How to Calculate Voltage Drop for NEC-Compliant Wiring

A 20 A kitchen circuit on 12 AWG copper doesn't lose much voltage across 25 feet. Stretch that same wire to 65 feet from a basement panel to a far-wall receptacle, and the drop hits 4.2 percent of 120 V. That's 5 volts gone before the load draws a watt. The toaster gets 115 V instead of 120, the microwave runs below rated power, and you've exceeded the NEC's 3 percent recommendation without breaking any hard code rule. The math is simple enough to do on a notepad.

The Formula

For DC and single-phase AC:

Vdrop = 2 × L × I × (R / 1000)

For three-phase AC:

Vdrop = √3 × L × I × (R / 1000)

Where:

• L = one-way wire length in feet
• I = load current in amps
• R = conductor DC resistance in ohms per 1000 feet, from NEC Chapter 9 Table 8 at 75 °C

The factor of 2 in the single-phase formula accounts for the round-trip current path: current travels out on the hot conductor and returns on the neutral, so the effective wire length is double the one-way run. In three-phase balanced circuits, the return current splits among the three phases, and the geometry produces √3 (about 1.732) instead of 2.

To convert volts to percentage:

Vdrop% = (Vdrop / Vsource) × 100

Worked Example: Kitchen Receptacle Circuit

You're wiring a 20 A small-appliance branch circuit per NEC 210.52(B) from a basement panel to the far end of a kitchen remodel. One-way run: 65 feet. Source: 120 V single-phase. Wire: 12 AWG copper, THWN-2 (75 °C column).

Step 1. Look up resistance. NEC Chapter 9 Table 8 lists 12 AWG copper at 1.93 Ω per 1000 ft (DC, 75 °C).

Step 2. Apply the formula.

Vdrop = 2 × 65 × 20 × (1.93 / 1000)
      = 2 × 65 × 20 × 0.00193
      = 5.018 V

Step 3. Convert to percentage.

Vdrop% = 5.018 / 120 × 100 = 4.18%

That exceeds the NEC 3% branch-circuit recommendation. The fix: upsize to 10 AWG copper (1.21 Ω/kft).

Vdrop = 2 × 65 × 20 × (1.21 / 1000) = 3.146 V → 2.62%

Now you're under 3% with room for the 5% combined feeder-plus-branch ceiling. Verify both numbers with the voltage drop calculator before pulling wire.

Worked Example: EV Charger on a Long Run

You're installing a Level 2 EV charger (NEMA 14-50, 40 A continuous) in a garage 80 feet from the panel. Source: 240 V, single-phase.

Step 1. Check ampacity first. Per NEC 310.16 at 75 °C, 8 AWG copper carries 50 A. For a 40 A continuous load, NEC 210.19(A)(1) requires conductors rated at 125% of the continuous load: 40 × 1.25 = 50 A. 8 AWG at 50 A passes exactly.

Step 2. Compute voltage drop. 8 AWG copper: 0.764 Ω/kft.

Vdrop = 2 × 80 × 40 × (0.764 / 1000)
      = 2 × 80 × 40 × 0.000764
      = 4.890 V

Vdrop% = 4.890 / 240 × 100 = 2.04%

Under 3%, but there's no margin for branch circuits downstream. If the garage also has a lighting subpanel, the combined feeder-plus-branch drop could breach 5%.

Stepping to 6 AWG (0.491 Ω/kft, 65 A at 75 °C):

Vdrop = 2 × 80 × 40 × 0.000491 = 3.142 V → 1.31%

The extra headroom costs roughly $0.80 per foot in material. On an 80-foot run that's $64 for the upgrade, cheap insurance against future load growth.

Why Low Voltage Amplifies the Problem

Voltage drop percentage scales inversely with source voltage. The same 3.07 V drop that represents 2.56% of 120 V is 25.6% of a 12 V landscape lighting system.

Consider a 12 V transformer powering 5 A of path lights through 100 feet of 14 AWG copper:

Vdrop = 2 × 100 × 5 × (3.07 / 1000) = 3.07 V → 25.6%

The lights at the end of the run receive 8.93 V. They'll be noticeably dim, or won't turn on at all if they have a minimum voltage threshold.

Upsizing to 10 AWG (1.21 Ω/kft):

Vdrop = 2 × 100 × 5 × 0.00121 = 1.21 V → 10.1%

Still high. Even 8 AWG (0.764 Ω/kft) only reaches 6.4%. Low-voltage landscape runs are where the "hub and spoke" wiring pattern earns its keep: instead of one long home run, wire short radial runs from a centrally placed transformer.

