I'm looking to wire up an off-grid system. What I'm stuck on is battery placement.

The likelihood is my batteries will need to be quite far apart (almost 2 metres). My question is how does this relate to the wire thickness I should use to add components to the circuit?

To flesh this out in case I'm not being clear: if I only use one battery, the fusebox can be 50cm away from both positive and negative terminal. Using a certain wire thickness, this could result in an X% voltage drop since the return distance of the current is 1m.

If I use the batteries in parallel, the positive terminal is now 50cm from the fusebox, whilst the negative terminal (on the other battery) is now 1.5m away.

  • Does this means the return distance is now 2m, resulting in a 2X% voltage drop?
  • Or does it include the parallel connections, meaning it's 3.5m or 5m?

2 Answers 2


Each wire has its own voltage drop and you have to compute it separately.

So you need to separately compute BAT1+ to BAT2+, BAT2+ to panel, panel to BAT2-, BAT2- to BAT1-. And those will have different currents on each.

I have to say, though, that if you are worried about voltage drop for 2 metres of length, you are probably going too thin with the wires. You should be pulling from the mains electric parts bin, and working off a table like NEC 310.15(B)(16).

Note a couple things about that table. First, it lists different thermal limits for different kinds of wire, based on their insulation quality. The thermal limits are based on current only; voltage isn't a factor. Second, it talks about aluminum wire, and that's something you should be thinking about. Obviously you've got some cost apprehension; aluminum wire is a great solution there. The moral panic about aluminum wire terminations applied to small wires at or below 10 AWG or 4mm.

So if you are pulling out of this table (or a Euro equivalent) you won't have any worries about either voltage drop or wire overheat.

The mains electrical parts bin has everything you need to cable for currents as high as 600 amps.

And by the way, when it comes to fuse boxes, Square D "QO" type breaker service panels, the run-of-the-mill types used in houses, are rated for DC power below 48 volts.


Yes, basically you have it right, more distance means greater voltage drop.

So, best thing to do is locate the batteries safely with proper venting, then consider your cable run.

To limit the voltage drop, which is due to the resistance of the cable ie proportional to length and inversely proportional to diameter, all you need to do is put in a nice thick cable which will compensate for the length.

So, if the maximum current from the batteries is 20A, then the cable is relatively cheap. If the current will be about 100A, then you are looking at the size of cable used for car battery to starter motor - which as it has more copper is more expensive.

Good cables and decent connections are cheaper than a fire...

  • \$\begingroup\$ Thanks Mike! As I'm looking at 300A discharge currents, I'm already looking at some of the thickest cable you can buy. My question really is how the parallel batteries work into the equation so I can size them right. Is it the total cable length (from positive, to device, to negative, AND the cable length connecting the two batteries), or are the parallel connections only considered in themselves since they're essentially supplying the other battery, or am I way off in understanding how this works? \$\endgroup\$
    – J. Carter
    Commented Dec 7, 2018 at 9:50
  • \$\begingroup\$ As the batteries are in parallel, you could size the cables so that the cable current carrying capacity is matched to the position that the cable is used, BTW never seen that done... So, all the cable links are normally sized for the maximum current to flow, so that the possibility of error is removed. \$\endgroup\$
    – Solar Mike
    Commented Dec 7, 2018 at 9:56

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