What is the input resistance of your meter?

If input impedance is extremely high, the meter draws virtually zero current. The meter may show a voltage, but there would be nearly zero "useful" current available.

When testing cells, you should always apply a load. For what you are measuring, something like 1/2-1mA (1k-2k Ohms/volt) between the two electrodes.

If you don't put a load on standard batteries, such as AA, they will almost always read as around 1.5V good or bad, even though bad cells drop to 0.9V when given a 10mA load.

Er... I set the resistance to 0, but hopefully I did it right. I don't know what a load is...

 

@Markd, you said the digital meter (with higher resistance) will probably show a higher voltage of water… I’m guessing this is because of Ohm’s law? I’m pretty sure I set my meter’s resistance to 0, though hopefully I did it right… no, wait a second… if my resistance was 0, if you put that in V=IR you would get V=0! Now I’m confused. Although, there could be an internal resistance in the electrolyte… There is a setting on my meter to measure DCmA… could I use that to get the current and calculate resistance?
I sort of see what you’re saying about the metals… like, that the difference in electrons wouldn’t even be there if the metals weren’t different... and the only reason the voltage is lower in weaker electrolytes is because the current is lower? Sorry if I'm not making much sense!

<Not MarkD>....
but...
All batteries have an internal resistance due to their chemical makeup. The lower the internal resistance, the more current the battery can source. The meter has relatively high internal resistance (usually 10k/volt+ for analog, 500k-1M/volt for digital), so the voltage drop across the battery's internal resistance is less, making the DMM(Digital Multimeter) read higher than the VOM (Analog Volt Ohm Meter).

AC Ohmmeters can be used as a non-destructive test of a battery's internal resistance, but they are more expensive than DMMs. Most are in the form of "ESR Meters" for measuring the parasitic resistance in electrolytic capacitors.

A New/Good AA Alkaline battery has about 0.08Ω, while a dead AA battery is around 0.5Ω or more. NiCd Chemistry are in the 0.001Ω range when fully charged. Knowing the voltage and current, as well as the exact value of the test resistor allows you to find the internal resistance. Car batteries are in the 0.0001Ω or less range, while a CR2032 cell can be over 2Ω.

The more current that is drawn from a battery, the lower the terminal voltage becomes. Battery testers are essentially "internal resistance meters" in a form, they apply a known resistance (which means a known current) and show the terminal voltage dropped across that resistance.

The internal resistance of a battery limits the maximum current a battery can source, and is a great indicator of the quality and health of the battery.

--ETA: A load is a resistor or any other circuit that draws current from the battery. Batteries source current to a load.

 

... Ah, that makes sense... so if you know the internal resistance of the meter, you could have an equation like V= (I)*(R of meter + X)
(where X is the resistance in the battery, of course)

but can I just get the current from the meter and plug that in to V=IR get the total resistance in the circuit?

 

You generally NEVER want to set the meter to measure current and put the leads across the battery. This will give a bit of a false value, and will blow the fuse in the meter, unless you are working with batteries that you are making in the first post. The pegging of the needle can actually bend it, in addition to blowing the fuse.

Never use the resistance range on a battery, since the meter tries to source a current through the leads to measure the voltage drop (hence resistance), the only thing that can really happen is pegging the needle or, again, blowing the fuse.

Rough way to get a "good guess", DMM is better for this due to accuracy:
Measure the battery's open terminal voltage and write it down.
get a resistor about 10 times the expected internal resistance (maybe 10 Ohms in your case).
Wire resistor between VOM + and - leads.
Switch meter to resistance mode.
Measure resistor's exact value and write it down.
Switch meter back to voltage.
Measure terminal voltage with the 10 ohm resistor, it will drop quickly as this is a "heavy load" to a lemon battery.
Subtract open circuit voltage from the highest (first) loaded voltage (with resistor).
Use ohms law to find current, and total resistance.
Subtract the exact value of the 10 Ohm resistor to get battery's internal resistance.

 

I will have to try that...

I feel many google searches coming on... I hope we start electricity in physics soon, haha.

Thank you so much.

 

I will have to try that...

I feel many google searches coming on... I hope we start electricity in physics soon, haha.

Thank you so much.

A video is worth a million words. Check the "Related Videos" as well.

 

@bleu This has been very interesting! Things I thought I knew and I was wrong. (I should have gone farther in chemistry.) Now I'm thinking, I have some almost completely iron nails, some what did they plate this quarter with, some I know it's not really a copper penny, and some deionized water, except I used my thumb to pour it slowly into the batteries. Then there is some table salt except it has iodine in it, a probably copper shield off a gigahertz transmitter, and a tree full of ripe grapefruit. Kitchen science can be so difficult without ANYTHING pure around the house!

Still, it's fun. Keep up the good work.