Superconducting energy storage and voltage 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.
 A: 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.
