jodiematthewrobinson Didyou take the voltage drop across the leads into account? It says ~.6A at .1V drop, so if you've got .2V difference at the battery, you would only need a 0.083 ohms in each lead to bring the voltage difference at the balancer down to .1V, and thus .6A will be drawn.
Antwort 23/01/2021mortenfj you wrote as an answer for s question that it did not work at all? so does it work slow or not at all?
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