---
title: "6-pole induction motor: slip, losses, efficiency"
subject: "Electrical Machines"
chapter: 5
key: "semester_4/electrical_machines:im-4pole-975rpm-slip-losses"
source: https://www.bctengineer.com/q/semester_4/electrical_machines/im-4pole-975rpm-slip-losses
site: BCT Engineering
---

# 6-pole induction motor: slip, losses, efficiency

Electrical Machines · Chapter 5: Three Phase Induction Machines

- Asked in: 2078 Poush, 2+2+2+2 marks
- Video: [Tutorial on how to calculate the efficiency of an induction motor](https://www.youtube.com/watch?v=1BukDI06z_s) — Marayati Marsadek
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/electrical_machines/im-4pole-975rpm-slip-losses
- Whole chapter: https://www.bctengineer.com/md/semester_4/electrical_machines/chapter-5

## Question

The power input to a 500 V, 50 Hz, 6-pole, 3-phase induction motor running at 975 rpm is 40 kW. The stator losses are 1 kW and friction loss is 2 kW. Calculate: (a) slip (b) rotor copper loss (c) output HP (d) efficiency.

## Short answer

Ns=120×50/6=1000 rpm; s=(1000-975)/1000=0.025 (2.5%). Air-gap power Pg=40-1=39 kW; rotor copper loss sPg=0.975 kW. Net mechanical power=39-0.975=38.025 kW; output=38.025-2=36.025 kW=48.29 HP. Efficiency=36.025/40×100=90.06%.

## Answer

**Given:** 500 V, 50 Hz, 6-pole, $N_r = 975$ rpm, input $P_{in} = 40$ kW, stator loss $= 1$ kW, friction loss $= 2$ kW.

#### (a) Slip
$$
N_s = \frac{120\times 50}{6} = 1000\ \text{rpm}, \qquad s = \frac{1000-975}{1000} = 0.025 = 2.5\%
$$

#### (b) Rotor copper loss
$$
P_g = 40 - 1 = 39\ \text{kW}
$$
$$
P_{cu2} = sP_g = 0.025\times 39 = 0.975\ \text{kW}
$$

#### (c) Output HP
$$
P_{m} = P_g - P_{cu2} = 39 - 0.975 = 38.025\ \text{kW}
$$
$$
P_{out} = 38.025 - 2 = 36.025\ \text{kW} = \frac{36025}{746} = 48.29\ \text{HP}
$$

#### (d) Efficiency
$$
\eta = \frac{36.025}{40}\times 100 = 90.06\%
$$

**Answer:** (a) $s = 2.5\%$; (b) rotor copper loss $= 0.975$ kW; (c) output $= 36.03$ kW $\approx 48.29$ HP; (d) $\eta = 90.06\%$.

## Check yourself

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

1. For 6 poles, 50 Hz and 975 rpm, slip is:
   A. 2.5 %
   B. 5 %
   C. 25 %
   D. 0.25 %
   Answer: A. N_s = 1000 rpm, s = 25/1000 = 0.025. (Key point: Slip)

2. Air-gap power is found by:
   A. Input plus stator loss
   B. Input minus stator loss: 40 - 1 = 39 kW
   C. Input minus friction
   D. Output plus 1 kW
   Answer: B. P_g = 39 kW. (Key point: Air-gap power)

3. Rotor copper loss is:
   A. 0.975 kW
   B. 1 kW
   C. 2 kW
   D. 0.39 kW
   Answer: A. s P_g = 0.025 x 39. (Key point: Rotor copper loss)

4. Efficiency (output 36.025 kW, input 40 kW) is about:
   A. 95 %
   B. 85 %
   C. 97.5 %
   D. 90.06 %
   Answer: D. 36.025/40 = 90.06 %. (Key point: Efficiency)

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