Single-phase and three-phase drop with the reactance term included, not just resistance. Set the power factor and the raceway and the calculator combines AC resistance and inductive reactance the way NEC Chapter 9 Table 9 says to.
Voltage drop calculator
1.39% is inside the 3% limit you set. About 1.39% of the 480 V supply is lost in the conductors under these conditions, leaving 473.34 V at the load. That is 6.66 V dropped across the run, line to line.
Your voltage drop appears here
Enter a supply voltage, a load current and a cable length. The result updates as you type.
One of the inputs needs attention. The figures below are the last complete calculation. Fix the highlighted field to update them.
Voltage dropWithin limit
1.39%
6.66 V lost against a3% limit
Supply
480.00V
At the load
473.34V
Lost as heat
1,275.8W
1.39% is inside the 3% limit you set. About 1.39% of the 480 V supply is lost in the conductors under these conditions, leaving 473.34 V at the load. That is 6.66 V dropped across the run, line to line.
Source480.00 V
350 ft one-way
At load473.34 V
3% limit
0%1%2%3%4%
Voltage falls from 480.00 volts at the source to 473.34 volts at the load, a drop of 1.39 percent against a limit of 3 percent.
Working, assumptions and sources
Conductor resistance
0.03274 Ω
Conductor reactance
0.03152 Ω
Impedance per 1000 ft
0.07329 Ω
Calculation details
Effective impedance per 1000 ft
Z_e = R × cos θ + X_L × sin θZ_e = 0.054 × 0.85 + 0.052 × 0.5268
Result: 0.07329 Ω per 1000 ft
R and X_L are for 250 kcmil copper in steel conduit, from NEC Chapter 9 Table 9. θ is the power factor angle.
Voltage drop
V_drop = √3 × I × Z_e × LV_drop = 1.732 × 150 A × 0.07329 Ω/1000 ft × 0.3500 (1000 ft)
Result: 6.66 V
A balanced three-phase circuit has no return conductor carrying full load current. The line-to-line drop is √3 times the drop along one conductor over the one-way length.
250 kcmil copper in steel conduit: AC resistance 0.052 Ω and reactance 0.052 Ω per 1000 ft. The last column is how much a resistance-only estimate understates the drop.
Effective impedance against power factor for 250 kcmil copper in steel conduit
Power factor
Effective Z
Resistance only
Understated by
1.00
0.054 Ω
0.054 Ω
—
0.95
0.06754 Ω
0.054 Ω
25.1%
0.90
0.07127 Ω
0.054 Ω
32.0%
0.85
0.07329 Ω
0.054 Ω
35.7%
0.80
0.0744 Ω
0.054 Ω
37.8%
0.70
0.07494 Ω
0.054 Ω
38.8%
Computed as Z_e = R cos θ + X_L sin θ from NEC Chapter 9, Table 9 values for 250 kcmil uncoated copper in steel conduit at 75 °C, 60 Hz.
A 350 ft feeder at 0.8 power factor calculated on resistance alone would come out roughly a third low. That is the difference between a run that passes a 3% target on paper and one that fails when it is commissioned. To calculate voltage drop for DC and AC circuits from one place, the main calculator handles all three system types.
Questions
Common questions.
When does conductor reactance actually matter?
When power factor is below about 0.9, when the conductors are large, or when they run in steel conduit, and most of all when all three are true at once. At unity power factor reactance contributes nothing to the drop. At 0.8 power factor on 250 kcmil copper in steel conduit, effective impedance is about 60% higher than the resistance-only figure, so a resistance-only calculation would tell you a run passes when it does not. On small conductors at good power factor the difference is a fraction of a percent and not worth chasing.
What is effective impedance and why is it not √(R² + X²)?
Effective impedance is Z_e = R × cos θ + X_L × sin θ, the projection of the conductor impedance onto the load-current direction. The magnitude √(R² + X²) would be right if you wanted the total impedance of the conductor, but voltage drop is about the component of that impedance in phase with the current. NEC Chapter 9 Table 9 states this formula in its notes and publishes an effective-Z column at 0.85 power factor computed with it.
Why does steel conduit increase voltage drop?
Steel is ferromagnetic, so it concentrates the magnetic field around the conductors. That raises inductive reactance, roughly 25% higher than in PVC, and adds eddy-current and hysteresis losses that show up as higher AC resistance. On 250 kcmil copper the reactance goes from 0.041 Ω per 1000 ft in PVC to 0.052 in steel. At poor power factor that difference is worth several percent of the total drop.
Does the three-phase result mean line-to-line or line-to-neutral?
Line-to-line. The √3 multiplier produces the drop between two phase conductors, which is what you compare against a line-to-line supply voltage like 480 V or 400 V. If you need the line-to-neutral drop on a balanced system, divide the result by √3, or equivalently calculate with the line-to-neutral voltage and a multiplier of 1. Mixing the two is a common way to get an answer that is out by 73%.
What frequency do these reactance values assume?
60 Hz. Inductive reactance is proportional to frequency, so at 50 Hz the reactance values are about 17% lower: multiply them by 50/60. Resistance is essentially unchanged at these frequencies for the conductor sizes involved. For anything above power frequency, skin effect becomes significant and this data no longer applies.