I have been using some old multimeters in a high school science lab. We did a current an voltage reading from a AA supply (two batteries) with a variety of resistors. In some cases a 1M resistor.

I noted that the only way to get current readings was to set it on the 10A circuit. OK, I thought, maybe the batteries put out that much current in toto.

But I saw this happen even when the 1M resistors were hooked up -- we would get readings of 2.7 V or so which makes sense (old batteries and all) and then when we measured current we would have to put it on the 10A and get readings like 2.0 A or 1.8. Ultimately, the digits made some sense in terms of Ohm's law, it was just the powers of 10 seemed off. So I wasn't sure if I was just reading the thing incorrectly. (Like, was it actually showing milliamps?)

To give one example: I put a resistor in line with a 3V supply. It's 46.4 Ohms. (The multimeter is an Extech instruments, found in a drawer). I touch the probes to the resistor and get 2.60 V, which seems reasonable enough. I connect the red probe to the 10A input and check current and I get ~1.9 A. But V = IR says that it should be

(2.6V) = I (46.4) --> I = 2.6 / 46.4 ~ 0.05A

We have gotten similar results with other meters, though my "good" one (a new Mestek DM91A) reads 0.04A when it's hooked up to the mA side (if I try to set it to mA range the alarm goes off, with the message "FUSE") and about 1.8 A on the 10A side.

In at least one multimeter I replaced the battery. I tried similar readings with a 965,000 Ohm resistor. I got 0.03-0.04 A with the newer Mestek, and 1.8 A with the Extech, which wouldn't read it when i used the milliamp connection.

I suspect there is some stupid mistake I am making in reading the things, and yes I checked that the positive was to positive and negative to negative. (I grew up with analog meters, and you could just read the scale :-) )

Anyhow I hope people about can help me figure this out.

Thanks in advance.

  • 2
    \$\begingroup\$ Please provide a circuit diagram of how you're hooking up the batteries, the resistor(s), and the multi-meter when doing these measurements. I suspect you're connecting the multi-meter incorrectly. \$\endgroup\$
    – marcelm
    Feb 14, 2019 at 21:11

3 Answers 3


Votage readings are taken with the meter in parallel with the load.

Current readings are taken with the meter in series with the load.

A voltmeter presents a very high resistance load to the circuit. The ammeter presents a very low resistance load to the circuit.


simulate this circuit – Schematic created using CircuitLab

Figure 1. Correct way to measure voltage and current.

By connecting the ammeter in parallel with the load you have drawn the full short-circuit current that the battery is capable of. This will have blown the mA fuse but the battery isn't powerful enough to blow the 10 A fuse. If you had tried it on a car battery it would have - and if there is no 10 A fuse (as is common in the cheap meters) it would have burned out the shunt or PCB tracks and possibly destroyed the meter.

Never short-circuit a battery with an ammeter.

  • 3
    \$\begingroup\$ "A voltmeter presents a very high resistance" ... Be aware this might influence your measurement when measureing across a 1M resistor. My meter e.g. has about 10M, so when measuring across 1M, the resistance the circuits sees is 1/(1/1+1/10) = 0.91M. \$\endgroup\$
    – Huisman
    Feb 14, 2019 at 21:25
  • \$\begingroup\$ Thanks! I think this is the problem. I haven't blown fuses by the way - this I know because I get correct ammeter readings elsewhere -- and I realized that the diagram on the lab sheets as published is a bit off (it's one of those old labs that uses the big 1-5A Ammeters, and in the pictures it looks like they are connected in parallel, and of course I forgot about the series rule because I was stupid). I will check the fuses, but I actually think they are OK since they give fine readings in other settings when I checked -- and realized I was connecting THOSE in series. D'OH! \$\endgroup\$
    – Jesse
    Feb 15, 2019 at 11:06
  • 1
    \$\begingroup\$ Good. Don't forget to upvote any useful answers and accept one if it answers your question. \$\endgroup\$
    – Transistor
    Feb 15, 2019 at 11:40

Guessing those meters have a blown fuse on the mA input, open it up and see if you find a 250mA-ish fuse that looks broken.

Which happened because of measuring the wrong way like the others suggest. And that also makes sense with a 2A short circuit current for AA batteries.


A fast way to check if the fuse is broken is to switch the multimeter to resistor measuring mode (having the probes connected to do resistor measurements).

Touch with the positive probe point (likely the red probe) the input of the 250mA on the meter. It should be very low ohmic, 0 to 1 ohm. If not, the fuse is broken.

NEVER replace the fuse with an AC fuse, use a DC fuse, same type, same ratings.

  • \$\begingroup\$ Yeah the resistance readings when I do the connection test (to make sure they go to zero when I do that) are fine, so I suspect the fuses are OK \$\endgroup\$
    – Jesse
    Feb 15, 2019 at 11:08

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