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Online, most household wiring guides suggest 14 gauge wire for 15 amp circuits, 12 for 20 amps, and 10 for 30 amps. Do these assume the voltage is 120, or does voltage not come into play when choosing wire gauge?

For example, I have an appliance that requires a 240 volt 15 amp power source. (Its plug is NEMA 6-15.) If I dedicate a circuit to this appliance, which is about 25' from the panel, will 14 AWG be sufficient? Why or why not?

When 14 gauge wire is appropriate, I've seen 12 used instead. (My home has several circuits like this.) I realize 12 AWG wire may be more difficult to work with, and it costs more than 14 AWG. Beyond these, are there reasons not to do so? Are there any advantages?

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    The voltage rating of wire is mainly dependent on the type of insulation around it. The risk of voltage is of a spark jumping to another conductor, and the insulation is what prevents that. Most household wire in the US is rated to 600V; it should be labeled on the side.
    – Hank
    Oct 29, 2014 at 14:41
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    Keep in mind the difference between peak and steady-state current (or power) draw. If your appliance draws 3600 W (roughly) all the time, you'll need heavier wire than if the steady-state is more like, say, 3 amps. As always, check your local building codes.
    – Carl Witthoft
    Oct 29, 2014 at 16:42
  • In addition too needing a wire large enough in diameter (of the copper) that it doesn't overheat and catch fire/melt due to the current, the wire needs to be large enough to prevent significant "voltage drop". Over about 50 feet wire size needs to be increased to minimize voltage drop. But the higher the supply voltage the less this increase needs to be.
    – Hot Licks
    Oct 29, 2014 at 18:38
  • (When you get into very high voltages -- tens of thousands of volts -- the wire diameter becomes important because a smaller wire results in more corona discharge.)
    – Hot Licks
    Oct 30, 2014 at 2:32
  • Just a few observations: 1) Usually 240v appliances specify a wire size in the installation instructions. 2) The difference in working with 12- and 14-gauge wire is negligible in most cases, especially a short run like 25'. 14 is slightly easier to connect devices to, but not that much easier - usually price is more of a concern than 10% more effort to bend the wire. 3) If the device specifies a 15 amp circuit, it probably is assuming a dedicated circuit. If panel space is an issue consider a double breaker; if you are short on panel capacity you need professional advice on what can be done.
    – brichins
    Oct 30, 2014 at 17:25

6 Answers 6

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The simple answer is you use the wire that your local code and the device manufacturer specifies for the situation.

But I think you are asking about how wire sizes are chosen for a particular application/current/voltage combination:

When choosing wire, current dictates the size of the conductor and voltage dictates the insulation.

Current causes the wire to heat up due to resistance. Metal expands and contracts when heated and cooled. This expansion and contraction, if too large, can loosen connections. Loose connections increase resistance, cause more heating, and will eventually allow a gap large enough to cause an arc or a high enough temperature to ignite surrounding materials or melt the insulation and cause an arc. A larger conductor reduces the resistance, which reduces these temperature changes. Therefore, using a large enough conductor keeps the expansion and contraction under the level that electrical fittings can tolerate without failing.

Similarly, voltage causes arcing so the covering (insulation) has to be designed to prevent arcing at the rated voltage. Usually you'll see common electrical wiring rated for 600 volts.

Another reason to use larger gauge wiring is to prevent voltage drop on long runs. This is generally not a problem when running wire in a house, but detached structures are a common place to see larger wire used.

