Voltage Drop Calculator
This voltage drop calculator estimates loss, load voltage, minimum listed size, and maximum length—not ampacity or code compliance.
Calculate voltage drop
Example result
120 V · Single-phase AC — two load conductors · 16 A · 50 ft one way · 12 AWG copper · K-factor estimate · 1 set
- Voltage drop
- 3.16 V
- Voltage drop percentage
- 2.63%
- Estimated load voltage
- 116.84 V
At or below the selected 3% target. This compares one modeled run, not wire safety or code compliance.
Approximate 75 °C conductor-resistance model (not ambient temperature). AC assumes power factor 1 and no reactance. No ampacity or code-compliance check.
Show the work
VD = 2 × K × I × L_ft / (CM × sets) VD = 2 × 12.9 × 16 × 50 / (6530 × 1) VD = 3.1607963246554363 V Reference-data cross-check (stranded, 75 °C DC resistance): R = 1.98 Ω per 1,000 ft VD_ref = 2 × 16 × 1.98 × 50 / (1,000 × 1) = 3.168 V The reference is the identified Table 8 reproduction, not a verified current-code edition. This cross-check does not change the primary K result. Drop percent = unrounded VD / 120 × 100 = 2.6339969372128635% Estimated load voltage = 120 − unrounded VD = 116.83920367534456 V Displayed outputs are rounded; target comparisons use unrounded values. A displayed 3.00% may be just above a 3% target.
Feet and meters use 1 ft = 0.3048 m exactly. A load’s current may change with voltage; this model holds the entered current fixed.
Which values should you enter?
| Input | Enter this | Common mistake |
|---|---|---|
| Length | One-way distance, source to load | Entering the round-trip length. The circuit multiplier is already included, so doubling the entered length doubles the result |
| Current | The load's operating current for the case you're checking | Assuming it's always the breaker rating. Use what your problem or design calls for |
| Voltage | The voltage across the two conductors feeding the load. A 240 V load uses 240; a 120 V load on a 120/240 V system uses 120. For three-phase, use line-to-line (208, 480) | Dividing a three-phase line-to-line drop by the line-to-neutral voltage (for example, 277 instead of 480) |
| Wire size | The conductor size on the run. For a quick "what if," change only the size and compare | Changing several inputs at once, then guessing which one mattered |
| Parallel sets | Equal current-sharing paths per phase or pole, joined at both ends; the entered current is the total before splitting | Treating this input as permission to parallel conductors, or dividing AC reactance without installation-specific data |
Voltage doesn't change the volts dropped. Push the same current through the same wire at 120 V or 240 V and you lose the same number of volts. The percent changes because it's measured against a bigger number. For example, 3.48 V is 2.90% of 120 V but only 1.45% of 240 V.
The calculator covers two-wire DC, two-wire single-phase, and balanced three-phase circuits with equal conductor properties on the modeled paths. It doesn't model unbalanced loads on shared neutrals, unequal return conductors, motor starting, harmonics, or voltage lost upstream of the circuit you enter. Supplied-data AC mode is limited to one modeled set; it does not automatically divide reactance for parallel conductors.
Voltage drop formulas
Single-phase and DC
VD = 2 × K × I × L ÷ CM
- VD = voltage drop, in volts
- K = a resistance constant: approximately 12.9 for uncoated copper, 21.2 for aluminum (ohm-circular mils per foot, around 75 °C conductor temperature)
- I = load current, in amps
- L = one-way length, in feet
- CM = conductor area, in circular mils, from the reference list below
Then: VD% = VD ÷ source voltage × 100, and voltage at load = source voltage − VD.
The 2 is there because current flows out on one conductor and back on the other. That's why you enter the one-way length and don't double it yourself. (IAEI Magazine, "Voltage Drop Formulas")
Three-phase: why 1.732 replaces the 2
VD = √3 × K × I × L ÷ CM
On a balanced three-phase circuit, the voltage drop between any two lines is √3 (about 1.732) times the drop on one conductor. So the √3 takes the place of the 2. Use line current for I and line-to-line voltage for the percent. Don't multiply by both 2 and 1.732.
