---
title: "Excitation emf of salient pole generator"
subject: "Electric Machines II"
chapter: 1
key: "semester_5/electric_machines_ii:salient-generator-excitation-500v"
source: https://www.bctengineer.com/q/semester_5/electric_machines_ii/salient-generator-excitation-500v
site: BCT Engineering
---

# Excitation emf of salient pole generator

Electric Machines II · Chapter 1: Three Phase Synchronous Generator

- Asked in: 2070 Chaitra, 6 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-generator-excitation-500v
- Whole chapter: https://www.bctengineer.com/md/semester_5/electric_machines_ii/chapter-1

## Question

A 50 Hz, 3 Phase, 500 V, star connected synchronous generator with salient pole rotor has Xd = 0.1 Ohm and Xq = 0.075 ohm. Armature winding resistance is 0.1 ohm. The generator supplies 100 A at 0.8 pf lagging. Calculate the excitation emf.

## Short answer

Using two-reaction theory: per-phase voltage 288.68 V, current 100 A at 0.8 pf lag. Compute E' = V + I(Ra+jXq) = 301.18 V, load angle 0°, Id=60 A, Iq=80 A. Excitation emf = E' + Id(Xd-Xq) = 302.68 V per phase (524.25 V line).

## Answer

Use the **two-reaction theory**. Given: star, 500 V, $X_d = 0.1\ \Omega$, $X_q = 0.075\ \Omega$, $R_a = 0.1\ \Omega$, $I = 100$ A at 0.8 pf lagging.

#### Step 1: Voltage behind $R_a + jX_q$
$$
\begin{aligned}
V &= \frac{500}{\sqrt3} = 288.68\ \text{V},\quad \bar I = 100\angle -36.87^\circ = 80 - j60\ \text{A} \\
\bar E' &= \bar V + \bar I(R_a + jX_q) = 288.68 + (80 - j60)(0.1 + j0.075) \\
&= 288.68 + (8 + 4.5) + j(6 - 6) = 301.18 + j0\ \text{V}
\end{aligned}
$$
So the load angle $\delta = 0^\circ$ (the q-axis lies along $V$ here) and $\psi = \phi + \delta = 36.87^\circ$.

#### Step 2: d- and q-axis currents
$$
\begin{aligned}
I_d &= I\sin\psi = 100\sin 36.87^\circ = 60\ \text{A} \\
I_q &= I\cos\psi = 100\cos 36.87^\circ = 80\ \text{A}
\end{aligned}
$$

#### Step 3: Excitation emf
$$
\begin{aligned}
E &= E' + I_d(X_d - X_q) = 301.18 + 60(0.1 - 0.075) \\
&= 301.18 + 1.5 = 302.68\ \text{V per phase} \\
E_L &= \sqrt3\times302.68 = 524.25\ \text{V}
\end{aligned}
$$

**Answer: excitation emf = 302.68 V per phase (524.25 V line), load angle ≈ 0°.**

## Check yourself

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

1. Which theory is used for a salient-pole generator with Xd and Xq?
   A. Single-reactance (synchronous impedance) theory
   B. Potier theory
   C. Two-reaction theory
   D. Transformer equivalent theory
   Answer: C. Blondel's two-reaction theory splits armature current into Id and Iq acting on Xd and Xq. (Key point: Theory used)

2. What is E' = V + I(Ra + jXq) here, and what does it show?
   A. 301.18 + j0 V, so delta = 0 degrees
   B. 288.68 V with delta = 36.87 degrees
   C. 314 + j20 V with delta = 10 degrees
   D. 500 V with delta = 0
   Answer: A. (80 - j60)(0.1 + j0.075) = 12.5 + j0, so E' = 301.18 V along V, giving delta = 0. (Key point: Voltage behind Ra + jXq)

3. What are Id and Iq with psi = 36.87 degrees and I = 100 A?
   A. Id = 80 A, Iq = 60 A
   B. Id = 60 A, Iq = 80 A
   C. Id = 100 A, Iq = 0
   D. Id = 60 A, Iq = 60 A
   Answer: B. Id = I sin(psi) = 60 A and Iq = I cos(psi) = 80 A. (Key point: Axis currents)

4. How is the excitation emf E obtained from E'?
   A. E = E' - Id(Xd - Xq)
   B. E = E' + Iq(Xd - Xq)
   C. E = E' + Id(Xd - Xq) = 302.68 V
   D. E = E' x Xd/Xq
   Answer: C. E = 301.18 + 60 x 0.025 = 302.68 V per phase. (Key point: Excitation emf)

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