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The more amperage a given appliance or device draws on a circuit, the thicker the wire needs to be, correct?

And this is to prevent the wire itself from becoming a heating element.

I understand that a 15A circuit needs a specific minimum gauge of wire, but how does the length of that gauge wire affect the situation?

If the wire is longer, despite the amperage rating of the circuit, would the wire need to be larger gauge?

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  • \$\begingroup\$ Yes. Typically, less than 3% voltage drop to wire is used. \$\endgroup\$ – StainlessSteelRat Dec 14 '20 at 22:40
  • \$\begingroup\$ The gauge is defining roughly the cross-section area of the cable, which affects its resistance per unit length - and will affect how much it is heating. The length of the cable will affect its overall resistance, and therefore the voltage drop on it - but won't affect the heating per unit length too much. There is no strict rule for required gauge vs current. It is depending on the application specifics. \$\endgroup\$ – Eugene Sh. Dec 14 '20 at 22:41
  • \$\begingroup\$ 1. yes, depending on what you mean by "needs". There's two problems (heat, voltage loss) that both stem from R, or resistance, which is the answer to the title question. 2. R/distance; longer wire = more resistance, thicker wire = less resistance, you have to balance it to keep R acceptably low for the application. 3. maybe. If 100 cubits of wire presented 10 ohms, and 5 ohms is your limit, you need thicker wire than you would to run 25 cubits. It's not just heat, as a longer wire has more surface area to dump out power, limiting the problem imposed by that aspect. \$\endgroup\$ – dandavis Dec 15 '20 at 0:05
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A wire that will carry 15A without overheating, will carry 15A whether it is 1 meter long, or 1 km long. What will change is the resistance, and resistance is directly related to voltage drop.

If you are powering a 120V load that draws, say 10A., if your wire has a resistance of 0.100\$\Omega\$ (half that in each direction), then there will be a voltage drop of 0.100\$\Omega\$ * 10A = 1V. Your load will not actually see a 120V supply, but a 120-1=119V supply.

Now, if your wire were the same gauge, but 10 times as long, there would be a voltage drop of 10V. Your load would see a 110V supply. Longer still, and you get the idea.

You may wonder what happens if the wire is 120 times as long. Will there be a voltage drop of 120V, and no voltage appearing at the load?

Only if the load acts like a short circuit when there is no voltage applied to it.

As the voltage on the load changes, the current through the load and the wire will probably also change. Some loads are resistive, like incandescent light bulbs. As the voltage across them drops, the current through them will also drop. Other loads behave differently. Some loads will attempt to compensate for a low supply voltage by drawing more current. An example would be most PCs. In such a case, a longer power cord, could cause the PC to draw more current. (Thanks to Bruce Abbott and JonRB for drawing attention to that phenomenon.)

I suppose in a worst case scenario, the current drawn by a load with a very long power cord could be above that for which the cord is rated. However, this would only happen if a fuse or circuit breaker is not present which is properly rated for the current carrying capacity of the wire.

To counteract the voltage drop of long wires, you can use a thicker gauge. That will decrease the resistance and voltage drop. But that is a separate issue from the ampacity or current carrying capacity of the wire. The current carrying capacity of a wire does not change with its length.

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    \$\begingroup\$ Note that the maximum allowable current (for RHW insulation, for example) doesn't scale with changes in cross-sectional area of the wire. For example, 4 gauge is rated at 85 A. But 000 gauge (which has more than 4 times the area) is limited to 200 A, barely more than twice the current. This has to do with the ratio of radiating surface area vs volume as it scales up and down over gauge. \$\endgroup\$ – jonk Dec 14 '20 at 23:33
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    \$\begingroup\$ "as the voltage drops, your device will probably pull less current" - probably, but not necessarily. If the device's output load is constant then it may draw more current at lower input voltage. This includes speed controlled motors and electronic equipment such as battery chargers, computers etc. Even a heater with thermostat control will draw more current on average. \$\endgroup\$ – Bruce Abbott Dec 15 '20 at 6:09
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    \$\begingroup\$ Note: if the load is a constant power load (computer, motor drive....) A drop in terminal voltage due to cable length will actually result in higher current being drawn and more volts drop - it's a quadratic to balance where it settles. Resistive/linear loads (heaters, cookers etc) will draw lower current for lower terminal voltage \$\endgroup\$ – JonRB Dec 15 '20 at 7:51
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    \$\begingroup\$ "The current carrying capacity of a wire does not change with its length": Correct as long as you assume it's not rolled up (like on a cable drum, or inside a vacuum cleaner, or the pile of cable under your workbench). \$\endgroup\$ – Peter - Reinstate Monica Dec 15 '20 at 9:57
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Yes. The longer the distance, the lower the the gauge (larger diameter), to maintain the service voltage at the point of load. If you are following electrical code guidelines, 3% is the magic number.

The National Electrical Code (NEC) recommends a maximum voltage drop of 3 percent for individual household circuits (known as branch circuits).

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Or use this voltage drop calculator:

enter image description here

Note the maximum current is set to 15A in this calculation, but for continuous loads it should be derated to 50%, and for unknown loads the budget should not exceed 80%. Both these cases will give you longer maximum lengths.

https://www.calculator.net/voltage-drop-calculator.html?material=copper&wiresize=8.286&voltage=120&phase=ac&noofconductor=1&distance=50&distanceunit=feet&amperes=15&x=55&y=23

https://www.thespruce.com/wire-size-underground-circuit-cable-length-1152899

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There is a lot to this question.

The first concept is ampacity. This is a thermal limit based on assumptions. In order to avoid having the wire get too hot for the insulation, you have to use a wire with the proper ampacity. Length doesn't matter. Insulation temperature rating of the insulation does matter. Also the details of how the wire is routed (in conduit or behind insulation, etc) matter.

https://www.cerrowire.com/products/resources/tables-calculators/ampacity-charts/

You should never use wire above its ampacity rating unless you know what you are doing. The melting point of copper is pretty high. So you may be able to run a lot of current through bare copper wire, or wire with high temperature insulation. I don't think there are tables for that. You have to do your own calculations or test and verify.

The second concern is voltage drop. This only depends on wire cross-section area and length. Longer length means more resistance and more voltage drop. small cross section also means more resistance and more voltage drop.

There are tables for this too. You can also calculate it if you know the resistance (Ohms per foot or Ohms per meter).

https://www.cerrowire.com/products/resources/tables-calculators/voltage-drop-tables/

One more thing. The resistance of copper (and almost all metals) increases with temperature. The increase is about 4 percent every 10 degrees C for copper. So if you ever do run copper hot, keep that in mind.

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I'm gonna ignore the description and simply answer the title:

  • Relationship between the length of a wire and it's AWG/amperage rating: none.
  • Relationship between AWG and amperage rating: basically, AWG is linked to wire diameter/cross-area. The higher is the cross-area, the higher is the ampering rating. You can find charts that tell you what is the max current a wire can carry based on its amperage, for exemple this one.
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