# Resistivity of Romex (NM) cable

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I'm interested in calculating some voltage drops for my shack service feed... Does anyone know where I can find a table of resistivities (ohms per unit length) for various Romex (NM or non-metallic sheathing) wire? Also for THHN? Please include the source for your data if possible so that I may follow up.

Thanks,

Jason

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Romex or THHN makes no difference as far as resistivity/voltage drop is concerned. The info you seek is available in any copper wire table, such as:

14AWG--2.575 ohms/M' 12AWG--1.619 ohms/M' 10AWG--1.018 ohms/M'

Be sure to remember the TOTAL wire length, not a one-way distance when computing voltage drop.

As far as allowable ampacities and derating factors for various types of cable, look in the National Electrical Code.

Bob Weiss N2IXK

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On Sat, 21 Feb 2004 00:56:51 GMT Bob Weiss wrote: | Jason Dugas wrote: |> I'm interested in calculating some voltage drops for my shack service |> feed... Does anyone know where I can find a table of resistivities (ohms |> per unit length) for various Romex (NM or non-metallic sheathing) wire? |> Also for THHN? Please include the source for your data if possible so that |> I may follow up. |> |> Thanks, |> |> Jason |> |> | | Romex or THHN makes no difference as far as resistivity/voltage drop is | concerned. The info you seek is available in any copper wire table, such as: | |

| | 14AWG--2.575 ohms/M' | 12AWG--1.619 ohms/M' | 10AWG--1.018 ohms/M'

That seems a little high. The figures I have are:

AWG Dia-mils TPI Dia-mm Circ-mils Ohms/Kft Ft/Ohm

10 101.90 9.8140 2.5881 10383 0.9989 1001.1 11 90.741 11.020 2.3048 8233.9 1.2596 793.93 12 80.807 12.375 2.0525 6529.8 1.5883 629.61 13 71.961 13.896 1.8278 5178.3 2.0028 499.31 14 64.083 15.605 1.6277 4106.6 2.5255 395.97 15 57.067 17.523 1.4495 3256.7 3.1845 314.02 16 50.820 19.677 1.2908 2582.7 4.0156 249.03 17 45.257 22.096 1.1495 2048.2 5.0636 197.49 18 40.302 24.813 1.0237 1624.3 6.3851 156.62

I'm also working on putting together a web page to calculate voltage drop. There is one out there that has been referred to before, but it produces wrong answers (such as large negative voltage drops).

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Probably the Number one site on the Internet for Voltage drop is

you find excellent voltage drop calculators and even a voltage drop slide show.

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|

Try

is my understanding that the advanced voltage drop calculator took over

10 years to write. Yours truly, Gerald Newton
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What about 6 and 8 AWG?

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| Try

| It is my understanding that the advanced voltage drop calculator took over | 10 years to write.

You mean this one?

That's the one that produces wrong answers!

Let's try it out. Put twenty 60 watt light bulbs (240 ohms for 120 volts each, for a net resistances of 12 ohms) at the end of a one mile run of

18 AWG copper wire. Power it with 120 volts That's a load of 10 amps.

Plug the numbers in:

1. Copper wire
2. 18 AWG
3. 120 volts 1-phase
4. 5280 feet
5. 10 amps

So you gonna believe that you will get -694.8 volts on those poor bulbs?

Maybe that's what is driving those "free electricity" crackpots :-)

I'm working on my own program. When I get it done, I'll post the URL here so you guys can see if I got it right or not.

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Well, actually, the answers are consistent, but you have given it some impossible values!.

The calculator assumes that the specified load current WILL FLOW. The negative numbers you reference arise because those sets of values would require additional voltage to achieve the specified load current. In other words, 120 VAC won't force 10 amps through a loop-mile of #18 even if the receiving end is short-circuited! But, if you connect an additional source of -694.8 volts at the receiving end, you will get your 10 amps.

It would have been nice to have a feature that flagged this situation, rather than giving the startling numbers, but the numbers are correct!

#18 @ 6.38 ohm/1000 ft x 10560 ft gives about 67 ohms. Add your 12 ohm load for a total of 79 ohms. Force 10 amps through it for a drop (including lamps) of 790 volts. Subtract the 120 volt source and you see that an additional potential of 670 volts is needed to develop a current of 10 amps.

I have rounded a few values, but the principle holds.

Not >

over

• posted

| Well, actually, the answers are consistent, but you have given it some | impossible values!.

The values are not impossible. They certainly are not practical, as no one in their right might would try to power 20 light bulbs that way.

| The calculator assumes that the specified load current WILL FLOW. The

The field on the form says load amps.

