On Mon, 30 May 2016 06:53:52 -0000 (UTC), Diesel
wrote:


Mon, 30 May 2016 06:09:42 GMT in alt.home.repair, wrote:

Voltage drop still depends on the load. If there is no load, there
is no voltage drop. If you have a wire with no load at all, there
will be full circuit voltage at both ends.

No, there won't. The wire isn't a super conductor. It takes a little
to push the electrons on it. The thinner and longer the length, the
more is lost in transit. If it were a super conductor, what you've
written would be absolutely true. As long as the wire has resistance
of it's own, we're subject to voltage drop. We have various ways in
which to minimize the voltage drop, though. Short of using a super
conductor however, we can't outright prevent it.

You are out of your area of expertise. With no current flow there is
NO voltage loss. Simple Ohm's Law. Nobody has been found crooked
enough to break that law.

Now, TECHNICALLY you are almost right. If you have a very low
impedence volt meter it will draw milliamps. Say it draws 5 milliamps
(0.005 amps) and the resistance of your cable is 100 ft of #10 copper
which means 200 feet of conductor. At 0..999 ohms per 1000 ft trhat
is 0.1998 ohms. Plug that into Ohm's law -E=IxR - and you get a
voltage drop of (.005 x .1998 )= a whopping 0.000999 volt drop across
the wire. Lets jump out on a limb and say you have a really rotten
voltmeter that draws 50 miliams (.050 amp) and do the math -
..050X.1998)=0.00999 volts.

That voltage drop will , in the real world, get "lost in the noise" -
well within the accuracy variance of all but the most expensive lab
type volt meters.

Watts is watts, man. Available voltage determines how many amps
it's going to take to get them, though. Lower voltage=more amps to
do the same job.

If you have a lower voltage, the watts will be lower. In our
example, that 1500 watt heater (at 240v) will be 1407 watts at
232v. Current will actually be less, not more.

Not exactly. You can have high voltage and next to no current behind
it (think of a taser) or low voltage and a considerable amount of
current behind it (think of your car battery) it's no chump and your
starter motor isn't exactly a low drain device...You can also lose a
few volts in our example, and, still be able to pull enough current
to make up the difference.

Yes, exactly. You are forgetting the "R" in E=IxR.
The starter draws a lot of current because it has a very low
resistance. A 12 volt starter drawing 120 amps has an effectice
resistance of 12/120= 0.1 ohms.and consumes 12X120= 1440 watts -
roughly 1 1/2 hp.

The taser is abour 26 watts. The open circuit voltage is about 50,000
vots. That means the current is (26/50,000)=0.00052 amps.. This means
the impedence or resistance of the tazer is something in the
neighbourhood of (50,000/.00052) = 96,153,846 ohms.

The easiest way I know of to explain the relationship is this:

It may be useful to consider the image of water in a hose. Voltage is
equivalent to pressure, water flow is equivalent to current and the
diameter of the pipe is equivalent ot the thickness of the wire - or
resistance.
The water theory of electrical theory is a very simplistic
explanation that "doesn't exactly hold water"