Cell (Battery) Testing

name         Larry
status       educator
grade        9-12
location     VA

Question -   A student has a basic multimeter and is trying to
read the strength of a AA battery.  The meter is set to 1.5V when
taking the reading.  The values for DC readings give three
possible strength of the current, which one is correct?

For a new battery it reads either 240 mA, 48 mA, or 9.5 mA.  I am
assuming mA is the proper unit of measurement for the
current...correct me if I am wrong.
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Hi Larry
I am not really sure what you have going on there. If the multimeter
is set to read voltage and it is measuring the open circuit voltage
across the battery terminals, then a reading around 1.5 v will be
correct. That alone, however, does not indicate if the battery is
fresh or spent. As the battery becomes expended, internal resistance
increases within the battery and that resistance will not necessarily
be revealed with open circuit voltage testing. For internal
resistance to become apparent, the battery must be connected to an
external resistance to cause a current to flow. If the internal
resistance is large, you will see a drop in voltage across the
battery terminals. The relationship will be V = (R*Voc)/(Ri + R)
where R is the resistance of the external load, Ri is the internal
resistance (which you cannot directly measure, but can infer from the
performance of the battery under load), V is the voltage across the
battery terminals, and Voc is the open circuit voltage across the
battery terminals. The higher the value of Ri, the less voltage your
battery will be able to supply under load.
Hope this helps.

Bob Froehlich
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Larry,

The confusion here is mistaking a voltage reading for an amperage 
reading.  A multimeter can read either volts or amps at a given 
time, depending on which option is selected on the multimeter's 
dial.  Since a battery is DC (direct current) you can set the dial 
to the 1.5V option and read the voltage of the battery.  A "AA" 
battery should run about 1.5 volts, depending on the specific brand 
of the battery as well as how old it is.

Amperage, however, depends on more than just the battery--it depends 
on the resistance through the circuit.  Ohm's Law states V = IR, 
where V is voltage, I is current (measured in amps) and R is 
resistance (measures in ohms).  Using a multimeter to measure amps 
is of no use because you would only be measuring the internal 
resistance of the multimeter itself.  You have to put the battery 
into a circuit and then take a reading before and after the resistor 
to measure the current across the resistor.

Matt Voss
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Larry,
You said the 1.5V range, so  the answer is in Volts not mA.
and I would guess you read the scale that has 1.5 at the high end.

On those old meters, if the switch is set to "1.5 volts",
you are looking to read the needle on the scale with 1.5
(or maybe 15) at the high end, and has evenly-spaced numbers all across.
The red-numbered scale with numbers near zero compressed
is for AC volts only.

Out of "240, 48, 9.5" the only one that fits a 15 scale well is 9.5.
So I am guessing your battery was reading 0.95v, pretty dead.
On the scale where it reads "9.5", is the highest number 15?
If not, maybe it meant 0.24v or 0.48v, even deader.

I like to use the 2.0v voltage range on digital voltmeters.
When the alkaline 1.5v battery is somewhat rested,
(not just finished driving large currents),
I simply read the voltage.
Brand new tends to be 1.55-1.65v.
Strong is over 1.45v.
Half done is about 1.3v.
   (In flashlights you can get some useable yellowish light out of those.)
Anything below 1.2v is playing with the dregs.
Anything below 0.5v is in danger of leaking corrosive ooze
in your equipment.

I have heard it said that accurate readings are of voltage
_with_substantial_load_current_, such as a resistor that draws
maybe 200mA from an AA battery (maybe a 7 ohm resistor).
(A constant-current load works fine too,
but you'd have to build your own electronic board to do it.
A resistor is so simple.)
It is true that the internal resistance of an alkaline battery increases as
it gets depleted.
Another way to check is put a load across it, say 10 ohms,
and see how far the voltage falls in 1 minute.
But I have never needed those methods,
my open-circuit voltage reading has always done pretty well for me.

Not sure how all that translates onto your analog multimeter.
The voltage ranges listed above should not be too hard to read on a 1.5v
scale.

My casual measurements a while ago showed me that
the internal resistance of a new D or AA battery,
measured after some seconds of loading, was about 1 ohm,
and the short-circuit current was just over an amp.
Perhaps this short-circuit current decreases with depletion,
but I have not plotted it.
I think measuring that is hard on both battery and meter,
and perhaps it is not what you want.

Jim Swenson
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