How to Calculate Voltage Drop: A Brief Answer
Voltage drop is calculated using Ohm’s law for a circuit section, taking into account the line length: ΔU = I · Rline. For a single-phase network, the line resistance is counted twice (phase and neutral conductors); for a three-phase network, it is calculated through the line voltage and the factor √3. The result is compared with the permissible drop: no more than 3 % for lighting and no more than 5 % for power lines (PUE RK). If the calculated drop exceeds the norm, the cross-section is increased to a value at which the condition is met.
The key check: ΔU% = (ΔU / Unom) · 100. For a 380 V line, the permissible drop of 5 % is 19 V. For a 220 V line, it is 11 V. Length and load current affect losses linearly; cross-section is inversely proportional.
What Data Is Needed for the Calculation
- Load current I (A) or power P (W), voltage U (V), cos φ.
- Line length L (m) — the distance from the source to the consumer along the cable route.
- Conductor cross-section S (mm²) — preliminarily selected by heating.
- Conductor material — resistivity ρ: copper 0.0175 Ω·mm²/m, aluminium 0.028 Ω·mm²/m (at +20 °C).
- Number of phases — single-phase (220 V) or three-phase (380 V) network.
- cos φ — load power factor.
- Installation type — open, in a pipe, in the ground, in a tray.
- Ambient temperature — if it differs from the rated value.
Formulas and the Principle of Calculation
Active Resistance of the Conductor
R = ρ · L / S
R — resistance of one conductor, Ω; ρ — resistivity, Ω·mm²/m; L — length, m; S — cross-section, mm².
Voltage Drop in a Single-Phase 220 V Line
ΔU = 2 · I · L · ρ / S · cos φ
The factor 2 accounts for the current in the phase and neutral conductors. I — current, A; L — length, m; ρ — resistivity; S — cross-section; cos φ — power factor.
Voltage Drop in a Three-Phase 380 V Line
ΔU = √3 · I · L · ρ / S · cos φ
√3 ≈ 1.732. The calculation is performed for one phase, but through the line voltage.
Voltage Drop in Percent
ΔU% = ΔU / Unom · 100
Unom — rated mains voltage: 220 V or 380 V.
Step-by-Step Example: From Power to Verification
Initial data. Three-phase asynchronous motor: Pnom = 15 kW at the shaft, U = 380 V, cos φ = 0.85, η = 0.88. Line length from the distribution board to the motor — 80 m. Open installation in a tray together with two other loaded cables. Air temperature +35 °C. Copper conductors, PVC insulation.
Step 1. Calculated load current.
I = Pnom / (√3 · U · cos φ · η) = 15,000 / (1.732 · 380 · 0.85 · 0.88) ≈ 30.4 A.
Step 2. Preliminary selection by heating.
For copper conductors with PVC insulation under open installation, a 4 mm² cross-section has a table permissible current of 36 A (Table 1.3.6 of the PUE RK). However, the table value is valid for +25 °C and a single cable.
Step 3. Correction factors.
Temperature +35 °C: factor 0.88 (Table 1.3.3 of the PUE RK for a rated conductor temperature of +65 °C and air temperature of +25 °C).
Group installation: three loaded cables in a tray — factor 0.85 (Table 1.3.26 of the PUE RK for installation in the ground; for trays, a similar approach is applied based on the number of cables laid together).
Adjusted permissible current: 36 · 0.88 · 0.85 ≈ 26.9 A.
The calculated current of 30.4 A exceeds 26.9 A — the 4 mm² cross-section does not pass.
Step 4. Repeat selection.
Cross-section 6 mm²: table current 46 A. Adjusted: 46 · 0.88 · 0.85 ≈ 34.4 A. The condition 30.4 ≤ 34.4 is satisfied.
Step 5. Voltage drop check.
ΔU = 1.732 · 30.4 · 80 · (0.0175 / 6) · 0.85 ≈ 12.5 V.
ΔU% = 12.5 / 380 · 100 ≈ 3.3 %.
For a power line, up to 5 % is permissible (PUE RK). The condition is satisfied.
Step 6. Result.
We adopt a 6 mm² cable with copper conductors. The initially selected 4 mm² cross-section had to be increased due to temperature and group installation. If the line length had reached 120–150 m, the voltage drop would have become the decisive factor. The construction and characteristics of power cable are described in a separate article.
