Try a full lesson: Voltage drop: sizing feeders and branch circuits for it

What this lesson covers

Voltage drop is where theory meets design. At the journeyman level you calculate the drop on a conductor someone else chose. At the master level you choose the conductor: you split a drop budget between a feeder and its branch circuits, then size each run to meet it. Theory and calculations carry about 8 of the 100 questions on our practice exam, and voltage drop also turns up inside feeder and grounding questions.

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

Step 1: the formulas

Use the circular mil method:

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

To size a conductor, solve for CM:

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

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.
• CM is the conductor's area in circular mils, from the conductor properties table in Chapter 9.
• VD is the allowed drop in volts.

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.

Step 2: check with the resistance method

Chapter 9 also lists each conductor's resistance per 1,000 ft. Drop = I × R, where R = resistance per foot × the circuit's conductor length (2 × L for single-phase). For three-phase, multiply the one-conductor drop by 1.732. The two methods agree within a percent or two, which is why one makes a good check on the other.

Step 3: the code status of 3% and 5%

The NEC suggests, in informational notes in Article 210 (branch circuits) and Article 215 (feeders), that a branch circuit or feeder drop of no more than 3%, and a combined drop of no more than 5%, gives reasonable efficiency. Informational notes are not requirements. Questions often state these figures as the design target; use whatever the question gives.

Two cautions:

• A few articles do set mandatory limits for specific equipment, such as some fire pump and sensitive electronic equipment installations. Read those articles when the question is about them.
• Voltage drop never lets a conductor go smaller than its ampacity rules allow. Size for ampacity first, then upsize for drop if needed. When you upsize ungrounded conductors for drop, the EGC grows in proportion (module 8).

Worked example 1: size a three-phase feeder for drop

A 480 V, three-phase feeder carries 150 A to a motor control center 320 ft away. The design limit is 2% for this feeder. Take these Chapter 9 values: 1/0 105,600 cmil, 2/0 133,100 cmil, 3/0 167,800 cmil.

Check with 2/0: VD = 1,072,454 ÷ 133,100 = 8.06 V, which is 1.68%, under 2%. With 1/0: 1,072,454 ÷ 105,600 = 10.16 V, which is 2.12%, over.

Worked example 2: split the budget

A 208Y/120 V panel is fed by a three-phase feeder with 2% drop. A 120 V, single-phase branch circuit from that panel carries 16 A to a load 140 ft away. What is the smallest copper conductor that keeps the total drop at or under 5%? Take these values: 10 AWG 10,380 cmil, 8 AWG 16,510 cmil.

• Total budget at the load: 5% of 120 V is 6.0 V. The feeder uses 2% of the panel's line-to-neutral voltage, 2.4 V.
• Left for the branch circuit: 6.0 − 2.4 = 3.6 V.
• CM = 2 × 12.9 × 16 × 140 ÷ 3.6 = 57,792 ÷ 3.6 = 16,053 cmil.
• 10 AWG is short. 8 AWG (16,510 cmil) meets it.

Check with 8 AWG: VD = 57,792 ÷ 16,510 = 3.50 V. Total 2.4 + 3.50 = 5.90 V, which is 4.9%, under 5%. With 10 AWG: 57,792 ÷ 10,380 = 5.57 V, total 7.97 V, 6.6%, over.

Worked example 3: the drop on an existing run

A 240 V, single-phase, 50 A load is 180 ft from its panel on 6 AWG copper. Take 6 AWG as 26,240 cmil and 0.491 ohms per 1,000 ft.

The two agree. The run exceeds 3%, so a designer wanting 3% would move to 4 AWG: 232,200 ÷ 41,740 = 5.56 V, or 2.3%.

Worked example 4: maximum length

How far can a 480 V three-phase, 60 A load be from its panel on 4 AWG copper (41,740 cmil) at 3% drop?

Check: at 448 ft, VD = 1.732 × 12.9 × 60 × 448 ÷ 41,740 = 600,571 ÷ 41,740 = 14.39 V. Same answer.

The traps

• Two-way length in the formula. L is one way; the 2 or 1.732 already accounts for the return.
• Using 2 on a three-phase circuit. The three-phase factor is 1.732.
• Rounding the conductor down. Pick the first size at or above the computed CM.
• Wrong base voltage. A 120 V branch circuit's 3% is 3.6 V, not 6.24 V.
• Forgetting the feeder's share. The 5% total covers feeder and branch circuit together.
• Treating 3% and 5% as requirements. They are recommendations unless a question or a specific article makes them binding.

Try it

Ten free practice questions on voltage drop follow this lesson. The full course has a lesson for each of the twelve modules, check questions after each one, and timed 100-question practice exams written to the areas state master electrician bulletins share, on the 2023 NEC.

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