The power is generated in the coil.
It's a separate, isolated circuit, having no relationship to plus, minus, or ground of the circuit that charged the coil.

Once the switch is opened, the original power supply is out of the picture.

Picture the coil replaced with an independent source, as a battery or generator.

 

The inductance is more like inertia than it is like a generator. The magnetic field did not store electrons, it stored energy. When you try to stop the current instantly, the current tends to continue flowing in the direction it was already flowing but it does not get generated inside the coil. It has to come from the positive supply. The collapsing magnetic field provides the energy but it does not provide the electrons. The energy wants to move some current but it doesn't supply he current. Moving the switch to the positive side of the circuit only changes which end of the current loop suddenly has a gap in it.

I am honestly trying to communicate, regardless of how difficult it seems to sound.

 

I guess my understanding is different.

I have never heard of current flow having inertia.

I have no training so I could be wrong.

I believe the energy is stored magnetically.

The field collapsing would be similar to a permanent magnet inside the coil being removed (near) infinitely quickly.
Generating a pulse.

 

I don't see any conflict or misunderstanding in your statements, inwo. We are merely thinking differently.
In fact, current flow does not have inertia. I only used that to try to explain what an inductor does. Inductance makes current flow persist until the magnetic field has collapsed. I am saying that the behavior of an inductor is analogous to inertia. Much like saying a capacitor is analogous to a coffee can. A capacitor can hold an electric field that will supply current between recharging pulses from a rectifier. In that scenario, you could use, "coffee can" instead of capacitor and maybe get a feel for the purpose of the capacitor.

 

Glad to hear that #12.

Thought a different explanation might help.
Whenever I here the phrase "Goes to grd" I cringe.
All my friends who mostly do automotive, are limited by the idea that all "electricity" seeks ground.

The original auto/electric ,common chassis ground conductor, design has done much to blur understanding of current flow. IMO

 

#12 I get the switch side, too much resistance, not a problem. I also understand induction, mutual induction, self induction and all the good stuff that happens when a wire is passed through a magnetic field and vise-versa. I am an automotive guy who likes to play with electronics and I stay on that side of my business because it is the most interesting and gratifying. I also understand that our term ground, simply means more negative than positive although many in the field really think it is ground. Perhaps I'm just not getting it. I'm just not grasping why it doesn't take the path of least resistance once that voltage is generated. I will have to think about it some more and read through the comments. Perhaps it will just hit me. Thanks for all the good feedback. Sounded like good questions?

 

The fine distinction I am working on is the difference between energy and coulombs. Waving a magnet next to a coil will induce coulombs to flow because energy was inflicted on the existing electrons in the wire. None of the electrons in the magnet went into the coil. If the ends of the coil are connected, electrons will move, and that is called current.

Pushing coulombs through an inductor with a battery will cause a magnetic field to appear around the wires of the coil. Generating this magnetic field requires energy. When you arrive at the point in time when the current is continuous and steady, the magnetic field will just stand there and the inductance will seem to disappear because you can only measure the effect of the ohms in the wire.

12 volts applied to a 12 ohm resistor will cause 1 amp (1 coulomb per second) to flow. 12 volts applied to a Henry of inductance that has 12 ohms of resistance in the wire it is wound with will arrive at 1 amp after the amount of time needed to arrive at steady state conditions. One Henry will allow 1 amp of change in current per second per volt if no resistance is involved. The resistance slows this down, but I don't remember the formula for how much it slows down the rate of increase of current.

The current ramps up until it gets to 1 amp. Now you have a magnetic field just standing there and the current is flowing just like it would if you only had a 12 ohm resistor. The time required for the current to change from zero to 1 amp is when the energy required to create the magnetic field was diverted into making the magnetic field. How many of the electrons are in the magnetic field? Zero. They didn't start flowing out of the battery at 1 amp and some of them never got back to the positive terminal (went missing). They simply didn't start at an amp, they ramped up to one amp as the ENERGY was diverted.

If you add an "ideal" switch that shorts out the battery, the stimulus that was keeping one amp flowing through the 12 ohm coil will stop. In the first instant of the shorted battery, the inductance will create 12 volts to push 1 amp through the 12 ohms of resistance and the current will diminish to zero across time and the energy in the magnetic field is converted to heat in the 12 ohms of resistance of the wire in the coil. How many electrons were provided by the battery during that time? Zero. The battery was shorted out by an ideal switch, so the voltage across the switch is zero and zero voltage means zero current. (The battery will start melting, but it will not contribute any electrons to the flow through the inductor.)

Now, back to steady state with 1 amp flowing: Instead of shorting the battery, we disconnect the battery. When you disconnect the circuit, the electrons from the battery are not available but the magnetic field is still standing there. The collapse of the magnetic field tends to make the coulomb per second continue to flow. If a diode is available, the inductance will create the 12.6 volts that is required to push 1 amp through a diode junction and a 12 ohm coil. One amp will flow in the first instant and the current will ramp down to zero. The battery is disconnected, so where are the electrons coming from? Nowhere. They were already in the wire. They are just moving.

