---
title: "Slip, rotor copper loss, shaft power and efficiency at 975 rpm"
subject: "Electrical Machines I"
chapter: 5
key: "semester_4/electrical_machines_i:im-500v-6pole-975rpm-40kw"
source: https://www.bctengineer.com/q/semester_4/electrical_machines_i/im-500v-6pole-975rpm-40kw
site: BCT Engineering
---

# Slip, rotor copper loss, shaft power and efficiency at 975 rpm

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

- Asked in: 2069 Poush, 8 marks
- Video: [IM Q#2 Numerical on power relation, slip in hindi](https://www.youtube.com/watch?v=YcOSAlNMctc) — Electrical paathshala
- 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-500v-6pole-975rpm-40kw
- Whole chapter: https://www.bctengineer.com/md/semester_4/electrical_machines_i/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 the friction and windage losses total 2 kW. Calculate: (i) The slip (ii) The rotor copper loss (iii) Shaft power and (iv) The efficiency.

## Short answer

Ns = 120×50/6 = 1000 rpm; s = (1000-975)/1000 = 0.025. Rotor input Pag = 40-1 = 39 kW; rotor copper loss = s×Pag = 0.025×39 = 0.975 kW. Shaft power = 39-0.975-2 = 36.025 kW; efficiency = 36.025/40 = 90.06%.

## Answer

**Given:** 500 V, 50 Hz, 6 poles, $N = 975$ rpm, input $P_{in} = 40$ kW, stator losses $= 1$ kW, friction and windage losses $= 2$ kW.

#### (i) Slip

$$
\begin{aligned}
N_s &= \frac{120 \times 50}{6} = 1000\ \text{rpm} \\
s &= \frac{1000 - 975}{1000} = 0.025 \ (2.5\%)
\end{aligned}
$$

#### (ii) Rotor copper loss

$$
\begin{aligned}
P_{ag} &= P_{in} - \text{stator losses} = 40 - 1 = 39\ \text{kW} \\
P_{cu2} &= sP_{ag} = 0.025 \times 39 = 0.975\ \text{kW} = 975\ \text{W}
\end{aligned}
$$

#### (iii) Shaft power

$$
\begin{aligned}
P_m &= P_{ag} - P_{cu2} = 39 - 0.975 = 38.025\ \text{kW} \\
P_{shaft} &= P_m - \text{friction and windage} = 38.025 - 2 = 36.025\ \text{kW}
\end{aligned}
$$

#### (iv) Efficiency

$$
\eta = \frac{P_{shaft}}{P_{in}} = \frac{36.025}{40} = 0.9006 = 90.06\%
$$

```text
 40 kW in --> [-1 kW stator] --> 39 kW air gap
          --> [-0.975 kW rotor Cu] --> 38.025 kW
          --> [-2 kW F&W] --> 36.025 kW shaft
```

**Answer: (i) s = 0.025; (ii) rotor Cu loss = 975 W; (iii) shaft power = 36.025 kW; (iv) η = 90.06%.**

## Check yourself

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

1. What is the slip for a 6-pole 50 Hz motor at 975 rpm?
   A. 0.025
   B. 0.975
   C. 0.25
   D. 0.0256
   Answer: A. Ns = 1000 rpm, so s = (1000 - 975)/1000 = 0.025. (Key point: Slip)

2. How is air-gap power found from the input?
   A. P_ag = P_in - friction and windage = 38 kW
   B. P_ag = P_in - stator losses = 39 kW
   C. P_ag = P_in x (1 - s)
   D. P_ag = P_in + stator losses = 41 kW
   Answer: B. Stator losses (copper and core) are lost before the power crosses the air gap. (Key point: Air-gap power)

3. What is the rotor copper loss?
   A. 1000 W
   B. 38.025 kW
   C. 975 W
   D. 1.0 kW
   Answer: C. P_cu2 = s x P_ag = 0.025 x 39 kW = 0.975 kW. (Key point: Rotor copper loss)

4. What are shaft power and efficiency?
   A. 38.025 kW and 95.06%
   B. 36.025 kW and 92.4%
   C. 39 kW and 97.5%
   D. 36.025 kW and 90.06%
   Answer: D. P_shaft = P_ag - P_cu2 - friction and windage = 36.025 kW, and 36.025/40 = 90.06%. (Key point: Shaft power and efficiency)

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## Related in other subjects

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