Understanding Resistivity in Superconductors

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SUMMARY

The resistivity of YBCO (Yttrium Barium Copper Oxide) in its superconductive state is zero for direct current (DC), confirming its superconducting properties. However, alternating current (AC) resistivity is not zero, leading to dissipative losses primarily at high frequencies, such as microwave frequencies. Heat generation in superconductors occurs under non-ideal conditions, such as excessive current causing weak-links across grain boundaries or flux movement between pinning sites, which are more related to materials science than superconductivity itself.

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What is the Resistivity of YBCO in superconductive state because it is not zero, or there would be not heat created so what is it or am I wrong? Explain to me where the heat from a current passing into a superconductor comes from.
 
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The DC resistivity for YBCO in the SC is indeed zero.
The AC resistivity is not equal to zero, meaning there will some dissipative losses, but this is mainly an issue at quite high frequencies, i.e. microwave frequencies and above (this is true for all superconductors).

I am not sure what "heat" you are referring to. There are situations where e.g. cable might warm up if you pass too much current through it, but this a "non-ideal" situation where the heat could be generated from e.g. weak-links formed across high-angle grainboundaries going normal; flux moving between pinning sites can also cause dissipation.
However, these effects don't really have anything to do with superconductivity as such; it is more of a materials science issue.
 

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