# Simple concept question: inductors

1. Jul 17, 2014

### ffp

How does an inductor works? I know that it stores energy in form of magnectic fields, and all the effects it does to a circuit, but i don't know how it really, physically works.

The electrons go through the coils and induce a magnectic field. Why it is stored? Why doesn't the magnectic field just vanish when there is no more current? Why does it works like a current source when there is no more current going into it?

2. Jul 17, 2014

### Averagesupernova

The field does vanish, or more correctly stated, "collapses" when there is no more current. The collapse is what causes the current to continue. The inductor does not store energy in the same way a capacitor does as when the source is removed on an inductor you will get the energy back immediately. The source can be removed on a capacitor and it will store the energy for a period of time after that.

3. Jul 17, 2014

### jim hardy

4. Jul 17, 2014

### ffp

So the inductor doesn't really store the energy, unlike a capacitor that keeps the energy even when disconnected from the circuit, the inductor just induces a current when the source is first connected (back EMF) and again when it is disconnected, right?

5. Jul 17, 2014

### Averagesupernova

Technically that is stored energy isn't it? If a person looks at it in another way you could say that you can store the energy in the magnetic field indefinitely. Super-cooled super-conducting coils can have current passing through them indefinitely when the are short circuited. This does not work with regular conductors since there are resistive losses in the conductor. As soon as the coil is un-short circuited the voltage builds across the coil and the field starts to fall.

6. Jul 17, 2014

### MrSparkle

it is storing energy, but it cannot retain it. A capacitor 'stores' voltage, but you don't need a closed circuit to have a voltage difference. Ergo it can hold the energy for a period of time, until the internal leakage slowly releases it. An inductor stores current, but you have to have a current flow, so it immediately releases energy to try to restore the old current. If there is an open circuit, there isn't a possibility for current flow, so the stored energy responds by increasing voltage until current does flow (a spark somewhere in the circuit).

7. Jul 17, 2014

### Okefenokee

Have you ever heard of the "electromechanical analogies"? These are analogies that equate electrical components to mechanical components. They are useful for understanding the behavior of circuits in a way that can be visualized. It's more than that however. Any circuit has a mechanical equivalent and a machine will have a circuit equivalent. You may have heard of analog computers. You can use the analogies to simulate nearly any physical system with a electronic circuit.

Inductance is the equivalent mass and current is the equivalent of motion. When current flows in a circuit with inductance it will have momentum. So you can see that inductance will oppose changes in current because it behaves like mass which has inertia and opposes changes in motion.

That analogy doesn't explain what inductance is. It merely gives you a way to understand how it behaves which is what you seem to be after. The math works out the same if you treat inductance like mass.

Every real circuit will have some inductance. It's unavoidable because current always create a magnetic field which stores some energy in space. For plain wires that field is spread around thinly so it doesn't store much energy. An inductor will have a tightly wound coil which creates a significant field inside. Basically, inductance is a property of circuits that converts the flow of current to magnetic fields.

8. Jul 17, 2014

### sophiecentaur

I think it would be better to say that Inductance is a property of a component that produces an emf which opposes an attempt to change the current flowing through it. (This 'back emf' is produced - induced - as a result of a changing magnetic field around an inductor.) The magnetic field around the inductor corresponds to the energy that was used in order to produce it - as has been written above.

9. Jul 17, 2014

### ffp

Thank you very much for all the replies!!

Another question then: If I charge a capacitor and then open a switch, the capacitor will store that energy in its electric field unless the voltage is great enough to break air resistance and create a spark. Would the same happen with an inductor?
What would happen if the voltage isn't great enough to break the air resistance? Where does the energy stored in the inductor go? I believe it will manifest in the form of voltage across the terminals of the switch, trying to build the current that the inductor would in the case of a closed circuit. In this case it works the same way a capacitor does?

10. Jul 17, 2014

### Okefenokee

Ahh, but the units of inductance are Henries which corresponds to Webers per Amp. If a circuit with 1 Henry experiences 1 Amp then there will be 1 unit of magnetic flux (measured in Webers) somewhere in the physical space of the circuit. That's steady state.

The Weber is equivalent to Volts times seconds so if the flux changes by 1 weber per second then there will be 1 Volt of EMF somewhere in the circuit. That's what you are describing. That's time dependant.

We're acting sort of like the parable of the blind men trying to describe the elephant. You guys are describing different aspects of inductance so I offered some more alternatives hoping to make a more complete picture. No one is incorrect.

11. Jul 17, 2014

### MrSparkle

yes.
yes.
The voltage increases as the field collapses until the voltage is enough to spark. it doesn't work the same as a capacitor because caps don't increase voltage. You can see this in action when you unplug something with an inductive load like a cheap vacuum cleaner. Often you see a small spark at the outlet.

Also, DC-DC converters are built around this principle. Current is pulsed through an inductor using a hi-power transistor. During the on time, the magnetic field charges up. During the off-time, the field collapses which raises the output voltage.

Last edited: Jul 17, 2014
12. Jul 17, 2014

### ffp

Thanks again!!
So the only difference between inductor and capacitor, in the case I posted, is that the capacitor have a fixed voltage(equals to the source that charged it), while the inductor would increase the voltage as the field inside it is changing and then stop.
Considering there is not enough voltage to create a spark, the energy of the inductor would be dissipated in form of voltage on switch terminals, right?
Also,considering a capacitor and an inductor (2 different circuitry) that have the same voltage on them, the inductor would discharge faster when the switch is open, since it only lasts as long as the field fades and it does even when there is an open circuit, while the capacitor would store its energy and discharge very slowly due to leakage in an open circuit, am i right?

