by 12Element
Tags: inductors
 P: 3 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.
 PF Patron P: 4,948 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?
 P: 3 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.
PF Patron
P: 4,948