On Tue, 1 Aug 2006 08:01:50 -0400, "Greg Neill"
wrote:

"Adam Funk" wrote in message ...
I recently tested a 9V alkaline battery by measuring its open-circuit
voltage (9.0 V) and then measuring it with a car headlight lamp (R = 1
Ohm) across the terminals (4.0 V). The lamp lit up brightly and got
warm, but from the significant voltage drop I conclude that the
battery is basically dead. Correct?


From those measurements I get a Thevenin model of the circuit as
follows, where Rb is the battery's internal resistance and Rl is the
load (lamp).

- Vb + Rb
-----|||||-----/\/\/\-----
| |
| |
o o
| |
| Rl |
-----------/\/\/\---------


With the load removed, and assuming the voltmeter is an open circuit,
Vb = 9.0 V. With Rl = 1 Ohm in place and the voltage across o-o
measured as 4.0 V, the loop current is 4 A. So Rb is 1.25 Ohm.
Correct?

Is there a rule of thumb for judging a battery as "still OK" or
"dead" based on the calculated Thevenin resistance?

I measured the headlamp as 1.0 Ohm, which in a 12 V car circuit
(assuming a negligeable series resistance) should have a power of
144 W. Does that sound reasonable?

If this is a typical 9V battery as in:

http://data.energizer.com/PDFs/522.pdf

then the usual load is in the neighborhood of a few tens
of mA up to maybe a couple of hundred mA at greatly
shortened service life. If your load current was really
as high as 4A, then if the battery wasn't finished before
it probably is now.


Shorting this kind of battery for some seconds won't do permanent
harm.


Also, the car headlight will increase in resistance as
its filament gets hotter, so your load might have started
at 1 ohm, but would have risen considerably very quickly.


At high loads, the ESR of the battery will change with time, too.
Connect a 9v alkaline battery to an ammeter. It will start at, say, 2
amps and drop off as internal polarization kicks in; the decay and
recovery time constants are very roughly in the area of a minute. At
lower currents, ESR is pretty steady.

John