---
title: "Aluminium cable for 25 kW single-phase load"
subject: "Industrial Power Distribution and Illumination"
chapter: 5
key: "semester_6/industrial_power_distribution_and_illumination:aluminium-cable-25kw"
source: https://www.bctengineer.com/q/semester_6/industrial_power_distribution_and_illumination/aluminium-cable-25kw
site: BCT Engineering
---

# Aluminium cable for 25 kW single-phase load

Industrial Power Distribution and Illumination · Chapter 5: Electrification of Industrial Building

- Asked in: 2072 Asoj, 4 marks
- Video: [How to select cable size? // AC cable size & Voltage drop calculation. // AC cable design for load.](https://www.youtube.com/watch?v=DYsRldg3sG0) — Solar Facts
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_6/industrial_power_distribution_and_illumination/aluminium-cable-25kw
- Whole chapter: https://www.bctengineer.com/md/semester_6/industrial_power_distribution_and_illumination/chapter-5

## Question

Calculate the cross-sectional area of Aluminum cable require to supply a single phase load of 25 KW at power factor of 0.7 lagging which is 40 m away from the voltage source so that voltage drop in the cable does not exceed 3% of the standard supply voltage of 220 V.

## Short answer

Using voltage drop formula ΔV = I × ρ × 2L / A with 3% limit (6.6 V), load current 162.34 A, loop length 80 m, aluminium resistivity 0.0283 Ω·mm²/m, minimum area is 55.69 mm²; select 70 mm² aluminium cable (2.39% drop).

## Answer

The cross-section is found from the permissible voltage drop of the go-and-return (2-wire) circuit.

Constants: resistivity of aluminium ρ = 0.0283 Ω·mm²/m (2.83 × 10⁻⁸ Ω·m); the whole I·R drop is taken (conservative, reactance neglected).

```text
Load current I = P / (V × cosφ)
               = 25000 / (220 × 0.7) = 162.34 A
Allowed drop   = 3 % of 220 = 6.60 V
Loop length    = 2 × 40 = 80 m
ΔV = I × R = I × ρ × 2L / A ≤ 6.60
A  = ρ × 2L × I / ΔV
   = 0.0283 × 80 × 162.34 / 6.60
   = 55.69 mm²
```
Next standard size: **70 mm²**.
```text
Check: ΔV = 0.0283 × 80 × 162.34 / 70
          = 5.25 V (2.39 %) < 6.60 V  OK
```

**Answer: minimum area = 55.69 mm²; use a 70 mm² aluminium cable** (drop 5.25 V). If only the in-phase component of drop (I·R·cosφ) is considered, A = 38.98 mm² and a 50 mm² cable would just satisfy the drop limit; the current rating of the chosen size must also be checked from the manufacturer's table for 162 A.

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. What is the current for 25 kW at 220 V, pf 0.7, single phase?
   A. 113.6 A
   B. 162.34 A (P / (V cosφ))
   C. 196 A
   D. 93.5 A
   Answer: B. I = 25000/(220 x 0.7) = 162.34 A. (Key point: Load current)

2. What loop length and formula give the area?
   A. Loop = 2 x 40 = 80 m; A = ρ x 2L x I / ΔV = 55.69 mm²
   B. Loop = 40 m; A = 27.8 mm²
   C. Loop = 80 m; A = 556 mm²
   D. Loop = 40 m; A = 100 mm²
   Answer: A. Go-and-return path is 80 m, allowed drop 6.6 V (3 % of 220 V), giving 55.69 mm². (Key point: Minimum area)

3. Which standard size is chosen and what is its drop?
   A. 50 mm², drop 5.9 V
   B. 35 mm², drop 8 V
   C. 70 mm², drop 5.25 V (2.39 %)
   D. 95 mm², drop 2 V
   Answer: C. Next standard size above 55.69 mm² is 70 mm²; drop 5.25 V is below 6.60 V. (Key point: Final size)

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