Voltage Drop

Compute single-phase and three-phase conductor voltage drop in volts and percent, the load-end voltage, and pass/fail checks against NEC and IEC 60364 limits, with a drop-versus-length plot.


NEC (NFPA 70) & IEC 60364

System

Phase
Conductor material

Circuit

m
A
V

Results

3.61V
Voltage drop
0.90%
Drop (% of V)
396.39V
End voltage
0.4630Ω/km
R used
NEC branch ≤ 3%PASS
NEC total ≤ 5%PASS
IEC 60364 ≤ 4%PASS

Drop vs. Run Length

0255075100Length (m)00.511.52%

About Voltage Drop Calculator

The voltage drop calculator determines how much voltage is lost along an electrical conductor carrying a given load current over a given distance. Excessive voltage drop causes dim lighting, nuisance equipment tripping, motor overheating, and wasted energy, so codes recommend keeping it within a few percent of the nominal supply voltage.

Choose single-phase or three-phase, pick a copper or aluminium conductor from the built-in resistance table (metric IEC 60228 and North-American AWG/kcmil sizes), and enter the run length, load current, power factor, and system voltage. The tool reports the drop in volts and percent, the voltage delivered at the load, and whether the circuit satisfies the common NEC and IEC limits.

How It Works

  1. Select the phase configuration (single or three) and the conductor material (copper or aluminium).
  2. Pick a conductor size; the tool looks up its AC resistance (and reactance) per kilometre from the built-in table.
  3. Enter the one-way run length (m), the load current (A), the system voltage (V), and the power factor.
  4. Optionally include conductor reactance and parallel conductors per phase; the calculator returns the drop, the end voltage, and PASS/EXCEEDS flags against the NEC and IEC limits.

Worked Example

A three-phase feeder uses 50 mm^2 copper conductors with an AC resistance of 0.463 ohm/km. It carries 100 A over a one-way run of 60 m at unity power factor on a 400 V system. The line voltage drop is Vd = sqrt(3) * I * L * R / 1000 = 1.732 * 100 * 60 * 0.463 / 1000 = 4.81 V, which is 4.81 / 400 = 1.20%. The voltage at the load is 400 - 4.81 = 395.2 V, comfortably within the NEC 3% branch and IEC 60364 4% limits.

Formulas

Single-phase voltage drop
Vd = 2 * I * L * (R*cos(phi) + X*sin(phi)) / 1000
Three-phase voltage drop
Vd = sqrt(3) * I * L * (R*cos(phi) + X*sin(phi)) / 1000
Resistive simplification (pf = 1, X = 0)
Vd_1ph = 2*L*I*R/1000 | Vd_3ph = sqrt(3)*L*I*R/1000
Drop percent and end voltage
Vd% = 100 * Vd / V_system | V_end = V_system - Vd

Standards & References

  • NEC (NFPA 70) 210.19(A) & 215.2(A) FPN — 3% branch, 5% total recommendation
  • IEC 60364-5-52 — 4% typical voltage drop limit
  • NEC Chapter 9, Table 8/9 — conductor resistance and reactance
  • IEC 60228 — standard conductor cross-sections

Frequently Asked Questions

Why does the single-phase formula use 2 and the three-phase formula use sqrt(3)?

Single-phase current flows out along the line conductor and back along the neutral, so the drop occurs over two conductor lengths, giving the factor of 2. In a balanced three-phase system the return currents cancel, and the line-to-line drop works out to sqrt(3) (about 1.732) times the per-conductor drop.

What voltage drop limit should I design to?

The NEC recommends a maximum of 3% on a branch circuit and 5% on the combined feeder plus branch. IEC 60364 commonly uses 4% for general circuits. These are recommendations for efficiency and equipment performance, not hard safety limits, but most specifications enforce them.

When should I include conductor reactance?

Reactance matters for larger conductors, longer runs, and lower power factors. At unity power factor the reactive term drops out. For lagging loads (motors, transformers) enable the reactance option so the drop is computed as R·cosφ + X·sinφ, which is more accurate than resistance alone.

Does the calculator account for temperature and parallel conductors?

The table uses AC resistance at about 75°C, a common design temperature. You can also specify several conductors per phase in parallel, which divides the effective impedance and reduces the voltage drop proportionally.