How to Calculate Voltage Drop
The voltage drop formula for a single-phase circuit is VD = 2 × K × I × L ÷ CM, where K is the conductor's resistivity (12.9 for copper, 21.2 for aluminum), I is the current, L is the one-way length in feet and CM is the wire's area in circular mils. For three-phase, replace the 2 with 1.732. Divide VD by the system voltage for percent drop.
Voltage drop is the voltage lost along a conductor because of its resistance. It is Ohm's law applied to the wire itself: current flowing through the conductor's resistance turns part of the supply voltage into heat before it reaches the load.
The two formulas you'll use
Single-phase (120 V, 240 V):
VD = 2 × K × I × L ÷ CM
Three-phase (208 V, 240 V, 480 V):
VD = 1.732 × K × I × L ÷ CM
| Symbol | Meaning | Where to get it |
|---|---|---|
| VD | Voltage drop in volts | The answer |
| K | Resistance of one circular-mil-foot of conductor | ≈ 12.9 for copper, ≈ 21.2 for aluminum (ohm-cmil/ft at 75°C) |
| I | Load current in amps | Actual or calculated load |
| L | One-way length in feet | Measured along the cable route |
| CM | Conductor area in circular mils | NEC Chapter 9, Table 8 |
Percent drop = VD ÷ system voltage × 100.
The single-phase formula has a 2 because current flows out on one conductor and back on the other. In a balanced three-phase circuit the currents share the return path, and the geometry gives √3 (1.732) instead.
How to calculate voltage drop by hand, step by step
- Find the current. Use the calculated load or nameplate current. For a quick check, the breaker rating is a conservative stand-in.
- Measure the one-way length along the actual route: up walls, across ceilings and around obstacles, not straight-line on the plan.
- Look up the circular mils for the wire size in Chapter 9, Table 8 (the wire gauge chart lists them).
- Pick K for the material: 12.9 for copper, 21.2 for aluminum.
- Plug into the formula for single-phase or three-phase, then divide by the system voltage and multiply by 100 for percent.
- Compare to your target, and if the result is over, solve for the circular mils you need and go up to the next standard size.
The K-factor explained
K is the resistance of a piece of conductor one circular mil in area and one foot long, in ohm-circular-mils per foot. It lets you calculate a wire's resistance from its size without looking up a separate resistance value for each gauge: resistance per foot = K ÷ CM.
You can back it out of NEC Chapter 9, Table 8. Stranded 12 AWG copper is listed at about 1.98 Ω per 1,000 feet at 75°C with an area of 6,530 cmil, so K = 1.98 × 6,530 ÷ 1,000 ≈ 12.9. Aluminum's higher resistivity gives about 21.2.
Two things change K in practice:
- Temperature. Resistance rises as the conductor heats. The 75°C values are conservative for lightly loaded circuits; some references use a lower K for conductors running cooler.
- Coating. Tinned (coated) copper has a slightly higher resistance than uncoated copper. Table 8 lists both; for coated conductors, use its resistance value directly.
Example 1: single-phase, 240 V
A 30 A, 240 V well pump circuit, 150 feet away, on 10 AWG copper (10,380 cmil):
VD = 2 × 12.9 × 30 × 150 ÷ 10,380 = 11.18 V
11.18 ÷ 240 = 4.7%, which is above the recommended 3% for a branch circuit.
Example 2: solve for the wire size
Rearrange the formula to find the minimum area for a target drop:
CM = 2 × K × I × L ÷ VD
For the same pump with a 3% target (7.2 V at 240 V):
CM = 2 × 12.9 × 30 × 150 ÷ 7.2 = 16,125 cmil
The next size up in Table 8 is 8 AWG (16,510 cmil). With 8 AWG, VD = 7.03 V, or 2.9%.
The same method works for feeders. A 100 A, 240 V subpanel feeder, 200 feet, in aluminum, at 3%: CM = 2 × 21.2 × 100 × 200 ÷ 7.2 = 117,778 cmil. Ampacity alone needs only 1 AWG aluminum (75°C), but the voltage-drop target calls for 2/0 aluminum (133,100 cmil): VD = 6.37 V, or 2.7%. Check that the breaker and panel lugs accept the larger conductor.
Example 3: solve for the distance
Rearranged for length: L = CM × VD ÷ (2 × K × I).
How far can a 20 A, 120 V circuit on 12 AWG copper run before it exceeds 3% (3.6 V)?
L = 6,530 × 3.6 ÷ (2 × 12.9 × 20) = 45.6 feet one way at the full 20 A. At a 16 A load it is 56.9 feet.
Maximum one-way distance for a 3% drop, copper
| Circuit | Wire | At 100% of rating | At 80% of rating |
|---|---|---|---|
| 15 A, 120 V | 14 AWG | 38.2 ft | 47.8 ft |
| 20 A, 120 V | 12 AWG | 45.6 ft | 56.9 ft |
| 30 A, 240 V | 10 AWG | 96.6 ft | 120.7 ft |
| 40 A, 240 V | 8 AWG | 115.2 ft | 144.0 ft |
| 50 A, 240 V | 6 AWG | 146.5 ft | 183.1 ft |
Beyond these distances, go up a size or check the real load current. Most general-purpose receptacle circuits never carry their full rating, which is why long 120 V runs at full rating look worse on paper than they perform.
