We are working with an electrician to install permanent wiring to a large trailer unit - it will be a 120/240 V split-phase system (the trailer itself uses a NEMA 14-50 dryer plug) delivering 50 amps at a distance of 300 ft.

I (think I) understand that this wiring set up effectively gives us three circuits - two 120 V/50 A circuits and a single 240 V/50 A circuit.

What I am struggling to understand is how I determine the proper wiring size for the system, given the parameters of the circuit and our voltage drop needs ( <3% ). Following Ohm's Law, I have been using the formula to calculate the expected voltage drop for particular wire gauges (4 AWG or 1 AWG):

Vd = 2 * L * R * A


Vd: voltage drop

L: one way distance (in thousands of ft)

R: resistance of wire (in Ohms per 1000 ft)

A: load current (in amperes)

Ex. 1 1 AWG wire, 300ft, 50A

Resistance of 1 AWG wire / 1000ft = .124 Ohms

Vd = 2 * 50 * (.3) * (.124)

Vd = 3.72 V

Ex. 2 4 AWG wire, 300ft, 50A

Resistance of 4 AWG wire / 1000 ft = .249 Ohms

Vd = 2 * 50 * (.3) * (.249)

Vd = 7.47 V

Based on this information, my numbers suggest that we would need to use 1 AWG wire, but the electrician has said that 4 AWG is more than sufficient. I have complete confidence in this electrician, so I am sure there is something that I am missing. How does the presence of the split-phase 240 V system affect voltage drop considerations (if at all?).

Any comments or help is greatly appreciated!

  • \$\begingroup\$ It would help if you showed your numbers and calculation. \$\endgroup\$
    – DSI
    Commented Apr 22, 2020 at 16:57

1 Answer 1


The 50A breaker pair feeding the NEMA 14-50 plug limits the current on any phase to ... 50A, period, no matter how you load your phases. The breakers are linked, so if you have a single-phase fault on L1, it will trip itself and L2.

Your electrician will use that breaker panel limit not only to calculate the IR drop, but also the allowable cable temperature rise given the ambient temperature, conduit type, wire material and insulation type, as these all affect the 'ampacity' of the wire.

As for the voltage drop, the NEC specifies a maximum drop of 5%, not 3%. See https://www.mikeholt.com/technnical-voltage-drop-calculations-part-one.php

At 300ft and 50A, #4 ga. copper will be less than that, about 3.5%. So, plenty.

Your electrician knows what they're doing.

BONUS: An IR drop calc, with a wire resistance chart: https://www.calculator.net/voltage-drop-calculator.html

tl; dr:

  • resistance of #4 ga. is 0.2485 ohm/1000ft
  • One-way IR drop for 300' at 50A = 50 * 300/1000 = 3.73V
  • Two-way IR drop is 7.46V, or 3.1%
  • \$\begingroup\$ Thank you for your reply! As far as the calculations go, I understand what you are doing. What is unclear to me is why the relative voltage drop is only calculated with the voltage across the two legs (240 V) and not also across the leg / neutral combinations (120 V). \$\endgroup\$ Commented Apr 22, 2020 at 18:40
  • \$\begingroup\$ Say you have two 120V loads, one 50A and the other, 25A. Both their returns go to neutral for 75A... or do they? Nope. Some of the return current goes to the other phase. Neutral only carries the imbalance of the two legs, so the neutral net current is 25A. \$\endgroup\$ Commented Apr 22, 2020 at 18:46
  • \$\begingroup\$ Let me see if I understand correctly. Lets say we have an extreme case where one 120V leg has a load of 50A and the other has a load of 0A. In this case, the neutral line would carry the difference of 50A. Would the voltage drop effectively double for the 120V circuit with a 50A load? \$\endgroup\$ Commented Apr 22, 2020 at 19:04
  • \$\begingroup\$ 50A in, 50A return on neutral, so it's a round-trip drop of 7.46V as I calculated. That's the very worst case. Bottom line, there is no need to upsize neutral compared to L1 and L2. \$\endgroup\$ Commented Apr 22, 2020 at 19:09

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