---
title: "3.5 core cable for 60 A over 200 m"
subject: "Industrial Power Distribution and Illumination"
chapter: 5
key: "semester_6/industrial_power_distribution_and_illumination:cable-60a-200m"
source: https://www.bctengineer.com/q/semester_6/industrial_power_distribution_and_illumination/cable-60a-200m
site: BCT Engineering
---

# 3.5 core cable for 60 A over 200 m

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

- Asked in: 2076 Bhadra, 4 marks
- 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/cable-60a-200m
- Whole chapter: https://www.bctengineer.com/md/semester_6/industrial_power_distribution_and_illumination/chapter-5

## Question

Determine the size of 3.5 core copper cables for MDB to SDB required to carry the maximum current of 60A. It is given that length of cable is 200m. The permissible voltage drop is 2% of supply voltage. The supply system is 3-phase, 400V, 50 Hz. Current ratings of copper conductor is given in table below.

| Conductor area (sq.mm) | 1.5 | 2.5 | 4 | 6 | 10 | 16 | 25 | 35 | 50 | 70 | 95 | 120 | 150 |
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
| 3.5 core cable current rating (A) | 21 | 27 | 36 | 45 | 60 | 77 | 99 | 120 | 145 | 175 | 210 | 240 | 270 |

## Short answer

For 60 A, 10 mm² meets current rating but voltage drop requires 46.25 mm². Use 3½ core 50 mm² Cu cable; it has 145 A rating and 7.4 V drop (1.85% < 2%).

## Answer

The cable is selected by current rating and then checked for voltage drop; the larger size governs. Power factor is not given, so cosφ = 1 is assumed (largest resistive drop, conservative). ρ (Cu) = 0.0178 Ω·mm²/m.

#### By current rating
I = 60 A → from the table, 10 mm² carries 60 A (exactly), so 10 mm² is the minimum by current.

#### By voltage drop
```text
Allowed drop = 2 % of 400 V = 8 V
ΔV = √3 × I × ρ × L / A
For 10 mm²: ΔV = √3 × 60 × 0.0178 × 200 / 10
              = 36.99 V (9.25 %)  → NOT OK
A_min = √3 × 60 × 0.0178 × 200 / 8
      = 46.25 mm² → next standard size 50 mm²
Check 50 mm²: ΔV = 369.97 / 50 = 7.40 V (1.85 %) → OK
```

**Answer: use 3½ core, 50 mm² copper cable** (rating 145 A, voltage drop 7.40 V = 1.85 % < 2 %). The long length makes voltage drop, not current, the deciding factor. (With pf 0.8 the minimum area would be 37 mm², still giving 50 mm².)

## Check yourself

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

1. What size does current rating alone give for 60 A?
   A. 6 mm²
   B. 10 mm² (60 A)
   C. 16 mm²
   D. 50 mm²
   Answer: B. The table lists 60 A for 10 mm². (Key point: By current)

2. What is the drop for 10 mm² over 200 m at 60 A, and the verdict?
   A. About 7.4 V, OK
   B. 36.99 V (9.25 %), exceeds the 8 V limit, so NOT OK
   C. About 8 V, exactly OK
   D. 3.7 V, OK
   Answer: B. ΔV = √3 x 60 x 0.0178 x 200 / 10 = 36.99 V, far above 2 % of 400 V = 8 V. (Key point: Voltage drop)

3. Which size is finally selected and why?
   A. 10 mm², since current is satisfied
   B. 25 mm², average of the two
   C. 95 mm², for safety
   D. 50 mm², as A_min = 46.25 mm² from voltage drop and the long length governs
   Answer: D. Minimum area for 8 V drop is 46.25 mm², next standard 50 mm² (drop 7.40 V). (Key point: Final size)

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