Cable Voltage Drop
Compute voltage drop ΔV = 2·ρ·L·I/A in cable.
Voltage drop in electrical cables
Voltage drop in a cable depends on its resistance, length and current. For DC or single-phase circuits (round trip), ΔV = (2 · ρ · L · I) / A, where ρ is resistivity (copper ≈ 1.68 × 10−8 Ω·m or 0.0172 Ω·mm²/m), L is length in meters, I is current in amperes and A is cross-section in mm². For three-phase: ΔV = √3 · ρ · L · I / A. The Brazilian NBR 5410 limits voltage drop to 4% for lighting and 7% for power in residential branch circuits, measured from the origin of the installation; 5% at the terminals of utilization equipment. Example: 30 m, 2.5 mm², 10 A at 127 V single-phase → ΔV ≈ 4 V (3.1%) — within limits.
Applications: residential, industrial and solar
Voltage-drop checks are mandatory for: residential installations under NBR 5410 (long circuits to outdoor outlets, shower units far from the panel); industrial installations under NBR 14039 (medium voltage); photovoltaic systems, where long DC string cables between modules and inverter often dominate losses; data centers and underground LV/MV feeders. When drop exceeds the limit, the practical fix is to increase the cross-section (next standard size: 2.5 → 4 → 6 → 10 mm²) rather than raising voltage.
FAQ
Why the factor of 2 for single-phase? Current flows through phase and neutral, so both wires contribute resistance — total length is 2 × L.
Aluminum versus copper? Aluminum has ρ ≈ 2.82 × 10−8 Ω·m, about 1.6× higher than copper, so for the same drop it needs a larger cross-section.
Does temperature matter? Yes — copper resistivity increases ~0.4% per °C. Sizing standards assume 70 °C conductor temperature for PVC insulation.
What about power factor? For inductive loads, the formula includes cos φ: ΔV ≈ 2 · L · I · (R · cos φ + X · sin φ) / A. For resistive loads, cos φ = 1 and the simple form suffices.
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