Weird voltage from switch single pole switch

[dkarst]

If you don't want to use a Wiggy or old VOM like Simpson 260, many modern multimeters have a "Low-Z" setting that presents something like 3k ohms and will drop any coupled voltage to give you a true reading. Another alternative is Fluke make a "stray voltage eliminator" which plugs into meter and does essentially same thing for $100.

Without knowing more about your tic-tracer and how it responds to coupled voltage, anything else would be speculation.

 

[gar]

230613-1029 EDT

The original Simpson 260 was somewhere in the 20 to 30 dollar range new, had a square corner case, and AC was measured with a copper oxide rectifier at 1000 ohms per volt. DC was 20,000 ohms per volt, and still is. Some time between about 1948 and 1955 a germanium rectifier replaced the copper oxide, and AC sensitivitv became 5000 ohms per volt.

Thus, on a 1000 V DC range the input impedance became 20,000,000 ohms per volt, and AC was 5,000,000 ohms
per volt.

I have a 260 from about 1947 that has the rounded corner case. So from that time period to now the case from the front looks the same.

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[Frank DuVal]

Standard DVOM volt meter and I was away from other wires to avoid capacitive voltage.

As others have said, capacitance is in the wires in the wall.

Fluke adapter is $100. You can make your own (less safe, I'm sure) from a Pamona MDP-02 and a 3 killohm resistor (3,000 Ohms, 1/2 watt). Use heat shrink to cover the leads and only use the adapter after you have determined there is no 120 or higher voltage on the wires to be tested.

 

[ptonsparky]

Augie? who besides the 3 or 4 here even own a wiggy, let alone know what it is?

Well, if they have been reading here for any length of time, they should know. It comes up often enough. A trip to a BB store should find them one.

 

[LarryFine]

I've been using the Knopp K60 for troubleshooting for decades.

I only use a voltmeter when I need to read the exact voltage.

 

[don_resqcapt19]

My scenario multiple circuit neutrals tied together at a random J-box, power returns on all possible paths, thru a resistive light bulb load, and splitting the load between all neutrals, then getting a feed back thru another bulb onto the switch leg then contact between the EGC and switch leg.
Same reason when having multiple fault return paths it requires a much higher fault current to trip a breaker when you have the multiple return paths than otherwise needed.
....

There are only two conditions that could result in that, the first is a voltage drop on the neutral of 30 volts, but that is not likely. The second is that the neutral is open between the point where the shock occurred and the service equipment.

 

[Jpflex]

Update: I confirmed this 15 to 20 volts AC I encountered was indeed due to inductance in the conduit containing the line wire and switch leg wire in close proximity.

I removed switch wire and ran another wire on switch screw with switch off and checked voltage to neutral potential. Voltage was 0

 

[Jpflex]

Update: I wrote an update to this post this morning but I’m not sure why it’s not showing on my end.

Anyway I did get a chance to confirm that the cause of voltage on switch leg up to 20 volts AC was due to induction from adjacent line power wire ran in same 1/2 conduit to switch box.

I removed switch leg and ran a separate wire outside of conduit and voltage was 0 volts to neutral

Also the reason I came back was because old knob and tube “cloth” insulation tore and created a short to blow old Edison plug fuse. Switch wires to box were replaced but I was not originally given time for wire replacement from panel

 

[gar]

230614-2049 EDT

Jpflex:

I believe is more correct to describe the type of coupling as capacitive, rather than inductive. You need current flow to produce inductive coupling.

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[Dsg319]

Fluke 117 dual impedance meter is best of both worlds. Great trouble shooting meter that won’t break the bank.

 

[Jpflex]

230614-2049 EDT

Jpflex:

I believe is more correct to describe the type of coupling as capacitive, rather than inductive. You need current flow to produce inductive coupling.

.

Yes I realized this. However I believe in order to show a voltage you also have a current, although small with voltmeters typical 10 mega ohms of impedance

Or capacitive charge on electrical cabinets with ungrounded systems

 

[Jpflex]

Actually it could be inductance with alternating magnetic field of hot line wire 60 hertz inducing into second wire separated by dielectric which acts like a secondary on a transformer

Or if as a capacitor with plates (wires) separated by insulation but second plate or switch leg not being connected to neutral “neutralized” this idea

 

[Dsg319]

Yes I realized this. However I believe in order to show a voltage you also have a current, although small with voltmeters typical 10 mega ohms of impedance

Or capacitive charge on electrical cabinets with ungrounded systems

How can an open circuit with 120volts have current?

 

[gar]

230614-2349 EDT

Dsg319:

Depends upon what you want to call an open circuit. For example if you have 120 V applied across a 1000 megohm resistance you will have a very small conductive current, but in most applications this would be considered an open circuit. However it may e possible to measure this current.


Jpflex:

If I take a 10 ft length of thinwall steel conduit, to have magnetic material, and run a single wire thru this conduit with 120 V 60 Hz applied to this wire, and place a remote 10 A load thru this wire, then ---

(1) Connect the conduit to earth and put a thin sheet of Mylar around a potion of the conduit, and over this some copper foil. From the copper foil to earth I connect a high impedance meter, then the meter will read zero volts. The conduit acts as an electrostatic shield. There is capacitance from the 120 V wire to the conduit, and current flows thru this. But the conductive conduit acts as an electrostatic shield, and no current flows from the internal wire to the external foil around the conduit.

(2) Same setup as (1) except I now use a 2000 turn coil connected to the voltmeter. With the coil outside of the conduit I see an induced voltage. The conduit, even though a magnetic material, is insufficient to shield the wire's magnetic field from getting outside the conduit.

(3) Put an isolated test wire inside the conduit, and you will get both magnetic and capacitive coupled signals to the sense wire.

. .
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[Joethemechanic]

(3) Put an isolated test wire inside the conduit, and you will get both magnetic and capacitive coupled signals to the sense wire.

I would think that this would be best illustrated by an oscilloscope. The capacitive portion would lead and the magnetic would lag

 

[Joethemechanic]

@Jpflex

What you experienced was more or less negligible, but here is an explanation of one of the ways it can become very undesirable

 

[Jpflex]

How can an open circuit with 120volts have current?

Through the DVOM

 

[ptonsparky]

Through the DVOM

And how much current is flowing through your meter at that time?

 

[Jpflex]

And how much current is flowing through your meter at that time?

120i / 10,000 ohms x power factor

I didn’t say it would be significantly large but it is present

 

[Jpflex]

120i / 10,000 ohms x power factor

I didn’t say it would be significantly large but it is present

At least 120V / 10,000 ohm if loads are at unity power factor

120V not 120i