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Assume DC. For a given amperage and distance, does the wire gauge need to be constant over the length of the cable?

For example, assume 3% acceptable voltage loss, 24Vdc, 10 meter, 50 amp. A cable gauge of 2 AWG would be needed to stay within the acceptable voltage drop.

However, over a 1 meter cable, 10 AWG is enough.

If the cable length is 10 meter, could the first meter be 10 AWG, and the remaining 9 meter 2 AWG? Similarly, does the required AWG change gradually as the distance increases?

This calculator was used to arrive at the required AWG: https://www.solar-wind.co.uk/info/dc-cable-wire-sizing-tool-low-voltage-drop-calculator

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  • \$\begingroup\$ I don’t think the 10 AWG can handle the 50 amp requirement you have. See here for ampacity ratings of insulated conductors. \$\endgroup\$ Commented Mar 8, 2020 at 18:03
  • \$\begingroup\$ No. If 1m 10G is 1 unit loss, and 10m 2G is 1 unit loss, your proposal gives 1.9 units loss. \$\endgroup\$
    – user16324
    Commented Mar 8, 2020 at 18:15
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    \$\begingroup\$ Similarly, does the required AWG change gradually as the distance increases? <<< aaah, you're not familiar with Kirchhoff's law, which is really the most basic law of physics for electronics. The current flowing in a cable will be the same, no matter at which point you look at it, and the current defines the loss per Ohm's law, which is the second most basic law. So, you need to brush up your school-physics knowledge of electricity! Closing this as too broad. \$\endgroup\$ Commented Mar 8, 2020 at 18:27
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    \$\begingroup\$ Ask yourself this: Does increasing the diameter of a water pipe at the far end reduce the pressure drop more than just increasing the diameter all the way through? \$\endgroup\$
    – DKNguyen
    Commented Mar 8, 2020 at 18:43
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    \$\begingroup\$ In distribution the cables are designed to get thinner as you go further out IF the load gets less. So if you have part of your load closer to the source then it may benefit from a heavier cable. Many things to consider but the cost and weight over 10m of cabling may make it pointless to try and economise on a bit of copper. \$\endgroup\$
    – KalleMP
    Commented Mar 8, 2020 at 19:53

3 Answers 3

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assume 3% acceptable voltage loss, 24Vdc, 10 meter, 50 amp. A cable gauge of 2 AWG would be needed to stay within the acceptable voltage drop.

If the cable length is 10 meter, could the first meter be 10 AWG, and the remaining 9 meter 2 AWG?

A given length of wire with some of it a smaller diameter will obviously have higher resistance, and therefore higher voltage drop. But will it be acceptable in this case?

The resistance of a 10 m run of 2 AWG wire is ~0.000512 Ω/m * 20 m = 0.01 Ω, * 50 A = 0.512 V, which is 2.1% of 24 V.

A run with 1 m of 10 AWG wire plus 9 m of 2 AWG wire is (0.00328 Ω/m * 2 m) + (0.000512 Ω/m * 18 m) = 0.0158 Ω, * 50 A = 0.79 V, which is 3.3% of 24 V.

So no, it would not meet the acceptable loss requirement.

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Simply add resistance of each segment

When you change wire sizes, you simply add the resistances of each wire segment. For instance, 10 AWG has 3.277 milliohms per metre for 3.277 milliohms. 2 AWG has 0.5127 milliohms per metre, so 9 metres is 4.6143 milliohms. Your total is 7.8913 milliohms.

It's really that simple.

As a practical matter you'll also have two unnecessary splices, with their own issues.

So if you wanted 8 milliohms of resistance, you could've just used a continuous 10 metre run of 4 AWG, giving 8.152 milliohms. That gets rid of the splices, and also saves a lot of copper.

Although, really, you should be using aluminum for this sort of thing. Aluminum is absolutely the correct material for heavy feeder. Despite what you may have heard about trouble in small-branch-circuits, feeder has always been a different deal, and aluminum metallurgy is improved now.

Simply add the voltage drop for each segment

If you're using a voltage drop calculator, you just make 2 passes through the calc.

  • For 1 metre of 10 AWG wire @ 50A
  • For 9 metres of 2 AWG wire @ 50A

Then you add up the two voltage drops.

