Try a full lesson: Voltage drop: the calculation and the code status

What this lesson covers

Voltage drop is one of the most common calculations on state journeyman exams, and one of the easiest to get wrong by a single factor of 2. Theory and calculations are the largest area on our 100-question practice exam, about 14 questions. This lesson gives you one formula in two forms, a habit for checking it, and the code status of the 3% and 5% figures.

We cite the 2023 NEC by article in our own words. Keep your own copy open beside this page.

Step 1: pick the formula

Use the circular mil method:

• Single-phase: VD = 2 × K × I × L ÷ CM
• Three-phase: VD = 1.732 × K × I × L ÷ CM

Where:

• K is about 12.9 for copper and 21.2 for aluminum (ohm-circular mils per foot, a trade approximation).
• I is the load current in amperes.
• L is the one-way length in feet, from source to load.
• CM is the conductor's area in circular mils, from the conductor properties table in Chapter 9.

The 2 in the single-phase formula is there because current goes out on one conductor and back on the other. The 1.732 in the three-phase formula replaces it, because the return current is shared among the phases.

Step 2: check it with the resistance method

Chapter 9 also lists each conductor's resistance in ohms per 1,000 ft. Then:

• R = ohms per 1,000 ft × total conductor length ÷ 1,000 (for single-phase, the total length is out and back, 2 × L)
• VD = I × R

The two methods agree within a few tenths of a volt. Working both is the habit that catches a missing 2.

Step 3: compare it with the target

Divide the drop by the source voltage for a percentage. The NEC's informational notes in Articles 210 and 215 suggest no more than 3% on a branch circuit or feeder and 5% total from the service to the farthest outlet. These are recommendations, not general requirements. A few articles do set mandatory limits for specific equipment; if a question names that equipment, read its article.

Worked examples

Example 1: single-phase drop. A 120 V circuit feeds a 16 A load 75 ft away on 12 AWG copper. Take 12 AWG as 6,530 cmil and 1.98 Ω per 1,000 ft.

Example 2: sizing up to meet 3%. Same circuit. 3% of 120 V is 3.6 V. Solve the formula for CM: CM = 2 × K × I × L ÷ VD = 30,960 ÷ 3.6 = 8,600 cmil. Take 10 AWG as 10,380 cmil: it is the smallest that meets 8,600.

• Check: 30,960 ÷ 10,380 = 2.98 V, or 2.49%. Under 3%.

Example 3: three-phase. A 480 V, three-phase load draws 65 A at 250 ft on 6 AWG copper. Take 6 AWG as 26,240 cmil and 0.491 Ω per 1,000 ft.

• CM method: 1.732 × 12.9 × 65 × 250 ÷ 26,240 = 363,071 ÷ 26,240 = 13.84 V, or 13.84 ÷ 480 = 2.88%.
• Resistance check: one conductor is 250 × 0.491 ÷ 1,000 = 0.1228 Ω, and 65 × 0.1228 = 7.98 V per conductor. Times 1.732 for three-phase line to line: 13.82 V. Agrees.

Example 4: the longest run that meets 3%. A 240 V, single-phase, 40 A load on 8 AWG copper (16,510 cmil). 3% is 7.2 V. Solve for L: L = VD × CM ÷ (2 × K × I) = 7.2 × 16,510 ÷ (2 × 12.9 × 40) = 118,872 ÷ 1,032 = 115 ft.

• Check: at 115 ft, 2 × 12.9 × 40 × 115 ÷ 16,510 = 7.19 V. Just under 7.2.

Example 5: aluminum. Repeat Example 1 with aluminum of the same area: use K = 21.2. 2 × 21.2 × 16 × 75 ÷ 6,530 = 50,880 ÷ 6,530 = 7.79 V. Check by ratio: 4.74 × (21.2 ÷ 12.9) = 7.79 V. Aluminum drops about 64% more than copper of the same size.

What a drop does to the load

A resistive load's power changes with the square of its voltage. A 240 V, 4,800 W heater has R = 240² ÷ 4,800 = 12 Ω. At 228 V (a 5% drop), P = 228² ÷ 12 = 4,332 W. Check by ratio: 0.95² × 4,800 = 4,332 W. Motors respond differently: lower voltage makes them draw more current and run hotter.

Common traps

• Doubling the length and keeping the 2. L is one way. The formula's 2 covers the return.
• Using 2 on three-phase. Use 1.732. Using 2 overstates the drop by about 15%.
• Using the wrong voltage for the percentage. Divide by the circuit's source voltage: 120 V for a 120 V circuit, 480 V for a 480 V three-phase circuit.
• Calling 3% a requirement. It is an informational note unless a specific article requires a limit.
• Rounding the conductor down. If the CM you need falls between sizes, take the larger.
• Forgetting the EGC. If you upsize the phase conductors for voltage drop, the equipment grounding conductor grows in proportion (Article 250; see Module 6).

Check yourself

Try the ten questions in the free practice set for this lesson. Work each one twice, once each way.

Where to look in the NEC

• Chapter 9: conductor properties (circular mils and resistance per 1,000 ft)
• Articles 210 and 215: voltage drop informational notes
• Article 250: proportional increase of the equipment grounding conductor