Keep in mind that the “typical” wiring diagram may not have been followed in your case, but it still may be wired “correctly”. I`m still leaning towards open nuetral until I hear more from you.
There is no code requirements for colors to be used on 3 and 4 way switches. Except, when using romex (or other premade cables), you can use the white wire as a hot (but not as the switchleg). Keep that in mind when you trouble shoot this. You may have a hot white wire that you think is supposed to be a nuetral.
Some tricky wiring layouts involve the use of the “Chicago 3 way” method of wiring where the nuetral is switched. This has been illegal to do for many years now and should (hopefully) not be used in your case. This would always raise hell and confusion when homeowners replaced switches or modified the ciruit in any way.
True, up to a point.
There are those who claim that, if you were completely isolated from everything else in the universe, you would be able to touch any ground-referenced “hot” conductor with no ill effect. This is more-or-less true with DC, but what about AC? With AC, you have the dreaded capacitance to deal with. If the capacitance between you and earth ground is high enough, you could conceivably draw enough current to zap you, even though you’re “isolated” from the earth. Just something to keep in mind.
Yea, me too. If you really had 64 VAC on the blub, I suspect it would be glowing dimly. Do what I suggested above… measure the voltage of the top plug with the lamp plugged in. (Make sure the switch on the lamp itself is on. Then flip one of the 3-way switches so the lamp turns off. Measure the voltage between white and red… is it still 64 VAC??)
Let me know if the voltage magically goes to zero when you do this. If so, case closed.
I suggest you think about what happens when that “hot” white wire is connected to the other two neutral white wires, and come up with a better example. 
Psssst. Think about current.
Hmmm. I’m not following you here, Q.E.D. What I am describing is how it is possible for a “shared neutral” to zap you…
Sounds like a good example to me. The third wire, for all you know, is connected to active via a low resistance lamp filament or transformer winding.
Q.E.D.:
Let me explain it a different way.
Let’s say I wire up a standard outlet. Both the hot and the neutral go directly to the breaker box. A very simple arrangement. There’s no “sharing” going on.
Someone plugs a clock radio into the outlet. The radio is turned on. Music is playing. So far so good.
Now let’s say someone is in the basement screwing around with the wiring between the outlet and breaker panel. for whatever reason he grabs a pair of wire cutters and decides to cut the neutral wire.
We now have two white “neutral” wires:
- The white wire that goes to the neutral prong in the outlet.
- The white wire that goes to the breaker box.
#2 is safe to touch. But what about #1??
Assuming the clock radio is still plugged in, we can all agree that the radio is no longer powered, since the current is now zero. * But the white wire comming from the outlet has 120 VAC on it.* If you were to touch this wire we would have a voltage divider situation; the voltage would be divided between the clock radio and you. And because you probably have much more resistance than the clock radio, almost all of the voltage would appear across you.
BTW: The above example assumes you are grounded…
Oh, I see what you meant, then. One of the three wires is a current-carrying return from some device on the line. Gotcha.
Yep. And it’s a white “neutral” wire. Can definitely get zapped by them under the right (wrong?) circumstances…
cmosdes FYI - if you look at the outlets closely you will notice that the tab that connects the two brass colored terminals on the side of the device should be broken (removed). This facilitates the half hot / half switched configuration while sharing the same nuetral.
Crafterman - Ive gotten shocked by more open nuetrals than I have by hot wires. Youre right, the nuetral can get you!
whuckfistle - Thanks for the tip about the tab although I did already understand that part of it.
I’ll have to think about your “Chicago 3 way” comment and see if my circuit ends up like that. I’m not really sure how a neutral can’t end up in a switch leg on a three way, but I’ll think about it a bit more and ask later if I’m still confused.
For the record, doing the measurement Crafter_Man talked about (partially pulling the plug and measuring) seems to indicate he was dead on with his assessment of what was happening.
Also, my switch wasn’t bad as I now realize I was thevanizing the circuit by trying to measure it with it still hooked up to the rest of the circuit.
Thanks again to all. I might post a picture of how this thing is wired just to see if anyone sees any problems with it.
