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

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.

It sure as heck does.

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.

Yes, it is, but a higher resistance draws less current - so the
current through the entire circuit is reduced.

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.

You need to look at ohms law first. Then Kirchoff's applies.

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.
And you are also wrong.

If a load has an effective resistance of 20 ohms on a 115 volt supply
the circuit draws 5.75 amps.

If you add a .5 ohm resistance in the wiring to the device, the
effective resistance is now 20.5 ohms, and the current is 5.609 amps.
The voltage drop across the .5 ohm resistance is(.5X5.609=) 2.8 volts,
so the device sees only (115-2.8=) 112.2 volts.

Power without resistance in the line is (5.75X115=) 661.25va (or watts
if DC or straight resistance load) while it drops to
(5.609X112.2=)629.33 va.
31.92 watts of power is wasted in the resistance of the wire.

That's true any way you slice it.

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On Mon, 31 Aug 2009 02:47:34 GMT, (Doug Miller)
wrote:

In article m, David Nebenzahl wrote:
On 8/30/2009 1:58 PM Doug Miller spake thus:

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.

Changes current, too. E.g., a current-limiting resistor in series with a
LED, sized so that the LED won't draw excess current and burn itself out.

Wrong.

If you want to limit the current passing through an LED, you put a resistor in
*parallel* with it, not series.

Nope. And an LED is a "spacial case" too, because it is NOT a
resistance. It is a fixed voltage drop.

Total current in the circuit remains the same.

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