If you are running a new circuit to a detached garage, wiring a long LED strip, or troubleshooting a dimming lights issue, you need to know exactly how much voltage is lost in the wire. The direct answer for single-phase AC and DC circuits is the basic voltage drop formula: VD = (2 × K × I × L) / CM. This equation allows you to calculate the exact voltage lost across a conductor based on its material, the current flowing through it, and the one-way distance of the run.
Below, we break down every variable, rearrange the formula for practical jobsite use, and walk through two fully worked examples with strict unit tracking so you never undersize a feeder or branch circuit again.
The Core Equation: Symbols, Units, and Assumptions
The basic voltage drop formula is derived directly from Ohm’s Law (V = I × R), substituting the physical resistance properties of a wire. For single-phase AC and DC systems, the formula is expressed as:
VD = (2 × K × I × L) / CM
This formula applies under specific assumptions: it assumes a unity power factor (or close to it) and ignores AC reactance. For standard residential and light commercial wiring (up to 60Hz and conductors smaller than 1/0 AWG), the resistance of the wire dominates the impedance, making this formula highly accurate. For large feeders or long 3-phase runs, you must use the approximate line-to-line voltage drop formula that incorporates reactance (X) and power factor.
| Symbol | Definition | Standard Unit / Value |
|---|---|---|
| VD | Voltage Drop | Volts (V) |
| 2 | Multiplier for the out-and-back current path (Line + Neutral) | Dimensionless constant (Use 1 for DC one-way or 3-phase) |
| K | Direct Current Resistance Constant | 12.9 for Copper, 21.2 for Aluminum (at 75°C) |
| I | Current (Load) | Amperes (A) |
| L | One-way length of the circuit | Feet (ft) |
| CM | Cross-sectional area of the conductor | Circular Mils (CM) - Found in NEC Chapter 9, Table 8 |
The K constant represents the resistance of a circular mil-foot of the conductor material. While 12.9 is standard for copper at 75°C, if you are calculating for a cold environment (20°C), the K value for copper drops to approximately 10.8. Always match your K value to the expected operating temperature of the wire.
Rearranged Forms: Solving for Wire Size, Current, or Distance
On the workbench or jobsite, you rarely need to find the voltage drop of a known circuit; usually, you are trying to size a wire for a specific load or determine how far you can run a specific gauge. By rearranging the basic voltage drop formula algebraically, we get three highly practical variations:
- Solving for Wire Size (CM): Use this when you know your load and distance, and need to find the minimum Circular Mils to stay within a 3% drop limit.
CM = (2 × K × I × L) / VD - Solving for Maximum Current (I): Use this when you have an existing wire run and want to know the maximum safe load before exceeding your voltage drop threshold.
I = (VD × CM) / (2 × K × L) - Solving for Maximum Distance (L): Use this when sizing low-voltage runs (like 12V LED strips or 24V irrigation valves) to find the absolute maximum run length.
L = (VD × CM) / (2 × K × I)
Once you calculate the required CM, you cross-reference the result with standard AWG tables. For instance, if your calculation demands 10,000 CM, you must step up to the next standard size, which is 8 AWG (16,510 CM), because 10 AWG only provides 10,380 CM.
Worked Example 1: Sizing a Branch Circuit for a Garage Subpanel
Scenario: You are wiring a dedicated 120V branch circuit from a main panel to a detached garage subpanel. The continuous load is 15A, and the one-way distance is 150 feet. You are using copper wire and want to limit the voltage drop to the NEC-recommended 3% maximum for branch circuits.
Step 1: Identify the known variables and target VD.
- Voltage = 120V
- Target VD = 3% of 120V = 3.6V
- I (Current) = 15A
- L (One-way distance) = 150 ft
- K (Copper at 75°C) = 12.9
Step 2: Select the rearranged formula for wire size.
CM = (2 × K × I × L) / VD
Step 3: Plug in the values and track units.
CM = (2 × 12.9 Ω·CM/ft × 15 A × 150 ft) / 3.6 V
CM = 58,050 / 3.6
CM = 16,125
Step 4: Match to standard AWG.
Looking at the wire gauge chart, 10 AWG is 10,380 CM (too small). 8 AWG is 16,510 CM. Therefore, you must pull 8 AWG copper wire to maintain a 3% or less voltage drop at 150 feet for a 15A load. This highlights how quickly distance forces you to upsize wire beyond standard ampacity requirements (which would normally allow 14 AWG for 15A).
Worked Example 2: Evaluating an Existing 120V Receptacle Circuit
Scenario: You are troubleshooting a workshop receptacle that causes lights to dim when a table saw starts. The circuit is wired with existing 12 AWG solid copper wire. The one-way run from the panel is 80 feet, and the saw draws a steady 12A under load. What is the actual voltage drop?
