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I'm working on Chapter 4 of Practical Electronics For Makers and I'm working the reader through some very simple circuits so we can see how resistors affect (lower) current.
So I set up the simplest circuit that is simply a 3V battery supply and one 10 Ohm resistor.
Fist I calculate the expected current value I will see using Ohm's law (E = IR).
3V / 10 Ohm = 0.3A (300mA)
I hook up the circuit to run through my multimeter so I can measure (and confirm) the value.
My meter displays:
157.7 mA
What?
Google...
Apparently meters have their own internal resistance.
I decide to turn Ohm's law around and calculate internal resistance.
3V / 0.157 = 19.10 (Ohms)
So it looks like my meter has around 9-10 Ohms of its own resistance.
Hmmm...
Measuring the measurements of a measuring device.
Of, course if I had higher resistance in my circuit I'd probably not notice this, because it is only 10 Ohms.
I posted a question about this on electronics stack exchange where you can see pictures and the circuit.
Multimeter, measuring current calculates less. Can I calculate meter's internal resistance this way? - Electrical Engineering Stack Exchange[^]
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Don't forget the internal resistance of the battery or power source.
9-10 Ohms sounds like a bit much for the multimeter.
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Next time you'll measure the voltage over the resistor. 
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Maybe you need a more accurate meter:
Fluke[^]
The difficult we do right away...
...the impossible takes slightly longer.
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I'm weighing 500 gr of rice. I put the scale to 0, and put the bowl on the scale. What's that, it weighs 153 grams?
It's called calibrating, and your measuring-device should know how to compensate for its own interference.
Bastard Programmer from Hell 
If you can't read my code, try converting it here[^]
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You also need to know the sensitivity of measurement.
So while you may have zeroed your scale you will also need to know that the scale is, for example, accurate to within 1 gram.
“That which can be asserted without evidence, can be dismissed without evidence.”
― Christopher Hitchens
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I'd assume the last digit on the scale to be a guess. That's still a lot less than the weight of the scale itself
Bastard Programmer from Hell
If you can't read my code, try converting it here[^]
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And how is about to replace your 10 Ohm resistor by e.g. a 100 Ohm? Can you confirm with 100 Ohm your Observation?
It does not solve my Problem, but it answers my question
modified 19-Jan-21 21:04pm.
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I think that is a great idea and will try that and let you know.
I also learned a bit more about this see : This thread the Lounge[^]
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Thank you for your Feedback.
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.
Best regards
Bruno
It does not solve my Problem, but it answers my question
modified 19-Jan-21 21:04pm.
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Don't forget the tolerance of your resistors. Look at the last color band. Tolerances of 10% or worse were not uncommon, but today resistors with low tolerances are not so expensive anymore.
I have lived with several Zen masters - all of them were cats.
His last invention was an evil Lasagna. It didn't kill anyone, and it actually tasted pretty good.
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Good point and I did measure resistor on meter a d it gets 9.8 to 10.
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It's no big deal, it's just a case of knowing the error bars for your measurement.
Every measurement in science is meaningless without knowing the precision of the measuring instrument.
“That which can be asserted without evidence, can be dismissed without evidence.”
― Christopher Hitchens
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AND the tolerances of the parts you are measuring, in this case the resistor.
I have lived with several Zen masters - all of them were cats.
His last invention was an evil Lasagna. It didn't kill anyone, and it actually tasted pretty good.
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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.
Will Rogers never met me.
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That's a great point and helps me to learn. However I am using 2 AA in series to obtain 3v.
I still thought 300mA might be a bit too much to pull but in the spirit of experimentation I rolled on. 
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Carry on! 
Electronics remains my first love, despite nearly 40 years since school! It's fascinating, always...
Will Rogers never met me.
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Quote: So it looks like my meter has around 9-10 Ohms
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)?
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By the way, did you check the battery voltage when it was actually connected to the resistor?
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I measure approx 2.2 volts across the 10 Ohm resistor in my circuit.
I'm assuming that would mean something like:
2.2 / 10 Ohm = 220mA
However, I know when I measure that I only see 157mA of current.
So...
220mA - 157mA = 63mA (which are missing somewhere).
I plug in the missing 63 like so:
.8V (voltage drop) / .063 = 12.69 Ohms
So I'm still seeing a extra ~12 Ohms that I can't account for. That is obviously higher than my 9-10 Ohm extra I was calculating but could be due to rounding.
Does that seem at least somewhat correct?
I'm just noodling my way through this to learn more. Thanks,
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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.
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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).
[update]
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).
[/update]
What's the power rating of your resistor?
With 2.2V you are still well out-of-range on a 1/4W resistor.
modified 5-Mar-18 2:40am.
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CPallini wrote: With 2.2V you are still well out-of-range on a 1/4W resistor.
All good points. Thanks for input.and you're especially right about power rating of resistor.
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