The direct answer for residential single-phase wiring is that the standard approximate AC voltage drop calculation is VD = (2 × K × I × L) / CM. While the National Electrical Code (NEC) does not strictly mandate a specific voltage drop limit for most branch circuits, it strongly recommends a maximum of 3% drop on branch circuits and 5% total drop from the service entrance to the furthest outlet to ensure reasonable efficiency and prevent motor burnout.
When you are sizing wire for a long run, ampacity tables only tell you what the wire can handle before the insulation melts. They tell you absolutely nothing about whether the equipment at the end of the wire will actually function. That requires the AC voltage drop calculation. Below is the complete derivation, the algebraic rearrangements you need for the field, and step-by-step worked examples with strict unit tracking.
The Core AC Voltage Drop Calculation Formula
In a purely theoretical DC circuit, voltage drop is just Ohm's Law (V = I × R). But in AC circuits, we have to account for the out-and-back path of single-phase power, the specific resistivity of the conductor material at operating temperature, and the cross-sectional area of the wire. For practical field calculations on single-phase residential and light commercial systems, we use the approximate formula derived from NEC Chapter 9, Table 8.
| Symbol | Variable | Standard Unit | Definition & Field Notes |
|---|---|---|---|
| VD | Voltage Drop | Volts (V) | The total voltage lost across the entire circuit loop (line and neutral/ground). |
| 2 | Multiplier | Dimensionless | Accounts for the out-and-back path in single-phase AC (Line + Neutral). Use 1.732 for 3-phase. |
| K | Specific Resistance | Ω·cmil/ft | Material resistivity. Use 12.9 for Copper at 75°C, 21.2 for Aluminum at 75°C. |
| I | Current | Amperes (A) | The actual continuous load current, not the breaker size. (e.g., a 15A load on a 20A breaker). |
| L | Length | Feet (ft) | The one-way distance from the breaker panel to the load. |
| CM | Circular Mils | cmil | Cross-sectional area of the wire. (e.g., 14 AWG = 4,110; 12 AWG = 6,530; 10 AWG = 10,380). |
Note: For highly precise engineering involving large feeders, the exact AC formula incorporates power factor and reactance: VD = 2 × I × L × (R cosθ + X sinθ). However, for standard residential wire sizing up to 400A, the K-factor approximation above is the industry standard and matches NEC guidance (EC&M: Basics of Voltage Drop).
Rearranged Forms: Solving for Wire Size, Distance, or Current
On the jobsite, you rarely need to find the voltage drop; you already know you need to keep it under 3%. What you actually need to find is the wire size (CM) or the maximum distance (L). Here are the algebraic rearrangements of the core formula:
- Solving for Wire Size (CM):
CM = (2 × K × I × L) / VD
Use this to find the minimum Circular Mils required, then look up the corresponding AWG in NEC Chapter 9, Table 8. - Solving for Maximum One-Way Length (L):
L = (VD × CM) / (2 × K × I)
Use this to determine how far you can run a specific wire gauge before exceeding your 3% threshold. - Solving for Maximum Current (I):
I = (VD × CM) / (2 × K × L)
Use this when auditing an existing long circuit to see what the safe continuous load limit is.
Worked Examples: Branch Circuits and 240V Feeders
Let's run the math with strict unit tracking to prove how the dimensions cancel out to leave us with Volts.
Problem 1: 120V Branch Circuit (Kitchen Appliance)
Setup: You are running a dedicated 120V, 20A continuous load for a high-end espresso machine. The one-way distance from the panel is 80 feet. You plan to use 12 AWG Copper NM-B. Is 12 AWG sufficient to maintain a 3% drop?
- Identify Variables:
K = 12.9 Ω·cmil/ft (Copper at 75°C)
I = 20 A
L = 80 ft
CM = 6,530 cmil (12 AWG)
Target VD = 3% of 120V = 3.6V - Apply Formula with Units:
VD = [ 2 × 12.9 (Ω·cmil/ft) × 20 (A) × 80 (ft) ] / 6,530 (cmil) - Calculate Numerator:
2 × 12.9 × 20 × 80 = 41,280 (Ω·cmil·A)
Note: The 'ft' in the denominator of K cancels with the 'ft' in Length. - Divide by CM:
41,280 / 6,530 = 6.32 V
The 'cmil' units cancel out, leaving Volts (since V = I × R). - Conclusion: 6.32V is a 5.2% drop. 12 AWG fails the 3% recommendation. You must step up to 10 AWG (10,380 cmil), which yields a 3.97V drop (3.3%), or 8 AWG for strict compliance.
Problem 2: 240V Subpanel Feeder
Setup: Feeding a detached garage subpanel with a 40A continuous load. Distance is 150 feet. Using 8 AWG Copper THHN in conduit.