Three-Phase Circuits

For balanced three-phase loads, substitute √3 for the factor of 2:

Vdrop = √3 × L × I × (R / 1000)

Example: a 480 V three-phase feeder running 200 feet to a 50 A motor on 6 AWG copper.

Vdrop = 1.732 × 200 × 50 × (0.491 / 1000)
      = 1.732 × 200 × 50 × 0.000491
      = 8.51 V → 1.77%

Compare to the same parameters on single-phase:

Vdrop = 2 × 200 × 50 × 0.000491 = 9.82 V → 2.05%

The three-phase circuit drops about 87% as much voltage as the single-phase equivalent, matching the theoretical ratio of √3/2 = 0.866. This efficiency advantage is one reason industrial facilities distribute power at three-phase 480 V: fewer amps for the same power, less drop, and smaller wire.

What the NEC Actually Says About Drop Limits

The NEC does not mandate a maximum voltage drop. The limits most electricians cite are recommendations, not enforceable code.

Specifically, NEC 2023:

• Article 210.19, Informational Note No. 4: recommends no more than 3% drop on a branch circuit
• Article 215.2, Informational Note No. 2: recommends no more than 3% on a feeder
• Both notes: recommend no more than 5% on the combined feeder and branch circuit

"Informational notes" in the NEC are explanatory material, not enforceable requirements. Most jurisdictions treat them as de facto standards, and the 3%/5% thresholds are embedded in engineering practice.

Why those numbers? Motors are the most drop-sensitive common load. An induction motor at 95% of rated voltage draws roughly 5% more current, heats the windings faster, and may fail to start under full-load torque. At 90% voltage, starting torque drops to about 81% of rated (torque scales with V²). The 3%/5% limits keep the voltage at the load above 95% of nominal under steady-state conditions, which keeps motors, heating elements, and electronics within their design envelopes.

Quick Reference: Maximum One-Way Length at 3% Drop

For 120 V single-phase copper at 75 °C. Formula: L_max = (0.03 × V) / (2 × I × R/1000).

For 240 V circuits at the same gauge and current, double these distances. The higher source voltage means the same absolute drop is a smaller percentage.

Use the wire size calculator to check any combination of gauge, material, length, and current, and the Ohm's Law calculator to verify individual resistance calculations.

Copper vs. Aluminum

Aluminum conductors have about 61% the conductivity of copper. For the same AWG gauge, aluminum's resistance per foot is roughly 1.6× higher.

At 12 AWG: copper is 1.93 Ω/kft, aluminum is 3.18 Ω/kft. If the copper drop was 3.15 V on a given run, the same gauge in aluminum drops 5.19 V. The standard practice is to step up two AWG sizes when switching from copper to aluminum. The NEC ampacity tables confirm this: 10 AWG aluminum at 75 °C carries 30 A, close to the 25 A rating of 12 AWG copper.

For long feeder runs where material cost dominates, aluminum is standard. 4/0 aluminum is rated 180 A at 75 °C with 0.100 Ω/kft, versus 2/0 copper at 175 A with 0.0967 Ω/kft. The aluminum has slightly lower ampacity and slightly higher resistance, but costs substantially less per foot. Most 200 A residential services use 4/0 aluminum service-entrance cable for exactly this reason.

When DC Resistance Is Not Enough

The formula above uses DC resistance from NEC Chapter 9 Table 8. For building wiring at 60 Hz with conductors up through about 1/0 AWG, this is accurate. Two effects reduce accuracy for larger conductors or higher frequencies.

Skin effect pushes current toward the outer surface of the conductor, reducing the effective cross-section and increasing resistance. At 60 Hz, skin depth in copper is about 8.5 mm. A 4/0 AWG solid conductor (11.7 mm diameter) is already large enough that the center carries measurably less current than the surface. For conductors above 4/0, use NEC Chapter 9 Table 9 (effective impedance at 0.85 PF) instead of Table 8.

Reactance adds a voltage drop component in quadrature with the resistive drop. The impedance-based formula is:

Vdrop = I × (R cos θ + X sin θ) × phase_factor × L / 1000

where R and X are per-1000-ft values from NEC Table 9, θ is the power factor angle, and phase_factor is 2 (single-phase) or √3 (three-phase). For typical building circuits with 0.85 PF and small conductors, the reactive component adds only 1 to 3 percent to the total drop. For large feeders on inductive loads, it can add 10 to 20 percent.

For most residential and light commercial work, the DC resistance formula is the right tool. When you need full impedance precision, the voltage drop calculator handles both approaches.