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Conductor Resistance — and the Reactance

The familiar K-constant voltage-drop formula models resistance only. That is exactly right at unity power factor — and increasingly wrong as conductors get large, because reactance starts to rival resistance. This tool uses Chapter 9 Table 8 resistance and Table 9 reactance, and computes the effective impedance that actually causes drop.

Calculate the run impedance

Set 1.0 for a resistive load — reactance then contributes nothing.

ft
0 ft2,000 ft

Voltage drop

2.72 V 0.57%

From the effective impedance, not R alone

Run resistance

0.00608 Ω

Run reactance

0.00510 Ω

See the breakdown
Table 8 resistance
Table 9 reactance
Over the run
Effective Z
Voltage drop
K-constant method
I²R heat

Reactance adds real drop at 0.85 power factor on a conductor this size.

The method, explained in plain English

# Resistance — Chapter 9, Table 8, corrected for temperature
R = R₇₅ × [1 + α(T − 75)] × length ÷ 1000 ÷ sets
α = 0.00323 copper · 0.00330 aluminum
# Reactance — Chapter 9, Table 9
X = X_table × length ÷ 1000 ÷ sets    # steel raceway is higher
# What actually causes drop
Ze = R·cos θ + X·sin θ    # NOT √(R² + X²)
VD = (1.732 or 2) × I × Ze
# The two limits worth remembering
PF = 1.0  → Ze = R exactly    # why the K formula works
bolted fault → use √(R² + X²)    # a different question

Table 8 reproduces the K constants

Multiply any Table 8 resistance by that size's circular mils and divide by 1000 and you get 12.9 for copper, 21.2 for aluminum — every size. The shortcut is the table, averaged.

Low power factor can help or hurt

On small conductors R dominates, so a lagging load slightly reduces drop. On large conductors X rivals R and it increases it. The crossover is near 1/0.

Hot conductors drop more

4/0 copper is 18% less resistive at 20 °C than at 75 °C. A cold-resistance calculation flatters the design; loaded conductors run hot.

I²R is why loose lugs burn

Half an ohm of contact resistance at 15 A dissipates 112 W inside a box. Nothing is overloaded — the connection is simply a heater.

Worked examples

The same 4/0 copper run at 200 A over 100 ft, three-phase 480 V, in steel conduit.

1

At 0.85 power factor — the defaults

R = 0.0608 × 100 ÷ 1000 = 0.00608 Ω · X = 0.051 × 100 ÷ 1000 = 0.00510 Ω
Ze = 0.00608 × 0.85 + 0.00510 × 0.527 = 0.00785 Ω
VD = 1.732 × 200 × 0.00785 = 2.72 V (0.57%)

Result: 29% more drop than resistance alone would give. On a 4/0 feeder in steel conduit the reactance is no longer a rounding error.

2

The same run at unity power factor

Ze = 0.00608 × 1.0 + 0.00510 × 0 = 0.00608 Ω — exactly R
VD = 1.732 × 200 × 0.00608 = 2.11 V (0.44%)
K-constant method: 1.732 × 12.9 × 200 × 100 ÷ 211,600 = 2.11 V

Result: the two methods agree to 0.3%. This is the proof that the K shortcut is a resistance-only model — and exactly right when the load is resistive.

3

500 kcmil, where reactance takes over

R = 0.0258 Ω/kFT · X = 0.048 Ω/kFT in steel — X is nearly 2× R
Ze at 0.85 PF = 0.0472 Ω/kFT, against 0.0258 from resistance alone

Result: 83% more drop than the K formula predicts. On large feeders serving motor loads, a resistance-only calculation is not conservative — it is simply wrong, and this is exactly why Chapter 9 Table 9 exists.

Chapter 9 Tables 8 and 9, per 1000 ft

Rendered from the same data the calculator uses. The K check column multiplies the copper resistance by the conductor's circular mils and divides by 1000. From 6 AWG up it lands between 12.84 and 12.90 — which is exactly where the familiar 12.9 constant comes from (21.2 for aluminum). The 8 through 14 AWG rows sit about 2% lower, so the single averaged constant is very slightly conservative on the smallest conductors.