- Identify Variables:
K = 12.9 Ω·cmil/ft
I = 40 A
L = 150 ft
CM = 16,510 cmil (8 AWG) - Apply Formula:
VD = [ 2 × 12.9 × 40 × 150 ] / 16,510 - Calculate:
Numerator: 154,800
154,800 / 16,510 = 9.37 V - Conclusion: 9.37V on a 240V system is a 3.9% drop. For a feeder, the NEC recommends a maximum of 3% for the feeder alone, but allows up to 5% total. This is borderline. Upgrading to 6 AWG (26,240 cmil) drops it to 5.9V (2.4%), which is the professional choice for a subpanel to leave headroom for future loads.
Real-World Scenario: The 120V Compressor Burnout
Formulas are clean; jobsites are not. Here is a scenario that plays out constantly in home workshops, illustrating what happens when ampacity is prioritized over voltage drop.
The Setup: A hobbyist woodworker installs a 1.5 HP air compressor (rated 120V, 15A running current, but draws near 20A under heavy continuous cycling) at the far end of a 150-foot long detached shed. The shed is wired with a standard 12 AWG NM-B branch circuit on a 20A breaker. The wire is perfectly legal for the breaker and the running ampacity.
The Numbers:
Let's calculate the voltage drop at the compressor's 20A heavy-cycle draw:
VD = (2 × 12.9 × 20 × 150) / 6,530
VD = 77,400 / 6,530 = 11.85 V
The Outcome: The voltage arriving at the compressor motor under load is 120V - 11.85V = 108.15V. AC induction motors draw more current as voltage drops to maintain their mechanical power output (Watts = Volts × Amps × Power Factor). The compressor motor begins pulling 23A+ to compensate for the low voltage. The 20A breaker doesn't trip immediately because thermal-magnetic breakers have a time-delay curve for slight overloads. Instead, the motor's internal thermal overload protector trips, or worse, the winding insulation degrades over a few months and the motor shorts out.
What Went Wrong: The installer sized the wire purely for ampacity (preventing the wire from catching fire) and ignored voltage drop (ensuring the equipment actually works). To fix this, the circuit should have been wired with 8 AWG copper (CM = 16,510), which would limit the drop to 4.7V (3.9%), keeping the motor happy and the current draw stable. For a comprehensive look at wire properties, refer to standard manufacturer data like the Cerrowire Voltage Drop Guide.
Assumptions, Unit Traps, and Realistic Magnitudes
To use the AC voltage drop calculation effectively, you need to understand its boundaries and where DIYers typically break the math.
When the Formula Applies (and its Assumptions)
- Steady-State AC: This formula calculates the drop under a continuous, steady-state load. It does not calculate the momentary voltage sag during motor startup (Locked Rotor Amps), which can be 5 to 7 times higher than running current.
- Temperature Assumption: The K-factor of 12.9 assumes the copper is operating at 75°C. If you are using NM-B cable (which is limited to the 60°C column by NEC 334.80), the actual resistance is slightly lower (K ≈ 12.6), meaning your real-world drop will be marginally better than the 12.9 calculation. We use 12.9 as a conservative safety buffer.
- Sinusoidal Waveform: It assumes a clean utility sine wave. Heavy harmonic distortion from cheap VFDs (Variable Frequency Drives) or solar inverters can increase effective RMS current and skew the drop.
Unit Mistakes That Break the Math
If your answer is wildly off by a factor of 10 or 1,000, you likely committed one of these errors:
- Forgetting the '2': Single-phase power requires an out-and-back path. If you forget to multiply by 2, you are only calculating the drop on the hot wire, ignoring the neutral return path.
- Using Breaker Size Instead of Load Current: If you put '20' for 'I' on a 20A breaker, but the actual load is a 2A LED lighting circuit, your calculated drop will be 10x higher than reality. Always use the actual anticipated load.
- Confusing mm² with Circular Mils: The standard US formula requires Circular Mils (cmil). If you are using metric wire (mm²), you must convert, or use the metric formula: VD = (2 × ρ × I × L) / A, where ρ is resistivity in Ω·m and A is area in mm².
What a Realistic Answer Magnitude Looks Like
When you punch the numbers into your calculator, use these sanity-check ranges. If your result falls outside these, double-check your decimal placement:
- 120V Branch Circuits: A well-designed circuit should show a drop between 1.0V and 3.5V. If you are seeing 15V+ on standard 12 or 14 AWG wire, your run is either absurdly long (over 200ft) or you forgot to divide by the CM.
- 240V Feeders / Appliances: Expect to see drops between 2.0V and 8.0V. Because the system voltage is doubled, the percentage drop is naturally halved for the same physical wire and distance.
Mastering the AC voltage drop calculation bridges the gap between passing an electrical inspection and building a system that actually performs. Always pull the tape measure, verify your continuous load, and let the Circular Mils dictate your wire gauge on long runs.
