The multiplier is different
The calculator uses two times one-way distance for DC or single-phase circuits and the square root of three times one-way distance for a balanced three-phase circuit. The three-phase input uses line-to-line system voltage and assumes a balanced load under the simplified resistance model.
Changing the selector is not a way to improve a result. It must match the actual circuit. An unbalanced system, motor, transformer, nonlinear load or circuit with meaningful reactance can require a more complete engineering method.
How do you calculate voltage drop?
Every metallic conductor has internal electrical resistance. As current flows through a run of copper or aluminum wire, part of the source voltage is lost as heat along the cable path. Using Ohm's law (V = I × R) and circular-mil resistivity at 75°C, the voltage lost across a two-wire circuit is:
where:
- VD – Total voltage drop between the supply panel and the load terminals, measured in volts (need a target gauge instead? Use the wire size calculator);
- K – Conductor resistance constant at 75°C (K = 12.9 for copper and K = 21.2 for aluminum in Ω·CM/ft);
- I – Load current flowing through the circuit in amps (estimate from wattage with our watts to amps calculator);
- L – One-way wire distance in feet (the multiplier of 2 already accounts for the return conductor); and
- CM – Conductor cross-sectional area in circular mils from NEC Chapter 9, Table 8.
Three-phase AC voltage drop equation
In a balanced three-phase system, the return currents cancel in phase opposition, so the round-trip factor of 2 is replaced by √3:
Subtract VD from the source voltage (V_load = V_source − VD) to find the actual voltage delivered to the equipment terminals. Explore all related tools in our electrical calculators hub.
Formula comparison
| Circuit | Simplified volts-lost formula | Voltage input | Load assumption |
|---|---|---|---|
| DC / single phase | Vd = 2 × K × I × L ÷ CM | System voltage | Single current path and return |
| Balanced three phase | Vd = √3 × K × I × L ÷ CM | Line-to-line voltage | Balanced phase currents |
Worked example: 480 V, 50 A, 200 ft, 6 AWG copper
Using the calculator's copper constant and 26,240 circular mils for 6 AWG, the simplified DC/single-phase result is about 9.83 V lost, or 2.05% of 480 V. With the balanced three-phase multiplier, the estimate is about 8.52 V, or 1.77%.
The difference comes from the circuit geometry in the formula. It does not prove that 6 AWG is suitable. Ampacity, continuous load, motor rules, terminals, correction factors, protection and the actual system still govern the installed conductor.
Inputs that deserve a second check
- 01
Line-to-line voltage
For this balanced three-phase model, enter the actual line-to-line system voltage.
- 02
Line current
Use the design current determined for the circuit, not total three-phase current added together.
- 03
One-way length
Enter source-to-load distance once; do not double it before entering the value.
- 04
Load behavior
Identify motors, starting current, power factor, harmonics and imbalance that the simple model omits.
- 05
Conductor conditions
Verify material, area, temperature, conductor count, terminals and installation method separately.
When to stop using the simplified result
- The three-phase load is materially unbalanced.
- Motor starting or equipment inrush is important to operation.
- Power factor or conductor reactance materially affects the circuit.
- Parallel conductors, transformers or multiple loads occur along the run.
- The system voltage or phase configuration is uncertain.
- The result will be used for construction, permitting or equipment procurement without professional review.