A member present a situation to which I need input.
Assume a scenario with 40 Ballard lights along a walkway each light drawing 45 watts,
It's 800 ft from panel to the 1st light and 800 additional feet the last light.
How do you address voltage drop ??
Short answer: for a 2-wire circuit, only one wire sized used the entire length, and all loads the same, just average the distances from the source to the fixtures, and then pretend all 40 fixtures are at that average distance.
Long answer below.
Cheers, Wayne
You haven't specified the wiring configuration so I'll assume a single 2-wire circuit. Using a 3-wire (split phase, L-N loads, or 3 phase, L-L loads) or 4-wire (3 phase, L-N loads) is an interesting but more complicated version of the problem.
You can, as has been suggested, always just divide the circuit in individual lengths where the current is constant and then calculate the voltage drop on each piece separately.
For the case where only a single wire size is used throughout, you can simplify that into one calculation in a couple different ways. Since the fixtures are all the same, the simplest is probably just to add up the distance from the source to each fixture and apply the voltage drop formula for a single fixture at that combined distance. [If each fixture current were different, then you could still just add up for every fixture its current times its distance from the source, and use that sum as I * D in your voltage drop formula.]
If your fixtures equally spaced, the sum of the distances is just the average distance times the number of fixtures. So 40 fixtures * (800' + 1600') / 2 = 40 fixtures * 1200' = 48,000 fixture-feet. That would mean you can do one voltage drop calculation with the using formula, using a single 45W fixture at 48,000' of 1-way circuit length. Equivalently (40) 45W fixtures all at the average distance away, or 1,200', if that point of view is more appealing.
