# Electric resistance and superconductors

1. Jun 7, 2007

### Mr_Bojingles

I was reading there that if you apply an electromotive force of 1 vole to a wire with a resistance of 1 ohm it will cause a current of 1 ampere to flow.

If I was to obtain a wire with no resistance at all and applied 1 volt to it would it still cause a current of 1 amp to flow or would the current be insanely high because theres nothing stopping it? Its impossible to use Ohms law when it comes to superconductors because dividing anything by 0 leaves the answer undefined.

I read a bit about the superconductor theory but I didnt get into the technical details like exactly what kind of current flows at a given voltage etc.

BTW whats that site that lets you host big files so people can download them? I have a good electronics ebook I'll upload.

Last edited: Jun 7, 2007
2. Jun 7, 2007

### ranger

Theres a limit as to maximum amount of current that can flow. This is called the critical current. You put too much, and the material loses its superconductive property and will obviously heat up. The thing about superconductors is their lack of resistance. This means you could theoretically have a current that could flow without any driving voltage or decay.

Heres a quote from wiki:
http://www.quantiki.org/wiki/index.php/Superconductivity

So its not that easy as simply applying ohms law to the superconductor alone to find the resulting current.

Heres a thinking question. What would happen if you short the leads of fully charged capacitor with a superconductor?
Try rapidshare.

Last edited: Jun 7, 2007
3. Jun 7, 2007

### Mr_Bojingles

Does the superconductive material lose its conductivity solely because it heats up as a result of current flowing through it or are there other factors involved?

You say theoretically current could flow through the superconductive without any voltage as a driving force. Is that just because the electrons have a tendancy to flow around in the electron pool?

I cant even comprehend that capacitator question. I've only started learning about electronics in the last couple of months and I've never used a capacitator so I know very little about them. I've only been playing around with circuits as simple as a power source, resistor, LED, and a buzzer. I bought a potentiometer today to see if it will work as a dimmer for the LED. Works like a charm. Im guessing regular house light dimmers are just potentiometers.

Heres the ebook I was talking about
<< URL deleted by berkeman -- copyright issues >>
Its fairly detailed.

Last edited by a moderator: Jun 8, 2007
4. Jun 7, 2007

### ice109

i want to say nothing because theres no voltage drop across it??? but theres probablymore to it than that?

5. Jun 8, 2007

### Staff: Mentor

Last edited: Jun 8, 2007
6. Jun 8, 2007

### ranger

Its mostly caused by heat. Referred to as critical temperature.
http://hyperphysics.phy-astr.gsu.edu/hbase/solids/scond.html#c2
Think about it for a sec. If theres no opposing force for the electrons (zero resistivity) and we somehow set them in motion, whats there to stop them?
Well with regards to that capacitor question, start from the basics with shorting it with a piece of wire with very little resistance and ponder what will happen.
Its good that you're experimenting. Keep it up. But domestic dimmers do not work using potentiometers. Do you kow why your dimmer/how your dimmer worked? Lets just say for a sec that we would use your dimmer circuit. We designed a circuit that can vary the voltage across a regular incandescent bulb via a simple voltage divider (resistor and pot). What can you say about power dissipation across the resistor network? Remember we're dealing 120V mains here!
One way do design such a dimmer would be to make use AC phase difference. This circuit would be rather complex for beginner to electronics. It requires knowledge capacitors, inductors, AC theory, triac/diac operation (solid-state theory), etc. So I won't confuse you. For now learn how capacitors work (both DC and AC)
I would feel very cheap if I hinted you the answer to my question.

Last edited: Jun 8, 2007
7. Jun 8, 2007

### ZapperZ

Staff Emeritus
There's something very wrong here, and unfortunately, since it is in the engineering section, those of us who have some knowledge in superconductivity didn't get to see this until recently.

It is incorrect to think that the supercurrent will cause the superconductor to heat up. This contradicts the notion that such supercurrent flows with zero resistance. If it flows with zero resistance, and the heat generated is due to such resistance, then there is no heat generated via the flow of such supercurrent. It doesn't bump into any of the material's lattice and thus, generate no resistance and no heat.

