Why am I measuring 16 volts on a wire that should be at 0 volts?

Question

I am replacing a power receptacle at my apartment. It has the standard four wires, with one of the two reds going to a nearby toggle switch.

I am concerned because when the toggle switch is off, I measure 16V between one of the red wires and a white. (The other red wire always measures 0 V, so I assume it doesn't go to anything.)

I don't want to wire up the red if it's going to sit at 16 V when it's supposed to be off. Any ideas why this is happening or what I can do? Or is 16 sufficiently close to 0?

Answer 1

The 16 V is likely an induced/stray/phantom voltage. That power line is "receiving" a voltage just like a radio would, since it is bundled along with a live wire (~120-130 V). The other red wire may be connected to ground (or neutral) somewhere, so that it is held at ~0 V.

If my assumption of it being an phantom voltage is true, the 16 V wouldn't be able to power any devices, and can be considered safe. It is mostly caused by the capacitance between the two wires. When you attach the multimeter to the system, you create a current path between the "open" wire and the neutral. The AC current then can flow between the wires (AC goes through capacitors), and then through the multimeter (which has a finite input impedance). The current flowing through your multimeter determines the voltage that you are measuring.

You may want to look at how the switch is wired in order to fully understand the circuit before replacing the power receptacle. Also, remember that the switch should connect/disconnect the LINE (~120 V) and not the neutral. The neutral and ground should always be connected to the outlet (and be unswitched).

One way to avoid this problem is to use a low-input-impedance voltmeter. Modern digital voltmeters usually have input impedances of around 10 MΩ. Using a meter with an input impedance of less than 500 kΩ will load the unconnected wire enough that it will not be able to develop a substantial phantom voltage. Adding a 500 kΩ - 1MΩ resistor in parallel with your voltmeter's input would be a reasonable way to dispell the phantom voltages (but be careful that you you are within the power rating of the resistor, power=V^2/R).

Old analog voltmeters often have a low enough input impedance that they will not be able to measure the phantom voltage. Also, there are some modern digital multimeters that are designed to have low enough input impedances that the phantom voltages cannot be measured. These multimeters often use PTC thermistors in parallel with their input.

Lab experiment

As an example, I've connected about 1 meter of NM 12/2 cable in a way similar to your situation. I connected neutral and line to the outside two conductors of the NM cable, and left the ground floating. I measured 31 V between neutral and the ground wire:

Theoretical calculation

Here's an example calculation (with many simplifications, worst case scenario, etc), showing that this "phantom" can be quite large, written in Matlab code. It assumes that the "red" connector is between the "hot" and grounded wires, that you're using 12-gauge wire, 19 mil of insulation on each wire, dielectric constant of PVC, your multimeter's input impediance is 10 Mohm, and no inductive coupling (only capacitive coupling). It uses the capacitance formula on Wikipedia for a pair of parallel wires. The assumed wire length is one meter. The result is that you see a phantom voltage of 33.4 V, similar to what I measured in "real life". This shows that 16 V is a "reasonable" phantom voltage that could be measured with modern, high input impedance, voltmeters.

This calculation is based on assuming that your 12/3 cable looks something like:

This would produce a voltage divider circuit (assuming no inductive coupling) something like:

The phantom voltage is the voltage across Rmm (on the right side of the diagram). For AC circuits, complex numbers can be used to represent the impedance of each element in the circuit. The impedance of a capacitor is 1/(jωC). Wikipedia has more information on voltage dividers. The magnitude of the output voltage is what a multimeter would measure, and its phase can be discarded.

% For NM 12/2 cable, approx....
% Assume flat NM cable, with Red-Line-Ground-Neutral

f = 60; % Hz
w = 2*pi*f; % rad
Vin = 120; % V(rms)

% wire diameter
a=2.053e-3; % m

% Insulation, 19 mil
t_ins = 0.019*2.54/100; %m

% Cable length
l = 1; % m

% Dielectric constant
e0 = 8.854e-12; % F/m
e = 3 * e0; % PVC has a dielectric constant of 3.

%Multimeter input resistance, value of Fluke 80 series V
Rmm = 1e7;

% Wire capacitance, formula from Wikipedia
C = pi*e*l/acosh((2*t_ins+a)/a); % F
% The impedance of a capacitor is 1/(j*w*C)
Z_C = 1./(1j*w*C);

% Impedance of Z_C in parallel with Rmm.
% Parallel impedances are combined as the inverse of the sum of the
% inverses.
Z_2 = 1/(1/Z_C + 1/Rmm);

% The phantom voltage is a voltage divider of Z_C is series
% with Z_2. The phantom voltage is the voltage over Z_2.
Vphantom = Vin * abs(Z_2/(Z_C + Z_2));

fprintf('Phantom voltage is %f V.\n', Vphantom);

Answer 2

Check the ground circuit in the building/house. Check its integrity, check its connection. If all are OK, proceed to inspect the wiring to the point where the 16 Volts appear.

If some segment of the wiring is too old, then the dielectric (insulation) may have deteriorated so much not to the point where it is completely bare (and therefore discharge may occur with the nasty consequences that we all know) but to a point where induction (just like a transformer) current may happen despite a switch is breaking the rest of the circuit. if you found such segments, then isolate it by cutting it at both sides and then put a parallel path of good conductor that is will insulated, your problem should go.