Originally Posted by
SCSInet
Obviously we shouldn't turn this thread into an entirely different topic, but your own research contradicts your conclusions from it.
You are saying that an overloaded circuit will consume more power to deliver the power to the device.
But on the other hand, you are posting, as evidence of this claim, NEC documentation which states that the net effect of voltage drop is performance related, not consumption related.
The NEC documentation is right. When the conductor is undersized, given it's load AND length, voltage drop results, nobody is disputing that. Where I am suggesting that you are not accurate in your conclusion in that voltage drop doesn't cause an increase in current consumed. It's simple Ohm's law (and trust me, Ohm's law applies in the real world and doesn't simply look good on paper). Ohm's law states that current (I) is equal to Volage (V) / Resistance (R). The reason that conductors are upsized over long distances is not due to heating (though that is also an effect), but rather resistance. A 1000ft conductor carrying 15 amps will not experience any rise in temperature above what a 1 foot conductor of the same size (gauge) carrying the same current. The difference is that over 1000 feet, the resistance adds up and voltage drop increases.
Ergo, if you have a motor (in this example, we'll ignore power factor, real vs. apparent power, etc concepts associated with magnetic loads) with a resistance of 100 ohms, at 120 volts it will draw 1.2 amps. You have it connected to a 15 amp circuit, wired with the standard 14 gauge wire. The total length of the circuit, from the break itself to the motor's windings is 100 feet. Chapter 9, table 8 of the 1999 NEC lists 14 gauge wire as having a resistance of .00314 ohms per foot, or 0.314 ohms per 100 feet. That makes the total circuit resistance 100.314 ohms, which makes the total circuit draw now equal to 1.19 amps. Ergo, the motor gets LESS power, and the total circuit draws LESS power. In this case, given Ohm's law again (in this case, Voltage (V) = Amps (I) * Resistance (R)) the voltage drop of 120 volts across .314 ohms of transmission loss is 0.3768 volts, or 0.314%.
Apply that same formula to a fully loaded 15 amp branch circuit and you get 4.71 volts, or 3.925%. In this case, you are exceeding the 3% rule for branch circuits, and upsizing might be appropriate. This is what I believe you are trying to say, and I'm certainly not disagreeing with you that the voltage drop can cause a performance issue. Where I disagree is that a circuit with these parameters loaded in this way will consume MORE power, when in fact it consumes LESS, and even LESS is delivered to the equipment (which causes a whole other host of problems, especially with motors). I also content that most residential circuits are not 100 feet long.
Ok so long (very long) story short, what I'm saying is that fully loading your circuits with them at the same time being excessively long causes power you would have consumed anyway in the equipment to be wasted in transmission, but it does not cause the circuit to draw more than it would have had it been properly sized. It only causes power which you want to end up being consumed by the equipment to be consumed in transmission instead, so what we are talking about here is efficiency, not power usage.
So in the above example of
the circuit never takes up 16 amps, it takes up the same 15 amps (actually slightly less), and 15 amps never makes it to the device.
I must say I love it when a debate contains hard facts and I dont mind being proved Wrong, I actually like the fact that you showed me information I over looked/mis-interpreted. I also appreciate it not degrading to an "I'm right your wrong" childish banter that happens so often on boards. I do feel I'm over looking something though, I may get back to you later on this topic SCI,
But in the meantime I conceed that you seem to have a more accurate perspective, Thanks for holding a sensible conversation about the off topic topic.
