In article , Josh wrote:
On Mon, 31 Aug 2009 02:45:56 GMT, (Doug Miller)
wrote:
In article , Metspitzer
wrote:
On Sun, 30 Aug 2009 20:58:38 GMT, (Doug Miller)
wrote:
In article
, JIMMIE
wrote:
It does not affect the current that the wire COULD carry or the
AMPacity. A long cable, compared to a short cable, will the current
in the total circuit because it adds resistance to the total circuit.
Adding resistance to a circuit changes voltage, not current.
I can't believe what I am hearing.
Oh? What's the problem?
If you add resistance in series to the destination load (which is what
higher wire resistance does), the *total* R seen by the source is
higher. Assuming a constant V at the source, this means I (current)
through the wire will be lower.
No, it doesn't. It means *voltage* on the other side of the resistance will be
lower. Current is the same at all points in a series circuit.
At the *load*, V will be lower also
^^^^
You misspelled "only".
(often called the "IR drop" of the
wire) -- the load is now part of a "voltage divider".
So yes, adding resistance changes both voltage (away from the source)
and current.
No, it doesn't.
Another way to think of it -- if the V across the final load (constant
R) is lower, the current (I) must also be lower.
Wrong again. Current is the same at all points in a series circuit.
Circuit analysis can
be kind of fun -- you can often approach it from multiple perspectives
and get the same answer.
Evidently you've found multiple ways to get the same wrong answer.
Back to Circuit Analysis 101 for you, and this time pay attention when the
instructor discusses Kirchoff's Current Law.
This assumes a constant resistance load, which a light bulb isn't
completely, but is sufficient for this purpose.
You're assuming a *lot* of things; unfortunately, almost none of them are
correct.