---
title: "Max-torque speed and efficiency at 1350 rpm"
subject: "Electrical Machines I"
chapter: 5
key: "semester_4/electrical_machines_i:slip-ring-380v-180v-500w-1350rpm"
source: https://www.bctengineer.com/q/semester_4/electrical_machines_i/slip-ring-380v-180v-500w-1350rpm
site: BCT Engineering
---

# Max-torque speed and efficiency at 1350 rpm

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

- Asked in: 2072 Asoj, 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-380v-180v-500w-1350rpm
- Whole chapter: https://www.bctengineer.com/md/semester_4/electrical_machines_i/chapter-5

## Question

A 380 V, 4-pole, 50 Hz, 3 phase, slip ring induction motor has a star connected stator winding and a star connected rotor winding. At standstill the voltage between the two slip rings is 180 V. The stator impedance is 0.5+j2.5 ohm. The rotor resistance and reactance at standstill are 0.06 ohm and 0.3 ohm respectively. The stator to rotor turn ratio is 1:1. The motor consumes 500 watts at no-load. Calculate: i) Speed at which the motor develops the maximum torque ii) Efficiency of the motor when it is running at 1350 rpm

## Short answer

Effective ratio K = 180/380, referred R2'=0.2674 Ω, X'=1.337 Ω, total X=3.837 Ω. Speed at max torque: sm = R2'/√(R1²+X²)=0.0691, Nm = 1396 rpm. At 1350 rpm, s=0.1: input 18.48 kW, output 13.51 kW, efficiency 73.1%.

## Answer

**Given:** $V_L = 380$ V (star), 4 poles, 50 Hz, slip-ring voltage at standstill $= 180$ V (star), $Z_1 = 0.5 + j2.5\ \Omega$, $R_2 = 0.06\ \Omega$, $X_2 = 0.3\ \Omega$, no-load input $= 500$ W, $N = 1350$ rpm.

**Assumption:** the stated 1:1 turns ratio conflicts with the measured 380 V / 180 V; the measured voltages are used to find the effective ratio $K = 180/380$, and the rotor values are referred to the stator. The approximate equivalent circuit is used (no-load current neglected in the series branch), and the 500 W no-load loss is taken as the constant (core + mechanical) loss.

#### Rotor values referred to stator

$$
\begin{aligned}
K &= \frac{180}{380} = 0.47368, \qquad K^2 = 0.22438 \\
R_2' &= \frac{R_2}{K^2} = \frac{0.06}{0.22438} = 0.2674\ \Omega \\
X_2' &= \frac{X_2}{K^2} = \frac{0.3}{0.22438} = 1.3370\ \Omega \\
X &= X_1 + X_2' = 2.5 + 1.337 = 3.837\ \Omega
\end{aligned}
$$

#### (i) Speed at maximum torque

$$
\begin{aligned}
s_m &= \frac{R_2'}{\sqrt{R_1^2 + X^2}} = \frac{0.2674}{\sqrt{0.25 + 14.723}} = \frac{0.2674}{3.8695} = 0.0691 \\
N_m &= (1 - 0.0691) \times 1500 = 1396.3\ \text{rpm}
\end{aligned}
$$

#### (ii) Efficiency at 1350 rpm

$$
s = \frac{1500 - 1350}{1500} = 0.1, \qquad \frac{R_2'}{s} = 2.674\ \Omega
$$

$$
\begin{aligned}
V_{ph} &= \frac{380}{\sqrt{3}} = 219.39\ \text{V} \\
Z &= (0.5 + 2.674) + j3.837 = 3.174 + j3.837 = 4.980\ \Omega \\
I_1 &= \frac{219.39}{4.980} = 44.06\ \text{A}
\end{aligned}
$$

$$
\begin{aligned}
P_{in} &= 3I_1^2(R_1 + R_2'/s) = 3 \times 44.06^2 \times 3.174 = 18\,483\ \text{W} \\
P_{cu1} &= 3 \times 44.06^2 \times 0.5 = 2912\ \text{W} \\
P_{ag} &= 3 \times 44.06^2 \times 2.674 = 15\,572\ \text{W} \\
P_{cu2} &= sP_{ag} = 1557\ \text{W} \\
P_m &= P_{ag} - P_{cu2} = 14\,014\ \text{W} \\
P_{out} &= P_m - 500 = 13\,514\ \text{W} \\
\eta &= \frac{13\,514}{18\,483} = 0.7312 = 73.1\%
\end{aligned}
$$

(Note: 1350 rpm lies beyond the maximum-torque speed, in the high-slip region, which is why the efficiency is low.)

**Answer: (i) maximum torque at about 1396 rpm ($s_m = 0.0691$); (ii) efficiency at 1350 rpm $\approx 73.1\%$.**

## Check yourself

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

1. The question states a 1:1 turn ratio, but the answer uses K = 180/380. Why?
   A. The measured standstill voltages (180 V rotor, 380 V stator) disagree with 1:1, so they give the effective ratio
   B. Because the rotor is star connected and the stator is delta
   C. Because K must always equal the rotor voltage divided by the slip
   D. Because the 1:1 ratio only applies at full load
   Answer: A. The measured open-circuit rotor voltage versus the stator voltage gives the real ratio K = 180/380, which is used to refer rotor values to the stator. (Key point: Effective turns ratio)

2. How is the rotor resistance referred to the stator in this answer?
   A. R2' = R2 x K = 0.0284 ohm
   B. R2' = R2/K^2 = 0.2674 ohm
   C. R2' = R2 x K^2 = 0.0135 ohm
   D. R2' = R2/K = 0.1266 ohm
   Answer: B. Impedances refer through the square of the ratio; since K = 0.4737, dividing 0.06 by K^2 gives 0.2674 ohm. (Key point: Referring rotor values)

3. Which expression gives the slip at maximum torque here?
   A. s_m = X/R2'
   B. s_m = R1/(X1 + X2')
   C. s_m = R2'/sqrt(R1^2 + X^2), with X = X1 + X2'
   D. s_m = R2'/(R1 + R2')
   Answer: C. Maximum power transfer to R2'/s occurs when R2'/s equals the magnitude of the rest of the impedance, giving s_m = 0.0691 and about 1396 rpm. (Key point: Slip at maximum torque)

4. At 1350 rpm (s = 0.1), how is the output power obtained?
   A. P_out = P_ag - 500 W
   B. P_out = P_in - 500 W
   C. P_out = s P_ag - 500 W
   D. P_out = P_m - 500 W, where P_m = P_ag - s P_ag
   Answer: D. Mechanical power developed is P_ag(1 - s); subtracting the 500 W constant (core + mechanical) loss gives the shaft output of about 13.5 kW. (Key point: Output power and efficiency)

## Close variants set in other exams

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