Post by email@example.com
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 firstname.lastname@example.org
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.
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
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