# Why one of my multimeter shows a small voltage when only one probe is connected to AC and the other dont?

I've recently acquired a Minipa ET-997 multimeter. I noticed that when I connected one probe to 127V phase and let the other disconnected, it will show around 12V.

But I have an older and cheaper multimeter which shows 0V in this same situation. Why is that?

Another thing, the older multimeter shows -127V when the black probe is on the phase wire and the red probe is on the neutral wire. And it shows +127 when the red probe is on phase and the black probe is on neutral. Therefore, I can easily tell which is phase and which is neutral. Minipa ET-997 always shows 127V, no matter how the probes are connected. Is there a way to "fix" this?

• What's with the close vote? This is a legitimate question, on topic, well asked, and can be reasonably answered. I don't get it. May 28, 2013 at 16:49
• @OlinLathrop Going by the slew of "I don't get it either" close-votes seen in the last 24 hours, somebody's having a bad day, and is taking their frustration out on questions beyond their ken. I agree, can't see any reason for closing this one. May 28, 2013 at 18:32
• Your old multimeter is junk. When you're measuring AC voltage, the polarity of the probes doesn't matter. The measured value is non-negative. It's either am RMS value, amplitude, or peak-to-peak, as the case may be, but never negative.
– Kaz
May 28, 2013 at 20:57
• Thanks for the answers. For the sake of completeness, I just like to add that the old multimeter is a Uni-T DT830B. May 28, 2013 at 22:36
• The -127 vs 127 measurement makes no sense. The live wire will swing between a positive offset and a negative offset compared to the neutral wire, not between + and 0. To know which wire is neutral, you can measure offset to protection earth/ground, or another absolute ground point. May 29, 2013 at 0:15

What you are seeing is a result of capacitive coupling and the very high input impedance of the multimeter. This is the same effect that causes hum in a audio line when you touch it with your finger. You're body isn't directly connected to the power line, but it picks up enough, which is then fed into the audio lead, so that it is quite audible after the amplifier makes it larger.

Your multimeter is like the amplifier, except that it displays the value instead of driving a speaker with it. There is enough difference in what is coupled to the two inputs of the meter with a wire dangling from one and none from the other such that the difference is measureable.

Such signal derived from capacitive coupling is very high impedance, meaning it diminishes quickly with even a slight load. However, a good voltmeter presents a high impedance, so this capacitively coupled signal is not attenutated to oblivion. Note that it is still significantly attenuated in your example. The power line is 127 V but after the capacitive coupling and then loading by the meters input impedance you only get 12 V.

In short, this is all expected, and actually shows you have a good meter.

What Olin said, but also you have capacitive coupling because any two conductors not touching is a capacitor. (It's no coincidence that the schematic symbol looks like two non-touching conductors). The capacitance gets bigger as the conductors get closer, but the house wiring and your unconnected meter probe are still a (small) capacitor. So when you don't have one a lead connected, you have this:

simulate this circuit – Schematic created using CircuitLab

The larger the capacitor, and the larger the meter impedance (represented by the very large resistor), the higher voltage you will measure.

(the ground, by the way, is just there so the simulator can give you numbers relative to something. It's not meant to imply that the meter, or the AC supply is actually earthed, and is not relevant to the operation of the circuit at all.)

• Just a minor note: The Minipa ET-997 has an input impedance of 10MΩ (like most digital multimeters seem to). May 28, 2013 at 21:34