Example 4: three-phase, 208 V
A 200 A, 208 V three-phase feeder, 150 feet, on 4/0 copper (211,600 cmil):
VD = 1.732 × 12.9 × 200 × 150 ÷ 211,600 = 3.17 V
3.17 ÷ 208 = 1.5%. Comfortably under the 3% guideline for the feeder, leaving room for branch-circuit drop downstream.
Example 5: three-phase, 480 V, solve for size
A 480 V three-phase feeder with a 60 A calculated load runs 400 feet. The 3% target is 0.03 × 480 = 14.4 V. (If the load were a single motor, Article 430 would first require conductors rated for 125% of its full-load current.)
CM = 1.732 × 12.9 × 60 × 400 ÷ 14.4 = 37,239 cmil
Ampacity alone allows 6 AWG copper (65 A at 75°C), but 6 AWG drops 20.44 V, or 4.3%. The next size with enough area is 4 AWG (41,740 cmil): VD = 12.85 V, or 2.7%. When you upsize the phase conductors for voltage drop, 250.122(B) requires a proportional increase in a wire-type equipment grounding conductor.
K-factor method vs. impedance method
The K-factor method uses DC resistance and ignores reactance. It's accurate for small conductors and resistive loads, which covers most residential and light-commercial work.
For large conductors (roughly 1/0 and larger), long feeders and motor loads, engineers use the impedance method with NEC Chapter 9, Table 9. Table 9 gives AC resistance and reactance per 1,000 feet for copper and aluminum in PVC, aluminum and steel conduit. The effective impedance depends on the power factor:
Ze = R × PF + X × sin(arccos PF)
Then VD (single-phase) = 2 × Ze × I × L ÷ 1,000, and VD (three-phase) = 1.732 × Ze × I × L ÷ 1,000.
Results from the two methods usually agree within about 10% for common building-wire sizes. When a design is close to the limit, use Table 9.
Recommended limits
- 3% for a branch circuit (informational note in NEC 210.19(A)).
- 5% combined for feeder plus branch circuit (informational note in NEC 215.2(A)).
Informational notes are recommended practice, not enforceable requirements, except where the NEC makes them mandatory for specific systems (for example, sensitive electronic equipment under Article 647 and fire pumps under Article 695) or where the project specifications do.
Why voltage drop matters
- Motors run hotter and draw more current at low voltage, which shortens their life, and they may fail to start.
- Electronics and LED drivers can drop out or flicker below their input range.
- Heating elements lose output with the square of the voltage: a 5% drop costs about 10% of the heat.
- Energy is wasted as heat in the conductor, every hour the load runs.
Common mistakes
- Doubling the length twice. The 2 in the single-phase formula already accounts for the return conductor. Use the one-way length.
- Taking the percentage against the wrong voltage. On a 208Y/120 V system, a three-phase or 208 V load is measured against 208 V; a 120 V branch circuit against 120 V.
- Using the breaker rating when you know the load. It is conservative, which is fine for a check, but it can oversize long runs. Use the calculated or nameplate load, and the continuous-load value where it applies.
- Stopping at the branch circuit. The drop on the feeder adds to it. A 2% feeder leaves 3% for the branch circuit under the 5% total.
- Forgetting the ground and the lugs when the conductors get larger.
Frequently asked questions
What is voltage drop in simple terms?
It's the voltage lost in the wire between the panel and the load. Long runs, high current and small wire make it worse.
How do you calculate voltage drop quickly?
For single-phase copper: multiply 25.8 × amps × one-way feet, then divide by the conductor's circular mils. That is 2 × 12.9 combined into one number.
What is K in the voltage drop formula?
K is the resistance of one circular-mil-foot of conductor: about 12.9 ohm-cmil/ft for copper and 21.2 for aluminum at 75°C. Dividing it by the wire's circular mils gives resistance per foot.
How far can I run 12 gauge wire on a 20 amp circuit?
About 45.6 feet one way at the full 20 A on 120 V before the drop passes 3%, or about 56.9 feet at a 16 A load. Beyond that, use 10 AWG or check the real load.
Does voltage drop depend on voltage?
The volts lost do not depend on the system voltage, but the percentage does. The same 6 V drop is 5% at 120 V and 2.5% at 240 V. That is why higher-voltage circuits tolerate longer runs.
How do I measure voltage drop on an existing circuit?
Measure the voltage at the panel and at the load while the load is running at its normal current. The difference is the drop on that circuit.
Related
Skip the hand math: use the voltage drop calculator, or let SparkQuote flag long runs automatically while you estimate the whole job.
Last updated: October 6, 2026. Based on the 2023 NEC. Check results against the code edition and local amendments your jurisdiction enforces and with the AHJ; electrical work should be done by qualified people.