 You may have found an idiot proofing error, but in 99.9 per cent of the
 voltage drop problems a user that is familiar with electrical work can use
 the program to find practical useful answers. The program is not written
 for idiots; it is written for electricians.
 hint: electrician.com
I'll believe physicists and engineers before I'll believe an electrician.
 If the answer is negative, then obviously all the volts are dropped in the
 conductor. Any 1st year electrician apprentice should know that. But
Really? You think there will be ZERO VOLTS at the load? You really have
no concept of physics, do you.
According to my tables, the 5280 foot run of 18 AWG copper wire, at a rating
of 6.3851 ohms per 1000 feet, will have 33.713328 ohms for the full 5280
feet, and 67.426656 ohms for the full loop. Adding in the 12 ohms of load
makes the total ohmage of 79.426656. So, at 120 volts that's a current of
1.5108278 amps. That puts a voltage difference of 18.1299336 volts across
the load, and 101.8700664 volts across the wiring. The total watts in the
load is 27.3912077 (divided across the 20 bulbs means each dissipates only
1.36956 watts). The remaining 153.9081283 watts is dissipated in the wires
at 0.0145746 watts per foot per conductor.
 more than that, the ampacity of No. 18 is not 10 amperes. Voltage drop
 alone does not solve all the problem, the ampacity of a conductor is solved
 using different equations and Tables. Additionally, for No. 18 NEC
 Section 726.23 requires overcurrent protection no larger than 7 amperes.
 Also, the advanced voltage drop calculator finds the approximate temperature
 of a conductor and finds the voltage drop for that temperature. Read the
 article
Did you mean 725.23?
I suggested looking at table 310.16 under the 90 degree insulation column
for copper and note the ampacity for 18 AWG wire there is 14. That's for
wiring in raceways or buried, up to 3 conductors. Now have a look at table
310.17 and see what the ampacity for 18 AWG wire in free air is 18.
For the purpose of calculating voltage drop, the ampacity is not really
relevant. The temperature would be, but the temperature coefficient does
not alter the above values radically.
And finally, with a total resistance of 79.426656 ohms at 120 volts, the
current is only 1.5108278 amps, anyway. That's well below virtually every
rating for 18 AWG wire. You might not have much light, but the wire is
definitely not going to burn up.
 http://www.electrician.com/articles/advanced_voltage_drop_calculator.html
 for some enlightenment.

 You appear to be more into electronics and ham radio than electrical work.
 Perhaps you should stay away from the electrician page until you serve an
 8,000 hour apprenticeship and an additional 8,000 hours as a Journeyman
 electrician and then get your electrical administrators license then you
 might be able to use electrical calculators without evaluating them as a
 technician.
If electricians use calculators or formulas that result in violations of
physics, then I sure as hell don't want to hire them to wire anything of
mine. I'll play it safe and wire my own.


 Phil Howard KA9WGN  http://linuxhomepage.com/ http://ham.org/ 
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