# In superconductivity, is resistance =0?

If a metal shows superconductivity of electricity at definite temperature, then during superconductivity can we consider resistance of metal = 0?

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The first sentence in the wikipedia page may help. Read on for details. en.wikipedia.org/wiki/Superconductivity –  BMS Mar 1 '14 at 7:31
according to wiki during superconductivity electricesistance but acording to ohms law V=IR then if resistance =o then voltage provided by any battery when metal showing superconductivity will be 0 but voltage of a battery cannot be changed. –  Murtuza Vadharia Mar 1 '14 at 7:42
@MurtuzaVadharia You don't have to keep a battery attached to a superconductive circuit. Just an impulse is sufficient to get the current going - the electric current will continue to flow through the circuit even without battery: voltage is zero, resistance is zero, current is non-zero. This is done in superconductive electromagnets, like those in LHC. –  mpv Mar 1 '14 at 7:59
Superconductivity is strictly a DC phenomenon. If you even ramp it you can get losses. Note the difference between Type I and Type II supercons in a magnetic field (e.g., caused by current flow. A straight wire is 1/2 of a winding). –  Uncle Al Mar 1 '14 at 19:28

Superconductivity is a phenomenon of exactly zero electrical resistance and expulsion of magnetic fields occurring in certain materials when cooled below a characteristic critical temperature. When you cool metal to its critical temperature it becomes a superconductor.

In Metal there are Cooper pairs and it doesn't have enough energy to break these pairs. If ΔE is larger than the thermal energy of the lattice, given by kT, where k is Boltzmann's constant and T is the temperature (In Kelvin), the fluid will not be scattered by the lattice. The Cooper pair fluid is thus a superfluid, meaning it can flow without energy dissipation.$$E = kT$$Where:$$k = 1.3806488 × 10^{-23}\,m^2\;kg\;s^{-2}\;K^{-1}$$ Superconductivity means that metal resistance has dropped to exactly 0. So answer is yes

source: Wikipedia http://en.wikipedia.org/wiki/Superconductivity

EDIT: What about battery's internal Resistance? If you added Battery in circuit then you must add battery's internal resistance too (It will be small (10 - 100Ohms) but not zero) V = IR = I(R_superconcuctor(which is zero) + R_battery). But after disconnecting battery, voltage and resistance will be zero but current will still be flowing

EDIT 2: Cooper pair or BCS pair is two electrons (or other fermions) that are bound together at low temperature. Cooper pairing is a quantum effect, the reason for the pairing can be seen from a simplified classical explanation. An electron in a metal normally behaves as a free particle. The electron is repelled from other electrons due to their negative charge, but it also attracts the positive ions that make up the rigid lattice of the metal. This attraction distorts the ion lattice, moving the ions slightly toward the electron, increasing the positive charge density of the lattice in the vicinity. This positive charge can attract other electrons. At long distances this attraction between electrons due to the displaced ions can overcome the electrons' repulsion due to their negative charge, and cause them to pair up. The rigorous quantum mechanical explanation shows that the effect is due to electron–phonon interactions. For more info: https://en.wikipedia.org/wiki/Cooper_pair

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The question is how would one disconnect the battery and without disrupting the flow of current? –  Bleeding Fingers Mar 1 '14 at 9:58
@BleedingFingers See e.g en.wikipedia.org/wiki/Superconducting_magnet#Persistent_mode –  nephente Mar 1 '14 at 10:48
This answer is right, but it isn't helpful unless you already know what a Cooper pair is. If you want a simple explanation of the basics, try these links. superconductors.org/oxtheory.htm youtube.com/watch?v=l-_lsG_K2B4 –  mmesser314 Mar 1 '14 at 13:26