Question about the best length of 12 ga low voltage wire for 174 watt system using a 300 watt transformer?
I’ve designed a low voltage layout using (12) 11 watt pathway lamps and (6) 7 watt deck step lamps. The run starts at the 300 watt transformer, and my first light starts (deck step light) 11" away and the run (a big oval) terminates 157′ thereafter. My questions is this: Considering voltage drop, would it be better to just leave the run terminated at the last light 157′ total wire run, or, since the transformer is only 19′ from the terminus, continue the 12 ga wire to the transformer for a "loop" configuration for a more balanced line load?
Thanks in advance.
3 Responses
billrussell42
13 Mar 2010
guru
13 Mar 2010
There are several ways of connecting this
1
If you make a single string, where you feed the power at one end and all the lights are distributed over the length of the cable, this will have the largest difference between the start light voltage and finish light voltage.
2
If you can make a loop where the start end and the finish end of the line are available at the transformer.
Assuming your cable has a black and white wire.
Feed the power start end at white wire and finish end at black wire.
This way all the lights will have the same voltage, regardless of the run length! The losses are still there, only the distribution is even, this is true parallel connection.
3
If you feed at both ends, then there will be fewer losses, but there will be some variations in voltage from start to center of the loop.
The variation will be ¼ that of the single string, method 1.
Also if you avoid cutting the main wire line, just skin and tap onto-it this will have less loss.
Hope this properly answers your question
Guru
Irv S
13 Mar 2010
The loop is by far the better idea.
Without it you’d have to increase the wire size to reduce the voltage drop.
Loops are not good as you can disconnect a wire and have it still live.
So you have a total of 121+42 = 163 watts. At 12 volts that is 14 amps.
#12 wire is 1.6 ohms per 1000 ft or 0.16 ohms per 100 ft
just as a guess, say your maximum drop you want is 2 volts, which means the lights will get 10 volts at least. By ohms law:
R = E/I = 2/14 = 0.14 ohms
That means you can have 50 feet of wire (out and back, 100 total) and keep the voltage drop below 2 volts.
But you have 3 times that at 157 ft. I’d divide the lamps into 2 more or less equal sections and run a separate #12 pair to each section. Of course breaking it into sections means the current is lower, so 2 sections should do it. It may be better to setup the lamps furtherest away as one section, and only have 1/3 the lamps in that section, and the nearest 2/3 in the second section.
edit: thinking about it, best may be two strings, one for each half of the oval. That means the distance is about 80 ft, the resistance is 0.128 ohms or 0.26 for the pair. At half the current, or 7 amps, that is a drop of 1.8 volts, which would hardly be noticable.
This is similar to your idea of completing the loop. But think about the loop. At some point there will be virtually zero current, as the current is getting supplied from both sides. Just cut the wire at that point, and you have my proposal above under "edit".
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