First, I will have an electrician do all of the electrical work, but I'm trying to understand how this works, so I don't get "oversold"...
I want to put five light sources (floodlights, driveway lamp post, etc.) fairly evenly spaced (every 300') along this driveway. I believe that 120 watts of 120V LEDs at each location would be more than enough, but it keeps the numbers "round". The wire will be run underground in conduit. Assume all the lights will be on at the same time.
I think that means I need 5 amps at the first location, 4 amps at the second location, and so on, and the fifth/last location will need 1 amp.
The Southwire voltage drop calculator says I need to use 4 awg if I need 5 amps at the end of the circuit, but I really only need 1 amp at the end.
Please bear with me...
Working backwards from the 5th/last location to the 4th location, 300' of 12 awg wire will carry 1 amp with only a 1V drop. The wire between the 4th and 3rd locations will need to carry 2 amps (4th and 5th lights). The calculator says 10 awg will carry 2 amps for 300' with a 1.25V drop. Between 3rd and 2nd, 10 awg will carry 3 amps with a 1.9V drop. Between 2nd and 1st, 10 awg will carry 4 amps with a 2.5V drop. Finally, between the 1st light and the breaker, the calculator says 8 awg will carry 5 amps with a 2V drop.
The above adds up to a total voltage drop of 8.65V at the farthest distance. I understand this is more than is allowed, but given that the total amps will be closer to 2.5 amps instead of 5 amps, that should bring the total voltage drop down to (I think) allowable levels.
Is the above a realistic design? One electrician suggested running 6 awg for the whole thing, but that seems like overkill.
Thanks,
JL
I want to put five light sources (floodlights, driveway lamp post, etc.) fairly evenly spaced (every 300') along this driveway. I believe that 120 watts of 120V LEDs at each location would be more than enough, but it keeps the numbers "round". The wire will be run underground in conduit. Assume all the lights will be on at the same time.
I think that means I need 5 amps at the first location, 4 amps at the second location, and so on, and the fifth/last location will need 1 amp.
The Southwire voltage drop calculator says I need to use 4 awg if I need 5 amps at the end of the circuit, but I really only need 1 amp at the end.
Please bear with me...
Working backwards from the 5th/last location to the 4th location, 300' of 12 awg wire will carry 1 amp with only a 1V drop. The wire between the 4th and 3rd locations will need to carry 2 amps (4th and 5th lights). The calculator says 10 awg will carry 2 amps for 300' with a 1.25V drop. Between 3rd and 2nd, 10 awg will carry 3 amps with a 1.9V drop. Between 2nd and 1st, 10 awg will carry 4 amps with a 2.5V drop. Finally, between the 1st light and the breaker, the calculator says 8 awg will carry 5 amps with a 2V drop.
The above adds up to a total voltage drop of 8.65V at the farthest distance. I understand this is more than is allowed, but given that the total amps will be closer to 2.5 amps instead of 5 amps, that should bring the total voltage drop down to (I think) allowable levels.
Is the above a realistic design? One electrician suggested running 6 awg for the whole thing, but that seems like overkill.
Thanks,
JL