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A current through an inductor cannot change abruptly, so what happens if I have an inductor with current passing through, and I suddenly open circuit it so that no current flows through?

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    $\begingroup$ You get an arc (hence the diodes protecting solenoids). $\endgroup$ – Jon Custer Oct 21 '16 at 17:16
  • $\begingroup$ I believe the circuit technically explodes in this case. $\endgroup$ – Lelouch Oct 21 '16 at 19:05
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So this physical fact is actually used in the engineering world all the time. I would suggest looking up buck-boost converters if you want to see how. Basically if you have a circuit that switches on and off (abruptly, I might add), an inductor will "smooth" the current. This happens because:

$$ v(t) = L\frac{di(t)}{dt} $$

Which simply implies that the slope of the current cannot be infinite (otherwise your circuit will have an infinite energy per coulomb, which is physically impossible). This physical restriction provides continuous current transitions on the circuit.

Side note: the application here for buck-boost converters is to utilize a DC voltage switching node, with pulse width modulation (PWM) and an inductor to generate a DC output voltage of variable magnitude.

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    $\begingroup$ It might be nice to provide the practical implications of the math. ie fast switching leads to voltage spike maybe leads to an arc. Also noting how the built up charge dissipates could be helpful. $\endgroup$ – user118047 Oct 21 '16 at 21:34

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