Voltage Drop Calculator
Calculate conductor voltage drop the way the NEC intends: the Chapter 9 Table 9 effective-impedance method for AC circuits (with per-raceway resistance and reactance, power factor, temperature, and parallel sets) and the resistance method for DC. Solve for the drop, the minimum wire size — checked against ampacity — or the maximum run length.
Inputs
Results
Enter voltage, current, and run length to calculate
How this calculator works
For AC circuits the calculator uses the effective-impedance method from NEC Chapter 9, Table 9, Note 2. Table 9 lists each conductor’s AC resistance R and inductive reactance XL in ohms per 1000 feet — at 75°C, 60 Hz, three single conductors in a raceway — with separate columns for PVC, aluminum, and steel raceways (steel raises reactance, and skin effect raises AC resistance on large sizes). The effective impedance at your load’s power factor is:
Ze = R·cos θ + XL·sin θ (θ = arccos PF)
and the voltage drop over a one-way run of L feet at I amps is:
VD = 2 × I × Ze × L / 1000 (single-phase) VD = √3 × I × Ze × L / 1000 (three-phase, line-to-line)
DC circuits use the resistance-only form VD = 2 × I × R × L / 1000, with R derived from the conductor’s circular-mil area (the Chapter 9 Table 8 basis: 12.9 Ω·cmil/ft for copper, 21.2 for aluminum at 75°C). Parallel sets divide R and XL by the number of sets (permitted for 1/0 AWG and larger, NEC 310.10(G)). Resistance is corrected to your conductor operating temperature per Table 8 Note 2 — reactance does not change with temperature. When sizing a conductor, the recommendation is also floored at the load’s base 75°C ampacity from Table 310.16 (before derating), so the tool never suggests a wire that meets the drop target but can’t legally carry the current.
Worked example
A 240 V single-phase, 32 A load at the end of a 150 ft run of 8 AWG copper in PVC conduit, power factor 0.85, conductors at 75°C:
- Table 9, 8 AWG copper in PVC: R = 0.78, XL = 0.052 Ω/1000 ft.
- sin θ = sin(arccos 0.85) = 0.527.
- Ze = 0.78 × 0.85 + 0.052 × 0.527 = 0.690 Ω/1000 ft.
- VD = 2 × 32 × 0.690 × 150 / 1000 = 6.63 V.
- 6.63 / 240 = 2.76% — inside the 3% recommendation. Voltage at the load: 233.4 V.
Enter the same numbers above to reproduce this result. The “values used” block in the results shows the R, XL, and Ze behind every answer so it can be checked against the code tables.
Voltage-drop limits: recommended vs. mandatory
| Limit | Applies to | NEC reference | Status |
|---|---|---|---|
| 3% | Branch circuit, at the farthest outlet | 210.19(A) Informational Note | Recommendation |
| 5% | Feeder + branch circuit combined | 215.2(A) Informational Note | Recommendation |
| 1.5% / 2.5% | Sensitive electronic equipment (Art. 647 systems), branch / combined | 647.4(D) | Mandatory |
| 15% start / 5% run | Fire pump circuits (starting; running at 115% FLA) | 695.7 | Mandatory |
| EGC upsizing | When conductors are upsized for voltage drop, the EGC scales by the same cmil ratio | 250.122(B) | Mandatory |
Informational Notes are not enforceable (NEC 90.5), but energy codes and local amendments in some jurisdictions do make the 3%/5% figures binding — confirm with your AHJ.
