Mains power electrical systems on Earth supply a fixed standardized voltage to a number of devices, which are all wired in parallel with each other.

However, on the planet I'm writng about, they do it the other way around: instead of getting a fixed voltage, their mains power grid supplies a fixed current, with all devices wired together in series, and all the current-voltage symmetries flipped. This is of course confusingly labelled as "alternating voltage", and instead of a constant 120 volts AC, you have a constant 120 amps AV.

A lot of the same ideas from our electrical system seem to have some kind of equivalent:

An appliance's power would be rated by how many volts it takes to run at the constant current - lower power devices might only draw a few millivolts from the grid while high-power domestic devices might take tens of volts.

A basic electrical receptacle could work by having a single slot with contacts consisting of a pair of leaf springs pressed tightly against each other. The plug has a single insulating blade with a contact on either side, and when inserted it makes contact with both sides of the outlet before separating the leaf springs.

Switches would work by shorting out the terminals on a device, allowing current to flow uninterrupted past the device without creating a voltage. Breaks in the circuit would be essentially the equivalent of a short in our voltage-based grid, and a household electrical panel will contain 'circuit keepers' that short across the entire loop if the voltage rises too high. In the context of smaller devices, stuff like MOVs and fuses would trade places.

Transformers would still work the same way, and the current would, magnifying current instead of voltage and allowing massive transmission.

Is there somewhere that I'm not thinking of where this symmetry breaks down, or reasons that a mains power electrical grid couldn't work like this?

Furthermore, would it be possible to build electronic devices on this sort of system? It's unclear to me if electronics could work with some kind of standard "1 amp DV system" due to the nature of the fixed voltage drops across semiconductor junctions

Also, what would be a reasonable value for the current standard to actually be?

Per @AlexP's request, I've made some basic schematics for how the distribution grid and house wiring might work.

Distribution grid:


House wiring:


  • 1
    $\begingroup$ If you have the whole grid in series, then 1) You have a far more complicated network, since you have to maintain a continuous path through all devices on the grid; and 2) Any break in the path brings down the whole grid. With a parallel grid, you can add & loads & generation pretty much at will, or disconnect whole parts of the grid. A parallel grid is fairly modular, as long as you maintain powerflow & stability criteria (not exactly a simple task, but doable). $\endgroup$
    – jamesqf
    Mar 1 '18 at 4:47
  • $\begingroup$ @jamesqf Just as well, a single short anywhere in a parallel grid will bring down the entire system... oh wait we have fuses for that. The only difference here is that the "fuses" are wired in parallel and blow closed in response to the over-voltage condition created by an open, rather than wired in series and blowing open in response to an over-current condition caused by a short. $\endgroup$ Mar 1 '18 at 5:16
  • 3
    $\begingroup$ Please supply a circuit schematic drawing showing two power stations, a main distribution line, a commune/town distribution branch and ten appliaces in a house, because I don't understand at all how this could be done; unless you mean to have all the generators and consumers in the country in series, in which case for any reasonable current you will need to generate a voltage of many billions of volts -- which is not only not practical, it's utterly impossible. $\endgroup$
    – AlexP
    Mar 1 '18 at 12:17
  • 1
    $\begingroup$ @AJMansfield: In a series circuit, what happens when a 'fuse' blows somewhere in it? The whole circuit goes dead, because electricity must follow a continuous path - that's what circuit means :-) $\endgroup$
    – jamesqf
    Mar 1 '18 at 18:15
  • 1
    $\begingroup$ @jamesqf I think you misunderstand how his "circuit keepers" work-- they aren't like normal fuses that open the circuit, since this would indeed cut power to everyone. To understand how they work, imagine you have a house in this world with a circuit keeper across the incoming and outgoing wires to the house, and a bulb in the house burns out, creating an open circuit. The entire distribution voltage will build across the terminals of the light bulb, and the circuit keeper senses this and shorts the ingoing and outgoing wires of the house, effectively removing it from the circuit. $\endgroup$ Mar 1 '18 at 19:31

All in all, this would be a total logistical nightmare, for multiple reasons. If I come up with additional ones I'll add them, but off the top of my head I can think of the following:

1) Since every single appliance in the grid is connected in a single loop, the distance of conductor that current flows through would be extremely long. This means a whole bunch of wasted energy from resistive losses.

2) If you run through Maxwell's equations, you find that the voltage across the terminals of a generator depend on how fast it is turned. Keeping a constant current would require changing the speed of rotation for the generator depending on load, which again is a practical nightmare since its much easier to design the generator to operate at a stable frequency. Likewise, solar panels operate at a fixed voltage, as do most means of electricity generation. If you take an electronics class, you'll see that current sources are really rather difficult to implement, and (to my knowledge at least) ultimately rely on a voltage source that drives a known current that is then mirrored using some semiconductor magic. Long story short: voltage sources are easy, current sources are hard.

