Calculating Joule losses in an electrical cable
Estimate in seconds the power dissipated by the Joule effect (P = R·I²), the energy lost over the year and the corresponding cost in CHF for a copper or aluminium cable. Enter R directly, or let the calculation derive R from length, cross-section and resistivity — the convention one-way length is respected and the out-and-return path is included automatically.
What is the Joule effect?
L'Joule effect, demonstrated by James Prescott Joule in 1841, reflects the irreversible conversion of electrical energy into heat when a current flows through a resistive conductor. Every real cable has a non-zero resistance R that opposes the current flow and dissipates a power P = R·I² as heat. This energy is permanently lost to the downstream load: it merely heats the insulation, the cable tray and the ambient air, and is the invisible line item on electricity bills.
In Swiss low-voltage installations, Joule losses typically represent 1 à 4 % of the energy distributed in the building. In a large commercial building supplied at 400 V three-phase, this can reach several MWh per year. The NIBT 2020 chapter 5.2 requires sizing that guarantees Iz >= I, and thus already caps heating at a safe threshold (70 °C for PVC, 90 °C for XLPE), but does not minimise economic losses. The designer has strong technical latitude here: going one cross-section size above the normative minimum is often profitable over the service life of the installation (30 to 40 years).
Detailed formulas
The Joule loss calculation rests on three inseparable equations. The dissipated power instantaneous value equals P = R·I², expressed in watts. The quadratic form is fundamental: doubling the current quadruples the losses for the same cable. On a real cable, the total circuit resistance (out + return for single-phase, out only for balanced three-phase) follows Pouillet's law R = ρ·L/S, where ρ is the material resistivity (0.0224 Ω·mm²/m for copper at 20 °C, 0.036 for aluminium), L the length in metres and S the cross-section in mm².
Length convention used in this tool. The "Length L" field expects the one-way length of the cable (physical length of the run, as measured on the plan). The out-and-back path is then built into the calculation: we use R = 2·ρ·L/S, where the factor 2 represents the return conductor; R is the effective loop resistance to plug directly into P = R·I² to obtain the cable's total Joule losses. This is the most intuitive convention for the designer — no need to double the length manually, the tool does it for you.
L'energy lost over a usage period h is written E = P·h in Wh, i.e. E = P·h/1000 in kWh. The annual cost follows directly: Cost = E · pricekWh. With an average price of 0.22 CHF/kWh in Switzerland (Romande Energie, Groupe E, SIG tariffs, 2026 basis), a continuous loss of 100 W costs 193 CHF/year. The voltage drop associated is ΔU = R·I in single-phase, or ΔU = √3·R·I/2 in balanced three-phase. The transmission efficiency of the run is written η = 1 - P/(U·I) = 1 - R·I/U.
Important: ρ increases with temperature. At 70 °C (nominal operating point of a PVC cable loaded at Iz), ρcopper rises from 0.0224 to about 0.0273 Ω·mm²/m, i.e. +22 % more losses compared with the cold calculation. The temperature slider above automatically applies the correction ρ(T) = ρ20 · (1 + α · (T - 20)).
Typical losses by cross-section and length
The table below gives the Joule losses of a single-phase copper cable continuously loaded at its nominal current for 2000 h/year, for various lengths one-way and cross-sections (out-and-return losses included, R = 2·ρ·L/S). Values computed at ρ = 0.0224 Ω·mm²/m, price 0.22 CHF/kWh. Multiply by 2 for 4000 h/year, by 4 for continuous operation.
| Cross-section | Current I | L = 20 m | L = 50 m | L = 100 m | L = 200 m |
|---|---|---|---|---|---|
| 1,5 mm² | 10 A | 60 W / 26 CHF | 149 W / 66 CHF | 299 W / 131 CHF | 597 W / 263 CHF |
| 2,5 mm² | 16 A | 92 W / 40 CHF | 229 W / 101 CHF | 459 W / 202 CHF | 917 W / 404 CHF |
| 2,5 mm² | 32 A | 367 W / 161 CHF | 917 W / 404 CHF | 1835 W / 807 CHF | 3670 W / 1615 CHF |
| 4 mm² | 25 A | 140 W / 62 CHF | 350 W / 154 CHF | 700 W / 308 CHF | 1400 W / 616 CHF |
| 4 mm² | 32 A | 229 W / 101 CHF | 573 W / 252 CHF | 1147 W / 505 CHF | 2294 W / 1009 CHF |
| 6 mm² | 40 A | 239 W / 105 CHF | 597 W / 263 CHF | 1195 W / 526 CHF | 2389 W / 1051 CHF |
| 10 mm² | 63 A | 356 W / 156 CHF | 889 W / 391 CHF | 1778 W / 782 CHF | 3556 W / 1565 CHF |
| 16 mm² | 80 A | 358 W / 158 CHF | 896 W / 394 CHF | 1792 W / 789 CHF | 3584 W / 1577 CHF |
| 25 mm² | 100 A | 358 W / 158 CHF | 896 W / 394 CHF | 1792 W / 789 CHF | 3584 W / 1577 CHF |
Key takeaway: on a 50 m feeder loaded at 32 A, moving from 2.5 to 4 mm² cuts the losses from 917 W to 573 W, i.e. 151 CHF/year saved for an additional material cost of around 30 to 40 CHF. Payback period of a few months, then pure savings over a 30-year service life. The logic applies a fortiori to long runs or to a heat pump running 4000 to 5000 h/year.
