---
title: "Slip at max torque and full load, starting current"
subject: "Electrical Machines"
chapter: 5
key: "semester_4/electrical_machines:im-slip-torque-ratios"
source: https://www.bctengineer.com/q/semester_4/electrical_machines/im-slip-torque-ratios
site: BCT Engineering
---

# Slip at max torque and full load, starting current

Electrical Machines · Chapter 5: Three Phase Induction Machines

- Asked in: 2070 Bhadra, 8 marks
- Video: [Starting torque, maximum torque, full load torque & their relation of 3 phase induction motor](https://www.youtube.com/watch?v=bUo0WLRtDsQ) — Edu fun zo
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/electrical_machines/im-slip-torque-ratios
- Whole chapter: https://www.bctengineer.com/md/semester_4/electrical_machines/chapter-5

## Question

A 3-phase, 50 Hz induction motor has starting torque which is 1.25 times full load torque and a maximum torque which is 2.5 times the full load torque. Neglecting stator resistance and rotational losses and assuming constant rotor resistance. Find
i) slip at maximum torque
ii) the slip at full load
iii) the current at starting in per unit full load current

## Short answer

Slip at maximum torque = 0.268. Full-load slip = 0.0559. Starting current per unit of full-load current = 4.73.

## Answer

**Given:** $T_{st} = 1.25\,T_{fl}$, $T_{max} = 2.5\,T_{fl}$, stator resistance and rotational losses neglected, constant $R_2$.

#### (i) Slip at maximum torque
$$
\frac{T_{st}}{T_{max}} = \frac{1.25}{2.5} = 0.5 = \frac{2s_m}{1+s_m^2}
\;\Rightarrow\; s_m^2 - 4s_m + 1 = 0
$$
$$
s_m = 2-\sqrt3 = 0.2679
$$

#### (ii) Slip at full load
$$
\frac{T_{fl}}{T_{max}} = \frac{1}{2.5} = 0.4 = \frac{2 s s_m}{s^2+s_m^2}
\;\Rightarrow\; s^2 - 5 s_m s + s_m^2 = 0
$$
$$
s = s_m\,\frac{5 \pm\sqrt{21}}{2}
$$
Taking the smaller root (the stable region, $s < s_m$):
$$
s_{fl} = 0.2679\times 0.2087 = 0.0559
$$

#### (iii) Starting current in per unit of full-load current
Rotor current: $I_2 = \dfrac{E_2}{\sqrt{(R_2/s)^2 + X_2^2}}$. With $R_2 = s_mX_2$:
$$
\frac{I_{st}}{I_{fl}} = \frac{\sqrt{(R_2/s_{fl})^2 + X_2^2}}{\sqrt{R_2^2 + X_2^2}}
= \frac{\sqrt{(s_m/s_{fl})^2+1}}{\sqrt{s_m^2+1}}
$$
$$
= \frac{\sqrt{(0.2679/0.0559)^2 + 1}}{\sqrt{0.2679^2 + 1}} = \frac{\sqrt{23.96}}{1.0353}= 4.73
$$

**Answer:** (i) $s_m = 0.268$; (ii) $s_{fl} = 0.0559$; (iii) $I_{st} = 4.73$ p.u. of full-load current.

## Check yourself

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

1. With stator resistance neglected, what relation links Tst/Tmax to the slip at maximum torque sm?
   A. Tst/Tmax = 2sm/(1+sm^2)
   B. Tst/Tmax = sm/(1+sm)
   C. Tst/Tmax = 2sm^2/(1+sm)
   D. Tst/Tmax = (1+sm^2)/2sm
   Answer: A. T/Tmax = 2 s sm/(s^2+sm^2) at s = 1 gives 2sm/(1+sm^2). (Key point: Torque-slip ratio)

2. Tst/Tmax = 0.5. What is sm?
   A. 0.5
   B. 0.2679
   C. 0.0559
   D. 3.732
   Answer: B. 0.5 = 2sm/(1+sm^2) gives sm^2 - 4sm + 1 = 0, so sm = 2 - sqrt3 = 0.268 (the root below 1). (Key point: Solving for sm)

3. Why is the smaller root chosen for the full-load slip?
   A. The smaller root gives a negative slip
   B. The motor runs in the stable region where s < sm
   C. The larger root gives zero torque
   D. The larger root is greater than 1 only
   Answer: B. Stable operation is on the part of the curve where torque rises with slip, i.e. s < sm. (Key point: Choosing the root)

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