AC Line Loss Calculator

I²R power loss in a feeder or transmission line, and loss as a percentage of transmitted power.
Loss from Current & Resistance
Loss Percentage vs Transmitted Power

I²R Line Loss

Ploss = n×I2×R  (n=2 for single-phase, n=3 for three-phase conductors)
50A, 0.5Ω/conductor, 1Φ
100A, 0.2Ω/conductor, 3Φ
20A, 1.2Ω/conductor, 1Φ
A
Enter values and press Calculate.

Loss Percentage vs Transmitted Power

Pxmit = V×I×PF (1Φ) or √3×VLL×I×PF (3Φ)  •  Loss% = Ploss/Pxmit×100
230V, 50A, PF0.9, 0.5Ω, 1Φ
400V, 100A, PF0.85, 0.2Ω, 3Φ
11kV feeder, 30A, 3Φ
V
A
Enter values and press Calculate.

How I²R Line Loss Works

Every conductor has resistance, so any current flowing through it dissipates power as heat, following P=I²R. In a real feeder or transmission line, this loss occurs in every current-carrying conductor: a single-phase circuit has an outgoing and return conductor (loss counted twice), while a balanced three-phase circuit has three current-carrying conductors (loss counted three times, with no current in a balanced neutral). Because loss scales with the square of current, doubling the load current quadruples the line loss — which is exactly why power is transmitted at high voltage (low current) over long distances.

QuantityFormula
Single-phase line lossPloss = 2×I²×R
Three-phase line lossPloss = 3×I²×R
Single-phase transmitted powerPxmit = V×I×PF
Three-phase transmitted powerPxmit = √3×VLL×I×PF
Loss percentageLoss% = Ploss/Pxmit × 100

Conductor resistance R depends on wire gauge, material (copper/aluminium), length, and temperature; see our Wire Gauge / Ampacity Calculator for typical resistance-per-length values.

Real-World Applications & Examples

Worked examples

1. Single-phase feeder. I=50A, R=0.5Ω/conductor: Ploss=2×50²×0.5=2500 W (2.5 kW dissipated as heat in the cable).
2. Three-phase feeder. I=100A, R=0.2Ω/conductor: Ploss=3×100²×0.2=6000 W (6 kW).
3. Small single-phase branch circuit. I=20A, R=1.2Ω/conductor: Ploss=2×20²×1.2=960 W — a significant loss for a lightly loaded, high-resistance (long/thin) run.
4. Loss percentage, single-phase. V=230V, I=50A, PF=0.9, R=0.5Ω: Pxmit=230×50×0.9=10.35kW, Ploss=2.5kW → Loss%=2500/10350≈24.2% — unacceptably high, indicating the conductor is undersized.
5. Loss percentage, three-phase. VLL=400V, I=100A, PF=0.85, R=0.2Ω: Pxmit=√3×400×100×0.85≈58.9kW, Ploss=6kW → Loss%≈10.2% — still high, but far better than example 4's undersized single-phase run.
6. Medium-voltage feeder. An 11kV three-phase feeder at 30A with 2Ω/conductor: Ploss=3×30²×2=5.4kW against Pxmit=√3×11000×30×0.95≈543kW, giving Loss%≈1% — illustrating why higher transmission voltages drastically cut relative losses for the same power delivered.

Frequently Asked Questions

What is I²R loss?

The power dissipated as heat in a conductor due to its resistance, given by P=I²R. It represents real energy wasted (not delivered to the load) in every current-carrying wire.

Why does line loss quadruple when current doubles?

Because loss is proportional to the square of current (P=I²R); doubling current means the loss term increases by 2²=4 times, all else being equal.

Why is power transmitted at high voltage?

For a fixed power P=V×I, raising voltage lowers the required current proportionally; since loss depends on I², even a modest voltage increase dramatically cuts transmission losses.

Why is single-phase loss "2×I²R" but three-phase is "3×I²R"?

A single-phase circuit has two current-carrying conductors (line and neutral/return); a three-phase circuit has three line conductors, each carrying the same current in a balanced system, so the I²R loss is counted once per current-carrying conductor.

Does the neutral conductor add loss in a balanced three-phase system?

In a perfectly balanced three-phase system, the neutral carries zero current, so it contributes no additional I²R loss; unbalanced loads do cause some neutral current and loss.

What is a typical acceptable line loss percentage?

Utility transmission lines often target well under 5% loss; building/facility feeders are commonly designed for 1–3% voltage drop (a related but distinct metric), with correspondingly low I²R loss percentages.

How does conductor resistance R depend on temperature?

Resistance increases with conductor temperature (roughly 0.4%/°C for copper); a heavily loaded, hot conductor has measurably higher resistance and thus higher loss than the same conductor at room temperature.

How do I reduce line loss?

Use a larger conductor (lower R), shorten the run, increase system voltage (lowering I for the same power), improve power factor (lowering current for the same real power), or balance three-phase loads to minimize neutral current.

Does power factor affect I²R loss?

Indirectly, yes: a lower power factor means more current is needed to deliver the same real power, and since loss depends on I², poor power factor directly increases I²R losses for a given real power delivered.

What is the difference between line loss and voltage drop?

Line loss (P=I²R) is the power wasted as heat; voltage drop (V=I×R) is the reduction in voltage seen at the load end. Both come from the same conductor resistance but are different metrics — voltage drop affects equipment performance, while loss affects energy cost.

How do I find R for a real cable?

Use the cable's resistance-per-length rating (from datasheets or standard wire tables) multiplied by the one-way conductor length; see our Wire Gauge / Ampacity Calculator for typical copper and aluminum values.

Is DC line loss calculated the same way?

Yes, DC line loss uses the same P=I²R per conductor (typically 2×I²R for a two-wire DC circuit), without the power-factor or three-phase considerations that apply to AC.

Why is loss percentage important, not just absolute loss?

Absolute loss (watts) tells you the wasted energy, but loss percentage relative to transmitted power tells you the relative efficiency of the feeder — a small loss can still be a large percentage on a lightly loaded, undersized circuit.

Can line loss cause a conductor to overheat?

Yes — I²R loss is exactly the heat generated in the conductor; if it exceeds what the cable's insulation and installation method can safely dissipate, the conductor overheats, which is why ampacity tables (not just loss calculations) set the maximum allowed current.

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