|Subject:||Re: 9V battery testing; Thevenin equivalent; car headlamps.|
|Date:||Tue, 01 Aug 2006 12:23:13 GMT|
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.
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