What Can Change the Result
| Parameter | How it affects the result |
|---|---|
| Line length | Losses increase proportionally to length. A 100 m line loses twice as much as a 50 m line at the same current and cross-section. |
| Conductor cross-section | Losses are inversely proportional to cross-section. Increasing the cross-section from 4 to 6 mm² reduces voltage drop by a third. |
| Conductor material | Aluminium has a resistivity about 1.6 times higher than copper. At equal cross-section, losses in an aluminium line are higher. Comparison of copper and aluminium. |
| cos φ | When cos φ decreases from 0.95 to 0.8, the current at the same active power increases, increasing losses. |
| Conductor temperature | Copper resistance increases with temperature: when heated from +20 to +65 °C, resistance increases by about 17 %. In voltage drop calculations for continuously loaded lines, this is accounted for through the temperature coefficient. |
| Installation method | Affects the permissible current, and therefore the preliminary cross-section. Installation method has no direct effect on voltage drop, but determines which cross-section passed the heating check. |
Typical Mistakes
- Calculating voltage drop without taking cos φ into account. For motors and transformers, cos φ = 0.7–0.9, and neglecting it underestimates the calculated current and therefore the losses.
- Forgetting about the return conductor in a single-phase network. A formula without the factor 2 gives a result underestimated by half.
- Using table resistivity at operating temperature. The resistivity of 0.0175 Ω·mm²/m is given at +20 °C. For a heated conductor, the resistance is higher.
- Checking voltage drop for only one load. If the line supplies several consumers, losses are calculated for the most remote point or summed by sections.
- Ignoring inductive reactance for large cross-sections. For cables with cross-sections greater than 25–35 mm², the inductive component becomes noticeable. For household and most industrial 0.4 kV lines, active resistance predominates.
- Applying the USSR PUE of the 6th edition instead of the current PUE RK. The current document is the PUE RK 2015, as amended by Order of the Minister of Energy of the RK No. 340 dated 31.10.2022.
A Brief Algorithm for Voltage Drop Verification
- Calculate the load current I from power, voltage, cos φ, and efficiency.
- Select a preliminary cross-section from the permissible current table, taking into account corrections for temperature and group installation.
- Calculate the active line resistance R = ρ · L / S.
- Calculate the voltage drop using the formula for a single-phase or three-phase network.
- Convert to percent and compare with the norm: 3 % for lighting, 5 % for power lines.
- If the norm is exceeded, increase the cross-section and repeat the calculation.
- When choosing between single-core and multi-core versions, take into account installation convenience and route conditions. Differences between single-core and multi-core cable.
FAQ
What voltage drop is considered permissible?
The PUE RK establishes: for lighting — no more than 3 % of the rated voltage, for power lines — no more than 5 %. For motors during start-up, up to 20 % is permitted. GOST 32144-2013 standardises voltage deviations at the consumer, but the design check is performed according to the PUE.
Is it necessary to take the inductive reactance of the cable into account?
For 0.4 kV cables with cross-sections up to 25–35 mm², the inductive reactance is small compared to the active resistance, and it is neglected. For large cross-sections and long lines, the inductive component is accounted for separately, but for most practical calculations in 220/380 V networks, active resistance is sufficient.
How will the voltage drop change if the cross-section is doubled?
Voltage drop is inversely proportional to cross-section. When the cross-section is increased from 4 to 8 mm², losses will decrease by about half at the same current.
Can an aluminium cable be used for a long line?
It is possible, but at equal cross-section, losses in an aluminium line are about 1.6 times higher than in a copper one. For long lines, this often makes it necessary to increase the aluminium conductor cross-section compared to copper in order to meet the voltage drop norm.
How to check voltage drop for a line with several loads?
The line is divided into sections. For each section, the current flowing through it is calculated, then the losses in that section. The total drop is the sum of losses in all sections from the source to the most remote consumer.
Sources Used
- Electrical Installation Rules of the Republic of Kazakhstan 2015, as amended by Order of the Minister of Energy of the RK No. 340 dated 31.10.2022 — Committee for Technical Regulation and Metrology of the MIR RK — https://base.spinform.ru/show_doc.fwx?rgn=71151
- GOST 32144-2013 Electric energy quality norms in general-purpose power supply systems — Interstate Council for Standardisation, Metrology and Certification — https://docs.cntd.ru/document/1200104301
- GOST R 50571.5.52-2011/IEC 60364-5-52:2009 Low-voltage electrical installations. Part 5-52. Selection and erection of electrical equipment. Wiring systems — Rosstandart — https://dokipedia.ru/document/5149532
- Technical Regulation of the Customs Union TR CU 004/2011 “On the safety of low-voltage equipment” — EAEU — https://docs.eaeunion.org/docs/ru-ru/0145005
