# Voltage drop calculations (single-phase and three-phase)

> Voltage drop is the voltage lost in the conductors. For single-phase, VD = 2 × K × I × L ÷ CM; for three-phase, use 1.732 instead of 2. K is 12.9 for copper and 21.2 for aluminum, L is the one-way length in feet and CM comes from Chapter 9, Table 8. Percent drop is VD ÷ source voltage × 100.

Exams: Journeyman Electrician, Master Electrician. Conductors and overcurrent.

## Formula

    Single-phase: VD (V) = 2 × K × I × L ÷ CM
    Three-phase: VD (V) = 1.732 × K × I × L ÷ CM
    Minimum CM = 2 (or 1.732) × K × I × L ÷ Allowed VD (V)

## Variables

- **VD**: Voltage drop in the circuit conductors (V)
- **K**: Resistance of a 1 cmil, 1 ft conductor: 12.9 copper, 21.2 aluminum (ohm-cmil/ft)
- **I**: Load current (A)
- **L**: One-way length from source to load (ft)
- **CM**: Conductor area, Chapter 9, Table 8 (cmil)
- **R**: DC resistance per 1,000 ft, Chapter 9, Table 8 (ohm)

## Open your 2026 NEC to

- Chapter 9, Table 8: Circular mil area and DC resistance per 1,000 ft of each conductor size
- 210.19 Informational Note (2023: 210.19(A) Informational Note No. 3): Suggests about 3% drop for a branch circuit and 5% total. Not a requirement
- 215.2 Informational Note (2023: 215.2(A)(1) Informational Note No. 2): Suggests about 3% drop for a feeder and 5% total. Not a requirement
- 647.4(D): One of the few places where a voltage drop limit is mandatory (sensitive electronic equipment)

Every conductor has resistance, so current flowing through it loses some voltage before it reaches the load. Long runs and heavy loads lose the most. Motors run hot, lights dim and electronics act up when the drop is too large.

The K formula comes from Ohm's law. K is the resistance of a conductor one circular mil in area and one foot long, so K × L ÷ CM is the resistance of one wire. Single-phase circuits have two current-carrying wires, so you multiply by 2. Balanced three-phase circuits use 1.732 (the square root of 3) instead.

You can also use the resistance in Chapter 9, Table 8 directly: VD = 2 × I × R × L ÷ 1,000, with R in ohms per 1,000 ft. Both methods give nearly the same answer; use the one the question names.

The 3% and 5% values are informational notes under 210.19 and 215.2. They are advice, not enforceable rules, except where a section such as 647.4(D) makes a limit mandatory. Exam questions state the percent to use, often as a job specification.

## Worked example

A 240 V single-phase load draws 40 A. It is 210 ft from the panel and is fed with 2 AWG copper conductors. Using K = 12.9, what is the voltage drop?

1. Single-phase formula: VD = 2 × K × I × L ÷ CM. The 2 covers the conductor out and the conductor back.
2. K for copper: 12.9.
3. Look up Chapter 9, Table 8 for 2 AWG: 66,360 cmil.
4. VD = 2 × 12.9 × 40 × 210 ÷ 66,360 = 3.27 V.

**Answer: 3.27 V**

## How to solve it on the exam

1. Identify single-phase (factor 2) or three-phase (factor 1.732), copper (K = 12.9) or aluminum (K = 21.2), and the one-way length.
2. Look up the circular mil area of the conductor in Chapter 9, Table 8.
3. Find VD with the formula. For percent drop, divide VD by the source voltage and multiply by 100.
4. To find a size: allowed VD = percent × voltage. Then CM = factor × K × I × L ÷ allowed VD. Pick the first size in Table 8 with at least that many circular mils.
5. Check that the size you picked also has enough ampacity for the load.

## Common mistakes

- Using the round-trip length and then also multiplying by 2.
- Using 2 for a three-phase circuit, or 1.732 for single-phase.
- Dividing by 120 V or 277 V when the drop is asked as a percent of 240 V, 208 V or 480 V.
- Using the copper K for an aluminum conductor.
- Treating 3% as an NEC requirement for every circuit.

## Exam tips

- Write the formula with blanks first, then fill in each number. It prevents the doubled-length slip.
- Tab Chapter 9, Table 8. It has both the circular mils and the resistance you need.

## Questions people ask

### What is the voltage drop formula?

For single-phase, VD = 2 × K × I × L ÷ CM. For three-phase, VD = 1.732 × K × I × L ÷ CM, with K = 12.9 for copper or 21.2 for aluminum and L the one-way length in feet.

### Does the NEC require a maximum 3% voltage drop?

No, not for general circuits. The 3% branch circuit and 5% total values are informational notes, which are advice. A few sections, such as 647.4(D) for sensitive electronic equipment, do set a mandatory limit.

### What K value should I use on the Texas exam?

Use the K the question gives. If none is given, 12.9 for copper and 21.2 for aluminum are the usual values.

### How do I find the wire size for a 3% voltage drop?

Find the allowed drop (0.03 × voltage), then solve CM = 2 × K × I × L ÷ allowed drop (1.732 for three-phase). Pick the smallest size in Chapter 9, Table 8 with at least that many circular mils, then check its ampacity.

## Sources

- [TDLR: Electrician examination information](https://www.tdlr.texas.gov/electricians/elecexam.htm)
- [IAEI Magazine: Voltage drop calculations](https://iaeimagazine.org/electrical-fundamentals/voltage-drop-calculations/)

Interactive practice: https://texaselectricianexam.com/calculations/voltage-drop/