---
title: "Regulation of 10 kVA round-rotor generator"
subject: "Electric Machines II"
chapter: 1
key: "semester_5/electric_machines_ii:regulation-10kva-zero-pf"
source: https://www.bctengineer.com/q/semester_5/electric_machines_ii/regulation-10kva-zero-pf
site: BCT Engineering
---

# Regulation of 10 kVA round-rotor generator

Electric Machines II · Chapter 1: Three Phase Synchronous Generator

- Asked in: 2078 Bhadra, 2+2+2 marks
- Video: [Synchronous Generators - Question no. 4](https://www.youtube.com/watch?v=2HOkrie9-1c) — Ayman El-Hag - Electrical Engineering for Everyone
- Answer checked by Claude Sonnet 5.5 on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/electric_machines_ii/regulation-10kva-zero-pf
- Whole chapter: https://www.bctengineer.com/md/semester_5/electric_machines_ii/chapter-1

## Question

A 3-phase, star-connected, round-rotor synchronous generator rated at 10 kVA, 230 V has an armature resistance of 0.5 ohm per phase and a synchronous reactance of 1.2 ohm per phase. Calculate the percentage voltage regulation at full load at power factors of (a) 0.8 lagging, (b) 0.8 leading, (c) determine the power factor such that the voltage regulation is zero on full load.

## Short answer

At rated load: (a) 0.8 lagging, regulation = +21.8%; (b) 0.8 leading, regulation = –3.08%. Zero regulation occurs at a leading power factor of 0.869, obtained by solving E = V for φ.

## Answer

**Given:** star, 10 kVA, 230 V, $R_a = 0.5\ \Omega$, $X_s = 1.2\ \Omega$ per phase.
$$
\begin{aligned}
V_{ph} &= \frac{230}{\sqrt3} = 132.79\ \text{V},\quad I = \frac{10000}{\sqrt3\times230} = 25.10\ \text{A} \\
V\cos\phi + IR_a &= 106.23 + 12.55 = 118.78\ \text{V} \\
V\sin\phi &= 79.67\ \text{V},\quad IX_s = 30.12\ \text{V}
\end{aligned}
$$

#### (a) 0.8 pf lagging
$$
\begin{aligned}
E &= \sqrt{118.78^2 + (79.67 + 30.12)^2} = 161.76\ \text{V} \\
\%\text{Reg} &= \frac{161.76 - 132.79}{132.79}\times100 = 21.81\%
\end{aligned}
$$

#### (b) 0.8 pf leading
$$
\begin{aligned}
E &= \sqrt{118.78^2 + (79.67 - 30.12)^2} = 128.70\ \text{V} \\
\%\text{Reg} &= \frac{128.70 - 132.79}{132.79}\times100 = -3.08\%
\end{aligned}
$$

#### (c) Power factor for zero regulation
Zero regulation means $E = V$. With $Z_s = R_a + jX_s = 1.3\angle 67.38^\circ\ \Omega$ and current $I\angle\phi$ ($\phi$ positive for leading):
$$
\begin{aligned}
E^2 &= V^2 + 2VIZ_s\cos(\theta_z + \phi) + I^2Z_s^2 = V^2 \\
\cos(\theta_z + \phi) &= -\frac{IZ_s}{2V} = -\frac{25.10\times1.3}{2\times132.79} = -0.1229 \\
\theta_z + \phi &= 97.06^\circ \\
\phi &= 97.06^\circ - 67.38^\circ = 29.68^\circ\ \text{(leading)} \\
\text{pf} &= \cos 29.68^\circ = 0.869\ \text{leading}
\end{aligned}
$$

**Answer: (a) 21.81%; (b) −3.08%; (c) zero regulation at pf = 0.869 leading.**

## Check yourself

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

1. Full-load current of the 10 kVA, 230 V star machine is
   A. 10000/230 = 43.5 A
   B. 10000/(3 x 230) = 14.5 A
   C. 10000/(sqrt3 x 230) = 25.1 A
   D. 25.1/sqrt3 A
   Answer: C. Line current equals phase current in star. (Key point: Full-load current)

2. The regulation at 0.8 lagging is +21.81 % and at 0.8 leading it is
   A. +3.08 %
   B. -21.81 %
   C. 0 %
   D. -3.08 %
   Answer: D. Leading reactive drop subtracts, so E < V. (Key point: Lagging vs leading)

3. Zero regulation requires
   A. E = V, which occurs at a leading pf (0.869)
   B. a lagging pf of 0.869
   C. unity pf
   D. pf of zero
   Answer: A. Only a leading load lowers E to V. (Key point: Zero regulation)

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