Voltage Drop Calculator

Voltage Drop Calculator

Calculate exact electrical voltage drop in volts and percentage, end-of-line load voltage, conductor power loss (watts), and maximum run distance using NEC Chapter 9 Table 8 and Table 9 resistance and reactance values. Supports single-phase AC, three-phase AC, and DC circuits across copper and aluminum conductors.

Scenarios:

Conductor & Circuit Inputs

12V, 120V, 240V, 480V
V
1,920 W
Amps
30.5 m (200 ft round-trip)
ft
0.50 to 1.00

Voltage Drop & Line Loss Results Z_eff: 1.240 Ω / 1,000 ft

Voltage Drop — Within 5% NEC Feeder+Branch Limit
3.97 V (3.31%)

Delivers 116.03 V at the load over 100 ft using 10 AWG Copper.

End Load Voltage
116.03 V
From 120.0 V source
Line Power Loss
63.5 W
I²R heat in wire
Max Run (3% Drop)
90 ft
Max for 3.60 V drop
Max Run (5% Drop)
151 ft
Max for 6.00 V drop
Voltage Profile from Source to Load & NEC 3% / 5% Thresholds 116.03 V at Load

Voltage Drop Across All Wire Gauges (Current Load & Distance)

Wire Gauge R (Ω/1k ft) Voltage Drop (V) Drop (%) Load Voltage Line Loss (W) Max Run (≤3%) NEC Status

NEC Voltage Drop Formulas (Chapter 9, Table 8 & Table 9)

For DC circuits and resistive AC loads (`PF = 1.0`), voltage drop is determined by the conductor resistance (R in ohms per 1,000 feet at 75°C). For AC circuits with inductive loads (`PF < 1.0`), NEC Chapter 9 Table 9 incorporates the conduit's inductive reactance (X_L in ohms per 1,000 feet) to compute effective impedance (Z_eff):

Effective Impedance (Z_eff) = (R × PF) + (X_L × √(1 − PF²))   (for AC; for DC, Z_eff = R)
Single-Phase / DC Voltage Drop (V_drop) = (2 × L_ft × Z_eff × I) / (1,000 × Parallel Sets)
Three-Phase (3ϕ) Voltage Drop (V_drop) = (1.73205 × L_ft × Z_eff × I) / (1,000 × Parallel Sets)
Voltage Drop (%) = (V_drop / V_source) × 100
Line Power Loss (Watts) = Conductors × I² × (R × L_ft / (1,000 × Parallel Sets))

Illustrative Example

Scenario (Illustrative): A 120V single-phase circuit delivering 16 Amps (`PF = 1.0`) over a 100 ft one-way run of 10 AWG Copper in PVC conduit (`R = 1.24 Ω / 1,000 ft`):

  • Voltage Drop (V): (2 × 100 ft × 1.24 Ω × 16 A) / 1,000 = 3.97 V
  • Percentage Voltage Drop: (3.968 V / 120 V) × 100 = 3.31%
  • End-of-Line Load Voltage: 120.00 V − 3.97 V = 116.03 V
  • Resistive Power Loss in Wire: 2 × 16² × (1.24 × 100 / 1,000) = 63.5 Watts
  • Upsizing Comparison: Stepping up one size to 8 AWG Copper (`R = 0.778 Ω / 1,000 ft`) reduces the voltage drop to 2.49 V (2.07%), meeting the ≤ 3.0% NEC branch-circuit informational note.

How to Use This Voltage Drop Calculator

  1. Select Material & Wire Gauge: Choose Copper or Aluminum and select the wire gauge (`18 AWG` to `500 kcmil`) you want to evaluate.
  2. Configure Circuit Phase & Conduit: Choose Single-Phase AC, Three-Phase AC, or DC. For AC runs, select whether the conductors run in PVC/nonmetallic raceway, aluminum conduit, or steel conduit.
  3. Enter Voltage, Load Current & One-Way Distance: Input the nominal source voltage, load current in Amps, and the one-way distance from source to load.
  4. Compare All Wire Sizes: Check the full reference table below the calculator to see the exact voltage drop, line power loss in watts, and maximum run distance (`≤ 3%`) for every AWG size.

Frequently Asked Questions

What is the NEC standard for maximum allowable voltage drop?

Under NEC 210.19(A) Informational Note No. 4 and NEC 215.2(A)(1) Informational Note No. 2, the National Electrical Code recommends that voltage drop not exceed 3% on a branch circuit or feeder individually, and not exceed 5% total across the combined feeder and branch circuit for reasonable efficiency of operation.

Why does conduit material (PVC vs. Steel) affect AC voltage drop on large conductors?

Alternating current generates an alternating magnetic field around the conductors. Magnetic steel conduit increases inductive reactance (X_L) compared to nonmagnetic PVC or aluminum conduit. On large feeders (`1/0 AWG` and larger) with power factors below `1.0`, reactance becomes a substantial part of total impedance.

How do parallel conductor sets reduce voltage drop?

Running N identical parallel conductor sets per phase divides the current per conductor by N (or equivalently divides total circuit impedance by N), cutting the voltage drop in half for 2 parallel sets or to one-third for 3 parallel sets. Note that NEC 310.10(G) generally permits parallel conductors only in sizes 1/0 AWG and larger.