Disconnecting a live inductor?

I am sorry, if the question seem stupid, I am new to this: What happens if you have a circuit with an inductor (that is saturated and passing current in one direction acting as a wire) and then you open switches on both sides and disconnect the inductor?

According to theory current can't change instantaneously though the inductor - so it can't go to zero - but if you have physically severed any conductor routes where it can flow though, what happens?

Also another question: suppose you have a voltage source connected to an inductor, but on one side there is a wire loop after the inductor is saturated a swithis opened in the circuit and the inductor is connected only to the loop.

Would the current start circulating in the closed loop then?

look up "inductive kick" there is a large spike of reverse voltage generated as the magnetic field colapses, then a little ringing, then nothing. you cant store energy for very long in an inductor. not like a capacitor or battery.

In your first question, you can't open both switches truly simultaneously from the point of view of the inductor. Even 1 nanosecond difference changes how the inductor interacts with the circuit it was connected to. But if you could isolate both ends at the same instant, the voltage across the inductor would increase rapidly (inductive kick) and then drop to zero, The energy stored in the inductor would be radiated as electric and magnetic fields, and dissipated as heat in the copper.

ak

Sylvia said:

I am sorry, if the question seem stupid, I am new to this: What happens if you have a circuit with an inductor (that is saturated and passing current in one direction acting as a wire) and then you open switches on both sides and disconnect the inductor?

According to theory current can't change instantaneously though the inductor - so it can't go to zero - but if you have physically severed any conductor routes where it can flow though, what happens?

Also another question: suppose you have a voltage source connected to an inductor, but on one side there is a wire loop after the inductor is saturated a swithis opened in the circuit and the inductor is connected only to the loop.

Would the current start circulating in the closed loop then?

Click to expand...

You can deduce a fair bit by simply remembering that a current carrying conductor has lines of magnetic flux around it - when you interrupt that current, those lines of flux collapse and induce an emf in the conductor.

There's a Tyco relay appnote (possibly on the element 14 website) that describes in some detail the nature of recirculating current when you use a catch diode to clamp the back emf from the coil, and gives some detail about how the collapsing lines of flux give back energy.

Sylvia said:

According to theory current can't change instantaneously though the inductor - so it can't go to zero - but if you have physically severed any conductor routes where it can flow though, what happens?

Click to expand...

It's kinda like the irresistible force meeting the immoveable object: it depends on your definitions.

But the physics dictate the voltage across is proportional to the rate of change of the current thru an inductor. So an infinite change in current leads to an infinite voltage.

In the real world you can't quite change the current instantaneously, and a voltage getting large enough WILL find some way to short.

(Sure you can cut both sides together, just put the wires next to each other.)

Sylvia said:

Also another question: suppose you have a voltage source connected to an inductor, but on one side there is a wire loop after the inductor is saturated a swithis opened in the circuit and the inductor is connected only to the loop.

Would the current start circulating in the closed loop then?

Click to expand...

Yes it would. The initial value of current is exactly the same as when you started. Something very similar happens when you put a "catch diode" across an inductor as is commonly done for a relay. The current "switched" from the switch (or transistor) to the diode, then the current decays.

Sylvia said:

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What happens if you have a circuit with an inductor (that is saturated and passing current in one direction acting as a wire) and then you open switches on both sides and disconnect the inductor?

According to theory current can't change instantaneously though the inductor - so it can't go to zero - but if you have physically severed any conductor routes where it can flow though, what happens?
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Click to expand...

Assuming you could instantly and completely break the circuit, then the current will keep flowing until the stray inductor capacitance is charged to a voltage where the capacitor stored energy of 1/2 CV\(^{2}\) equals the initial inductor energy of 1/2 I\(^{2}\)R. If this high voltage doesn't cause an insulator breakdown and arcing then the tank circuit consisting of the inductor inductance and stray capacitance will oscillate from this energy until all the energy is dissipated by electromagnetic radiation and IR losses in the inductor series resistance.

you dont have to short the coil for current to flow when disconnected, the coil its self is conductive.

crutschow said:

Assuming you could instantly and completely break the circuit, then the current will keep flowing until the stray inductor capacitance is charged to a voltage where the capacitor stored energy of 1/2 CV\(^{2}\) equals the initial inductor energy of 1/2 I\(^{2}\)R. If this high voltage doesn't cause an insulator breakdown and arcing then the tank circuit consisting of the inductor inductance and stray capacitance will oscillate from this energy until all the energy is dissipated by electromagnetic radiation and IR losses in the inductor series resistance.

Click to expand...

Well said, but I'd like to add that in most real world cases it is not an aircored inductor but has some other metal, like a ferrite core or relay's steel armature. In most cases the big power loss to the released inductor energy is by the eddy current (iron losses) in the other metals, not the copper losses which are small in comparison.

I've scoped a lot of relays on coil turnoff and the iron losses from a simple steel armature are large enough that the peak back EMF is only about 100 to 150v on a 12v relay coil.

alfacliff said:

you dont have to short the coil for current to flow when disconnected, the coil its self is conductive.

Click to expand...

Only if you connect one end of the coil to the other end.

crutschow said:

Assuming you could instantly and completely break the circuit, then the current will keep flowing until the stray inductor capacitance is charged to a voltage where the capacitor stored energy of 1/2 CV\(^{2}\) equals the initial inductor energy of 1/2 I\(^{2}\)R. If this high voltage doesn't cause an insulator breakdown and arcing then the tank circuit consisting of the inductor inductance and stray capacitance will oscillate from this energy until all the energy is dissipated by electromagnetic radiation and IR losses in the inductor series resistance.

