Do you have two multimeters? Put the one in battery test mode, and measure resistance across the two relevant jacks of the meter, whatever you measure is the built in load resistor.
When you use a 100 ohm resister across the probes the voltage flows between the positive and negative probe around the resistor and the battery (see image).
To be more accurate, it is current that flows. In simple terms, voltage is just the pressure that makes current want to flow.
But if you don't use the resistor and you just use the probes, I assume it goes back to the meter? If so, why does it go through the resistor when connected as in the picture, and not just right back to the multi-meter and bypass the resistor?
Remember this: Electricity will flow in each and every part of a circuit that it is able.
In your example, when you have the resistor and meter connected, some current will flow through the resistor and some will flow through the meter. How much flows through each will depend on the nature of each part.
If we just use simple resistance** in our calculations, the resistor is 100 ohms and a typical digital multimeter is at least 10,000,000 ohms. Use ohms law to work out how much current flows through each.
** Actual circuit parameters in the real world can get really, really tricky, especially as the frequency of things increase - but it is not necessary to worry about that with the example you are trying to understand.
I do get that. But how do the meter know what is passing between the resistor and probes?
It doesn’t know and it doesn’t care. The meter simply measures the voltage differential between any two points in a circuit.
When you apply a load to a battery, the battery’s internal resistance is also in series with the battery’s voltage source. So in essence, your load and the internal resistance create a resistive voltage divider, and your meter is measuring the center tap, so to speak. That’s why the voltage drops under load. The different internal resistance values of different battery types is why you need a different load resistor.
See the attached sketch. (The section in blue is for the pedants who will otherwise remark that a thorough equivalent circuit also shows parasitic inductance and capacitance. You can completely ignore that for your purposes.)
I am fine with someone who doesn't know that's what we here for. But as you said I suspect that the OP is trying to pull our legs.
"pull our legs. . . ." Like a conspiracy theory or something? I'm wondering why you would think that?
I've never really thought about how a multi-meter measures voltage. The image I posted made me curious. I didn't really mean how a MM theoretically measures voltage, but, rather, how does it measure voltage when the circuit seems to be independent of the meter, seems to just bypass it. In the image, the green line actually doesn't go through the battery. I think maybe that is what threw me off.
An old-fashioned meter (e.g., Simpson 260 VOM) measures the voltage between the two input terminals by connecting a largish resistor in series with a microammeter.
"20,000 ohms/volt" means a 50 uA meter with 200 k\$\Omega\$ total resistance to make a 10 V full-scale.
New-fangled meters (the usual 3.5-digit DMM) connect the two input terminals to a switchable voltage divider to obtain, say, 200 mV at the full-scale setting of the voltmeter.
The internal ADC is scaled to obtain 1999 counts at 199.9 mV at the divider output: it's an electronic circuit.
Nothing in the usual hand-held meter (of either type) is connected to ground, and "voltage" is always the difference between two points.
At higher frequencies, consider a coaxial transmission line. Cut the cylindrical line with an imaginary plane perpendicular to the wire. At a given position along the line, the voltage is the difference in potential between the center conductor and the outer shield. The current is the flow along the wire through that plane. At high frequencies, this is a local measurement.
Do you have two multimeters? Put the one in battery test mode, and measure resistance across the two relevant jacks of the meter, whatever you measure is the built in load resistor.
It reads out 13.5 ohms. That can't be right?
I tested a dead AA battery that read .8v in normal voltage mode. Then I tested in using the meter's internal battery test function, which replied with .0v. I then connected the resistor to the leads and it also read .0v.
So even though the readings were a little different between a fully charged AA when measured with the MM's internal battery test function and the resistor using normal voltage function, both the resistor and the internal battery function measured 0v (actually blipped on both .01v). So that tells me the 100 ohm resistor method and the internal battery test function are working.
I'd still like to test them again with a battery that isn't so dead, such as one that has around 1.0v -1.1v measured with the internal battery tester and then the resistor.
That sounds about right, probably a 10 ohm resistor (measuring low resistances to high precision is hard). You need to load the battery down to show the voltage dip if it's nearing end of life, 10 ohms is only 150mA
@1.5V. I have one of those generic testers with the little analog meter, with the specs molded into the case, the stated load is 4 ohms on the 1.5V position. A good AAA (or larger) will accelerate the needle to the green zone hard enough it overshoots and makes an audible click against the stop, a handy feature/bug.
That sounds about right, probably a 10 ohm resistor (measuring low resistances to high precision is hard). You need to load the battery down to show the voltage dip if it's nearing end of life, 10 ohms is only 150mA@1.5V, I have one of those generic testers with the little analog meter with the specs molded into the case, the stated load is 4 ohms on the 1.5V position, a good AAA (or larger) will accelerate the needle to the green zone hard enough it overshoots and makes an audible click against the stop, a handy feature/bug.
I'm just wondering what the different readings would be between a 100 ohm resistor and the built in 15 ohm resistor. Everyone says use a 100 ohm resistor for AAA AA 9V and C and D batteries.
Checking the 9V side I get a reading of 1500 ohms?
A higher resistor will load the battery less, who is "everyone"? What's their logic behind 100 ohms?
Then 1500 ohms is what they picked, my analog tester uses 220 on the 9V contacts, my cheap harbor freight meter with similar function probably uses yet another set of values.
There's no standard for this sort of thing, at the end of the day it's a go/no go sort of test. That's why I have the little analog battery tester. I think it's silly to measure such a thing to such high precision and I don't care to drag a DMM out every time I change a battery.
I am fine with someone who doesn't know that's what we here for. But as you said I suspect that the OP is trying to pull our legs.
"pull our legs. . . ." Like a conspiracy theory or something? I'm wondering why you would think that?
I've never really thought about how a multi-meter measures voltage. The image I posted made me curious. I didn't really mean how a MM theoretically measures voltage, but, rather, how does it measure voltage when the circuit seems to be independent of the meter, seems to just bypass it. In the image, the green line actually doesn't go through the battery. I think maybe that is what threw me off.
The current flow into all things connected to the battery in parallel that including the resistor and the DMM. But let assume the DMM is an ideal voltmeter that no current goes thru it still there is a voltage accross the 2 battery terminal whether you have current goes thru something or not (like a resistor).
A higher resistor will load the battery less, who is "everyone"? What's their logic behind 100 ohms?
Then 1500 ohms is what they picked, my analog tester uses 220 on the 9V contacts, my cheap harbor freight meter with similar function probably uses yet another set of values.
There's no standard for this sort of thing, at the end of the day it's a go/no go sort of test. That's why I have the little analog battery tester. I think it's silly to measure such a thing to such high precision and I don't care to drag a DMM out every time I change a battery.
Oh right I forgot about that. Higher resister lets less power through so less of a load?
And that explains the discrepancy between using the 100 ohm resistor and using the battery test function, as the battery test function resulted in less voltage.
I wonder why such a high resistor on the 9V side?
The 100 ohm resistor seems to be a thing on the internet. Search string: test AA battery multimeter