A very low current with very high voltage can travel through high resistant wires without much loss. Could this type of current be used on magnetic energy storage without superconducting materials?

Superconducting magnetic energy storage systems work by making an electromagnetic field on a superconducting coil, which in turn self-induces a current that produces an electromagnetic field. Since the superconducting material have almost no resistance at all, it has almost no losses and keeps self-inducing the current until discharge.

The logic goes that since very high voltages can travel through wires with very high resistance without many losses, then an induction of such current on an electromagnet could produce a very low loss inductance.

Or I'm incorrect also?

  • 1
    \$\begingroup\$ Since the superconducting material have almost no resistance at all - Not almost. Superconducting means literally zero resistance, which is qualitatively different from low resistance like a really wide copper bus-bar. \$\endgroup\$ Jan 16 at 14:01

4 Answers 4


You have misunderstood how "high voltage" is beneficial, in the context of conduction losses.

The first part of this answer doesn't not address your question directly, but it will help you must understand why "high voltage" can help, and the context in which this principle is valid.

In power delivery systems, loss occurs in the conductors themselves, due to electrical resistance of those conductors. As you know, Ohm's law says that any conductor will develop a potential difference between its two ends proportional to the current flowing through the conductor, due to its resistance. Here's the scenario:


simulate this circuit – Schematic created using CircuitLab

RW1 and RW2 represent resistance of the wires carrying current to and from the load, represented by RL. Current is 1A, the same everywhere in the loop, but wire resistance develops a total of 2V, for a dissipation of \$2 \times 1A \times 1V = 2W\$ of power, lost in the wires.

The load receives only 10V of the total 12V available from source V1, after the wires have "eaten" 2 of them, and therefore dissipates \$1A \times 10V = 10W\$.

Now we'll increase the supply voltage V1 by a factor of 10, to become 120V, and adjust RL so that it still consumes only 10W:


simulate this circuit

Now we have only 14mW of power lost in the wires (a factor of 140 improvement), and still 10W being delivered to the load. It may seem contrived that we had to change the load resistance, but the point here is that we are delivering the same power to the load, but with vastly reduced conduction losses in the wires, and that reduced loss is due to the reduction of loop current that we asking the wires to carry.

This principle is used in all national power grids, to minimise power dissipation in the wires, and reduce the diameter of wires needed to carry current. While we don't directly "change the resistance" of the load (that's determined by the households, factories and whatever else is connected), an equivalent function is performed by transformers.

It's important to understand that voltage of the the 12V or 120V source isn't "travelling through" anything. It's being shared between whatever elements exist in the loop. It's the current that's flowing through things, in combination with the elements' resistance, that is causing power to be delivered to them.

Now back to your energy storage system. In such a system, a zero-resistance coil has a current passed through it, which develops a magnetic field in which the energy is stored. Current is allowed to continue flowing around the loop, and the absence of resistance in the loop means that no power is delivered to anything. By conservation of energy, the energy of the magnetic field is preserved.

To retrieve that energy, a resistance (or more generally, an "impedance") is introduced into the loop. This new element does not have zero impedance, and must therefore develop a voltage across it, by Ohm's law, which means that it receives \$P = I \times V\$ Joules of energy per second. That energy comes from the magnetic field.

Evidently, any resistance in the loop will receive power, causing the magnetic field to diminish over time, until it's completely exhausted, whether this is during "charging" of the storage system, or during "discharge". If the coils themselves have any resistance, they will dissipate power, and drain the energy of the magnetic field, by heating.

This is why it's necessary to use super-conducting coils; the loss of energy to non-zero resistances in the loop, including the coil itself, will occur regardless of the voltages present anywhere in the system. The success of such an energy storage system depends entirely on the resistances present, and not on what voltage you use to "charge up" the store.

In fact, since the idea is to establish a "semi-permanent" current in the loop, it doesn't make sense to talk of reducing current there. The current that flows is necessarily as high as you can make it, because it's intimately linked to the strength of magnetic field.

I hope you can see why the principle of using "high voltage" to improve efficiency is not applicable in this scenario, and it doesn't matter what voltage you use to "charge up" the energy store. The goals for the storage system are chiefly:

  1. Develop a strong magnetic field, which necessitates high current

  2. Have zero (or as close to zero as is possible) resistance in the current loop, to avoid depleting the energy of the field

Neither of these are going to be helped by using high voltages. As a further talking point, this might surprise you; when there's nothing receiving energy in the loop (no resistance present), the voltage across the coils must be zero, even when thousands of amps are flowing.