13. Jul 18, 2014

### sophiecentaur

I think you are seeing a parallel here where there is not one - apart from the mention of Volts.The voltage across a Capacitor can be static in a static situation. The voltage across an Inductor would need the current to be increasing at a steady rate - so definitely not a static situation. You might as well say that a resistor is like a Capacitor because, if you pass the appropriate current through it, there will be a similar voltage across it.
To sum up:
Capacitor Volts: depends of Charge.
Resistor Volts: depends on rate of charge flow (current).
Inductor Volts: depends on rate of change of current.

14. Jul 18, 2014

### ffp

I think I'm not being very clear, sorry about my english. I'll try again with questions:

1- If an inductor is charged and after a while one of its terminals is open, where does the energy "stored" in the inductor go? What will be the voltage across the open terminal? Consider that that is not enough voltage to create a spark.

2- An inductor is charged with a voltage source and then the circuit is open. The same is done with a capacitor, with a source of same voltage. Which one would discharge first?

@sophiecentaur:
Doesn't the inductor just need an increase in current (don't need to be steady) Vl=L di/dt. Any variation of currrent will create a voltage across the inductor.

15. Jul 18, 2014

### jim hardy

into whatever is between those terminals.

Whatever is necessary to maintain the current.

16. Jul 18, 2014

### sophiecentaur

The word "charged" does not apply to an inductor. In the context of electricity, the word 'charged' has a specific meaning.
I think you may mean that a current is set up and circulates for ever in the absence of any resistance.
This is 'ideal' and you must continue with an ideal model. Disconnecting (breaking) the circuit must be instantaneous and complete. Ldi/dt will be infinite so the volts developed will be infinite. This is not a real scenario. The nearest to it is an inductor with finite resistance somewhere in the circuit (and a top-up current from a supply) and a switch that takes a finite time to break the circuit. As the current falls to zero, there will be a (possibly large) emf induced. This is an everyday situation and can be dealt with - as mentioned previously.

Perhaps you should use one of the free circuit simulators that are available (hate them, myself) on the net and see what happens under the conditions you are contem[plating.

If you want a steady emf from your experiment then the rate of change of current must be constant. That's all I meant.

17. Jul 18, 2014

### sophiecentaur

There will be a finite Capacitance between the switch contacts. You have a parallel tuned circuit, in effect. Your 'missing energy' will be stored in the charge difference across the switch contacts. Once the L current has dropped to zero (volts at a maximum) the situation will reverse and the capacitor will discharge back through the coil. You will have a free oscillation. There is another twist to this. The oscillating currents in this circuit will actually radiate energy in the form of an electromagnetic wave so, even if you don't include an explicit Resistive element, there will still be a radiation resistance to cause the energy to leak away and the oscillations will die down.

You could work out order of magnitude of the frequency of this oscillation if you think of a, say 10mH inductor and a parasitic Capacitance of, say, 10pF.

18. Jul 18, 2014

### jim hardy

That thought entered my mind and i considered saying it, but that necessitates either ideal or superconducting wire

That's the scenario i had in mind.

quite agreed.
ffp needs to grasp the concept of an ideal inductor's behavior then tone his thought experiment down to realty;; at least that's how i learn basics.

You and i have both 've had our fingers on automobile points i know, so our understanding of inductance is firmly ingrained to the point it's intuitive now. We can barely remember when it wasn't.

ffp is struggling up his learning curve and sadly probably will not ever encounter a battery and coil or magneto ignition , which places the concept so viscerally at your fingertips

so I was searching for a terse phrase that conveys the concept.

if an ideal inductor is interrupted by an ideal switch
the energy must go into the capacitance across the contacts of that switch
and the voltage on that capacitance will rise until it's absorbed all the energy
hence my phrase "to whatever is necessary.."

hence also my phrase "whatever is between those [STRIKE]contacts[/STRIKE] terminals" - even if it's just free space (which has capacitance)

Ahhh i see you've already posted whilst i was typing
and you expressed my points more succinctly

sorry if i butted-in, just wanted to nudge the discussion a little .

regards as always - old jim

Last edited: Jul 18, 2014
19. Jul 18, 2014

### sophiecentaur

Cross posting is a sign of an interesting thread!
regards back to you my boy.

20. Jul 18, 2014

### ffp

Got it, the open terminals would behave like a capacitor. If there's enough voltage across them we got a spark, otherwise the energy oscilates back to inductor and we got kind of a LC oscilator.
Yes I'm considering everyting ideal. Didn't know that an open terminal could work as a capacitor (even on ideal circuits). That's why i was so curious.
Thanks!

21. Jul 19, 2014

### sophiecentaur

Glad the 'penny dropped'. Gives you a good feeling dunnit?
There is a more common version of the 'where did the energy go?' conundrum. This is often given to A Level students who are starting to feel too damned confident about their knowledge.
Take a Capacitor value C and charge it to Voltage V. Disconnect the supply.
The stored charge is CV and the Stored Energy is now CV2/2

Connect an identical capacitor in parallel. The total Capacity is now 2C and the charge is the same as you started with so the new voltage is V/2

The Energy stored is now CV2/4
Half of the energy has 'gone missing'. Explain: Dissipated in any resistance in the circuit or radiated due to the implied Inductance of the loop of connecting wire.