---
title: "Regulation of delta connected 1600 kVA alternator"
subject: "Electric Machines II"
chapter: 1
key: "semester_5/electric_machines_ii:regulation-delta-1600kva"
source: https://www.bctengineer.com/q/semester_5/electric_machines_ii/regulation-delta-1600kva
site: BCT Engineering
---

# Regulation of delta connected 1600 kVA alternator

Electric Machines II · Chapter 1: Three Phase Synchronous Generator

- Asked in: 2079 Bhadra, 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-delta-1600kva
- Whole chapter: https://www.bctengineer.com/md/semester_5/electric_machines_ii/chapter-1

## Question

A 3-phase delta connected alternator is rated at 1600 kVA, 13.5 kV having per phase armature effective resistance and synchronous reactance of 1.5 Ω and 3 Ω per phase respectively. Calculate the voltage regulation for a load of 1.28 MW at (i) 0.8 pf lagging (ii) 0.8 pf leading.

## Short answer

Phase voltage = 13.5 kV (delta), phase current = 39.5 A. At 0.8 lagging pf: generated emf = 13619 V, regulation = +0.88%. At 0.8 leading pf: emf = 13477 V, regulation = –0.17%. Calculated using E = √[(Vcosφ+IR_a)² + (Vsinφ ± IX_s)²].

## Answer

**Given:** delta connected, so $V_{ph} = V_L = 13.5$ kV; $R_a = 1.5\ \Omega$, $X_s = 3\ \Omega$ per phase; load 1.28 MW at 0.8 pf, i.e. $1.28/0.8 = 1.6$ MVA (full load).

$$
\begin{aligned}
I_{ph} &= \frac{1.6\times10^6}{3\times13500} = 39.51\ \text{A} \\
V\cos\phi + IR_a &= 13500\times0.8 + 39.51\times1.5 = 10800 + 59.26 = 10859.26\ \text{V} \\
V\sin\phi &= 8100\ \text{V},\quad IX_s = 39.51\times3 = 118.52\ \text{V}
\end{aligned}
$$

#### (i) 0.8 pf lagging
$$
\begin{aligned}
E &= \sqrt{10859.26^2 + (8100 + 118.52)^2} = \sqrt{10859.26^2 + 8218.52^2} \\
&= 13618.65\ \text{V} \\
\%\text{Reg} &= \frac{13618.65 - 13500}{13500}\times100 = 0.879\%
\end{aligned}
$$

#### (ii) 0.8 pf leading
$$
\begin{aligned}
E &= \sqrt{10859.26^2 + (8100 - 118.52)^2} = \sqrt{10859.26^2 + 7981.48^2} \\
&= 13476.93\ \text{V} \\
\%\text{Reg} &= \frac{13476.93 - 13500}{13500}\times100 = -0.171\%
\end{aligned}
$$

The regulation is small because the given impedance (3 Ω) is very small compared with the phase voltage.

**Answer: (i) +0.88% at 0.8 pf lagging; (ii) −0.17% at 0.8 pf leading.**

## Check yourself

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

1. For a delta alternator at 13.5 kV, phase voltage is
   A. 13.5/sqrt3 kV
   B. 13.5 x sqrt3 kV
   C. 13.5 kV
   D. 13.5/3 kV
   Answer: C. In delta, phase voltage equals line voltage. (Key point: Delta)

2. Phase current at 1.6 MVA is
   A. 1.6e6/(sqrt3 x 13500) = 68.4 A
   B. 1.28e6/13500
   C. 1.6e6/13500 = 118.5 A
   D. 1.6e6/(3 x 13500) = 39.51 A
   Answer: D. In delta, S = 3 Vph Iph. (Key point: Phase current)

3. Regulation at 0.8 lagging is
   A. +0.88 %
   B. -0.17 %
   C. +5 %
   D. +22 %
   Answer: A. E = 13618.7 V vs 13500 V. (Key point: Lagging)

4. Regulation at 0.8 leading is small negative because
   A. the resistance is zero
   B. the reactive term is V sinφ - I Xs, giving E slightly below V
   C. the machine is star connected
   D. delta gives lower voltage
   Answer: B. E = 13476.9 V is less than 13500 V; impedance is small relative to voltage. (Key point: Leading)

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