# Charging a bank of batteries in series with a resistor

Few days ago I was trying to recharge a bank of ten batteries connected in parallel. I connected the charger directly to the battery terminals at 13.75 V as specified in battery manufacturer datasheet.

The problem: the batteries were about 11.4 V at that time and my DC supply transformer was supplying a the maximum current of 50 A. Well ... 50 A for 10 $\times$ 9 Ah batteries is way too much, not to mention that the wires started to melt due to high current so I realised that I need something to limit the current and quickly figured it out that I need a resistor.

I've seen numerous youtube videos and documentation about how a resistor works and it's limitations, I ended up using this calculator.

13.75 V at 4 A = 3.4375 $\Omega$ resistor, but I found a 3.3 $\Omega$ resistor which (based on the calculator) should be capable to limit current at 4.16 A, dissipating with 57.29 W of heat.

Big surprise when I connected it on positive terminal (tried on negative also) and only about 0.6 A of current was flowing to the batteries, despite the fact that the batteries were discharged and without any kind of resistor they alone would draw > 30 A.

So my question is: why is that? Why does the 3.3 $\Omega$ resistor limit the current to 0.6 A and not to 4.16 A as calculated. Is there a possible difference in resistance which doesn't allow the current to flow at max capacity?

I've also tested the resistor with a multimeter and the resistance is fine. I've also connected it to positive and negative of a single battery and about 4.3 A of current was flowing - so it works as it should but not when connected with the charger.

Thank you!

• NB: Make sure each of the batteries voltage levels are about the same, or you'll have possibly high currents running between the batteries themselves, as well! – Steve Mar 15 '17 at 18:40

You have to use the voltage difference to calculate the resistor. $$13.75~V - 11.4~V = 2.35~V$$