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This seems like a googleable question, but I simply cannot come across anything.

I'm working on a 12v system (aluminum boat), trying to supply power to a motor 12ft away from the circuit breaker. Through online sources, I was able to determine that at the 45amp max, 6awg wire could handle it, with a 5% voltage drop.

The confusion enters when I try to calculate the voltage drop with this 6awg wire, plus the potentially 8awg wire that runs 1.5 feet from the battery to the circuit breaker.

How does one correctly calculate the voltage drop over two different gauged pieces of wire?

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  • \$\begingroup\$ It might be a good idea to replace the 1.5 feet of 8AWG wtih 6AWG while you are at it. Then you have a continuous 6AWG path from battery to motor. \$\endgroup\$
    – user57037
    Aug 27, 2018 at 3:59
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    \$\begingroup\$ Copper wire, plus batteries, plus an aluminum hull in water can lead to some pretty interesting failures. Be careful! \$\endgroup\$
    – user57037
    Aug 27, 2018 at 4:00
  • \$\begingroup\$ @mkeith - Is there a significant advantage there? Without getting into detail, the 8awg wire saves me quite a bit of money. \$\endgroup\$
    – xtraorange
    Aug 27, 2018 at 19:25
  • \$\begingroup\$ If the motor current flows through the 8AWG wire (which is what I think based on your question) then it will produce a voltage drop between battery and breaker. I am envisioning that the breaker is in a small panel with other stuff. Everything connected to that panel will experience that voltage drop. If the only consumer of power is the motor, then it probably doesn't matter as long as the 8AWG can handle 45 Amps without overheating. \$\endgroup\$
    – user57037
    Aug 28, 2018 at 0:27
  • \$\begingroup\$ @mkeith - that circuit breaker is separate from my main fuse panel, which has its own connection to the battery. However, they do both share a common ground... would that be an issue at all? If not, it sounds like I can use the 8awg for the short run, which would be great new for me, haha. \$\endgroup\$
    – xtraorange
    Aug 28, 2018 at 1:45

4 Answers 4

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Each piece of wire is a resistor. Add up the resistance of all the wire that the current to the motor runs thru. Remember to consider both directions.

You find the resistance of each piece of wire by looking in a wire chart to get the resistance per unit length. Multiply that by the actual length of the wire segment to get the resistance of that segment.

Add up all the individual resistances of the individual wire segments. That's the total wire resistance in your installation. Multiply that by the current, and you have the voltage drop caused by the wire.

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    \$\begingroup\$ How did this get downvoted? This is correct. \$\endgroup\$
    – K H
    Aug 27, 2018 at 3:33
  • \$\begingroup\$ If the motor is not a balanced multiphase load, make sure you take into account the resistance both on the way to and back from the motor. \$\endgroup\$
    – K H
    Aug 27, 2018 at 3:40
  • \$\begingroup\$ @KH The OP is talking about a 12 volt boat. There’s little doubt it’s a 2 wire DC motor. \$\endgroup\$
    – DoxyLover
    Aug 27, 2018 at 6:21
  • \$\begingroup\$ Figured it was possible anyway tiny hobby motors go to 3 phase and boats sometimes vary on the side of fancy. \$\endgroup\$
    – K H
    Aug 27, 2018 at 6:32
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From google: https://www.calculator.net/voltage-drop-calculator.html

You can just first calculate the drop over the 6AWG wire: 45A, (assume DC) 12V, 6AWG, 12ft = 11.57 V.

Then for the extra: 45A, (assume DC) 11.57V, 8 AWG, 1.5ft = 11.517 V.

If your load is not DC you can recalculate.

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  1. You calculate the first drop over 6AWG: 45A, 12V, #6, 12ft = 11.57V (3.58% drop)
  2. Then calculate the remaining on 8AWG: 45A, 11.57V, #8, 1.5ft = 11.517V (0.46% drop)

Total drop is calculated as such: 1 - [(1-0.0358) * (1-0.0046)] = 4.02% drop or 11.517/12

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$$R = \frac {\rho \ell} {A} $$

$$R_T = \frac {\rho \ell_1} {A_1} + \frac {\rho \ell_2} {A_2} = 10.37 \Omega \cdot CM /ft \times \left( \frac {12ft \times 2} {26,250CM} + \frac {1.5ft \times 2} {16,509CM} \right) = 11.37m \Omega$$

Worse case scenario at 45A, will give 0.5115V voltage drop or 4.26%. This is at 20°C for copper.

Actual voltage drop will depend upon the actual current and temperature.

The \$\times 2\$ comes from having to get to the load and get back to the source.

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