Where K = 12.9 and 21.2 come from
K is resistance in ohms per foot multiplied by conductor area in circular mils. For 12 AWG copper: 1.98 Ω per 1,000 ft × 6,530 circular mils ÷ 1,000 = 12.93. The familiar K values approximate these resistance–area products; they are not exact for every size and construction. (IAEI Magazine)
The default example differs by about 0.007 V between the K estimate and the resistance-data cross-check. That difference is not a universal tolerance: it scales with current and length and also depends on the selected resistance. The "Show the work" panel prints both for the wire-size estimate so you can see it. A circular mil, by the way, is a unit of area, not diameter. It's the area of a circle 0.001 inch across. (Blue Sea Systems, "Circular Mils Explained")
Conductor areas used here
The 26 selections below use the areas in the inspected Table 8 reproduction, page 1. Keep this rounding convention when reproducing these answers. These selections are calculation inputs, not a list of permitted material/size/application combinations. The resistance cross-check uses that reproduction's uncoated-copper or aluminum values at 75 °C; selecting solid changes only the cross-check, not K.
| Size | Area (cmil) | Size | Area (cmil) |
|---|---|---|---|
| 18 AWG | 1,620 | 3/0 AWG | 167,800 |
| 16 AWG | 2,580 | 4/0 AWG | 211,600 |
| 14 AWG | 4,110 | 250 kcmil | 250,000 |
| 12 AWG | 6,530 | 300 kcmil | 300,000 |
| 10 AWG | 10,380 | 350 kcmil | 350,000 |
| 8 AWG | 16,510 | 400 kcmil | 400,000 |
| 6 AWG | 26,240 | 500 kcmil | 500,000 |
| 4 AWG | 41,740 | 600 kcmil | 600,000 |
| 3 AWG | 52,620 | 700 kcmil | 700,000 |
| 2 AWG | 66,360 | 750 kcmil | 750,000 |
| 1 AWG | 83,690 | 800 kcmil | 800,000 |
| 1/0 AWG | 105,600 | 900 kcmil | 900,000 |
| 2/0 AWG | 133,100 | 1000 kcmil | 1,000,000 |
Minimum size and maximum length
Start with allowed drop = source voltage × target percent ÷ 100, in volts. Let F = 2 for the supported DC/single-phase models or F = √3 for balanced three-phase, and n be the number of equal current-sharing sets.
- Required area per set: CM = F × K × I × L ÷ (allowed drop × n). Choose the first listed area at least as large as the unrounded result. This is voltage-drop sizing only.
- Maximum one-way feet: L = allowed drop × CM × n ÷ (F × K × I). Round down to a whole foot, or convert using 0.3048 m/ft and round down to 0.1 m, so the displayed distance does not exceed the computed limit.
These rearrangements use the same K-factor estimate. Custom R/X mode calculates voltage drop only. At zero current, this model does not establish a finite maximum length; zero-current or zero-distance inputs do not select a minimum wire size.
Using resistance, reactance, and power factor
AC conductors have reactance (X) as well as resistance (R); the reactance term can matter when interpreting a resistive-only estimate. When your problem or your cable data gives you R, X, and the load's power factor, use:
- Single-phase: VD ≈ 2 × I × L × (R × PF + X × sin φ)
- Balanced three-phase: VD ≈ √3 × I × L × (R × PF + X × sin φ)
Here R and X are per conductor, per unit length, PF is the lagging power factor, and sin φ = √(1 − PF²). AC current is RMS line current. For DC, use R alone: VD = 2 × I × L × R. Keep the length and the R and X units matched: if R is in ohms per 1,000 ft, L is in thousands of feet. This is an approximation for steady loads, and it's only as good as the R and X values you enter. (Schneider Electric Electrical Installation Guide, "Calculation of voltage drop in steady load conditions")
Worked examples
These are original learning examples, not official exam questions. Keep intermediate values unrounded; the displayed answers are rounded.
Example 1: A 100-foot, 120 V circuit, three wire sizes
Inputs: 120 V single-phase · 15 A · 100 ft one way · copper · 12 AWG (6,530 cmil) · 3% target
VD = 2 × 12.9 × 15 × 100 ÷ 6,530 = 38,700 ÷ 6,530 = 5.93 V
Percent = 5.93 ÷ 120 × 100 = 4.94%
Voltage at load = 120 − 5.93 = 114.07 VAnswer: 5.93 V, 4.94%, 114.07 V at the load. That's above the 3% target, which is 3.60 V on a 120 V circuit.
Now change only the wire size:
| Copper size | Area (cmil) | Voltage drop | Drop % | Voltage at load |
|---|---|---|---|---|
| 12 AWG | 6,530 | 5.93 V | 4.94% | 114.07 V |
| 10 AWG | 10,380 | 3.73 V | 3.11% | 116.27 V |
| 8 AWG | 16,510 | 2.34 V | 1.95% | 117.66 V |
10 AWG is still just over 3%. 8 AWG gets under it. The table compares voltage drop only. It doesn't pick a cable or a breaker.
