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Voltage Drop Calculator

Calculate voltage drop, percentage loss, and remaining load voltage for DC, single-phase, and 3-phase circuits over distance.

Calculates AC Effective Impedance (Z) with conduit reactance per NEC Ch. 9 Table 9.

V
A

Typical: 0.85 motors, 0.95 lighting, 1.0 resistive

Voltage Drop Analysis

Voltage Drop (ΔV)
5.185 V (4.32%)

⚠ Acceptable for total feeder + branch circuit (≤ 5%)

Voltage at Load Terminal
114.81 V

From 120V supply under 15A load

Power Dissipated in Conductors
90 W

Conductor thermal I²R energy loss

Effective Loop Impedance (Z)
0.34569 Ω

Includes conduit magnetic reactance

Step-by-Step Engineering Calculation

  1. 1Method: NEC Chapter 9 Table 9 (AC Effective Impedance at 75°C, 60Hz)
  2. 2Conduit: STEEL Raceway | Conductor: COPPER 12 AWG
  3. 3Code Resistance: R=2 Ω/kftR = 2\ \Omega/\text{kft} | Reactance: XL=0.054 Ω/kftX_L = 0.054\ \Omega/\text{kft}
  4. 4Power Factor: PF=cos⁡θ=0.85\text{PF} = \cos\theta = 0.85 (sin⁡θ=0.5268\sin\theta = 0.5268)
  5. 5Effective Impedance: Zeff=(2×0.85)+(0.054×0.5268)=1.7284 Ω/kftZ_{\text{eff}} = (2 \times 0.85) + (0.054 \times 0.5268) = \mathbf{1.7284\ \Omega/\text{kft}}
  6. 6Circuit Length: L=100.0 ftL = 100.0\text{ ft} (30.5 m) | Parallel sets: 1
  7. 7Voltage Drop: ΔV=2×15 A×1.7284 Ω/kft×100.0 ft1000×1=5.185 V\Delta V = 2 \times 15\text{ A} \times \frac{1.7284\ \Omega/\text{kft} \times 100.0\text{ ft}}{1000 \times 1} = \mathbf{5.185\text{ V}}
  8. 8Voltage Drop Percentage: 5.185 V120 V×100=4.32%\frac{5.185\text{ V}}{120\text{ V}} \times 100 = \mathbf{4.32\%}
  9. 9Voltage at Terminal Load: Vload=120 V−5.185 V=114.815 VV_{\text{load}} = 120\text{ V} - 5.185\text{ V} = \mathbf{114.815\text{ V}}
  10. 10Total Conductor Thermal Power Loss: Ploss=90.00 WP_{\text{loss}} = \mathbf{90.00\text{ W}}

Formula & Theory

ΔV_1φ = 2 × I × (ρ × L / A) | ΔV_3φ = √3 × I × (ρ × L / A)

Calculates the voltage reduction along a conductor run caused by the internal electrical resistance of copper or aluminum wiring.

Variable Definitions
ΔV:
Voltage Drop (Volts (V)) – Voltage lost across the conductor run
%ΔV:
Voltage Drop % (Percentage (%)) – Proportion of source voltage lost
L:
One-Way Length (Meters / Feet) – Distance between power source and load
ρ:
Resistivity (Ω·m) – Conductor resistivity at operating temperature

Worked Engineering Example

100-Foot 120V Branch Circuit

A 120V 15A load powered over 100 feet of 12 AWG copper wire at 20°C.

Inputs:
  • Voltage: 120 V
  • Current: 15 A
  • Length: 100 ft
  • Wire: 12 AWG Copper
Result: Voltage Drop = 4.76 V (3.97%), Load Voltage = 115.24 V

Key Assumptions

  • •Conductor resistance is calculated at specified temperature (standard 20°C / 75°C).
  • •Power factor is assumed to be 1.0 (purely resistive) for DC and basic AC calculations.

Limitations & Safety Scope

  • •Inductive cable reactance (X_L) becomes significant in large conductors (> 4/0 AWG) or steel conduit runs and should be factored in for large commercial feeders.
  • •Does not guarantee regulatory electrical code compliance. Always verify with applicable local codes (NEC, IEC, CEC).

Frequently Asked Questions

What is the maximum allowable voltage drop under the NEC?

The NEC recommends a maximum 3% voltage drop on branch circuits and 5% total voltage drop combined across feeder and branch circuits for optimal efficiency.

Why does copper have less voltage drop than aluminum?

Copper has a lower electrical resistivity (1.724 × 10⁻⁸ Ω·m) compared to aluminum (2.82 × 10⁻⁸ Ω·m), meaning an aluminum conductor must have roughly 61% greater cross-sectional area to match the resistance of copper.

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Educational & Reference Notice: This calculator provides technical estimates based on standard physical formulas. Actual electrical installation, protective device sizing, and conductor selection must be verified against the applicable electrical code (NEC, IEC, CEC) by a licensed electrician or qualified professional engineer.