If a diode is not available when you disconnect the battery, the current still has to change from 1 amp to zero during some time period. One Henry will allow 1 amp of change in current per second per volt. At the first instant of time, one amp is going to continue flowing. If the resistance in the circuit was still 12 ohms, the inductance would produce 12 volts to get 1 amp through the resistance. Disconnecting makes the resistance change from 12 ohms to nearly infinity and almost all of that resistance is in the gap between the contacts in the switch. At this point, the inductance will generate nearly infinity volts to get an amp to flow. This is where the air ionizes and lets current flow through the switch, in spite of the fact that is is opening at the same time. Where is that energy coming from? The magnetic field. Where are the electrons coming from? They were already in the wire. They are just moving.

What if you had a magic switch that opened the contacts by 100 feet in zero time at all? Then the current would, "go straight to ground" because ground is less than 100 feet away.

If you used a transistor to open the circuit, the inductance would create enough voltage to get that amp through the transistor, even if it has to ionize the transistor (and they do).

A simple circuit has been described to show how the math works and how much voltage can be generated when you try to instantly stop the current through an inductor.

http://forum.allaboutcircuits.com/blog.php?b=478

 

..............................
I'm just not grasping why it doesn't take the path of least resistance once that voltage is generated.
..................

It does take the path of least resistance. But remember that a complete current path always needs two connections from the source of the voltage, whatever they are.

 

It does take the path of least resistance. But remember that a complete current path always needs two connections from the source of the voltage, whatever they are.

depends on the voltage. Also there are often hidden capacitances (even if only small). The higher the frequency, the more they can conduct.

 

I'm just not grasping why it doesn't take the path of least resistance once that voltage is generated. I will have to think about it some more and read through the comments. Perhaps it will just hit me. Thanks for all the good feedback. Sounded like good questions?

Once you open the switch, no matter if it is to ground or B+, there simple is no path thru b+ to ground.

With the switch is open that is now the path of most resistance.

The path of least resistance is thru the diode.

 

Would this help explain my separate source explanation?

The polarity is different, but the idea of energy being stored either in a coil or capacitor, can still illustrate the circuit path taken by discharge.

If C1 is charged and looking for a place to dump its energy, plugging it in will discharge thru lamp 1. Not go to ground, lamp 2.

This could actually be demonstrated to your students with a large capacitor and a couple leds.

It would have to be explained that this is only a demonstration of circuit paths relation to ground, not as the way energy is stored in a coil.

As it is reversed but analogous.

It may also work with a large coil, and two neon lamps.
But the capacitor method will be slower to discharge and easier to study.

 

Neon diagram...............................................

 

Good answers and I am getting a better sense of how it dissipates but what happens when the switch is located on the B+ side and not the ground side. Again, why does it not head for the ground post of the battery?

I think the thing that is hanging you up is what I and others have said, there has to be a complete circuit for current to flow back to the supply or ground, whatever the case may be.
Just connected to one or the other when switched off will not in itself complete a path.
Max.

 

Max, I think you are right. #12 you have made it very understandable and the others have made comments similar to Max. I think I have forgotten the part where there has to be a complete path. Something so simple and yet I made it so complicated. It's funny how you can go day to day without having to think about something until someone challenges you. I guess that's why I love to teach. There are those who dare to challenge and that makes my day all worthwhile. You guys have been a great help and I will pay it forward. There is hope for our younger generation. For what it's worth, I spoke to several of my colleagues and none of them could answer this either. I like to think I'm pretty good with DC electrical but I just got humbled. Very cool.

 

This simulation might help understanding:

I model two standard automotive relays (12V coil, 85Ω coil resistance, 100mH coil inductance) with both high-side and low-side switching. For the simulation, the relay coils were energized for a long time and then the switch contacts open, de-energizing the coils at time=1ms.

Note that the coil currents I(L1) and I(L2) and the diode currents I(D1) and I(D2) are the same in both cases. The only difference is the voltage across the coil. Note the current that flows in the coils just before 1ms, and just after 1ms. Obviously, after the dust settles, the coil current tends toward zero after a few ms.

Notice when the currents in diodes flow. Notice that the diode current is exactly the same as the coil current after the switch contacts open. This has to be the case by Kirchoff's current law.

Now look at the voltages at the relay coil V(a) and V(b). In the high-side switched example, V(a) goes one-diode drop (~0.65V) below 0V (ground) while the coil current continues to flow through the diode, finally settling to 0V because the other end of the relay coil is connected to 0V.

In the low-side switched example, V(b) goes one-diode drop (~0.65V) above the +12V rail while the coil current continues to flow through the diode, finally settling to 12V because the other end of the relay coil is connected to the +12V rail.

There is Zero current in-to/out-of the 12V battery after the switches open! Note that the primary reason for adding the snubber diodes to the relay coils is to constrain the voltage across the opening switch so as to eliminate the potential (pun) for creating an arc and possibly burning or welding the switch contacts.

 

There is hope for our younger generation.

Thank Dog! I have been pessimistic about that for several years.

 

It does take the path of least resistance. But remember that a complete current path always needs two connections from the source of the voltage, whatever they are.

depends on the voltage. .......................

What does voltage have to do with the current going through the path of least resistance?