The Single-Phase Voltage Drop Formula (and What Every Symbol Means)
When you run a branch circuit more than 50 feet from the panel, the wire itself becomes a resistor in series with your load. If you ignore this, your 120V receptacle might only deliver 108V under load, causing motors to overheat and heaters to underperform. The voltage drop formula single phase is your primary defense against undersized feeders and long-run branch circuits.
Here is the standard NEC-derived formula for single-phase AC (and DC) circuits:
VD = (2 × K × I × D) ÷ CM
Before you plug in numbers, you need to know exactly what each variable represents and where it comes from. The '2' in the numerator accounts for the round-trip path of single-phase current (out on the ungrounded conductor, back on the grounded neutral).
| Symbol | Definition | Standard Units | Notes & Bench Context |
|---|---|---|---|
| VD | Voltage Drop | Volts (V) | The absolute voltage lost in the wire, not a percentage. |
| 2 | Multiplier | Dimensionless | Accounts for the out-and-back path of single-phase current. (Use 1.732 for 3-phase). |
| K | Resistivity Constant | Ohms-cmil/ft | Use 12.9 for Copper and 21.2 for Aluminum at 75°C. |
| I | Current (Load) | Amperes (A) | The actual continuous current draw of the load, not the breaker size. |
| D | Distance | Feet (ft) | The one-way physical distance from the panel to the load. |
| CM | Circular Mils | cmil | The cross-sectional area of the wire. (e.g., 12 AWG = 6,530 CM). |
When This Formula Applies (and Its Assumptions)
This formula is highly accurate for DC circuits and single-phase AC circuits where the power factor is close to 1.0 (like resistive heating elements or incandescent lighting). It assumes steady-state current and ignores AC reactance ($X_L$). For small wire sizes (14 AWG to 1/0 AWG) and standard 60Hz residential power, the reactance is negligible (under 2%), making this formula the industry standard for everyday sizing.
What does a realistic answer magnitude look like? NEC Informational Note 210.19(A) recommends a maximum of 3% voltage drop on branch circuits and 5% total for feeder plus branch. On a 120V circuit, 3% is 3.6V. On a 240V circuit, 3% is 7.2V. If your math spits out a 15V drop on a 120V line, your wire is drastically undersized.
Rearranging the Equation: Solving for Wire Size, Distance, or Load
On the jobsite, you rarely solve for VD. Usually, you know your maximum allowable drop (e.g., 3.6V) and need to find the right wire size (CM) or the maximum distance you can run. Here are the rearranged forms:
- Solving for Wire Size (CM):
CM = (2 × K × I × D) ÷ VD
Use this to find the minimum circular mils, then look up the next largest AWG in NEC Chapter 9, Table 8. - Solving for Maximum Distance (D):
D = (VD × CM) ÷ (2 × K × I)
Use this when a client asks, 'How far can I run my 10 AWG wire for a 20A compressor?' - Solving for Maximum Current (I):
I = (VD × CM) ÷ (2 × K × D)
Use this to determine if an existing buried cable can handle a new load addition.
Worked Examples: Tracking Units from Paper to Panel
Abstract formulas are useless if you drop a unit or misread a table. Let us walk through two solved problems, tracking every unit to ensure the math holds up.
Problem 1: Finding the Voltage Drop on an Existing Run
Setup: You are powering a 120V, 15A resistive wall heater. The one-way distance from the panel is 100 feet. The installed wire is 12 AWG copper. What is the voltage drop?
Knowns:
- K = 12.9 (Copper at 75°C)
- I = 15 Amps
- D = 100 Feet
- CM = 6,530 (from NEC Chapter 9, Table 8 for 12 AWG)
Step-by-Step Math:
- Numerator: 2 × 12.9 × 15 × 100 = 38,700
- Denominator: 6,530
- VD = 38,700 ÷ 6,530 = 5.92 Volts
Outcome: A 5.92V drop on a 120V circuit is a 4.9% drop. This exceeds the NEC 3% recommended limit. The heater will only see 114.08V. To fix this, you must upsize to 10 AWG (10,380 CM), which would drop the VD to a compliant 3.7V.
