# Determining wire gauge/AWG for DC

I have 15 kW electrical motor that will be connected to a 50 V battery. I'm trying to determine the wire gauge/AWG. Looking at AWG created confusion from as every chart has different recommendation. I would like to know how can I determine the wire gauge ?

• Length: 3 meters
• Voltage: 50 V
• Max. Current: 315 A
• what electrical frequency (ie skin depth) how much voltage drop can you tolerate
– user16222
Feb 8 at 12:48
• Please link the chart(s) that caused you confusion so that maybe an explanation can be found. Feb 8 at 12:49
• Take any gauge resistance per meter, multiply by 3 meters, then by two since two conductors travel this distance. $P=\Omega \cdot A^2$ --> power wasted in cables. Bigger cable = lower loss. Feb 8 at 13:03
• If both pieces of equipment are mounted on a heavy metal chassis you may be able to use this as the ground return, for about half the loss (per rdtsc's comment). But stay safe. Feb 8 at 14:34

## How much voltage drop do you want?

You're gonna get some. How much is acceptable?

If you decided in percentage, convert that to volts by multiplying the percentage x your voltage. Now, apply Ohm's Law:

  (Tolerable) Voltage (drop) = current x resistance

X volts tolerable = 315A x resistance
X /315A = max resistance per 3m


Now divide both sides by 3

  X / 315A / 3 = max resistance per m


now multiply both sides x 1000m since most wires are rated in resistance per 1000m

  X / 315A / 3 * 1000 = max resistance per 1000m
X / 315A / 3 * 1000 / 3.28 = max resistance per 1000 feet


## Don't go smaller than legal minimums, though!

Generally your local electrical codes, or appliance design standards, will specify a maximum current for a particular size of wiring. So for instance, when I did your calculation, I got this result based on resistance alone:

for 4% (2 volt drop) I get 0.0063 ohms. 2.116 ohms per 1km. 0.645 ohms per 1000'. That gives 6 AWG aluminum (17 mm2) or 8 AWG copper (8.3 mm2).

That won't work! Those wires are only allowed 50A! (at 75C thermal). In fact I once saw an installation of #6 copper where the insulation had melted so completely that it extruded into the strands of wire. Don't do that.

In your case, your wire run is so short that your question will be decided by the thermal/safety minimums of the wire.

For instance in North America: for in-wall wiring, for 315A you'd need 500 kcmil aluminum (250 mm2). However for outdoor pole line wiring, 250 kcmil aluminum (125 mm2) is acceptable at 315A. (because it will cool better in free air).

## Aluminum is your very best friend.

If you have a huge wire, or any significant distance to go, don't blow money on copper conductors. Go over into AC mains distribution parts bin and use aluminum wire. America has long experience with aluminum, and has it dialed in.

The crux is two things: First, always use lug connectors whose bodies are made of aluminum to connect aluminum wires. They work great on copper too -- thermal characteristics work favorably there. Second, (and America found this to be an issue for copper also, in fact it was discovered in a copper context!), set torques with a torque wrench. Use of antioxidant can help too, but it's not important enough to be a Code requirement.

• Good answer. It would also be helpful to mention that sometimes the limitation is current carrying capacity of the wire. You may be able to tolerate a large voltage drop, but if there’s too much power loss per unit length, it will burn up. This is affected by things like number of wires, stranded vs solid core, conductor material, insulation material, convection, and ambient temperature. Not always trivial to calculate which is probably why the tables OP mentions give different values.
– Ryan
Feb 10 at 5:55
• Holy smoke @Ryan what was I thinking? Yeah, good save. Fixed! Feb 10 at 6:17