Is Voltage Drop Greater in Weak Batteries Compared to Good Ones?

[Whakataku]

Relevant equations:

V = I*R
ε = I*(R+r)
Then,
V = {ε/(R+r)}*r

V = potential difference, ε = electromotive force, R = resistance of external load resistor,
r = internal resistance of the battery.

So say the external load resistor's resistance is a constant for two batteries of the same make, difference being that one is good and the other is weak. Then since by intuition the internal resistance will be different, the good battery at room temperature having less r than the weak battery. The ε is constant.

Then is it correct to say that the voltage or potential difference will be smaller, or there will be voltage drop for the weak battery compared to the good battery?

 

[berkeman]

Relevant equations:

V = I*R
ε = I*(R+r)
Then,
V = {ε/(R+r)}*r

V = potential difference, ε = electromotive force, R = resistance of external load resistor,
r = internal resistance of the battery.

So say the external load resistor's resistance is a constant for two batteries of the same make, difference being that one is good and the other is weak. Then since by intuition the internal resistance will be different, the good battery at room temperature having less r than the weak battery. The ε is constant.

Then is it correct to say that the voltage or potential difference will be smaller, or there will be voltage drop for the weak battery compared to the good battery?

It depends on the type of battery, but often you will not see much difference in the open circuit voltage for much of the discharge cycle. Most of the droop comes near the end of the discharge cycle.

You can look at the battery's datasheet to see what its discharge curves look like. This is a curve for a deep-discharge battery:

http://hamwaves.com/qrp/portable/dm12-7.2_discharge.gif

 

[mrspeedybob]

What do you mean by a "weak" battery? Do you mean a battery that is nearly discharged or do you mean a rechargable battery that has nearly lost its ability to store a charge? My knowledge and experience pertains mostly to lead-acid batteries but I can tell you the following about them...

Open circuit voltage of a fully charged lead-acid cell is about 2.1 volts, it will deplete about 75% of its capacity before dropping to 2.0 volts. The same applies to a cell that is nearly worn out, though the capacity will be lower, the internal resistance will be higher, and the recovery time will be longer. (recovery time is how long it takes the open circuit voltage to come back up after a load is removed)

 

[Whakataku]

Sorry I didn't specify a battery. Let's just say a D cell alkaline battery with 1.5V. And by weak, I mean, nearly discharged, or nearly depleted in chemical charge due to the extensive oxidation reaction at the anion.

You see what I'm getting confused is that I ask someone about it and they say one thing, and ask another person they say the totally different thing. I just want to know if the voltage is dependent on the internal resistance, or totally independent like the electromotive force which is due to the chemical reaction within the cell. The equations I mentioned seems to indicate the former where the potential difference is dependent on the internal resistance.

The chart was helpful Berkeman, but are there any universal mathematical statement that relates the internal resistance to the voltage of an electromotive cell?

Regarding mrspeedybob "Open circuit voltage of a fully charged lead-acid cell is about 2.1 volts, it will deplete about 75% of its capacity before dropping to 2.0 volts. "
Was what you stated a phenomenon, or can it be mathematically explainable?
I crave mathematical proof regarding the potential difference between the anode and cathode of a battery.