Calculating power in a superconductor

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  • #26
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.

All,

What if you take a superconducting loop with a known current density. While the current circulates, not energy is transferred since the loop resistance is ZERO. But, for sake of discussion, flip a 1 ohm resistor into the loop. Measure the current drop across the resistor... if current is flowing in the loop, then Ohm's law states E =IXR... does this work?

-e
 
  • #27
CWatters
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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?

It would help if you posted a link to the article.

You appear to be confusing the voltage drop down the wire with the voltage delivered to the load.

In a super conductor the voltage drop down the wire is zero. The voltage delivered to the load need not be zero.

For example... Suppose you have a 9V battery connected to a 3 Ohm light bulb by super conducting wires. There will be no voltage loss in the wires so all 9V will arrive at the light bulb. The power delivered to the light bulb is V2/R = 92/3 = 27W

However in an experiment operating at 20,000A they are very unlikely to be using a 9V battery as the source and a 3 ohm light bulb as the load!
 
  • #28
All,

New member here.

I am just learning how to post and read the forums & topics. I am very interested in this post. We known SC materials in the superconducting state have zero resistance and when it's exposed to an external magnetic field will produce a counter field on the surface of the superconductor.The process is very complex. In this example the SC is a loop, then the current will circulate on the surface of the SC.

If you have a copy of Michael Tinkham "Introduction to Superconductivity, 2 nd addition", Fig 5.3, page 164. We see a circulating J current in his example. If no change in the external magnetic field takes place and the Tc is maintained, this J current will continue forever, since there is no resistance in the SC loop. The circulating J current represents a potential energy in this state.

So my question is what happens if?

Now for the sake of discussion we want to measure the J current of the SC loop. In this case, a 1 Ohm resistor is placed in series with the loop. For the sake of discussion, its done instantaneously. We can now reduce the complex nature of the superconductor loop to a simple Ohm's law computation. Ohm's law states that E = I x R. Since the resistor placed into the SC loop value is known, we can now measure the voltage response and from this compute and confirm the J current.

Will this experiment produce a valid J current response or will the measured response be zero?
 
  • #29
Drakkith
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First, remember that current is found by dividing V by R. (or E by R depending on your nomenclature)
In a superconductor R is zero, and dividing by zero is undefined, so you cannot find the current using Ohm's law. (Obviously)
Placing a resistor in series with the superconductor loop, you now have 1 ohm of resistance. Based purely off of my basic understanding of circuits, when you increase the resistance in a circuit, you reduce the current flow through that circuit. I would expect the current in the superconductor + resistor to be drastically less than before the resistor was introduced, and to decay like an inductor + resistor where the inductor is providing the voltage to the circuit.
However, since I have only a basic understanding of circuits, I'm not completely certain about that.
 
  • #30
The key to the example is to find a simple means to measure the unknown max current flowing on the surface of the SC. Since the SC is a closed loop, you can't measure the current without some special effort and techniques. We can measure the current as suggested by inserting a 1 ohm resistor. Now the current will flow out of the SC loop and with a oscilloscope we can measure the voltage drop across the resistor. If the SC surface has any current flowing, then it must flow thru the 1 ohm resistor to complete the circuit. Yes, I agree that this voltage will roll off just like an inductor discharge curve since you're introducing resistance into the SC loop, but it will have some peak, rolling off and decay to zero, i.e. some time constant based on the dimensions and specifics of the SC . It's the peak signal that is of interest and then we can integrate the total energy under the curve for the total energy. The main question was; "Will a voltage signal be greater than zero?"
 

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