# Is it safe to measure AAA battery current with a multimeter directly? [duplicate]

Just bought a consumer grade multimeter, read carefully the instructions and tried some things. I know some very basic things about electricity, but I am unsure whether what I did was safe and didn't damage my multimeter or make it not work properly whatsoever.

After reading many forums I know it is relatively safe to measure voltage. As for current, I know you must be cautious.

• connect the red probe to the 10A plug (2)
• put the dial switch of the multimeter to 200m (3)
• turn it on
• touch the red probe to the positive and black - to the negative of the AAA battery Nothing blew up. :)

So, here are my questions:

1. Is it safe what i did for the multimeter?
2. What does the reading 54.0 on the display mean?
3. The red probe is in the 10A plug (2), so how am I supposed to interpret the scale to the switch? I mean, there is a position that says 20m/10A (4), so I assume that is 10A. But what about the next position 200m (3)? Is it 20A? Because, you know, they say when you don't know what you're measuring it's best to start with the largest scale and work your way down. So, what is the largest scale? Is it 20m/10A (4) or 200m (3)?
4. What about the measurement when the dial switch is in position (4)?
5. After all, I can't seem to understand how many miliamps is the current that my battery gives.
6. Is it safe to measure if the probe is in port (1)?
7. I know the multimeter didn't blow any fuse, but my OCD wants to know whether the tool suffers from these measurements.

EDIT:

1. If that measurement is so bad for the battery and the multimeter, as many people say, then is it also bad (for the battery and for the tester) to measure batteries with this battery tester that I bought for a couple of bucks:

EDIT 2:

This concrete meter has a function "Battery test". See in the photos - the dial switch has a "BATT" section for 1.5V and 9V and the red probe must be in plug (1). However, when I test this same battery, it shows 1.27 - just the same as when I measure its voltage. The manual says (copy/paste): "In the measuring ranges BATT 1.5V and BATT 9V, the battery to be measured is charged by an internal resistance, thus you obtain practical information on the condition and functionality of the tested battery". I don't know if the translation of the manual to English is correct, but that's what it says. I still wonder if it does anything different, though. May be it just doubles the voltage meter for people who couldn't figure it out.

That being said, I wonder why don't all multimeters costing more than a few dollars have the same features that the Battery Tester 3 (costing \$5) has.

## marked as duplicate by Chris Stratton, mkeith, W5VO♦Sep 18 '16 at 17:04

• You don't measure current across a battery like that because an ammeter setting is effectively short circuiting the battery. In this case you were lucky it was only an AA cell. Had it been a car battery you would certainly have blown the fuse and/or destroyed the meter. Current is measured in series with a load. Voltage is measured across. – JIm Dearden Sep 18 '16 at 15:29
• It means that little cell is supplying 5.38 Amps. Which it won't do for very long... In any other scale the reading is meaningless. And for any battery larger than AA, it's dangerous for both the meter and the battery. Don't do this with ANY size of lithium battery, for example! – Brian Drummond Sep 18 '16 at 15:35
• I shorted an NiMH cell with a wire once accidentally, and the result was unbelievably energetic. There was a giant blue spark, and I was so startled I dropped the cell. A huge chunk of the positive terminal completely melted away. I believe short-circuiting an eneloop can produce 20A or more. Luckily your meter must have around 200mOhm resistance which limits the short circuit current to around 5A. – mkeith Sep 18 '16 at 16:08
• I agree with Chris Stratton that this is a duplicate. I even made the same comment on the linked question that I made on this one. LOL. However, the OP did put a lot of effort into this question. – mkeith Sep 18 '16 at 16:13
• If the pictured battery tester performs a loader measurement at all, and if it is well designed, it applies a "reasonable" load rather than a short circuit. Of course what is a reasonable load depends on the intended capability of the cell - a load that challenges an AAA won't be much for a D cell. – Chris Stratton Sep 18 '16 at 16:25

No, it is not safe. An ideal current meter is a dead short. An ideal battery has zero internal resistance. So, in an ideal world measuring a battery by directly connecting it to a current meter will create an infinite amount of current.

In the real world, there is some resistance in just about everything. So the current will be limited. But most of the time this will either blow a fuse in the meter or damage the meter. In some rare cases the battery may destroy it's self. Which, for some battery chemistries, would be very bad.

That said, let's look at the multimeter's settings and what is displayed by the multimeter.

