---
title: "Star-delta starting torque ratio and full-load slip"
subject: "Electrical Machines"
chapter: 5
key: "semester_4/electrical_machines:im-star-delta-starting-torque-ratio"
source: https://www.bctengineer.com/q/semester_4/electrical_machines/im-star-delta-starting-torque-ratio
site: BCT Engineering
---

# Star-delta starting torque ratio and full-load slip

Electrical Machines · Chapter 5: Three Phase Induction Machines

- Asked in: 2067 Mangsir (old course), 8 marks
- Video: [IM 18 Star Delta Starting of 3 phase Induction Motor](https://www.youtube.com/watch?v=EVIWkrtI1Fk) — Lectures in Electrical Engineering
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/electrical_machines/im-star-delta-starting-torque-ratio
- Whole chapter: https://www.bctengineer.com/md/semester_4/electrical_machines/chapter-5

## Question

A 3-ph induction motor has a ratio of maximum torque to full load torque as 2.5:1. Determine the ratio of actual starting torque to full load torque for star delta starting. Also calculate full load slip. [Given: rotor resistance per phase = $0.4\ \Omega$ and rotor reactance per phase at stand still = $4\ \Omega$]

## Short answer

Slip at maximum torque = 0.1. Direct start torque ratio to full load = 0.495. With star-delta starter, ratio = 0.165. Full-load slip = 0.0209.

## Answer

**Given:** $T_{max}/T_{fl} = 2.5$, $R_2 = 0.4\ \Omega$, $X_2 = 4\ \Omega$ at standstill.

#### Slip at maximum torque
$$
s_m = \frac{R_2}{X_2} = \frac{0.4}{4} = 0.1
$$

#### Starting torque with direct switching
$$
\frac{T_{st}}{T_{max}} = \frac{2s_m}{1+s_m^2} = \frac{0.2}{1.01} = 0.198
$$
$$
\frac{T_{st}}{T_{fl}} = 0.198\times 2.5 = 0.495
$$

#### Star-delta starting
In star, the phase voltage is $1/\sqrt3$ of the line voltage, and torque $\propto V^2$, so the starting torque is one third of the direct (delta) value:
$$
\frac{T_{st(Y\Delta)}}{T_{fl}} = \frac{0.495}{3} = 0.165
$$

#### Full-load slip
$$
\frac{T_{fl}}{T_{max}} = 0.4 = \frac{2 s s_m}{s^2+s_m^2}
\;\Rightarrow\; s^2 - 0.5 s + 0.01 = 0
$$
$$
s = \frac{0.5 \pm\sqrt{0.25-0.04}}{2} = 0.0209 \ \text{(stable root)}
$$

**Answer:** starting torque with star-delta starter $= 0.165\,T_{fl}$; full-load slip $= 0.0209$ (2.09 %).

## Check yourself

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

1. How is the slip at maximum torque found from the rotor data?
   A. sm = X2/R2
   B. sm = R2/X2 = 0.1
   C. sm = R2 x X2
   D. sm = R2/(R2+X2)
   Answer: B. Maximum torque occurs when R2 = sX2, so sm = 0.4/4 = 0.1. (Key point: Slip at max torque)

2. What is Tst/Tfl for direct switching?
   A. 0.495
   B. 0.165
   C. 2.5
   D. 0.198
   Answer: A. Tst/Tmax = 2sm/(1+sm^2) = 0.198, times Tmax/Tfl = 2.5 gives 0.495. (Key point: Direct starting torque)

3. Why is the starting torque with a star-delta starter one third of the direct value?
   A. Rotor resistance triples
   B. Phase voltage is 1/sqrt3 and torque is proportional to V^2
   C. Frequency drops by 3
   D. Slip is divided by 3
   Answer: B. In star each phase gets V/sqrt3, so torque falls by (1/sqrt3)^2 = 1/3, giving 0.165 Tfl. (Key point: Star-delta effect)

4. Which equation is solved to find the full-load slip?
   A. s^2 - 0.5s + 0.01 = 0 from Tfl/Tmax = 0.4 = 2 s sm/(s^2+sm^2)
   B. s = sm x 2.5
   C. s^2 + 0.5s + 0.01 = 0
   D. s = 0.4 x 0.1
   Answer: A. Tfl/Tmax = 0.4 gives s^2 - 5 sm s + sm^2 = 0, i.e. s^2 - 0.5s + 0.01 = 0, with stable root 0.0209. (Key point: Full-load slip equation)

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