### Author Topic: Battery voltage drop.  (Read 1264 times)

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#### McBryce

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##### Battery voltage drop.
« on: October 31, 2017, 02:40:06 pm »
Hi all,
I was just messing about with some old batteries, setting up my DC Load for a battery capacity test. I have two old 9V batteries, both read about 7.5V with no load. If I connect either one to my DC Load and try to pull 500mA, the voltage drops to about 3V, but supplies the 500mA. All as expected so far. However, if I connect the two batteries in series (total no load voltage now around 14.8V) and try to pull 500mA the voltage drops to 0.027V and can only manage around 280mA. I wasn't expecting this, but I'm no battery expert.
I would have expected somewhere around 6V and the 500mA still being supplied, but obviously I was wrong. Is this due to the two ESRs being added together? The maths don't add up for me. Can anyone explain exactly what's happening here?

Thanks,
McBryce.
30 Years making cars more difficult to repair.

#### IanMacdonald

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##### Re: Battery voltage drop.
« Reply #1 on: October 31, 2017, 02:59:32 pm »
The internal resistance per battery is (7.5-3)/0.5 or 9 ohms.

With two in series the volts drop across the two internal resistances (18 ohms total) with 500mA flowing would be 9v, so the terminal voltage should be (7.5x2)-9 or 6v.

I suspect you have some kind of measurement error with the result you get.

#### McBryce

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##### Re: Battery voltage drop.
« Reply #2 on: October 31, 2017, 03:20:22 pm »
Thanks for confirming that my maths wasn't wrong.

After some experimentation I found the answer: One of the batteries seems to have an issue. It only supplies the 3V / 500mA for about 10 seconds and then drops to 0V. I suspect I didn't wait long enough when checking the batteries individually.

McBryce.
30 Years making cars more difficult to repair.

Smf