Tom Biasi 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?


Nice experiment and good math practice but that battery was not designed to
supply that much current. You really can not tell much about the battery
from that test. Get the test specs on the battery and load it properly and
repeat your experiment.
BTW: The resistance of the lamp filament will not stay the same as it heats
up.

By the way, some 9 volt batteries are made up using cells that are not
welded together, and only touching each other, which can lead to higher
ESR and unreliability. I don't know if there is another battery which
has been made like this. Beware.

GS