A discharged battery pack reads 20 Volts and the charger reads 24 Volts. The battery pack has an internal resistance of 4 ohms, which voltage is used to find out the current going to the battery? The power supply, battery pack, or difference between them?

24/4 = 6 amps
20/4 = 5 amps
4/4 = 1 amps

When measured with a meter the amps appear to be close to 1 amp. I want to understand this and be sure how it works.

Any help is greatly appreciated,

Sandy

 

A discharged battery pack reads 20 Volts and the charger reads 24 Volts. The battery pack has an internal resistance of 4 ohms, which voltage is used to find out the current going to the battery? The power supply, battery pack, or difference between them?

24/4 = 6 amps
20/4 = 5 amps
4/4 = 1 amps

When measured with a meter the amps appear to be close to 1 amp. I want to understand this and be sure how it works.

Any help is greatly appreciated,

Sandy

The current flowing through the battery is the supply voltage minus the battery voltage and then divide them by the battery's internal resistance.

But this is true at the beginning of the charging because then the battery voltage increases

 

Mik3, thank you for the help!
Sandy

 

Apologies, Mik3, but I do not understand what you are saying in your response to Sandy. What is the numerical value of "...them" using Sandy's example? If "them" is the difference voltage (4 volts), then dividing "them" by battery internal resistance is incorrect.

Let's call whatever circuit exists between the charger output and the battery terminal "R?" (shorthand for unknown R), the internal resistance of the battery "Rb", the charger output voltage Vch, and the battery terminal voltage Vb. The current in the series circuit of charger, R?, and the battery would be (Vch-Vb)/R?, not (Vch-Vb)/Rb.

Ohm's law applies to both voltage and current across the same circuit. You can't take voltage across some circuit element and divide it by the resistance somewhere else in the series circuit and determine current.

Circuit current would also be Vb/Rb, depending upon how Rb was measured. I don't know enough about batteries to know if Rb is the same for charging current as it is for discharge current, that is, if RB is properly modelled as a simple resistor in series with a pure voltage source, but I suppose it would be.

I have no explanation for Sandy's observed voltage and current, since we don't know how he determined Rb or what circuit constitutes R?.

I agree (and you will see immediately when charging a battery from a metered constant voltage power supply) that current changes rapidly if you are using a constant voltage source and a series resistor. Additionally, the effective internal resistance of a battery will change rapidly as the battery charges up.

Maybe Sandy can fill in some details of his measurements.

awright