Calculating power in a superconductor

I see in the main page an article touting a record setting 20,000 amps flowing in a superconductor.

Normally, with resistance in the circuit, it is a simple matter to multiply volts times amps to come up with watts or joules/second. But in a superconductor the resistance is zero so 20,000 amps times zero resistance would be in that case zero watts. So how do you calculate the power of those 20,000 amps when there is zero resistance?

I am thinking it would have to do with counting the number of electrons passing by a point in a given time or something but don't know how you would do that.

Anyone help with this?

UltrafastPED
Gold Member
Power is volts x amps.

The current is number of units of charge per second; in amps it is coulombs per second.

ZapperZ
Staff Emeritus
I see in the main page an article touting a record setting 20,000 amps flowing in a superconductor.

Normally, with resistance in the circuit, it is a simple matter to multiply volts times amps to come up with watts or joules/second. But in a superconductor the resistance is zero so 20,000 amps times zero resistance would be in that case zero watts. So how do you calculate the power of those 20,000 amps when there is zero resistance?

I am thinking it would have to do with counting the number of electrons passing by a point in a given time or something but don't know how you would do that.

Anyone help with this?

Have you ever used a clip-on ammeter? It is basically a loop measuring the amount of magnetic field induced by a current moving through it. This measurement tells you directly the amount of current in the wire without having to know power, potential difference, and resistance.

Zz.

Khashishi
Superconductors don't dissipate power, so power is zero.

UltrafastPED
Gold Member
Superconductors don't dissipate power, so power is zero.

The question is how much power is being carried; power dissipated is something else - typically the joule heating of the wire/resistors, but there are other possibilities.

Khashishi
In that case, the superconductor is acting as an inductor. The energy stored in an inductor is
##U=\frac{1}{2} LI^2 = \int \frac{1}{2}\frac{B^2}{\mu} dV##

The inductance will depend on the shape of the superconductor. If it is coiled into a solenoid, then the field is
##B = \mu \frac{NI}{l}##
so
##U = 2 \pi r^2 l \left( \mu \frac{NI}{l} \right)^2 /\mu##
##=2 \pi r^2 \mu I^2 \frac{N^2}{l}##

Have you ever used a clip-on ammeter? It is basically a loop measuring the amount of magnetic field induced by a current moving through it. This measurement tells you directly the amount of current in the wire without having to know power, potential difference, and resistance.

Zz.
Trust me, I know all about clamp on ammeters, AC and DC. That wasn't needed in this case because the current was already given. I just wanted to know how much energy was in that current, how many joules.

In that case, the superconductor is acting as an inductor. The energy stored in an inductor is
##U=\frac{1}{2} LI^2 = \int \frac{1}{2}\frac{B^2}{\mu} dV##

The inductance will depend on the shape of the superconductor. If it is coiled into a solenoid, then the field is
##B = \mu \frac{NI}{l}##
so
##U = 2 \pi r^2 l \left( \mu \frac{NI}{l} \right)^2 /\mu##
##=2 \pi r^2 \mu I^2 \frac{N^2}{l}##
Thanks, I think that answers the question, the energy is in the form of the magnetic field. The only problem I have with that is the magnetic field is all held inside the superconducting material, right? It is not allowed to flow freely as a field like in an iron core electromagnet. Does that change things? I guess it would not but the field strength internally would be a lot stronger since the lines of force would be all squished inside the superconductor and not spread out over a large volume, right?

Thanks, I think that answers the question, the energy is in the form of the magnetic field. The only problem I have with that is the magnetic field is all held inside the superconducting material, right? It is not allowed to flow freely as a field like in an iron core electromagnet. Does that change things? I guess it would not but the field strength internally would be a lot stronger since the lines of force would be all squished inside the superconductor and not spread out over a large volume, right?

Can you explain the units in the equation? I don't know what the u is for instance. I think N is the number of turns squared but then there is the little l under that. Can you expand on that? Thanks again. Oh, I think the lower case l is the length of the circle in the case of a looping superconductor? I also assume I is the current. Of course Pi and such is obvious.

