Voltage Drop Calculator — Wire Gauge, Distance & Drop Percent
Long wire runs steal voltage. Enter your circuit’s voltage, current and one-way distance, pick a wire gauge, and see how many volts survive the trip — and whether the drop stays inside NEC guidance.
Your circuit
Advanced options
The core answer uses the fields above; these refine it.
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At the load
Your wire run
Voltage at the panel on the left, voltage that reaches the load on the right. The wire changes color with the guidance check.
Find your wire size
Give the calculator a target and it tests every wire size with your current method and circuit details, smallest first.
The finder sizes for voltage drop only. Ampacity, temperature ratings and local code still govern the final choice — have the run designed by a licensed electrician.
- Every conductor resists current a little, and that resistance turns some of your voltage into heat along the run — the drop grows with current and distance, and shrinks with a bigger conductor.
- At the defaults — 120 V, 15 A, 100 ft of 12 AWG copper — 5.93 V is lost on the run (4.94%), leaving 114.07 V at the load.
- The NEC’s informational notes suggest no more than 3% drop on a branch circuit and 5% total across feeder and branch combined.
Keep exploring
Three companion tools for electrical and math work.
How it works
Three steps from panel to verdict.
Enter the circuit
The source voltage and the current the load draws — from the device label or the breaker size feeding it.
Measure the run
One-way distance — feet, meters, miles or kilometers — panel to load. The formula doubles it, because current travels out and back.
Pick wire and material
Choose the AWG gauge and copper or aluminum, then read the drop percent against the 3% and 5% guidance lines.
The math behind voltage drop
Every conductor resists current a little, and that resistance turns some of your voltage into heat along the run. For single-phase AC and DC circuits, electricians use the circular-mil formula: the drop grows with current, distance and the material constant K (12.9 for copper, 21.2 for aluminum, at operating temperature), and shrinks as the wire’s cross-section — its circular mils (CM) — grows.
For a three-phase run the constant changes from 2 to √3 and the result is the line-to-line drop. DC and single-phase AC share the 2× form above.
Worked examples:
| Scenario | Calculation | Result |
|---|---|---|
| 120 V, 15 A, 100 ft, copper 12 AWG | 2 × 12.9 × 15 × 100 ÷ 6,530 | Drop 5.93 V · 4.94% · 114.07 V at load |
| 240 V, 20 A, 150 ft, aluminum 10 AWG | 2 × 21.2 × 20 × 150 ÷ 10,380 | Drop 12.25 V · 5.11% · 227.75 V at load |
| 208 V three-phase, 20 A, 150 ft, copper 8 AWG | √3 × 12.9 × 20 × 150 ÷ 16,510 | Drop 4.06 V · 1.95% · 203.94 V at load |
| 240 V, 30 A, 200 ft, copper 6 AWG | 2 × 12.9 × 30 × 200 ÷ 26,240 | Drop 5.90 V · 2.46% · 234.10 V at load |
Circular mils by gauge: 18 AWG 1,620 · 16 AWG 2,580 · 14 AWG 4,110 · 12 AWG 6,530 · 10 AWG 10,380 · 8 AWG 16,510 · 6 AWG 26,240 · 4 AWG 41,740 · 3 AWG 52,620 · 2 AWG 66,360 · 1 AWG 83,690 · 1/0 105,600 · 2/0 133,100 · 3/0 167,800 · 4/0 211,600 · 250 kcmil 250,000 · 300 kcmil 300,000 · 350 kcmil 350,000 · 400 kcmil 400,000 · 500 kcmil 500,000 · 600 kcmil 600,000 · 750 kcmil 750,000 · 1000 kcmil 1,000,000.
Resistance method: if you already know the conductor’s resistance per unit length — from a datasheet or NEC Chapter 9 Table 8 — switch the calculator’s method to resistance and enter it in Ω/km, Ω/m, mΩ/m, Ω/1000 ft, Ω/ft or mΩ/ft. The drop is the phase constant (2, or √3 for three-phase) × current × resistance per km × length in km, and parallel conductor sets divide it. Table 8 resistances differ slightly (about ±2–3% on small gauges) from the circular-mil figures above, which is why every result names the method that produced it.
