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04172019, 09:19 PM  #1 
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Voltage Drop Question
Hi All,
I'm helping a friend build a chicken coop that sits about 200 feet from his garage where his service panel is. Initially he was just wanting to put a light out there and our plan was to use a 12/2 direct burial conductor with a GFCI breaker in the panel (running the conductor through the wall just beneath the panel and out through the vinyl siding inside a piece of 3/4 or 1 inch conduit elbowed down to the ground and then buried about 2 feet deep all the way out to the chicken coop). He has now decided that it might be a good idea to be able to put a 1500 watt heater out there in case he wants to warm it for chicks so now we're talking about a duplex outlet in addition to a 60 watt light bulb. My question is this, will the 12/2 conductor be sufficient (I know it will handle the amperage draw) and should we be concerned about voltage drop for a 200 ft run? Is voltage drop a concern for something like a heat lamp or light or is it only something that should be considered when powering appliances that have motors that rotate? I appreciate any advice. Thanks for reading! 
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04172019, 09:33 PM  #2 
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Re: Voltage Drop Question
You can use this calculator to figure the voltage drop. Normally, no more than a 3% drop is desirable. For your load of 13A, that would require 8 AWG copper cable. If all you have is a heater and light bulb, you could probably get away with 10 AWG with a 5% drop.
Things with motors tend to be more sensitive to voltage drop than pure resistive loads. If you want to do it right, use the heavier cable. 
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04172019, 09:34 PM  #3 
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Re: Voltage Drop Question
There will be a voltage drop of about 6.6% or 8V, assuming good connections. You'll get less heat, and a slightly dimmer light. it'll otherwise run fine. It would be a greater concern with a motor. (some heaters do have fans in them.) IIRC, the general rule is a 3% max voltage drop, if you want to be more proper. (that would require 8awg cooper, or 6awg aluminum.) I'm sure the electricians will correct me.
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04172019, 09:58 PM  #4 
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1500 + 60 watts is about 13 amps at 120 volts so it will be about an 8 volt drop over the 200 feet using #12 copper. Maxing out the full 20 amps is nearly a 13 volt drop. I'm not a code expert but I think I remember NEC listing 5% as the maximum acceptable voltage drop at the end of a circuit. To be under 5% voltage drop at 13 amps you need to use #10 copper. You would be slightly over 5% voltage drop at 16 amps (80% capacity) which is what you should max out a 20 amp circuit at. For me, using #10 would be acceptable. A 250 ft roll of #10 uf at your local big box store is about $60 more than #12 uf.

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04172019, 11:16 PM  #5 
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Re: Voltage Drop Question
The resistance of 12 gauge copper wire is 1.588 ohms per 1000 ft or .001588 per ft.
.001588 ohms x 400 ft (200 ft there and 200 ft back) is .6352 ohms. That is less than a 100 watt incandescent light bulb (if you could find one) 100W / 120V = .83 amps. (R=E/I) therefore (120/.83= 144.5 ohms) therefore 100 watt light bulb has 144.5 ohms resistance Total Load: 1500 watt heater = 12.5 amps 100 watt light bulb = 0.83 amps 200 ft cable (400 ft wire) = 0.64 amps Total amperage draw = 13.97 amps 12 gauge cable will carry 20 amps. Don't worry about it. MTN REMODEL LLC likes this.
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Harold, The Left Handed Widget Maker Last edited by hkstroud; 04172019 at 11:25 PM. 
04182019, 12:32 AM  #6 
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Re: Voltage Drop Question
I stand corrected. I forgot my 5th grade math. To divide by a number less than one you multiply.
Therefore, a 100 watt light bulb which uses .83 amps of current has 120V divide by .83 amps or (120 X .83) = 9.96 ohms resistance. To make sure I wasn't lying to you I just measured one with my volt meter. Actually came out to be 10.1 ohms resistance. The bulb, the meter or I must be defective. Its probably me but don't tell the chickens.
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Harold, The Left Handed Widget Maker Last edited by hkstroud; 04182019 at 12:38 AM. 
04182019, 08:40 AM  #7 
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Re: Voltage Drop Question
Different experts will differ but IMHO voltage drop should not exceed 5% from pole transformer to the furthest outbuilding where the equipment is being used. Three percent from main panel to the furthest outbuilding is a reasonable number to shoot for.
Note: The resistance of the lamp (bulb) when lit (the filament is hot) is somewhat greater compared when when the lamp is unscrewed and cold. Always disconnect power before touching ohmmeter probes to a circuit. The quick method of measuring the resistance of the lamp in operation is to measure the current flowing through and the voltage across the lamp power connections (or fixture terminals). So 120 volts divided by 0.83 amps is approx. 144 ohms for the white hot filament. Dividing by a number less than one by employing multiplication is superficially incorrect. The confusion arises from dealing with decimals versus fractions. When you want to divide a quantity X (any number) by a quantity Y (expressed as a proper or improper fraction and not as a mixed number) you may elect to invert the fraction Y (example: 3/4 becomes 4/3) and then multiply the two quantities. Dealing with the 0.83 amps here (not expressed as a fraction), you do not multiply instead of divide.
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Heat went off in the dead of winter? Excellent idea to be late or absent for work and to spend the day winterizing your house to keep pipes from freezing. Every house is different. For those of you not hit yet, think of what things you need to do. Last edited by AllanJ; 04182019 at 09:04 AM. 
04252019, 02:53 PM  #8 
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Re: Voltage Drop Question
A motor when it starts up and charges the windings will have 3x the load and so wire sizing is important although there are options for adding a capacitor as is often done with air conditioning compressor motors.
For long distances I would rather run a 240v line and split off 120v for lights and have 240v for space heaters, arc welders, etc. at little additional cost. https://www.electricalestimating101....ropCharts.pdf 
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