If you're wanting to calculate how much voltage drop you get with a fixed current, you're calculating an impractical and unrealistic situation, AND not one the form items describe.

| negative numbers you reference arise because those sets of values would | require additional voltage to achieve the specified load current. In other | words, 120 VAC won't force 10 amps through a loop-mile of #18 even if the | receiving end is short-circuited! But, if you connect an additional source | of -694.8 volts at the receiving end, you will get your 10 amps. | | It would have been nice to have a feature that flagged this situation, | rather than giving the startling numbers, but the numbers are correct!

It doesn't need to be flagged. It needs to give the correct answer for what people expect.

| #18 @ 6.38 ohm/1000 ft x 10560 ft gives about 67 ohms. Add your 12 ohm load | for a total of 79 ohms. Force 10 amps through it for a drop (including | lamps) of 790 volts. Subtract the 120 volt source and you see that an | additional potential of 670 volts is needed to develop a current of 10 amps.

I don't want to force 10 amps through it. I want to know how much voltage I get at the other end (or lose in the wire) when connecting it to the normal supply voltage of 120 volts. This is what the form working says it will do (but doesn't). I know what you are describing, but that isn't the way the form makes it out to be, and it isn't what most people need.

Sure, if you're got a step-up variac and want to get those light bulbs up to full brightness by adjusting the supply voltage, this could tell you how much you have to go up to. But this isn't a form for that, and isn't what most people go about doing. Sure, in some cases you probably want to figure out whether to install a boost transformer or not, or how much. And a form to do just that would be helpful. This one isn't that.

In real life, people want to find out what gauge to run power to the barn with and still get a reasonable voltage out there. A form that would show a matrix of choice wiring and varying loads would probably be most useful.

| I have rounded a few values, but the principle holds. | | Not intuitive, but not wrong either!!

Based on what you are describing, neither intuitive, nor useful.

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over

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OK, I stand corrected.

What IS the correct answer for what people expect, using your figures?

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| You may have found an idiot proofing error, but in 99.9 per cent of the | voltage drop problems a user that is familiar with electrical work can use | the program to find practical useful answers. The program is not written | for idiots; it is written for electricians. | hint: electrician.com

I'll believe physicists and engineers before I'll believe an electrician.

| If the answer is negative, then obviously all the volts are dropped in the | conductor. Any 1st year electrician apprentice should know that. But

Really? You think there will be ZERO VOLTS at the load? You really have no concept of physics, do you.

According to my tables, the 5280 foot run of 18 AWG copper wire, at a rating of 6.3851 ohms per 1000 feet, will have 33.713328 ohms for the full 5280 feet, and 67.426656 ohms for the full loop. Adding in the 12 ohms of load makes the total ohmage of 79.426656. So, at 120 volts that's a current of

1.5108278 amps. That puts a voltage difference of 18.1299336 volts across the load, and 101.8700664 volts across the wiring. The total watts in the load is 27.3912077 (divided across the 20 bulbs means each dissipates only 1.36956 watts). The remaining 153.9081283 watts is dissipated in the wires at 0.0145746 watts per foot per conductor.

| more than that, the ampacity of No. 18 is not 10 amperes. Voltage drop | alone does not solve all the problem, the ampacity of a conductor is solved | using different equations and Tables. Additionally, for No. 18 NEC | Section 726.23 requires overcurrent protection no larger than 7 amperes. | Also, the advanced voltage drop calculator finds the approximate temperature | of a conductor and finds the voltage drop for that temperature. Read the | article

Did you mean 725.23?

I suggested looking at table 310.16 under the 90 degree insulation column for copper and note the ampacity for 18 AWG wire there is 14. That's for wiring in raceways or buried, up to 3 conductors. Now have a look at table

310.17 and see what the ampacity for 18 AWG wire in free air is 18.

For the purpose of calculating voltage drop, the ampacity is not really relevant. The temperature would be, but the temperature coefficient does not alter the above values radically.

And finally, with a total resistance of 79.426656 ohms at 120 volts, the current is only 1.5108278 amps, anyway. That's well below virtually every rating for 18 AWG wire. You might not have much light, but the wire is definitely not going to burn up.

|

| for some enlightenment. | | You appear to be more into electronics and ham radio than electrical work. | Perhaps you should stay away from the electrician page until you serve an | 8,000 hour apprenticeship and an additional 8,000 hours as a Journeyman | electrician and then get your electrical administrators license then you | might be able to use electrical calculators without evaluating them as a | technician.

If electricians use calculators or formulas that result in violations of physics, then I sure as hell don't want to hire them to wire anything of mine. I'll play it safe and wire my own.

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| What IS the correct answer for what people expect, using your figures?