DrDu
The only problem I have with that is the magnetic field is all held inside the superconducting material, right?
No, on the contrary. The main problem in achieving high currents in superconductors is that there is no magnetic field allowed inside a superconductor as it will lead to a breakdown of superconductivity. There are materials where superconductivity breaks down only along some isolated flux lines within the superconductor, so called type 2 superconductors. They allow for higher field strength, nevertheless the field is also there concentrated in regions which are not superconducting.

UltrafastPED
Gold Member
Thanks, I think that answers the question, the energy is in the form of the magnetic field. The only problem I have with that is the magnetic field is all held inside the superconducting material, right? It is not allowed to flow freely as a field like in an iron core electromagnet. Does that change things? I guess it would not but the field strength internally would be a lot stronger since the lines of force would be all squished inside the superconductor and not spread out over a large volume, right?

While there is a magnetic field around the superconducting wire, it is the same magnetic field that you would find outside of any wire carrying the same current - the physics here is the same.

The equations shown above were for magnetic induction ...

Once again, the power transmitted is:
volts x amps = (joules/coulomb) x (coulombs/second) = (joules/second) = watts.

Resistive power losses can be calculated as (current)^2 x (resistance) = I^2 x R = 0.

The electric company bills you for watt-hours = (joules/second) * (3600 seconds) = 3600 joules.
Thus they bill you for total energy used.

f95toli
Gold Member
So how do you calculate the power of those 20,000 amps when there is zero resistance?

You don't. You can't calculate the power delivered to a load if you only know the current in the leads; nor is it very relevant since the dissipation in a conductor only depends on the current it is carrying, not on the power it is delivering.

This has nothing to do with superconductors as such, the situation is exactly the same for a normal conductor: the power dissipated in a properly sized conductor will be negligeble compared to the power delivered to the load.

Also, the energy carried by that current could in theory be arbitrarily close to zero since the voltage across the load can be tiny; and the only energy stored in the lead itself will be due to its self inductance(*) which depends on the geometry.

(*) Superconductors also have kinetic inductance(which is temperature dependent); meaning there are two contributions to L, but this has nothing to do with the question as such.

Drakkith
Staff Emeritus
Am i correct in saying that power is a rate of energy transfer, so the superconductor has no power to measure since no energy is being transferred anywhere?

UltrafastPED
Gold Member
Am i correct in saying that power is a rate of energy transfer, so the superconductor has no power to measure since no energy is being transferred anywhere?

Yes, power is energy per unit of time [watts = joules/second].

But superconducting wires can and do carry power. This power, like any electrical power, is volts x amps - and the amount delivered depends upon the load.

Why would you think that a superconducting wire could not carry any power?

Drakkith
Staff Emeritus
Why would you think that a superconducting wire could not carry any power?

I was under the assumption that the OP was talking about the current in a superconductor with no load.

DrDu
Of course you could use also the poynting vector ExH to calculate the energy transported. The magnetic field strength is proportional to the current in the SC, and E is proportional to the voltage drop on the load. This should give you a more concrete idea where the energy really is transported.

sophiecentaur
Gold Member
Trust me, I know all about clamp on ammeters, AC and DC. That wasn't needed in this case because the current was already given. I just wanted to know how much energy was in that current, how many joules.

Current doesn't have 'energy'. It is arguable that a flow of electrons could be said to have Kinetic Energy, in that they have a (very small) mass and travel at a (very low) velocity. So I guess that could be regarded as Power. But the way Current is regarded in electrical matters doesn't involve this idea. Along with the water analogy, it tends to lead to more misconceptions than 'understanding'.

I may be completely wrong here, but this is what I believe with my very limited knowledge of superconductors, and a little experience with Ohm's Law.

A superconductor has zero resistance. Therefore, since V = IR, the voltage drop across the superconductor is given by V = 20000*0 = 0V. So 0 volts, and since Power is given by P = VI, P = 0*20000 = 0W. This, as stated previously, is the power dissipated in the superconductor.

