---
title: "Full-load torque, rotor current and copper loss of 440 V motor"
subject: "Electrical Machines I"
chapter: 5
key: "semester_4/electrical_machines_i:im-440v-1425rpm-rotor-impedance"
source: https://www.bctengineer.com/q/semester_4/electrical_machines_i/im-440v-1425rpm-rotor-impedance
site: BCT Engineering
---

# Full-load torque, rotor current and copper loss of 440 V motor

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

- Asked in: 2081 Chaitra (new course), 2+2+2 marks
- Video: [Numerical on full load torque & Rotor Copper Loss](https://www.youtube.com/watch?v=T-KdAuHvBtY) — #Electrical Academy
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/electrical_machines_i/im-440v-1425rpm-rotor-impedance
- Whole chapter: https://www.bctengineer.com/md/semester_4/electrical_machines_i/chapter-5

## Question

A 440 V 3-phase 50 Hz, 4-pole Y-connected Induction motor has a full load 1425 rpm. The rotor has an Impedance of (0.4 + j 4) Ω per phase and rotor/stator turn ratio is 0.8. Calculate: (i) Full-load torque (ii) Rotor current and (iii) Full-load rotor Cu loss.

## Short answer

Slip=0.05, rotor standstill E2=203.23 V/phase. Rotor current I2=22.72 A. Rotor Cu loss=619.5 W. Air-gap power=12.39 kW, torque=78.88 N·m.

## Answer

**Given:** $V_L = 440$ V (Y), $f = 50$ Hz, $P = 4$, $N = 1425$ rpm, rotor impedance at standstill $= 0.4 + j4\ \Omega$/phase, rotor/stator turns ratio $K = 0.8$. Stator impedance is neglected.

#### Slip and rotor emf

$$
\begin{aligned}
N_s &= \frac{120 \times 50}{4} = 1500\ \text{rpm}, \quad s = \frac{1500 - 1425}{1500} = 0.05 \\
E_1 &= \frac{440}{\sqrt{3}} = 254.03\ \text{V/phase} \\
E_2 &= K E_1 = 0.8 \times 254.03 = 203.23\ \text{V/phase (standstill)}
\end{aligned}
$$

#### (ii) Rotor current

$$
\begin{aligned}
I_2 &= \frac{sE_2}{\sqrt{R_2^2 + (sX_2)^2}} = \frac{0.05 \times 203.23}{\sqrt{0.4^2 + (0.05 \times 4)^2}} \\
&= \frac{10.161}{\sqrt{0.16 + 0.04}} = \frac{10.161}{0.4472} = 22.72\ \text{A}
\end{aligned}
$$

#### (iii) Full-load rotor copper loss

$$
P_{cu2} = 3 I_2^2 R_2 = 3 \times 22.72^2 \times 0.4 = 619.5\ \text{W}
$$

#### (i) Full-load torque

$$
\begin{aligned}
P_{ag} &= \frac{P_{cu2}}{s} = \frac{619.5}{0.05} = 12\,390\ \text{W} \\
\omega_s &= \frac{2\pi \times 1500}{60} = 157.08\ \text{rad/s} \\
T &= \frac{P_{ag}}{\omega_s} = \frac{12\,390}{157.08} = 78.88\ \text{N·m}
\end{aligned}
$$

Check: $P_m = (1-s)P_{ag} = 11\,771$ W and $T = P_m/\omega = 11\,771/149.23 = 78.88$ N·m, which agrees. (This is the gross developed torque; mechanical losses are not given.)

**Answer: (i) Full-load torque = 78.88 N·m; (ii) rotor current = 22.72 A; (iii) rotor copper loss = 619.5 W.**

## Check yourself

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

1. What is the full-load slip?
   A. 0.05
   B. 0.95
   C. 0.0525
   D. 0.5
   Answer: A. (1500 - 1425)/1500 = 0.05. (Key point: Slip)

2. How is the standstill rotor emf per phase found?
   A. K x 440
   B. K x 440/sqrt(3) = 203.2 V
   C. 440/(K sqrt(3))
   D. 440 x sqrt(3)
   Answer: B. Stator phase emf is 254.03 V, times 0.8. (Key point: Rotor emf)

3. What is I2?
   A. 22.72 A
   B. 45.4 A
   C. 10.2 A
   D. 50.8 A
   Answer: A. I2 = sE2/sqrt(R2^2 + (sX2)^2) = 10.161/0.4472. (Key point: Rotor current)

4. How is torque found from rotor copper loss?
   A. Pag = Pcu2/s, T = Pag/ws = 78.88 N·m
   B. T = Pcu2/ws
   C. T = Pcu2 x s / ws
   D. T = Pcu2/wr
   Answer: A. Pag = 12390 W; ws = 157.08 rad/s. (Key point: Torque)

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