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    It may be worth mentioning that the practical significance of a particular amount of voltage drop may depend substantially upon the starting voltage, especially when wiring is used for low voltage. For example, if one were using a 200-foot cable to power a 240V device that required 10 amps, the cable would drop about 10 volts, leaving 230 for the device. If instead the supply was 12 volts, having the cable drop ten volts would leave only two volts available for the load.
    – supercat
    Oct 29, 2014 at 17:47
  • @supercat - Correct. In effect, a higher voltage permits a thinner wire, for a given current. (But only where a long run would require that the wire be oversized in the first place.)
    – Hot Licks
    Oct 29, 2014 at 18:28
  • @supercat Which is why long distance voltage transport makes use of extra high voltages (up to several hundred kV).
    – Mast
    Oct 30, 2014 at 13:10
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    Add I said above, current defines the size of the conductor. To oversimplify, voltage can be easily converted to current, and vice versa. So by increasing the voltage, you can reduce the conductor size. And reducing the conductor size makes the wires cheaper.
    – longneck
    Oct 30, 2014 at 13:21
  • So, if I were to sum up your answer, you're in favor of using bigger conductors, true? ;-) It does seem likely that the instructions with the OP's 3,600 Watt appliance (whatever it is--it wasn't specified) indicate what size wire should be used with the appliance?
    – Craig Tullis
    Oct 4, 2015 at 7:42
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As others have stated, the voltage of the appliance/circuit has no bearing on the size (gauge) of the wire. Voltage dictates the quality of the insulation of a wire and most (power) wire we encounter will be rated for 600 Volts.

The gauge should be primarily selected by determining the current draw - in Amperes - of all the devices to be connected to the circuit...

Given 

Amps = Watts / Volts

Then

TotalAmps = [Device1(Watts) + Device2(Watts) + Device3(Watts)] / CircuitVolts

Or

TotalAmps = Device1(Amps) + Device2(Amps) + [Device3(Watts)/CircuitVoltage]

... and then referencing a chart which can be traced back to the NEC recommendations. Herein lies the problem. There is no one official NEC "if X then Y" chart for all situations. The actual NEC charts are for engineers/contractors to reference when designing for an application and are not very easy reads. Here is what the NEC has to say: http://www.fs.fed.us/database/acad/elec/greenbook/3_basicdesigns.pdf

Fun, right? What we normies have to do is rely on charts that interpret those recommendations and those charts vary - sometimes wildly - in easy of readabilty. Compare my favorite chart http://www.cerrowire.com/ampacity-charts to this one http://www.usawire-cable.com/pdfs/nec%20ampacities.pdf They are both technically accurate from a rule of thumb basis but the latter requires more in depth evaluation such as Note4 which indicates a derating of the wire's maximum ampacity if the conduit fill (number of wires in the raceway/cable) is greater than 3.

Amperes is not the only factor for wire size, but we are working with rule of thumb here. The other MAJOR factors that contribute to selection are (A) the wiring installation application type (THHM, UF, etc...) and temp ratings, (B) the length of the circuit feeder which increases resistance, voltage losses and ultimately, unacceptable heating of the feeder wire and notably it's connections, (C) single- vs multi-phase applications (we are only concerned with single in household systems), (D) whether the load is inductive or not (big motor/compressor in the appliance?) and a couple of other more obscure factors we won't address here.

Item (A) in residential applications is typically NM/NMC class wiring for Romex-style, THWM for BX or conduit-style wiring and UF for cable buried in your yard. Item (B) is actually quite important. If the wiring run is very long, the resistance of the wire (all wire is resistive to a degree) and thusly the temperature of the wire will increase. If that temperature rises above a wire's insulation rating, it could melt causing a short or worst case, start a fire in the surrounding building materials. This is where my second favorite chart comes in: http://www.cerrowire.com/voltage-drop-table EDIT: longnecks's top-rated answer above is a better explanation of temperature's effects on circuits especially regarding the wire/fixture interface where most fires begin.

Knowing what we do now after referencing those two charts from Cerro we can answer:

If I dedicate a circuit to this appliance, the length of which is about 50' (including the return), will 14 AWG be sufficient? Why or why not?

with YES because you indicated that the device will be the only one on the circuit and because the run is actually 25' by the definition of the rules which do not calculate the total length of WIRE, rather the length of the CIRCUIT which is comprised of both conductors. In 240 land, there is no Return or Neutral. This allows 240 circuits to often use a gauge that would seem to be too small! In 120 land, the neutral of a given circuit is allowed to be (and nearly always) shared amongst the numerous branches of said circuit which introduces some derating. But mostly because circuits of a higher voltage introduce less voltage drop then an equivalent circuit at a lower voltage. E = R * I ... where E = voltage drop (volts, V); R = electrical resistance (ohms, Ω); I = current (amps, A) This is not intuitive because the supply voltage is not used in the calculation. However, if you have two loads which are both rated at 2400 Watts, one of which runs at 120V and another at 240V, the former will draw 20 amps, the latter 10. Half the current draw will introduce only half the voltage drop, reducing that element of the calculation for a wire's guage.