Problem 2: Sizing Wire for a 240V Subpanel Feeder
Setup: You need to feed a 240V, 30A subpanel located 200 feet away. You want to limit the voltage drop to exactly 3%. What size copper wire do you need?
Knowns:
- VD = 7.2V (which is 3% of 240V)
- K = 12.9 (Copper)
- I = 30 Amps
- D = 200 Feet
Step-by-Step Math:
- Numerator: 2 × 12.9 × 30 × 200 = 154,800
- Denominator: 7.2
- CM = 154,800 ÷ 7.2 = 21,500 Circular Mils
Outcome: You need a wire with at least 21,500 CM. Checking Southwire's wire tables or NEC Chapter 9 Table 8, 8 AWG is 16,510 CM (too small). 6 AWG is 26,240 CM. You must pull 6 AWG copper THHN for this feeder.
Real-World Autopsy: The 150-Foot Space Heater Disaster
Let us look at a real-world failure where ignoring the voltage drop formula single phase resulted in a callback and a melted neutral pigtail.
The Setup: A homeowner plugged a 1500W, 120V portable space heater into a receptacle in a detached workshop. The workshop was wired with 14 AWG copper, and the one-way distance from the main house panel was 125 feet. The heater ran constantly during a freezing snap.
The Numbers:
- Load Current (I): 1500W ÷ 120V = 12.5 Amps
- Distance (D): 125 feet
- Wire (CM): 14 AWG = 4,110 CM
The Calculation:
VD = (2 × 12.9 × 12.5 × 125) ÷ 4,110
VD = 40,312.5 ÷ 4,110 = 9.8 Volts
The Outcome: The voltage at the receptacle under load was only 110.2V (120V - 9.8V). Because a resistive heater's power output drops with the square of the voltage ($P = V^2 ÷ R$), the 1500W heater was only outputting about 1,269 Watts of heat. The homeowner complained the workshop was freezing.
What Went Wrong: To compensate for the lack of heat, the homeowner plugged a second 1500W heater into the same 14 AWG circuit using a power strip. The total current spiked to 25A. The 15A breaker eventually tripped, but not before the 14 AWG wire heated up significantly in the walls, and the neutral pigtail in the receptacle box softened and deformed due to the sustained thermal stress. The fix required pulling a dedicated 10 AWG 20A circuit for the workshop heaters.
Three Unit Mistakes That Will Break Your Math
If your calculator is giving you answers that suggest you need 500 MCM wire for a 15A lighting circuit, you have fallen victim to one of these three unit traps.
1. Using Metric Cross-Section Instead of Circular Mils
The formula demands Circular Mils (CM). If you are working with metric wire (like 2.5 mm² automotive or solar cable), you cannot plug '2.5' into the CM slot. You must convert. The conversion factor is roughly 1,973.5 CM per square millimeter. A 2.5 mm² wire is actually 4,933 CM. Plugging in 2.5 will result in a calculated voltage drop that is nearly 2,000 times higher than reality.
2. The 'Double-Counting the 2' Trap
The '2' in the numerator exists solely to account for the return path (the neutral wire). 'D' is strictly the one-way physical distance from source to load. If your load is 100 feet away, D = 100. A common mistake is measuring the total wire pulled from the spool (200 feet for hot + neutral) and plugging 200 into 'D' while leaving the '2' in the formula. This artificially quadruples your calculated voltage drop. According to Fluke's electrical testing guidelines, always use the one-way physical run distance for D.
3. Using the 20°C DC Resistance Constant
If you look up the raw resistivity of copper in a physics textbook, you will find 10.4 ohms-cmil/ft. This is the DC resistance at 20°C (68°F). However, AC current experiences the skin effect, and wires in a bundled conduit easily reach 50°C to 75°C under load. Copper's resistance increases by about 0.4% per degree Celsius. Using 12.9 (the 75°C effective AC constant) builds a necessary safety margin into your math. Using 10.4 will yield an optimistically low voltage drop that fails in the heat of summer.
Bench Tip: Always verify your math with a physical test when possible. After pulling a long run, measure the voltage at the panel under load, then measure it at the receptacle under the exact same load. The difference is your real-world VD. If it deviates wildly from your formula result, check your connections—a loose lug at the breaker adds resistance that the wire formula cannot predict.