The 3-way switching circuit, which is composed of two SPDT wall switches & accompanying wiring, is supposed to be in series with the top plug’s “hot” connection. (In electric wiring applications, you’re supposed to switch on the hot side.) The neutral for the top plug (and the bottom plug, for that matter) is supposed to go directly to the breaker box.
Don’t feel bad; even us EE’s get “bit” by this all the time.
The funny thing is that, when you were measuring 64 V and 36 V, the meter was not lying to you; those voltages were really there. The problem is that we naturally assume these voltages are “stiff,” i.e. we assume these voltages will still be present when the load is changed. As the experiment proved, these voltages are highly dependent on the load – the higher the load impedance, the higher the voltage. If your digital voltmeter were “ideal,” i.e. it had infinite impedance, it would have measured close to 120 VAC with the switches off!
As you can surmise, the presence of leakage current / capacitance has lead many folks down the wrong path during diagnostics of home wiring. The problem is that meter manufacturers have given us too much of a good thing; in a quest to make their meters more “ideal,” they have made the meter’s input impedance very high (1 to 10 Mohms). While this is desirable when working on electronic equipment, it is often not a good thing when troubleshooting home wiring (as you learned). If you had used an old Simpson analog meter, neon bulb, or solenoid indicator, all of which have much lower impedance than a digital voltmeter, you would not have seen a problem.
I hate to admit it, but I am an EE. I just never would have thought of the capacitance in the wires as something to be concerned about.
What I’m still unsure about, however, is how this shakes out now. It sounds like you are saying I have a capacitor in series with my load. In the case of my digital multi-meter, that load is about 10M. Using a simple resistor divider,
Vload = Vsource * Zload/(Zload + Zcap)
Zcap = 1/(jwC)
Vload = 120 * 10/(10 + 8.8) = 67v (as was shown before).
But if I put a low impedance load on there, the voltage across that load would be far less than 120v.
Take a 40w bulb.
R = V^2/P = 120^2/40 = 360ohms
Vload = 120 * 360 / (360 + 8,800,000) =~ 0!
So I must be missing something really basic here.
Cmosdes: No, you’re not missing anything. Your calculations are correct.
In both cases the top outlet is “off”. Also in both cases, you have a resistor divider composed of a capacitor (8.8M ohms) and load.
In the first case the load resistance is the digital meter (10M ohms). Under this arrangement the voltage across the meter is 67 V.
In the second case the load resistance is the light bulb (360 ohms). Under this arrangement the voltage across the meter is extremely low (roughly zero). This is why you measured zero volts when you plugged the lamp in with the top outlet “off.”
Remember, the capacitances across the switches are only present when the top outlet is off. When you turn the outlet on, the capacitances across the switches go away.
You’re right; when doing diagnostics & testing on home wiring, you shouldn’t have to worry about capacitance! That’s why I’m not a big advocate of using a fancy digital voltmeter when troubleshooting home wiring problems (as I mentioned in a previous post). While there’s nothing inherently wrong with them, using digital voltmeters can cause you to make incorrect conclusions due to high input impedance. One solution: stick a 220K resistor in parallel with your meter leads.
Thanks for the info Crafterman - I had never come across this problem. Probably because I use a very basic meter with low resistance for my troubleshooting.
This is a handy thing to keep in mind if I ever use a high impedence meter sometime down the road.
I think the thing that threw me was the idea the capacitance is caused by the wiring and outlet. I guess the open switch is the capacitor?
I suppose this is easily tested by pulling the switch and separating the wires to many inches. That would change the capacitance of the open switch to approximately 0 and should leave the load impedence (as seen by the breaker box) near infinite. Putting a 10M meter in that circuit should make the voltage close to 0 across the outlet.
Capacitance is everywhere: in and around the switches, between adjacent parallel conductors, between conductors and earth ground, in and around the outlet, outlet to earth ground, etc. And keep in mind we’re talking about very little capacitance, around 0.0001 uF or so. That ain’t much. But at 60 Hz it’s enough impedance to create resistor divider scenerios with high impedance voltmeters.
You can’t get rid of it without a lot of effort. And there’s no reason to get rid of it; the impedance is high enough that it doesn’t create a safety problem.
Also keep in mind (as mentioned in my original post) that, in addition to capcaitance, there might be leakage due to simple resistance.