Voltage Drop Calculator (AC)

NEC 210.19 Informational Note No. 4 recommends: branch circuits — maximum 3% voltage drop at the farthest outlet. Feeders — maximum 2%. Combined feeder + branch — maximum 5%. These are Informational Notes, meaning they are guidance, not enforceable code requirements (enforceable requirements use the word 'shall'). However, they are universally followed in design practice and often written into project specifications, making them contractually binding. Energy codes such as ASHRAE 90.1-2019 §8.4.1 do make feeder voltage drop limits enforceable (maximum 2% for feeders) as part of mandatory energy efficiency provisions. Fire pump circuits under NEC 695.7 carry a mandatory and specific 15% maximum during motor starting (locked-rotor current) and 5% at running load — these are enforceable 'shall' requirements. Additionally, IEC 60364-5-52 for international projects specifies 3% for lighting and 5% for other uses from the origin of the installation.

Four options, in order of typical cost-effectiveness: (1) Increase conductor size — the most common solution. Going from #12 to #10 AWG increases cross-sectional area by 59% (6,530 to 10,380 CM) and reduces VD proportionally. (2) Increase system voltage — doubling the voltage halves the current for the same power, reducing VD by 50%. For a 10 kW motor, 208V requires 27.8 A vs. 480V requiring 12.0 A — same kW, half the VD. (3) Reduce distance — relocate the panel or transformer closer to the load. For outdoor or campus installations, adding a pad-mounted transformer near the load building can reduce feeder length from thousands of feet to hundreds. (4) Parallel conductors — running two identical cables per phase halves the current per conductor, reducing VD by 50%. NEC 310.10(H) requires paralleled conductors to be the same length, material, cross-sectional area, insulation type, and terminated in the same manner. For existing installations where recabling is impractical, options include: installing a boost transformer at the load end, adjusting transformer primary taps, or using power conditioning equipment (e.g., constant-voltage transformers for sensitive single-phase loads).

Copper resistivity (K=12.9) is about 39% lower than aluminum (K=21.2). This means an aluminum conductor needs approximately 64% more cross-sectional area (21.2 ÷ 12.9 = 1.64) to achieve the same voltage drop as a copper conductor. For equal ampacity per NEC 310.16, aluminum conductors are typically 1-2 AWG sizes larger than copper — but aluminum is roughly 50% lighter and 30-50% cheaper per ampere-foot, dominating utility transmission and distribution, large service entrance cables (residential 4/0 Al for 200 A services), and commercial feeders above 400 A. In commercial building branch circuits (15-60 A), copper is overwhelmingly preferred due to smaller conduit fill, better termination reliability (aluminum oxide is an insulator that must be wire-brushed and coated with anti-oxidant compound), and avoidance of the thermal expansion mismatch that can loosen aluminum terminations over repeated load cycling. For large commercial feeders (400-4000 A), the cost advantage of aluminum often makes it the default choice, with careful attention to AL/CU-rated lugs and proper torque procedures per manufacturer instructions.

The simple K-factor formula (DC resistance method) assumes unity power factor and gives acceptable results for conductors #2 AWG and smaller where inductive reactance is negligible. For larger conductors, power factor significantly affects voltage drop. The exact AC formula is: VD = I × (R × cos φ + X × sin φ) × L × √n, where R is AC resistance (Ω/ft), X is inductive reactance (Ω/ft), and φ is the power factor angle. At lagging power factor (inductive loads: motors, transformers), the voltage drop is HIGHER than the unity-PF calculation because the inductive reactance component adds to the drop — a 0.85 PF motor circuit experiences roughly 10-15% more VD than the simple DC calculation predicts. At leading power factor (capacitive loads, lightly loaded long cables with capacitance), voltage RISE can occur (Ferranti effect). For a 500 A, 480V feeder with 0.80 PF at 750 ft using 750 kcmil copper: DC method → VD = (1.732 × 750 × 12.9 × 500) ÷ 750,000 = 11.2 V (2.33%). AC exact (R=0.0172, X=0.0340 Ω/1000 ft): VD = 1.732 × 500 × [(0.0172 × 0.80) + (0.0340 × 0.60)] × (750/1000) = 22.2 V (4.63%) — nearly double the DC estimate. For precision work on conductors #2/0 and larger, consult IEEE 141 AC impedance tables with actual load power factor.

NEC Table 8 conductor properties are listed at 75°C because this is the assumed operating temperature for most building wire types under full ampacity loading. Copper's resistance increases with temperature: R_T = R_25°C × [1 + 0.00393 × (T − 25)]. At 75°C, resistance is approximately 19.7% higher than at 25°C. Using the 75°C value is conservative for steady-state voltage drop because the conductor will be at or near 75°C when carrying full rated current. For cold-start conditions (e.g., checking voltage for motor starting when the conductor is at ambient temperature), the actual VD may be slightly lower than calculated. Conversely, for rooftop conduits in direct sun where ambient plus solar gain can reach 60-70°C, the conductor temperature may exceed 75°C and resistance increases further — NEC Table 310.15(B)(1) provides temperature correction factors that can be applied to the resistance values for installations above 30°C ambient.

For a feeder serving both three-phase and single-phase loads, calculate VD separately for each portion. The three-phase portion uses VD_3Φ = (1.732 × L × K × I_3Φ) ÷ CM. The single-phase portions are served line-to-neutral and use VD_1Φ = (2 × L_1Φ × K × I_1Φ) ÷ CM for each phase conductor + neutral. The worst-case phase voltage (line-to-neutral) is the critical value. If the single-phase loads are reasonably balanced across all three phases, use the phase with the highest current for the VD check. For a mixed panel (100 A 3Φ motor + 40 A miscellaneous 1Φ on each phase), the design approach is: (a) calculate feeder VD using 3Φ formula at the total phase current (100 + 40 = 140 A per phase), (b) for each 1Φ branch circuit, calculate its VD from the panel to the load using the 1Φ formula, (c) sum the feeder VD% + worst branch VD% to verify the 5% combined limit.