How to reduce Joule losses
Five technical levers are available to a designer, ranked by economic efficiency:
1. Increase the cross-section
It is the dominant lever: R varies as 1/S, so going from 2.5 to 4 mm² cuts losses by 37 %, and from 4 to 6 mm² by a further 33 %. The golden rule : on any run longer than 30 m and loaded above 50 % of its Iz, check the economic calculation for the next cross-section up.
2. Shorten the cable
Since R is proportional to L, optimising the routing and placing sub-distribution boards at the load barycentre mechanically divides the losses. A well-placed floor sub-board can divide the cumulative length of socket and lighting feeders by 3. This lever is often forgotten at preliminary/detailed design stage because it depends on the architecture, not on the sizing.
3. Balance the phases
In three-phase, an imbalance drives current through the neutral, producing Joule losses with no benefit. Distributing single-phase loads across L1/L2/L3 brings the neutral current close to 0. The NIBT 5.5.3 requires neutral sizing suited to unbalanced and harmonic loads.
4. Limit harmonics
Harmonic currents (orders 3, 5, 7) increase the effective RMS current and therefore the losses (P = R·I²rms). In a commercial building with many switched-mode power supplies (LED, IT, drives), provide passive filters or an oversized neutral. See IEC 61000-3-2 and NIBT 5.5.
5. Switch to solid aluminium
Contre-intuitif mais valide sur les départs principaux longs : à coût matière équivalent, l'aluminium permet une section géométrique 1,6× plus grande, donc des pertes équivalentes ou inférieures au cuivre, avec un poids divisé par 2. Pratique sur les colonnes montantes des immeubles collectifs et les liaisons transformateur-TGBT, sous réserve de connectique adaptée (cosses bimétal, graisse contact).
FAQ
How do you calculate Joule losses?
The Joule losses of a conductor are P = R·I², in watts, where R is the total circuit resistance in ohms and I the current in amperes. If R is not known directly, it is estimated using Pouillet's law applied to the loop: R = 2·ρ·L/S, where ρ is the resistivity (0.0224 Ω·mm²/m for copper at 20 °C) and L the one-way length in metres (the factor 2 accounts for the out-and-return path) and S the cross-section in mm². The energy lost over time is E = P·h in watt-hours, and the corresponding cost is Cost = E·pricekWh.
Why do Joule losses increase with the square of the current?
The dissipated power is P = UR·I, where UR = R·I est la chute de tension aux bornes du conducteur. En substituant, P = R·I². Cette dépendance quadratique a un impact massif : doubler le courant quadruple les pertes, alors que la puissance utile transportée ne double que de manière linéaire. C'est précisément pourquoi le transport d'électricité longue distance se fait en très haute tension (HT 220 kV, THT 380 kV) : à puissance constante, monter U permet de baisser I et donc de diviser les pertes par un facteur considérable.
What is the difference between Joule losses and voltage drop?
The two quantities are related but distinct. The voltage drop ΔU = R·I (single-phase) or √3·R·I/2 (three-phase) measures the potential difference lost between the feeder and the point of use; it is expressed in volts or as a percentage of U, and NIBT 5.2 limits it to 3 % for lighting and 5 % for sockets/power. The Joule losses P = R·I² = ΔU·I is the corresponding thermal power in watts. An installation can comply with the voltage-drop limit while still wasting hundreds of CHF/year in Joule losses.
Should ρ be taken at 20 °C or at 70 °C?
It all depends on the load profile. For a sizing NIBT safety (max voltage drop, Iz), ρ is taken at the maximum insulation temperature (70 °C PVC, 90 °C XLPE), i.e. ρCu ≈ 0,0273 Ω·mm²/m at 70 °C. For a calculation realistic economic over the service life, take ρ at the average operating temperature (often 40-50 °C in commercial buildings), i.e. ρCu ≈ 0,024 Ω·mm²/m. The temperature slider above automatically applies the correction.
How much do Joule losses represent on a bill?
En résidentiel suisse, les pertes Joule internes au bâtiment représentent typiquement 1 à 2 % de la consommation, soit 50 à 150 CHF/an pour une villa de 5000 kWh. En petit tertiaire (cabinet, commerce), elles montent à 2 à 4 %, soit 300 à 800 CHF/an pour 20 000 kWh consommés. En grand tertiaire ou industrie légère avec colonnes montantes longues et charges déséquilibrées, on peut dépasser 5 %, plusieurs milliers de CHF/an. Ces pertes sont totalement invisibles sur la facture mais bien réelles : un audit énergétique sérieux les chiffre toujours.
Is the Joule effect always undesirable?
Non. L'effet Joule est exploité dans tous les appareils résistifs où la chaleur est l'effet recherché : chauffage électrique, plaques vitrocéramiques, fours, chauffe-eau résistifs, sèche-serviettes, planchers chauffants, fers à souder, ampoules à incandescence. Dans ces cas, le rendement électrique vers chaleur est de 100 %. Dans les câbles, par contre, les pertes Joule sont purement parasites et doivent être minimisées dans la mesure économiquement raisonnable.
Going further
To check that the selected cross-section complies with the NIBT 5.2 voltage drop, use the voltage drop calculator. To size the initial cross-section from the current and the installation method, see the NIBT 2020 cable cross-section calculator. The complete list of free ElectroCAD tools also covers EN 12464-1 illuminance, panel layout and short-circuit current calculation. Users of the AutoCAD plugin ElectroCAD Tools benefit from these calculations integrated directly into the schematic's bill of materials, with automatic annotation updates whenever a cross-section changes.