So, it has been established that the power loss is 0W in the super conductor. However, I'm not sure if this is the case or not, but if the superconductor was not the only part of the circuit, then the only part that would affect the power of the circuit is everything other than the superconductor.
The superconductor in this case would be like one of those ideal wires you find on Physics tests, where they do not affect the circuit at all apart from to allow the movement of charge.

This means that (assuming what I have said is correct), the voltage and current, and therefore the power, are governed only by all of the circuit elements excluding the superconductor.

So, for your case, where you only have the current in the superconductor, it would not be possible to determine any voltage or power through the rest of the circuit.

But a battery can store energy and so can a loop of superconducting material. It has been tested as a technique for storing energy, so there is energy in it. The fact that at any given time it is not being tapped does not mean there is no energy to tap. It's just like any other inductor but with normal conductors, the energy is dissipated because of internal resistance turns the energy into heat.

A battery not hooked up to a circuit still has X amount of potential energy stored as electric charges across an ionic barrier or a capacitor charged up to some voltage stores energy.

If you have a resonant circuit you have energy swapping between the inductor and the capacitor which would continue for as long as the energy took to heat up the normal conductors and then the energy is gone.

You could see, for instance, a capacitor and inductor but made of superconductor material below its superconductive temperature maintaining a certain frequency depending on the specific inductance and capacitance, maintaining that frequency forever till it was tapped somehow, by another inductor nearby, for instance.

f95toli
Gold Member
But a battery can store energy and so can a loop of superconducting material. It has been tested as a technique for storing energy, so there is energy in it. The fact that at any given time it is not being tapped does not mean there is no energy to tap. It's just like any other inductor but with normal conductors, the energy is dissipated because of internal resistance turns the energy into heat.

But the energy depends on the inductance, which is determined by the geometry. Saying that a conductor can carry x amount of current does not tell you anything about how much energy it could potentially store if you made an inductor out of it.

Also, superconducting LC resonators (which is what you are describing) have very low losses compared to resonators made from normal metals; but the losses are nowhere near zero, meaning they are not useful for storing energy. This is partly because you will always have losses due to other mechanisms (in e.g. nearby dielectrics) but also -more fundamentally- because the losses in a superconductor are only zero at DC; as the frequency goes up so does the losses. Since resonators are by their very nature AC/RF devices they will always be lossy.

UltrafastPED
Gold Member
I may be completely wrong here, but this is what I believe with my very limited knowledge of superconductors, and a little experience with Ohm's Law.

If there is no voltage drop along the conductor - then there are no losses.

This is actually what you assume during simple modeling of circuits; wires are treated as having no resistance, they are just paths for the current. In your "limited experience" with Ohm's law this is almost certainly how you would have treated every wire.

The voltage drop will occur at the _load_, which is where the power is delivered. The voltage applied at one end of the superconducting wire will be unchanged at the other end ... an ideal wire ... and so that voltage is available to do work.

The power delivered will then be Power = Volts x amps.

If there is no voltage drop along the conductor - then there are no losses.

This is actually what you assume during simple modeling of circuits; wires are treated as having no resistance, they are just paths for the current. In your "limited experience" with Ohm's law this is almost certainly how you would have treated every wire.

The voltage drop will occur at the _load_, which is where the power is delivered. The voltage applied at one end of the superconducting wire will be unchanged at the other end ... an ideal wire ... and so that voltage is available to do work.

The power delivered will then be Power = Volts x amps.

And even if you had a micro ohm meter and included the wires, the difference would probably be one in ten thousand so setting the wires to zero resistance is a big help in calculating currents and energy in a circuit.

But there still has to be a way to calculate the energy stored in a looping superconductor, like that one guy said, the energy would be in the magnetic field and the equations can be used for that.

Drakkith
Staff Emeritus
And even if you had a micro ohm meter and included the wires, the difference would probably be one in ten thousand so setting the wires to zero resistance is a big help in calculating currents and energy in a circuit.

But there still has to be a way to calculate the energy stored in a looping superconductor, like that one guy said, the energy would be in the magnetic field and the equations can be used for that.

Yes, but the energy stored in the superconductor has nothing to do with power since the energy is just being stored, not transferred.

UltrafastPED