Voltage Drop: NEC Rules, Formula, and Exam Calculations Explained

Voltage drop is one of the most common calculation questions on journeyman and master electrician exams — and one that many candidates lose points on unnecessarily. The formula looks intimidating, but it follows a consistent pattern. Once you understand the NEC recommendations, the single-phase and three-phase formulas, and the wire sizing procedure, voltage drop problems become straightforward.

This guide covers everything the exam tests: the NEC recommendations, both formulas, worked single-phase and three-phase examples, and how to size wire to keep voltage drop within limits.

Why Voltage Drop Matters

When current flows through a conductor, the conductor's resistance causes a voltage loss along the length of the wire. The load receives less voltage than the source provides. This matters for several reasons:

• Equipment performance: Motors run hotter and inefficiently at low voltage. Lighting is dim. Electronic equipment may malfunction or fail.
• Code compliance: While the NEC's voltage drop recommendations are not strictly mandatory, they represent the minimum for "satisfactory" operation — and many AHJs and customers treat them as requirements.
• Exam questions: Voltage drop calculations appear on virtually every journeyman and master licensing exam.

The NEC Voltage Drop Recommendations

Here is what the NEC actually says — note the language carefully.

NEC 210.19(A) Informational Note No. 4 (Branch Circuits):

"Voltage drop on branch circuit conductors should not exceed 3 percent, and the combined voltage drop on both feeders and branch circuit conductors should not exceed 5 percent for satisfactory performance."

NEC 215.2(A)(1) Informational Note No. 2 (Feeders):

"Conductors for feeders as defined in Article 100, sized to prevent a voltage drop exceeding 3 percent at the farthest outlet of power, heating, and lighting loads, or combinations of such loads, and where the maximum total voltage drop on both feeders and branch circuits to the farthest outlet does not exceed 5 percent, will provide reasonable efficiency of operation."

The Key Point: "Recommended" vs. "Required"

The NEC calls these informational notes — they are recommendations, not mandatory requirements (the NEC uses "shall" for requirements). In practice:

• Many AHJs and job specifications adopt the 3%/5% recommendations as enforceable requirements
• Licensing exams almost universally treat the 3% and 5% limits as the standard to apply in calculation problems
• For your exam, use 3% for branch circuits, 3% for feeders, and 5% combined

The 3% / 5% Summary

If a feeder has 2% voltage drop, the branch circuit should have no more than 3% (totaling 5%). If the feeder has 1%, the branch can have up to 4%. The combined total should not exceed 5%.

The Voltage Drop Formula

Single-Phase Formula

VD = (2 × K × I × L) / CM

Where:

• VD = Voltage drop (volts)
• K = Resistivity constant of the conductor (12.9 for copper, 21.2 for aluminum, both at 75°C)
• I = Current in amperes
• L = One-way length of the conductor run in feet
• CM = Conductor area in circular mils (from NEC Chapter 9 Table 8)
• 2 = Factor for two conductors (current travels out and returns)

Three-Phase Formula

VD = (1.732 × K × I × L) / CM

The only difference is replacing 2 with 1.732 (√3). Three-phase systems are more efficient because the return path is shared — the voltage drop is lower for the same wire size, current, and distance compared to single-phase.

The K Values

Some older references use K = 11.9 for copper (calculated at 25°C room temperature). Most licensing exams use 12.9 for copper because conductors operate at elevated temperatures under load. Use 12.9 unless the exam problem states otherwise.

Circular Mil (CM) Values — NEC Chapter 9 Table 8

You need circular mil areas for the voltage drop formula. These come from NEC Chapter 9 Table 8. Memorize the most common ones:

Note: kcmil = thousands of circular mils. 250 kcmil = 250,000 CM.

Worked Examples

Example 1: Calculate Voltage Drop (Single-Phase)

Problem: A 20A, 120V single-phase branch circuit is wired with 12 AWG copper THHN. The circuit run is 75 feet (one-way). What is the voltage drop, and does it meet the 3% recommendation?

Solution:

• K = 12.9 (copper)
• I = 20A
• L = 75 feet
• CM = 6,530 (12 AWG copper, Chapter 9 Table 8)

VD = (2 × 12.9 × 20 × 75) / 6,530

VD = (38,700) / 6,530

VD = 5.93 volts

Percentage: 5.93 / 120 = 4.9%

This exceeds the 3% recommendation (3% of 120V = 3.6V). You would need a larger wire.

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Example 2: Size Wire to Limit Voltage Drop (Single-Phase)

Problem: Same circuit as above — 20A, 120V, 75 feet. What is the minimum wire size to keep voltage drop at or below 3%?

Solution: Rearrange the formula to solve for CM:

CM = (2 × K × I × L) / VD

Maximum allowable VD = 3% × 120V = 3.6 volts

CM = (2 × 12.9 × 20 × 75) / 3.6

CM = 38,700 / 3.6

CM = 10,750 circular mils minimum

Looking up Chapter 9 Table 8: 10 AWG copper = 10,380 CM (too small), 8 AWG copper = 16,510 CM (meets requirement).

Minimum wire size for voltage drop = 8 AWG copper.

Check ampacity: 8 AWG at 75°C = 50A. Circuit is 20A. Ampacity is fine. Use 8 AWG copper.

