---
title: "Delta induction motor: slip, rotor emf, turns ratio"
subject: "Electrical Machines"
chapter: 5
key: "semester_4/electrical_machines:im-delta-440v-1440rpm-rotor-emf"
source: https://www.bctengineer.com/q/semester_4/electrical_machines/im-delta-440v-1440rpm-rotor-emf
site: BCT Engineering
---

# Delta induction motor: slip, rotor emf, turns ratio

Electrical Machines · Chapter 5: Three Phase Induction Machines

- Asked in: 2077 Chaitra, 8 marks; 2071 Magh, 4 marks (asked 2 times)
- Video: [07 Solved Problems - Slip of Induction Motor #04](https://www.youtube.com/watch?v=hfshBTtiK18) — SP Engg. ACADEMY
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/electrical_machines/im-delta-440v-1440rpm-rotor-emf
- Whole chapter: https://www.bctengineer.com/md/semester_4/electrical_machines/chapter-5

## Question

A three-phase delta-connected 440 volts, 50 Hz, 4-pole induction motor has a rotor standstill emf per phase of 130 volts. If the motor is running at 1440 rpm, calculate the slip, frequency of rotor induced emf, the value of rotor induced emf per phase and stator to rotor turn ratio for same speed.

## Short answer

Synchronous speed is 1500 rpm, so slip is 0.04. Rotor frequency is 2 Hz and rotor emf per phase is 5.2 V. Stator to rotor turns ratio is 440/130 = 3.385.

## Answer

**Synchronous speed:**

$$
N_s = \frac{120f}{P} = \frac{120\times 50}{4} = 1500\ \text{rpm}
$$

#### Slip

$$
s = \frac{N_s - N}{N_s} = \frac{1500 - 1440}{1500} = 0.04 \ (4\%)
$$

#### Frequency of rotor emf

$$
f_r = sf = 0.04\times 50 = 2\ \text{Hz}
$$

#### Rotor induced emf per phase at 1440 rpm

The rotor emf per phase is proportional to slip: $E_{2r} = sE_{2}$, where $E_2 = 130$ V is the standstill value.

$$
E_{2r} = 0.04\times 130 = 5.2\ \text{V}
$$

#### Stator to rotor turns ratio

The stator is delta connected, so the stator phase voltage equals the line voltage, 440 V. At standstill the induction motor behaves like a transformer, so

$$
\frac{N_1}{N_2} = \frac{E_1}{E_2} = \frac{440}{130} = 3.385
$$

The turns ratio is independent of speed, so it is the same at 1440 rpm.

**Answer:** slip $= 0.04$ ($4\%$); rotor frequency $= 2\ \text{Hz}$; rotor emf per phase $= 5.2\ \text{V}$; turns ratio $N_1/N_2 = 3.385$ (stator phase turns : rotor phase turns, about 3.38:1).

## Check yourself

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

1. The slip at 1440 rpm for 4 poles, 50 Hz is:
   A. 0.06
   B. 0.4
   C. 0.04
   D. 0.4%
   Answer: C. Ns = 1500; s = 60/1500 = 0.04. (Key point: Slip)

2. Rotor emf frequency is:
   A. 50 Hz
   B. 2 Hz
   C. 1.2 Hz
   D. 4 Hz
   Answer: B. fr = s f = 0.04 x 50. (Key point: Rotor frequency)

3. Rotor emf per phase at 1440 rpm is:
   A. 130 V
   B. 3.25 V
   C. 5.2 V
   D. 0.04 V
   Answer: C. E2r = s E2 = 0.04 x 130 = 5.2 V. (Key point: Running rotor emf)

4. The stator-to-rotor turns ratio uses which stator voltage?
   A. Line voltage of 440 V equals the phase voltage for delta
   B. 440/sqrt(3)
   C. 440 x sqrt(3)
   D. 220 V
   Answer: A. In delta, phase voltage = line voltage; ratio = 440/130 = 3.385, independent of speed. (Key point: Turns ratio)

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