*Post by t***@optonline.net*I agree it's hard to make the case that the voltmeter becomes

a ammeter because it's not measuring amps. But with some

finite resistance, it does exist and it's effects on what kind of

circuit you connect it too can't be ignored in all cases. Plug

it into a powered AC outlet and it reads 120V because the

effects can be ignored. Connect it as you did and the

effects can't be ignored. Again, if you understand AC and

DC circuit fundementals, what's going on is easily understood.- Hide quoted text -

Theoretically, everything is a circuit. Even the air is a conductor,

and since there an infinite number of paths in parallel, and you have

to allow for resistive, capacitive, and inductive coupling, there is

alwasy current flow.

But in real world practical circumstances any sufficiently large

resistance is an open.

Except of course when you're using an instrument that

has a high enough impedance in a way that you will

see those effects. That is exactly what you are doing.

The proof? Simply use two meters in series, one set to voltmeter and

the other to ammeter. Put them between the terminals of an open

switch, or an electrical outlet, and read them both. Especially read

the one set to ammeter.

I have no idea that point you're trying to make here

besides using instruments in very strange ways.

*Post by t***@optonline.net*Again, if you understand AC and

DC circuit fundementals, what's going on is easily understood.-

Oh really? Explain it to me then. Explain why I consistently get 80

V on that circuit.

Because it's likely capacitive coupling on the

cable run. I think that has been pointed out many

times now, now? There is enough capacitance

there in that open run of wire that together with the high impedance

meter it forms an RC circuit.

Not why I might get some phantom voltage reading,

with a high impedance meter; I understand that part. I don't

understand why THAT circuit will always read 80 V when open.

What exactly do you think a "phantom voltage" is?

Something that can't be explained by electrical circuits

and instead by withcraft? It's only a phantom if

you don't understand circuit basics.

Show me

the math. Feel free to use Norton or Thevenin equivalents.

It's most simply modeled as a circuit with one

resistor, that being the multimeter, one

capacitor, that being the distributed capacitance

of the cable, and a voltage source, that being

120V AC.

I think that you are simply wrong. What is going on is NOT current

flow in a circuit, as you allege, but measurement of voltage potential

difference between two different conductors.

But unless you accept that the one conductor is

acting as a small capacitor in an AC circuit, then

you have no potential on that unconnected wire

going to the empty light bulb socket. If it's just

a perfect conductor with no capacitve effect,

no part of a complete circuit, etc.

then there is no potential difference and you

have no explanation for the 80V you see. When

you start to model it by what it really looks like,

which is a distributed model with R, C, and L,

then the voltage you see is no longer some

mysterious "phantom" voltage, but is instead

explainable by the circuit dynamics and the

meter.