0.1376
Ω
594
W
0.594
kW
11.88
V
0.1376
Ω
594
W
0.594
kW
11.88
V
The Power Loss Calculator (Joule Effect) computes the I²R power dissipated as heat in electrical conductors, combining any specified resistance with a cable resistance calculated from conductor geometry. The Joule effect — heat generation by electrical current flowing through resistance — is the fundamental mechanism behind conductor heating, cable sizing, and energy loss in distribution systems.
James Prescott Joule established in 1841 that the heat generated per second in a conductor is proportional to I² × R, where I is current and R is resistance. This law, fundamental to electrical engineering, explains why power transmission is conducted at high voltage (low current) to minimize I²R losses. Doubling current quadruples losses — a powerful incentive for power factor correction, right-sizing cables, and minimizing cable runs.
Cable resistance is calculated as R = ρ × L / A, where ρ is resistivity (1.72×10⁻⁸ Ω·m for copper, 2.82×10⁻⁸ for aluminum), L is length in meters, and A is cross-sectional area in m². The factor of 2 for two-way (round-trip) accounts for both supply and return conductors. This formula, combined with the I²R loss formula, gives direct calculation of conductor power losses from physical cable parameters.
Voltage drop (V = I × R) is the directly measurable consequence of I²R loss. At the load end, voltage is reduced by V_drop = I × Rcable. For three-phase systems, voltage drop = √3 × I × Rcable × cos(φ) (considering both resistive and reactive drop). NEC and IEC recommend maximum voltage drop of 3% from service entrance to final outlet (5% total including feeder and branch).
Energy loss quantification drives cable sizing economics. A cable running at 50A with 200W of I²R loss for 8000 hours/year wastes 1600 kWh = $192/year at $0.12/kWh. Upsizing from 6mm² to 10mm² copper reduces resistance by 40%, saving $77/year. Payback on the larger cable investment is typically 1-3 years, making conductor upsizing an excellent long-term investment for high-current, long-run circuits.
Cable resistance: R = 2ρL/A (round trip), where A is converted from mm² to m² (÷10⁶). Total resistance = input resistance + cable resistance. Power loss P = I² × R_total (watts). Voltage drop V = I × R_total. The input resistance field allows including terminal, joint, or additional resistance beyond the cable itself.
Power loss percentage = P_loss / P_load × 100%. Design target: cable losses below 1-2% of load power. Voltage drop should be below 3% (IEC recommendation) or 5% (NEC maximum). High voltage drop indicates undersized cable. For long runs, upsizing cable by one or two sizes significantly reduces both losses and voltage drop.
Inputs
Results
344W loss in cable, 6.88V drop. At 230V single-phase, this is 3% voltage drop — at the NEC/IEC limit. Consider 35mm² for longer runs.
Inputs
Results
12.9V drop on a 230V circuit = 5.6% — exceeds NEC/IEC recommendations. Upsize to 25mm² cable.
Copper: ρ = 1.72×10⁻⁸ Ω·m (at 20°C). Aluminum: ρ = 2.82×10⁻⁸ Ω·m. Aluminum is 64% more resistive than copper by cross-section. However, aluminum is lighter and cheaper — aluminum cables used at 1.5× the cross-section of copper provide equivalent resistance while saving cost and weight. Utility power lines are mostly aluminum (ACSR).
Resistance increases with temperature: R_T = R_20 × (1 + α × ΔT), where α (temperature coefficient) = 0.00393/°C for copper. A cable at 90°C maximum operating temperature has 27% higher resistance than at 20°C: R_90 = R_20 × (1 + 0.00393 × 70) = 1.275 × R_20. Current-carrying capacity tables in NEC/IEC already account for this.
P = I²R. If current doubles: P_new = (2I)² × R = 4I²R = 4 × P_original. This quadratic relationship is why high-current applications require careful conductor sizing — small increases in current cause disproportionately large increases in heating. It is also why power is transmitted at high voltage (low current) over long distances.
IEC 60364-5-52: maximum 4% voltage drop from origin of installation to any point. NEC (NFPA 70) FPN recommends: 3% for branch circuits, 5% total (feeder + branch). For sensitive equipment (PLC, VFD), more stringent limits (1-2%) are common in specifications. Voltage drop is additive through cascaded circuits.
Reduce current through the cable: power factor correction (reduces reactive current component), load rebalancing (distributes load across phases), load reduction or shifting, adding parallel cable runs, or installing a local transformer to reduce impedance. Replacing conductors is most effective but expensive for long runs.
Roboculator Team
The Roboculator Team explains calculations, planning tools, and practical formulas in clear language for real-life situations.
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