I learned many years ago that according to Ohm's law, current is equal to voltage divided by resistance. Now if superconductors have zero resistance then the current should be infinite. Moreover the power should be zero too as $P=I^2R$.

In superconductors, current can pass without applied voltage. Since current is flowing so there is definitely some energy flowing in the wires but VIt=energy and V is zero so there is no energy ?

How to solve this discrepancy?

  • $\begingroup$ It's probably better to think of superconductors as materials where current can freely flow without an applied voltage. $\endgroup$ Apr 30, 2015 at 22:13
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    $\begingroup$ @SumitSingh Better to remember that electric potential is potential energy per unit charge. To move charge to higher potential, work must be done on it; when the charge moves back to lower potential the energy is released. At constant potential energy, without friction, a mass can move with constant momentum indefinitely. $\endgroup$
    – rob
    Apr 30, 2015 at 22:33
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    $\begingroup$ They have zero static resistance, but they still have inductance. So applying a voltage does not cause infinite current to appear. $\endgroup$
    – BowlOfRed
    Apr 30, 2015 at 22:55
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    $\begingroup$ possible duplicate of How can Ohm's law be correct if superconductors have 0 resistivity? or physics.stackexchange.com/q/31646 $\endgroup$
    – Danu
    May 1, 2015 at 7:33
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    $\begingroup$ Division by zero is undefined, its not equal to infinity, so Ohm's law apparently doesn't specify what the current will be when there is zero resistance. $\endgroup$
    – bdsl
    May 1, 2015 at 9:03

5 Answers 5


In a superconductor, the current can keep flowing "forever" since there is no resistance. But since conductors have inductance (in fact, superconductors are used most often to create magnets like for an MRI scanner), applying a voltage would not (immediately) cause an infinite current to flow.

It is instructive to see how an MRI magnet is "ramped" (turned on). From that link, I reproduce a circuit diagram here:

enter image description here

By turning on a very small (mW) heater, you can locally heat the superconducting loop so it becomes resistive. Now you can apply a voltage to the circuit - and it will preferentially send current around the superconducting part of the loop. The current will not be infinite though - for a given inductance $L$, the magnet will "ramp" as the current increases according to $$V = -L\frac{dI}{dt}$$.

When the desired field strength is reached, you turn off the heater - the loop will be closed and the current will keep flowing. At this point you turn off the voltage (current) supply. And you have a (strong) magnet that will persist for a very long time - as the current keeps flowing without resistance.

As was pointed out by J... And hexafraction in the comments, there is another important concept in superconductivity: that of a critical current. This is the largest current (density) that a particular superconductor can carry without becoming resistive. This is a function both of the magnetic field, and the temperature of the wire. This has a number of implications:

1) the current will never be "infinite" - you can only get up to some finite current before superconductivity stops

2) if you have a local imperfection in your wire, it might have a lower critical current and will therefore become the source of the transition to normal resistivity

3) designers of superconducting wire spend a great deal of effort on stability; for example they surround the superconducting core with normal conductor (silver or copper) to allow small temporary fluctuations in resistivity to dissipate (current temporarily diverted to the normal conductor). This greatly improves the stability of a superconducting system

4) because critical current goes down as magnetic field goes up, there is a non linear relationship between the magnetic field achieved and the amount of superconductor needed- one reason why higher field MRI systems are disproportionately more expensive.

5) sometimes a superconducting magnet will "quench" - local heat dissipation causes a runaway effect as more wire becomes resistive. You end up with all the stored magnetic energy being dissipated as heat. This will boil off the cryogen (usually liquid helium) which needs venting safely to the outside world (very expensive event!)

It is quite a complex subject...

  • $\begingroup$ What happens if you don't heat a part but still apply a voltage? $\endgroup$ May 1, 2015 at 8:32
  • $\begingroup$ @Joshua: I'm guessing something bad happens to the part of the circuit outside the cryostat (or, more likely, safety features prevent anything at all interesting happening). $\endgroup$ May 1, 2015 at 9:17
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    $\begingroup$ @JoshuaLin In that case, the superconductor will quench. The coil will create too large a magnetic field which will cause a portion of the superconductor to cease to superconduct. Energy is then converted as heat as in a normal wire. $\endgroup$
    – nanofarad
    May 1, 2015 at 10:11
  • $\begingroup$ @JoshuaLin the voltage source itself has a finite impedance so the current you generate is limited by that: $I = \frac{V}{R_{int}}$ $\endgroup$
    – Floris
    May 1, 2015 at 10:28
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    $\begingroup$ @hexafraction I think quenching is the key concept that none of these answers really touch. Superconductors can conduct current with no resistance, but only up to a point. There is a maximum current carrying capacity and beyond that value the superconductor collapses and fails to be a superconductor. $\endgroup$
    – J...
    May 1, 2015 at 12:27

The voltage is zero. That's the point.

The main way current gets started, like in an NMR magnet, is by inductive coupling.


If they have 0 resistance then I (V/R) should be infinite?

According to Ohm's law, the voltage and current associated with a conductor are proportional:

$$V = R \cdot I$$

where the resistance $R$ is the constant of proportionality. This equation holds for an (ideal) ohmic material. We can rearrange this equation to be

$$I = \frac{V}{R}$$

except for the case that $R = 0$ since division by zero is undefined. In that case, we have that $V = 0$ for any value of $I$ (for physical superconductors, there is an upper limit on $I$).

Also, as you've noted, there is no power dissipated in the conductor when the resistance is zero.

Finally, as a commenter noted, there will be a voltage across when the current is changing since there is a magnetic field associated with the current and thus, an associated inductance.


...and V is zero so there is no energy ?

The other answers all touch on this part of your question, but none of them explicitly says, that there is no energy dissipated in the superconductor itself.

Ohm's Law applies to a conductor. The $I$ in Ohm's Law refers to the current flowing in the conductor, the $V$ refers to the voltage difference between two points along the length of the conductor, and the $R$ is the resistance between those same two points.

The power dissipated between those same two points, can be expressed as $I^{2}R$, as $\frac{V^2}{R}$, or as $IV$; all of which are zero when $V=0$ and $R=0$.


@Jerry Schirmer is right, the best way to think of a superconductor is like a skier being able to ski at a constant speed without a slope. The power output is zero since no power is dissipated through any heating affects of the conductor so no energy is transferred except when resistance is met at the other end (like a bulb).


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