very low voltage -- dangerous?

Jan 05, 2008 42 Replies

I'm doing some rewiring and have run into some confusion. I have a



240V circuit, which, when on shows about 240V on my multimeter. When I flip it's breaker, though, it shows some number of mV (instead of
0). This confused me.

I mentioned this to someone else, and they asked what the amperage was when the breaker was off. Seemed like a good question. So I went and picked up a clamp meter that detects amperage. It showed 0. This seemed strange, so I called the company (Ideal) and they said I had to put a load on the circuit to detect any amperage. Ok, but that defeats the point...



Or does it? Maybe that it says 0A when the breaker is off, combined with a few mV means it's not a problem at all?


Hi, If you are using DVM, it's OK. DVM has too high an input impedance, it'll pick up suck stray voltage. That is why I still keep and use old Simpson 260 analog meter.

If you have no load, you have no current, hence zero amperage.

Think of voltage as water pressure, and current as water flow. If there is no water flowing (no load), then a water flow meter (ampmeter) will read nothing.

mv (milivolts) is not current. It is voltage.

I suggest you try reading a book on elementry electricity. You don't have to be such an ignoramus.

Hmmm, Basic AC/DC theory is taught in high school physics class!

The wires on the shut off circuit can pick up some induced voltage from nearby live wires in the panel or anywhere along the length of those wires where they are in close proximity to any live wires. It's not a problem and not dangerous. Turn off the main and see if it goes away. Kevin

This is fairly common with high impedance solid state DMM's. You are reading a "phantom" or "ghost" voltage that does not really indicate anything. A low impedance meter is a good thing to have in addition to a DMM because it will apply a load to a circuit and eliminate "ghost" voltages. Wiggy, Knopp K-60, Ideal Vol-Test, or Fluke T+Pro all fit the bill. Try to stick with at least a Category 3 rated meter if you are using it at the Service Entrance.

Yes, but the effects of measuring millivolts with a high impedance probe is not often understood until people get some hands on experience in the field, after taking classes at a significantly higher level than what is taught in High Schools.

To expand on Floyd's post: all materials leak current. There is no 'zero'. There is only a near zero that is so low as to measure zero. And then those wires are also acting as an antenna. Just another source of some millivolts of voltage. As Floyd notes, this becomes apparent from experience. But understanding why means learning the many sources of 'leakage'.

To better appreciate the c> Yes, but the effects of measuring millivolts with a high

Put a load on the circuit, turn off the breaker, then measure -- it will almost certainly show you 0 volts as it should. If you see voltage under these circumstances _then_ you have a problem of some sort. As others have pointed out, you can not blindly rely upon a high-impedance meter in home-wiring situations since they can measure capacitive leakage across open contacts and between adjacent conductors.

easiest way to check is use a night light, 5 watts or so as a tester, or even a 25 watt light.

no light? no glow? no problem. if you test the voltage at the light it will be zero.

digital multi meters pick up all kinds of stray fields, just carry one under a above ground power line, like the power drop to your home.

It used to be anyway. And it is still available to anybody capable of entering a library.

That is not necessarily true. A "load", could for example be provided by switching on a few lights some appliance plugged into that particular circuit.

It might, or might not, affect the voltage reading, depending on the wiring between the "load" and the point where a reading is taken.

It isn't "capacitive leakage", either across open contacts or between adjacent conductors. It's induction.

And note that regardless of the load, the amount of induction is exactly the same. That is, the power induced into a given circuit by another is not going to change when the load is changed. It's just a matter of E = I * R, and reducing the load means the current goes up and the voltage goes down. Since the meter is reading voltage, it show a lower reading.

Hey, It is E = I * Z since it is AC circuit. Your formula is for DC circuit.

Don't worry, that's the exact *same* thing.

Hi, Not really. If they are same, why R and Z? My understanding is Z has two component, pure resistance plus reactance by inductance and capacitance combined. No? Are there many non-reactive AC load?

Thanks for the info.

Interestingly, today I was working on an existing circuit that was shut off at the main panel. I had tested it earlier and found no voltage. I was then handling the wires and felt enough of something that it made me recheck the voltage (e.g. the 'something' felt like a bit of current, but it didn't hurt and was very slight). The meter showed fluctuated between .9 and 1.3 volts. This is more than the phantom voltage I'd be seeing and mentioned above.

I continued to work on it and had no ill effects. But it seemed strange.

Yeah, really. It's actually E = I * (R + j)

And of course j is usually very close to zero for 50/60 Hz power frequencies.

Oh... how about the resistors that have been recommended as a "load" to make the meter reading more reasonable? How about the impedance of the meter itself! How about just about everything (lights, heaters, etc) except for motors.

The simple fact is that unless you have a large electric motor as the main part of the load, the reactance will be negligible. But even with a high reactance load it is still insignificant for measuring voltage to determine if a circuit is hot or not... after all, plus or minus 20-30 volts is close enough!

Hmmm, Then why is there a big discrepancy between Watt meter reading and VA reading? You mean typical residential electrical load has a PF of near 1? I thought my original answer to the poster was good enough.

How big is a "big discrepancy"?

Close enough.

Apparently not.

I agree with about everyone that for "phantom" voltage the impedance of the circuit you are measuring has to be very high. A light bulb connected anywhere between the wires you measure the "phantom" voltage would kill the voltage.

You get induction from a magnetic field. You get a magnetic field from current. To get induction you need a very near conductor with relatively high current.

Capacitive leakage you need a very near conductor with voltage, producing an electric field. I agree with John that the source of phantom voltage is likely capacitive.

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