What Happens When an Inductor is Open Circuited?

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SUMMARY

When an inductor is open-circuited, it generates a spark due to the sudden interruption of current flow, with the energy of the spark being equivalent to the energy stored in the inductor. For non-superconducting inductors, this energy dissipates as heat due to intrinsic resistance. In the case of superconducting inductors, while the charges can remain at each end, the absence of a current path leads to a potential violation of energy conservation if the superconductivity condition is not maintained. Thus, open-circuiting an inductor without proper clamping will always result in sparking.

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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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