Quick Answer: Single-phase: VD = 2×K×I×D / CM | Three-phase: VD = 1.732×K×I×D / CM (K = 12.9 Cu, 21.2 Al; D = one-way distance in ft; CM = circular mils). NEC targets: branch circuit ≤3%, feeder ≤3%, total ≤5% (Informational Notes — NEC 210.19/215.2). Example: 20A on 12 AWG Cu (CM=6,530), 75ft → VD = (2×12.9×20×75)/6,530 = 5.93V = 4.9% — over limit; upsize to 10 AWG (CM=10,380) → 3.1% ✓. → Voltage Drop Calculator
Quick Formula Reference
| Circuit Type | CM Formula | Resistance Formula |
|---|---|---|
| Single-Phase | VD = 2×K×I×D / CM | VD = 2×I×R×L |
| Three-Phase | VD = 1.732×K×I×D / CM | VD = 1.732×I×R×L |
| Solve for CM | CM = 2×K×I×D / VD_max | CM = 1.732×K×I×D / VD_max |
Where K = 12.9 (copper at 75°C), 21.2 (aluminum at 75°C); D = one-way conductor length (ft); CM = conductor area in circular mils.
NEC Voltage Drop Guidelines
Recommended Maximum Voltage Drop
| Application | Branch Circuit | Feeder | Total (Feeder + Branch) |
|---|---|---|---|
| NEC Recommendation | 3% max | 3% max | 5% max |
| Sensitive Equipment | 2% max | 2% max | 3% max |
| Motor Circuits | 3% max | 3% max | 5% max |
Note: NEC 210.19(A) and 215.2(A) Informational Notes recommend these limits for "reasonable efficiency of operation."
Voltage at Load
V_load = V_source - V_drop
Example: 120V source, 3% drop
V_drop = 120V × 0.03 = 3.6V
V_load = 120V - 3.6V = 116.4V
Single-Phase Voltage Drop Formulas
Standard Formula
V_drop = 2 × I × R × L
Where:
- V_drop = Voltage drop (Volts)
- I = Current (Amperes)
- R = Resistance per unit length (Ohms/foot or Ohms/meter)
- L = One-way length of conductor (feet or meters)
- 2 = Factor for round-trip (supply and return)
Using Circular Mils
V_drop = (2 × K × I × D) / CM
Where:
- K = Resistivity constant (12.9 for copper, 21.2 for aluminum)
- I = Current (Amperes)
- D = One-way distance (feet)
- CM = Wire area in circular mils
Percentage Voltage Drop
V_drop (%) = (V_drop / V_source) × 100
Three-Phase Voltage Drop Formulas
Standard Formula
V_drop = √3 × I × R × L
Or equivalently:
V_drop = 1.732 × I × R × L
Using Circular Mils
V_drop = (1.732 × K × I × D) / CM
Three-Phase Percentage
V_drop (%) = (V_drop / V_L-L) × 100
Wire Resistance Reference
Copper Wire Resistance (Ohms per 1000 feet)
| AWG | DC Resistance | AWG | DC Resistance |
|---|---|---|---|
| 14 | 3.14 Ω | 2 | 0.194 Ω |
| 12 | 1.98 Ω | 1 | 0.154 Ω |
| 10 | 1.24 Ω | 1/0 | 0.122 Ω |
| 8 | 0.778 Ω | 2/0 | 0.0967 Ω |
| 6 | 0.491 Ω | 3/0 | 0.0766 Ω |
| 4 | 0.308 Ω | 4/0 | 0.0608 Ω |
| 3 | 0.245 Ω | 250 kcmil | 0.0515 Ω |
Circular Mil Areas
| AWG | Circular Mils | AWG | Circular Mils |
|---|---|---|---|
| 14 | 4,110 | 2 | 66,360 |
| 12 | 6,530 | 1 | 83,690 |
| 10 | 10,380 | 1/0 | 105,600 |
| 8 | 16,510 | 2/0 | 133,100 |
| 6 | 26,240 | 3/0 | 167,800 |
| 4 | 41,740 | 4/0 | 211,600 |
Quick Reference: Maximum Distance Tables
Single-Phase 120V (3% Drop, Copper)
| Load (A) | 14 AWG | 12 AWG | 10 AWG | 8 AWG | 6 AWG |
|---|---|---|---|---|---|
| 10 | 45 ft | 70 ft | 115 ft | 180 ft | 290 ft |
| 15 | 30 ft | 45 ft | 75 ft | 120 ft | 190 ft |
| 20 | 22 ft | 35 ft | 55 ft | 90 ft | 145 ft |
| 25 | — | 28 ft | 45 ft | 70 ft | 115 ft |
| 30 | — | — | 38 ft | 60 ft | 95 ft |
Single-Phase 240V (3% Drop, Copper)
| Load (A) | 14 AWG | 12 AWG | 10 AWG | 8 AWG | 6 AWG |
|---|---|---|---|---|---|
| 10 | 90 ft | 140 ft | 230 ft | 360 ft | 580 ft |
| 15 | 60 ft | 90 ft | 150 ft | 240 ft | 385 ft |
| 20 | 45 ft | 70 ft | 115 ft | 180 ft | 290 ft |
| 30 | — | — | 75 ft | 120 ft | 190 ft |
| 40 | — | — | — | 90 ft | 145 ft |
Three-Phase 480V (3% Drop, Copper)
| Load (A) | 10 AWG | 8 AWG | 6 AWG | 4 AWG | 2 AWG |
|---|---|---|---|---|---|
| 20 | 175 ft | 280 ft | 440 ft | 700 ft | 1,115 ft |
| 30 | 115 ft | 185 ft | 295 ft | 465 ft | 745 ft |
| 50 | 70 ft | 110 ft | 175 ft | 280 ft | 445 ft |
| 75 | — | 75 ft | 120 ft | 185 ft | 295 ft |
| 100 | — | — | 90 ft | 140 ft | 225 ft |
Worked Examples
Example 1: Single-Phase Residential
Given: 20A load, 120V, 75 feet one-way, copper wire
Find: Required wire size for 3% max drop
Solution:
- Calculate allowable voltage drop:
V_drop (max) = 120V × 0.03 = 3.6V
- Using the CM formula, solve for CM:
CM = (2 × K × I × D) / V_drop
CM = (2 × 12.9 × 20A × 75ft) / 3.6V
