Electrical wiring: the "last inch"

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The sub-thread up yonder about using 15-amp switches on 20-amp circuits got me to thinking. Actually, have been wondering about this for a long time, so here goes.
The thing is, we have circuits where the wiring and devices on that circuit are designed to safely carry a certain amount of current, for example a 15-amp circuit using devices rated for that amount and wired with #14 wire. All well and good.
But our discussions here inevitably leave out what I'm calling the "last inch". By that I mean such things as the wires that feed a lighting fixture, attached inside the wall or ceiling box, or the wires connecting a dimmer switch. These wires are always a *lot* smaller than the cable used to wire the circuit; often they're around 16 or 18 gauge.
This would seem to violate the integrity of the circuit, because now you have weak point. In the worst case, a short circuit at the device, you'd have a lot of current going through these smaller wires, until the breaker trips. Isn't there a greater chance of fire in that case?
So how does the NEC reconcile this apparent violation of the integrity of the circuit? How do folks like us who install and work on such wiring rationalize it? Is it simply a matter of practicality? (It's obviously not practical to use 14-gauge wire all the way up to every device on a lighting circuit.)
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Lets say you have a light fixture with a 100 watt bulb. Inside the fixture they use the smaller wires and something shorts out. Those wires are encased in the fixture and the box. The short circuit lowers the resistance and the amps shoot up and the breaker pops. The excess heat from the short is contained and dissipated in the metal fixture long before it causes mischief.
Now consider a different proposition. The circuit has a penny in the fuse box and the circuit is overloaded. The wires inside the walls heat up and catch the house on fire.
See the difference?
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On 8/29/2009 2:38 PM Roger Shoaf spake thus:

>

That makes sense; the idea is to confine any potential fires within boxes, where they presumably won't burn the damn house down.
Which is why I use metal boxes instead of plastic ones, and pay attention to properly clamping cables going into the box (rather than just sticking the cable through a hole in the box).
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David Nebenzahl wrote:

Hmmm, Is there anyone who does not do that when wiring? Then it'll fail inspection. Every thing is simple math and cool head. Actually life is.
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Typical dimmer is rated a 600 watts. The wiring attached to it, is of substantial size to handle its rating. If you get a 1000, 1500, or 2000 watt dimmer, you will find that it has larger conductors attached to it, to handle it's respective load. A switching device is not necessarily designed to carry the entire load of a circuit, unless it's going to carry the entire load of a circuit.
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Also, just as a long wire needs to be a thicker gauge, compared to a normal length of wire, to carry a fixed amount of amperages, a very short length of wire can be rated to carry a larger amperage at a smaller gauge than normally used. This is the rational used on appliance cords, and the internal wiring in appliances. I have seen a formula somewhere that will determine the exact gauge needed for a given length at a specific amperage.
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On 8/29/2009 2:49 PM EXT spake thus:

Saying "a fixed amount of *amps*" would do.

That's not true. Conductors are rated at a certain current regardless of their length.

I was going to bring up the aspect of cords too, as our 20-amp circuits have cords plugged into them that are rated at far less than that, creating another potential source of fire.
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David Nebenzahl wrote: ...

... Not exactly so--look at the voltage drop tables; at a given voltage the drop becomes excessive at a minimum conductor size and a larger conductor is required.
All these points have been considered--in essence, the answer is that the individual appliance/light/whatever has conductors sized specifically for the load.
A 100W bulb, for example, on the 15A 14ga circuit doesn't need 14ga because it draws only 1A (in round numbers)--the circuit wiring is required to be larger to account for the loading of all devices in simultaneous usage on the circuit.
Also, again, NEC specifically covers the wiring not the end devices; they're under other guidelines such as UL, etc., ...
In the end, there's no increased risk in common usage as long as you don't do something that is in obvious contravention to intended use--put a 300W bulb in a 25W rated fixture or 3 1000W hair driers on a 25-ft 16ga light-duty extension cord, say. Sure, one _can_ do stupid, it's presumed the Darwin rule will take care of that...
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Yes, exactly so. Voltage and current are not the same. A long conductor *does* cause voltage to drop, but it does *not* affect current.
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On Aug 30, 7:50am, snipped-for-privacy@milmac.com (Doug Miller) 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. Voltage E, Current I and Resistance R are not independent of each other under normal conditions.
E=IR I=E/R R=E/I
Jimmie
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I WROTE WHAT!!!!! Sorry answered two phone calls and three questions from my wife while trying to write this. Please ignore my dribble.
Jimmie
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I normally do anyway... <g>
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Adding resistance to a circuit changes voltage, not current.
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On Sun, 30 Aug 2009 20:58:38 GMT, snipped-for-privacy@milmac.com (Doug Miller) wrote:

I can't believe what I am hearing.
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Oh? What's the problem?
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On Mon, 31 Aug 2009 02:45:56 GMT, snipped-for-privacy@milmac.com (Doug Miller) 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.
At the *load*, V will be lower also (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.
Another way to think of it -- if the V across the final load (constant R) is lower, the current (I) must also be lower. Circuit analysis can be kind of fun -- you can often approach it from multiple perspectives and get the same answer.
This assumes a constant resistance load, which a light bulb isn't completely, but is sufficient for this purpose.
Josh
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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.

^^^^ You misspelled "only".

No, it doesn't.

Wrong again. Current is the same at all points in a series circuit.

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're assuming a *lot* of things; unfortunately, almost none of them are correct.
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On Mon, 31 Aug 2009 03:47:35 GMT, snipped-for-privacy@milmac.com (Doug Miller) wrote:

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...

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"

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).

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

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

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
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On Mon, 31 Aug 2009 03:47:35 GMT, snipped-for-privacy@milmac.com (Doug Miller) wrote:

It sure as heck does.

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

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

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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Josh wrote:

Hi, R is lower and I is lower? Where does it come from?

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