- 15 Apr '14 14:45 / 1 edithttp://phys.org/news/2014-04-cern-world-record-current-superconductor.html

This piece is showing 20,000 amperes going through this 20 meter cable.

My question: A superconductor has zero resistance and the usual way you calculate power is to multiply current times voltage which gives you an answer in watts, or joule/seconds. So if you just do the math you go 20,000 times zero which of course is zero.

So how do you figure out how much power they are talking about in these cables? How many watts are transmitted?

Maybe you have to calculate using the number of electrons passing a point in a given time. Don't know how to do that however. - 15 Apr '14 15:23

The number of electrons passing through a point per unit of time is directly proportional to the current in Ampère. That's what the "current" is - a current of electrons.*Originally posted by sonhouse***http://phys.org/news/2014-04-cern-world-record-current-superconductor.html**

This piece is showing 20,000 amperes going through this 20 meter cable.

My question: A superconductor has zero resistance and the usual way you calculate power is to multiply current times voltage which gives you an answer in watts, or joule/seconds. So if you just do the mat ...[text shortened]... sing the number of electrons passing a point in a given time. Don't know how to do that however. - 15 Apr '14 16:23

Not sure but perhaps, in the context of superconductors, you have to think in terms of the power not being in the superconductor cable itself but rather the power when it final comes out of that superconductor and into a device such as a lamp or an electric motor which should have a voltage drop across it while it is switched on and drawing current?*Originally posted by sonhouse***http://phys.org/news/2014-04-cern-world-record-current-superconductor.html**

This piece is showing 20,000 amperes going through this 20 meter cable.

My question: A superconductor has zero resistance and the usual way you calculate power is to multiply current times voltage which gives you an answer in watts, or joule/seconds. So if you just do the mat ...[text shortened]... sing the number of electrons passing a point in a given time. Don't know how to do that however. - 15 Apr '14 16:43

Exactly. A superconductor cable uses zero power.*Originally posted by humy***Not sure but perhaps, in the context of superconductors, you have to think in terms of the power not being in the superconductor cable itself but rather the power when it final comes out of that superconductor and into a device such as a lamp or an electric motor which should have a voltage drop across it while it is switched on and drawing current?** - 15 Apr '14 19:07 / 1 edit

It 'uses' zero power but for instance, in a superconducting loop, you have charged it up to X amount of amperes and it is just looping the same current forever.*Originally posted by KazetNagorra***Exactly. A superconductor cable uses zero power.**

How do you calculate how much power is stored in that loop? How many Joules? - 15 Apr '14 20:13 / 2 edits

The "real" power would be whatever machine was drawing neglecting any power losses from the power transmission. The Super conductor just removes line losses. Btw, The Joule is a unit of energy. the Watt is a unit of power (the time rate of the energy consumption). You are asking different things, neither seem to be answerable to me.*Originally posted by sonhouse***It 'uses' zero power but for instance, in a superconducting loop, you have charged it up to X amount of amperes and it is just looping the same current forever.**

How do you calculate how much power is stored in that loop? How many Joules? - 15 Apr '14 20:28 / 3 edits

You are mixing current and resistance up in your calculation. If you have "zero" resistance, you have "infinite" current in a loop ( A short circuit, no matter what the voltage across that loop is). What you get is:*Originally posted by sonhouse***http://phys.org/news/2014-04-cern-world-record-current-superconductor.html**

This piece is showing 20,000 amperes going through this 20 meter cable.

My question: A superconductor has zero resistance and the usual way you calculate power is to multiply current times voltage which gives you an answer in watts, or joule/seconds. So if you just do the mat ...[text shortened]... sing the number of electrons passing a point in a given time. Don't know how to do that however.

P = V*I = Infinite ( short circuit conditions )

Look at it this way:

P = V^2/R ( the power used by the super conductor is equal to the square of the voltage divided by the resistance of the superconductor, which is 0)

In the case above R = 0, and P is once again infinite ( division by zero ), (giving short a short circuit, and very bad things).

If the Resistance "R" were infinite ( a non conductor), the power dissipated by line would be zero ( open circuit conditions) - 16 Apr '14 03:54

The power loss from a current in a superconductor is negligible, so if the material stays in a superconducting state then there is nothing to stop these current loops persisting for huge periods of time.*Originally posted by sonhouse***It 'uses' zero power but for instance, in a superconducting loop, you have charged it up to X amount of amperes and it is just looping the same current forever.**

How do you calculate how much power is stored in that loop? How many Joules?