3) A common voltage is easier to work with than a common current, since there is so much variability in current required by different electrical devices. You could theoretically get around this by putting transformers in every device, but this would be incredibly bulky, costly, and in some cases just straight up impractical since the winding ratios would have to be so gigantic.

4) Many electrical devices, especially electronics, require DC rather than AC. But since all devices are in series, we can't rectify the AC for one device without doing so for all devices down the line. So, we either need no DC devices or we need to insert an inverter after the DC device, which again adds a lot of unnecessary costs and inefficiencies.

5) Your form of circuit protection fundamentally relies on closing switches when a voltage gets too high. The problem is that building high voltage switches is much harder than low voltage ones, since you have to worry about arcing wearing out the contacts. For a typical house, you might have a voltage drop of a couple thousand volts (which as Samuel pointed out, is a huge safety liability), so your over-voltage protection would already have to be pretty heavy duty and expensive. To protect larger portions of the grid, the voltage requirements could easily cross into the realm of functional impossibility.

6) See Samuel's great answer about the absurd supply voltage requirements.

7) Not to harp too much on the circuit followers, since I think they're a cool idea, but I'm also pretty sure that they wouldn't be able to close fast enough to be useful. You said in the comments that the voltage rise time is mitigated by large capacitance of the system, but I don't think this is the case:

Reactances of other loads in series shouldn't matter if your current source is truly a current source, since it will push constant current through them regardless. If it significantly differs from a true current source, this will cause problems as your standard current will always be fluctuating as open circuits are momentarily created in the grid.

Shunt capacitance (Ie inherent capacitances formed between transmission wires) also wouldn't limit voltage rise, as far as I can tell. To see why, take a look at the Telegrapher's equations for transmission lines: $$\frac{\partial V(x,t)}{\partial x}=-L\frac{\partial I(x,t)}{\partial t}-RI(x,t)$$ $$\frac{\partial I(x,t)}{\partial x}=-C\frac{\partial V(x,t)}{\partial t}$$ Where C is the shunt capacitance per unit length, L is the series inductance per unit length, and R is series resistance per unit length. Solving these for the exact boundary conditions would be rather involved, but isn't really necessary, as my point can be proven simply by looking at the equations themselves.

Now, picture we have a transmission line where an open circuit is created from x to x+l at time t. From the second equation, $\frac{\partial V(x+l,t)}{\partial t}$ could be arbitrarily large, indicating the voltage drop would build very quickly. To see why, note that the current at x+l should pretty immediately drop from a large number to zero, since otherwise you would have an arc across the open circuit which is a Bad Thing. But due to the finite propagation speed of current waves which you see if you solve those equations for simpler boundary conditions, the current just after x+l will still be non-zero at time t. So, $\frac{\partial I(x+l,t)}{\partial x}$ will be incredibly large, and when you divide it by the already small C (which is usually on the order of $10^{-12} F/m$), you will see $\frac{\partial V(x+l,t)}{\partial t}$ can be made arbitrarily large. So, shunt capacitance also doesn't help you mitigate the voltage rise across the open circuit.

So, really the only capacitance we're left with is the tiny series capacitance formed by the break in the line itself, which would be on the order of picofarads. With the current source pushing tens of amps, it would take only nanoseconds for the voltage to build up to thousands of volts. There isn't a switch in the world fast enough to close before your burnt out lightbulb becomes an impromptu arc furnace.

As to your question about whether electronics would work with constant current: It depends, but most likely no, at least with familiar logic design. See, most logic design depends on the assumption that you have a rail of high voltage and a rail of low voltage with lots of transistors and whatnot in between. You might be able to come up with a way to implement logic using constant current, but it would be pretty alien to any of our electrical engineers looking at it (my guess is it would use current controlled BJTs as opposed to the more prevalent voltage controlled MOSFETs used in contemporary electronics). In addition, to stress point 3 above, currents in integrated circuits are often in the range of picoamps unless you're trying to design a stove shaped like a computer. This is a far, far cry from the tens of amps needed for large, powerful appliances.

