Are you using a digital voltmeter?
If so, then I think I might know what the “problem” is. I say “problem” because, if you’re using a digital meter, you probably don’t have a problem.
Most digital meters have a FET input stage, and typically have an input impedance of 10M ohms. This impedance, in conjunction with series impedances in a circuit, forms a voltage divider. This can cause erroneous readings. It has “bitten me” several times.
This explanation is best performed with an example. Let’s say you flip the switches so that the top outlet is off. This means that the red wire (which goes to the top plug’s hot terminal) is isolated from the rest of the world. Or at least you think it is. In reality, there is a finite amount of impedance between:
a) The red wire and “hot”.
b) The red wire and “neutral”.
In both cases, the impedance comes from two sources:
- Material resistivity (bulk and surface).
- Capacitance.
#1 is usually very high (over 50M ohms). In properly wired installations, most of the impedance is due to capacitive coupling.
Red-to-White Measurement
In this case, the circuit looks like this:
hot -> capacitor (wall switches) -> 10M ohms (digital meter) -> neutral.
Let’s say that, with the top outlet off, there exists 0.0003 uF of capacitance between the red wire and “hot”. (This is not much capacitance, and is easily created via the two switches and associated cabling.) At f = 60 Hz this capacitance has an impedance of 8.8M ohms. So what happens when you stick a digital meter (Z = 10M ohms) across 120 VAC with a source impedance of 8.8M ohms? You get a voltage divider. And the voltage across the digital meter is… 64 VAC.
Black-to-Red Measurement
In this case, the circuit looks like this:
hot -> 10M ohms (digital meter) -> capacitor (outlet) -> neutral.
Let’s say that, with the top outlet off, there exists 0.000114 uF of capacitance between the red wire and “neutral”. (This is not much capacitance, and is easily created via the outlet and associated cabling.) At f = 60 Hz this capacitance has an impedance of 23.3M ohms. So what happens when you stick a digital meter (Z = 10M ohms) across 120 VAC with a source impedance of 23.3M ohms? You get a voltage divider. And the voltage across the digital meter is… 36 VAC.
To prove the above stuff, do the following:
Plug in a lamp to the top outlet and turn it on. Verify the voltage across the lamp (while it is on) is 120 VAC. (To do this, partially pull out the plug to expose the bare terminals, then connect across them using your meter leads. Be careful!) Then toggle one of the wall switches to turn the lamp off. What is the voltage now? With the lamp still connected to the outlet, I’m confident you’ll find that it’s zero. If it is, you’ll know the 64 V and 36 V readings were due to finite source impedances as described above.
Keep in mind, however, that while the analysis above is correct, it is based on an assumption that there’s nothing wrong with your wiring and that you are using a digital meter with high input impedance. If either one of these is not true then I’ll have to come up with something else.