CM = 38,700 / 3.6 = 10,750 CM
- Select wire: 10,750 CM → 10 AWG (10,380 CM is close, use next size up)
Verify:
V_drop = (2 × 12.9 × 20 × 75) / 10,380 = 3.73V (3.1%)
Answer: Use 10 AWG copper (or 8 AWG for margin)
Example 2: Three-Phase Industrial
Given: 100A load, 480V, 200 feet, copper, PF = 0.85
Find: Wire size for 3% max drop
Solution:
- Allowable drop:
V_drop (max) = 480V × 0.03 = 14.4V
- Using resistance method:
V_drop = √3 × I × R × L
14.4V = 1.732 × 100A × R × 200ft
R = 14.4 / (1.732 × 100 × 0.2) = 0.416 Ω/1000ft
- From resistance table: 2 AWG = 0.194 Ω/1000ft ✓
Verify:
V_drop = 1.732 × 100A × 0.000194 × 200 = 6.7V (1.4%)
Answer: Use 2 AWG copper
Example 3: Long Distance Motor Feeder
Given: 50 HP motor, 480V 3-phase, 500 feet, PF = 0.87
Find: Wire size for 3% drop
Solution:
-
Calculate motor FLC (from NEC table): 65A
-
Allowable drop: 480V × 0.03 = 14.4V
-
Required CM:
CM = (1.732 × K × I × D) / V_drop
CM = (1.732 × 12.9 × 65A × 500ft) / 14.4V
CM = 726,063 / 14.4 = 50,421 CM
- Select: 1 AWG (83,690 CM) or 1/0 AWG (105,600 CM)
Answer: Use 1 AWG minimum, recommend 1/0 AWG for motor starting
Factors Affecting Voltage Drop
Temperature Correction
Wire resistance increases with temperature:
R_T = R_25 × [1 + α × (T - 25)]
Where:
- R_25 = Resistance at 25°C
- α = Temperature coefficient (0.00393 for copper)
- T = Operating temperature (°C)
Power Factor Effect
For AC circuits with significant reactance:
V_drop = I × L × (R × cos(θ) + X × sin(θ))
Where X = reactance (significant for larger conductors)
Conductor Material
| Material | K Factor | Relative Resistance |
|---|---|---|
| Copper | 12.9 | 1.00 (baseline) |
| Aluminum | 21.2 | 1.64× copper |
Voltage Drop vs Wire Size Trade-offs
Cost-Benefit Analysis
| Factor | Smaller Wire | Larger Wire |
|---|---|---|
| Initial cost | Lower | Higher |
| Voltage drop | Higher | Lower |
| Energy loss | Higher | Lower |
| Heat generation | More | Less |
| Future capacity | Limited | Available |
When to Exceed Minimum
Consider upsizing wire when:
- Long distance runs
- Motor circuits (starting voltage)
- Sensitive electronic equipment
- Expected load growth
- High duty cycle operations
Common Mistakes to Avoid
| Mistake | Why It's Wrong | Correct Approach |
|---|---|---|
| Using one-way distance for single-phase | Misses return conductor | Use 2×L factor |
| Ignoring 3-phase √3 factor | Result off by 15% | Use 1.732 factor |
| Using DC resistance for large AC conductors | Ignores skin effect | Use AC resistance tables |
| Forgetting to check both ampacity AND drop | May still be undersized | Verify both requirements |
Related Calculators
| Calculator | Use When... |
|---|---|
| Voltage Drop Calculator | Quick sizing calculations |
| Wire Size Calculator | Complete wire selection |
| Conduit Fill Calculator | Selecting conduit size |
| Motor Circuit Calculator | Motor feeder sizing |
Summary
Key Formulas:
- Single-Phase: V_drop = 2 × I × R × L
- Three-Phase: V_drop = √3 × I × R × L
- CM Method: V_drop = (2 × K × I × D) / CM
NEC Recommendations:
- Branch circuit: 3% max
- Feeder: 3% max
- Total: 5% max
K Factors:
- Copper: 12.9
- Aluminum: 21.2
FAQ
What is acceptable voltage drop per NEC?
NEC recommends (but doesn't require) maximum 3% drop for branch circuits and feeders, with 5% total from source to load. These are informational notes, not mandatory requirements.
Why does voltage drop matter?
Excessive voltage drop causes: reduced equipment performance, motor overheating, dimming lights, tripped breakers, and wasted energy as heat in conductors.
How do I reduce voltage drop?
Options include: using larger wire, shortening the distance, increasing voltage (208V vs 120V), using higher power factor, or running multiple parallel conductors.
Do I need to consider voltage drop for short runs?
For runs under 50 feet, voltage drop is usually not a concern for typical residential loads. Always verify for high-current circuits regardless of distance.
Should I use copper or aluminum for long runs?
Copper has lower resistance but costs more. For long runs, the cost difference may favor aluminum with larger gauge. Consider: initial cost, installation labor, terminal compatibility, and conduit fill.