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Thinking again (and again and again ...) about this: In case you have a second multimeter available, I think the most easy Thing is to measure the voltage over the ammeter with the second meter. Having the current displayed by the ammeter and the voltage over the ammeter makes it easy too judge wheter the ammeter's internal shunt is the reason
I like your experiments and the articles.
It does not solve my Problem, but it answers my question
There are several factors affecting the measurement; the meter most certainly does not have an internal impedance anything close to 10 ohms. Most likely is that you're far overloading the battery - 300 mA is a lot for a button cell or even a small alkaline battery. They're designed for a discharge rate much lower, and their internal resistance increases when you try to pull more current than they can deliver. Even a cheap multimeter is expected to measure 10A without disturbing the circuit - perhaps a 100 mV drop = .01 ohm. Try again using a larger resistor.
That's difficult to believe.
A quick and dirty test with my cheap multimeter gave a internal resistance of about 0.2 Ohms while measuring about 0.1 A (2.2 V lab power supply on 22 Ohm resistor).
By the way, is the resistor properly sized (common 1/4W ones cannot stand the 0.9 W of your test)?
Combining a cheap multimeter and an overloaded battery may lead to all kinds of silly conclusions. If you want to comment on a multimeter, make sure you use a lab-grade power supply and resistor both capable to act nominally in your circuit; and likewise when checking a battery, make sure to use a lab-grade multimeter and resistor. And make sure none of the components involved heats up (which indicates operation outside the nominal operating range), as currently your resistor probably and your battery most likely does.
That's strange. I would espect about 3.0V in a circuit with only the batteries and the resistor (the multimeter providing negligible effect with its high parallel resistance).
According to the answer to your post on electronics stack exchange it looks you are asking too much to your batteries (I'm not an expert). So I would suggest you to make a test within limits of both resistor and the batteries (e.g. use a 100 Ohms resistor).
What's the power rating of your resistor?
With 2.2V you are still well out-of-range on a 1/4W resistor.
In addition to the other sources of error (I'd wager battery internal resistance is the biggest), you've run into ammeter burden[^]. In most modern digital multimeters, the maximum burden is approximately the full scale on your lowest voltage range, probably something like 200mV. (It measures current by the voltage drop across a shunt.)
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