The critical current, which is temperature dependence (i.e. at what temperature below Tc you set it determine what the critical current is), is more of a reflection of the supercurrent density, i.e. how much charge carrier is there at a given temperature. If you are at T<Tc but just below, then there is very little supercurrent density and you could easily "exhaust" all of them and start to now use the normal charge careers and thus, you'll detect resistance. If T is well-below Tc, then you have a larger supercurrent density and the critical supercurrent will be a lot larger.

Administrative note: please keep in mind that we do not allow the sharing of copyrighted material without permission of the author/publisher. This is an unlawful activity and it is something we do not tolerate.

Zz.

8. Jun 8, 2007

### cabraham

Here is what will happen if you place a superconductor directly across the leads of a charged capacitor. Current will gradually increase in accordance with the cap initial voltage and the inductance of the circuit. Since the closed path has a finite area, some inductance is present. Thus a cap is terminated by an inductance with zero resistance. The current starts at zero and builds up while the cap discharges. When the cap has transferred all of its energy into the inductance, the current is at its maximum. Then, the energy now stored inductively will be returned to the capacitor with reversed polarity. An L-C oscillator is what I would expect with radian frequency given by 1/sqrt(LC). That is my off the cuff educated guess.

9. Jun 8, 2007

### ice109

you think the supercon wire will act as an inductor forming a resonant lc circuit? i think not, theres no $${A} \cdot {d \phi}= 0$$ , i don't think there will be any inductance

Last edited: Jun 8, 2007
10. Jun 8, 2007

### cabraham

Rest assured that there will indeed be a finite inductance value. Take a superconductor and loop into a 5 turn coil and connect it across the cap. You do acknowledge the inductance don't you? Now reduce the number of turns to one. The inductance is equal to the permeability (free space value of 0.4*pi uh/m) times the turns squared (one) times the loop area (pi*r^2) divided by the magnetic path length. There has to be an inductance. Since the cap and inductor are directly connected, the inductor current must gradually ramp upward starting at zero since current in an inductor cannot change instantly. The cap discharges as the current increases. The cap's electrical energy is transferring to the inductance as magnetic energy. When the cap is at 0 volts, the current is at max value. The inductance current continues, as energy now is tranferring from L over to C (magnetic to electric) but the voltage on the cap is increasing with polarity opposite to its original state. When the inductor is out of current, the cap voltage is at the same magnitude as its starting state, but with reversed polarity. The cap then discharges back into the inductance, and the current polarity is opposite before. Once discharged, the cap is at 0 volts, and the inductance recharges it to its original polarity. The process repeats indefinitely. An L plus a C with no R and initially energized will be a lossless LC resonant tank circuit.

Yes, there is an "A dot d(phi)". As soon as the cap starts discharging into the one turn superconducting loop, a current exists and a magnetic flux, phi, increases from 0 upward. Magnetic flux, phi, is indeed present. This flux divided by the current equals the inductance, since 1 henry = 1 weber-turn/amp.

Best regards.

Claude

11. Jun 8, 2007

### Averagesupernova

You are assuming that there is no resistance in the plates of the capacitor. With a normal cap and a superconducting wire the oscillation will decay similarly to a cap and plain old copper wire.

12. Jun 8, 2007

### ice109

does a straight wire have inductance?

13. Jun 8, 2007

### Averagesupernova

Yes .

14. Jun 9, 2007

### ice109

i don't understand how the theory yields that? theres no area, theres no closed surface through which to make flux

15. Jun 9, 2007

### ranger

16. Jun 9, 2007

### cabraham

But the problem made no mention of cap plate resistance, or "esr", so I neglected it. However, if such resistance exists, then of course, the oscillations will decay, just like any damped RLC network. I have no argument there.

As far as a straight piece of wire goes, inductance is indeed present. As the cap discharges, the current path consists of the + plate, through the superconductor, to the - plate. Although the plates are separated by insulation (vaccuum or otherwise), a "closed" path does exist. This is called "displacement current". A magnetic field surrounds any current carrying conductor, including a straight one. Any attempt to change the value of current requires that work be done, since the stored energy must be changed. Hence, a straight wire carrying current stores energy in the magnetic field, and tends to oppose changes in current. Work is needed to change current value, and its associated mag field value. This is what "inductance" is all about. Best regards.

C;aude

Last edited: Jun 9, 2007