---
title: "Torque at 1450 rpm from starting torque"
subject: "Electrical Machines I"
chapter: 5
key: "semester_4/electrical_machines_i:im-4pole-150nm-1450rpm-0p5-2ohm"
source: https://www.bctengineer.com/q/semester_4/electrical_machines_i/im-4pole-150nm-1450rpm-0p5-2ohm
site: BCT Engineering
---

# Torque at 1450 rpm from starting torque

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

- Asked in: 2073 Magh, 8 marks
- 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-4pole-150nm-1450rpm-0p5-2ohm
- Whole chapter: https://www.bctengineer.com/md/semester_4/electrical_machines_i/chapter-5

## Question

A 4 pole, 50 Hz, 3 phase induction motor develops starting torque of 150 N-m. Calculate the torque developed by the motor when running at a speed of 1450 rpm. Given that rotor resistance and reactance at stand still are 0.5 Ω and 2 Ω respectively and rotor EMF at stand still is 400 V per phase.

## Short answer

Slip s = (1500-1450)/1500 = 0.0333. Torque ratio T/Tst = s(R2²+X2²)/(R2²+(sX2)²) = 0.0333×(0.25+4)/(0.25+(0.0667)²) = 0.5568. Developed torque at 1450 rpm is 0.5568×150 = 83.5 N·m.

## Answer

**Given:** 4 poles, 50 Hz, $T_{st} = 150$ N·m, $R_2 = 0.5\ \Omega$, $X_2 = 2\ \Omega$ at standstill, $E_2 = 400$ V/phase, $N = 1450$ rpm. Stator impedance neglected.

#### Slip

$$
\begin{aligned}
N_s &= \frac{120 \times 50}{4} = 1500\ \text{rpm} \\
s &= \frac{1500 - 1450}{1500} = 0.03333
\end{aligned}
$$

#### Torque ratio

$$
T = \frac{K s E_2^2 R_2}{R_2^2 + (sX_2)^2}, \qquad T_{st} = \frac{K E_2^2 R_2}{R_2^2 + X_2^2}
$$

Since $E_2$ is the same in both, it cancels:

$$
\begin{aligned}
\frac{T}{T_{st}} &= \frac{s(R_2^2 + X_2^2)}{R_2^2 + (sX_2)^2} \\
&= \frac{0.03333 \times (0.25 + 4)}{0.25 + (0.03333 \times 2)^2} \\
&= \frac{0.14167}{0.25 + 0.00444} = \frac{0.14167}{0.25444} = 0.5568
\end{aligned}
$$

$$
T = 0.5568 \times 150 = 83.52\ \text{N·m}
$$

Rotor current at this speed (for information): $I_2 = \dfrac{sE_2}{\sqrt{R_2^2 + (sX_2)^2}} = \dfrac{13.33}{0.5044} = 26.4$ A.

**Answer: Torque at 1450 rpm $\approx 83.5$ N·m.**

## Check yourself

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

1. What is the slip at 1450 rpm for a 4-pole 50 Hz motor?
   A. 0.0345
   B. 0.0333
   C. 0.967
   D. 0.05
   Answer: B. s = (1500 - 1450)/1500 = 0.0333. (Key point: Slip)

2. Why can E2 be ignored in finding T/Tst?
   A. Because E2 is zero at standstill
   B. Because E2 is the same in T and Tst and cancels
   C. Because E2 is proportional to slip
   D. Because E2 is replaced by R2
   Answer: B. Both torques use the same standstill rotor emf E2, so it cancels in the ratio. (Key point: Torque ratio method)

3. With T/Tst = 0.5568 and Tst = 150 N-m, what is the torque at 1450 rpm?
   A. 150 N-m
   B. 26.4 N-m
   C. 269 N-m
   D. 83.5 N-m
   Answer: D. T = 0.5568 x 150 = 83.5 N-m. (Key point: Result)

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