Method
Voltage drop from resistance/reactance per kilometre for single-phase or balanced three-phase circuits.
It covers two cases: three phase and single phase. Equations that apply to only one of them are tagged.
Equations
k = 2single phase
k = √3three phase
sin φ = √1 − PF2
ΔU = k · I · L · (R · PF + X · sin φ)1000
ΔU% = 100 · ΔUUn
Ur = Un − ΔU
Symbols and units
| Symbol | Quantity | Unit |
|---|---|---|
| Inputs | ||
| I | Load current | A |
| L | Route length (one way) | m |
| R | Conductor resistance | Ω/km |
| X | Conductor reactance | Ω/km |
| PF | Load power factor | 0–1 |
| Un | Nominal voltage | V |
| Intermediate values | ||
| k | Circuit factor | – |
| sin φ | Sine of the load phase angle | – |
| Results | ||
| ΔU | Voltage drop | V |
| ΔU% | Voltage drop | % |
| Ur | Receiving voltage | V |
Assumptions and limits
- No conductor selection or permissible-drop limit is applied.
In practice
- On three-phase circuits use the line-to-line voltage for Un; on single-phase circuits the phase-to-neutral voltage.
- Take R and X from the cable maker for the conductor size and operating temperature. The permitted drop comes from the wiring regulations and the specification.
Worked example
The input values are the calculator's own verification test, run through the calculator. They show the working; they are not design values.
Method: Three phase, balanced.
Inputs
| I | Load current | 100 A |
| L | Route length (one way) | 100 m |
| R | Conductor resistance | 0.5 Ω/km |
| X | Conductor reactance | 0.08 Ω/km |
| PF | Load power factor | 0.8 0–1 |
| Un | Nominal voltage | 415 V |
Results
| ΔU | Voltage drop | 7.75959 V |
| ΔU% | Voltage drop | 1.86978 % |
| Ur | Receiving voltage | 407.24 V |
A preliminary check for engineer review, not an engineering approval. The project calculation to the applicable codes, authority requirements and specification governs.