It should be noted, that the answer would still be "YES, 14awg will do" if the run were actually 50' according to the Cerro charts....BUT just on the edge. After browsing a few other charts that are popular, some indicate 12awg, others 14awg. YMMV. That's why we have the really in depth NEC findings to fall back on and take into account EVERY factor.

As for:

I realize 12 AWG wire may be more difficult to work with, and it costs more than 14 AWG. Beyond these, are there reasons not to do so? Are there any advantages?

The answer is a judgement call for the contractor/homeowner. Take this example: I'm running a new 240 circuit for a new window air conditioner. The unit I can fit in the window opening can be handled by a 14awg/15amp circuit BUT is right near the maximum rating. Suppose the unit is barely able to meet my cooling needs and suddenly, the market introduces a higher BTU output unit that fits in the opening but it's going to require a 12awg/20amp circuit. This would be a future-proofing judgement call.

And remember the most important thing: Your local building codes supercede NEC's. If it's your property, the work you do along the way may impact your ability to sell the property down the road.

Hope I've answered all your questions. Disclaimer: I do not work for Cerro cable, just a tired old HVAC/R pro that deals with a lot of crappy wiring, residential and commercial. And the links are munged because this site only allows two links for noobs.

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The National Electrical Code (NEC) determines the required minimum size for conductors. Under the NEC three broad categories cover most installations: low voltage, less than 600 volts, more than 600 volts.

Keep in mind that code requirements specify the worst legally allowable construction. One common reason to increase conductor size above code minimums is to reduce voltage drop by reducing conductor resistance.

Seasoned professionals often exceed code based on judgement built on years of experience.

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  • ...and another does not answer the question.
    – Ecnerwal
    Oct 29, 2014 at 15:24
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    @Ecnerwal So answer the question ...
    – bib
    Oct 29, 2014 at 15:43
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    @Ecnerwal Not quite. What is the answer to the question about the specific 240v/15a/14awg 50 ft run?
    – bib
    Oct 29, 2014 at 15:57
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    @Ecnerwal longneck's answer doesn't solve an actual construction problem. It's textbook chat about electricity. A person is no closer to wiring their house after reading it. Even worse, there is nothing that points out that conductor sizes come from tables, not from electrical theory. Steering people toward model codes, particularly in regard to life safety issues is the useful answer. Implying, even by ommission that there is another prudent course is either made irresponsibly or from lack of knowledge.
    – user23752
    Oct 29, 2014 at 16:14
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    @CarlWitthoft In my case, I would be rendering a professional opinion. Doing so in this case would not be consistent with an ordinary standard of care due to the lack of relevant details, first hand knowledge or reliable second hand knowledge from a person generally qualified to do electrical work or design of the actual field conditions, and because I do not have a contractual relationship (this is minor from a personal standpoint but insurance companies care). From experience on both sides of the building department counter, the answer is "it depends on what the code says".
    – user23752
    Oct 29, 2014 at 17:12
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Yes and no (or is it no and yes?)

The minimum wire gauge at household and light commercial voltages (less than 600V) indeed does not depend on the voltage -- the first entry in NEC table 310.106(A) specifies that 14AWG copper or 12AWG aluminum is usable all the way up to 2000V when suitably insulated.

However, in high-voltage work (upwards of 2kV), the wire must be upsized as per table 310.106(A). However, this only matters in heavy commercial and industrial systems where high-voltage feeders are used to avoid excessive losses, as well as the occasional load that is powered by high voltage (such as an electrode-type industrial steam boiler, or an extremely large motor).

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For AC power wiring, you just follow your electrical codes.

Here's my cheat sheet for NEC.

enter image description here

This is just the standard NEC table with the "no go zones" grayed out.

Note that the green zone (aluminum NM and UF) is found in old work, and apparently is starting to be sold again, go figure. Someone makes a Copper Clad Aluminum NM cable. Wish they would offer it in SER instead, so we could use the higher amp rating.

If your wire and all your terminals are rated for 75C thermal rating, you can use the 75C column (barring 240.4(D)), otherwise you use the 60C column.

