Coil Thought Experiment

Hi, a coil is like a capacitor or battery when charged correct? So if I took a large coil, ran a large current through it, and quickly disconnected the leads from the power source (assume this is possible) the coil should retain its inductance/capacitance within its large EM field. I could like walk around with it and then attach a lightblub to the leads and the light bulb would light up for a bit.

True or false?

False.

I suppose the math implies that a large enough inductor as a sole component would have current available for use on such human time scales, maybe even a coil smaller than a few prarie provinces - but certainly not a mobile one.

This is true for capacitors (they store energy in an electric field). You can charge one up by applying a voltage across it's terminals. When disconnected, an ideal cap will retain that charge forever. Real life caps however slowly discharge due to leakage currents through the dielectric and through moisture in the air.

In equipment that hasn't been used for a long time, the caps could still be charged and can deliver a lethal shock when touched.

Utterly false.

The energy stored in an inductor is in the form of flux in the magnetic core or in the air if it is an air coil. As soon as the current through the coil is interrupted by opening the path from the current source, the magnetic field starts to collapse generating a voltage in the coil.

If there is a conductive path allowing current to flow through the coil, that current supports the magnetic flux which decreases as the current through the coil decays due to energy dissipated in the resistance of the coil winding and the external conductive path. If the closed circuit through the coil and the external path is superconducting, the magnetic flux will be sustained for a very long time, but as soon as you introduce resistance into the circuit in an effort to extract and use that energy, the current will decay and the field will collapse. For common copper conductors, the magnetic field collapses fairly rapidly due to the finite resistance of the copper.

This phenemenon is noticeable in the slowing down of the release of a solenoid or relay that is protected from arcing by putting a diode across the ends of the coil. In the absence of the diode the device or contacts interrupting the current through the relay coil arcs, dissipating the energy of the collapsing magnetic field and allowing the relay or solenoid to release rapidly. A capacitor accomplishes a similar function by providing a reservoir to absorb the energy of the collapsing field. To the transient energy pulse, the capacitor looks like a short circuit.

If there is no conductive path allowing current to flow through the coil, the voltage generated by the collapsing magnetic field builds up until something blocking current flow fails, thereby allowing curent to flow to dissipate the energy of the collapsing field. The failure can be in the insulation between coil windings or in the air between the end of the coil and the conductor that is breaking the circuit. If it is in the insulation you will usually not be aware of it until enough damage accumulates to cause an obvious problem. If it is in the air, it is often perceptable in the form of a tiny or large (depending upon the size of the inductor) arc between the breaking contacts. Note that this is the principle of the Kettering engine ignition system in which the points interrupt the current and the voltage builds up until the gap of the spark plug breaks down and you get an ignition spark.

awright

Ever unplug a motor while it's running, and see the big spark? That's what happens when you quickly disconnect an inductor.

I used to have trouble with this one too.
For some reason, it's easy to grasp how a capacitor stores charge at a voltage such that you can pluck it out of the circuit and put it in your pocket. And for some reason, it is equally mystifying that an inductor, too, stores energy. I think of it like this: plucking a charged inductor from a circuit and putting it in your pocket is just the same as shorting out a charged capacitor (ie, plugging it into a circuit). Examining the equations that describe the behaviors of capacitors and inductors respectively shows this. In particular, it shows how the two are sort of mirror images of each other.

For instance: i = C dV/dt means that if you change the voltage across a capacitor, you get a current running through it (ie, you're either charging or discharging it). By contrast, V = L di/dt means that if you change the current flowing through an inductor, you get a voltage across the inductor...which opposes the voltage that is driving the current (and that's why people like to say that inductors "oppose changes in current")

If you look at V = L di/dt, and integrate both sides over time, you get int[V dt] = L i, then C dV/dt = (int[V dt] / L) and it becomes apparent that these elements have a mirror-image relationship (the product of capacitance and the derivative of voltage is equal to the quotient of inductance and integral of voltage).

I don't know how meaningful that is, just some musings, is all

-Adam O.

it is true for coils too, except that in the case of a coil, the moment you disconnects it, it will generate a voltage high enough to discharge itself - the reason for arcing.

Ok, thanks for the only valid answer. Its a thought experiment. I know a large super charged coil will arc across the air to make sure it gets itself discharged before someone is able to pull the leads away from the power source.

But if someone did pull the leads away fast enough your saying it would act like a capacitor and simply hold the EM field in a suspended state. So I'm correct.

Can anyone confirm? I mean someone that doesn't dwell on the physical technicality of it. I know its 99.9% impossible to do this.

I think you were given an answer in awright's post, digest it a bit more.

Hamlet

Umm, wrong. Go back to what Awright said.

He explained both extreme situations. One, the coil is a closed loop. Two, the coil is an open loop.

If it's an open loop, (you disconnected the coil really really fast, even infinitely fast) the voltage rises until it is high enough that it is no longer an open loop. That is, the coil will arc between windings or to the two end terminals. Then, it does as he explained in the closed loop scenario.

So, even if you pulled it away, and even if you had awesome insulation between the windings and pulled the end terminals way far away from each other, they would still very soon arc. Once they've arced it eventually it drains all its energy. This is true even if it is a superconductor, because even if the resistance of the coil itself is zero, you've tried to make resistance as high as possible with insulation and separation of the terminals.

The only way it would stay charged is if the coil is both shorted to itself instantly after being instantly removed from the circuit, and is also perfectly superconducting, as explained.

An inductor is a capacitor with V and I reversed.

You charge up a capacitor by applying a current; the voltage increases over time.

You charge up an inductor by applying a voltage; the current increases over time.

To maintain a charge, the capacitor current must be zero (open circuit).

To maintain a charge, the inductor voltage must be zero (short circuited!).

If you had a really, really large inductor, large enough that you had a few seconds of charge to play with, you could charge it with a DC voltage, open the circuit and observe the large, stretchy arc you pull from it, then quickly short it together, maintaining the charge. Then you can pull more charge out by opening the circuit and arcing some more.

This is analogous to a capacitor, where you get sparks when shorting, rather than opening, the circuit.

Tim

to answer your question, yes the bulb would light up a bit...

but only if you can connect the bulb within pico seconds after you disconnect the inductor.

your inductor also has a self resonant frequency, in the case of a tesla coil secondary this can be say 100KHZ*, so common sense would say that if you connect the light bulb within one tenth of a cycle (say 500 nano seconds) most of the energy would end up in the light bulb.
The problem is that the energy stored in the inductor (connected to a steady state dc supply in real life), when converted into voltage (charging its self capacitance) will always be enough to arc over.
so you either have to accept a rather low current, or artificially increase the self capacitance of the inductor.

i don't waste my time with theoretical questions involving inductors that don't have self capacitance.

*this is exceedingly low for real life inductors designed to be inductors btw.

Why not use a battery in the first place?

Not sure if someone else already mentioned this. A coil is the "opposite" of a capacitor. A capacitor needs to remain open (nothing connected to it) for it to hold a charge, a coil needs to be closed (both wires connected) for it to hold a charge. So if you put a current through it, and then quickly disconnected one wire and connected it to the other side, your charge would remain there a little longer. It would dissipate quickly due to resistance in the wire though.

TheComet

define opposite, please.

this would be like saying a negative resistance resistor is the opposite of a resistor, which is not entirely correct.

Opposite in the sense that voltage and current are reversed.

The fundamental equation of a capacitor is:
I = C * dV/dt

The fundamental equation of an inductor is:
V = L * dI/dt

Tim