So far I've found that the thickness of the wire influences the resistance. A thicker wire has less resistance. Also, the length of a wire has a linear correlation with the resistance. 2x length = 2x resistance. So your wire resistance will be length / thickness, but that doesn't explain why the thickness should increase with length. What am I missing?


I can think of two consequences of what I've explained above: first, with a high wire resistance, your voltage will drop and you might not get the voltage you need. Second, having a thin wire with lots of amps will potentially melt the isolation and be a hazard.

So let's say we would accept any voltage loss and just don't want a hazardous situation. We take a wire of 1m. Then we run a certain amount of current though it which will produce heat in the wire. Now we double the wire length to 2m, so double the resistance. We run the same current through it which will produce double the amount of heat, but over double the amount of length. Unless you're coiling your wire I don't see how it should make for a more hazardous situation.

And the other way around, according to this calculator http://www.solar-wind.co.uk/cable-sizing-DC-cables.html if you have a 1cm wire and running 100 amps @ 12 volt you can do with 0.5 mm2. How is that possible?

  • 3
    \$\begingroup\$ You say you don't care about the voltage drop, but most people do. That's the explanation for all of your questions. \$\endgroup\$
    – Dave Tweed
    Apr 7, 2014 at 14:15
  • \$\begingroup\$ I thought back about this problem and realized the problem really is with non-linear loads. A light bulb requiring X watts will have lower resistance if the voltage across it isn't ideal. The lower resistance allows higher current. To summarize: longer wire = higher resistance = lower voltage at load = lower resistance load = more current through wires = greater chance of starting fire. If you have only linear loads, then your assumption is correct: longer wires with the same or less current = no increase in fire danger. \$\endgroup\$
    – horta
    Jun 5, 2014 at 20:27
  • \$\begingroup\$ ".. according to this calculator if you have a 1cm wire and running 100 amps @ 12 volt you can do with 0.5 mm2. How is that possible?" - It's not; the wire would melt. (a 1cm 0.5mm2 copper wire has a resistance of about 84µΩ; at 100A it would dissipate 0.84W, far too much for such a small wire) \$\endgroup\$
    – marcelm
    Aug 20, 2016 at 18:14

2 Answers 2


The wire gauge you need is a function of several things.

  • The acceptable voltage drop or power loss (that appears to be the only thing considered in the website you linked). The voltage drop (and power loss) is proportional to wire length and inversely proportional to the cross-sectional area of the wire- in other words inversely proportional to the square of the wire diameter (assuming constant current).
  • The acceptable temperature rise. This is a function of the number of current-carrying wires bundled together, the environment (maximum ambient temperature and air pressure or altitude, for example), the insulation type, the wire type (some types of wire are plated to withstand higher temperatures than bare copper without corroding).
  • Regulatory requirements and other considerations- for example, the wire may be rated for 200°C insulation, but you might not want the wire to run that hot.
  • Fusing- the fuse or circuit breaker should protect the wire in the case of faults such as overload or short circuit.

Very short lengths of wire can depend on heat sinking through the ends (indeed, in a vacuum, that may be the main heat loss mechanism), but usually that's not taken into account.

Normally you'd run through a checklist such as the above to make sure ALL the requirements are satisfied simultaneously, so you might find that using PTFE insulated wire allows you to use AWG 18 wire, but because of the voltage drop limitation you'll have to use AWG 12 wire.

  • \$\begingroup\$ Reading your answer and looking again at the calculator I linked, I think it only calculated voltage loss and you would have to then figure out if the resulting wire doesn't heat up to much. \$\endgroup\$ Apr 7, 2014 at 14:58

I think you misunderstand the relation. It's not length * thickness = resistance. It's more like length / thickness = resistance. If your length gets longer you need the thickness to go up proportionally if you want the resistance to stay the same.

Read up on it here and you'll understand that it's the length/(the cross-sectional area) is proportional to resistance. http://en.wikipedia.org/wiki/Electrical_resistivity_and_conductivity

You can think of it like a pipe with a thick fluid. The longer the pipe, the more resistance to movement you have. The larger the pipe, the less resistance to movement you have.

To answer your last question, 100 amps at 12V is possible because the wire length is so short. A short wire length results is a very low resistance.

  • \$\begingroup\$ Thanks for correcting the relationship. I wrote it down incorrectly, but sshouldn't affect the reast of my reasoning. \$\endgroup\$ Apr 7, 2014 at 14:20
  • \$\begingroup\$ Then yes, your reasoning would be correct. Heat per unit length won't increase. It won't get any hotter. The problem then is the voltage drop which would affect the performance of whatever is plugged into it. \$\endgroup\$
    – horta
    Apr 7, 2014 at 14:28
  • \$\begingroup\$ Also, regarding the 100 amps @ 12volt. let's say my wire is 0.3ohm/meter and we're running 100 amps through it. Then the amount of heat per cm of wire would be 30 watts/cm which looks like a lot to me. \$\endgroup\$ Apr 7, 2014 at 14:35
  • \$\begingroup\$ I'm afraid it's 30watt, you forgot to square in I2 * R = P. \$\endgroup\$ Apr 7, 2014 at 15:05
  • \$\begingroup\$ You're right, it's 30 watts out of 1 cm of wire. That's going to get toasty. That calculator is pretty messed up. \$\endgroup\$
    – horta
    Apr 7, 2014 at 15:22

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