Voltage drop calculator

Feeder voltage drop from current, length, conductor R and X, and power factor.

Circuit
V
A
Conductor impedance, per conductor
Ω/km
Ω/km
Estimate R from the conductor size
°C

DC resistance: 0.08276 Ω/km

DC resistance only. At 60 Hz, large conductors have a higher AC resistance from skin effect; use the manufacturer's AC value when you have it.

Result

Voltage drop
3.674V
Of nominal
0.919%
R over the length
0.02Ω
X over the length
0.008Ω
Load current lagging the receiving-end voltage by 31.8 degreesV = 1.00∠0.0°. I = 0.70∠−31.8°.φVI

The formula

The drop along a feeder is the component of the I·Z drop that lies along the receiving-end voltage, where Z is the conductor impedance. With R and X per unit length of one conductor and L the one-way length:

ΔV1ϕ=2 I L (Rcos⁡φ+Xsin⁡φ)\Delta V_{1\phi} = 2\,I\,L\,(R\cos\varphi + X\sin\varphi)
ΔV3ϕ=3 I L (Rcos⁡φ+Xsin⁡φ)\Delta V_{3\phi} = \sqrt{3}\,I\,L\,(R\cos\varphi + X\sin\varphi)

The single-phase formula counts the conductor out and the conductor back, hence the 2. The three-phase formula gives the line-to-line drop of a balanced load, hence the square root of 3. Both leave out the small component of the drop at right angles to the voltage, which is why they are called approximate; for drops of a few percent the error is negligible.

Getting R

Use the resistance from the cable manufacturer or from the impedance table of the code edition you design to. If you only know the conductor size, the helper estimates DC resistance from the resistivity at 20 °C, corrected to the operating temperature:

RT=ρ20 (1+α (T−20))AR_T = \dfrac{\rho_{20}\,\bigl(1 + \alpha\,(T - 20)\bigr)}{A}

It uses 0.017241 Ω·mm²/m and 0.00393 per °C for copper, 0.028264 Ω·mm²/m and 0.00403 per °C for aluminum, and 1 kcmil = 0.506707 mm².

Worked example

A balanced three-phase load draws 100 A at 0.85 power factor, lagging, from a 400 V source through 100 m of cable with R = 0.2 Ω/km and X = 0.08 Ω/km. The length is 0.1 km, cos φ = 0.85 and sin φ = 0.5268:

ΔV=3×100×0.1×(0.2×0.85+0.08×0.5268)=3.674 V\Delta V = \sqrt{3} \times 100 \times 0.1 \times (0.2 \times 0.85 + 0.08 \times 0.5268) = 3.674\ \text{V}
3.674400×100%=0.919%\dfrac{3.674}{400} \times 100\% = 0.919\%

Switch the power factor to leading and the reactive term changes sign: the drop falls to 2.215 V. The result turns negative, a voltage rise toward the load, only when the load leads and X sin φ is larger than R cos φ: a strongly leading load on a cable whose reactance is large next to its resistance.

Questions

Is this the exact voltage drop?
It is the standard approximate formula, which leaves out the small quadrature component of the drop. For drops of a few percent the difference from the exact phasor solution is negligible.
Why does power factor change the voltage drop?
The drop is I times (R cos φ + X sin φ). At a lagging power factor the reactance term adds to the drop. At a leading power factor it subtracts, and on a reactive line the voltage can even rise toward the load.
Where do the R and X values come from?
From the cable manufacturer's data or the impedance table in the code edition you design to, for your conductor, raceway and temperature. The helper on this page estimates DC resistance from the conductor's area and temperature; it leaves out skin effect, which matters for large conductors at 60 Hz.

For learning, not for design

PhasorPrep is an independent study resource. It is not affiliated with, endorsed by, or sponsored by NCEES. Questions are original and written for practice; they are not actual exam questions. Calculators are for learning, not for engineering design.

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