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I've been reading about self-inductance in circuits in M. Purcell's Electricity and Magnetism and came across an interesting fact:

"What happens if we open the switch after the current $I0$ has been established, thus forcing the current to drop abruptly to zero? That would make the term $L dI/dt$ negatively infinite! The catastrophe can be more than mathematical. People have been killed opening switches in highly inductive circuits. What happens generally is that a very high induced voltage causes a spark or arc across the open switch contacts, so that the current continues after all."

What do they mean exactly by highly inductive circuits? If I have a small circuit with a coil as an inductor, a battery with 2.3V and some resistance R, and if I open the switch after a long time, will there be an electric arc? Or this applies only to circuits with really high voltages and currents?

I mean, the mathematical equations still apply here and $dI/dt$ will be very negative when the switch is opened.

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  • $\begingroup$ The energy in an inductor is $W=LI^2/2$. If you make L=1H and I=10A, which is neither a very large technical inductor nor a very large current, then you are storing 50J of energy in it. That's plenty to kill a person. $\endgroup$ – CuriousOne Jun 25 '15 at 20:41
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As you slide the switch open, it doesn't instantly transition from zero resistance to infinite resistance. As the contact area decreases the resistance rises, which acts upon the current to produce a voltage operating against the current. Even when the contact resistance becomes zero, there is capacitance across the gap, which produces a rapidly rising voltage, again operating against the current. If this voltage rises fast and high enough (which depends on the current and induction) then you will get a spark.

You can get a spark from a 1.5V alkaline D cell if you connect and disconnect a reasonable inductor to it.

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