Click to expand...

I guess you meant

\(\text{\frac{1}{2}I^{2}L}\)

rather than

\(\text{\frac{1}{2}I^{2}R}\)

t_n_k said:

I guess you meant

\(\text{\frac{1}{2}I^{2}L}\)

rather than

\(\text{\frac{1}{2}I^{2}R}\)

Click to expand...

I certainly did.

why would you have to connect the ends of a coil for current to flow? coils are made of cinductive wire. and the colapse of the magnetic field induces a reverse voltage across the coil, which then flows back through the coil, and causes another reversal. thats what causes the ringing of the coil, with the voltage decreasing each time due to the winding resistance and core losses.

alfacliff said:

why would you have to connect the ends of a coil for current to flow? coils are made of cinductive wire. and the colapse of the magnetic field induces a reverse voltage across the coil, which then flows back through the coil, and causes another reversal. thats what causes the ringing of the coil, with the voltage decreasing each time due to the winding resistance and core losses.

Click to expand...

That ringing current flows between the inductance and stray capacitance of the coil. You can't have any current flow in an ideal inductor with no stray capacitance if the winding is open.

not in the entire coil if the coil is opened, but in each half, as the magnetic field colapses. just as in the entire coil if the circuit is opened. there has to be current flow if the satuating current is interupted. coils conduct.

alfacliff said:

not in the entire coil if the coil is opened, but in each half, as the magnetic field colapses. just as in the entire coil if the circuit is opened. there has to be current flow if the satuating current is interupted. coils conduct.

Click to expand...

That doesn't make sense to me. There aren't two halves to an inductor, it's just one contiguous inductor. Anything that happens, happens in the complete inductor at the same time (subject to the speed of light). An ideal inductor can have no current flow if it is open circuit and there is no stray capacitance. If you try to open such an inductor while current is flowing an arc will form across the coil contacts until the inductive energy is dissipated.

Sylvia said:

I am sorry, if the question seem stupid, I am new to this: What happens if you have a circuit with an inductor (that is saturated and passing current in one direction acting as a wire) and then you open switches on both sides and disconnect the inductor?

According to theory current can't change instantaneously though the inductor - so it can't go to zero - but if you have physically severed any conductor routes where it can flow though, what happens?

Also another question: suppose you have a voltage source connected to an inductor, but on one side there is a wire loop after the inductor is saturated a swithis opened in the circuit and the inductor is connected only to the loop.

Would the current start circulating in the closed loop then?

Click to expand...

There is a very good representation of what happens when an inductor is disconnected from its source in Allied's Electronics Data Handbook, pub. 1962. pgs 18 & 19. I still use this book today as these principles never go out of date. When your inductor is suddenly disconnected from its source, the inductor now becomes the source and it wants to keep the current flowing in the same direction it was going; ie the current does not reverse! You can see this visually by connecting a scope across the inductor. Note; a free wheeling diode should be placed across the inductor to avoid the inductive kick and subsequent high voltage (arc).

Cheers, DPW [Everything has limitations...and I hate limitations.]

Sylvia said:

I am sorry, if the question seem stupid, I am new to this: What happens if you have a circuit with an inductor (that is saturated and passing current in one direction acting as a wire) and then you open switches on both sides and disconnect the inductor?

According to theory current can't change instantaneously though the inductor - so it can't go to zero - but if you have physically severed any conductor routes where it can flow though, what happens?

Also another question: suppose you have a voltage source connected to an inductor, but on one side there is a wire loop after the inductor is saturated a swithis opened in the circuit and the inductor is connected only to the loop.

Would the current start circulating in the closed loop then?

Click to expand...

Its interesting to note that the little mopeds usually used what is called "energy transfer" ignition - to produce the spark, the points close at the point where the rotor magnets are giving the coils fastest changing flux.

This is the exact opposite to the more usual Kettering ignition where the points open to break the coil current and produce a large back emf.

If you short a coil that's surrounded by changing flux - the energy has to go somewhere.

ian field said:

Its interesting to note that the little mopeds usually used what is called "energy transfer" ignition - to produce the spark, the points close at the point where the rotor magnets are giving the coils fastest changing flux.

This is the exact opposite to the more usual Kettering ignition where the points open to break the coil current and produce a large back emf.
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Click to expand...

An "energy transfer" ignition is a type of magneto and magnetos are still commonly used in many small engines, such as in lawn mowers and chain saws, that don't have a battery.

They do not operate the opposite of a Kettering ignition. The main difference is that the primary current is generated by moving a magnet relative to a coil, instead of by a battery, but the primary current, in both cases, is interrupted by the points to generate the back EMF spark.

Agreed, I have a 500cc single cylinder motorcycle (batteryless ignition) that opens the points at the time when the 1 phase coil on the crankshaft reaches max current. The spark still occurs at points-break.

crutschow said:

An "energy transfer" ignition is a type of magneto and magnetos are still commonly used in many small engines, such as in lawn mowers and chain saws, that don't have a battery.

They do not operate the opposite of a Kettering ignition. The main difference is that the primary current is generated by moving a magnet relative to a coil, instead of by a battery, but the primary current, in both cases, is interrupted by the points to generate the back EMF spark.

Click to expand...

I've certainly encountered a moped ignition with the points closing on the timing mark - at first I thought it was a miss-print in the workshop manual, the engine was running (after a fashion) and I verified when and where the points opened and closed before wading in with a screwdriver.