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I'm looking for the right copper wire gauge for my project. I've found several charts listing the maximum current that various gauges, solid and stranded, can transmit. The charts often disagree. One says an 8 gauge wire with a stranding of "7" has an allowable ampacity of 95A, another says 40A, another says 150A, and another says 75A for a solid wire, but 20A for 43 strands. I've found some charts that say stranded wire can handle much, much less current, and others that say that stranded wire "might not" be able to handle as much. How do I know what the right answer is? Is there a formula I can use to right the right gauge?

Specifically, here's what I'm doing:

  • Running a heater
  • 12 volts (needs to be low voltage for safety reasons)
  • around 40 amps
  • maybe 10 meters of wire
  • on a mobile application, so must be stranded for durability
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  • \$\begingroup\$ One source of discrepancies is "chassis wiring" vs cabling. Current limits are due to heat dissipation, and when you're running a single wire you can dissipate a lot more than one conductor in a cable can handle. \$\endgroup\$
    – Nick Johnson
    Commented Jul 1, 2015 at 14:20
  • \$\begingroup\$ I'm willing to bet the charts don't disagree so much as they are for different situations. As Nick says, a single wire can handle a different max current than a wire in a cable bundle. The charts should have a section explaining what specific situation/wire type they are relevant for. You simply need to find the chart that corresponds to your situation. \$\endgroup\$
    – I. Wolfe
    Commented Jul 1, 2015 at 14:29
  • \$\begingroup\$ 0.5 ohm per mtr dissipates power of about 80 W. Isn't something wrong? Is the source of heat from wires is the heater? \$\endgroup\$
    – User323693
    Commented Jul 1, 2015 at 14:31
  • \$\begingroup\$ There's a variety of factors - voltage drop, heating, sheathing (insulation), is it in free air or bundled together in a sealed conduit... that's why there is not a single "right" wire. \$\endgroup\$
    – John U
    Commented Jul 1, 2015 at 15:36

2 Answers 2

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There is no single formula which will do what you want to do. Rather, it is a series of steps.

  1. What is your total power? - We already know the answer: 12 volts @ 40 amps, or 480 watts.

  2. How much power are you willing to waste in your wire? - Let's use 5% as an illustration. As a check, this is 12 watts. Dissipated over a 10-meter cable, this is about 1.2 watts / meter, which seems reasonable. The wire will get warm but not hot, as long as the cable is not contained in an enclosed, insulating cable run.

  3. 5% of 12 volts is .6 volts. Since the cable runs to the heater and also back from the heater, the voltage drop on a single conductor is .3 volts in 10 meters, or .03 volts/meter.

  4. Since the voltage drop in the cable occurs at 40 amps, the resistance of the wire must follow Voltage = Current x Resistance, and Resistance = Voltage/Current = .03/40 = .00075 ohms/meter. Note that this is the maximum resistance allowable. Any greater value produces more drop in the wire and more power dissipated.

  5. From a standard table http://www.daycounter.com/Calculators/AWG.phtml you can see that a 3-guage wire will have a resistance of .000647 ohm/meter, while a 4-guage wire has .000815 ohms/meter.

Is this pretty heavy wire? Yup. Can you do better? Well, only if you're willing to heat up the cable more. The problem here is that your supply voltage is so low that you can't afford much resistance. As illustration of this, consider 480 watts at 120 volts. Then the current is 4 amps while the allowable (5%) voltage is 6 volts. Then you can repeat the calculations above and see that you only need 23 guage wire.

Oh yes, and at 3 ga, you want stranded, not for durability, but just so you can bend the suckers.

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  • \$\begingroup\$ I had a similar question, but was wondering if it was worth replacing all AWG20 wires in my PC power supply by a single AWG10 not stranded. I was wondering if it would reduce the noise in the DC voltage as bigger wire should have lower inductance. \$\endgroup\$
    – Loïc Faure-Lacroix
    Commented Dec 19, 2017 at 11:44
  • \$\begingroup\$ @LoïcFaure-Lacroix - Not a great idea, but not for the reason you think. If anything, the increase in inductance will tend to reduce noise at the load, since high-frequency components will be resisted by the inductance. No, the problems are that a) 10 gauge is a lot stiffer than 20, and b) if you go to a single wire all current will pass through a single set of connector contacts. And finding a connector which will handle a single 40 amp contact is much more expensive than multiple, smaller lines. \$\endgroup\$
    – WhatRoughBeast
    Commented Dec 19, 2017 at 16:11
  • \$\begingroup\$ Ah right, I was planning to connect the wires to binding posts in a closed enclosure (well ventilated) with unregulated output + a current / voltage regulated output. But if you say that more wire should reduce the noise compared to one wire I might not bother unsoldering/soldering AWG10 wire. I guess the wire will sink most of the heat making it a pain to do. \$\endgroup\$
    – Loïc Faure-Lacroix
    Commented Dec 19, 2017 at 17:00
  • \$\begingroup\$ @LoïcFaure-Lacroix - Nah. Soldering is easy. You just have to use the right tool. homedepot.com/p/… (Grin) \$\endgroup\$
    – WhatRoughBeast
    Commented Dec 19, 2017 at 19:28
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check out the 16th ed. of the wiring regs,BS1716,cable selection,and cable calculations. Of course this is a British publication,so it doesn't deal in 'gauges',but deals in conductor cross-sectional-area,in millimeters. hope this helps...

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    \$\begingroup\$ It would be helpful if you provide a link to the standard. It would also help legibility if you capitalise and space properly. (There should be a space after every punctuation mark.) \$\endgroup\$
    – Transistor
    Commented Aug 20, 2016 at 11:22

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