"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?
No. A 1-ohm load is way too low a resistance to use with a 9V battery; the
excessive current that flowed (over 4 amps) will damage the battery. The
battery may be dead *now* because you killed it. Or, if the abuse was only
momentary, it may be OK.
I would have tried maybe a 100-ohm load, drawing 90 mA, which is a typical
(rather heavy) load on a 9V battery.
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?
If you're sure the load was 1.0 ohm, then the total resistance in the
circuit is 9/4 ohm, so indeed, Rth is 1.25 ohms.
That is a rather low Rth for a 9-volt battery. As much as 5 or 10 ohms
might be OK.
Is there a rule of thumb for judging a battery as "still OK" or
"dead" based on the calculated Thevenin resistance?
Indirectly, yes. At the current that the battery is designed to supply, the
voltage should be at least 90% of full voltage.
Actually, in this era of digital voltage meters, I measure Vth, not Rth.
Vth drops from well over 9.0 V for a new battery to 8.8 V for one that is
showing its age.
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?
Yes. Headlamps normally draw something like 10 amps at 12 V, which would
imply about 1.2 ohms. And that is approximate.