---
title: "Rotor copper loss and efficiency of 35 kW induction motor"
subject: "Electrical Machines I"
chapter: 5
key: "semester_4/electrical_machines_i:im-4pole-1440rpm-35kw-losses"
source: https://www.bctengineer.com/q/semester_4/electrical_machines_i/im-4pole-1440rpm-35kw-losses
site: BCT Engineering
---

# Rotor copper loss and efficiency of 35 kW induction motor

Electrical Machines I · Chapter 5: Three-Phase Induction Machines

- Asked in: 2082 Kartik (new course), 6 marks
- Video: [losses and efficiency - Three Phase Induction Motor - Electrical Machines 3](https://www.youtube.com/watch?v=sKS5d40MEic) — Ekeeda
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/electrical_machines_i/im-4pole-1440rpm-35kw-losses
- Whole chapter: https://www.bctengineer.com/md/semester_4/electrical_machines_i/chapter-5

## Question

A 4 pole, 3 phase, 50 Hz, 1440 rpm induction motor is drawing 35 kW. The stator and friction losses are 1 kW and 1.5 kW respectively. Find: (i) rotor copper losses (ii) efficiency of motor

## Short answer

Synchronous speed 1500 rpm, slip 0.04. Air-gap power 34 kW, rotor copper loss = 0.04×34 = 1.36 kW. Output = 34 - 1.36 - 1.5 = 31.14 kW, efficiency = 88.97%.

## Answer

**Given:** $P = 4$, $f = 50$ Hz, $N = 1440$ rpm, input $P_{in} = 35$ kW, stator losses $= 1$ kW, friction (mechanical) losses $= 1.5$ kW.

**Power flow:** input → (minus stator losses) → air-gap power → (minus rotor Cu loss) → gross mechanical power → (minus friction) → output.

#### Synchronous speed and slip

$$
\begin{aligned}
N_s &= \frac{120 f}{P} = \frac{120 \times 50}{4} = 1500\ \text{rpm} \\
s &= \frac{N_s - N}{N_s} = \frac{1500 - 1440}{1500} = 0.04
\end{aligned}
$$

#### (i) Rotor copper loss

$$
\begin{aligned}
P_{ag} &= P_{in} - \text{stator losses} = 35 - 1 = 34\ \text{kW} \\
P_{cu2} &= s \, P_{ag} = 0.04 \times 34 = 1.36\ \text{kW}
\end{aligned}
$$

#### (ii) Efficiency

$$
\begin{aligned}
P_{m} &= P_{ag} - P_{cu2} = 34 - 1.36 = 32.64\ \text{kW} \\
P_{out} &= P_m - \text{friction} = 32.64 - 1.5 = 31.14\ \text{kW} \\
\eta &= \frac{P_{out}}{P_{in}} = \frac{31.14}{35} = 0.8897 = 88.97\%
\end{aligned}
$$

| Item | Power (kW) |
|---|---|
| Input | 35.00 |
| Stator losses | 1.00 |
| Air-gap power | 34.00 |
| Rotor Cu loss | 1.36 |
| Gross mechanical power | 32.64 |
| Friction loss | 1.50 |
| Output | 31.14 |

**Answer: (i) Rotor copper loss = 1.36 kW; (ii) efficiency = 88.97%.**

## Check yourself

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

1. What is the slip at 1440 rpm for a 4-pole 50 Hz motor?
   A. 0.04
   B. 0.06
   C. 0.4
   D. 0.03
   Answer: A. Ns = 1500 rpm, s = 60/1500 = 0.04. (Key point: Slip)

2. How is air-gap power found?
   A. Input minus stator losses = 34 kW
   B. Input minus friction loss
   C. Input plus stator losses
   D. Input times efficiency
   Answer: A. 35 - 1 = 34 kW. (Key point: Air-gap power)

3. What is the rotor copper loss?
   A. s x Pag = 1.36 kW
   B. (1 - s) x Pag = 32.64 kW
   C. s x input = 1.4 kW
   D. 1.5 kW
   Answer: A. Pcu2 = 0.04 x 34 = 1.36 kW. (Key point: Rotor copper loss)

4. What is the efficiency?
   A. 88.97%
   B. 93.3%
   C. 96%
   D. 85%
   Answer: A. Output = 34 - 1.36 - 1.5 = 31.14 kW; 31.14/35 = 88.97%. (Key point: Efficiency)

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