On Mon, 31 Aug 2009 03:47:35 GMT, (Doug Miller)
wrote:

In article , Josh wrote:


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.

Yes, I agree. I could have worded that better -- I meant "lower than
it would have been without the extra resistance in series", not "lower
at this point in the circuit than at some other point".

The earlier post I was responding to claimed that the current would
not change if the resistance of the wire changed. That's what I was
trying to refute. You and I both agree that the current at any point
in one circuit will be the same as any other point in that (series)
circuit.

I suspect a lot of the next comments stem from that unclear wording;
here goes...


At the *load*, V will be lower also
^^^^
You misspelled "only".

Yes -- again, I meant "The V across the load will be lower than it
would have been without the extra series resistance, just as the I is
lower than it would have been without the extra series resistance"


(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.

Adding resistance to the wire changes the voltage across the load, and
thus the current through the load (and the rest of the circuit). It
doesn't change the voltage across the source (until you consider
battery internal resistance, regulator quality, and other factors).



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.

I agree. But it's not the same as it would be in a *different*
circuit, one with more or less resistance in series.


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.

I've never been accused of getting things right the first time :-)


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.

Hopefully I've reworded my statements in a way that Kirchoff would
agree with -- it's been 18 years since my Circuit Analysis classes; I
have an ECE degree but do logic design/architecture for a living, so
rarely get to break out the Ohms law these days; the Fourier
transforms scared me away from analog design for good :-)

Josh