- Solar Electrics
- Battery Charging
- Battery Resistance

# Battery Resistance

### For lead Acid Batteries

There are two processes where the battery resistance will change.

a) The normal operation of charging a battery will increase it's resistance, whilst removing sulfation, and the resistance will drop as sulfation occurs. This follows the formula that V=I × R.

With a constant current (I) the battery charges the voltage (V) rises. Is simple terms as an equation must balance then if one side increases then so must the other. So if V rises then I × R must also rise and of I is constant then R must rise. The reverse is the case when the battery discharges.

b) Although sulfation and de-sulfation are the normal reactions in a lead acid battery sulfation can become ingrained and not subject to being processed by normal charging. In this case the battery appears charged and has a high resistance. To overcome this, or at least to try to, a high voltage can be applied in an attempt to force the sulfate to return to the electrolyte.

Created by roger • Last edit by roger on 22 Mar 2019

See https://chemistry.stackexchange.com/questions/29179/how-come-diluting-a-cell-decreases-the-voltage

My assertion the voltage in a cell is decreased by diluting the cell is not necessarily true.

According to the Nernst Equation: