According to the National Electrical Code the current-carrying capacity of any wire is determined by the temperature rating of its insulation. Common ratings for 12 gauge wire are 60C (21 amps), 75C (28 amps), and 85C (30 amps). But wait, that is not the final answer.
Again, according to the NEC, overcurrent protection for 12 gauge wire shall not exceed 20 amps. Furthermore, good design practice demands that no circuit be intentionally loaded to more than 80% of its overcurrent protection. In the case of a 20 amp circuit, that means 16 amps.
So even though 85 degree C wire can safely carry 30 amps, you can't use a circuit breaker larger than 20 amps and you can't design the system to carry more than 16 amps continuously (over 3 hours).
Your original question was about maximum distance. Voltage drop does play a part here, but it is proportional to the actual load you put on the circuit. The heavier the load, the more the voltage will drop. Conversely, with no load on the system there will be no voltage drop. The worst-case scenario would be to start a large motor (such as an air conditioner) at the end of a long wire run. Motor starting current is usually 6X the normal running current. On a hot summer day when you are most likely to need the A/C, the power company is most likely to be experiencing brownout conditions because everybody wants A/C at that time. In that case your supply voltage will likely be low to start with and the voltage drop at the end of a very long wire may just make it impossible to start the compressor. Even if you do get it started, the ongoing low voltage situation will cause the compressor to draw higher amps. The result will be overheating of the compressor and likely early burnout.
Read more: What is the maximum safe distance for a residential 20 amp 120 volt circuit when using 12/2 AWG NM Romex wire? | Answerbag
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