---
title: "Speed and value of maximum torque of 2-pole motor"
subject: "Electrical Machines I"
chapter: 5
key: "semester_4/electrical_machines_i:im-2pole-2ohm-8ohm-max-torque"
source: https://www.bctengineer.com/q/semester_4/electrical_machines_i/im-2pole-2ohm-8ohm-max-torque
site: BCT Engineering
---

# Speed and value of maximum torque of 2-pole motor

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

- Asked in: 2076 Baisakh, 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-2pole-2ohm-8ohm-max-torque
- Whole chapter: https://www.bctengineer.com/md/semester_4/electrical_machines_i/chapter-5

## Question

A 3 phase, 400 V, 50 Hz, 2 pole induction motor has rotor circuit resistance of 2 Ω and rotor circuit reactance of 8 Ω at stand still. It develops a starting torque of 10 N-m. At which speed, the motor develops maximum torque and calculate the value of maximum torque.

## Short answer

At slip 0.25, corresponding to 2250 rpm, the motor develops maximum torque. Maximum torque is 2.125 times starting torque, giving 21.25 N·m.

## Answer

**Given:** 2 poles, 50 Hz, $R_2 = 2\ \Omega$, $X_2 = 8\ \Omega$ (standstill), $T_{st} = 10$ N·m. Stator impedance neglected.

#### Speed at maximum torque

$$
\begin{aligned}
N_s &= \frac{120 f}{P} = \frac{120 \times 50}{2} = 3000\ \text{rpm} \\
s_m &= \frac{R_2}{X_2} = \frac{2}{8} = 0.25 \\
N_m &= (1 - s_m)N_s = 0.75 \times 3000 = 2250\ \text{rpm}
\end{aligned}
$$

#### Maximum torque

$$
T_{st} = \frac{K E_2^2 R_2}{R_2^2 + X_2^2}, \qquad T_{max} = \frac{K E_2^2}{2X_2}
$$

$$
\begin{aligned}
\frac{T_{max}}{T_{st}} &= \frac{R_2^2 + X_2^2}{2R_2X_2} = \frac{2^2 + 8^2}{2 \times 2 \times 8} = \frac{68}{32} = 2.125 \\
T_{max} &= 2.125 \times 10 = 21.25\ \text{N·m}
\end{aligned}
$$

**Answer: The motor develops maximum torque at 2250 rpm (slip 0.25), and $T_{max} = 21.25$ N·m.**

## Check yourself

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

1. With R2 = 2 Ω and X2 = 8 Ω, what is the slip at maximum torque?
   A. 4
   B. 0.25
   C. 0.5
   D. 0.0625
   Answer: B. sm = R2/X2 = 2/8 = 0.25. (Key point: Slip at max torque)

2. For a 2-pole 50 Hz motor (Ns = 3000 rpm), what is the speed at maximum torque?
   A. 750 rpm
   B. 3000 rpm
   C. 2250 rpm
   D. 1500 rpm
   Answer: C. N = (1 - 0.25) x 3000 = 2250 rpm. (Key point: Speed at max torque)

3. Which ratio gives Tmax/Tst?
   A. (R2^2 + X2^2)/(2 R2 X2) = 2.125
   B. 2 R2 X2/(R2^2 + X2^2)
   C. X2/R2 = 4
   D. (R2 + X2)/(2 R2) = 2.5
   Answer: A. Dividing Tmax = KE2^2/(2X2) by Tst = KE2^2R2/(R2^2+X2^2) gives 68/32 = 2.125, so Tmax = 21.25 N-m. (Key point: Tmax from Tst)

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