Adam Funk wrote:
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?

Your calculations are correct but they are on paper and your testing
arrangement is in real life with real components so:
a) R1 can and does change depending on brightness produced by ~1-10 or
20. (Read the Watts rating at 12- 13.7V -- car battery on charge).
b) Battery has I dependant on rate where the production of current has
some upper limit and then the voltage drops independant of Rb

So to test your measurment introduce in this circuit an ampermeter and
do your calculations again.

Have fun

Stanislaw
Slack user from Ulladulla.