Voltage drop is the loss of electrical potential along a conductor due to its inherent resistance. When current flows through a wire, some energy is converted to heat, resulting in a lower voltage at the load than at the source. To calculate voltage drop in wire for a standard single-phase AC or DC branch circuit, use the foundational formula:
VD = (2 × K × I × L) / A
While the National Electrical Code (NEC) does not strictly mandate voltage drop limits for most residential branch circuits, it strongly recommends a maximum of 3% drop on branch circuits and 5% total drop (feeder plus branch) for reasonable efficiency (NEC Article 210.19(A) Informational Note). Exceeding these limits leads to dimming lights, overheating motors, and nuisance tripping of sensitive electronics.
The Core Voltage Drop Formula and Symbol Definitions
The formula above is derived from Ohm’s Law (V = I × R), substituting the specific resistance formula for a wire based on its material, length, and cross-sectional area. The multiplier "2" accounts for the complete circuit path: the current must travel out to the load and return to the source.
| Symbol | Definition | Standard Unit (US Customary) |
|---|---|---|
| VD | Voltage Drop (the absolute loss in volts) | Volts (V) |
| 2 | Multiplier for the out-and-back return path (single-phase/DC) | Dimensionless |
| K | Specific Resistivity of the conductor material at a given temperature | Ohm-cmil / ft |
| I | Current (the continuous load drawn by the device) | Amperes (A) |
| L | One-way length of the circuit (source to load) | Feet (ft) |
| A | Cross-sectional area of the conductor | Circular Mils (cmil) |
Reference Data: Wire Areas and Resistivity Constants
To use the formula accurately, you need exact values for A (wire area) and K (resistivity). The NEC Chapter 9, Table 8 provides baseline DC resistance at 75°C (167°F), which is the standard operating temperature column for most modern THHN/XHHW-2 building wire. Using the 75°C column is critical because wire resistance increases as it heats up under load.
| AWG Size | Area (A) in cmil | Copper K (75°C) | Aluminum K (75°C) |
|---|---|---|---|
| 14 AWG | 4,110 | 12.9 | 21.2 |
| 12 AWG | 6,530 | 12.9 | 21.2 |
| 10 AWG | 10,380 | 12.9 | 21.2 |
| 8 AWG | 16,510 | 12.9 | 21.2 |
| 6 AWG | 26,240 | 12.9 | 21.2 |
| 4 AWG | 41,740 | 12.9 | 21.2 |
Bench Note: If you are sizing feeders using the 60°C column (common for older NM-B Romex or terminals rated only for 60°C), the K value for copper jumps to roughly 14.8. Always match your K value to the lowest temperature rating in your circuit chain.
Rearranged Forms for Sizing and Distance Limits
In the field, you rarely just solve for voltage drop. Usually, you know your maximum acceptable drop and need to find the right wire size, or you have a fixed wire size and need to know how far you can run it. Here are the algebraically rearranged forms of the core equation:
- Solve for Wire Size (A):
A = (2 × K × I × L) / VD
Use this to find the minimum circular mils required, then round UP to the next standard AWG size. - Solve for Maximum One-Way Length (L):
L = (VD × A) / (2 × K × I)
Use this when extending a circuit to a detached garage or well pump to ensure you don't exceed the 3% threshold. - Solve for Maximum Current (I):
I = (VD × A) / (2 × K × L)
Use this to determine if an existing buried cable can handle a new load addition without excessive drop.
Worked Examples: Calculating Drop and Sizing Wire
Let’s apply the math to two common jobsite scenarios. Pay close attention to the unit tracking—dropping a unit is the most common reason DIY calculations fail.
Example 1: Finding the Voltage Drop on an Existing Branch Circuit
Scenario: You are wiring a 120V dedicated outlet for a high-end espresso machine in a kitchen island. The circuit is 12 AWG Copper, protected at 20A, but the continuous load is 15A. The one-way distance from the panel to the island is 85 feet. What is the voltage drop, and does it pass the 3% NEC recommendation?
