---
title: "6-pole motor: rotor loss and efficiency"
subject: "Electrical Machines"
chapter: 5
key: "semester_4/electrical_machines:im-6pole-440v-950rpm-rotor-loss"
source: https://www.bctengineer.com/q/semester_4/electrical_machines/im-6pole-440v-950rpm-rotor-loss
site: BCT Engineering
---

# 6-pole motor: rotor loss and efficiency

Electrical Machines · Chapter 5: Three Phase Induction Machines

- Asked in: 2066 Magh (old course), 8 marks
- Video: [Electrical Machines | Lec 6C | Three-Phase Induction Motor | Rotor Quantities: Slip, Torque, etc](https://www.youtube.com/watch?v=s-wCI5jDu3g) — Engineering Devotion
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/electrical_machines/im-6pole-440v-950rpm-rotor-loss
- Whole chapter: https://www.bctengineer.com/md/semester_4/electrical_machines/chapter-5

## Question

A 3-phase, 440 V, 50 Hz, 6-pole induction motor draws 50 kW at 0.85 lagging p.f. from the source when connected to rated supply. If rotor rotates at 950 rpm, calculate (i) rotor loss and (ii) overall efficiency of the motor. The friction loss and stator loss are 2 kW and 1.5 kW respectively at this running condition.

## Short answer

Synchronous speed 1000 rpm, slip 0.05. Air-gap power = 50 kW – 1.5 kW = 48.5 kW. Rotor loss = slip × Pg = 2.425 kW. Out put = Pg – rot or loss – 2 kW friction = 44.075 kW; efficiency = 88.15 %.

## Answer

**Given:** 440 V, 50 Hz, 6-pole, input $P_{in} = 50$ kW at 0.85 lagging p.f., $N_r = 950$ rpm, friction loss $= 2$ kW, stator loss $= 1.5$ kW.

$$
N_s = \frac{120\times 50}{6} = 1000\ \text{rpm}, \qquad s = \frac{1000-950}{1000} = 0.05
$$

#### Power flow
Air-gap power (rotor input):
$$
P_g = 50 - 1.5 = 48.5\ \text{kW}
$$
#### (i) Rotor loss
$$
P_{cu2} = sP_g = 0.05\times 48.5 = 2.425\ \text{kW}
$$
#### (ii) Efficiency
Mechanical power developed: $P_m = 48.5 - 2.425 = 46.075$ kW
Output: $P_{out} = 46.075 - 2 = 44.075$ kW
$$
\eta = \frac{44.075}{50}\times 100 = 88.15\%
$$
(The power factor is not needed.)

**Answer:** (i) rotor copper loss $= 2.425$ kW; (ii) efficiency $= 88.15\%$.

## Check yourself

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

1. What is the slip of the 6-pole, 50 Hz motor at 950 rpm?
   A. 0.05
   B. 0.0526
   C. 0.95
   D. 0.5
   Answer: A. Ns = 1000 rpm, so s = 50/1000 = 0.05. (Key point: Slip)

2. How is the air-gap (rotor input) power obtained from the input?
   A. Subtract the stator loss from the input power
   B. Subtract the friction loss
   C. Multiply by the power factor
   D. Multiply by the slip
   Answer: A. Pg = Pin - stator loss = 50 - 1.5 = 48.5 kW. (Key point: Air-gap power)

3. Which formula gives the rotor copper loss?
   A. s x Pg = 2.425 kW
   B. (1-s) x Pg
   C. Pg/s
   D. s x Pin
   Answer: A. Rotor copper loss = s x air-gap power = 0.05 x 48.5 = 2.425 kW. (Key point: Rotor copper loss)

4. What is the efficiency and why is the power factor not needed?
   A. 88.15 %; input power in kW is already given, so pf is not needed
   B. 94 %; pf multiplies output
   C. 88.15 %; pf is needed to find output
   D. 96.5 %; pf equals efficiency
   Answer: A. Pout = 46.075 - 2 = 44.075 kW and eta = 44.075/50 = 88.15 %; the given input is real power. (Key point: Efficiency)

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