Step 1: Identify the known variables.
- Wire Size = 12 AWG. From standard tables, 12 AWG = 6,530 CM.
- I (Current) = 12A
- L (One-way distance) = 80 ft
- K (Copper) = 12.9
Step 2: Select the base voltage drop formula.
VD = (2 × K × I × L) / CM
Step 3: Calculate the drop.
VD = (2 × 12.9 × 12 × 80) / 6,530
VD = 24,768 / 6,530
VD = 3.79V
Step 4: Evaluate the result.
A 3.79V drop on a 120V system represents a 3.16% drop (3.79 / 120 = 0.0316). While this is slightly over the strict 3% NEC recommendation for a single branch circuit, it is well within the 5% total maximum (feeder + branch combined). The dimming lights are likely caused by the motor's inrush current (which can be 5x to 7x the running 12A), not the steady-state voltage drop. For inrush issues, you would need to calculate using the locked-rotor amperage (LRA) rather than the full-load amperage (FLA).
Common Unit Traps and Realistic Magnitudes
When using the basic voltage drop formula, a single unit error will result in a wire size that is either dangerously undersized or comically oversized. Watch out for these specific traps:
- The "2" Multiplier Trap: The "2" in the formula accounts for the out-and-back path (Line and Neutral). Therefore, L must be the one-way distance. If you measure 100 feet of cable pulled from the spool, the one-way distance is 50 feet. If you use 100 for L and still multiply by 2, you will double-count the distance and oversize your wire by two AWG steps.
- Circular Mils vs. Square Mils: The formula requires Circular Mils (CM), not square mils or square millimeters. CM is calculated by squaring the diameter of the wire in mils (1 mil = 0.001 inches). If you accidentally use the cross-sectional area in square mils (Area = 0.7854 × d2), your calculated voltage drop will be off by roughly 27%.
- Metric Conversions: The K constant (12.9) is specifically derived for feet and circular mils. If you are working in meters and square millimeters (the IEC standard), you must use the metric formula: VD = (2 × ρ × I × L) / A, where ρ is resistivity in Ω·m and A is area in mm2.
What does a realistic answer magnitude look like?
For a standard 120V residential branch circuit, a realistic VD answer should be between 0.5V and 3.6V. If your calculation spits out 45V or 120V, you have made a math error—most likely forgetting to divide by the CM value, or inputting the wire diameter in inches instead of circular mils. Conversely, if you calculate a drop of 0.001V for a 100-foot run, you likely multiplied by CM instead of dividing by it.
Frequently Asked Questions
How does the basic voltage drop formula change for 3-phase systems?
For balanced 3-phase systems, the out-and-back path geometry changes, altering the multiplier. The basic formula becomes VD = (√3 × K × I × L) / CM. The multiplier changes from 2 to the square root of 3 (approximately 1.732). This reflects the phase-to-phase voltage relationship in a 3-phase wye or delta configuration. Always ensure you are calculating line-to-line voltage drop when using this variant.
Can I use the basic voltage drop formula for low-voltage DC LED strips?
Yes, but the margins for error are much tighter. On a 12V DC LED strip, a 3% drop is only 0.36V. Because the current (I) for long LED runs can be quite high (e.g., 10A for a 120W strip), the voltage drop will be severe over long distances. You will frequently use the rearranged distance formula (L = VD × CM / 2 × K × I) to find that you can only run 18 AWG wire for a few feet before needing to inject power from both ends or switch to 24V strips to halve the current.
Why does my calculated voltage drop differ from my multimeter reading?
If your multimeter reads a higher voltage drop than your calculation, the most common culprit is terminations and connections. The basic formula only calculates the drop across the continuous length of the conductor. It does not account for the contact resistance at breakers, wire nuts, terminal lugs, or receptacle backstabs. A loose or corroded connection can easily add 1 to 2 volts of drop. Additionally, if the wire is routed through a hot attic, the ambient temperature rises, increasing the K value and thereby increasing the actual resistance of the copper.
What K value should I use if the wire is aluminum instead of copper?
If you are sizing aluminum conductors (common for service entrance feeders and subpanels to save money), you must change the K constant. At 75°C, the K value for aluminum is 21.2. Because aluminum has a higher resistance than copper, you will find that you need to step up roughly one to two AWG sizes when switching from copper to aluminum to achieve the exact same voltage drop profile. Always verify that your terminations are rated for aluminum (CO/ALR or marked AL/CU) and use anti-oxidant paste.
