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I might be in the wrong place but I want to try and get a decently credible answer.

I have just bought a brand new 12v AGM car battery. The voltage drop in the battery when a load is applied appears to be too large. It's voltage when disconnected is 12.8V. I know my car draws very close to 1amp when a battery is first connected and my clamp meter confirmed it did again this time, but the voltage of the battery dropped to 12.7V at the posts. R=V/I, thus the internal resistance is apparently 0.1 ohms. Is this not an appropriate way to measure internal resistance, as I believe 0.1 ohms is way too high?

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    \$\begingroup\$ Suspect measurement issues first, e.g. 12.76 rounds up to 12.8, 12.74 rounds down... \$\endgroup\$ Jul 16 '21 at 0:41
  • \$\begingroup\$ I would suggest using a meter with at least 4.5 digits, that gives you a decade above your ref voltage. Connect your load and let it operate for at least 10 minutes on a fully charged battery. At that point the voltage readings should be stable, then you can calculate the internal resistance and be close. \$\endgroup\$
    – Gil
    Jul 16 '21 at 2:16
  • \$\begingroup\$ In principal what you are doing is right . You might need a bigger load to do this measurement with a reasonable precision, maybe a set of DC bulbs acting as load. Note that the internal resistance of a battery changes with it's state of charge and cell life. \$\endgroup\$ Jul 16 '21 at 5:25
  • \$\begingroup\$ The meter has 4 digits.It read 12.80 and then went down to 12.70. It then slowly creeps back up to 12.8V 1/100 of a volt at a time... \$\endgroup\$ Jul 16 '21 at 6:32
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That's not how you measure primary ESR. That's the secondary double layer ESR which you seem to be measuring.

The Primary ESR must drop high current to measure accurately or after a sustained drain to equalize the secondary charge down after charging to a steady voltage.

from ratings if CA = 900 A for a 5V drop then ESR = 5V/900A is what you expect. But that is high current. 1A current is not accurate.

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In real life, a battery does not act as a voltage source in series with a resistor - if it did then a graph of voltage vs current would be a straight line. As you have observed, this is not the case and there is a significant voltage gradient where the current is relatively small. At larger currents the gradient, which corresponds to the internal resistance, will be much flatter.

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  • \$\begingroup\$ So a battery isn't ohmic? \$\endgroup\$ Jul 16 '21 at 6:43
  • \$\begingroup\$ Well no it doesn’t follow Ohm’s law; in real life nothing does exactly, some things far from it. \$\endgroup\$
    – Frog
    Jul 16 '21 at 11:23

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