Common mistake: entering 200 ft and also keeping the 2. That counts the return conductor twice and doubles your answer.
Example 2: Minimum wire size for a three-phase run
Inputs: 208 V three-phase · 60 A · 300 ft one way · copper · 3% target
- Allowed drop = 208 × 0.03 = 6.24 V
- Required area: CM = √3 × 12.9 × 60 × 300 ÷ 6.24 = 64,452 cmil
- Find the first listed size with at least that many circular mils. 3 AWG has 52,620, which is too small. 2 AWG has 66,360, which is enough.
- Check it: VD = √3 × 12.9 × 60 × 300 ÷ 66,360 = 6.06 V = 2.91%.
Answer: 2 AWG copper, for voltage drop.
Choose the first listed area at least as large as the unrounded requirement; do not round to a smaller area. If you used 2 instead of √3 here, you'd get 74,423 cmil and pick 1 AWG, one listed size larger than this voltage-drop calculation requires.
Example 3: Adding reactance and power factor
Inputs: 480 V three-phase · 200 A · 250 ft one way · R = 0.063 Ω per 1,000 ft · X = 0.050 Ω per 1,000 ft · power factor 0.85 lagging. These R and X values are made-up inputs for practice, not data for a specific cable.
sin φ = √(1 − 0.85²) ≈ 0.526783
R × PF + X × sin φ = (0.063 × 0.85) + [0.050 × √(1 − 0.85²)]
≈ 0.0798891 Ω per 1,000 ft
L = 250 ft = 0.250 thousand feet
VD ≈ √3 × 200 × 0.250 × 0.0798891 ≈ 6.92 V
Percent = 6.92 ÷ 480 × 100 = 1.44%Answer: about 6.92 V, 1.44%, 473.08 V line-to-line at the load.
With resistance alone (power factor 1, no reactance), the same run shows 5.46 V. That is about 21% lower, but it changes both power factor and reactance. To isolate reactance, keep power factor at 0.85 and set X to zero: the result is 4.64 V, about 33% lower than 6.92 V. Compare the same assumptions before attributing a difference to one input.
How much voltage drop is acceptable?
Use the limit specified for your actual task. The calculator's 3% default is an example comparison, not a rule that approves a circuit. A receiving voltage must suit the load as well as the applicable installation requirements. (Schneider Electric, "Determination of voltage drop")
The familiar 3% branch-circuit / 5% combined feeder-and-branch guidance appears in Schneider Electric's reproduction of the 2017 NEC branch-circuit informational note, where it describes efficient operation. That historical note is not a current, universal acceptance test for every circuit. (Schneider Electric, 2017 NEC 210.19 excerpt)
| Comparison | What it does—and does not—tell you |
|---|---|
| One run against the entered target | Whether that run's calculated drop is at or below your chosen percentage |
| Feeder plus branch circuit | Requires both parts of the path and a consistent voltage basis; this calculator does not silently add a run you have not entered |
| An actual installation | Requires the applicable code edition, local requirements, equipment limits, and the full design—not just this result |
Do not apply this steady-load calculator to fire-pump starting or treat its default target as a limit for special systems. For an actual installation, check the applicable requirements with the authority having jurisdiction and the equipment documentation.
Voltage drop vs. ampacity: check both
Ampacity and voltage drop answer different questions. Ampacity concerns the current a conductor can carry under its temperature and installation conditions. Voltage drop is how much voltage the load loses along the way. A wire can pass one check and fail the other. On long runs, voltage drop is often what pushes you to a bigger size, but it never lets you use a size smaller than ampacity requires. (Schneider Electric, conductor-sizing context)
This calculator checks voltage drop only.
Its minimum-size result is the smallest listed size meeting the voltage-drop target under the selected model. It does not size overcurrent protection or equipment grounding conductors, check terminations, or establish whether parallel conductors are permitted. Changing circuit-conductor size does not complete those separate checks.
Why voltage drop calculators give different answers
When two calculators disagree, it's usually one of these:
- One-way vs. total length. Some tools want round-trip length. Enter the same distance into a tool that expects the other convention and one answer doubles or halves.
- K rounding vs. supplied resistance. 12.9 is an approximation. The size of the difference depends on the conductor data, current, and distance; it is not always a few hundredths of a volt.
- DC resistance vs. AC impedance. The reference table supplies DC resistance. An AC estimate can use resistance, reactance, and power factor together. Its result is not automatically larger; compare the actual inputs (see Example 3).
- Temperature. The resistance reference is at 75 °C conductor temperature. For the same copper or aluminum conductor and other inputs, a lower resistance temperature gives less resistive drop. Ambient temperature and conductor temperature are not interchangeable.
