Worked problem

Voltage Drop on a 240 V Single-Phase Feeder

Calculate percent voltage drop for a 50 A load 150 ft from the panel and choose a conductor that meets the 3% guideline.

Electrical & Power · Last reviewed

Given

240 V single-phase, 50 A load, 150 ft one way, copper conductors. 6 AWG is 26,240 circular mils; 4 AWG is 41,740. Use K = 12.9 ohm-cmil/ft (copper at 75 °C).

Find

The voltage drop in percent, and whether it meets the 3% guideline in the informational note to NEC 210.19.

Solution

VD = 2 × K × I × L ÷ CM
  1. 6 AWG: VD = 2 × 12.9 × 50 × 150 ÷ 26,240 = 7.37 V.
  2. Percent = 7.37 ÷ 240 = 3.07%. Slightly over 3%.
  3. 4 AWG: VD = 193,500 ÷ 41,740 = 4.64 V, which is 1.93%. This passes.

Answer

6 AWG: 3.07% (just over) | 4 AWG: 1.93%

Common mistakes

  • Using the one-way length without the factor of 2 for the return conductor.
  • Treating 3% as a code limit. It is a recommendation: 3% on a branch circuit, 5% for feeder plus branch together.
  • Checking only ampacity. A conductor can carry the current and still drop too much voltage on a long run.

Try a variation: 480 V three-phase

A balanced three-phase 480 V feeder carries 80 A for 200 ft one way, with 4 AWG copper (41,740 circular mils).

VD = √3 × K × I × L ÷ CM
  1. VD = 1.732 × 12.9 × 80 × 200 ÷ 41,740 = 8.56 V.
  2. Percent = 8.56 ÷ 480 = 1.78%, which passes the 3% guideline.
  3. With 3 AWG (52,620 circular mils) the drop falls to 1.42%.

Work backward: the smallest conductor for a 3% drop

  1. For the original 240 V circuit, 3% of 240 V is 7.2 V.
  2. Required area = 2 × K × I × L ÷ VD = 2 × 12.9 × 50 × 150 ÷ 7.2 = 26,875 circular mils.
  3. 6 AWG is 26,240 circular mils, just short. The next size, 4 AWG, is the smallest that meets 3%.

Sanity check

Voltage drop is proportional to current and to length, and inversely proportional to the conductor area. Doubling the length doubles the drop. Each step up in AWG size changes the area by about 26%, so one size larger cuts the drop by roughly a fifth.

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Questions

Why is K 12.9?

It is the approximate resistivity of copper at 75 °C in ohm-circular mils per foot. Some references use 12.4 (copper at 25 °C) or 21.2 for aluminum.

How does three-phase change the formula?

Replace the factor 2 with √3 (1.732) for a balanced three-phase circuit.

Does power factor change the voltage drop?

For AC circuits with large conductors, reactance matters as well as resistance, and the drop depends on the load power factor. The simple resistance-only formula is accurate for small and medium conductors. For large conductors use impedance values from the NEC tables or software.

Where does the 3% figure come from?

The informational notes to NEC 210.19 and 215.2 recommend a drop of no more than 3% on a branch circuit or feeder and 5% overall. They are recommendations, though some energy codes and project specifications make them requirements.

What about the dip while a motor starts?

A motor draws several times its running current for a moment, so the voltage at the motor and elsewhere on the circuit dips briefly. That is a separate check from the steady-state drop.

Educational material. Examples use simplified, hypothetical inputs and are not a design, a construction document or professional engineering advice. Confirm requirements with the codes and standards adopted where the work is built and with a licensed engineer.

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