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ceecee avatar image
ceecee asked

Cable size for 2 X 180ah batteries in parralel

Hi, I'm trying to calculate the needed cable size for 2 x 180ah batteries in parralel.

So the question is should I match the 180ah (power from one to the other) or double it 360ah a wire accordingly. The wiring from the batteries will supply my campervan with 12v 3000ka multiplus along with 12v lights, USB sockets etc. the wiring for these has all ready been installed and is correct to spec. So 70mm x 2 to multiplus.


This last bit I'm finding hard to find the correct information. If you have links to further reading that will help that would also be helpful.


Many thanks



battery
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3 Answers
seb71 avatar image
seb71 answered ·

You size the wires based on the current you want them to carry, not based on battery capacity.


Another important thing is to protect all wires with fuses. A fuse should blow before the current reaches the maximum current a certain wire can safely carry.


It would be best to use busbars. Don't connect everything directly at the battery terminals.

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myhomebuzz avatar image
myhomebuzz answered ·

@ceecee

Batteries in parallel, the intensity (A) is added. Conversely, batteries in series, the voltage (V) adds up.

In your case, the two batteries give 12V 360Ah. Your Multiplus 3000VA according to information to be confirmed: 120V or 220V (important for cables from the Multiplus in AC)? Which model ?

Take the example with a Multiplus 3000VA / DC 12V / AC 220V / 3000W / 6000W peak is positioned 1.5 meters away from the battery bank.

So the intensity A calculated on W peak: 6000W / 12V = 500A

The comments of seb71/JohnC are important to understand, as well as their comments to increase the voltage of the batteries (12-> 24) which will reduce the sections and heating of the cables.

The following section rule can be used and explained in another of my posts here, but cable length is also important to consider due to voltage loss:

Ω = copper 0.0179 (but be careful with aluminum: 0.028!)

S = ρ x A / 0.03 x V or

S = (Ω x (L x 2)) x A / 0.03 x V or

S = (Ω x (L x 2)) x (W / V) / 0.03 x V

S = 0.0179 x 3m x 500A / 0.36 = 74.58 mm2 -> 100 mm2

For battery-battery connection cables, the larger the cable, the more Coulomb (electric charge passing through a conductor) can pass. Prefer at least 35/50 or even better 70 mm2 than 25 mm2 informed everywhere. This will avoid heating the cables. This comes at a cost, but so does security.

I hope I have brought some help.

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ceecee avatar image
ceecee answered ·

Thank you for the replies

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