Internal Resistance of a Battery and Temperature?

[TimNJ]

I’m working on a real simple lab for my circuit analysis class, but I’m having a hard time sourcing information to make sense of my results. Part of the lab was to hook up a 2 AAs, and a 0.5ohm resistor in series and measure the voltage across the battery terminals every minute for 15 minutes. Initially we measured 3.0V open circuit, and about 1.8V across the battery, and by the end, it had decreased to 0.37V. BUT, after unloading the batteries, the open circuit voltage went back up to 2.56V. It makes sense that the battery had discharged so it lost some of its voltage. However, I don’t understand why the voltage across the load kept dropping. By the 15th minute, the calculated internal resistance was about 3ohms, while it had been about 0.3ohm at the very start. I thought it had to do with the battery heating up like a PTC which caused the resistance to go up…but after reading about batteries, it seems that the internal resistance of a battery goes down with increased temperature. What am I missing!? Thanks so much!

[sacherjj]

Two factors comprise the internal resistance of a battery: electrical resistance and ionic resistance.  Electrical resistance is the normal resistance of electron flow.  Ionic resistance is due to electrochemical factors like electrolyte, etc.

Usually, internal resistance of the battery rises as it discharges, due to the use of active materials in the cell.  The rate of change is not always consistent.

Ionic resistance increases with a decrease in temperature, due to lower mobility of the chemicals in the battery.  I believe that electrical resistance would increase with temperature, but I'm not sure of that.

[TimNJ]

Thanks! So with a heavy load, internal resistance is going to increase due to the higher current and power dissipation. If the battery could dissipate more power it could probably handle the high current better.

[IanB]

I’m working on a real simple lab for my circuit analysis class, but I’m having a hard time sourcing information to make sense of my results. Part of the lab was to hook up a 2 AAs, and a 0.5ohm resistor in series and measure the voltage across the battery terminals every minute for 15 minutes. Initially we measured 3.0V open circuit, and about 1.8V across the battery, and by the end, it had decreased to 0.37V. BUT, after unloading the batteries, the open circuit voltage went back up to 2.56V. It makes sense that the battery had discharged so it lost some of its voltage. However, I don’t understand why the voltage across the load kept dropping. By the 15th minute, the calculated internal resistance was about 3ohms, while it had been about 0.3ohm at the very start. I thought it had to do with the battery heating up like a PTC which caused the resistance to go up…but after reading about batteries, it seems that the internal resistance of a battery goes down with increased temperature. What am I missing!? Thanks so much!

When you calculate internal resistance you cannot continue to assume the open circuit EMF remains at the initial 3.0 V during the discharge. So your estimate that the "internal resistance" increased to 3 ohms is really not the case, not unless you periodically disconnected the load resistor and remeasured the open circuit voltage during the discharge. The internal resistance most likely did increase, but I doubt your exact number.

As to why the voltage across the load kept dropping, this is the only thing that can happen, right? The voltage across the load is given by V = IR, where R is constant at 0.5 ohms. As the cell depletes and loses energy the current it can produce decreases towards zero, meaning that the voltage similarly must decrease.

There are various factors inside the cell that cause its output to decrease during a discharge test like this. Firstly, the output current represents a rate of chemical reaction going on inside the cell. As the chemicals get used up their concentration decreases and reaction rates decrease with decreasing concentration ("the law of mass action").

There are temporary factors too. For instance the chemicals get used up faster next to the electrodes where the electrochemical reactions are occurring and it takes time for fresh reactants to diffuse in and fill the gap. Also the reactions in the cell produce hydrogen gas. Normally this gas is catalytically removed by some of the chemicals but in a heavy discharge gas may be formed faster than it can be removed. This will cause gas bubbles to accumulate next to one of the electrodes and reduce the available surface area for reaction.

[TimNJ]

Ian to the rescue!...as usual. Alright so I suppose this is more chemically related than electrically/thermally...though its probably a mixture of all. But the simple ohm's law analysis makes a lot of sense. Thank you.