I want to build a system where I have a solar charger connected to one 12V battery, which is then wired in parallel by a long cable (30 meter, 2.5 mm2) to the other battery.

Will both batteries eventually be fully charged, or will the battery on the long end have a lower voltage/charge due to voltage drop in the cable?

I am trying to understand the relevance of current in the resistance of the cable. As the difference in voltage between battery A and B is being reduced, the current on the cable should drop - and this will reduce the voltage drop due to resistance in the cable, I am assuming?

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    \$\begingroup\$ Having the storage so far from the charger is a bad idea. You may end up damaging the charger. \$\endgroup\$ Jul 10, 2016 at 12:06
  • \$\begingroup\$ This. Can't you just keep both batteries and the charger in the same place, possibly 30m away? Not the best solution, but better than what you have in mind now. \$\endgroup\$ Jul 10, 2016 at 13:33
  • \$\begingroup\$ I want more battery capacity in total, and the other house is much more suitable location wise for the batteries. I was thinking that one battery close to the charger, and another in the extra house would in general cause fewer amps on the wire which is more efficient resistance wise. \$\endgroup\$
    – frodeborli
    Jul 10, 2016 at 13:45
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    \$\begingroup\$ Will both batteries eventually be fully charged, or will the battery on the long end have a lower voltage/charge due to voltage drop in the cable? Yes both will fully charge, as a battery charges, the current eventually drops to 0 (zero) so also the voltage drop across the cable will be zero and both batteries will end up equally charged. \$\endgroup\$ Jul 10, 2016 at 15:36
  • \$\begingroup\$ @FakeMoustache You should post that as an answer so that I can accept it. It seems that it is a quite good solution to have two battery banks, one close to the charger, and another one in the far end. Especially since we will be using most electricity from the battery close to the charger. \$\endgroup\$
    – frodeborli
    Jul 10, 2016 at 17:46

1 Answer 1


2.5mm copper wire has a resistance of about 3.5mΩ per meter, so your 30 meter cable is about 0.0035 * 30 * 2 = 0.21Ω. How much effect this will have depends on the type of battery, their capacities, the available charging current, and what charging algorithm is used.

For example, if both batteries are 100Ah flooded lead-acid type, and the charger is delivering a constant 20A up to a float voltage of 13.8V, the initial situation (both batteries less than 50% charged) is something like this:-


simulate this circuit – Schematic created using CircuitLab

R1 and R2 are the internal resistances of the batteries. The charging current will split between the batteries, but not equally because the cable resistance has increased the effective resistance of BAT2 to 260mΩ, 5 times higher than BAT1. Therefore BAT1 will get 5 times more of the current than BAT2 (17A vs 3A), and charge up 5 times quicker.

As BAT1 is charging faster at higher current, its internal voltage will rise faster and reduce the difference in charging currents. The relationships between state of charge, voltage, current and resistance are not linear so calculating the exact charging currents is difficult, but BAT2 will always get significantly less current and lag behind.

If the float voltage is maintained for long enough then BAT2 will eventually reach full charge. But how long will it take? As well as charging slower during the 'bulk' phase, it will also take 5 times longer to become to become fully charged after reaching float voltage (the 'absorption' phase). Depending on how many sunlight hours your solar panels get per day, this may take several days or even weeks.

Meanwhile you may use the batteries to power some devices. Provided the connection is made to BAT1 the discharge current will also split unevenly (with BAT1 providing the most) so even if BAT2 hasn't reached full charge it should be safe to run them both down together. However you should not connect devices directly to BAT2, because then it would be discharged faster than BAT1 and require an even longer recharge time.

At high discharge current this arrangement will be less efficient than having both batteries in the same location, because there is loss in the cable and the unevenly split current reduces the effective capacity of BAT2 (since you must stop drawing current when BAT1 reaches cutoff voltage, but BAT2 was discharging at a lower rate so it may still have charge left in it).

  • \$\begingroup\$ I guess BAT1 will help with charging BAT2 until they are equalized - while at the same time as the charger is primarily charging BAT1 due to less resistance. \$\endgroup\$
    – frodeborli
    Jul 20, 2016 at 20:21

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