  • $\begingroup$ The system doesn't actually run at a constant speed, it's a little slower under high load and a little faster under low load. Tuned such that a mains powered clock is accurate first thing in the morning. $\endgroup$
    – Separatrix
    Mar 1 '18 at 8:30
  • $\begingroup$ @Separatrix True, but the difference is small enough that we're still able to reliably call it 60 Hz. Which is another problem: as you change the generator speed to adjust voltage, you'd also be changing the AC frequency, which could wreak havoc on appliances tuned for a specific frequency (most notably AC motors). $\endgroup$ Mar 1 '18 at 14:13
  • $\begingroup$ @el duderino: Small differences in frequency are part of how the grid reacts to changes in load. It's similar to letting in the clutch on your car: if you let it in without increasing power by feeding more gas, the engine lugs and might stall. If you add a large load, the frequency drops in response, until generators can react to pick up the load. The real problem is that all the generators on the grid have to be frequency synchronized: en.wikipedia.org/wiki/Synchronization_(alternating_current) $\endgroup$
    – jamesqf
    Mar 1 '18 at 18:28
  • $\begingroup$ Could the issue with actuation speed be solved by intentionally adding additional capacitance along side each circuit keeper? Or as another option (possibly used together with an extra capacitor), add a MOV or something similar that can start conducting faster before the mechanical switch is able to make contact. $\endgroup$ Mar 2 '18 at 16:59
  • 2
    $\begingroup$ At this point, you're adding more complexity just to avoid using parallel circuits and using constant volts. It wouldn't take very long before the engineers (and electricians) on your planet realized there was a far simpler solution. $\endgroup$ Mar 2 '18 at 17:45

You're unlikely to be able to produce enough voltage.

Every device you connect in series is another load. Each load dissipates some power, since the whole thing is in series, that means each load represents a voltage drop. In order to drive 120 amps through a nation's worth of appliances in series you would need millions and millions of volts. You would quickly reach the dielectric breakdown strength for all know materials. The power grid would arc across a mile of air rather than go through all those appliances.

Even two appliances in your own home could be at dangerously different voltages. A built in problem like ungrounded AC, but without an easy solution.

I'm guessing your assumption of the symmetries comes from the Norton/Thévenin equivalents. But those are only truly equivalent on paper and even then only in specific cases. If you actually build a Norton circuit it will be warmer than its Thévenin equivalent, because its resistor is constantly dissipating power.

This demonstrates the other problem you'll have. Say I have a 1200W clothes dryer and a 12W coffee cup heater. The dryer will drop 10V, at 120 amps to generate its rated output. To make sure my coffee cup heater doesn't burn up in series is to make the resistive element 0.0008 Ohms (yes, 800 micro-ohms, about 1 foot of 9 gauge copper wire).

Perhaps you just mean Ohm's law for symmetries. It's worth noting that just because there is an equation that relates two phenomena, like voltage and current (assuming a fixed resistance), they are very different animals. Even just describing what each is makes that obvious. Voltage is the electric potential difference between two points while current is a quantity of charge passing through a cross section of a volume in a given time. There is no reason to think that you can exchange one quantity for the other in a physical system and expect nothing to change.

  • 1
    $\begingroup$ That's a really good point, and something that I completely forgot to include in my answer. The only way I could see it being anything remotely close to workable is if there were individual generators and circuits for every few houses, which would be pretty inefficient and impractical. $\endgroup$ Mar 1 '18 at 13:51
  • 1
    $\begingroup$ For those who are not familiar with gauges of wire, 9 gauge wire is commonly used in chain link fences. That should help give a sense of scale. $\endgroup$
    – Cort Ammon
    Mar 1 '18 at 20:38

Can't be done once you get a significant number of devices hooked up, and here "significant" means "really low".

Forget the grid for the moment, and just consider your house. If all the electrical devices, everything in the house has to be in series, what happens to all the food in your refrigerator when someone forgets to turn off a light before going to work and the bulb burns out? And if you live in a colder climate, that's also the power to your heating system, which in winter means the possibility of a frozen home, burst pipes, and all other sorts of joys.

Now moving up a little, what happens when your neighbour has a bulb burn out, or someone a few hundred kilometers away? I don't see many customers for a system where someone in Chicago can suffer extensive damage to their home because someone in Detroit had a burned out light bulb.

What's more, how do you identify where the problem is? In the system we have now, if a grid goes down, say due to a storm, you can bring parts of it up and the parts that don't come up isolate where the remaining problems are. But with a system of thousands, or millions, of devices, how do you find what the single one causing the grid to go down because it broke the circuit is?


One issue separate from all of the above is how does such a system develop in the first place? Even if people can develop super-fast (on the order of nanosecond) switches that's not going to happen--barring Alien Space Bats--for quite some time long after people have figured out parallel and series electrical systems and using them. Without those super-fast switches, parallel systems using constant voltage are so much easier to deal with for all the reasons given in other answers that there's going to be so much infrastructure in place that there's going to be no advantage to switching to a more complicated system with greater points of failure requiring more complicated and expensive wiring for no benefit.