Ampacity doesn't care about voltage

Heat rise (which is what all this is about) is caused by voltage drop. Let's consider a 12 volt installation running 10 amps. The wire is quite long and has 0.1 ohm resistance. Ohm's law says voltage drop will be

 Vdrop = I (current) x R (resistance) 
 Vdrop = 10 amps     x 0.1 ohms 
 Vdrop = 1 volt

How much heat is that in watts?

 P = V (volts) x I (current)
 P = 1 volt    x 10 amps 
 P = 10 watts

So at our 12 volt system at 10 amps, we have 1 volt of voltage drop (8%) and 10 watts of wire heating.

So let's look at the situation at 120V. 10 amps, 0.1 ohm wire.

 Vdrop = 10 amps     x 0.1 ohms 
 Vdrop = 1 volt 

 P = 1 volt    x 10 amps 
 P = 10 watts

Wait. That is exactly the same formula and answer! Yeah. Nothing in the voltage drop and wire heating formulas gives even the slightest care what the system voltage is. At 600V at the same amps, the wire will heat up the same.

Since heat is the deciding factor in the above tables, they don't care about system voltage either. (It matters only as a percentage; 1 volt drop is a concerning 8% for 12V, a nothingburger 0.8% for 120V, and 0.16% for 600V).

Now you know why Edison lost the war of the currents. Westinghouse could just keep bumping distribution voltage to reduce voltage drop to nothing. Edison couldn't, and had to pay for every inch in copper.

Voltage drop in practice

For example, I have an appliance that requires a 240 volt 15 amp power source. (Its plug is NEMA 6-15.) If I dedicate a circuit to this appliance, which is about 25' from the panel, will 14 AWG be sufficient? Why or why not?

I don't know. Plug it into a Voltage Drop Calculator online and see what you're looking at, and figure out that percentage based on your system voltages. I can tell you this, below 75' for 120V or 150' for 240V, no need to even bother running the calculation, as it will be below any number anyone cares about.

If you're shoveling biblical quantities of power on a regular basis, e.g. a large solar array or an F150 Lightning E-truck, you might bother to do the number-crunch of whether the cost is justified to pay for larger feeder wire. I mean if the bigger wire costs $50 and reduces voltage drop by 1%, you need to buy or sell $5000 of power before that makes sense. Aluminum wire is cheap.

Why do more than the bare minimum?

When 14 gauge wire is appropriate, I've seen 12 used instead. (My home has several circuits like this.) I realize 12 AWG wire may be more difficult to work with, and it costs more than 14 AWG. Beyond these, are there reasons not to do so? Are there any advantages?

The NEC electrical code specifies the absolute slumlord bare minimum beneath which a house is unfit for human habitation.

Some people confuse that for a "best practice".

I think you can see the folly in that thinking.

In practice, 20A circuits where 15A is allowed mean you can run more stuff before the breaker trips. So a heater on high and a PC.

How well does "#14 is cheaper" really work?

There's another reason: The SKU tax. Owning two different kinds of thing is more expensive than owning one. Here, price the following:

  • a 250' spool of 12/2
  • a 100' spool of 12/2 + 100' of 14/2 + 50' of 14/2.

Wowza! The second one costs twice as much even though it has less copper. So realistically to get sane wire prices, you'll buy

  • a 250' spool of 12/2 + a 250' spool of 14/2

Well now, that's a lot of capital tied up, isn't it? That's what I mean by "SKU tax". Your money is tied up from having to own more things. And this out-of-pocket "carrying cost" blows away the theoretical "savings" on #14 being cheaper per-foot. So owning #14 is a net lose for most people.

Obviously it works for builders who wire 3 houses a week, but they buy it by the mile.

I don't own any #14 myself. My smallest wire is #12.

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V = I x R (Voltage = Current x Resistance)

Voltage drop across the wire is equal to the current times the wire's resistance.

P = V x I

Power loss across the wire is equal to the voltage drop due to the wire's resistance times the current.

Now plug equation 1 into equation 2:

P = (I x R) x I

Simplify:

P = I^2 x R (I squared times R)

As you see there is no voltage in this equation!

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