Every cable is a resistor you paid for, and it bills you every hour it carries current.
Enter the load and the run, get the I²R loss, the voltage drop, and what the wasted
kilowatt-hours cost per year — with conductor resistance corrected to operating temperature.
Runs entirely in your browser.
How it works
DC & 1-phase: I = P / (V·pf) · loss = 2·I²·r·L (out + return)
3-phase: I = P / (√3·V·pf) · loss = 3·I²·r·L (three conductors)
r = ρ₂₀/A × (1 + α·(T − 20)) Ω/m — same resistance model as the size converter
Annual cost = loss × hours/1000 × price
L is the one-way run length; the loop factor (2 or 3) is applied automatically.
Resistance is corrected to the conductor temperature you set — the same loaded cable that
carries the loss is the one being measured, not a cold 20 °C table row.
Worked example
11 kW, three-phase 400 V, pf 0.95, 120 m one-way of 16 mm² copper at 70 °C:
I ≈ 16.7 A, r ≈ 1.29 Ω/km, loss = 3 × 16.7² × 0.00129 × 120 ≈ 130 W —
about 1.2 % of the load, 519 kWh a year at 4000 h, ≈ $78/yr at $0.15/kWh.
Step the same run up to 25 mm² and the loss falls to 83 W — saving about $28 a year, so
the 25 mm² premium pays for itself in a few years of continuous duty. Whether that is worth
it is now a two-number comparison instead of a guess.
Limitations
- DC resistance model — no skin/proximity correction. At power frequency this matters only above roughly 25 mm²; treat big-feeders results as slightly optimistic.
- Current assumed steady at the entered value. Real duty cycles average out; enter the mean load, not the nameplate, for energy cost.
- No reactive-power or harmonic losses, no neutral loss in unbalanced single-phase loads.
- Ampacity (can this cable safely carry the current?) is a code-table question — not covered here.
Why does halving the voltage quadruple the loss?
Loss is I²R. Same power at half the voltage means double the current, and double current squared is four times the loss. Long low-voltage runs are loss burners; the same power at a higher voltage is almost free.
Is voltage drop the same as power loss?
No. Drop is impedance arithmetic; loss is I²R in the copper. On pf≈1 circuits they track closely; with motors or inverters the loss fraction is usually lower than the drop fraction. Both are shown — they answer different questions.
What temperature should I enter?
The temperature the conductor reaches under load. If you don't know, 70 °C is a fair default for loaded feeders; the worst legal case for 90 °C insulation is 90 °C.
Does it work for solar DC strings?
Yes — pick DC, enter Vmp as the voltage and string current as implied by the power field. Remember a PV string only produces its share of hours per day; the annual energy figure scales accordingly.