Voltage drop formula, with worked examples

The formula

Single-phase:  VD = 2 × K × I × D ÷ CM
Three-phase:   VD = 1.732 × K × I × D ÷ CM

VD = voltage drop (V)   K = 12.9 copper, 21.2 aluminum
I = load current (A)    D = one-way length (ft)
CM = conductor area in circular mils (NEC Chapter 9, Table 8)

The 2 in the single-phase version accounts for the current going out on one conductor and back on the other. In a balanced three-phase circuit the line-to-line drop is √3 (1.732) times the drop in one conductor, so 1.732 replaces the 2. D is always the one-way distance from the source to the load.

K is the resistance of a conductor one circular mil in area and one foot long. Multiplying a conductor's resistance per 1,000 ft from Chapter 9, Table 8 (given at 75 °C) by its circular mils and dividing by 1,000 gives about 12.9 for copper; aluminum, with higher resistivity, comes out near 21.2. That is why an aluminum run drops roughly 64% more voltage than a copper run of the same size.

Percent drop is simply VD ÷ source voltage × 100. A 6 V drop is 5% of 120 V but only 2.5% of 240 V, which is one reason long runs are fed at the higher voltage.

How much drop is acceptable

The NEC does not set a mandatory limit for ordinary branch circuits and feeders. Informational notes to 210.19(A) and 215.2(A) recommend sizing conductors so the drop is no more than 3% on a branch circuit or feeder and no more than 5% for the feeder and branch circuit together. Inspectors and engineers widely use those figures as the design target.

Some equipment instructions and some specific code sections are stricter or mandatory, so check the installation manual for motors, pumps and electronics on long runs.

Solving for wire size

Rearrange the formula to get the smallest conductor area that keeps the drop at your limit, then pick the next size up from Table 8:

CM = 2 × K × I × D ÷ VD allowed        (single-phase)
CM = 1.732 × K × I × D ÷ VD allowed    (three-phase)

VD allowed = limit % × source voltage

Example: 30 A at 240 V, 150 ft, copper, 3% (7.2 V): CM = 2 × 12.9 × 30 × 150 ÷ 7.2 = 16,125 cmil. 10 AWG has 10,380 cmil (too small) and 8 AWG has 16,510 cmil, so 8 AWG.

Voltage drop never replaces the ampacity check. Size the conductor for ampacity first (Table 310.16, the terminal temperature rating, any derating), then check the drop, and use whichever conductor is larger.

Worked examples

1. A 20 A, 120 V circuit, 100 ft one way, copper. 12 AWG (6,530 cmil): VD = 2 × 12.9 × 20 × 100 ÷ 6,530 = 7.90 V, 6.58%. 10 AWG: 4.97 V, 4.14%. 8 AWG: 3.13 V, 2.60%. Ampacity alone allows 12 AWG; holding 3% takes 8 AWG.

2. A 30 A, 240 V single-phase load, 150 ft, copper. 10 AWG is enough for the ampacity (30 A in the 60 °C column) but drops 11.18 V, 4.66%. 8 AWG drops 7.03 V, 2.93%, inside 3%.

3. A 50 A, 208 V three-phase feeder, 200 ft, copper. 6 AWG (65 A at 75 °C) drops 8.51 V, 4.09%. 4 AWG drops 5.35 V, 2.57%. If the branch circuits fed from it add their own drop, 4 AWG keeps the total within 5% more comfortably.

4. A 100 A, 240 V aluminum feeder to an outbuilding, 250 ft. 1/0 AL carries 120 A at 75 °C, but drops 10.04 V, 4.18%. 3/0 AL drops 6.32 V, 2.63%. Long feeders to detached garages and barns are the classic case where voltage drop, not ampacity, picks the conductor.

When the simple formula is not enough

The K-factor method uses DC resistance. For large conductors, and for loads with a low power factor such as motors, the conductor's reactance matters too. Chapter 9, Table 9 gives AC resistance and reactance per 1,000 ft for each size and raceway type, and its notes show how to combine them into an effective impedance at a given power factor. Engineers use that method for big feeders and service conductors.

Temperature matters as well: resistance rises as the conductor heats up. K = 12.9 is based on 75 °C resistances, which is a reasonable working value for loaded building wire.

Motor starting is a separate question. A motor can draw several times its running current for a moment, so the momentary drop at start-up can be much larger than the running drop this formula gives.

Common mistakes

• Doubling the distance and then using the single-phase formula, which doubles it again.
• Using the wrong voltage for the percentage: a 240 V circuit's drop is divided by 240, not 120.
• Using the breaker size instead of the real load current. The formula wants the current the load actually draws; using the breaker rating gives a conservative answer.
• Treating the 3% recommendation as permission to skip the ampacity check, or the reverse.
• Forgetting that the feeder's drop and the branch circuit's drop add up toward the 5% total.

Questions people ask