Graphene’s sleeping superconductivity awakens

[jessriedel]

The article broke the DOI number by dropping the decimal. It's 10.1038/ncomms14024 . You can get the PDF from the journal webpage since it's open access. http://www.nature.com/articles/ncomms14024

[ythn]

I don't get how superconductors fit in with V = IR. Assuming that superconductors literally have zero resistance, doesn't that imply there will always be zero voltage on the line as well? And without voltage, how can you push current through?

[ghkbrew]

The V = IR equation relates the voltage _difference_ across a resistor to it's resistance and the amount of current flowing though it. So a super conductor in a circuit will have 0 drop in voltage across it, while still having an "absolute" voltage (or rather voltage difference between it and a ground).

In theory this it's unrelated to the amount of current passing through the superconductor, but in reality a material will stop acting like a superconductor if you try to push too much amperage through it.

[pjc50]

Ohm's law is like Newtonian gravity: an approximation that ignores quantum effects. V=IR is not precise; across every resistor there is also a quantity of Johnson-Nyquist noise. This nearly always doesn't matter except when building high-precision instrumentation amplifiers.

(What I don't know: are superconductors also free of thermal noise? That could have useful properties of its own..)

[marcosdumay]

No voltage drop means, when you push an electron on the superconductor with some energy, another one is pushed out somewhere else with the same energy.

There is no voltage difference on the line, but unless your entire circuit is a superconducting line, there will be some voltage drop somewhere determining your current. And if the entire circuit is superconducting, you set your current by magnetic means.

[scentoni]

Ohm's Law is not relevant for superconductors and is not fundamental. Ohm's Law is a simple phenomenological relationship that is a useful approximation for many everyday situations where V/I is approximately constant. cf https://en.wikipedia.org/wiki/Friction and https://en.wikipedia.org/wiki/Hooke's_law

[hawkice]

It doesn't sound like you _would_ be able to maintain a voltage across a superconductor. [Think about what that would mean, in terms of electric charge... very odd. Doesn't sound like something that would happen. It's like saying there's a equilibrium top-heavy-ness to a weighted vacuum tube. The weights should fall, unabated, no matter how many you add at the top.] Doesn't mean you can't have current, though, and modulate the current. Just that you can't use [the superconductor's] voltage to do so.

[posterboy]

Your V is the Voltage drop over the superconductor, but for the complete picture consider a voltage source that adds its inherent resistance (output impedance) to the loop and limits the overall current. A superconductor, to my limited understanding is equivalent to the idealized resistance free wires in schematics.

Voltage isn't on the line but it's the difference of two electric potentials. Without Voltage drop there is no energy loss for conducting any current, as P=IV and E(t)=Pt.

Edit: OTOH you might argue that then there is no Energy gradient for the current to follow, but that's only the macroscopic picture where the Voltage drop is on average zero. Another comment pointed at quantum mechanics, but I couldn't explain lossy resistors on that level either.

[IshKebab]

Well, R and V are zero, so I can have any value.

[jobu]

Flip the equation and think of it as I = V/R

As resistance nears zero, the current becomes infinite, regardless of the voltage.

[mdorazio]

This article doesn't say, but the paper on Nature does. 4.2K. So, no, it's not any higher temperature than existing superconductors.

[careersuicide]

I saw the title and immediately said to myself (out loud even), "Yeah, but at what temperature?"