"mm" wrote in message
...
On Thu, 01 Mar 2007 07:53:56 -0500, Terry
wrote:


Not sure which situation you are referring to, but if you mean the hot
wire is connected to the bulb, and you are measuring voltage on the
other, the cold side of the bulb when the neutral is switched and the
neutral is open, there cannot be 90 volts there for that reason. There
would be a full 117 volts or whatever is measured at the receptacle or
at the hot side of the lightbulb.

E=IR. When I (the current) = zero, then E, the voltage drop accross
the resistor (R) is also zero. It doesn't matter how narrow the
filament in the light bulb is, with the neutral open and no current**
flowing through the lightbulb, there will be no voltage drop accros
the lightbulb or accross the wire that feeds it from the receptacle.

**But there will be some current when you attach a meter***. How much
current depends a lot on what kind of meter you use.
***I think this is a good example of the Heisenberg (sp?) Uncertainty
Principle, which says iirc that when one attempts to measure
something, he changes it. Not sure why that would always be true, but
physics was a long time ago.

But this is a good time to talk about the difference between a meter
that has 20,000 ohms per volt and one that has 11 million ohms (per
volt?, I don't know) For house current, 20,000 is plenty, but the
meters don't have the advantages of auto-polarity, and even auto
ranging might be useful sometimss. The change to digital meters might
have been started by the space program, but is driven by the fact that
20,000 ohms is not so good for electronics.

If you have a transistor that is connected by a 10 meg ohm resistor
to ground, and you want mesure the voltage on the non-grounded end of
the transistor, and you connect a 20,000 ohm meter across the
resistor, you now have two resistors in parallel, one the original 10
megs, and the other the 20,000 ohm meter. So the total resistance is
now about 20,000 ohms and current that might be flowing through the
transistor has, barring other factors, increased tremendously, up to a
factor of 500 since the resistance is now 1/500th of what it used to
be. IF otoh, you used a 11 meg meter, the resistance (impedance might
be a better term but they are both measured in ohms) of the two
resistances in parallel is about 5 megs, not such a big change.

So people in electronics, especially with transistors and not tubes
(but tubes too), needed higher impedance meters.


What is an advantage in electronics can be a disadvantage in
electricity. If you have a pretty high voltage for a home (like 90
volts) but one with a very low maximum current, and you connect a
digital 11 meg meter across it, the currrent draw will be very low,
and the voltage will not drop much because the voltage drop is
determined by E=IR and the resistance of the circuit is low compared
to the resistance of the meter. So it's like a big pipe feeds a very
teeny pipe, adn there is almost no current through the big pipe and no
pressure drop.

If on the other hand one uses a 20,000 ohm meter, that will still be
quite a restriction to current flow, but only 1/500th and there will
be 500 times as much current, so the voltage drop will be 500 times as
much.


While your description of what is hapning is ok, much of what you are saying
is not accurate. The 20,000 ohm meter is really that per volt of the full
scale setting. That is if you have the standard Simpson 260 meter
(concidered the best analog meter for electricians for years) and set it for
250 VAC there is about 5 meg of resistance in the circuit which is only
about half the resistance of the digital meter. In the problem above if it
is leakage or induced voltage it will read about half of what the digital
meter will. Also if you click to the next lowest scale you will see the
relative position of the meter will not change much if it is very low
current leakage.

Also many of the transistor circuits are high current low voltage circuits
and the tube circuits are high voltage low current circuits. The analog
meter will usually be more accurate in reading the voltages there than they
will be in the tube circuits. A 10 meg resistor in the circuit in parallel
with the 11 meg meter will register a very large change, not a small one
like you state. Anything over about 10 % is usually noticable and you have
a 50 % differance .