NEC impedance method: for AC runs the NEC’s own model (Chapter 9, Table 9) splits the conductor’s opposition to current into resistance R and inductive reactance XL, and both depend on the conduit — steel raises the reactance, and on large conductors it raises the resistance too. The calculator looks up R and XL for your size, material and conduit, combines them at your power factor with the Table 9 Note 2 formula above, and multiplies by current and length (× 2 single-phase, × √3 three-phase, divided by parallel sets). Worked example, in the style of the NEC handbook: 240 V three-phase, 100 A, 150 ft of 1 AWG copper in aluminum conduit at 0.85 power factor — Table 9 gives Ze = 0.16 Ω/1000 ft, so the drop is √3 × 0.16 × 100 × 150 ÷ 1,000 = 4.16 V (1.73%), leaving 235.84 V at the load. On DC the reactance does nothing, so the NEC method works from the Table 9 resistance alone and the conductor line says so.
Temperature correction (computed): the K constants and the Table 9 resistances are 75 °C figures, but resistance climbs as a conductor warms. Set the conductor temperature in Advanced options and the resistance is rescaled with the standard temperature relation — (T + 234.5) ÷ (75 + 234.5) for copper, with 228 in place of 234.5 for aluminum, using the inferred absolute-zero constants. The correction is computed from the table values rather than read from a table, and the conductor line flags it whenever the temperature is not 75 °C.
| Size | XL — PVC or aluminum conduit (Ω/1000 ft) | XL — steel conduit (Ω/1000 ft) | R — copper in steel conduit (Ω/1000 ft) | R — aluminum in steel conduit (Ω/1000 ft) |
|---|---|---|---|---|
| 14 AWG | 0.058 | 0.073 | 3.1 | — |
| 12 AWG | 0.054 | 0.068 | 2.0 | 3.2 |
| 10 AWG | 0.05 | 0.063 | 1.2 | 2.0 |
| 8 AWG | 0.052 | 0.065 | 0.78 | 1.3 |
| 6 AWG | 0.051 | 0.064 | 0.49 | 0.81 |
| 4 AWG | 0.048 | 0.06 | 0.31 | 0.51 |
| 3 AWG | 0.047 | 0.059 | 0.25 | 0.4 |
| 2 AWG | 0.045 | 0.057 | 0.2 | 0.32 |
| 1 AWG | 0.046 | 0.057 | 0.16 | 0.25 |
| 1/0 AWG | 0.044 | 0.055 | 0.12 | 0.2 |
| 2/0 AWG | 0.043 | 0.054 | 0.1 | 0.16 |
| 3/0 AWG | 0.042 | 0.052 | 0.079 | 0.13 |
| 4/0 AWG | 0.041 | 0.051 | 0.063 | 0.1 |
| 250 kcmil | 0.041 | 0.052 | 0.054 | 0.086 |
| 300 kcmil | 0.041 | 0.051 | 0.045 | 0.072 |
| 350 kcmil | 0.04 | 0.05 | 0.039 | 0.063 |
| 400 kcmil | 0.04 | 0.049 | 0.035 | 0.055 |
| 500 kcmil | 0.039 | 0.048 | 0.029 | 0.045 |
| 600 kcmil | 0.039 | 0.048 | 0.025 | 0.038 |
| 750 kcmil | 0.038 | 0.048 | 0.021 | 0.031 |
| 1000 kcmil | 0.037 | 0.046 | 0.018 | 0.025 |
Extract from NEC Chapter 9, Table 9 — ohms to neutral per 1000 ft at 75 °C. The calculator carries the full table, with resistance for all three conduit types; aluminum has no 14 AWG row in Table 9.
Data: conductor circular-mil areas and the NEC impedance method’s resistance and reactance values from NFPA 70, National Electrical Code, Chapter 9, Tables 8 and 9 (600-volt cables, three single conductors in conduit, 75 °C); effective impedance per Table 9, Note 2. Temperature corrections computed with the standard resistance–temperature relation (inferred absolute zero −234.5 °C for copper, −228 °C for aluminum). Checked October 2026.
Go deeper
Why the formula looks the way it does — and what a bad drop does.
Current makes a round trip: out to the load on one conductor and back on the other. Both legs resist it, so the wire that matters is twice the run:
- One-way input: you enter the panel-to-load distance; the 2 in the formula adds the return leg for you.
