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# How do I calculate what gauge wire I need to use over very long distance runs - like 500 ft on 3 conductor solid copper wire at 1.5 amps and 120V-formula-or way to calculate for various lengths-amps?

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If this is an electrical installation that needs to comply with the National Electrical Code, I'd advise you to read the code book article 310. It's not exactly easy to summarize, there are many ins and outs. For NEC purposes you would likely size according to Table 310.16. If you're experimenting, or just want some math to figure out what you need, check out NEC Chapter 9, Table 8. It lists the specific resistance per length of wire for different wire sizes. You can use this to determine how much resistance your circuit will have to overcome. As far as formulas, everything is based on Ohm's law. It is: Voltage (in Volts)=Current (in Amps) x Resistance (in Ohms). also: Current = Voltage divided by Resistance Resistance = Voltage divided by Current For example, NEC rules aside: If you have a 1.5 amp load at 120 volts, 500 feet away from your voltage source, that gives you 1000 feet round trip. If you were to use #18 copper, it's 8.08 ohms per 1000 feet. Dividing 120 (volts) by 8.08 (ohms) gives you 14.85 (amps). This wire's resistance will allow 14.85 amps to flow @ 120v, and your circuit is only 1.5 amps @ 120v. Of course, this would most likely not be allowed in your application by the NEC, as the smallest wire permissible for general use is #14. #18 is commonly used for low current loads, such as a fluorescent lighting ballast, and in this application it is permissible. Here is a way you can calculate the answer: First, You must determine how much voltage drop your device can tolerate. The NEC suggests no more than 3% drop for branch circuits. I will use 3% in the example. 3% of 120V is 3.6V, so the wiring run must drop no more than 3.6V. Now calculate the maximum resistance of the wiring run using R=E/I (resistance = voltage divided by current). Your example uses 1.5A, so let's use that. R = 3.6V / 1.5A R = 2.4 ohms Remember, use the voltage dropped across the wiring divided by the current through the wiring to get resistance of the wiring. Now, consult a table of wire resistance per size, such as NEC Chapter 9. The resistance in the table will be for a standard length, such as 1000 feet, so you will need to factor for your length. Remember to add both wires because the current has to make the round trip! Example: My distance from breaker to load is 350 feet. Double this because there are two wires = 700 feet. The 700 feet of wire can have no more than 2.4 ohms of resistance, so 1000 feet of the same wire can have no more than 3.43 ohms: 2.4 * (1000 / 700) = 3.43 Check your table for a wire that has 3.43 ohms per 1000 feet or less of resistance. According to my table, #14 copper has a resistance of 3.07 ohms per thousand, and will do nicely. I did not use your example of a 500 foot run, because it worked out to 1000 feet of wire, too easy and nothing to learn! #14 is normally good for 15 amps, so in this case we are not using it because we need the current capacity, but because we need to minimize voltage drop over a huge distance. Now, just for fun, let's reverse engineer your example using #18 wire to see what performance we could expect: #18 = 8.08 ohms per 1000 feet. Wiring run = 1000 feet (500 * 2) Voltage drop (E=IR): E = 1.5A * 8.08ohms E = 12.12V Voltage delivered to the load = 120 - 12.12 = 107.88V

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