---
title: "5 MVA alternator: regulation and pf for zero regulation"
subject: "Electrical Machines"
chapter: 6
key: "semester_4/electrical_machines:alt-5mva-11kv-regulation-zero"
source: https://www.bctengineer.com/q/semester_4/electrical_machines/alt-5mva-11kv-regulation-zero
site: BCT Engineering
---

# 5 MVA alternator: regulation and pf for zero regulation

Electrical Machines · Chapter 6: Three Phase Synchronous Machines

- Asked in: 2066 Magh (old course), 8 marks
- Video: [Numerical on Voltage Regulation by MMF Method - Synchronous Machine - Electrical Machines - IV](https://www.youtube.com/watch?v=pFZPbi0rrNQ) — Ekeeda
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/electrical_machines/alt-5mva-11kv-regulation-zero
- Whole chapter: https://www.bctengineer.com/md/semester_4/electrical_machines/chapter-6

## Question

A 3-phase star connected, 5 MVA, 11 kV synchronous generator has armature resistance of $0.12\ \Omega$ and synchronous reactance of $2\ \Omega$ per phase. Calculate voltage regulation if the generator delivers full load at rated terminal voltage at 0.9 lagging p.f. Also find out generator power factor at which the voltage regulation is zero.

## Short answer

Voltage regulation at full load 0.9 lagging is 4.30%. Zero regulation occurs at a leading power factor of 0.995.

## Answer

**Given:** 5 MVA, 11 kV, star, $R_a = 0.12\ \Omega$, $X_s = 2\ \Omega$ per phase, full load at 0.9 pf lagging.

$$
V_{ph} = \frac{11000}{\sqrt3} = 6350.85\ \text{V}, \qquad I = \frac{5\times10^6}{\sqrt3\times11000} = 262.43\ \text{A}
$$
$$
IR_a = 31.49\ \text{V}, \qquad IX_s = 524.86\ \text{V}
$$

#### Voltage regulation at 0.9 lagging
$\cos\phi = 0.9$, $\sin\phi = 0.4359$:
$$
E = \sqrt{(V + IR_a\cos\phi + IX_s\sin\phi)^2 + (IX_s\cos\phi - IR_a\sin\phi)^2}
$$
$$
= \sqrt{(6350.85 + 28.34 + 228.78)^2 + (472.38 - 13.73)^2} = \sqrt{6607.97^2 + 458.65^2}
$$
$$
E_{ph} = 6623.9\ \text{V}
$$
$$
\text{Regulation} = \frac{6623.9-6350.85}{6350.85}\times100 = 4.30\%
$$

#### Power factor for zero regulation
For zero regulation $E = V$ at full load current. The current must lead. Let $\vec I = I\angle\theta$ with $\theta > 0$ leading, $Z_s = 2.003\angle86.57^\circ$ ($|Z_s| = \sqrt{0.12^2+2^2}= 2.0036\ \Omega$):
$$
|\vec V + \vec IZ_s|^2 = V^2 \;\Rightarrow\; 2VI|Z_s|\cos(\theta + 86.57^\circ) + I^2|Z_s|^2 = 0
$$
$$
\cos(\theta+86.57^\circ) = -\frac{I|Z_s|}{2V} = -\frac{262.43\times2.0036}{2\times6350.85} = -0.0414
$$
$$
\theta + 86.57^\circ = 92.37^\circ \;\Rightarrow\; \theta = 5.81^\circ \text{ (leading)}
$$
$$
\cos\theta = 0.9949
$$

**Answer:** regulation at 0.9 lagging $= 4.30\%$ (E = 6623.9 V per phase). Regulation is zero at **0.995 leading** power factor.

## Check yourself

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

1. What are Vph and full-load current for the 5 MVA, 11 kV star generator?
   A. 6350.85 V and 262.43 A
   B. 11000 V and 262.43 A
   C. 6350.85 V and 454.5 A
   D. 11000 V and 454.5 A
   Answer: A. Vph = 11000/sqrt3 = 6350.85 V; I = 5e6/(sqrt3 x 11000) = 262.43 A. (Key point: Full-load current)

2. What is the regulation at 0.9 lagging?
   A. About 10.2 %
   B. About 4.30 %
   C. About 0 %
   D. About 18.6 %
   Answer: B. E = 6623.9 V per phase, so regulation = (6623.9 - 6350.85)/6350.85 = 4.30 %. (Key point: Regulation at 0.9 lagging)

3. What condition gives zero regulation?
   A. E = V at full-load current, which needs a leading current
   B. E = 0
   C. A lagging current of 0.8 pf
   D. I = 0
   Answer: A. Zero regulation means E equals V; the magnetising armature reaction of a leading current is needed to achieve this. (Key point: Zero regulation condition)

4. What power factor gives zero regulation?
   A. 0.9 lagging
   B. Unity
   C. 0.995 leading
   D. 0.995 lagging
   Answer: C. Solving |V + I Zs| = V gives cos(theta + 86.57) = -0.0414, theta = 5.81 degrees leading, so pf = 0.995 leading. (Key point: Zero regulation result)

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