If you need to know how to calculate voltage drop for a single-phase AC or DC circuit, the direct answer is the standard approximate formula: VD = (2 × K × I × D) / CM. This equation tells you exactly how many volts will be lost as heat across a wire run before the electricity reaches your load. Sizing wire purely by ampacity (the breaker size) is a rookie mistake; if the run is long enough, the voltage at the receptacle or motor will sag, causing poor performance, overheating, and tripped breakers.
Below, we will break down every symbol in this formula, rearrange it to solve for the variables you actually need on the jobsite, and walk through two fully tracked worked examples so you can see the math in action.
The Core Voltage Drop Formula & Symbol Definitions
The standard formula used by electricians and engineers for single-phase voltage drop is derived directly from Ohm’s Law (V = I × R), adapted to use the physical dimensions of standard American Wire Gauge (AWG) conductors.
VD = (2 × K × I × D) / CM
| Symbol | Definition | Units & Standard Values |
|---|---|---|
| VD | Voltage Drop (the actual volts lost across the wire pair) | Volts (V) |
| 2 | Multiplier for the return path (accounts for both the hot and neutral/ground wire in single-phase) | Dimensionless constant |
| K | Specific resistance of the conductor material | Ω·CM/ft. Copper: 12.9 (at 75°C), Aluminum: 21.2 (at 75°C) |
| I | Current (the actual load draw, not the breaker size) | Amperes (A) |
| D | One-way physical distance from the source to the load | Feet (ft) |
| CM | Circular Mils (the cross-sectional area of the wire) | CM (Found in NEC Chapter 9, Table 8) |
When This Formula Applies (and Its Assumptions)
This formula is highly accurate for single-phase AC circuits, DC circuits, and 3-phase circuits (with a modified multiplier) under steady-state loads. It assumes a power factor near 1.0 (unity) and ignores AC reactance (the inductive/capacitive opposition of the wire). For standard residential and light commercial wiring (up to 1/0 AWG in standard raceways at 60Hz), reactance is negligible, making this formula the industry standard for practical field calculations.
Rearranged Forms: Solving for Wire Size, Distance, and Current
On the bench or in the field, you rarely just want to find the voltage drop. Usually, you know your maximum allowable drop (e.g., 3% of 240V = 7.2V) and need to find the right wire size. Here are the algebraically rearranged forms:
- Solving for Wire Size (CM):
CM = (2 × K × I × D) / VD - Solving for Maximum Distance (D):
D = (VD × CM) / (2 × K × I) - Solving for Maximum Current (I):
I = (VD × CM) / (2 × K × D)
Note: For 3-phase systems, replace the '2' in all these formulas with the square root of 3 (1.732).
Worked Example 1: Sizing Wire for a 240V Subpanel Feeder
Scenario: You are running a 240V single-phase feeder to a detached garage subpanel. The continuous load is 50A. The one-way trench distance is 200 feet. You are using copper THHN wire and want to keep the voltage drop under the NEC-recommended 3% maximum for feeders.
Step 1: Identify the known variables.
- I = 50 A
- D = 200 ft
- K = 12.9 (Copper at 75°C operating temperature)
- VD = 7.2 V (Calculated as 240V × 0.03)
Step 2: Select the rearranged formula to solve for CM.
CM = (2 × K × I × D) / VD
Step 3: Plug in the numbers and track the units.
CM = (2 × 12.9 Ω·CM/ft × 50 A × 200 ft) / 7.2 V
CM = 258,000 / 7.2
CM = 35,833.33
Step 4: Convert CM to AWG.
Looking at NEC Chapter 9, Table 8, 6 AWG copper is only 26,240 CM (too small). The next size up is 4 AWG copper, which is 41,740 CM. Therefore, to maintain a 3% or lower voltage drop over 200 feet at 50A, you must pull 4 AWG copper conductors.
