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Diesel

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.

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.

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.

--
MID:
Hmmm. I most certainly don't understand how I can access a copy of a
zip file but then not be able to unzip it so I can watch it. That
seems VERY clever!
http://al.howardknight.net/msgid.cgi?ID=145716711400