  • $\begingroup$ As I've pointed out in a number of other places, this open-circuit fault condition is dealt with in the same way that our electrical grid deals with short circuits. $\endgroup$ Mar 1 '18 at 16:04
  • $\begingroup$ @AJMansfield: The problem is that there isn't any way of dealing with an open circuit and still maintaining your series grid condition. Indeed, there really isn't any way of turning anything off. (AFAIK, anyway: you might find more expertise if youask on the electronics site) $\endgroup$
    – jamesqf
    Mar 1 '18 at 18:31
  • $\begingroup$ Except that with the ridiculously high voltages necessary in order to keep constant current, your switches and fuses would need to huge in order to provide sufficient distance that the current doesn't simply arc across the contacts when the switch/fuse is open (ie, when everything is supposed to be "normal"). $\endgroup$ Mar 1 '18 at 18:31
  • $\begingroup$ @KeithMorrison A switch would only need to be able to provide as much isolation as would be needed to make the next breaker up the chain fire, the same way that in our power grid a regular lightswitch on a 15 amp circuit here only needs to be able to interrupt 15 amps (not counting safety factors, lightning protection, etc). $\endgroup$ Mar 1 '18 at 21:54
  • 1
    $\begingroup$ "I unplugged my coffee maker and suddenly there was a nation wide blackout the outlet failed and didn't complete the circuit." $\endgroup$ Mar 2 '18 at 1:13

Won't work...

Losses inside a conductor are proportional to R * I^2.

Therefore, all other things being equal (insulation melting temperature, etc) the conductor cross-section which determines R has to be inversely proportional to I^2.

Problem #1: If your house had a "120A" circuit instead of "120V" then all wires would have to be rated for 120A, which would make them impractically large and expensive. In fact, everything would have to be rated for this, even the switches, lampcord, etc.

Here (France) the average wall socket allows 230V/16A (ie, 3500W nominal). So let's keep the wires rated at 16A, and the constant current supply at 16A too.

If the house draws 10kW maximum, then it will have 625V across its supply. Manageable.

However, it is in series with the neighbor's houses unless the utility provides a transformer winding for each subscriber. Since voltages add up, the wiring will be at potentially several kV relative to Protection Earth. Therefore:

Problem #2: Unsafe voltages.

Problem #3: Three phase induction motors would be a problem. This means no industry.

Problem #4: The efficiency for low loads would be catastrophic. Say a cellphone charger, drawing 5W = 0.32V @ 16A. The losses in wiring would no longer be insignificant relative to delivered power.

  • $\begingroup$ With #3, I'm not sure three-phase would be completely out, although it would probably be much more cumbersome. A motor could be wired with three independent coils rather than using a star point or sharing phases. The plug and socket would have to be a six-pole connector that connects each coil into a completely separate circuit. $\endgroup$ Mar 1 '18 at 19:24
  • $\begingroup$ When supplied from fixed voltage a motor draws current according to torque. So using an AC current source as supply would limit torque, I wonder even if an induction motor would work at all. $\endgroup$
    – bobflux
    Mar 1 '18 at 19:38

Is there somewhere that I'm not thinking of where this symmetry breaks down, or reasons that a mains power electrical grid couldn't work like this?

Not only does your design work, we use its basis all the time.

I am not going to give a full physics class here, you can find the whole thing easily if you have the disposition. The short version of it is that we alternate the voltage in order to alternate the current. You can't dissociate one from another, because V = ri. Invert the voltage and you invert the current, and vice-versa. The way the grid is wired does not change that.

Another way to see this relationship is by looking at this graphic from Wikipedia:

types of current

Notice how voltage and current are represented by the same axis in all cases.

Last but not least, a curiosity: look at any appliance power data label and you'll see a frequency in the data, practically always either 50 or 60 hz. That's how fast the appliance expects the voltage to alternate for it to work.

  • 3
    $\begingroup$ I know this type of system is used e.g. for powering LED lighting systems. But is this actually used at the scale of the entire electrical grid for an area, with everything in series rather than in parallel? The symmetry I'm interested in isn't so much the voltage-current one as the series-parallel one. $\endgroup$ Mar 1 '18 at 4:16
  • 1
    $\begingroup$ Also, alternating vs direct current is irrelevant here, the point is that instead of acting as a voltage source, the mains grid is designed to act as a current source, with devices operating at the complete opposite side on the load line compared to how they do here. $\endgroup$ Mar 1 '18 at 5:31

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.