I've been reading up on superconducting magnetic energy storage, and while I know how the energy is stored ($E=\frac12LI^2$) I have no idea how the discharging works. Specifically, as per this paper 1 a voltage is applied across the superconducting coil, and the power output is given by $VI$, where $I$ is the current circulating in the superconductor. But I thought the notion of applying a voltage across a superconductor didn't make sense, as per this physics SE question.

Furthermore, I'm a bit uncertain as to how you just apply a voltage across something. For example, you could presumably put a 12V battery across the coil, but wouldn't that just result in your battery being charged? What if you wanted to deliver electrical power elsewhere?

1 P. Tixador, "Superconducting Magnetic Energy Storage: Status and Perspective", IEEE/CSC & ESAS EUROPEAN SUPERCONDUCTIVITY NEWS FORUM, No. 3, January 2008.

  • $\begingroup$ Where in the paper [1] does it say "voltage is applied"? Voltage is created after the SC circuit is disconnected (or some element of non-zero resistance is introduced) by mobile charges due to inductor pushing current into shorted terminals or resistive element. $\endgroup$ Commented Dec 15, 2020 at 16:11
  • $\begingroup$ @JánLalinský Page 3, "The product of the magnet current ($Io$) by the maximum allowable voltage($Vmax$) across it gives the power of the magnet ($Io*Vmax$). $\endgroup$ Commented Dec 15, 2020 at 20:48
  • $\begingroup$ Where in that statement do you see "voltage is applied"? The statement says that maximum usable power of the device is determined by maximum allowed voltage on its terminals (by its designer or by other analysis). This maximum allowed voltage does not have to be applied by anybody, voltage will appear spontaneously and the operator has to make sure it isn't too great. $\endgroup$ Commented Dec 15, 2020 at 21:27
  • $\begingroup$ For example, if we use the device to generate heat and so put in parallel a big resistive element with resistance $R$ and current $I$ goes through it instead of through the short, then voltage $RI$ will appear spontaneously on the resistive element as a result of large current passing through. In order to be safe from reaching the maximum allowed voltage, the ohmic resistance must not be too big. $\endgroup$ Commented Dec 15, 2020 at 21:28

1 Answer 1


Normally a superconducting magnet (a coil) is closed. If a resistor is connected at two points on the coil and the coil is opened between those points, current will flow through the resistor. The back EMF caused by the current through the resistance will cause a voltage between the ends of the coil. This in turn causes the current in the coil to drop. When current in the coil changes, self-inductance of the coil induces a back EMF that opposes the change. So, the rate of energy transfer to the resistor is controlled by the coil geometry, the amount of current in the coil, and the value of the resistor.

In practice, current from the superconducting coil can be switched rapidly to drive an AC current in an external coil, allowing energy to be transferred out inductively instead of resistively. The portion of the system that deals with the switching, then converting the power to a useful form, is called a power conditioning unit. There is a lot of information online about power conditioning units for superconducting systems, such as this.

  • $\begingroup$ The authors name of the linked document is Superczynski :-) $\endgroup$ Commented Dec 20, 2020 at 19:17
  • $\begingroup$ „a resistor is connected at two points on the coil and the coil is opened between those points“: I know it a bit different. The supercooled coil will be heated a bit between these two connections and it is no longer superconducting in between. That allows the current from the source to flow through the whole coil. After charge the heater is switched off. $\endgroup$ Commented Dec 23, 2020 at 4:36
  • $\begingroup$ That's a slightly different, probably better approach. $\endgroup$
    – S. McGrew
    Commented Dec 23, 2020 at 5:31

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