---
title: "Max torque speed and Tmax/Tst of slip ring motor"
subject: "Electrical Machines I"
chapter: 5
key: "semester_4/electrical_machines_i:slip-ring-0p4-4ohm-max-start"
source: https://www.bctengineer.com/q/semester_4/electrical_machines_i/slip-ring-0p4-4ohm-max-start
site: BCT Engineering
---

# Max torque speed and Tmax/Tst of slip ring motor

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

- Asked in: 2074 Bhadra, 8 marks
- Video: [Numerical problems on 3-phase induction motors: Maximum Torque & Corresponding slip](https://www.youtube.com/watch?v=14-83DFgA4Q) — Gargi Basu
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/electrical_machines_i/slip-ring-0p4-4ohm-max-start
- Whole chapter: https://www.bctengineer.com/md/semester_4/electrical_machines_i/chapter-5

## Question

The rotor resistance and reactance of a 4-pole, 50 Hz, 3-phase slip ring induction motor are 0.4 and 4 ohm/phase respectively at stand still. Calculate the speed at maximum torque and the ratio (max torque)/(starting torque). What value should the resistance per phase have so that the starting torque is half of maximum torque?

## Short answer

Speed at maximum torque is 1350 rpm (slip 0.1). The ratio of maximum to starting torque is 5.05. To obtain starting torque half of maximum, rotor resistance per phase must be 1.072 Ω, requiring an extra 0.672 Ω.

## Answer

**Given:** 4 poles, 50 Hz, $R_2 = 0.4\ \Omega$, $X_2 = 4\ \Omega$ per phase at standstill. Stator impedance neglected.

#### Speed at maximum torque

$$
\begin{aligned}
N_s &= \frac{120 \times 50}{4} = 1500\ \text{rpm} \\
s_m &= \frac{R_2}{X_2} = \frac{0.4}{4} = 0.1 \\
N_m &= (1 - 0.1) \times 1500 = 1350\ \text{rpm}
\end{aligned}
$$

#### Ratio of maximum to starting torque

$$
\begin{aligned}
\frac{T_{max}}{T_{st}} &= \frac{R_2^2 + X_2^2}{2R_2X_2} = \frac{0.16 + 16}{2 \times 0.4 \times 4} = \frac{16.16}{3.2} = 5.05
\end{aligned}
$$

#### Rotor resistance for $T_{st} = \tfrac{1}{2} T_{max}$

Let total rotor resistance per phase be $R$ and $a = R/X_2$:

$$
\frac{T_{st}}{T_{max}} = \frac{2RX_2}{R^2 + X_2^2} = \frac{2a}{1 + a^2} = \frac{1}{2}
$$

$$
\begin{aligned}
a^2 - 4a + 1 &= 0 \\
a &= \frac{4 \pm \sqrt{16 - 4}}{2} = 2 \pm \sqrt{3}
\end{aligned}
$$

Take the smaller root (the larger one, $a = 3.73$, also works but wastes much more power and gives lower running speed):

$$
\begin{aligned}
a &= 2 - \sqrt{3} = 0.2679 \\
R &= 0.2679 \times 4 = 1.072\ \Omega
\end{aligned}
$$

External resistance to be added per phase $= 1.072 - 0.4 = 0.672\ \Omega$.

Check: $2 \times 1.072 \times 4/(1.072^2 + 16) = 8.574/17.149 = 0.5$.

**Answer: Speed at $T_{max}$ = 1350 rpm; $T_{max}/T_{st}$ = 5.05; rotor resistance needed = 1.072 Ω per phase (i.e. 0.672 Ω extra per phase).**

## Check yourself

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

1. R2 = 0.4 Ω, X2 = 4 Ω, 4-pole 50 Hz. What is the speed at maximum torque?
   A. 1500 rpm
   B. 1650 rpm
   C. 1350 rpm
   D. 150 rpm
   Answer: C. sm = 0.1 so N = 0.9 x 1500 = 1350 rpm. (Key point: Speed at max torque)

2. What is Tmax/Tst?
   A. 10
   B. 0.2
   C. 2.5
   D. 5.05
   Answer: D. (R2^2 + X2^2)/(2 R2 X2) = 16.16/3.2 = 5.05. (Key point: Tmax/Tst)

3. Which equation gives the required total rotor resistance R (a = R/X2)?
   A. 2a/(1 + a^2) = 1/2, giving a^2 - 4a + 1 = 0
   B. a = 1/2
   C. 2a/(1 + a^2) = 2
   D. a^2 + 4a + 1 = 0
   Answer: A. Tst/Tmax = 2RX2/(R^2 + X2^2) = 2a/(1+a^2); setting it to 1/2 gives a^2 - 4a + 1 = 0. (Key point: Condition for Tst = Tmax/2)

4. Taking a = 2 - sqrt3 = 0.2679, what external resistance per phase must be added?
   A. 1.072 Ω
   B. 0.4 Ω
   C. 0.672 Ω
   D. 0.268 Ω
   Answer: C. R = 0.2679 x 4 = 1.072 Ω total; subtracting the 0.4 Ω rotor resistance leaves 0.672 Ω. (Key point: External resistance)

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