---
title: "Unarmoured cable for 15 hp motor at 79 m"
subject: "Industrial Power Distribution and Illumination"
chapter: 5
key: "semester_6/industrial_power_distribution_and_illumination:cable-15hp-motor-79m"
source: https://www.bctengineer.com/q/semester_6/industrial_power_distribution_and_illumination/cable-15hp-motor-79m
site: BCT Engineering
---

# Unarmoured cable for 15 hp motor at 79 m

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

- Asked in: 2068 Jestha (old course)
- Video: [कितने Hp की मोटर पर कितने mm की कैबल लगाए | Motor cable size calculation formula](https://www.youtube.com/watch?v=S6CWUja5M5k) — Electrical Help (Rakesh Hooda)
- 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-15hp-motor-79m
- Whole chapter: https://www.bctengineer.com/md/semester_6/industrial_power_distribution_and_illumination/chapter-5

## Question

A 3-phase 15 hp, 400V, 50 Hz induction motor is to be installed in a workshop. Assuming the efficiency of the motor to be 85% and power factor 0.8, calculate the size of the unarmoured copper cable to be used if the distance between SDB and motor is 79m.

| 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 | 31 | 45 | 60 | 77 | 99 | 120 | 145 | 175 | 210 | 240 | 270 |

## Short answer

Motor full-load current is 23.75 A; design current (125%) is 29.7 A. From the table, 4 mm² unarmoured copper cable rated 31 A is selected. Voltage drop is 11.57 V (2.89%), within 5%. Cable size: 4 mm².

## Answer

**Given:** P_out = 15 hp, V = 400 V (3-phase), η = 85 %, pf = 0.8, length SDB → motor = 79 m.

**Assumptions:** 1 hp = 746 W; cable chosen for 125 % of motor full-load current (motor circuit rule of IS 732/NEC); Cu resistivity 0.0178 Ω·mm²/m; permissible voltage drop 5 % (20 V). Table ratings are used for the unarmoured multicore Cu cable.

#### Step 1: Full-load current

```text
Output P_out = 15 × 746        = 11,190 W
Input  P_in  = 11190 / 0.85    = 13,164.7 W
I_FL = P_in / (√3 × V × pf)
     = 13164.7 / (√3 × 400 × 0.8)
     = 23.75 A
```

#### Step 2: Design current and cable from table

```text
Design current = 1.25 × 23.75 = 29.69 A
```

From the table, the next rating above 29.69 A is **31 A → 4 mm²**.

#### Step 3: Voltage drop check

```text
R  = ρL/A = 0.0178 × 79 / 4 = 0.3515 Ω
ΔV = √3 × I × R × cos φ
   = √3 × 23.75 × 0.3515 × 0.8 = 11.57 V
%ΔV = 11.57/400 × 100 = 2.89 %  (< 5 %, OK)
```

**Answer: 4 mm² Cu unarmoured (3.5/4-core) cable** — full-load current 23.75 A, voltage drop 11.57 V (2.89 %). Protect the motor with a 32 A TP MCB (D-curve) or MPCB and a thermal overload relay set at about 24 A.

## Check yourself

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

1. How is the motor input power found from the 15 hp output?
   A. P_in = 15 x 746 x 0.85
   B. P_in = (15 x 746)/0.85 = 13,164.7 W
   C. P_in = 15 x 746
   D. P_in = 15 x 1000
   Answer: B. Output is 11,190 W; dividing by efficiency 0.85 gives input 13,164.7 W, then I = P_in/(sqrt3 V pf) = 23.75 A. (Key point: Full-load current)

2. Why is the cable chosen for 1.25 x I_FL = 29.69 A?
   A. To allow a 125 % margin per motor circuit rule, so 31 A (4 mm2) is selected
   B. To reduce the voltage drop to zero
   C. Because efficiency is 125 %
   D. The cable is chosen for 23.75 A exactly, giving 2.5 mm2
   Answer: A. The motor-circuit rule uses 125 % of FLC; the next table rating above 29.69 A is 31 A (4 mm2). (Key point: Design current)

3. How is voltage drop checked?
   A. R = rho L/A = 0.3515 Ω, dV = sqrt3 I R cos phi = 11.57 V (2.89 %) < 5 %
   B. dV = I R only, equal to 8.3 V, with no limit
   C. dV = V/I
   D. It is not needed for long cables
   Answer: A. Using R = 0.0178 x 79/4, drop is 11.57 V = 2.89 %, within the 5 % limit. (Key point: Voltage drop check)

4. What protection is suggested for the motor?
   A. No protection
   B. 32 A TP MCB (D curve) or MPCB with a thermal overload relay near 24 A
   C. A 100 A fuse
   D. A single-pole 6 A MCB
   Answer: B. A D-curve 32 A TP MCB handles starting current, and the overload relay is set near the 23.75 A full-load current. (Key point: Final protection)

## Close variants set in other exams

- [Cable types and cable for 15 hp motor](https://www.bctengineer.com/question.md?key=semester_6%2Findustrial_power_distribution_and_illumination%3Acable-types-15hp-motor) — Industrial Power Distribution and Illumination, chapter 5, asked 2×

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