- Circular mils: CM measures the conductor’s cross-section. Bigger wire, more circular mils, less resistance, less drop.
- The K constant: 12.9 for copper and 21.2 for aluminum are the standard resistivity figures at typical operating temperature — resistance rises as wire heats up, which is why these are not the cold-lab values.
- Gauge steps: each AWG size multiplies CM by about 1.26, so one size up cuts the drop by roughly a fifth.
A load designed for 120 V does not shrug off 105 V. The effects show up fast:
- Motors run slower, draw more current to compensate, and overheat — the classic killer of well pumps and compressors on long runs.
- Lights dim and shift color; a 10% drop is plainly visible.
- Heating elements lose output with the square of the voltage — a 5% drop is nearly a 10% heat loss.
- Electronics may reset, misbehave, or trip their own protection.
- The fixes: upsize the wire, shorten the run, split the load across circuits, or step up to 240 V where the equipment allows it — the same watts at double voltage means half the current.
Frequently asked questions
Quick answers about voltage drop and wire sizing.
What is an acceptable voltage drop?
The NEC’s informational notes suggest keeping branch-circuit drop to 3% or less, and the combined feeder-plus-branch total to 5% or less, for reasonable efficiency. At 120 V, 3% is 3.6 V. These are recommendations rather than hard limits, but equipment runs noticeably better inside them.
How do I calculate voltage drop?
Use VD = 2 × K × I × L ÷ CM: twice the material constant (12.9 copper, 21.2 aluminum), times the current in amps, times the one-way length in feet, divided by the wire’s circular mils. A 15 A load 100 ft away on copper 12 AWG: 2 × 12.9 × 15 × 100 ÷ 6,530 = 5.93 V.
Do I enter one-way or round-trip distance?
One-way. Enter the distance from the panel to the load only. Current physically travels out and back, and the 2× in the formula accounts for the return conductor — doubling the distance yourself would count the trip twice and overstate the drop.
Is copper or aluminum better for long runs?
Copper drops less voltage at the same gauge — its K is 12.9 against aluminum’s 21.2 — so it suits long runs where drop is the binding constraint. Aluminum needs roughly two AWG sizes larger for a similar drop, but costs and weighs less, which is why it is common for large feeders.
What happens if voltage drop is too high?
Motors run hot and slow, lights dim, heating elements lose nearly twice the percent of output as the voltage percent lost, and electronics can reset or fault. Sustained undervoltage also shortens motor life — it is an equipment problem, not just an efficiency one.
Can I calculate three-phase drop, or use a known resistance per length?
Yes to both. Set the circuit type to AC three-phase and the calculator swaps the 2× constant for √3, giving the line-to-line drop, in either method. If you already know the conductor’s resistance per unit length — from a cable datasheet or NEC Chapter 9 Table 8 — switch the method to resistance and enter it in Ω/km, Ω/m, mΩ/m, Ω/1000 ft, Ω/ft or mΩ/ft. A third method, NEC impedance, models AC runs from Chapter 9 Table 9 — the resistance and reactance for your conduit type, combined at your power factor. Every method also handles 1–9 parallel conductor sets, dividing the drop by the number of sets.
How do I fix a voltage drop problem?
The standard fixes, in order of usual cost: shorten the run if the layout allows, upsize the wire (each AWG step up cuts the drop by roughly 21%), split the load across two circuits, or move the load to 240 V if the equipment supports it — the same power then needs half the current. The Find your wire size tool under the results does the upsizing maths for you: give it a target percent and it names the smallest gauge that meets it under your current method.
How does conductor temperature change the drop?
Resistance climbs as a conductor warms, so a loaded run drops more volts hot than cold. The K constants (12.9 copper, 21.2 aluminum) and the NEC Table 9 resistances are 75 °C figures. If your conductor runs cooler or hotter, set its temperature in Advanced options: the calculator rescales the resistance with the standard temperature relation — computed from the 75 °C values using the inferred absolute-zero constants, −234.5 °C for copper and −228 °C for aluminum — and the conductor line flags the correction.
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Disclaimer
This calculator is for information only and is not a substitute for a licensed electrician or your local electrical code. The 3% and 5% figures are NEC informational-note guidance; conductor ampacity, temperature ratings and local amendments also govern real installations. When in doubt, have the run designed by a professional.