---
title: "Slip ring motor: speed at max torque, resistance"
subject: "Electrical Machines"
chapter: 5
key: "semester_4/electrical_machines:im-slipring-max-torque-ratio"
source: https://www.bctengineer.com/q/semester_4/electrical_machines/im-slipring-max-torque-ratio
site: BCT Engineering
---

# Slip ring motor: speed at max torque, resistance

Electrical Machines · Chapter 5: Three Phase Induction Machines

- Asked in: 2074 Bhadra (old course)
- Video: [SPEED TORQUE CHARACTERISTICS OF SLIP RING MOTOR, HOW TO USE EXTERNAL RESISTANCE IN SLIP RING MOTOR,](https://www.youtube.com/watch?v=yDlzN4R9y08) — Vijay kumar jain Tech Electrical
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/electrical_machines/im-slipring-max-torque-ratio
- Whole chapter: https://www.bctengineer.com/md/semester_4/electrical_machines/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 max torque = 1350 rpm (slip 0.1). T_max / T_st = (1+0.1²)/(2×0.1) = 5.05. For T_st = 0.5 T_max, required rotor resistance per phase = 1.072 Ω (add 0.672 Ω external).

## Answer

**Given:** 4-pole, 50 Hz, $R_2 = 0.4\ \Omega$, $X_2 = 4\ \Omega$ per phase (standstill).

#### Speed at maximum torque
$$
N_s = \frac{120\times 50}{4} = 1500\ \text{rpm}, \qquad s_m = \frac{R_2}{X_2} = \frac{0.4}{4} = 0.1
$$
$$
N_m = N_s(1-s_m) = 1500\times 0.9 = 1350\ \text{rpm}
$$

#### Ratio $T_{max}/T_{st}$
$$
\frac{T_{st}}{T_{max}} = \frac{2s_m}{1+s_m^2} = \frac{0.2}{1.01} = 0.198
\;\Rightarrow\;
\frac{T_{max}}{T_{st}} = \frac{1+s_m^2}{2s_m} = \frac{1.01}{0.2} = 5.05
$$

#### Resistance for $T_{st} = \frac12 T_{max}$
Let the new total rotor resistance be $R_2'$, so $s_m' = R_2'/X_2$:
$$
\frac{2s_m'}{1+s_m'^2} = 0.5 \;\Rightarrow\; s_m'^2 - 4s_m' + 1 = 0
$$
$$
s_m' = 2 - \sqrt3 = 0.2679 \quad (\text{the root} < 1)
$$
$$
R_2' = s_m' X_2 = 0.2679\times 4 = 1.0718\ \Omega
$$
Additional resistance to be added per phase $= 1.0718 - 0.4 = 0.6718\ \Omega$.

**Answer:** speed at $T_{max}$ = 1350 rpm; $T_{max}/T_{st} = 5.05$; rotor resistance per phase $= 1.072\ \Omega$ (external $0.672\ \Omega$ added).

## Check yourself

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

1. R_2 = 0.4, X_2 = 4, 4 poles: speed at max torque is:
   A. 1500 rpm
   B. 1350 rpm
   C. 1200 rpm
   D. 150 rpm
   Answer: B. s_m = 0.1, so 1500 x 0.9 = 1350. (Key point: Speed at Tmax)

2. T_max/T_st =
   A. 0.198
   B. 10
   C. 5.05
   D. 2
   Answer: C. (1 + 0.01)/0.2 = 5.05. (Key point: Torque ratio)

3. For T_st = T_max/2, the equation for s_m' is:
   A. s^2 - 4s + 1 = 0
   B. s^2 + 4s + 1 = 0
   C. s^2 - 2s + 1 = 0
   D. s^2 - s + 4 = 0
   Answer: A. 2s/(1 + s^2) = 0.5. (Key point: Condition)

4. Total rotor resistance required is about:
   A. 0.4 ohm
   B. 0.67 ohm
   C. 1.07 ohm
   D. 2 ohm
   Answer: C. R_2' = 0.2679 x 4 = 1.072 ohm. (Key point: Resistance)

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