# Opening switch on inductor circuit in absolute vacuum

1. Jun 21, 2011

### osnarf

So, a professor in class a about a year ago mentioned that opening a switch in a circuit that an inductor is providing power to would cause a breakdown in the air and a spark would jump across it because the current can't change instantaneously to zero in an inductor. Something just made me think about that. But anyways, what would happen if it was in a perfect absolute vacuum?

2. Jun 21, 2011

### phinds

Very likely it would have little effect since the dielectric effect of a tiny amount of air as the circuit opened would only make it a tiny bit easier for the spark to fly across the gap than it would do it without the air there.

3. Jun 21, 2011

### RedX

I thought it was the opposite, that having an inductor prevents an instantaneous change of current through the circuit which would be disastrous, since an instantaneous change in current would create an infinite voltage which would lead to sparks.

4. Jun 21, 2011

### phinds

You have it right but seem to misunderstand the effect. The inductance in the circuit prevents an instantaneous change in current so what happens, as the OP described, is that a spark occurs when the circuit opens because the voltage goes WAY up in order to provide a current across the gap for a VERY brief time so that the current can go to zero quickly but NOT instantaneously. That is, the current CAN'T change instantaneously, and the voltage in the circuit is NOT going to be enough to drive a spark across a gap, so the inductive effect causes the voltage to rise dramatically very quickly so that the current can decrease very quickly. It doesn't take infinite voltage to create a gap spark.

5. Jun 21, 2011

### RedX

I think I see what you mean, but in an LR circuit, upon disconnection, the current goes as:

$$I(t)=I_0e^{-Rt/L}$$

so that the voltage across the gap with high resistance R is:

$$V(t)=-L*I'(t)=RI_0e^{-Rt/L}$$

so it seems that this effect doesn't depend on the value of L at all, just the value of the gap resistance R. The maximum voltage drop across the gap will always be $$RI_0$$ at time zero, regardless if you have an inductor or just have the self-inductance of the wires.

6. Jun 21, 2011

### 256bits

By opening the switch the gap width is not constant.
The value of resistance across the gap is thus not a constant value of R, but is increasing from 0 at the time the switch is thrown to a greater and greater resistance.
Subsequently, the voltage across the gap increases to a higher and higher value.

From your second equation V(t)=−L∗I′(t)=RI0e−Rt/L , with the R being variable and increasing, the voltage itself increases alongside the gap becoming wider if an inductance is included in the circuit..

7. Jun 21, 2011

### RedX

That certainly makes sense, and is very perceptive. If you include a time-varying gap resistance, then the equation for the voltage drop across the gap:

$$V(t)=I_0 R(t) e^{-(1/L) \int R(t) dt}$$
If R(t) is increasing with time, then a huge inductance L will allow the factor of R(t) in front to increase with time without much drop in the exponential as time passes, so the voltage attains higher value with increasing L.

That is strange though as I thought inductors were good things and prevented sparks, but were bad for things like signal processing as they smooth out waveforms.

8. Jun 26, 2011

### Gfellow

A slight segue here. An absolute vacuum? Really? Really? Do you mean a partial vacuum? Absolute vacuums are the realm of theoretical physics. Example: Present definition of a vacuum is "http://www.thefreedictionary.com/vacuum" [Broken]" Since mass/energy, space and time all seem to be intertwined, perhaps a better definition might be that an absolute vacuum is "A volume devoid of space." (The volume only being defined and described by space surrounding the absence.)
This is frontier stuff, mostly dealt with in far-fetched scientific papers and http://www.youtube.com/watch?v=2fdhyhPu6PY".

Last edited by a moderator: May 5, 2017