Magnetic field occupies space. To intercept that field, you need to fill some fraction of that space (the winding space) with conductor.

It's a nearly completely free choice, whether you fill the winding space with a single solid conductor, or millions of finely divided ones. And it's a free choice whether those strands are connected in series or parallel; thus you can design for any V/I ratio, for a given discharge rate.

The conductor arrangement has no effect on time constant: say we take half the strands and connect them in series with the other half. We've quadrupled the inductance, but we've also doubled the winding length and halved its area, thus quadrupling its resistance. Time constant is independent of how we divide the conductor.

Practical considerations (which constrain the freedom of that choice) relate to the density of conductor you can pack in there (finer-divided conductors are harder to pack), how fast you want to interact with the field (thick conductors and fast rates are susceptible to skin effect), and knock-on effects, like how you are able to cool the conductor, whether immersing finely-divided ones in liquid permeating the spaces between, or using solid conductors with passages for coolant (drilled holes, tubing, perforations).

We can make the time constant larger by making the inductor itself (physically) larger -- using much more material, or by using lower resistance material, or higher permeability material filling the space around the conductor (to a limit; real permeable materials saturate at some field strength, reducing energy capacity). For a given combination of those variables, there is a maximum time constant, given by an optimal arrangement of those materials (most likely a toroidal or poloidal geometry).

A related concern:

"Energy storage" usually relates to time constants of days at least; for magnetic storage, this necessitates superconductors. In which case, the choice is forced: superconductors are 100% skin effect by definition, so must be very finely divided. The ratio of voltage to amperage is therefore merely a matter of how those strands are arranged, in series or parallel. Since resistance is zero (at least in a perfectly closed superconducting loop), time constant of the inductor itself is N/A, though practical considerations (cryocooler power consumption, heat entry due to connections at room temperature, switching losses) limit an overall solution to finite times.

Superconductors, by the way, have a similar energy density limitation as permeable materials: a critical flux density beyond which they cannot operate, at any temperature. This is why superconducting magnets top out at about 20T.

As far as I know, typical designs are in the low kA range; superconducting magnets have indeed been used for specialty energy backup/storage applications.

  • \$\begingroup\$ "Finer divided conductors are harder to pack" -- as are very low turn-count coils. In between is easy, though. At the high turn-count end, you run into problems with losing more and more space to insulation. At the low turn-count end, you run into problems getting just a few turns to fit neatly into the space (which is why there's high-power inductors out there with "flat spring" coil designs). \$\endgroup\$
    – TimWescott
    Jan 15 at 1:03
  • \$\begingroup\$ @TimWescott It depends -- in the extreme, manufacturers of large motors, generators, etc. use solid copper bar. A fill factor of 90% is achievable! \$\endgroup\$ Jan 15 at 1:48
  • \$\begingroup\$ @TimWilliams Also square wire is a thing. \$\endgroup\$
    – user4574
    Jan 15 at 3:54
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    \$\begingroup\$ fun fact: Strong electromagnets will increasee their fill factor during operation. The inwards mechanical compression due to Lorentz force, pushes round wires into a close packed honeycomb over time. This is a reliability issue, so the voids are either filled with fiber reinforcement in the first place or non-round voidless wire is used \$\endgroup\$
    – tobalt
    Jan 15 at 5:06

There are inductive energy storage systems used with normal copper wire. The main application for inductive energy storage is in flyback converters. Here the time constants are in the microseconds to milliseconds.

Regardless of whether you want to connect your storage system to a high or low voltage supply or load, when an inductive energy storage system is in the 'store' mode, it is short-circuited, to allow the current to continue to circulate as long as possible. The only voltage the current sees is the IR of the copper itself.

Unfortunately, the L/R ratio depends on the material, the copper, and the magnetic volume it can surround, rather than how many turns you divide it into, or what shapes you wind the coils.

Contrast that with energy storage in capacitors, where the insulating dielectric material can achieve CR time constants in thousands of seconds, or much more.

There was one interesting case of using inductive energy storage, used to charge one of Sandia's early Z-pinch mega-volt water-insulated transmission line capacitors, where inductive storage was just possible, and was cheaper than capacitive storage. (I'll see if I can find a link to that article)

  • \$\begingroup\$ Small correction -- SMPS inductors have a time constant in the ~ms, so that only a small fraction of energy (some percent) is lost at the 1s to 10s µs switching rate. \$\endgroup\$ Jan 14 at 14:49

No. The energy stored in an inductor is 0.5LI^2. So it depends mainly on current. Voltage is irrelevant


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