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Electronics Repair (sci.electronics.repair) Discussion of repairing electronic equipment. Topics include requests for assistance, where to obtain servicing information and parts, techniques for diagnosis and repair, and annecdotes about success, failures and problems. |
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#1
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Posted to sci.electronics.basics,sci.electronics.repair
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![]() "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? 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? Nice experiment and good math practice but that battery was not designed to supply that much current. You really can not tell much about the battery from that test. Get the test specs on the battery and load it properly and repeat your experiment. BTW: The resistance of the lamp filament will not stay the same as it heats up. Tom |
#2
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Posted to sci.electronics.basics,sci.electronics.repair
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![]() Tom Biasi wrote: "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? 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? Nice experiment and good math practice but that battery was not designed to supply that much current. You really can not tell much about the battery from that test. Get the test specs on the battery and load it properly and repeat your experiment. BTW: The resistance of the lamp filament will not stay the same as it heats up. By the way, some 9 volt batteries are made up using cells that are not welded together, and only touching each other, which can lead to higher ESR and unreliability. I don't know if there is another battery which has been made like this. Beware. GS |
#3
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Posted to sci.electronics.basics,sci.electronics.repair
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On 2006-08-01, Tom Biasi wrote:
Nice experiment and good math practice How about the ASCII art? ;-) |
#4
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Posted to sci.electronics.basics,sci.electronics.repair
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![]() "Adam Funk" wrote in message ... On 2006-08-01, Tom Biasi wrote: Nice experiment and good math practice How about the ASCII art? ;-) I hate ascii art :-) |
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