# Understanding inductance coil.

I know that an inductor opposes a change in current through it. So in the circuit diagram attachment, When the switch is closed, current starts increasing in the circuit and reaches a maximum value. And When in steady state, if the switch is suddenly turned off, The inductor opposes the sudden drop in current, and hence, the current slowly drops to zero.

But if this has to happen, there must be a closed path for the current to flow. When the switch is turned off, the path for the electric current is opened. So why doesn't the current drop to zero immediately? Where is the path for the current to slowly drop to zero?

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• Inductor.PNG
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In the circuit you post. You don't have any closed path for the current when the switch is turned off.
So current drop to zero immediately after the switch was turned off. This immediately change in current will cause a very large voltage to be produced. This produces a very short, very high voltage spike across the coil.
And I highly recommend you to read this pdf from page 22 "Understanding the Inductor".
http://www.elsevierdirect.com/samplechapters/9780750679701/9780750679701.PDF

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jim hardy
Gold Member
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I know that an inductor opposes a change in current through it. So in the circuit diagram attachment....

....there must be a closed path for the current to flow. When the switch is turned off,.....
.. Where is the path for the current to slowly drop to zero?

In real simple terms:

As you correctly stated, the inductor opposes a change in current.
It does so by changing itself from a current sink into a current source. Lenz's law at work.
It reverses its voltage and raises it to whatever is necessary to maintain current , until it has expended the energy of its magnetic field.

The key phrase here is "whatever is necessary".
As your switch contacts begin to move apart, a spark will form between them through which current flows until the inductor has spent its energy down to the point it can no longer maintain the arc voltage.

That is why switch and relay contacts that are made for switching inductive loads are made of hard metal that is resistant to arcing.

The condenser in 1960's automobile ignitions is there to prevent the points from quickly burning away. It also extends the duration of the spark produced by the ignition coil.

That's how a pure ideal inductor behaves.
A real one has limits of how much voltage it can make because of effects like core eddy currents and winding capacitance.

But that's the basic idea.
Now put some of your math to it and you'll have the concept forever.

old jim

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To be precise you should say that the inductor opposes a change in magnetic field (magnetic flux linkage).
When you close the switch the current starts to flow and the magnetic field increases. An emf is induced in the inductor to oppose the increasing field, ie the increasing current. This is known as a back emf and the effect is that the current increases slowly at switch on.
When the switch is opened the current cannot continue to flow, it decreases instantaneously and an emf is produced to try to maintain the magnetic field, ie this emf tries to keep the current flowing.
As explained above this causes a large emf to be generated.

In real simple terms:

As you correctly stated, the inductor opposes a change in current.
It does so by changing itself from a current sink into a current source. Lenz's law at work.
It reverses its voltage and raises it to whatever is necessary to maintain current , until it has expended the energy of its magnetic field.

The key phrase here is "whatever is necessary".
As your switch contacts begin to move apart, a spark will form between them through which current flows until the inductor has spent its energy down to the point it can no longer maintain the arc voltage.

That is why switch and relay contacts that are made for switching inductive loads are made of hard metal that is resistant to arcing.

The condenser in 1960's automobile ignitions is there to prevent the points from quickly burning away. It also extends the duration of the spark produced by the ignition coil.

That's how a pure ideal inductor behaves.
A real one has limits of how much voltage it can make because of effects like core eddy currents and winding capacitance.

But that's the basic idea.
Now put some of your math to it and you'll have the concept forever.

old jim

So to dissipate the energy stored in the inductor (LI2/2), during the interval of the spark being produced, the large emf produced in the inductor will constitute a current which will cause the energy stored to be dissipated through the switch (due to the arc produced) and the winding resistance? Am i on the right path of understanding the concept now?

jim hardy
Gold Member
Dearly Missed
Am i on the right path of understanding the concept now?

Sounds like it to me.

We might wordsmith it a little bit:
the large emf produced in the inductor will [STRIKE]constitute[/STRIKE] cause a current

Tech is right about Lenz's law - induced EMF attempts to produce a current which will oppose the change in magnetic flux.

In a pure inductor the voltage will become whatever is necessary to establish that current.
Note that if there are other coils linking the same flux, current can flow in them to satisfy Lenz.
So a coil that is designed to produce sparks must have a good core that doesn't allow induced currents to flow there instead of in the hv winding.

You can 'feel' this effect with a AA battery and a small transformer. It'll shock your fingers.
Don't try it with a big one though they can be quite painful. Palm size only, like a doorbell transformer.

old jim

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Note that if there are other coils linking the same flux, current can flow in them to satisfy Lenz.
So a coil that is designed to produce sparks must have a good core that doesn't allow induced currents to flow there instead of in the hv winding.

So if i connect a circuit as attached in my post, and if the "inductor" is a plain iron core inductor, when i open the switch, can i expect the LI2/2 energy stored in the inductor to be dissipated through eddy currents, and the main current in the circuit to suddenly drop to zero?

jim hardy
Gold Member
Dearly Missed
So if i connect a circuit as attached in my post, and if the "inductor" is a plain iron core inductor, when i open the switch, can i expect the LI2/2 energy stored in the inductor to be dissipated through eddy currents, and the main current in the circuit to suddenly drop to zero?

No, unless the iron core is a really poor one.
The eddy currents in the iron will dissipate some of the energy and the remainder of the energy will behave as we discussed above in posts 3, 4 and 5.. So the spark will be smaller, maybe too small to notice.
I'm sorry if I confused you.

It is good to get set in your mind how the ideal inductor works, which is as you described earlier,
and be aware of the difference between an ideal and a practical component.
Real inductors are incapable of producing infinite volts like the formula says an ideal one should.

An automobile spark coil will make tens of kilovolts when its current is interrupted
the little doorbell transformer will make only maybe a few hundred volts.
I've measured 2500 volts from AC relay coils about the size of an egg.

So, if you just wrap a lot of turns around a hardware store c-clamp to make an iron core inductor
it will not make as much voltage as one with a better core.
Clearly the core designer will work to keep eddy current(and other) losses to minimum so his inductor's behavior will approach ideal.

In your early studies your classroom work will likely involve ideal components.
I mentioned this only to plant the seed of "non ideal-ness"so that when you hit the real world and find out components are not ideal, it won't be such a surprise.
I worked on some inductors that had a solid rod for an iron core. We measured their inductance at various frequencies. Above 400 hz the iron core was no longer detectable. They were far from ideal.
That had us scratching our heads for a while because there's no frequency term in the textbook equation for inductance.

I was glad to see you questioning 'how it works' at the basic level. You'll do fine.
Magnetics is an interesting sub-field. Check out www.mag-inc.com.

old jim

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Wow!!! Now I will be convinced with whatever "weird" far-from-ideal things that might happen while working with inductors. Now I am confident enough that i will be able to explain the reason for such kind of behavior, when i start working with different type of non-ideal components. Thanks a lot jim!