On 2006-08-01, Joop van der Velden <pe1dna@xxxxxxxxx> wrote:
> 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.
Interesting. I though a 44% voltage drop sounded like a lot, but as
you and others have pointed out, the load resistance I've used is very
low. What sort of resistance do I really need for this sort of test?
>> - 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.
> Yep. But remember that the 1 ohm of the lamp is measured in cold state.
> At 9V it is probably a lot more.
Right. I measured the lamp's resistance with an ohmmeter, which of
course puts very little current through it.
But I took the measurements by clipping the voltmeter (actually it's
the same meter) leads onto the battery terminals, reading the
open-circuit voltage, then pressing the lamp's terminals against the
battery terminals (the spacing was convenient --- that's where I got
the idea from) and immediately reading the loaded voltage (before the
lamp heated up).
>> 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.