Hong Life CalculatorsUnderstanding the voltage drop calculator — concept and principle
Long cables waste some voltage before power reaches the load, which can dim lights, slow motors or trip equipment. This calculator uses the method of IEC 60364-5-52 to find the voltage drop for DC, single-phase or balanced three-phase circuits from the current, cable length, cross-section and conductor material, and checks it against your allowed limit.
How it is calculated
u = b × (ρ₁ × L ÷ S × cos φ + λ × L × sin φ) × I
Drop (%) = 100 × u ÷ U₀
- b = 2 for DC and single-phase (outgoing and return conductor), b = 1 for three-phase (phase-to-neutral drop; the line-to-line drop is √3 × u).
- ρ₁ = 0.0225 Ω·mm²/m for copper and 0.036 Ω·mm²/m for aluminium, values at normal service temperature.
- λ = 0.08 mΩ/m is the cable reactance per metre; L is the one-way length in m, S the cross-section in mm², I the current, cos φ the power factor.
- U₀ is the phase-to-neutral voltage. Cable power loss = (2 or 3) × I² × R.
The principle
A conductor has resistance proportional to its length and inversely proportional to its cross-section, R = ρL ÷ S. Ohm's law gives the voltage lost along it, I × R. In a single-phase or DC circuit, current flows out and back, so two conductors' worth of drop occurs. In a balanced three-phase circuit, the neutral carries no current, so only one conductor is counted per phase. With AC, the cable's small inductance also matters, and how much it adds depends on the phase angle of the load, which is why cos φ and sin φ appear.
Worked example
Default inputs: single-phase 230 V, 16 A, 30 m, 2.5 mm² copper, cos φ = 1, limit 3 %.
- R of one conductor = 0.0225 × 30 ÷ 2.5 = 0.27 Ω.
- With cos φ = 1, sin φ = 0, so u = 2 × 0.27 × 16 = 8.64 V.
- Drop = 100 × 8.64 ÷ 230 = 3.76 %.
- Voltage at the load = 230 − 8.64 = 221.4 V.
- 3.76 % is above 3 %, so the answer is no — use a larger cross-section or a shorter run.
- Power lost = 2 × 16² × 0.27 = 138.2 W.
| Supply voltage (V; line-to-line for three-phase) | Voltage drop (%) |
|---|---|
| 57.5 | 15.03% |
| 115 | 7.51% |
| 172 | 5.01% |
| 230 | 3.76% |
| 288 | 3.01% |
| 345 | 2.5% |
| 460 | 1.88% |
How to read the result
IEC 60364-5-52 recommends at most 3 % for lighting and 5 % for other uses when supplied from a public low-voltage network, with higher limits from a private supply. Korea's KEC applies the same 3 % and 5 % limits for public LV supply. If you exceed the limit, a larger cross-section is usually the simplest fix; in this example 4 mm² would bring the drop to about 2.3 %.
Common mistakes
- Entering the round-trip length instead of the one-way length.
- Using the 20 °C resistivity, which understates the drop at operating temperature.
Frequently asked questions
Why is the drop doubled for single-phase?
The current flows through both the outgoing and the return conductor.
What voltage drop is acceptable?
Typical limits are 3 % for lighting and 5 % for other loads from a public supply; check your local rules.
Does a larger cable save energy?
Yes. Lower resistance reduces the I²R loss as well as the voltage drop.
Sources: IEC 60364-5-52:2009, Annex G (voltage drop in consumers' installations) · Korea Electro-technical Code (KEC) 232.3.9