Reference: AC resistance & reactance (copper)
Excerpt from NEC Chapter 9, Table 9 — ohms to neutral per 1000 ft at 75°C, uncoated copper, three single conductors in a raceway. The calculator holds the full table for copper and aluminum in PVC, aluminum, and steel raceways, 14 AWG through 750 kcmil.
| Size | R — PVC | R — steel | X∟ — PVC | X∟ — steel |
|---|---|---|---|---|
| 12 AWG | 2.0 | 2.0 | 0.054 | 0.068 |
| 10 AWG | 1.2 | 1.2 | 0.050 | 0.063 |
| 8 AWG | 0.78 | 0.78 | 0.052 | 0.065 |
| 6 AWG | 0.49 | 0.49 | 0.051 | 0.064 |
| 4 AWG | 0.31 | 0.31 | 0.048 | 0.060 |
| 2 AWG | 0.19 | 0.20 | 0.045 | 0.057 |
| 1/0 AWG | 0.12 | 0.12 | 0.044 | 0.055 |
| 4/0 AWG | 0.062 | 0.063 | 0.041 | 0.051 |
| 250 kcmil | 0.052 | 0.054 | 0.041 | 0.052 |
| 500 kcmil | 0.027 | 0.029 | 0.039 | 0.048 |
Frequently asked questions
Is the NEC 3% voltage drop a code requirement?
Generally no. The 3% branch-circuit and 5% total (feeder plus branch) figures come from Informational Notes to NEC 210.19(A) and 215.2(A), and Informational Notes are not enforceable code. But there are real exceptions: NEC 647.4(D) makes 1.5% branch / 2.5% total mandatory for sensitive-electronics circuits, fire pump circuits have hard limits in 695.7, some energy codes and local amendments make 3%/5% binding, and if you do upsize conductors for voltage drop, NEC 250.122(B) requires the equipment grounding conductor to be upsized proportionally — that part is enforceable.
Do I enter the one-way length or the round-trip length?
One-way. Enter the circuit length from the source to the load; the calculator applies the return path itself — a factor of 2 for single-phase and DC circuits, and √3 (1.732) for balanced three-phase circuits.
What power factor should I use?
Use 1.0 for purely resistive loads such as electric resistance heat and incandescent lighting. Use around 0.85 — the calculator default, and the value NEC Table 9 itself tabulates effective impedance at — for typical mixed commercial loads and motors. Motor nameplates and equipment cut sheets often state the actual power factor; if you have it, enter it.
Why does this calculator give a different answer than a simple K-factor calculator?
Simple calculators use VD = 2·K·I·L/cmil with K ≈ 12.9 (copper), which is a DC resistance model. It ignores conductor reactance and AC skin effect, both of which NEC Chapter 9 Table 9 includes. For small conductors the difference is a few percent, but for large conductors at typical power factors the resistance-only method can understate voltage drop by 50% or more — on a 500 kcmil copper feeder in steel conduit at 0.85 PF, the true effective impedance is roughly double the K-method figure. This calculator uses the Table 9 values and its Note 2 formula, which is what the code intends for AC circuits.
Can I run conductors in parallel to reduce voltage drop?
Yes — paralleling divides both resistance and reactance by the number of sets, which divides the voltage drop by the same factor. NEC 310.10(G) only permits parallel installations for conductors 1/0 AWG and larger (with matched length, material, size, insulation, and termination), and the calculator enforces that limit.
Copper or aluminum for long runs?
Aluminum has about 61% of copper’s conductivity, so an aluminum conductor needs roughly 1.6× the circular-mil area for the same voltage drop. Aluminum is still often the economical choice on feeders because it costs and weighs less per ampacity — you simply size it larger. Compare both in the calculator: switch the material and check the recommended size and the drop.
Method: NEC Chapter 9, Table 9 (AC effective impedance, Note 2) and Table 8 basis (DC resistance), with temperature correction per Table 8 Note 2 and parallel sets per 310.10(G). Table 9 values assume 60 Hz, three single conductors in a raceway, at 75°C; single-phase results apply the standard ×2 use of the to-neutral values, and cable/direct-burial runs use the PVC column. Sizing results screen against the base 75°C column of Table 310.16 only — run the full ampacity derate separately. Verify against the governing edition of the NEC and any local amendments for your jurisdiction.
Pair with the wire ampacity chart to confirm the size you land on can carry the load.