How do I know how much charge a battery has left?

I have a box full of batteries (AA, AAA, 9V, and CR2032). I know some are empty because I chucked them in there together with full ones, which was probably not a smart thing to do.

Is there an easy way to know the (approximate) charge of these batteries? For the AA & AAA I don't have any device that accepts just a single battery, and for the 9V & CR2032 I have just one device where changing the battery is comparatively a lot of work (smoke alarm & light switch).

I have a multimeter, but I remember from a few years ago that measuring the volts doesn't seem to give a good indication of charge...

Measuring the voltage with a multimeter gives a perfectly fine indication of charge. You just need to know how to convert the voltage into a useful form.

For most 1.5V alkaline batteries, (voltage-1)*300 will give you the approximate percentage remaining.

Same formula works for 9V alkalines, only it's voltage-6.

Note this is only for alkaline. Other battery chemistries have different formulas.

Lithium batteries are more difficult to do this with, as they don't lose much voltage until near the end of their lifespan, as shown on the graph below though 2.5-2.6V is a reasonable dead/not dead cutoff.

• This is the right method for me. I suggested it in another question, but it was downvoted, despite being the best answer. – vladiz Dec 26 '14 at 14:27
• While this is a valid answer, it's not all that much of a lifehack. To make it a better answer on this site, you might consider adding a homemade alternative to create something similar using commonly found household goods – Zach Saucier Dec 26 '14 at 15:43
• @ZachSaucier - aside from the drop trick Door Handle suggests, I don't really think there is any kind of "lifehack" method of doing this. – Compro01 Dec 29 '14 at 9:16
• "For most 1.5V alkaline batteries, (voltage-1)*300 will give you the approximate percentage remaining." So, for an alkaline cell... So if it measures 1.4 volts, that minus 1 is .4, .4 x 300 is 120... so it has 120% charge remaining? I don't think so! ... Anyway, a multimeter puts almost no load on the cell, and so is not a valid test. It's perfectly possible for a cell to be unable to provide any significant current to a load while still measuring at, or almost at, its "nominal" voltage on a multimeter. – Jamie Hanrahan Jan 30 '15 at 8:01
• It's still not a valid test with no load. and btw, if you test a brand new "1.5V" alkaline AA cell with no load, it will read 1.6V. Subtract 1 and multiply what's left by 200 and you're back to 120% silliness. – Jamie Hanrahan Dec 19 '18 at 9:23

For the A-type batteries, drop them with the flat (negative) side down. If they land and "stick" (stay on negative side, they don't fall over), the battery is good.

This YouTube video explains this trick much better.

For your 9V and CR2032, testing with multimeters is actually pretty accurate. When new, they should be slightly greater than 9 volts and 3 volts, respectively.

• Also in addition for Li-Ion batteries that you get for torches and such (18650s, 18350s), multimeter is also a good way to estimate, as they run from 4.2v down to around 3.6v – Andrew Williams Dec 10 '14 at 10:46
• does this work for rechargeable batteries? – Angelo Fuchs Dec 10 '14 at 10:58

My take is that the multiplier in the formula is incorrect. For 1.5 volt alkaline batteries it is (voltage-1)*200. For 9 volt alkaline batteries it is (voltage-6)*33.3. A 1.5V battery is exhausted at 1V and a 9V battery is exhausted at 6V. A 1.5V battery has .5V of life and a 9V battery has 3V of capacity. You need to find the percentage of remaining life of that capacity.

• Hi, your formula is interesting. Can you please give a source article so as to understand its' details ? – SebMa Dec 19 '18 at 9:07

For 9V batteries you may touch the contacts with the tip of your tongue and if you feel a light tingle that means there's still charge left. If you feel nothing, the battery is dead

• I know it sounds scary but it does work – L.B. Mar 27 '17 at 13:55
• this only differentiates between some charge/no charge, does not indicate with any accuracy how much charge is left. – Hobbes Sep 18 '17 at 13:07
• Man, 9V batteries sting my tongue like hell if they're new. New 4.5V batteries make tolerable tingle and wet my mouth plenty if I keep it more than a second. – Matija Nalis Sep 25 '17 at 0:41

You could create a simple device using an LED, a suitable resistor and some crocodile clips.

Something like the following circuit:

The value of the resistor could be changed depending upon the voltage battery you were attempting to test. The values you choose will depend upon the LED chosen, various utilities such as this will help you calculate the correct one

If you are feeling adventurous you could change the resistor used using a switch. Something like the following circuit:

The brightness of the LED would be an indication of charge.

If this is a common problem for you then you could mount this on pretty much anything ready to use when you needed it.

• Don't think this is actually practical - how much would the LED brightness change from a battery that's brand new vs one that's 80% spent? would it even be larger than the manufacturing tolerance between two different LED's, so that you could compare them side-by-side? – user2813274 Jul 25 '15 at 0:41