---
title: "Rotor copper loss, frequency, current and efficiency of 11 kW motor"
subject: "Electrical Machines I"
chapter: 5
key: "semester_4/electrical_machines_i:squirrel-cage-1450rpm-11kw-losses"
source: https://www.bctengineer.com/q/semester_4/electrical_machines_i/squirrel-cage-1450rpm-11kw-losses
site: BCT Engineering
---

# Rotor copper loss, frequency, current and efficiency of 11 kW motor

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

- Asked in: 2071 Magh, 2+2+2+2 marks
- Video: [Relation between rotor copper loss and rotor input- Three Phase Induction Motor](https://www.youtube.com/watch?v=63Sx6-Gcw_g) — Ekeeda
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/electrical_machines_i/squirrel-cage-1450rpm-11kw-losses
- Whole chapter: https://www.bctengineer.com/md/semester_4/electrical_machines_i/chapter-5

## Question

A 4 pole, 400 V, 3-phase, 50 Hz squirrel cage induction motor runs at 1450 RPM at 0.8 power factor lagging developing 11 kW. The stator losses are 1100 watt and mechanical losses are 400 watt. Determine: i) Rotor copper loss ii) Rotor frequency iii) Line current iv) Efficiency

## Short answer

Synchronous speed 1500 rpm, slip 0.0333. Rotor copper loss = 379.3 W, rotor frequency = 1.667 Hz. Input power = air‑gap power + stator losses = 12 479.3 W, line current = 22.52 A. Output = 10 600 W, efficiency = 84.94%.

## Answer

**Given:** 4 poles, 400 V, 50 Hz, $N = 1450$ rpm, pf $= 0.8$ lag, mechanical power developed $P_m = 11$ kW, stator losses $= 1100$ W, mechanical losses $= 400$ W.

Assumption: "developing 11 kW" is the gross mechanical power developed by the rotor (before mechanical losses).

$$
N_s = \frac{120 \times 50}{4} = 1500\ \text{rpm}, \qquad s = \frac{1500 - 1450}{1500} = 0.03333
$$

#### (i) Rotor copper loss

$$
P_{ag} : P_{cu2} : P_m = 1 : s : (1 - s)
$$

$$
\begin{aligned}
P_{ag} &= \frac{P_m}{1 - s} = \frac{11\,000}{0.96667} = 11\,379.3\ \text{W} \\
P_{cu2} &= sP_{ag} = 0.03333 \times 11\,379.3 = 379.3\ \text{W}
\end{aligned}
$$

#### (ii) Rotor frequency

$$
f_2 = sf = 0.03333 \times 50 = 1.667\ \text{Hz}
$$

#### (iii) Line current

$$
\begin{aligned}
P_{in} &= P_{ag} + \text{stator losses} = 11\,379.3 + 1100 = 12\,479.3\ \text{W} \\
I_L &= \frac{P_{in}}{\sqrt{3}\,V_L\cos\phi} = \frac{12\,479.3}{\sqrt{3}\times 400 \times 0.8} = 22.52\ \text{A}
\end{aligned}
$$

#### (iv) Efficiency

$$
\begin{aligned}
P_{out} &= P_m - \text{mech. losses} = 11\,000 - 400 = 10\,600\ \text{W} \\
\eta &= \frac{10\,600}{12\,479.3} = 0.8494 = 84.94\%
\end{aligned}
$$

| Item | Power (W) |
|---|---|
| Input | 12 479.3 |
| Stator losses | 1 100 |
| Air-gap power | 11 379.3 |
| Rotor Cu loss | 379.3 |
| Mechanical power developed | 11 000 |
| Mechanical losses | 400 |
| Output | 10 600 |

**Answer: (i) 379.3 W; (ii) 1.667 Hz; (iii) 22.52 A; (iv) 84.94%.**

## Check yourself

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

1. What is the ratio P_ag : P_cu2 : P_m used for the rotor copper loss?
   A. 1 : s : (1 - s)
   B. 1 : (1 - s) : s
   C. s : 1 : (1 - s)
   D. 1 : s^2 : (1 - s)
   Answer: A. Rotor copper loss is s times the air-gap power and mechanical power developed is (1 - s) times it. (Key point: Air-gap power ratio)

2. With s = 0.03333 and P_m = 11 kW, what is the rotor copper loss?
   A. 366.7 W
   B. 379.3 W
   C. 1100 W
   D. 11 379 W
   Answer: B. P_ag = 11000/0.96667 = 11379.3 W, and P_cu2 = s P_ag = 379.3 W. (Key point: Rotor copper loss)

3. What is the rotor frequency at 1450 rpm on a 4-pole 50 Hz supply?
   A. 50 Hz
   B. 3.33 Hz
   C. 1.667 Hz
   D. 48.3 Hz
   Answer: C. f2 = s f = 0.03333 x 50 = 1.667 Hz. (Key point: Rotor frequency)

4. How is input power found, and how is output obtained?
   A. P_in = P_m + stator losses; P_out = P_ag - mechanical losses
   B. P_in = P_ag + rotor Cu loss; P_out = P_m
   C. P_in = P_m + mechanical losses; P_out = P_m - stator losses
   D. P_in = P_ag + stator losses; P_out = P_m - mechanical losses
   Answer: D. Stator losses are added to air-gap power to get input (12 479.3 W); mechanical losses are subtracted from P_m to get output (10 600 W), giving 84.94%. (Key point: Input and efficiency)

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