Pretend you had a 120v load, where the hot (ungrounded conductor) was say only 5 feet in length from the source to the load, but the neutral was say, 1000 feet. Excluding things such as voltage drop (we use a thick enough wire, or whatever), what would happen when you turn on that light bulb? Would it pause for a fraction of a second while the electricity passed through the entire circuit traveling on the neutral?

Would it matter if it was a 240v load (ie. a motor). Would it the hurt the motor, pause before turning on, or maybe the motor would turn on despite one phase (say 'ungrounded conductor #1') being 0.1 second behind since the wire was so much longer than the other?

  • 2
    \$\begingroup\$ The speed of light delay for a signal to travel 1000 ft (~300 m) is about 1 us. Compare that to the time it takes a light bulb filament to heat up enough to glow. \$\endgroup\$ – The Photon Apr 24 '17 at 16:43
  • \$\begingroup\$ I'm more interested in what will happen when neutral and hot don't 'meet' at that load at the same time. So, if pretending that it's 100 miles long for the neutral (pretend there's no voltage drop), and 1 inch for the hot, that should create enough disparity to have a difference, even if fractional. And also allows for the discussion of what happens when those 2 parts of the circuit aren't even and whether things operate or not, etc. \$\endgroup\$ – Tony DiNitto Apr 24 '17 at 16:55
  • 2
    \$\begingroup\$ @TonyDiNitto For purposes of this musing, consider that the "power" only flows from the "hot" wire. Neutral wires are connected to earth ground anywise, so it's not as inaccurate an assumption as you might think. - With US 110/120V hot/neutral wiring, the electronig lengths of hot/neutral are completely immaterial to function (not counting wire resistance & EM radiation losses), because there's no phasing to match; only a 110-120V AC source & GND. ... \$\endgroup\$ – Robherc KV5ROB Apr 24 '17 at 17:23
  • 1
    \$\begingroup\$ ... On the other hand, with 220/240V "Split Phase" US wiring, you have 2 110-q20V "legs" that are both AC sources & 180 degrees out-of-phase with each other. As you electrical lengths there become different (on a scale of miles more than feet), the phase-delay on the longer wire means that the two waveforms will be less than 180 degrees out of phase at most lengths, and even 0 degrees at specific lengths. At 0 degrees phase difference there would be no power transferred, even if you shorted the two wires together. \$\endgroup\$ – Robherc KV5ROB Apr 24 '17 at 17:27
  • \$\begingroup\$ Try looking up reflection (or bounce or lattice) diagrams. \$\endgroup\$ – Andy aka Apr 24 '17 at 18:01

In a 2 wire system the only thing you can do is make the total wire so long that it either acts inductively or resistively against your system. In the case of resistive it makes absolutely no difference where the wire is and in the case of inductive the total length or discrepancy therein is of much smaller interest than the path those wires take. For there to be a mile-long difference, you're likely coiling one up, which is much more interesting in that respect.

For three phase systems that directly use the phase synchronicity, such as fixed armature three phase motors, you can disrupt it all if you can delay one of the phases by enough.

Again, however, the speed of light being 3*10^8 m/s and the speed of electrons electricity [credit:Andrew] in free hanging copper being approximately 95% of that, you would need a huge deal of length to get enough displacement, say 10%-ish (rough guesstimate - too lazy to really think heavily on that number) of the 60Hz wave (or in my case 50Hz), to start noticing it. 10% of a 60Hz wave would take about 1.68μs 1.68ms [credit: Jon].

In 1.68μs 1.68ms the wave in copper would have travelled (approximate speed of electricity in copper) * (time) = (3*10^8 * 0.95)m/s * 1.68ms = 478m 478km.

Seems feasible, right? quite ridiculous already...

Well, to go on... Let's, for the ease of it, assume that my 10% was a correct guesstimate and it isn't 30%. Let's also assume that the other wires are all connected directly to the source and thus have no losses at all (good luck with that one), then still:

A 3 phase system usually needs at least 10A per phase (and that's tiny!), if you'd want to limit the total resistive losses to 5%, you'd need a resistance smaller than, if we assume an RMS voltage of 240V for your 120VAC 60Hz situation, of less than 12V/10A = 1.2Ohm. And losing 12V of your RMS is quite a lot to power engineers! To get 1.2Ohm at 478km of wire, and I won't throw around another formula, as when looking for the exact resistivity of copper I found this, which shows me for 47897000cm (= the value I got a bit more exactly, which I rounded to 478km) I'd need a wire diameter of 92.335mm. That's an area of (0.92335 / 2)^2 * pi = 0.67 dm2. I'm using dm, or 1/10th of a meter, for a reason. 47897000cm = 4789700dm. So the volume of copper you need is 0.67 dm2 * 4789700dm ~= 3207235 dm3.
1dm3 is the same as 1 liter. So that's 3207235 liters of copper. Copper weighs about 8.93kg per liter, so that's about 28640616 kg of copper. At currently about €5.30 per kg, that comes to € 152 million!! With only about €10 you can buy an inductor to screw things up just as bad.

Which brings me to the next problem. 478m of copper wire, diameter 2.9201mm, laid out in a perfect loop, without coiling it, from a nice site with a calculator which I did not verify as I am still lazy, is close to 1mH. So in the most efficient way you could place that wire you already get 1mH, assuming the calculator is right. Placed on a normal type of floor which is sort-of-grounded I suspect it to be higher. This may, depending on the motor, already be enough of an inductor to put things out of whack. But if you then want to put that wire in an orientation so that you don't need a HUGE amount of space to make a perfect circle, the inductance will increase significantly.

The calculator stops working at about 5km, but with the numbers I entered I'd say it's safe to say we get in the range of actual Henris, not mili or micro.

At the time of writing €1 is approximately $1.08

Note: I made a mistake in my simplification steps to get 5% of RMS 240V in a three phase means 1.2 Ohm, but since it's all blue-sky I'm not going to re-do it.

Another note: Just calculated "back-of-envelope"-style, because my brain wouldn't let it go, that if you'd want the wire to be coiled such that it could only just stand up in my living room you'd need 77 windings, making an inductor of 74mH, assuming no metals interfere. (And it's a largely concrete living room). you'd need many of my living rooms for the immense length the coil would take, making that a futile exercise.

  • \$\begingroup\$ I've heard of people almost starting their house on fire by hanging a coiled 100ft cord on a door knob, the knob was almost glowing. \$\endgroup\$ – laptop2d Apr 24 '17 at 18:11
  • 2
    \$\begingroup\$ @laptop2d if you can bring up the patience to coil nearly 500m of 2,9mm diameter wire tight enough to couple significantly into a doorknob without at any time wondering if that might be a good idea, I'm going to call that natural selection. \$\endgroup\$ – Asmyldof Apr 24 '17 at 18:14
  • 1
    \$\begingroup\$ Can you cite a source for "the speed of electrons in free hanging copper"? I was under the impression that the speed of electricity is quite fast, but the underlying velocity of the free electrons is about 0.1 mm/s. \$\endgroup\$ – Andrew Morton Apr 24 '17 at 18:25
  • \$\begingroup\$ 10% of a 60 Hz wave is 1.67 milliseconds, not microseconds. Thus, your calculations of cost go up by a factor of 1000. \$\endgroup\$ – Jon Watte Apr 24 '17 at 18:30
  • \$\begingroup\$ @AndrewMorton you are correct, but it doesn't make a difference to this scenario. \$\endgroup\$ – BeB00 Apr 24 '17 at 18:30

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.