---
title: "Regulation of 100 kVA alternator from tests"
subject: "Electric Machines II"
chapter: 1
key: "semester_5/electric_machines_ii:regulation-100kva-test-data"
source: https://www.bctengineer.com/q/semester_5/electric_machines_ii/regulation-100kva-test-data
site: BCT Engineering
---

# Regulation of 100 kVA alternator from tests

Electric Machines II · Chapter 1: Three Phase Synchronous Generator

- Asked in: 2076 Chaitra, 6 marks
- Answer checked by Claude Sonnet 5.5 on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/electric_machines_ii/regulation-100kva-test-data
- Whole chapter: https://www.bctengineer.com/md/semester_5/electric_machines_ii/chapter-1

## Question

The data obtained on 100 kVA, 1100 V, Y-connected, 3-phase alternators are:
DC resistance Test: Voltage between lines 6 V DC, current in lines = 6 A
Open Circuit Test: Field Current = 12.5 A DC, line voltage = 420 V AC
Short Circuit Test: Field Current = 12.5 A DC, line current = rated current
Calculate the voltage regulation of alternator at 0.8 p.f. lagging. Take effective resistance to be 1.667 times of DC resistance.

## Short answer

DC resistance per phase is 0.5 Ω, effective Ra=0.8335 Ω. From OC and SC tests at same field current: Zs=4.62 Ω, Xs=4.544 Ω. At full load 0.8 pf lagging, generated emf per phase E=829.7 V, voltage regulation=30.64%.

## Answer

Regulation is found by the **synchronous impedance (EMF) method**.

#### Armature resistance
DC test between two lines (star: two phases in series):
$$
\begin{aligned}
R_{dc,ph} &= \frac{1}{2}\times\frac{6}{6} = 0.5\ \Omega \\
R_a &= 1.667\times0.5 = 0.8335\ \Omega
\end{aligned}
$$

#### Synchronous impedance (same field current 12.5 A)
$$
\begin{aligned}
I_{rated} &= \frac{100\times10^3}{\sqrt3\times1100} = 52.49\ \text{A} \\
E_{oc,ph} &= \frac{420}{\sqrt3} = 242.49\ \text{V} \\
Z_s &= \frac{242.49}{52.49} = 4.620\ \Omega \\
X_s &= \sqrt{4.620^2 - 0.8335^2} = 4.544\ \Omega
\end{aligned}
$$

#### Regulation at full load, 0.8 pf lagging
$$
\begin{aligned}
V_{ph} &= \frac{1100}{\sqrt3} = 635.09\ \text{V} \\
V\cos\phi + IR_a &= 508.07 + 52.49\times0.8335 = 551.82\ \text{V} \\
V\sin\phi + IX_s &= 381.05 + 52.49\times4.544 = 619.56\ \text{V} \\
E &= \sqrt{551.82^2 + 619.56^2} = 829.67\ \text{V per phase} \\
\%\text{Reg} &= \frac{829.67 - 635.09}{635.09}\times100 = 30.64\%
\end{aligned}
$$

**Answer: $Z_s$ = 4.62 Ω, $X_s$ = 4.544 Ω, regulation at 0.8 pf lagging = 30.64% (full load assumed).**

## Check yourself

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

1. DC test line-to-line 6 V, 6 A in a star machine gives Ra
   A. 1.667 x 1 = 1.667 Ω
   B. 1.667 x 0.5 = 0.8335 Ω
   C. 0.5 Ω
   D. 1.667 x 0.333
   Answer: B. Rdc per phase is half of line-to-line resistance. (Key point: Ra)

2. Zs is
   A. 420/52.49 = 8 Ω
   B. (1100/sqrt3)/52.49 = 12.1 Ω
   C. (420/sqrt3)/52.49 = 4.62 Ω
   D. 420/(3 x 52.49)
   Answer: C. Open-circuit phase emf over short-circuit current at the same field current (rated current). (Key point: Zs)

3. Xs is
   A. Zs - Ra
   B. Zs + Ra
   C. Zs/Ra
   D. sqrt(Zs^2 - Ra^2) = 4.544 Ω
   Answer: D. Zs^2 = Ra^2 + Xs^2. (Key point: Xs)

4. Regulation at 0.8 lag is found with E = sqrt((Vcosφ+IRa)^2+(Vsinφ+IXs)^2) giving
   A. 30.64 %
   B. -30.64 %
   C. 19.6 %
   D. 0.8 %
   Answer: A. E = 829.7 V vs V = 635.1 V. (Key point: Regulation)

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