What Happens When an Inductor is Open Circuited?

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When an inductor is open-circuited, it generates a spark due to the sudden interruption of current, releasing energy equivalent to what was stored in the inductor. This occurs because an inductor cannot maintain current flow in an open circuit, leading to a rapid discharge of energy. In the case of a superconductor inductor, while it may initially store energy without dissipating it as heat, the same principle applies: it will also spark if opened without clamping. The discussion clarifies that the assumption of charging an inductor with voltage is incorrect; it is charged with current. Overall, the behavior of both types of inductors during an open circuit is fundamentally similar, emphasizing the importance of proper circuit design to manage energy release.
12Element
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Hello everybody,
First of all I want to thank you all for this great forum.

My question is if a superconductor inductor is charged with a certain voltage and then it is
open circuited what will happen.
My very basic understanding tells me that if the inductor is not a superconductor (it has
intrinsic resistance) the induced EMF will first accumulate opposite charges at each end
of the inductor and then the charges will move from the higher potential end to the lower
potential end in a decaying manner (a simple RL circuit) and all the energy will be dissipated
as heat due to the inductor intrinsic resistance, am I right?

And for a superconductor inductor, the same will take place however, the charges will remain
at each end (acting like a capacitor) storing the energy in the form of electric field as long as the temperature is low enough to maintain the superconductivity condition otherwise, the
law of conservation of energy will be broken because energy have nowhere to go, again am I
right?


Thank you again for all your efforts.

Regards.
 
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First, you don't "charge" an inductor with a voltage, you do it with a current.

When inductors are "open-circuited" they spark across the gap as the circuit opens, with the energy of the spark equivalent to the energy in the inductor. Why would a superconducting inductor act any differently in principle?
 
Thank for replying.
What I have understood from your reply is that when the inductor
is open-circuited it will have to spark (giving that no clamping method
is provided), and therefore all of my assumption (which is based on
the assumption that you could open-circuit an inductor without having
it to spark) don't even have a reason to occur.
 
12Element said:
Thank for replying.
What I have understood from your reply is that when the inductor
is open-circuited it will have to spark (giving that no clamping method
is provided), and therefore all of my assumption (which is based on
the assumption that you could open-circuit an inductor without having
it to spark) don't even have a reason to occur.

Yes, that is correct, otherwise it would be possible to have a the same amount of current flowing through an open circuit as through the closed circuit, which just doesn't make any sense. How could you have current in an open circuit?
 
Thank you very much, that clarified a lot for me.
 
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