Why Ampacity Alone Isn't Enough
Ampacity tables (NEC Table 310.16) tell you the maximum current a conductor can carry without overheating its insulation. They say nothing about how much voltage is lost along the way. A 12 AWG conductor might be rated for 20A all day long, but run it 250 ft to a load and the voltage at the far end can drop low enough to trip motor protection, dim lighting noticeably, or cause VFDs and electronic loads to fault on undervoltage. Voltage drop is a separate calculation that has to be checked on every long run, regardless of whether ampacity passes.
The Voltage Drop Formula
Voltage drop occurs because every conductor has resistance, and current flowing through that resistance creates a voltage loss (Ohm's Law: V = I x R). The standard formulas used in the industry:
| Circuit Type | Formula |
|---|---|
| Single-phase (1Φ) | VD = (2 × K × I × D) / CM |
| Three-phase (3Φ) | VD = (1.732 × K × I × D) / CM |
Where:
- K = resistivity constant: 12.9 for copper, 21.2 for aluminum (at 75°C; use 11.2 / 18.4 at 20°C for more conservative numbers)
- I = load current in amps
- D = one-way distance from source to load, in feet
- CM = conductor cross-sectional area in circular mils (from AWG/kcmil tables)
The factor of 2 in the single-phase formula accounts for the round trip (out and back) through both conductors. Three-phase uses 1.732 (√3) because the phase relationship between conductors reduces the effective voltage drop compared to a straight round trip.
Worked Example 1: Single-Phase 240V Circuit
A 30A load is located 180 ft from the panel, wired with 10 AWG copper (CM = 10,380):
VD = (2 × 12.9 × 30 × 180) / 10,380 = 139,320 / 10,380 = 13.4 volts
As a percentage: 13.4V / 240V = 5.6% drop - this exceeds the 3% branch circuit recommendation. Upsizing to 6 AWG copper (CM = 26,240) brings it down to:
VD = (2 × 12.9 × 30 × 180) / 26,240 = 5.3 volts (2.2%) - now within limits.
Worked Example 2: Three-Phase 480V Motor Feeder
A 75 HP motor (FLA 96A) is fed from a panel 300 ft away with 1/0 AWG copper (CM = 105,600):
VD = (1.732 × 12.9 × 96 × 300) / 105,600 = 644,134 / 105,600 = 6.1 volts
As a percentage: 6.1V / 480V = 1.3% drop - well within the 3% recommendation, so 1/0 AWG is adequate for voltage drop even though a smaller conductor might still pass straight ampacity tables.
NEC Recommended Limits
| Circuit Segment | Recommended Max Drop |
|---|---|
| Feeder conductors | 3% |
| Branch circuit conductors | 3% |
| Total (feeder + branch combined) | 5% |
These figures come from the informational notes to NEC 210.19(A) and 215.2(A) — they are recommendations for reasonable efficiency, not mandatory code rules in most jurisdictions. That said, many engineering specs, utility interconnection agreements, and AHJs treat them as hard requirements, and sensitive equipment (VFDs, PLCs, lighting controls) often has its own tighter undervoltage tolerance that makes the 3%/5% guidance a practical necessity regardless of code status.
Practical Consequences of Excessive Voltage Drop
- Motors: Run hotter, draw more current to produce the same torque, and can trip thermal overloads under load
- Lighting: Visibly dims, and HID/LED drivers may flicker or fail to start reliably
- VFDs and electronics: Undervoltage faults, especially on startup when inrush current spikes the drop further
- Heating cable and resistive loads: Output power drops with the square of voltage - a 10% voltage drop cuts heating output by roughly 19%
How to Fix a Voltage Drop Problem
- Upsize the conductor - the most common fix; larger CM area directly reduces VD proportionally
- Increase distribution voltage - doubling voltage (e.g., 240V to 480V) for the same power load quarters the current, cutting voltage drop to roughly a quarter for the same conductor
- Shorten the run - relocate the panel, transformer, or motor control center closer to the load
- Switch to copper from aluminum - copper's lower resistivity (K=12.9 vs 21.2) reduces drop by roughly 39% for the same CM area; see our copper vs aluminum comparison for the full trade-off analysis
Frequently Asked Questions
What is an acceptable voltage drop percentage for industrial cable runs?
NEC recommends a maximum of 3% on branch circuits and 5% combined (feeder plus branch circuit), per the informational notes to NEC 210.19(A) and 215.2(A). These aren't hard violations in most jurisdictions, but exceeding them risks equipment malfunction and some specs/AHJs enforce them as hard limits.
What is the formula for calculating voltage drop?
Single-phase: VD = (2 × K × I × D) / CM. Three-phase: VD = (1.732 × K × I × D) / CM. K is 12.9 for copper or 21.2 for aluminum, I is load current, D is one-way distance in feet, and CM is conductor area in circular mils.
Does upsizing the conductor always fix a voltage drop problem?
Yes, larger conductors have lower resistance and less voltage drop for the same load. But upsizing has limits - larger conduit, larger terminals, and higher cost. Beyond a certain run length it's often more economical to raise distribution voltage or relocate the source closer to the load.
Source Correctly Sized Cable for Long Runs
Shanghai Unicorn manufactures copper and aluminum conductor cable in a full range of AWG and kcmil sizes, UL/CSA listed. Tell us your load, distance, and voltage class and we'll help you size for both ampacity and voltage drop.