---
title: "Maximum torque and torque at 1430 rpm of slip ring motor"
subject: "Electrical Machines I"
chapter: 5
key: "semester_4/electrical_machines_i:slip-ring-400v-200v-10nm-1430rpm"
source: https://www.bctengineer.com/q/semester_4/electrical_machines_i/slip-ring-400v-200v-10nm-1430rpm
site: BCT Engineering
---

# Maximum torque and torque at 1430 rpm of slip ring motor

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

- Asked in: 2076 Bhadra, 4+4 marks; 2069 Bhadra, 8 marks (asked 2 times)
- 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-400v-200v-10nm-1430rpm
- Whole chapter: https://www.bctengineer.com/md/semester_4/electrical_machines_i/chapter-5

## Question

A 400 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 200 V. The stator impedance is 0.8+j3 ohm. The rotor resistance and standstill reactance are 0.08 ohm and 0.25 ohm respectively. The motor produces a torque of 10 Nm at starting. Calculate: (i) Maximum torque that can be developed by the motor. (ii) Calculate the torque developed by the rotor at 1430 rpm.

## Short answer

Refer stator impedance to rotor side (K=0.5 gives R₁'=0.2 Ω, X=1 Ω). Using given starting torque as reference, Tmax/Tst=5.525, so Tmax≈55.25 Nm (slip 0.0784). At 1430 rpm (slip 0.04667), torque ratio=4.954, so torque≈49.5 Nm.

## Answer

**Given:** $V_L = 400$ V (star), 4 poles, 50 Hz, slip-ring voltage at standstill $= 200$ V (star), $Z_1 = 0.8 + j3\ \Omega$, $R_2 = 0.08\ \Omega$, $X_2 = 0.25\ \Omega$, $T_{st} = 10$ N·m.

**Method:** Refer the stator impedance to the rotor side and use torque ratios (the supply voltage cancels). The 10 N·m starting torque is used as the reference value.

#### Step 1: Transformation ratio and referred impedance

$$
\begin{aligned}
K &= \frac{E_2}{E_1} = \frac{200}{400} = 0.5 \\
R_1' &= K^2 R_1 = 0.25 \times 0.8 = 0.2\ \Omega \\
X_1' &= K^2 X_1 = 0.25 \times 3 = 0.75\ \Omega \\
X &= X_1' + X_2 = 0.75 + 0.25 = 1.0\ \Omega
\end{aligned}
$$

Torque (with $V'$ = stator voltage referred to rotor):

$$
T = \frac{3}{\omega_s}\cdot\frac{V'^2 (R_2/s)}{(R_1' + R_2/s)^2 + X^2}
$$

$N_s = 120 \times 50/4 = 1500$ rpm.

#### (i) Maximum torque

Maximum torque occurs when $R_2/s_m = \sqrt{R_1'^2 + X^2}$:

$$
\begin{aligned}
s_m &= \frac{R_2}{\sqrt{R_1'^2 + X^2}} = \frac{0.08}{\sqrt{0.04 + 1}} = \frac{0.08}{1.0198} = 0.0784
\end{aligned}
$$

$$
T_{max} = \frac{3}{\omega_s}\cdot\frac{V'^2}{2\left(R_1' + \sqrt{R_1'^2 + X^2}\right)}
$$

Ratio with starting torque ($s = 1$):

$$
\begin{aligned}
\frac{T_{max}}{T_{st}} &= \frac{(R_1' + R_2)^2 + X^2}{2R_2\left(R_1' + \sqrt{R_1'^2+X^2}\right)} \\
&= \frac{(0.28)^2 + 1^2}{2 \times 0.08 \times (0.2 + 1.0198)} \\
&= \frac{1.0784}{0.19517} = 5.525
\end{aligned}
$$

$$
T_{max} = 5.525 \times 10 = 55.25\ \text{N·m}
$$

(It occurs at $N = 1500(1 - 0.0784) \approx 1382$ rpm.)

#### (ii) Torque at 1430 rpm

$$
s = \frac{1500 - 1430}{1500} = 0.04667, \qquad \frac{R_2}{s} = \frac{0.08}{0.04667} = 1.714\ \Omega
$$

$$
\begin{aligned}
\frac{T}{T_{st}} &= \frac{R_2/s}{R_2}\cdot\frac{(R_1'+R_2)^2 + X^2}{(R_1' + R_2/s)^2 + X^2} \\
&= \frac{1}{0.04667}\times\frac{1.0784}{(0.2 + 1.714)^2 + 1} \\
&= 21.43 \times \frac{1.0784}{4.6645} = 4.954
\end{aligned}
$$

$$
T = 4.954 \times 10 = 49.54\ \text{N·m}
$$

Note: if the stator impedance were neglected, $T_{max}/T_{st} = (R_2^2+X_2^2)/(2R_2X_2) = 1.72$ only; including $Z_1$ (as given) changes the result a lot, so it has been included.

**Answer: (i) $T_{max} \approx 55.25$ N·m (at slip 0.0784); (ii) torque at 1430 rpm $\approx 49.5$ N·m.**

## Check yourself

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

1. What is K and how is the stator impedance referred to the rotor?
   A. K = 0.5; multiply by K^2 = 0.25
   B. K = 2; multiply by 4
   C. K = 0.5; multiply by 0.5
   D. K = 0.5; divide by 0.25
   Answer: A. K = 200/400 = 0.5, so R1' = 0.2 ohm and X1' = 0.75 ohm. (Key point: Transformation ratio)

2. Why can Tmax be found from the torque ratio with Tst?
   A. The supply voltage and constant 3/ws cancel
   B. Slip is the same
   C. The rotor is locked
   D. Resistance is zero
   Answer: A. Both torques share V'^2 and 3/ws, so only impedance terms remain. (Key point: Why use ratios)

3. What is the slip at maximum torque?
   A. R2/X = 0.08
   B. R2/sqrt(R1'^2 + X^2) = 0.0784
   C. R2/R1' = 0.4
   D. 1
   Answer: B. With the stator impedance included, sm = 0.08/1.0198 = 0.0784. (Key point: Slip at Tmax)

4. What is the result for Tmax and the 1430 rpm torque?
   A. 55.25 N·m and 49.5 N·m
   B. 10 N·m and 5 N·m
   C. 17.2 N·m and 10 N·m
   D. 55.25 N·m and 5.5 N·m
   Answer: A. Tmax/Tst = 5.525 gives 55.25 N·m; at s = 0.0467 T/Tst = 4.954 gives 49.5 N·m. (Key point: Max torque)

## Close variants set in other exams

- [Slip ring motor max-torque slip and full-load output](https://www.bctengineer.com/question.md?key=semester_4%2Felectrical_machines_i%3Aslip-ring-380v-180v-60nm-starting) — Electrical Machines I, chapter 5, asked 4×

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