NEC Voltage Drop Calculator
AWG and kcmil conductors. Uses effective impedance at your actual power factor, not a fixed 0.85.
120, 240 or 277
Actual load, not the breaker size
Board to load, one direction
Steel raises reactance
1.0 for heating and resistive loads
Advanced
Corrected value per conductor
Table 8 is published at 75 °C. Battery and solar runs often sit cooler.
- Drop
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- Voltage at load
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- Effective Z
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- Power lost
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AC results use effective impedance, Z = R × PF + XL × sin(arccos PF), from the 600 V table values at 75 °C. The 3% and 5% figures are recommendations in the Code, not enforceable limits. Ampacity, overcurrent protection and terminal ratings are separate checks.
How the calculation works
The NEC gives you two sets of conductor data, and which one you use depends on the circuit.
Chapter 9, Table 9 covers alternating current. It lists resistance and inductive reactance for 600 V conductors at 75 °C, broken out by raceway. Chapter 9, Table 8 covers direct current. It lists plain DC resistance at 75 °C with no reactance at all, because DC has none.
This calculator picks the right table when you choose the circuit type. Everything below explains what happens after that.
Effective impedance
On AC, resistance alone undersizes the answer. The magnetic field around the conductors opposes current too, and how much it opposes depends on the phase angle of the load. The Code handles this with effective impedance:
Z = R × PF + X_L × sin(arccos PF)
That single number replaces resistance in the drop calculation. On a resistive load at unity power factor the reactance term vanishes and you are back to plain resistance. On a motor at 0.8, reactance can add a fifth to the total.
Why 0.85 shows up everywhere
Table 9 publishes a ready-made effective Z column, but only at 0.85 power factor. Plenty of calculators just read that column and hand you the number, whatever your load actually is.
This one recalculates Z from the raw R and X_L values for the power factor you enter. On a 0.7 lagging motor feeder the difference is worth several percent of the result — enough to change what cable you buy.
The raceway changes the answer
Table 9 has separate columns for PVC, aluminum conduit, and steel conduit, and they are not the same. Steel raises reactance by roughly a quarter. Aluminum conduit raises resistance slightly through induced currents in the raceway wall.
On small conductors the difference is buried in rounding. On a 500 kcmil feeder in steel it is real money. The raceway selector picks the correct column instead of assuming PVC.
The multiplier
Once you have Z per 1000 feet, the rest is arithmetic:
Vd = multiplier × Z × I × (L ÷ 1000)
The multiplier is √3 for three-phase and 2 for single-phase or DC. Length is one way. The 2 accounts for the return conductor; balanced three-phase does not need one.
Percent is that drop over the system voltage as entered. For a 480 V three-phase feeder the drop is compared to 480 V, not 277 V. That is NEC convention, and it is the opposite of the metric approach, which references line to neutral. Two correct methods that give different percentages for the same cable.
3% and 5% are recommendations, not rules
This surprises people. The 3% figure for branch circuits and the 5% combined figure for feeder plus branch appear in the Code as informational notes. Informational notes are explanatory. They are not enforceable requirements, and no inspector can red-tag a circuit for exceeding them on voltage drop alone.
Specific rules elsewhere in the Code do mandate voltage drop calculations for certain installations, and plenty of design specs and utility agreements make 3% contractual. But the general branch-circuit figure is guidance. Treat it as a design target you should have a reason to exceed, not a pass-fail line.
DC runs and temperature
Table 8 is published at 75 °C. A battery bank in a cool room or a solar string on a mild day is nowhere near that, and a cooler conductor has less resistance.
The correction is straightforward:
R = R₇₅ × [1 + α × (T − 75)]
α is 0.00323 for copper and 0.00330 for aluminum. On a 6 AWG copper run at 30 °C, the drop comes out at 1.40% instead of 1.64%. The calculator applies this in DC mode, where the assumption matters most.
A worked example
250 kcmil copper in PVC, 300 ft, 200 A, 480 V three-phase, power factor 0.9.
From the table: R = 0.052 Ω/kft, X_L = 0.041 Ω/kft. sin(arccos 0.9) = 0.436, so Z = (0.052 × 0.9) + (0.041 × 0.436) = 0.0647 Ω/kft. Vd = 1.732 × 0.0647 × 200 × 0.300 = 6.72 V. Against 480 V that is 1.40%. Well inside a 3% target.
Parallel conductors
Sets in parallel split the current, so the drop falls roughly in proportion. Two 4/0 runs carry half the current each and drop about half what one would.
That only holds if the sets are identical: same size, same material, same insulation, same length, terminated the same way at both ends. Mismatch any of those and the current divides unevenly, one set runs hot, and the calculation stops describing reality.
What voltage drop does not tell you
A conductor that passes this check can still be wrong. It has to carry the current after correction for ambient temperature and conductor count. The overcurrent device has to clear a fault at the far end. The terminals at both ends have to be rated for the temperature column you sized from — a 90 °C conductor landing on 75 °C lugs is a 75 °C circuit.
Long runs usually fail voltage drop first. Short runs usually fail ampacity first. Both have to pass.
FAQ
Is the 3% voltage drop limit required by the NEC?
Not as a general rule. It appears as an informational note, which is explanatory text rather than an enforceable requirement. Certain installations do have mandatory calculation rules, and many design specs and utility agreements make 3% contractual, so check your project documents. As a design target it is sound practice either way.
Why does Table 9 assume 0.85 power factor?
It is a reasonable average for a mixed commercial load, and publishing one effective Z column is simpler than publishing a dozen. The table also gives raw resistance and reactance so you can work out effective impedance for any power factor, which is what this calculator does.
Does the conduit type really change the answer?
Yes, more than most people expect. Steel raises inductive reactance by about a quarter compared with PVC because the steel concentrates the magnetic field. Aluminum conduit adds a little resistance through induced currents in the raceway. On 14 AWG it is lost in the rounding. On 500 kcmil it changes the conductor you buy.
Should I use Table 8 or Table 9?
Table 9 for anything alternating current, because it includes reactance. Table 8 for DC circuits, where reactance does not exist. Using Table 8 for AC ignores reactance entirely and gives you a drop figure that is too low.
Why is the percentage based on 480 V and not 277 V?
Because that is how the NEC states it. The drop is compared against the nominal system voltage as it appears on the one-line. Metric practice references line-to-neutral instead, so the same cable shows a higher percentage under the IEC method. Neither is wrong. Just do not compare a number from one method against a limit from the other.
Do parallel conductors change the calculation?
They divide the current between the sets, so the drop drops roughly in proportion. The catch is that the sets have to be identical in size, material, insulation and length, and terminated the same way at both ends. Unequal sets do not share current equally and the math stops holding.