---
title: "Synchronous reactance and regulation from test data"
subject: "Electric Machines II"
chapter: 1
key: "semester_5/electric_machines_ii:regulation-1500kva-2300v-test"
source: https://www.bctengineer.com/q/semester_5/electric_machines_ii/regulation-1500kva-2300v-test
site: BCT Engineering
---

# Synchronous reactance and regulation from test data

Electric Machines II · Chapter 1: Three Phase Synchronous Generator

- Asked in: 2082 Baisakh, 6 marks
- Video: [viii) Alternator : Armature Resistance & Synchronous Reactance | Math Problem | Synchronous Gen](https://www.youtube.com/watch?v=JQxACd0wA20) — 4 Years Engineering BD
- Answer checked by Claude Sonnet 5.5 on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/electric_machines_ii/regulation-1500kva-2300v-test
- Whole chapter: https://www.bctengineer.com/md/semester_5/electric_machines_ii/chapter-1

## Question

A 3-phase, 1500 kVA, star connected, 50 Hz, 2300 V alternator has a resistance between each pair of terminals as measured by direct current is 0.16 Ω. Assume that the effective resistance is 1.5 times the ohmic resistance. A field current of 70 A produces a short-circuit current equal to full-load current of 376 A in each line. The same field current produces an emf of 700 V on open circuit. Determine the synchronous reactance of the machine and its full load regulation at 0.8 pf leading.

## Short answer

Effective resistance per phase = 0.12 Ω. Using O.C. 700 V line and S.C. 376 A at the same field current, Z_s = (700/√3)/376 = 1.0749 Ω, so X_s = √(1.0749² – 0.12²) = 1.068 Ω. At full-load 0.8 pf leading, generated emf = 1175.8 V/ph (2036.6 V line), regulation = –11.45%.

## Answer

#### Effective resistance
DC resistance between two terminals = 0.16 Ω, which is two phases in series (star).
$$
\begin{aligned}
R_{dc,ph} &= \frac{0.16}{2} = 0.08\ \Omega \\
R_a &= 1.5 \times 0.08 = 0.12\ \Omega
\end{aligned}
$$

#### Synchronous impedance and reactance
The same field current (70 A) gives 700 V (line) on open circuit and 376 A on short circuit:
$$
\begin{aligned}
Z_s &= \frac{700/\sqrt3}{376} = \frac{404.15}{376} = 1.0749\ \Omega \\
X_s &= \sqrt{Z_s^2 - R_a^2} = \sqrt{1.0749^2 - 0.12^2} = 1.068\ \Omega
\end{aligned}
$$

#### Regulation at full load, 0.8 pf leading
$$
\begin{aligned}
V_{ph} &= \frac{2300}{\sqrt3} = 1327.91\ \text{V},\quad I = 376\ \text{A} \\
V\cos\phi + IR_a &= 1062.32 + 376\times0.12 = 1107.44\ \text{V} \\
V\sin\phi - IX_s &= 796.74 - 376\times1.068 = 796.74 - 401.62 = 395.12\ \text{V} \\
E &= \sqrt{1107.44^2 + 395.12^2} = 1175.82\ \text{V per phase} \\
\%\text{Reg} &= \frac{1175.82 - 1327.91}{1327.91}\times100 = -11.45\%
\end{aligned}
$$

**Answer: $X_s$ = 1.068 Ω per phase; full-load regulation at 0.8 pf leading = −11.45% (E = 1175.8 V/phase, 2036.6 V line).**

## Check yourself

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

1. DC resistance between terminals is 0.16 Ω for a star machine. Ra is
   A. 1.5 x 0.16 = 0.24 Ω
   B. 0.16 Ω
   C. 0.08 Ω
   D. 1.5 x 0.08 = 0.12 Ω
   Answer: D. Two phases are in series, so per phase is 0.08 Ω; effective is 1.5 times. (Key point: Effective resistance)

2. Xs is found from
   A. Zs + Ra
   B. sqrt(Zs^2 - Ra^2) = 1.068 Ω
   C. Zs - Ra
   D. sqrt(Zs^2 + Ra^2)
   Answer: B. Zs^2 = Ra^2 + Xs^2. (Key point: Xs)

3. At 0.8 pf leading full load, regulation is
   A. +11.45 %
   B. +25 %
   C. -11.45 % because E < V
   D. 0 %
   Answer: C. E = 1175.8 V is below V = 1327.9 V for a leading load. (Key point: Regulation)

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