Read the instructions. That should the the last word. But an interpretation of the picture of the setup shows you have plugged in the probes to the 10A socket. And the dial is set to 10A. Guessing, the alternative label 20m is for when the probe is connected to the mA port.

The display is showing 5.38. Guessing, this is based on a 10A scale and is telling us that the battery is producing 5.38A. That is really good for a 1.5 volt cell. This is good news as we see on the multimeter's labeling next to the 10A port that the multimeter's fuse should blow when exposed to 10A for 15 seconds. Good thing you were only testing a small battery!

Even though what you did is not recommended, let's not let the information go to wast!

We assume the battery chemistry is designed to produce 1.5 volts. And you measured 5.4 amps. We can then calculate the total resistance the current is running through.

V = I x R
1.5 = 5.38 x R
R = 0.28 ohms


So now we know the resistance of your 10A current meter plus the internal resistance of the battery. If you used a second meter to measure the resistance of the 10A current meter, we could subtract that from 0.28 ohms and find the internal resistance of the battery.

Internal battery resistance is a whole other (interesting) area with respect to your question which indicates how efficient your battery will perform.

As stated earlier, an ideal current meter has zero ohms of resistance. As a current meter is really a voltage meter (ideally infinite resistance) in parallel with a shunt resistor, we pick the shunt resistor to be a small as practical.

This arrangement makes for a good current meter. But is a bad way to test a power source.

When testing a power source we use a "load". Generally a resistor which has the capacity to dissipate the power we are about to put into it from the power source. We chose the resistance of the load to be the equivalent of what we normally use the power source for. For example, a 1.5 volt incandescent light might be 100 ohms. So we can pick a 100 ohm "load" or "shunt" resistor to test a 1.5 volt battery. And we expect to see about 1.5 volts for a good battery. So we create a graphic for our voltage meter such that when it is deflected by a 1.5 volt potential the needle is over the color green.

• I believe that is an NiMH cell. – mkeith Sep 18 '16 at 16:02
• If you search on NiMH you will find they have very low internal resistance. If your NiMH battery had 25mohms resistance, that would infer your 10A meter was using an about 250mohms resistor to sense current. Which sounds plausible. – st2000 Sep 18 '16 at 16:13
• Yes. My point was that the 1.5V you are assuming for V may not be realistic for an NiMH cell that is delivering 5.38A to a load. Maybe 1.1V would be more reasonable. This doesn't materially change anything, though. Maybe the Ohmeter is closer to 200 mOhm than 280. – mkeith Sep 18 '16 at 16:19
• Yes, that is a good observation. Which does effect the calculations. Fortunately, you have the means in your hands to (safely) verify the voltage the battery is producing. – st2000 Sep 18 '16 at 16:47
• Not me. The OP. But he would need two meters. – mkeith Sep 18 '16 at 16:58

With the positive lead in the 10 Amp socket, the meter reading is only valid with the switch in the 10 Amp position.

While measuring battery current as you did is officially a Bad Thing, I have often done it to estimate how dead used AA or AAA cells are, but I wouldn't do it on any larger cells as they would be able to provide enough current to damage the meter (or me! ).

The point you need to understand is that in an ideal circuit, the current is proportional to the load resistance. This means that the battery does not have an inherent current to measure. The battery will "attempt" to supply however much current that the stuff connected to its terminals (the "load") demand.

In practice, the battery has a limit to how much current it can supply which is characterised most simply as an "internal" resistance, so the current delivered is not infinite into a dead short (a zero load). But with many types of battery this is rather low and thus a very large current will flow through the short circuit (in this case, your multimeter), causing sparks and bangs.

This indeed is why short circuits make big bangs.

As I said you cannot measure the battery's current on its own, it has no meaningful value.

• Thank you for the answer! the battery does not have an inherent current to measure. you cannot measure the battery's current on its own, it has no meaningful value These two sentences really did help me grasp the concept of current which was kind of unclear to me for years. Wish I knew it before I tried measuring. :) very large current will flow through the short circuit (in this case, your multimeter), causing sparks and bangs Hmm, don't they put any resistors in those multimeters? I don't think it would be a true short circuit, no? – Pavel Tankov Sep 18 '16 at 19:38
• @Pavel, it depends on the meter, but most will just have a fuse or electronic equivalent. Anything measuring amps has to have a resistance as low as possible so that it does not limit the current itself (thus giving a false reading). – Ian Bland Sep 18 '16 at 20:25