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Typical car battery, 12V. While connected to the car, It read 12.38V on my multimeter. Using the absolutely same multimeter, I measured it once I disconnected it from the car - voltage went to 12.56V. This is in the span of 5 minutes. First measure was taken outside at 1C, second at home at 18C.

Car sat for 1 week straight before disconnecting it. I charged it and reconnected, after sitting for one day voltage was back to 12.4V, thus I eliminate some sort of crazy parasitic draw as else after 1 week it would be dead, not same as after one day.

Is that type of drop to be considered normal?

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  • \$\begingroup\$ The State of Charge (SoC) of a lead-acid battery is dependent on many variables. And voltage measurement alone of LA batteries is insufficient to gauge the health of them. What is the real question here? Is the battery not starting the automobile? Most car batteries last from about 3-7 years, then must be recycled. \$\endgroup\$
    – rdtsc
    Commented Jan 20, 2020 at 13:39
  • \$\begingroup\$ Car battery is starting the car, no problem. I just think it starts it a bit slower than I would like, especially now in the winter. I had the battery load tested and it came positive, returning about 85% of it's stated CCA. The battery is only 2 years old and a couple of months. \$\endgroup\$
    – Tisho
    Commented Jan 20, 2020 at 13:43
  • \$\begingroup\$ Only things which will help: a) Use a lighter-weight engine oil in the winter (consult owner's manual), b) replace the battery, c) replace the car's starter, and/or d) relocate to a warmer climate. :) \$\endgroup\$
    – rdtsc
    Commented Jan 20, 2020 at 18:13
  • \$\begingroup\$ Already on 5W-40, starter is fast as as long as it's not the first start in the morning, even if it's 5-6 hours after, it cranks super fast and starts up fast too. I'm already in the warmest part of my country. Guess I can either replace the battery or deal with the slightly slower cranking time. I mean, it still fired at -5C with only one glow plug cycle, a diesel that is. It can be better though! \$\endgroup\$
    – Tisho
    Commented Jan 20, 2020 at 19:07

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Yes, this is perfectly normal. A battery has an internal resistance. When you add a load on to it, it becomes a resistor divider.

For example, assume you have a 12V battery and it read 12V exactly when measuring it. Imagine this battery has a 0.2 Ohm internal resistance. You then add a load of 10 Ohms.

Use the voltage divider equation [Rload/(Rbat+Rload)]*Vbat = 11.76V

This is why you get a variation in voltage measurement when you measure it with and without a load. Other environmental factors can play a part, such as temperature, but the main culprit in many cases I think you will find is the internal resistance.

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  • \$\begingroup\$ Makes sense. Is there really that much resistance in a car electrical system, a car that's turned off? Which value is the right one to look at in terms of SoC of the battery? 12.56V is very nearly full, while 12.4V is around 65-70%. \$\endgroup\$
    – Tisho
    Commented Jan 20, 2020 at 13:46
  • \$\begingroup\$ As the battery gets drained over time, the internal resistance increases, hence the voltage drop becomes more. And yeah, there is likely going to be something loading it when the car is off, even if it is only a small amount. \$\endgroup\$
    – MCG
    Commented Jan 20, 2020 at 13:48
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    \$\begingroup\$ @Tisho No, you can't assume battery SOC from a single voltage reading. If you could, there would be no need for 'gas gauge' ICs. Wht you can say about lead acid is <12v needs charging, >14v does not need charging. For middling voltages and SOCs, the voltage depends heavily on charge/discharge history, aka polarisation. \$\endgroup\$
    – Neil_UK
    Commented Jan 20, 2020 at 13:48

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