- Voltage basis on three-phase. Line-to-line vs. line-to-neutral changes the percent.
Compare like with like before deciding a tool is wrong. The assumptions behind this page's result are listed right under it.
Voltage drop practice problems
Original practice problems, not official exam questions. Use the K method (12.9 copper, 21.2 aluminum) and the conductor areas in the reference list above. Assume equal conductor properties on the modeled current paths and balanced loads for three-phase questions. Round only the final answer; √3 is not truncated to 1.732 in the calculations.
Question 1 (VD-01)
A 120 V single-phase circuit supplies 16 A through 12 AWG copper. The load is 100 ft from the panel. What is the voltage drop?
- A. 3.16 V
- B. 6.32 V
- C. 12.64 V
- D. 5.27 V
Reveal answer
B. 6.32 V VD = 2 × 12.9 × 16 × 100 ÷ 6,530 = 6.32 V. Choice A drops the 2, so it counts only one conductor. Choice C doubles the length and keeps the 2, which counts the return path twice. Choice D is the percent (6.32 ÷ 120 × 100 = 5.27%), not volts.
Sources: K-factor equations; conductor reference, page 1.
Question 2 (VD-02)
A 120 V single-phase load draws 7 A. It is 200 ft from the source and wired with 10 AWG copper. What percentage voltage drop does it have?
- A. 2.90%
- B. 3.48%
- C. 1.45%
- D. 5.80%
Reveal answer
A. 2.90% VD = 2 × 12.9 × 7 × 200 ÷ 10,380 = 3.48 V, and 3.48 ÷ 120 × 100 = 2.90%. Choice B is the drop in volts written as a percent, a very common slip. Choice C is the percent on a 240 V circuit. Choice D doubles the correct answer by doubling the length.
Sources: K-factor equations; conductor reference, page 1.
Question 3 (VD-03)
A 240 V single-phase load draws 40 A and sits 200 ft from the panel. What is the smallest listed copper conductor that keeps voltage drop at or below 3%, considering voltage drop only?
- A. 8 AWG
- B. 6 AWG
- C. 4 AWG
- D. 3 AWG
Reveal answer
C. 4 AWG Allowed drop = 240 × 0.03 = 7.2 V. CM = 2 × 12.9 × 40 × 200 ÷ 7.2 = 28,667 cmil. 6 AWG is only 26,240 cmil, so it falls short; 4 AWG (41,740 cmil) is the first size that meets it, giving 4.94 V (2.06%). 8 AWG is far too small. 3 AWG works, but it isn't the smallest size that does.
Sources: K-factor equations; conductor reference, page 1.
Question 4 (VD-04)
A 208 V three-phase load draws 60 A. The one-way distance is 300 ft. What is the smallest listed copper conductor that keeps voltage drop at or below 3%, considering voltage drop only?
- A. 3 AWG
- B. 2 AWG
- C. 1 AWG
- D. 1/0 AWG
Reveal answer
B. 2 AWG Allowed drop = 208 × 0.03 = 6.24 V. CM = √3 × 12.9 × 60 × 300 ÷ 6.24 = 64,452 cmil. 3 AWG (52,620) is too small; 2 AWG (66,360) is the first that works. Choice C is what you get if you use the single-phase 2 instead of √3 (74,423 cmil). Choice D is larger than this voltage-drop calculation requires.
Sources: K-factor equations; conductor reference, page 1.
Question 5 (VD-05)
A 208 V three-phase feeder carries 150 A over 300 ft of 4/0 AWG aluminum. What is the percentage voltage drop?
- A. 1.88%
- B. 3.75%
- C. 4.34%
- D. 2.28%
Reveal answer
B. 3.75% VD = √3 × 21.2 × 150 × 300 ÷ 211,600 = 7.81 V, and 7.81 ÷ 208 × 100 = 3.75%. Choice C uses 2 instead of √3 (9.02 V). Choice D uses the copper K of 12.9 (4.75 V). Choice A is half the correct answer.
Sources: K-factor equations; conductor reference, page 1.
Question 6 (VD-06)
A 120 V single-phase circuit uses 12 AWG copper and carries 20 A. What is the longest one-way run that keeps voltage drop at or below 3%, rounded down to the whole foot?
- A. 45 ft
- B. 91 ft
- C. 56 ft
- D. 22 ft
Reveal answer
A. 45 ft Allowed drop = 120 × 0.03 = 3.6 V. L = 3.6 × 6,530 ÷ (2 × 12.9 × 20) = 45.6 ft, which rounds down to 45 ft. Round down, because 46 ft would go over the target. Choice B forgets the 2. Choice C uses 16 A instead of 20 A. Choice D counts the return conductor twice.
