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.
| Quantity | Formula |
|---|---|
| Single-phase line loss | Ploss = 2×I²×R |
| Three-phase line loss | Ploss = 3×I²×R |
| Single-phase transmitted power | Pxmit = V×I×PF |
| Three-phase transmitted power | Pxmit = √3×VLL×I×PF |
| Loss percentage | Loss% = 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.
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.
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.
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.
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.
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.
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.
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.
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.
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.
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.
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.
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.
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.
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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