Current in superconductors when a voltage is applied.

AI Thread Summary
In a superconducting loop connected to a 9V battery, the theoretical current calculation using Ohm's law suggests an infinite current due to zero resistance. However, the actual current is limited by the battery's internal resistance, which must be considered. Assuming an internal resistance of 1.5Ω, the current can be calculated as 6A, acknowledging that superconductors can only carry currents below a critical threshold. If the current exceeds this limit, the superconductor transitions to a normal conducting state. This discussion highlights the importance of factoring in non-ideal conditions when analyzing superconducting circuits.
JackA7X
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Suppose you a have superconducting loop connected to a 9V battery. What is the current through it? With a resistance of 0Ω, V = IR gives ∞. I don't think that can be right... If it were, V=\frac{I}{nAe} would give an infinite velocity... Does that equation even apply to superconductors?
 
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Current will be limited by the internal resistance of the battery. This happens for all batteries connected to low impedance loads.
 
So if we assume the internal resistance is 1.5Ω how do you calculate the current through the superconductor?
 
I=V/R, just like always. In this case, 9A since a superconductor has 0 resistance.
 
Marcus
There seems to be an arithmetical error here.
 
Superconductors can only carry currents lower than a critical current value. If the current rises above that limit, they become normal conducting.
 
syhprum said:
Marcus
There seems to be an arithmetical error here.
Quite so, I hit the wrong key on my Blackberry. 6A.
 
Thank you, very helpful.
 
This is a "irresistible force and immovable object" type of question, if you don't allow something to be non-ideal.
 
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