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In the above figure, the circuit is closed for a very long time and is then opened. It is stated that "the emf across the inductor just after opening will be very high so as to oppose the change of current". I am aware of the mathematical expression of an inductor during charging and discharging, is there any similar expression to justify this behaviour?

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  • $\begingroup$ Are you aware of this? $$v(t)=L\frac{di(t)}{dt}$$ $\endgroup$ – Bob D May 7 at 10:29
  • $\begingroup$ @BobD . Yes I am, so you mean to say that since rate of change of current is high, due to the equation, the voltage across the inductor will also be high, Thanks. $\endgroup$ – Vaishakh Sreekanth Menon May 7 at 11:05
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    $\begingroup$ Exactly. An attempt to change current in zero time theoretically results in an infinitely high voltage. In reality it doesn’t because opening an air gap switch results in an arc ionizing the air allowing the current to continue. The breakdown of air is evidence that thousands of volts were generated. Hope this helps. $\endgroup$ – Bob D May 7 at 11:49
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The appearance of high emf is due to current changing with high rate. This in turn is due to fact that conducting path was interrupted so current can't go on unchanged. It charges up the surface of the wires, especially at S, and the resulting voltage may or may not be high enough to electrically breakdown the air in between. If it is, a spark will occur. In any case, the current will be decreasing and so emf has to appear in the coil.

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