Voltage drop formula
A long conductor has resistance, and current through resistance drops voltage. The circular-mil form of the formula puts that in terms you can work with a calculator:
- Single-phase: VD = 2 × K × I × L ÷ CM. The 2 covers the trip out and the trip back.
- Three-phase: VD = √3 × K × I × L ÷ CM, with √3 = 1.732.
- Percent drop = VD ÷ source voltage × 100. For three-phase, use the line-to-line voltage.
- K is the conductor material constant in ohm-circular mils per foot, I is the load current in amps, L is the one-way length in feet, and CM is the conductor area in circular mils.
This tool starts K at 12.9 for copper and 21.2 for aluminum, which are typical figures. Type another K and the tool uses it, because exam problems sometimes hand you their own.
Where circular mils come from
A circular mil is the area of a circle one mil (0.001 in) across, so a round conductor's area in circular mils is its diameter in mils, squared. AWG sizes follow a fixed geometric rule for diameter, and ASTM B258 gives the standard nominal diameters and areas. This calculator computes each AWG area from that rule, and takes kcmil sizes at face value, since 500 kcmil is 500,000 circular mils. On the exam, use the circular mil area your book's Chapter 9 table gives you. Small differences in rounding will not change which answer choice is closest.
Worked example 1: a long single-phase branch circuit
120 V, 20 A, 150 ft one way, 10 AWG copper (10,383 circular mils), K = 12.9. VD = 2 × 12.9 × 20 × 150 ÷ 10,383 = 7.45 V. Percent drop = 7.45 ÷ 120 × 100 = 6.21%. To get under 3%, 8 AWG (16,510 circular mils) gives 3.91%, which still misses, and 6 AWG (26,251 circular mils) gives 2.46%, which clears it. A conductor that fixes the drop may need a larger equipment grounding conductor as well, so check that part of your book too.
Worked example 2: a three-phase feeder in aluminum
480 V three-phase, 80 A, 300 ft one way, 3/0 AWG aluminum (167,806 circular mils), K = 21.2. VD = 1.732 × 21.2 × 80 × 300 ÷ 167,806 = 5.25 V, which is 1.09% of 480 V. These are original arithmetic examples, not exam questions.
What the NEC says about the limit
The NEC does not set a mandatory voltage drop limit for ordinary branch circuits and feeders. The recommended percentages sit in informational notes under 210.19 for branch circuits and 215.2 for feeders. An informational note explains and recommends, and it is not a rule you must enforce. When a question asks what the NEC requires, that is a different answer from what it suggests. Read the notes in your own book to see the numbers. For the full method and the traps, see the NEC voltage drop guide.
Common voltage drop mistakes
Why is my single-phase answer half what the book says?
You probably used the one-way length without the 2, or the round-trip length with the 2. The 2 in the formula already counts the return path, so L is always one way.
Which voltage do I divide by?
The source voltage the problem gives. For a three-phase circuit that is the line-to-line voltage.
Does the smallest size here mean it is the right conductor?
No. It only meets your drop target. Ampacity, terminations and the rest of the design still have to work. For that, see the conductor ampacity guide.
Voltage drop shows up on the Calculations portion of the Texas journeyman exam. See the Calculations portion guide for how that portion is built.