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Example 3: Three-Phase Voltage Drop Calculation

Problem: A 50A, 208V three-phase circuit is run with 6 AWG copper conductors for a one-way distance of 120 feet. What is the voltage drop?

Solution:

• K = 12.9 (copper)
• I = 50A
• L = 120 feet
• CM = 26,240 (6 AWG copper)
• Three-phase: use 1.732 multiplier

VD = (1.732 × 12.9 × 50 × 120) / 26,240

VD = (1.732 × 12.9 × 6,000) / 26,240

VD = 134,258.4 / 26,240

VD = 5.12 volts

Percentage: 5.12 / 208 = 2.46%

This is within the 3% branch circuit recommendation. ✓

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Example 4: Combined Feeder + Branch Circuit

Problem: A 277V branch circuit serves lighting at the far end of a building. The feeder has 1.5% voltage drop. What is the maximum allowable voltage drop on the branch circuit?

Solution:

• Combined limit: 5%
• Feeder uses: 1.5%
• Maximum for branch circuit: 5% − 1.5% = 3.5%
• Maximum VD in volts: 3.5% × 277V = 9.70 volts

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Example 5: Sizing Wire for a Long 240V Run (Single-Phase)

Problem: A 30A, 240V single-phase circuit feeds a subpanel 200 feet away (one-way). Size the copper conductors to keep voltage drop within 3%.

Solution: Maximum VD = 3% × 240V = 7.2 volts

CM = (2 × 12.9 × 30 × 200) / 7.2

CM = 154,800 / 7.2

CM = 21,500 circular mils minimum

Looking up Chapter 9 Table 8:

• 8 AWG = 16,510 CM — too small
• 6 AWG = 26,240 CM — meets requirement ✓

Check ampacity: 6 AWG copper at 75°C = 65A. Circuit is 30A. Fine.

Use 6 AWG copper.

Voltage Drop vs. Ampacity: Two Separate Calculations

This is a point that trips up many exam candidates. Voltage drop and ampacity sizing are independent calculations — you must satisfy both.

A 12 AWG conductor at 75°C has an ampacity of 25A — it won't overheat at 25A. But if the run is 150 feet at 20A on a 120V circuit, voltage drop is 11.9% — completely unacceptable. You would need to upsize for voltage drop even though the ampacity is adequate.

Rule: When a voltage drop calculation requires a larger wire than ampacity alone, use the larger wire. The conductor must satisfy both requirements.

The Alternative Formula (Using Table 9 Resistance)

NEC Chapter 9 Table 9 gives conductor resistance in ohms per 1,000 feet. You can use this instead of the CM/K formula:

VD = (2 × R × I × L) / 1,000

Where:

• R = Resistance in ohms per 1,000 feet (from Table 9, at 75°C)
• L = One-way length in feet
• 2 = Two conductors (single-phase; use 1.732 for three-phase)

Example: 12 AWG copper, 20A, 75 feet, 120V, single-phase. R for 12 AWG copper at 75°C = 2.0 Ω/1000 ft (Table 9) VD = (2 × 2.0 × 20 × 75) / 1000 = 6,000 / 1000 = 6.0 volts (5%)

(Slightly different from the K formula due to rounding in Table 9.)

The K formula and Table 9 formula give similar results. Use whichever the exam question provides data for. If the exam gives you circular mils, use the K formula. If it gives you resistance in Ω/1000 ft, use Table 9.

Common Exam Traps

One-way vs. total length: The formula uses L as the one-way circuit length. The factor of 2 (single-phase) accounts for both conductors. Do not double L in addition to using the factor of 2 — that counts the conductors twice.

Three-phase vs. single-phase: The only change is the multiplier: 2 for single-phase, 1.732 for three-phase. A three-phase circuit with the same current and wire size has about 13% less voltage drop than a single-phase circuit over the same distance.

Voltage drop percentage: Always express voltage drop as a percentage of the source voltage at that point in the system (e.g., 120V for a 120V branch circuit), not the load voltage or a different system voltage.

Copper vs. aluminum: A question that swaps copper for aluminum (K = 21.2 vs. 12.9) nearly doubles the voltage drop for the same wire size. Read carefully.

Neutral conductor: For single-phase 2-wire circuits and single-phase 3-wire circuits, count the neutral as carrying return current — this is why the factor of 2 appears. For balanced 3-phase circuits, the neutral carries negligible current and is not included.

Using the Voltage Drop Calculator

Working through these calculations by hand on a timed exam is slow. Use our Voltage Drop Calculator to practice quickly checking your hand calculations and to build intuition for how wire size, run length, and current interact. Being fluent with the relationship between these variables is what lets you eliminate wrong answers quickly on the actual exam.

Key Takeaways

• The NEC recommends (not requires) 3% max for branch circuits, 3% for feeders, 5% combined total
• Single-phase formula: VD = (2 × K × I × L) / CM
• Three-phase formula: VD = (1.732 × K × I × L) / CM
• K = 12.9 for copper, K = 21.2 for aluminum (at 75°C)
• To find minimum wire size: solve for CM = (2 × K × I × L) / VD, then look up in Chapter 9 Table 8
• Voltage drop and ampacity are separate requirements — the conductor must satisfy both
• Three-phase circuits have lower voltage drop than single-phase for the same wire size and current (by a factor of √3/2 ≈ 0.866)
• L in the formula is the one-way distance — the factor of 2 handles the return conductor