What is the voltage drop formula for single phase?

VD = 2 × I × R × L ÷ 1000 using resistance in ohms per 1000 ft, or VD = 2 × K × I × L ÷ CM using the K-factor method. L is the one-way length in feet.

What is the voltage drop formula for three phase?

VD = 1.732 × I × R × L ÷ 1000, or VD = 1.732 × K × I × L ÷ CM. Compare the result with the line-to-line voltage.

What is K in the voltage drop formula?

K is the resistance of a 1 circular mil, 1 ft conductor in ohms. Common values are 12.9 for copper and 21.2 for aluminum at 75 °C. Some references use 10.8 for copper at 25 °C.

Which method is more accurate?

Using NEC Chapter 9 Table 8 resistance is slightly more accurate because it reflects the actual stranding of each size. The K method is quicker by hand and usually lands within a few percent.

How do I find the wire size for a given voltage drop?

Rearrange the K formula: CM = 2 × K × I × L ÷ VD (single phase). Then choose the first size whose circular mil area is at least that value.

Voltage Drop Formula: Single & Three Phase (Examples)

The two formulas electricians use to calculate voltage drop, the resistance method and the K-factor (circular mil) method, with single- and three-phase worked examples.

Method 1: resistance (NEC Chapter 9 Table 8)

Single phase: VD = 2 × I × R × L ÷ 1000
Three phase: VD = 1.732 × I × R × L ÷ 1000

VD: voltage drop in volts
I: load current in amps
R: conductor resistance in ohms per 1000 ft (e.g. 12 AWG copper = 1.93)
L: one-way length of the circuit in feet

The 2 accounts for current flowing out and back on two conductors. In a balanced three-phase circuit the factor is √3 = 1.732. Percent drop is VD ÷ source voltage × 100.

Example: 120 V branch circuit

16 A, 100 ft, 12 AWG copper:

VD = 2 × 16 × 1.93 × 100 ÷ 1000 = 6.18 V → 6.18 ÷ 120 = 5.15%

That exceeds the recommended 3%. 10 AWG (1.21 Ω) gives 3.87 V or 3.23%, and 8 AWG (0.764 Ω) gives 2.44 V or 2.04%.

Method 2: K-factor (circular mils)

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

K: resistivity in ohm-circular mils per foot: 12.9 for copper, 21.2 for aluminum (at 75 °C)
CM: conductor area in circular mils (12 AWG = 6,530; 10 AWG = 10,380; see the wire gauge chart)

Example: the same circuit by the K method

VD = 2 × 12.9 × 16 × 100 ÷ 6,530 = 6.32 V → 5.27%

Slightly higher than the Table 8 method because K is an average and Table 8 reflects the coated, stranded conductor for each size. Either answer leads to the same decision.

Sizing a conductor for voltage drop

Rearrange the K formula to solve for the minimum circular mil area:

CM = 2 × K × I × L ÷ VD_allowed

Example: 40 A at 240 V, 75 ft, 3% limit

Allowed drop is 240 × 0.03 = 7.2 V:

CM = 2 × 12.9 × 40 × 75 ÷ 7.2 = 10,750 cmil

10 AWG (10,380 cmil) is just short, so use 8 AWG (16,510 cmil). By the Table 8 method, 8 AWG gives 1.91%. The wire size calculator does this and checks ampacity too.

Three-phase example

A 480 V feeder carries 100 A over 250 ft on 1/0 copper (105,600 cmil, 0.122 Ω/kft):

Table 8: VD = 1.732 × 100 × 0.122 × 250 ÷ 1000 = 5.28 V → 1.10%
K method: VD = 1.732 × 12.9 × 100 × 250 ÷ 105,600 = 5.29 V → 1.10%

For more three-phase cases, see the 3 phase voltage drop calculator.

How much voltage drop is allowed?

In the US, NEC informational notes to 210.19(A) and 215.2(A) recommend no more than 3% on a branch circuit or feeder and 5% total. They are recommendations, not requirements, except where codes such as energy codes or NEC 647.4(D) make them mandatory. In Canada, CEC Rule 8-102 sets the 3% and 5% limits as requirements. See the Canadian voltage drop calculator.

Common mistakes

Using the round-trip length for L; the formula's 2 already counts the return path.
Using 2 instead of 1.732 for three-phase circuits.
Using the breaker rating instead of the actual load current.
Comparing a three-phase drop with line-to-neutral voltage.
Ignoring reactance on large conductors in steel conduit at low power factor, where Table 9 impedance is more accurate.

Voltage Drop Formula