- Identify the variables:
- K = 12.9 (Copper at 75°C)
- I = 15 A
- L = 85 ft
- A = 6,530 cmil (from 12 AWG table above)
- Plug into the formula:
VD = (2 × 12.9 × 15 × 85) / 6,530 - Calculate the numerator (out-and-back resistance factor):
2 × 12.9 × 15 × 85 = 32,895 - Divide by the area:
32,895 / 6,530 = 5.03 Volts - Calculate the percentage:
(5.03V / 120V) × 100 = 4.19%
Verdict: At 4.19%, this exceeds the recommended 3% branch circuit limit. The espresso machine will see roughly 115V instead of 120V, which may cause its internal heating element to run longer or its pump motor to draw excess current. Fix: Upgrade to 10 AWG wire.
Example 2: Sizing Wire for a Long-Run 240V Load
Scenario: You are running a 240V circuit to a detached workshop to power a 30A air compressor. The one-way distance is 150 feet. You want to strictly limit the voltage drop to 3% to protect the compressor motor. What size copper wire do you need?
- Identify the variables:
- VD = 240V × 0.03 = 7.2 Volts (Maximum allowed drop)
- K = 12.9 (Copper)
- I = 30 A
- L = 150 ft
- Use the rearranged formula for Area (A):
A = (2 × K × I × L) / VD - Calculate the numerator:
2 × 12.9 × 30 × 150 = 116,100 - Divide by the max voltage drop:
116,100 / 7.2 = 16,125 cmil
Verdict: You need a wire with at least 16,125 circular mils of cross-sectional area. Looking at our reference table, 8 AWG is 16,510 cmil. While 8 AWG technically clears the 16,125 threshold by a razor-thin margin, standard electrical practice and manufacturer voltage drop calculators dictate stepping up to the next standard size for long runs to account for voltage sag at the utility transformer. Best practice: Pull 6 AWG THHN (26,240 cmil) in conduit.
Assumptions, Unit Traps, and Realistic Magnitudes
The formula VD = (2 × K × I × L) / A is highly reliable, but it operates under specific physical assumptions. Misunderstanding these will yield dangerously incorrect wire sizes.
When the Formula Applies (and When It Doesn't)
This exact formula applies to single-phase AC and DC circuits operating near unity power factor. It assumes a steady-state continuous load and a uniform ambient temperature. If you are calculating for a balanced three-phase circuit (like a large commercial HVAC unit or industrial mill), the out-and-back multiplier changes. Because the phases share the return current vectorally, you replace the "2" with the square root of 3 (approximately 1.732).
Unit Mistakes That Break the Math
The most catastrophic errors in voltage drop calculations come from unit mismatching:
- The "L" Trap: "L" is the one-way physical distance from panel to load. If you measure the total length of wire pulled from the spool (which includes both the hot and the neutral), you must drop the "2" from the numerator, or you will double-count the distance and oversize the wire by two full AWG steps.
- Metric vs. Imperial: This formula relies on Circular Mils (cmil) and feet. If your wire is specified in square millimeters (mm²), you cannot plug that number directly into "A". You must convert first: 1 mm² ≈ 1,973.5 cmil. Alternatively, use the metric voltage drop formula: VD = (2 × ρ × I × L) / A, where ρ is in Ω·mm²/m and A is in mm².
- Forgetting the Temperature Column: Using the 20°C K-value (10.8 for copper) for a wire that will operate at 75°C under load will understate your voltage drop by nearly 20%.
What a Realistic Answer Magnitude Looks Like
Developing a "feel" for the numbers prevents decimal errors. On a standard 120V residential branch circuit, a 3% drop is exactly 3.6 Volts. On a 240V feeder, a 3% drop is 7.2 Volts.
If you run your math and the result says your 120V circuit has a 45-volt drop, you have made a mistake—likely forgetting to divide by the circular mil area, or misplacing a decimal point in the wire area. Conversely, if your calculation yields a 0.05V drop on a 100-foot run, you likely multiplied by the area instead of dividing by it. Always sanity-check your final VD against the 3% to 5% thresholds before buying wire.
