inductor back-EMF and open circuits

[csdave]

Hi all,
I am trying to understand inductors better and I have a question that I can't get an answer for.

Assume I have a circuit with a generator, an ideal switch, an inductor and a resistor all in series (nothing in parallel with the inductor).

I turn on the switch and the current increases exponentially as the voltage across the inductor decreases from the generator's voltage to 0.

I leave the circuit on for a while,

then I open the switch. Current would want to stop, but it continues for a while because of the inductor, right? This generates an EMF (opposite to the one we had when closing the switch) across the inductor.

Now, what happens to this EMF if the switch remains open (and it's ideal, infinite R)? Does this EMF grow indefinitely? Does it generate a current across the inductor?

I thought of this, because I was thinking of capacitors and of the fact that if you remove a charged capacitor from a circuit, it will remain charged (ideally forever, in practice until the charge escape through parasitic resistances). Is there an equivalent phenomenon for an inductor? Or is an inductor without a generator simply a piece of wire?

Thanks to anyone who can answer.

 

[KlausST]

Hi,

the inductor has limited magnetic capacity to store the energy.
If you open an ideal switch, the voltage becomes very high. Usually you will see a spark at the just opening switch contacts. the most energy is dissipated with the spark.
And in very short time the inductor is "empty".

But you said it is an ideal switch. Maybe you say that the swich is opened so fast, that no spark can happen. For sure the voltage becomes very high. The voltage rise rate is limited by the inductors stray capacitance and the core material characteristics. And the voltage is limited by the isolation. Maybe inside the inductor a spark happens, maybe the air or other material gets ionized... what else? i don´tknow. But the energy is lost within us.

If you want to keep the energy in the inductor for a longer time you have to short circuit the inductor. (This is how superdonducting magnets work.)
In a real inductor the power is dissipated because of heating by the current in the wires. Soalso here the power is lost witin a few ms, maybe a few 10ms.

Hope this helps.

klaus

 

[D.A.(Tony)Stewart]

An ideal generator will be constant voltage and the current in inductor will rise LINEARLY until it saturates magnetically ( B Field => limit in Tesla's) then the current rises much faster as you noted until the inductance is almost zero for some times or the generator current becomes limited by the series resistance and thus the voltage begins to decay to Vmin=Imax*R(L).

The voltage increases as store more energy in the inductor (time but before it saturates and reduces its L)
The voltage increases 10x as you reduce the switch capacitance from 1pF to 0.1 pF ( smaller faster moving contacts with a big gap. )
the voltage increases as you switch the inductor faster at rate, 1/t ( more force, velocity )

When the switch opens it must arc , but there are ways to reduce the arc voltage with snubbers, MOV, SF6 gas etc.

For an exercise prove to yourself what is the threshold where there is no arc using short durations of charging the inductor and do experiments like Faraday did in a lab.

Use an old microwave oven transformer and a LiPo cell of 3.7V.. This can generate lethal arcs up to 50kV using a very fast knife switch with a big gap after a few seconds of charging... if you happen to be in the path of the current or touching an ungrounded switch .... but with proper understanding, you can learn to make it safe.

Full explanation of arcing is beyond scope of this question.