I copied this from another followup in this same thread:

According to my tables, the 5280 foot run of 18 AWG copper wire, at a rating of 6.3851 ohms per 1000 feet, will have 33.713328 ohms for the full 5280 feet, and 67.426656 ohms for the full loop. Adding in the 12 ohms of load makes the total ohmage of 79.426656. So, at 120 volts that's a current of

1.5108278 amps. That puts a voltage difference of 18.1299336 volts across the load, and 101.8700664 volts across the wiring. The total watts in the load is 27.3912077 (divided across the 20 bulbs means each dissipates only 1.36956 watts). The remaining 153.9081283 watts is dissipated in the wires at 0.0145746 watts per foot per conductor.
• posted

I ran this on another commercial calculator that I use, and it gave "vd=100%" as the answer. These calculators do assume a constant load, since they are designed for real world application where this is a good estimate of actual operation, not precise physics. They give good results when used in more realistic calculations, with voltage drops up to perhaps 15 or 20%.

Of course, you are also assuming a constant resistance for the bulbs, which is not real physics either. The problem you gave would result in no lights, regardlss of the actual voltage drop number. Not a very useful installation by any calculation result.

Ben Miller

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Certainly it is not a useful configuration. It was chosen so that the current would not be so excessive as to melt the wire, yet still show results that are obviously in error without having to manually run the formula through on a hand calculator to be sure.

I think what you are using is something that would be more appropriately called a "voltage configuration analyzer". If your "vd=100%" is meant to say "this situation cannot be practically compensated for", that I could agree with that. Another poster suggested the purpose was really to see how much drop was present in a constant current situation where in the case of 20 60-watt light bulbs we want a constant current of 10 amps. But the only case where that would be a real world concern is if you are going to be raising the voltage to correct it. That still could be done in the example I gave, but quite impractical since the feed voltage needed would be 794 volts (which would also exceed the rating of 600v wiring for a big part of the path, and 18 AWG wire generally is rated well less than that).

If you are a distribution line engineer calculating what transformer taps to use to ensure that a given drop ends up with the correct voltage at the far end, then you do need a calculator that deals with a constant load situation. You also need to do that kind of calculation over the full range of loads so you don't end up with excessive voltages when the load is reduced. The calculator on that web page would still be inappropriate for that case.

The case more people end up needing to work with is where changing the supply voltage is not generally an option (other than to jump up to the next higher voltage increment to reduce the I^2R losses). Marginal cases deal with trying to understand how much voltage is actually going to be present at the far end, and selecting the correct wire size to get things into an acceptable range. That was how this and other threads like this have started. The calculator on that web page would still be just as inappropriate in this case as well.

A more practical calculator for the distribution line engineer would consider the choices available for transformer voltages, drop wiring, and the full load range, then plot a graph showing voltage swings with changes in load for each of the choices. A smarter application could dismiss any choices considered out of range, and even advise exactly the best choice, including the costs of equipment, materials, and the energy loss.

A more practical calculator for someone trying to decide what wiring to go with to light up the barn would take the supply voltage, distance, load figures, and show the end result voltage under various wire choices across the load ranges. These cases generally don't involve voltage change other than choosing 120, 208, 240, 277, 347, 480, or 600. They rarely involve the use of a transformer or variac. And of course they don't invole a one mile distance, either.

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I am not sure which ones you have used, but the one I use the most takes into account conductor material, raceway material, conductor temperature, circuit configuration (1ph, 3ph, conductors/phase, etc.), type of load (constant current or constant VA), type of current (AC or DC). It will solve for any one of the following, given the other two: Conductor size, conductor length, and voltage drop. It also calculates series voltage drop for a string of loads. I am at a loss as to what to call it if it is not a voltage drop calculator (which is what the program calls it).

Ben Miller

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FYI the VD calculator has the idiot proof code put in as follows:

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I fixed the program so that you don't get a negative voltage. It took four lines and 15 minutes:

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Voltage drop happens, but that may not be what people need to know. Instead, people need to know what to change to make the load voltage be correct. For what you do as an engineer, you need to know all those things. And for most people most of the time, that's what gauge wire to use. You could call the program by any name of any data is gives you. So if you want to call it by the name voltage drop, then go ahead. I'll call what I would use to solve a problem by what it is that defines the name of the problem. But we're probably not talking about the same program. The program on the web site is definitely not the one I would use to determine whether AWG 8 wire is enough to power my barn 200 feet away. The program you described is probably overkill (does it model the cooling level of airflow).

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Why not? It doesn't directly calculate it, so you need to plug in a few wire sizes to determine the correct one, which only takes a few seconds. I am not saying it is perfect, but it seems to work OK based on a few quick trials that I ran. I don't use it regularly, and this is not an endorsement, just an observation.

Actually, it is quite simple to use. It does not calculate the actual conductor temperature. It allows you to select the temperature of the wire (60, 75, or 90 C) and then calculates the voltage drop at that temperature. In many cases this gives a conservative solution, and further precision is not necessary.

Ben Miller

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