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No, it’s the input that’s disconnected



Irrelevant. If it's not an open circuit, then the inductor is connected to things in parallel, and the impedance increase creates a voltage spike. If the load impedance is significantly lower than the thing being disconnected, then you're just disconnecting something that doesn't matter to the circuit and it's silly to be that pedantic about an irrelevant situation. You're bending the statement from "disconnecting an inductor" to "disconnecting something from an inductor (while something else is still connected)"


A voltage spike between what two points in the circuit?




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