Voltage Drop Calculation: NEC 3% and 5% Guidance With Examples

To calculate voltage drop on a single-phase circuit, use VD = 2 × L × R × I ÷ 1,000. R is the wire resistance from NEC Chapter 9 Table 8. The NEC suggests keeping drop to 3% on a branch circuit or feeder and 5% for both together. Those numbers sit in informational notes, so they're advice, not enforceable rules.

Voltage drop calculator

Enter the circuit. The calculator gives the drop in volts and percent, plus the smallest wire size that stays under your limit.

Is NEC voltage drop of 3% and 5% required?

No. The informational note under 210.19(A) covers branch circuits: no more than 3% drop at the farthest outlet, and 5% total with the feeder. It says that gives "reasonable efficiency of operation." The note under 215.2(A)(1) says the same for feeders.

Section 90.5(C) makes informational notes informational only, so an inspector can't enforce them. A job spec or the engineer of record can still require them, and exam questions may use 3% as the target.

The ampacity tables ignore voltage drop too. An informational note in 310.14(A)(1) points back to these notes.

How to calculate voltage drop

1. Look up R for the wire size and metal in Chapter 9 Table 8, in ohms per 1,000 feet.
2. Multiply: 2 × one-way feet × R × amps ÷ 1,000 for single-phase, or use 1.732 instead of 2 for three-phase. The answer is the drop in volts.
3. Divide by the circuit voltage and multiply by 100 for the percent. Over your target? Try the next larger size, as the examples below do.

The voltage drop formula, piece by piece

• 2 (single-phase): current flows out on one wire and back on the other, so the length counts twice.
• 1.732 (three-phase): the square root of 3, from three-phase math, not from an NEC table.
• L: one-way length in feet.
• R: ohms per 1,000 feet from Chapter 9 Table 8 (uncoated copper or aluminum, 75°C).
• I: load current in amps.
• ÷ 1,000: converts the per-1,000-foot resistance to your run length.
• Percent drop: VD ÷ circuit voltage × 100.

Where K = 12.9 comes from

Many books write VD = 2 × K × I × L ÷ CM, with CM the wire's area in circular mils. K isn't in the NEC. It's Table 8 resistance times circular mils, divided by 1,000.

• Copper, 12 AWG stranded: 1.98 × 6,530 ÷ 1,000 = 12.9
• Aluminum, 4/0: 0.100 × 211,600 ÷ 1,000 = 21.2

Both forms give the same answer. The K form helps when solving for wire size.

Table 8 or Table 9?

Table 8 lists DC resistance, which the simple formula uses. Table 9 lists AC resistance and reactance in conduit, plus an "effective Z" valid only at 0.85 power factor.

For small wire they're close: 12 AWG copper is 1.98 ohms per 1,000 feet in Table 8 and 2.0 in Table 9. Use the table the question names.

Voltage drop calculation, worked example: 240-volt load

The job: a 24-amp, 240-volt single-phase load, 150 feet from the panel, on 10 AWG stranded copper. Is it within 3%?

1. R from Table 8: 10 AWG stranded, uncoated copper is 1.24 ohms per 1,000 feet.
2. Formula: 2 × 150 × 1.24 × 24 ÷ 1,000 = 8.93 volts.
3. Percent: 8.93 ÷ 240 = 3.72%. Over the 3% guidance.
4. Try 8 AWG: R = 0.778. 2 × 150 × 0.778 × 24 ÷ 1,000 = 5.60 volts, or 2.33%. That works.

Or solve for wire size: 3% of 240 volts is 7.2 volts. CM = 2 × 12.9 × 24 × 150 ÷ 7.2 = 12,900 circular mils. Table 8 lists 10 AWG at 10,380 (too small) and 8 AWG at 16,510, so pick 8 AWG.

Worked example: three-phase feeder

The job: a 50-amp load on a 208-volt, three-phase feeder, 200 feet long, on 6 AWG copper.

1. R from Table 8: 6 AWG uncoated copper is 0.491 ohms per 1,000 feet.
2. Formula: 1.732 × 200 × 0.491 × 50 ÷ 1,000 = 8.50 volts.
3. Percent: 8.50 ÷ 208 = 4.09%. Over 3%.
4. Try 4 AWG: R = 0.308. 1.732 × 200 × 0.308 × 50 ÷ 1,000 = 5.33 volts, or 2.56%. That works.

Common voltage drop exam traps

• Doubling twice. L is the one-way length. The 2 already covers the return trip.
• Wrong voltage for the percent. Divide by the circuit voltage: 240 for a 240-volt load, 120 for a 120-volt load.
• Forgetting the ground. Under 250.122(B), upsizing the hots for voltage drop upsizes a wire-type ground by the same circular mil ratio. In the 240-volt example, 16,510 ÷ 10,380 = 1.59, so a 10 AWG ground becomes 8 AWG.
• Ignoring temperature. Table 8 is at 75°C. Its Note 2 gives the formula for other temperatures.

Size for ampacity first. Voltage drop never replaces it. See JCR's ampacity derating guide and dwelling service load calculation.

Voltage drop questions

Is 3% voltage drop required by the NEC?

No. The 3% and 5% figures are informational notes, which 90.5(C) says are not enforceable. A project spec or engineer can still require them.

Where is voltage drop covered in the NEC?

In informational notes: 210.19(A) for branch circuits and 215.2(A)(1) for feeders. Chapter 9 Table 8 has the resistance values for the math.

What is K in the voltage drop formula?

K is the resistance of a conductor one circular mil in area and one foot long. It works out to about 12.9 for copper and about 21.2 for aluminum.

How far can you run 12 AWG on a 20-amp circuit at 3%?

It depends on the load. At 120 volts with a 16-amp load, 3% is 3.6 volts. 3.6 × 1,000 ÷ (2 × 1.98 × 16) = about 57 feet one way, and a smaller load can go farther.

Does the equipment ground get bigger if I upsize wire for voltage drop?

Yes, if it's a wire-type ground. Under 250.122(B), hots upsized for any reason other than ampacity correction or adjustment take the ground up by the same circular mil ratio. Voltage drop is one of those reasons.

Related guides: Ampacity derating step by step · Equipment grounding conductor size (Table 250.122) · Dwelling service load calculation

More worked examples: conduit fill calculation, service wire size (Table 310.12) and all NEC Answers.