Voltage drop

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propower34

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If you had 15 incandescent light fixtures on the same branch circuit equally spaced 10' apart, with the first fixture 10' from the source and the last 150', how would you calculate the voltage drop for each light?

Each light is 100W, 120V, 75 deg.C max. ambient temp., #12 AWG CU.

How is the voltage for each fixture affected by the other fixtures in the circuit, and how is this represented mathematically?
 
Re: Voltage drop

Real world or testing?

I'd worry about the last light in the circuit. It's going to have it the worst, so why wonder about the others along the way?

15 x 100W = 1500W / 120 = 12.5 amps

Vd = 2 x R x I x D
Vd = 2 x .00198 x 12.5 x 150
Vd = 7.425 V
Vd = 120 - 7.425 = 112.575 V
6.6%

Not excruciating, but not great.
 
Re: Voltage drop

George,
But that is really a lot greater voltage drop than what you will really have because you did the calculation will all of the load at the far point. A common method, but not exact, is to make the calculation with all of the load at the midpoint in the total length. To do it in an exact manner, you need to do calculations at each light using the length and load at that point.
Don
 
Re: Voltage drop

I am looking for the best method to find the real voltage drop for this circuit at each point of delivery, i.e. each light.

Thanks
 
Re: Voltage drop

Originally posted by don_resqcapt19:
A common method, but not exact, is to make the calculation with all of the load at the midpoint in the total length.
Click to expand...
That makes sense. Thanks! :)

I imagine there's a formula out there for this. This seems achingly similar to something I vaguely recall from Algebra II. I'll betcha Rattus knows. :)
 
Re: Voltage drop

:D

You know, you could graph it, too. Them graphing calculators do cool stuff.

I can just see my old math teacher shaking his head right now. Whenever there's a slow, drawn out method of finding an answer to a problem like this, there's a super-duper formula that would give you the voltage at any point in the circuit in an instant. He always had it instantly, and I'd just stare at the blackboard and utter a barely audible, "Duuuuuuh..." :D
 
Re: Voltage drop

George kidding aside and maybe I will be showing my limits but I can not see how there could be a fit all formula to figure VD at any point on a typical circuit.

The example used in this thread was text book circuit.

15 - 100 watt lamps, equally spaced apart.

How about a real circuit with different loads and unequal spacing?

IMO we are stuck with point to point or Don's suggestion.
 
Re: Voltage drop

Try this example...

100' to first light @ 5amps, next light 150' @7amps, next light 200' @3amps

find load center by:

100' x 5 amps = 500 (node 1)
250' x 7 amps = 1750 (node 2)
450' x 3 amps = 1350 (node 3)
--------------------------------------
15 amps 3600 3600 / 15 = 240 feet

Use this in your normal VD calcs as the total circuit length.
The wire size calculated will be the same throughout the circuit.

#6 thhn would be the calculated wire size @ 2.94% VD
 
Re: Voltage drop

The exact solution of this problem is messy, however there is an approximate formula for N uniformly spaced lights, but you may not like it.

1) Figure the "gain" of the very last section, call it alpha which will be less than unity.

alpha = Rlamp/(Rlamp + Rwire)

The "gain" of the remaining sections is,

2*alpha, 3*alpha, ..... N*alpha

2) Multiply these terms together, then

total "gain" = N!*alpha^N

N! is called N factorial, that is,

N! = N*(N-1)*(N-2).......3*2*1
 
Re: Voltage drop

If you want a really messy voltage drop problem, try to figure the voltage at the end of a heat trace line where the current is different at every point along the line.


Steve
 
Re: Voltage drop

I got ambitious and ran the calculations for voltage drops from the source to each 100 watt lamp and these are the results starting with the lamp closest to the soruce.
0.119v, 0.924v, 1.287v, 1.584v, 1.815, 2.078v, 2.374v, 1.98v, 1.81v, 1.58v, 1.287v, 0.921v, 0.495v
 
Re: Voltage drop

I got ambitious and ran the calculations for voltage drops from the source to each 100 watt lamp and these are the results starting with the lamp closest to the soruce.
0.119v, 0.924v, 1.287v, 1.584v, 1.815, 2.078v, 2.374v, 1.98v, 1.81v, 1.58v, 1.287v, 0.921v, 0.495v
Click to expand...
I think the VD should be the highest at the last lamp. Isn't it?
 
Re: Voltage drop

Here is what I came up with. Voltage at the lamp, all lamps burning. If any lamps quit burning, it all changes of course.

1st = 119.52 volts
2nd = 119.07 volts
3rd = 118.65 volts
4th = 118.26 volts
5th = 117.91 volts
6th = 117.59 volts
7th = 117.30 volts
8th = 117.04 volts
9th = 116.82 volts
10th = 116.62 volts
11th = 116.46 volts
12th = 116.33 volts
13th = 116.24 volts
14th = 116.17 volts
15th = 116.14 volts
 
Re: Voltage drop

Gentlemen: My apologies, looks like I calculated the VD with only one lamp lit at each location. The corrected calculated voltages at each lamp are as follows for number 12 copper conductors at 75 degrees C:
119.8v, 118.97, 118.46, 117.95, 117.44, 116.41, 115.9, 115.39, 114.87, 114.36, 113.85, 113.34, 112.82, 112.31. Sorry for the wrong info.
 
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