Ghost Voltage Brain Teaser

Jerramundi

Senior Member
Location
Chicago
Occupation
Licensed Residential Electrician
Hey guys,

I just came across a bit of a brain teaser that I thought you all might find interesting and I wanted to get some input on it...
Here is a simple diagram of the situation:

Sump Outlet (II) ---------- Mid Outlet (II) ---------- Mid Outlet (II) ---------- Homerun Outlet* (II) ---------- UNKNOWN ---------- Panel

*N.B. What I'm referring to as the Homerun Outlet technically IS NOT the Homerun Outlet as it's not the first opening in the circuit and there is likely some unknown circuitry between this box and the panel, but relative to the room I'm working in, it is where the power comes from. So for simplicity sake, I'm just calling it the Homerun Outlet.

It is fed by a whip containing a 15a Network. Downstream from this box, the Red Hot is dedicated for the Sump and the Black Hot feeds the Mid Outlets.

I don't want to get into ALL the details as this will be pages long, but after some work I ended up with a Red Hot and Neutral from the Homerun Outlet to the Sump Outlet, completely disconnected from everything (see below)...

Sump Outlet (II) ---------- Mid Outlet (II) ---------- Mid Outlet (II) ---------- Homerun Outlet* (II) ---------- UNKNOWN ---------- Panel

Red Hot X -------------------------------------------------------------------------------- X Red Hot
Neutral X --------------------------------------------------------------------------------- X Neutral

These are two wires completely independent of EVERYTHING. Capped off on both ends, no receptacles, NOTHING. Just two wires in the pipe from box to box. When I connect the Red Hot to its' Feed at the Homerun Outlet I get 20v from Neutral to Ground at the Homerun (see below)...

Sump Outlet (II) ---------- Mid Outlet (II) ---------- Mid Outlet (II) ---------- Homerun Outlet* (II) ---------- UNKNOWN ---------- Panel

Red Hot X -------------------------------------------------------------------------------- Red Hot ------------- UNKNOWN ------------ Red Hot
Neutral X --------------------------------------------------------------------------------- X Neutral (20v Neutral to Ground)

Keep in mind, there is NOTHING connected. The Wires are capped off at the Sump Outlet. There is NOTHING connected in between. The ONLY connection is the Red Hot to it's local source in the room. This 20v ONLY appears on the Neutral when the Red Hot is livened up. There is ZERO potential connection between the Hot and Neutral. No potentially failed receptacles, NOTHING.

The only thing I can think of is that somewhere in between the two boxes the wires are scarred and maybe crossing. Thoughts?
 
Last edited:

GoldDigger

Moderator
Staff member
Location
Placerville, CA, USA
Occupation
Retired PV System Designer
We need to know what kind of meter you are using (mechanical with low input impedance or electronic with high input impedance.)
If the latter, the voltage you read is almost certainly the result of capacitive coupling between wires and cannot deliver more than microamps of current.
If there were any kind of actual insulation piercing short I would expect to see full line voltage when there is no load connected.

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LarryFine

Master Electrician Electric Contractor Richmond VA
Location
Henrico County, VA
Occupation
Electrical Contractor
A floating wire will exhibit some voltage, as read by a high-impedance meter, when run close to a wire that is energized, due to inductive coupling. That's one reason we prefer to use a solenoid tester (low impedance) for determining the presence of "real" power, and use a voltmeter when we need to know actual voltage.

If you connect even a low-wattage light bulb in parallel with your meter leads, I'll bet you'll read zero.
 

GoldDigger

Moderator
Staff member
Location
Placerville, CA, USA
Occupation
Retired PV System Designer
A floating wire will exhibit some voltage, as read by a high-impedance meter, when run close to a wire that is energized, due to inductive coupling. That's one reason we prefer to use a solenoid tester (low impedance) for determining the presence of "real" power, and use a voltmeter when we need to know actual voltage.

If you connect even a low-wattage light bulb in parallel with your meter leads, I'll bet you'll read zero.
We may refer to "induced" voltages, but the coupling involved is almost always capacitive. Inductive coupling would produce a voltage proportional to the current in the source wire(s). No current would mean no voltage.

Sent from my XT1585 using Tapatalk
 

Jerramundi

Senior Member
Location
Chicago
Occupation
Licensed Residential Electrician
We need to know what kind of meter you are using (mechanical with low input impedance or electronic with high input impedance.)
If the latter, the voltage you read is almost certainly the result of capacitive coupling between wires and cannot deliver more than microamps of current.
If there were any kind of actual insulation piercing short I would expect to see full line voltage when there is no load connected.

Sent from my XT1585 using Tapatalk
When you say "Mechanical" or "Electrical" do you mean "Analog" or "Digital?" If so, it is Digital Multimeter.
 

steve66

Senior Member
So if we understand correctly, you are trying to measure the voltage on a wire that isn't connected to anything?

That basically amounts to waving one meter probe around in the air. Most of the time it will read 0 volts, but if someone keys up a radio nearby, you will probably measure something besides 0. The meter probe is basically working as an antenna.

Now when you connect this same probe to an unconnected neutral wire, in a way you have a very long antenna. It may also read 0 volts. But then when you connect the red wire to power, your antenna is running right beside that voltage for 20 feet, 30, or maybe even 100 feet. You are going to start reading voltage.

So the basic lesson is don't try to measure voltage on an unconnected wire with a digital meter. If you use a wiggy, or an analog meter, it will usually have low enough internal resistance to connect two open wires together, and give you a zero reading.

Otherwise you are just getting sidetracked. Is there a problem you are trying to fix?
 

LarryFine

Master Electrician Electric Contractor Richmond VA
Location
Henrico County, VA
Occupation
Electrical Contractor
We may refer to "induced" voltages, but the coupling involved is almost always capacitive. Inductive coupling would produce a voltage proportional to the current in the source wire(s). No current would mean no voltage.
Yeah, that's what I meant. :rolleyes:
 

GoldDigger

Moderator
Staff member
Location
Placerville, CA, USA
Occupation
Retired PV System Designer
When you say "Mechanical" or "Electrical" do you mean "Analog" or "Digital?" If so, it is Digital Multimeter.
For what it is worth, some digital meters offer a specific range or set of ranges that have a low input impedance, to allow them to give a similar reading to what you would get from a wiggy or analog meter under the same conditions.
Also, some analog meters, like the old vacuum tube volt meter (VTVM), combined an analog meter with an amplifier to allow high input impedance.
So the true distinction, regardless of appearance, is based on the listed input impedance of the meter in question.
 

Hv&Lv

Senior Member
Location
-
Occupation
Engineer/Technician
Just throwing this out there... what if the red is twisted(or wrapped) in a circle in the boxes and the unconnected neutral is along with it somewhere basically making a crude transformer by induction...:unsure:
 

GoldDigger

Moderator
Staff member
Location
Placerville, CA, USA
Occupation
Retired PV System Designer
Just throwing this out there... what if the red is twisted(or wrapped) in a circle in the boxes and the unconnected neutral is along with it somewhere basically making a crude transformer by induction...:unsure:
True, it becomes two air core coils with mutual inductance. But, unlike a designed voltage transformer where the full line voltage is applied to the primary, the situation you describe is more of a current transformer in series with a load. A voltage will be induced into the "secondary" wire which is proportional to the current flowing in the primary wire.

My suspicion is that the effect will not be particularly noticeable in terms of phantom voltage.
 
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