---
title: "3-phase induction machine at slip 0.05"
subject: "Basic Electrical and Electronics Engineering"
chapter: 2
key: "semester_1/basic_electrical_and_electronics_engineering:induction-machine-3phase-numerical"
source: https://www.bctengineer.com/q/semester_1/basic_electrical_and_electronics_engineering/induction-machine-3phase-numerical
site: BCT Engineering
---

# 3-phase induction machine at slip 0.05

Basic Electrical and Electronics Engineering · Chapter 2: Electrical Machines

- Asked in: 2081 Baishakh (new course), 6 marks
- Video: [(B)EM (5th ed) || Example 6.5 || Slip, Power, and Torque of Induction Motor](https://www.youtube.com/watch?v=z9j23rttgWY) — Electrical Engineering Academy
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_1/basic_electrical_and_electronics_engineering/induction-machine-3phase-numerical
- Whole chapter: https://www.bctengineer.com/md/semester_1/basic_electrical_and_electronics_engineering/chapter-2

## Question

A 3-phase, 400 V, 80 kW, 50 Hz, 4-pole induction machine delivers rated output power at a slip of 0.05. Determine the: a) Synchronous speed, b) Motor speed, c) Frequency of the rotor current, d) Rotor induced voltage at the operating speed, if the stator-to-rotor turns ratio is 1.5 and e) Rated torque.

## Short answer

a) Synchronous speed Ns = 120×50/4 = 1500 rpm. b) Motor speed = 1500(1-0.05) = 1425 rpm. c) Rotor frequency = 0.05×50 = 2.5 Hz. d) Stator phase V = 400/√3 = 230.9 V; standstill rotor emf = 230.9/1.5 = 153.9 V; running emf = 0.05×153.9 = 7.70 V per phase. e) Torque = 80000 / (2π×1425/60) = 536.1 N·m.

## Answer

**Given:** $V_L=400$ V, output $=80$ kW, $f=50$ Hz, $P=4$ poles, $s=0.05$, stator : rotor turns $=1.5$. The stator is taken as star-connected.

**a) Synchronous speed:**

$$
N_s=\frac{120f}{P}=\frac{120\times50}{4}=1500\ \text{rpm}
$$

**b) Motor speed:**

$$
N=N_s(1-s)=1500\times0.95=1425\ \text{rpm}
$$

**c) Rotor current frequency:**

$$
f_r=sf=0.05\times50=2.5\ \text{Hz}
$$

**d) Rotor induced voltage at running speed.** Stator phase voltage:

$$
V_1=\frac{400}{\sqrt3}=230.9\ \text{V}
$$

Standstill rotor emf per phase $E_2=\frac{V_1}{1.5}=153.9$ V. At slip $s$ the rotor emf is $sE_2$:

$$
E_{2r}=0.05\times153.9=7.70\ \text{V per phase}
$$

(Line value $= \sqrt3\times7.70 = 13.33$ V.)

**e) Rated torque** (shaft output 80 kW at 1425 rpm):

$$
\omega=\frac{2\pi\times1425}{60}=149.23\ \text{rad/s}, \qquad T=\frac{P_{out}}{\omega}=\frac{80000}{149.23}=536.1\ \text{N·m}
$$

**Answer: a) 1500 rpm, b) 1425 rpm, c) 2.5 Hz, d) 7.70 V per phase, e) 536.1 N·m.**

## Check yourself

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

1. For 4 poles, 50 Hz and slip 0.05, what are the synchronous speed and motor speed?
   A. 1500 rpm and 1575 rpm
   B. 3000 rpm and 2850 rpm
   C. 1500 rpm and 1425 rpm
   D. 1425 rpm and 1500 rpm
   Answer: C. Ns = 120 x 50/4 = 1500 rpm and N = Ns(1 - s) = 1425 rpm. (Key point: Motor speed)

2. What is the rotor current frequency?
   A. 50 Hz
   B. 47.5 Hz
   C. 25 Hz
   D. 2.5 Hz
   Answer: D. fr = s f = 0.05 x 50 = 2.5 Hz. (Key point: Rotor frequency)

3. How is the rotor induced emf at the running speed found?
   A. Standstill emf E2 = V1/1.5 = 153.9 V per phase, then multiplied by slip: 7.70 V
   B. 400 x 1.5 x 0.05 = 30 V
   C. V1 x 1.5 = 346 V
   D. Equal to 153.9 V at all speeds
   Answer: A. Rotor emf at slip s is s x E2, with E2 = (400/sqrt(3))/1.5 = 153.9 V, giving 7.70 V. (Key point: Rotor emf)

4. How is rated torque calculated from the shaft output?
   A. T = P/Ns in rpm = 53.3
   B. T = P/omega with omega = 2 pi N/60 at 1425 rpm: 536.1 N·m
   C. T = P/omega at 1500 rpm: 509 N·m
   D. T = P x omega
   Answer: B. Shaft torque is output power divided by actual angular speed: 80000/149.23 = 536.1 N·m. (Key point: Rated torque)

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