Worked Example 2: Calculating Drop on a 120V Branch Circuit
Scenario: You have an existing 120V branch circuit wired with 12 AWG copper. It powers a 15A continuous load (like a space heater or server rack). The outlet is 80 feet from the panel. What is the actual voltage drop, and is it acceptable?
Step 1: Identify the known variables.
- I = 15 A
- D = 80 ft
- K = 12.9 (Copper)
- CM = 6,530 (12 AWG from NEC Table 8)
Step 2: Use the base formula to solve for VD.
VD = (2 × K × I × D) / CM
Step 3: Calculate and track units.
VD = (2 × 12.9 × 15 × 80) / 6,530
VD = 30,960 / 6,530
VD = 4.74 V
Step 4: Check the realistic magnitude.
To find the percentage: (4.74V / 120V) × 100 = 3.95%.
The NEC (Informational Note to 210.19(A)(1)) recommends a maximum of 3% for branch circuits. At nearly 4%, this 12 AWG run is technically undersized for a full 15A continuous load at this distance. If this were a critical motor or sensitive electronics, you would need to upgrade to 10 AWG (10,380 CM), which would drop the loss to a highly efficient 2.48%.
Common Unit Mistakes That Break the Math
When electricians get the wrong answer from this formula, it is almost always due to one of three unit or assumption errors:
- Doubling the Distance Manually: The '2' in the numerator already accounts for the out-and-back loop (hot and neutral). If you measure a 100-foot trench and plug in '200' for D because 'the wire goes there and back', you will accidentally double your voltage drop calculation. D is always the one-way physical distance.
- Using the Wrong 'K' Constant: Many old textbooks list K = 10.8 for copper. That is the resistance of copper at a room temperature of 20°C (68°F). Under load, wire heats up. The NEC and modern engineering practice use K = 12.9 to reflect copper at a 75°C operating temperature. Using 10.8 will result in undersized wire that sags more than your math predicted.
- Confusing AWG with mm²: The CM (Circular Mils) unit is strictly an Imperial/AWG measurement. If you are working with metric wire (e.g., 4mm²), you cannot plug '4' into the CM slot. You must either convert mm² to CM (1 mm² ≈ 1,973.5 CM) or switch to the metric voltage drop formula:
VD = (2 × ρ × I × D) / A, where ρ is resistivity in Ω·m and A is area in mm².
Frequently Asked Questions
How do I calculate voltage drop for a 3-phase motor?
For balanced 3-phase circuits, the current flows through three conductors, and the vector math changes the multiplier. Replace the '2' in the standard formula with the square root of 3 (approximately 1.732). The formula becomes: VD = (1.732 × K × I × D) / CM. Because 1.732 is smaller than 2, 3-phase systems inherently suffer less voltage drop over the same distance and wire size compared to single-phase systems, which is why industrial facilities use them for heavy machinery.
What is a realistic voltage drop magnitude for home wiring?
In a properly designed residential system, your voltage drop should rarely exceed 1% to 2% on standard branch circuits. The National Electrical Code (NEC) recommends a maximum of 3% on the farthest outlet of a branch circuit, and a maximum of 5% total combined drop from the utility transformer to the appliance (feeder + branch). If your calculations show a drop of 8% or 10%, you will notice incandescent lights dimming when the HVAC kicks on, and induction motors (like in well pumps or table saws) will draw excess amperage to compensate for the low voltage, leading to premature thermal failure.
How do I calculate voltage drop if I only know the wire resistance per 1000 feet?
If you have a spec sheet that lists the wire's resistance in Ohms per 1,000 feet (Ω/kft) instead of Circular Mils, you can bypass the 'K' and 'CM' variables entirely. Use this alternative Ohm's Law derivation:
VD = 2 × (R_per_1000 / 1000) × D × I
For example, if 10 AWG copper has a resistance of 1.24 Ω/kft, and you are running 12A over 150 feet:
VD = 2 × (1.24 / 1000) × 150 × 12 = 2 × 0.00124 × 1800 = 4.46V. This method is highly accurate as it uses the manufacturer's exact tested resistance rather than the theoretical 'K' constant.