Size R copper R aluminum X steel X PVC/Al Ze Cu, 0.85 PF K check
1 AWG 0.154 0.253 0.057 0.046 0.1609 12.89
2 AWG 0.194 0.319 0.057 0.045 0.1949 12.87
3 AWG 0.245 0.403 0.059 0.047 0.2393 12.89
4 AWG 0.308 0.508 0.06 0.048 0.2934 12.86
6 AWG 0.491 0.808 0.064 0.051 0.4511 12.88
8 AWG 0.764 1.26 0.065 0.052 0.6836 12.61
10 AWG 1.21 2 0.063 0.05 1.0617 12.56
12 AWG 1.93 3.18 0.068 0.054 1.6763 12.60
14 AWG 3.07 0.073 0.058 2.6480 12.62
1/0 AWG 0.122 0.201 0.055 0.044 0.1327 12.88
2/0 AWG 0.0967 0.159 0.054 0.043 0.1106 12.87
3/0 AWG 0.0766 0.126 0.052 0.042 0.0925 12.85
4/0 AWG 0.0608 0.1 0.051 0.041 0.0785 12.87
250 kcmil 0.0515 0.0847 0.052 0.041 0.0712 12.88
300 kcmil 0.0429 0.0707 0.051 0.041 0.0633 12.87
350 kcmil 0.0367 0.0605 0.05 0.04 0.0575 12.85
400 kcmil 0.0321 0.0529 0.049 0.04 0.0531 12.84
500 kcmil 0.0258 0.0424 0.048 0.039 0.0472 12.90
600 kcmil 0.0214 0.0353 0.048 0.039 0.0435 12.84

Sources & standards: NEC (NFPA 70) 2023 — Chapter 9 Table 8 (direct-current resistance at 75 °C, uncoated stranded conductors, with the temperature-correction note) and Chapter 9 Table 9 (alternating-current reactance for three single conductors in a raceway at 60 Hz). Effective impedance is computed as Ze = R·cos θ + X·sin θ from those two tables rather than read from Table 9's own effective-Z column, so the power factor is yours to set. Voltage drop targets of 3% and 5% appear only in Informational Notes to 210.19(A) and 215.2(A) and are not enforceable requirements. Local amendments override the model code, and the AHJ has final say.

Frequently asked questions

Common questions about conductor resistance, reactance and impedance.

Why not just use the K-constant voltage drop formula?

Because it models resistance only. At unity power factor that is exactly right — the effective impedance collapses to R and the two methods agree to a fraction of a percent. But on a lagging load, reactance contributes real drop, and above about 1/0 in a steel raceway the reactance is comparable to the resistance. On 500 kcmil copper in steel conduit the reactance is nearly twice the resistance, and the K formula simply cannot see it.

What is effective impedance and why is it not √(R² + X²)?

The impedance magnitude √(R² + X²) is the right quantity for a bolted fault, where current is limited only by impedance. For a normal load the drop along the line is Ze = R·cos θ + X·sin θ, because only the component of current in phase with the voltage produces IR drop and only the quadrature component produces IX drop. Using the magnitude overstates drop on a real load — sometimes substantially.

Does a low power factor always make voltage drop worse?

No, and this catches people out. Look at the formula: Ze = R·cos θ + X·sin θ. On a small, highly resistive conductor where R greatly exceeds X, a lagging power factor reduces line drop, because the R·cos θ term shrinks faster than the X·sin θ term grows. On a large conductor where X is comparable to R, it increases it. The crossover sits around 1/0.

Which resistance value does Table 8 give me?

DC resistance at 75 °C, in ohms per 1000 ft, listed separately for coated and uncoated copper and for aluminum, solid and stranded. This calculator uses the uncoated stranded values. A good sanity check: multiply the Table 8 resistance by the conductor's circular mils and divide by 1000, and you get back the familiar K constants — 12.9 for copper and 21.2 for aluminum, for every size in the table.

How do I correct resistance for temperature?

Chapter 9 Table 8's note gives R₂ = R₁[1 + α(T₂ − 75 °C)], with α of 0.00323 for copper and 0.00330 for aluminum. It is worth doing: 4/0 copper is 0.0608 Ω/kFT at 75 °C but only 0.0500 at 20 °C — 18% lower. Conductors run hot when loaded, so voltage drop at full load is genuinely worse than a cold-resistance calculation suggests.

Why does a steel raceway increase reactance?

Because steel is magnetic, so it concentrates the magnetic field around the conductors and raises their inductance. Chapter 9 Table 9 lists separate reactance columns for steel and for PVC or aluminum raceway — roughly 25% higher in steel. It only matters on large conductors, where the reactance is a meaningful share of the total impedance.

What is I²R heat useful for?

Two things. It is real energy lost as heat — on the default 200 A run it is about 730 W, dissipated continuously into the raceway and the building. And it is the reason a loose connection is dangerous: the same formula applied to a half-ohm of contact resistance at 15 A gives 112 W inside a device box, which is a small heater with nothing overloaded.

Feeder sized? Price the run.

Conductor, raceway, terminations, supports, and the labor to pull it — feeders are where material takeoffs get expensive. TradesQuote turns the scope into a detailed line-item estimate with quantities and totals.

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