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?

No, 1,25 ohm ESR (Equivalent Series Resistance) for a 9V battery is
quite good. For a 1,5V "D" cell i would consider it too high.

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?

Yep. But remember that the 1 ohm of the lamp is measured in cold state.
At 9V it is probably a lot more.

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

It depends of the size, technology and voltage. A large "D" cell will
have in its new state an Rb of about 0,1 ohm or even less.
A new 9V battery might give you somathing like 1 ohm.

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?

No, the resistance will increase by a factor of 2 or 3 as the lamp
gets hotter. Connect it to a decent power supply (or car battery) and
measure the current. That will give you the corrent operating wattage
and resistance.

--
Joop van der Velden -
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