Voltage Drop Calculator
By Andrew ·
Quick answer
Single-phase voltage drop is 2 × K × I × L ÷ CM: K is 12.9 for copper or 21.2 for aluminum, I the current, L the one-way length in feet and CM the wire's circular mils. Three phase uses 1.732 in place of 2. A 20 A load 100 ft away on 12 AWG copper at 120 V loses 7.9 V (6.6%); 8 AWG keeps it within 3%.
Calculator
- Results
- Voltage drop
- 7.9 V
- Drop as a percent
- 6.6%
- Voltage at the load
- 112.1 V
- Smallest wire for a 3% drop
- 8 AWG
The K-factor formula, with K = 12.9 for copper and 21.2 for aluminum and the wire areas in NEC Chapter 9, Table 8. It ignores reactance, which matters on large wire and long runs.
The K-factor formula
VD = 2 × K × I × L ÷ CMVD = 1.732 × K × I × L ÷ CM- K is the resistance in ohms of a wire one circular mil in area and one foot long: about 12.9 for copper and 21.2 for aluminum, from the resistances in Chapter 9, Table 8 at 75 °C.
- I is the load current in amps.
- L is the one-way length from the source to the load in feet, measured along the route the wire takes.
- CM is the wire’s area in circular mils, from Chapter 9, Table 8.
| Wire | Circular mils |
|---|---|
| 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 AWG | 105,600 |
| 2/0 AWG | 133,100 |
| 3/0 AWG | 167,800 |
| 4/0 AWG | 211,600 |
A kcmil size is its own number of thousands: 250 kcmil is 250,000 circular mils. Divide the drop by the supply voltage for the percent.
Worked example: 20 A, 100 ft away
- System
- Single phase, 120 V
- Load current
- 20 A
- One-way length
- 100 ft
- Wire
- 12 AWG copper, 6,530 CM
- Voltage drop
- 2 × 12.9 × 20 × 100 ÷ 6,530 = 7.9 V
- Drop as a percent
- 7.9 V ÷ 120 V = 6.6%
- Voltage at the load
- 112.1 V
- Smallest wire for 3%
- 8 AWG
To find the wire for a target drop, turn the formula around and solve for the circular mils.
CM = 2 × K × I × L ÷ allowed voltsChecked the other way: 10 AWG drops 4.97 V (4.1%) on this circuit and 8 AWG drops 3.13 V (2.6%).
Three phase, and copper against aluminum
- System
- Three phase, 208 V
- Load current
- 40 A
- One-way length
- 150 ft
- Wire
- 8 AWG copper, 16,510 CM
- Voltage drop
- 1.732 × 12.9 × 40 × 150 ÷ 16,510 = 8.12 V
- Drop as a percent
- 8.12 V ÷ 208 V = 3.9%
- Voltage at the load
- 199.88 V
- Smallest wire for 3%
- 6 AWG, which drops 5.11 V (2.5%)
On three phase the percent is of the line-to-line voltage, 208 V here.
- System
- Single phase, 240 V
- Load current
- 60 A
- One-way length
- 200 ft
- 6 AWG copper
- 11.8 V, 4.9%
- 4 AWG aluminum
- 12.19 V, 5.1%
- Smallest copper for 3%
- 3 AWG
- Smallest aluminum for 3%
- 1 AWG
Aluminum’s K is about 64% higher than copper’s, so for the same drop it takes about two sizes bigger.
3% and 5% are recommendations
The figures everyone quotes come from informational notes in the NEC, in 210.19(A) for branch circuits and 215.2(A) for feeders. They suggest no more than 3% on a branch circuit or a feeder, and no more than 5% for the feeder and branch circuit together, for reasonable efficiency of operation.
Informational notes are not enforceable, so for ordinary circuits the NEC sets no voltage drop limit. Some specific articles, engineers’ specifications and local rules do, and your AHJ has the final say.
| One-way length | 15 A load | 20 A load |
|---|---|---|
| 50 ft | 12 AWG | 10 AWG |
| 75 ft | 10 AWG | 8 AWG |
| 100 ft | 8 AWG | 8 AWG |
| 150 ft | 8 AWG | 6 AWG |
| 200 ft | 6 AWG | 4 AWG |
Where the simple formula falls short
- It uses DC resistance and ignores reactance. On large conductors, long feeders and loads with a low power factor, such as motors, the real drop can be noticeably different, and an AC calculation that includes reactance is more accurate.
- K is for wire at 75 °C. Lightly loaded wire runs cooler and drops a little less, so the answer leans conservative.
- L is the one-way length. Don’t double it: the 2, or the 1.732, already accounts for the return path.
- Use the current the load actually draws. Using the breaker rating instead gives a conservative answer.
- A motor draws several times its running current while it starts, so the drop at starting is much bigger than the steady figure.
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Common questions
What is the maximum voltage drop allowed by the NEC?
For ordinary branch circuits and feeders there is no enforceable maximum. Informational notes in 210.19(A) and 215.2(A) recommend 3% for a branch circuit or feeder and 5% for both together, and some specific articles, job specifications and local rules do set limits.
What size wire do I need for 20 amps at 100 feet?
For a 3% drop at 120 V on copper, 8 AWG, which drops 3.13 V. At 240 V, 10 AWG is enough. It can never be smaller than the circuit's ampacity requires, which is 12 AWG copper for an ordinary 20 A circuit.
How do you calculate voltage drop?
Multiply 2 × K × amps × one-way feet and divide by the wire's circular mils, with K = 12.9 for copper or 21.2 for aluminum. For three phase use 1.732 instead of 2, and divide the drop by the supply voltage for the percent.
Is voltage drop calculated with the one-way or round-trip length?
Enter the one-way length from the source to the load. The 2 in the single-phase formula, or the 1.732 for three phase, already accounts for the return path.
What is K in the voltage drop formula?
K is the resistance of a conductor one circular mil in area and one foot long: about 12.9 ohms for copper and 21.2 ohms for aluminum at 75 °C, worked out from the resistances in Chapter 9, Table 8.