out just how insistent arcs are, even in specialized materials. The ideal insulator you need for #2 in your question is just not a thing that happens in reality. We work very hard to approximate it.

Since, for an ideal inductor, you can't change its current instantaneously (i.e., in zero time), the current before disconnecting the inductor will continue flow regardless of what interrupts the circuit, be it an open circuit involving air, or by insertion of some insulator (dielectric) as in your second case.

So for the second case there are two possibilities. Either (1) the dielectric survives the high induced emf and permits current to flow per Ohm's law through a very high impedance or (2) the dielectric electrically breaks down creating a low impedance path for current to flow, just as in the case of an open circuit involving the dielectric breakdown of air, with resulting arcing.

Hope this helps.

• Why do you always write "Hope this helps" at the end of your every answer? Aug 2, 2021 at 15:17
• Because I do hope it helps and because it is has been my signature since joining the Exchange. Aug 2, 2021 at 15:34

Although it may not be immediately obvious in effect you have and LCR circuit with the open switch / break acting as a capacitor.

So it boils down to how much resistance, inductance and capacitance there is in the circuit as to "What happens next?"

If the electric field strength across the open switch / break becomes too large the once insulator (air/dielectric) becomes a conductor and the energy stored in the inductor is dissipated as heat in the resistance within the circuit.
This is possibly more likely to happen with the open switch as the dielectric strength of air may well be lower than that of the dielectric inserted in the break.
Also that dielectric will also probably have a reduced the electric field electric field within it due to it being polarised.
The LCR system is most likely over-damped and so the current will continuously fall to zero.

If the air/dielectric does not break down then the energy being lost by the inductor starts to be stored in the capacitor as the potential across the open switch / break increases.
Subsequently the capacitor discharges and there is a current flowing in the circuit which in turn means that the energy which was stored in the capacitor now becomes energy stored in the inductor.
So the current in the circuit oscillates at a frequency determined by the capacitance and inductance value - this is damped (due to the resistance in the circuit) simple harmonic situation.