Sources: K-factor equations; conductor reference, page 1.
Question 7 (VD-07)
A 12 V DC load draws 10 A. It is 25 ft from the battery and wired with 10 AWG copper. What is the percentage voltage drop?
- A. 2.59%
- B. 5.18%
- C. 0.62%
- D. 10.36%
Reveal answer
B. 5.18% DC uses the same two-conductor formula: VD = 2 × 12.9 × 10 × 25 ÷ 10,380 = 0.621387… V, and 0.621387… ÷ 12 × 100 ≈ 5.18%. Choice C is the drop in volts. Choice A forgets the 2. Choice D doubles the length. Less than a volt looks small, but on a 12 V system it's over 5%.
Sources: K-factor equations; conductor reference, page 1.
Question 8 (VD-08)
The calculator reports a voltage drop below the selected 3% target. Which conclusion does that result support?
- A. The selected conductor has enough ampacity for the installation.
- B. The modeled run is below the selected voltage-drop target under the entered assumptions.
- C. The feeder and every downstream branch circuit together are below 5%.
- D. The installation complies with every applicable electrical rule.
Reveal answer
B. The modeled run is below the selected voltage-drop target under the entered assumptions. The result compares one modeled run with the target you entered. Choice A needs a separate ampacity calculation; current-carrying capacity is not calculated here. Choice C adds upstream or downstream runs that were never entered. Choice D treats a numerical comparison as a complete installation review. None of those conclusions follows from this result alone.
Sources: voltage-drop interpretation; conductor-sizing context.
Your answers on these eight problems tell you how you did on these problems only. They are not an exam score or a prediction. For more calculation practice, try the electrical load calculation practice problems.
Voltage drop FAQ
Can I use this for 12 V or 24 V DC, like solar, RV, or low-voltage lighting?
Yes. Choose DC and enter the one-way length. At low voltages the percentage climbs fast, as Question 7 shows. This page doesn't check the separate installation rules for automotive, marine, or photovoltaic systems.
Does it handle parallel conductors?
For the K estimate and supplied-resistance DC mode, set the number of equal parallel sets under More options. The drop is divided by the number of sets, which assumes the sets are the same material, area, length, and resistance and share the total entered current evenly. For DC and single-phase, the return path must have the same modeled resistance.
This is an electrical-model assumption, not permission to use a particular parallel installation. Supplied-data AC mode requires one modeled set: use installation-specific effective R/X data rather than assuming reactance divides by the set count.
Is this the method used on electrician exams?
These exercises use the K-factor method, and it's what the "Show the work" panel prints for the wire-size estimate. No particular exam or NEC edition is implied. Follow your candidate bulletin's references and calculator rules, and use the data and method specified in the problem.
Sources
- IAEI Magazine, "Voltage Drop Formulas", April 5, 2017: K-factor derivation and equations. Used for the calculation method, not current code requirements.
- Table 8, "Conductor Properties," reproduced by NECA-IBEW, page 1: the listed circular-mil areas and 75 °C DC resistance. The reproduction does not identify its edition; it is the fixed reference used here, not a claim of current NEC table verification.
- Schneider Electric Electrical Installation Guide, "Calculation of voltage drop in steady load conditions", Figure G29: resistance/reactance and power-factor equations. Its IEC-specific installation rules are not implemented here.
- Schneider Electric, "Determination of voltage drop": interpretation in relation to the load and installation requirements.
- Schneider Electric, 2017 NEC 210.19 excerpt: historical 3%/5% efficiency guidance, not a current-edition code determination.
- Schneider Electric, conductor-sizing context: current-carrying capacity is a separate installation-dependent calculation.
- Blue Sea Systems, "Circular Mils Explained": the circular-mil area unit.
- NIST, "U.S. Survey Foot": the international foot is exactly 0.3048 meter.
By Castleport Test Prep Editorial Team
Last checked: September 30, 2026 — calculation formulas, the identified reference-table entries, unit conversion, and the original worked answers. These checks do not establish current NEC compliance or constitute a professional review of an electrical installation.
AI tools assisted with drafting and checking this resource. Source checks and calculation tests are not credentialed subject-matter review. See our editorial standards.
Castleport Test Prep is an independent exam prep publisher, not affiliated with, endorsed by, or approved by NFPA, any licensing board, or any testing vendor. Code and exam names identify their subjects; trademarks belong to their respective owners. These are original, unofficial learning exercises, not actual exam questions. They do not guarantee an exam result or license.