Although, you should not be running 50A on 10 AWG, no way, no how. You're required to breaker 10 AWG at 30A for copper or 25A for aluminum, unless you have 75C rated terminations, then +5 amps.

A continuous single length really is the best

As you'll see, a continuous run of the same size simply gets better numbers than a mixed-size run. Remember changing sizes midstream also means you'll have splices and their resistance.

But there's another reason splices are a bad thing. You can't just splice anywhere, wrap it in a booger of electrical tape, and call it good. Even in low voltage DC at these power levels, the splices must have proper execution of work. That means they must be inside an approved junction box. At 5 AWG or larger, a bunch of complicated rules about wire bending radius also enter the fray, and the junction box ends up needing to be quite large. It's just a bad idea.

That said, there's nothing wrong with pigtailing a heavy wire run to a smaller size that will fit on the lugs of the equipment. Just keep the pigtail workably short, e.g. 100-200mm.

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  • \$\begingroup\$ Thank you for all the details, it was very helpful! I have a solar setup with two separate cabins connected with a 10 meter thick gauge copper cable. The battery bank was going to move from one cabin to the other, but was missing one meter. I was wondering if I had to stick with the gauge, but it is clear I do to stay within the amps I want to target more or less. It will have to be a same gauge splice. \$\endgroup\$
    – Karl
    Commented Mar 9, 2020 at 16:12
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Your concept is flawed.

Because the math is so small, let's make it 1000m (cause I have the Tables for copper Ω/100m). So 1000m of #2 vs 100m of #11AWG, 100m of #10, .... to 100m of #2.

1000m of #2 has a resistance of 0.512Ω. 100m of #11 has a resistance of 0.413Ω, 100m of #10 0.327Ω. This is already greater than 1000m of #2. So your voltage drop would be greater than 3% (at 10m).

Actual math:

\begin{array} {|r|r|}\hline Length & \#2 & \#11 to \#2 & AWG \\ \hline 100 & 0.0512Ω & 0.0512Ω & \#2 \\ \hline 100 & 0.0512Ω & 0.0646Ω & \#3 \\ \hline 100 & 0.0512Ω & 0.0815Ω & \#4 \\ \hline 100 & 0.0512Ω & 0.1027Ω & \#5 \\ \hline 100 & 0.0512Ω & 0.129Ω & \#6 \\ \hline 100 & 0.0512Ω & 0.163Ω & \#7 \\ \hline 100 & 0.0512Ω & 0.206Ω & \#8 \\ \hline 100 & 0.0512Ω & 0.26Ω & \#9 \\ \hline 100 & 0.0512Ω & 327Ω & \#10 \\ \hline 100 & 0.0512Ω & 0.413Ω & \#11 \\ \hline \hline 1000m & 0.512Ω & 1.798Ω & Total \\ \hline \end{array}

So your #11 to #2 cable would have 351% more resistance.

To achive #2 equivalency, you can't start (end) with #2, but #2 must be somewhere in the middle.

Now same approach but going from 4/0 to #6 will get the resistance down to 0.5636Ω. This is bigger than 0.512Ω, but it illustrates the problem.

\begin{array} {|r|r|}\hline AWG & Ω/100m & Area (CM) & Area (CM) \\ \hline 4/0 & 0.016Ω & 211,600 & 66,373 \\ \hline 3/0 & 0.02Ω & 167,810 & 66,373 \\ \hline 2/0 & 0.026Ω & 133,080 & 66,373 \\ \hline 1/0 & 0.032Ω & 105,530 & 66,373 \\ \hline \#1 & 0.0406Ω & 83,694 & 66,373 \\ \hline \#2 & 0.0512Ω & 66,373 & 66,373 \\ \hline \#3 & 0.0646Ω & 52,634 & 66,373 \\ \hline \#4 & 0.0815Ω & 41,742 & 66,373 \\ \hline \#5 & 0.1027Ω & 33,102 & 66,373 \\ \hline \#6 & 0.129Ω & 26,250 & 66,373 \\ \hline Total & 0.5636Ω & 921,815CM & 663,730CM \\ \hline \end{array}

So this approach uses 38.8% more copper. Aside from cost of manufacturing, more copper means more cost.

So more resistance or greater cost. Either way, your concept is flawed.

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