In the ideal case of a perfect conductor, the absence of resistance means that no electo-motive force (voltage) is required to sustain a current loop. So if I = current; R = resistance = 0; V = I*R = 0. The power loss is V*I = 0. The energy stored in the loop is therefore constant. The current cannot be deduced from the resistance and the voltage drop because I = V/R = 0/0...

To get a handle on the energy stored I think you'd need to know about the specific physics of the superconductor and it will depend on the geometry of the current loop. I don't think it's a simple calculation. - 16 Apr '14 10:33 / 1 edit

I asked that question in Phys.org physic forum and one answer was the energy stored was in the magnetic field and it was the energy of the solenoid and gave an equation of a loop of current equated to the magnetic field strength and it is not a simple equation, not hard either but not like P=E*R squared.*Originally posted by joe shmo***The "real" power would be whatever machine was drawing neglecting any power losses from the power transmission. The Super conductor just removes line losses. Btw, The Joule is a unit of energy. the Watt is a unit of power (the time rate of the energy consumption). You are asking different things, neither seem to be answerable to me.**

I'll look for the post and try to port it over here.

I know they look at superconducting loops as an energy storage device but how do they get the energy out? If it is basically a DC field it is unchanging so how do you couple energy out of such a system? - 16 Apr '14 11:43

My bad. I was talking about using a perfect conductor in a regular circuit across a voltage source. "Superconductive loops" is apparently something very specific, that I know nothing about. ( Sorry for any confusion)*Originally posted by sonhouse***I asked that question in Phys.org physic forum and one answer was the energy stored was in the magnetic field and it was the energy of the solenoid and gave an equation of a loop of current equated to the magnetic field strength and it is not a simple equation, not hard either but not like P=E*R squared.**

I'll look for the post and try to port it over her ...[text shortened]... If it is basically a DC field it is unchanging so how do you couple energy out of such a system? - 16 Apr '14 13:39 / 1 edit

I think they would be very similar, one just a linear conductor, I think it was 20 meters long in the article but provided with a continuous flow of current. Of course they would have the energy supplied because it would have to start out in a non-superconductor so there would be the normal measurements of power applied and so forth.*Originally posted by joe shmo***My bad. I was talking about using a perfect conductor in a regular circuit across a voltage source. "Superconductive loops" is apparently something very specific, that I know nothing about. ( Sorry for any confusion)**

In a loop, I would assume you would have to couple power in magnetically, not totally sure about that but you can see a variable strength field inducing a current into the loop, I think that would work, then using the equation which goes like this: Field strength U=2pir^2 uI^2 * N^2/l

so if you used 1 meter as r, the first part would just be 6.28...times uI^2, not sure what the u stands for but I must be the loop current, then N^2/l I think N would be the number of turns divided by l, probably the loop circumference.

That is the equation given to me on the phys.org general physics forum. My handle there is Litup. It took me a long time to find a handle not used already! Sonhouse was already in use, as was Sonhouse1,2,3 and so forth! Go figure. - 16 Apr '14 21:02 / 1 edit

As far as the field strength goes I'm not sure that your using those variables quite correctly.*Originally posted by sonhouse***I think they would be very similar, one just a linear conductor, I think it was 20 meters long in the article but provided with a continuous flow of current. Of course they would have the energy supplied because it would have to start out in a non-superconductor so there would be the normal measurements of power applied and so forth.**

In a loop, I would a ...[text shortened]... dle not used already! Sonhouse was already in use, as was Sonhouse1,2,3 and so forth! Go figure.

the little "u" may be the energy density factor.

The "2" should be "1/2"

the little "r" is the radius of the conductor

"l" is the arc then of the entire solenoid ( or the superconducting loop in your case). - 17 Apr '14 12:25

That should be (pi r)/2? Radius of the conductor sounds right. You wouldn't need two descriptors of the size of the whole loop. But it looks to me like the pi r^2 thing would be the area of the cross section of the conductor which would be one part of the description of the volume of the entire assembly.*Originally posted by joe shmo***As far as the field strength goes I'm not sure that your using those variables quite correctly.**

the little "u" may be the energy density factor.

The "2" should be "1/2"

the little "r" is the radius of the conductor

"l" is the arc then of the entire solenoid ( or the superconducting loop in your case).

Like pi r^2 H for the volume of a cylinder.