---
title: "Excitation emf and maximum power of 1.5 MW motor"
subject: "Electric Machines II"
chapter: 2
key: "semester_5/electric_machines_ii:salient-motor-1-5mw-max-power"
source: https://www.bctengineer.com/q/semester_5/electric_machines_ii/salient-motor-1-5mw-max-power
site: BCT Engineering
---

# Excitation emf and maximum power of 1.5 MW motor

Electric Machines II · Chapter 2: Three Phase Synchronous Motor

- Asked in: 2083 Baisakh (new course), 3 marks
- Video: [#machines Question 24: Excitation emf in salient pole synchronous machines. #ee #gate](https://www.youtube.com/watch?v=2rRoVImgOvQ) — Electrical Questions
- Answer checked by Claude Sonnet 5.5 on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/electric_machines_ii/salient-motor-1-5mw-max-power
- Whole chapter: https://www.bctengineer.com/md/semester_5/electric_machines_ii/chapter-2

## Question

A 300 V, 1.5 MW, 3 phase synchronous motor has Xd = 4 Ω/phase and Xq = 3 Ω/phase, neglecting all losses, calculate the excitation emf when motor supplies rated load at unity power factor. Calculate the maximum mechanical power which the motor would develop for this field excitation.

## Short answer

With 3300 V star connection, excitation emf per phase is 2162 V (3744 V line). Maximum mechanical power is approximately 3.21 MW at load angle 75.2°.

## Answer

**Reading of data:** at 300 V a 1.5 MW motor would take about 2900 A, and $I_aX_q$ would be many times the supply voltage, so "300 V" is a misprint for **3300 V** (the standard version of this problem). Star connection, losses neglected.

$$
V = \frac{3300}{\sqrt3} = 1905.26\ \text{V},\quad I_a = \frac{1.5\times10^6}{\sqrt3 \times 3300} = 262.43\ \text{A}\ (\text{upf})
$$

#### Excitation emf (two-reaction method)

$$
\begin{aligned}
\vec E' &= V - jX_qI_a = 1905.26 - j787.30 = 2061.51\angle -22.45^\circ \Rightarrow \delta = 22.45^\circ \\
I_d &= I_a\sin\delta = 100.22\ \text{A} \\
E_f &= E' + (X_d - X_q)I_d = 2061.51 + 1 \times 100.22 = 2161.7\ \text{V per phase}
\end{aligned}
$$

Line value $= 3744$ V.

#### Maximum mechanical power

$$
P = 3\left[\frac{E_fV}{X_d}\sin\delta + \frac{V^2}{2}\left(\frac{1}{X_q}-\frac{1}{X_d}\right)\sin2\delta\right]
$$

With $a = E_fV/X_d = 1029665$ W and $b = V^2(1/X_q - 1/X_d) = 302500$ W:

$$
\begin{aligned}
\cos\delta_m &= \frac{-a + \sqrt{a^2 + 8b^2}}{4b} = 0.2554 \Rightarrow \delta_m = 75.20^\circ \\
P_{max} &= 3a\sin\delta_m + \tfrac{3b}{2}\sin2\delta_m = 2.987 + 0.224 = 3.21\ \text{MW}
\end{aligned}
$$

**Answer:** $E_f = 2162$ V per phase (3744 V line); maximum power $\approx 3.21$ MW at $\delta = 75.2^\circ$.

## Check yourself

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

1. Why is 300 V taken as 3300 V?
   A. At 300 V a 1.5 MW motor would draw about 2900 A and Ia Xq would greatly exceed the supply voltage
   B. Because 300 V is not a standard unit
   C. Because the motor is delta connected
   D. Because Xd is zero
   Answer: A. The numbers are impossible at 300 V; the standard version uses 3300 V. (Key point: Data correction)

2. What is Ef per phase?
   A. 1905 V
   B. 2062 V
   C. 2162 V
   D. 3744 V
   Answer: C. E' = 2061.5 V, Id = 100.22 A, so Ef = 2061.5 + (4 - 3) x 100.22 = 2162 V (3744 V line). (Key point: Excitation emf)

3. What are the maximum power angle and power for this excitation?
   A. 90 degrees and 1.5 MW
   B. 22.45 degrees and 1.5 MW
   C. 75.2 degrees and about 3.21 MW
   D. 75.2 degrees and 0.224 MW
   Answer: C. cos(delta_m) = 0.2554 gives 75.2 degrees; Pmax = 2.987 + 0.224 = 3.21 MW. (Key point: Maximum power)

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