A 20 A, 120 V single-phase circuit with 100 ft (one way) of 12 AWG copper drops 7.9 V (6.6 %), leaving 112.1 V at the load; 8 AWG copper keeps it within 3 %.
Calculator
Voltage Drop
Over the recommendation
6.6%
Drop
7.9V
Voltage at Load
112.1V
Exceeds 5% combined recommendation
Use #8 AWG CU for ≤3% drop
Method: K factor (copper 12.9), Chapter 9 Table 8.
Per NEC 2023 210.19 Informational Note, NEC 2023 215.2(A)(2) Informational Note No. 2, NEC 2023 Chapter 9, Table 8
How it’s calculated
VD = m × K × I × L ÷ A (K method)
VD = m × I × (R × PF + X × sin θ) × L ÷ 1000 (Table 9 method)
Percent = VD ÷ V × 100; voltage at the load = V − VD
m
2 for single-phase, 1.732 for three-phase
K
12.9 for copper or 21.2 for aluminum (ohm-circular mils per foot), from the conductor resistance at 75 °C in Chapter 9 Table 8
I
load current, amperes
L
one-way length, feet
A
conductor area in circular mils (Chapter 9 Table 8)
R, X
AC resistance and reactance per 1000 ft for the chosen raceway (Chapter 9 Table 9)
PF
power factor; sin θ = √(1 − PF²)
V
system voltage (line to line for three-phase)
Table 9 is used whenever the power factor is below 1 and the table lists the size and metal. At a power factor of 1 it is used only with the switch on and a size of 1/0 AWG or larger; otherwise the K method runs.
The suggested size is the first size, from 14 AWG up, whose drop, computed the way it would be for that size, is 3 % or less.
Assumptions
The conductors run at 75 °C. A lightly loaded conductor runs cooler and drops a little less.
A power factor below 1 is lagging (an inductive load).
One circuit, all conductors the same size. Single-phase has two conductors carrying the load; three-phase is balanced.
The percentage is of the voltage you pick, line to line for three-phase.
K is one value per metal, not per size. Sizes without a Table 9 row use the K method.
3 % and 5 % are recommendations; the calculator flags them and enforces nothing.
Ampacity is not checked here; the wire size calculator does that.
Results are rounded to 0.1 V and 0.1 %.
Worked example
A 208 V three-phase feeder, 150 A, 225 ft
2/0 AWG copper in steel conduit, power factor 0.85.
The power factor is below 1, so Table 9 applies. For 2/0 AWG copper in steel conduit:R = 0.100 Ω and X = 0.054 Ω per 1000 ft
Effective impedancesin θ = √(1 − 0.85²) = 0.527; Z = 0.100 × 0.85 + 0.054 × 0.527= 0.1134 Ω per 1000 ft
Voltage dropVD = 1.732 × 150 × 0.1134 × 225 ÷ 1000= 6.6 V, 3.2 % of 208 V, 201.4 V at the load
That is above the 3 % the notes recommend for a feeder, though within 5 % for feeder and branch circuits together. The first size that meets 3 %, computed the same way:3/0 AWG copper= 5.5 V, 2.7 %
For comparison, the K method on 2/0 AWG ignores reactance and the power factor:1.732 × 12.9 × 150 × 225 ÷ 133,100= 5.7 V (2.7 %), which looks better than it is
Answer: 3.2 % (6.6 V) with 2/0 AWG; 3/0 AWG copper brings it to 2.7 %.
This calculator uses K = 12.9 for copper, the resistance at 75 °C. Tools that use copper’s resistance at room temperature (a K near 10.4 to 10.8) show roughly a sixth to a fifth less drop.
Check: which K, or which conductor temperature, the other tool uses.
One-way or round trip
The formula already doubles the length for single-phase. Entering the round trip here doubles the drop.
Check: what the other tool’s length field means.
Reactance left out below a power factor of 1
A resistance-only formula, or a K result multiplied by the power factor, understates the drop on larger conductors, where reactance carries part of it. The worked example above shows the gap.
Check: whether and how the other tool uses the power factor.
Three-phase conventions
A balanced three-phase drop uses 1.732 (√3) where a single-phase drop uses 2; 2 would overstate it by about 15 %. Table 9’s impedance gives a drop to neutral, so it is multiplied by 1.732 before it is compared with the line-to-line voltage.
Check: the multiplier and which voltage the percentage is taken of.
Recommendation or requirement
The NEC figures are recommendations. Energy codes, local amendments and job specifications can make a limit mandatory, or split it between feeder and branch circuit.
Check: the job specification and your AHJ.
DC and automotive calculators
They assume a DC loop, no reactance and other limits. They do not fit AC building wiring.
Check: that the tool is meant for AC building wiring.
Questions
How much voltage drop does a 20 A, 120 V circuit have over 100 ft of 12 AWG copper?
About 7.9 V, or 6.6 %, with the K method (single-phase, 100 ft one way). 10 AWG gives about 4.1 %, and 8 AWG brings it to about 2.6 %.
Is a 3 % voltage drop a code requirement?
Not in the NEC itself. The 3 % figure (a branch circuit, or a feeder on its own) and the 5 % figure (feeder and branch circuits together) sit in informational notes, under 210.19 and after 215.2(A)(2); section 90.5(C) makes such notes explanatory, not enforceable. A local amendment, an energy code or the job specification can make a limit mandatory.
Do I enter the one-way or the round-trip distance?
One way, from the source to the load. The 2 (single-phase) or 1.732 (three-phase) in the formula already accounts for the return path.
When does the calculator use Chapter 9 Table 9?
Whenever the power factor is below 1 and the table lists the conductor. At a power factor of 1, only with “Use Ch 9 Table 9 impedance” switched on and a size of 1/0 AWG or larger.
Correct an ampacity for heat and for conductors in one raceway.
About this calculator
Written and maintained by Yusuf Can Samiloglu: MEP/HVAC design background, mechanical-engineering degree, not a licensed electrician.
How it’s checked: the calculation is the method of the Electric Master app, ported to the web and tested case by case against the app’s source code. Every table value was checked against the NEC 2023 values in at least two independent published sources.
The App Store version of Electric Master (2.2.0) predates the NEC corrections this page follows; until its next update, its result can differ from this page for some inputs. Version 2.2.0 multiplies the K result by the power factor and reaches Table 9 only from 1/0 AWG up, so below a power factor of 1 it shows less drop than this page.
: First version, NEC 2023. The K method uses Chapter 9 Table 8; Chapter 9 Table 9 impedance is used whenever the power factor is below 1, and on request from 1/0 AWG up; the suggested size is found with the same method as the result.