---
title: "Regulation and power angle of 1650 kVA generator"
subject: "Electric Machines II"
chapter: 1
key: "semester_5/electric_machines_ii:regulation-power-angle-1650kva"
source: https://www.bctengineer.com/q/semester_5/electric_machines_ii/regulation-power-angle-1650kva
site: BCT Engineering
---

# Regulation and power angle of 1650 kVA generator

Electric Machines II · Chapter 1: Three Phase Synchronous Generator

- Asked in: 2078 Kartik, 7 marks
- Video: [How to Calculate Max Power and Torque Angle of Synchronous Machines (Electrical Power PE Exam)](https://www.youtube.com/watch?v=FUko4Yn0jRQ) — Electrical PE Review
- Answer checked by Claude Sonnet 5.5 on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/electric_machines_ii/regulation-power-angle-1650kva
- Whole chapter: https://www.bctengineer.com/md/semester_5/electric_machines_ii/chapter-1

## Question

A three phase star-connected synchronous generator has effective resistance and synchronous reactance of 1.6 ohm and 35 ohm respectively per phase. Determine the percentage regulation and power angle for a load of 1320 kW at power factor of (a) 0.8 lagging and (b) 0.8 leading when the rating of the machine is 1650 kVA and 13.8 kV.

## Short answer

Per-phase Vph=7967.4 V, I=69.03 A. (a) 0.8 lagging: E=9687 V, regulation=21.58%, power angle δ=11.11°. (b) 0.8 leading: E=6902 V, regulation=−13.37%, δ=16.84°.

## Answer

**Given:** star, 1650 kVA, 13.8 kV, $R_a = 1.6\ \Omega$, $X_s = 35\ \Omega$; load 1320 kW at 0.8 pf (= 1650 kVA, full load).
$$
\begin{aligned}
V_{ph} &= \frac{13800}{\sqrt3} = 7967.43\ \text{V} \\
I &= \frac{1320\times10^3}{\sqrt3\times13800\times0.8} = 69.03\ \text{A} \\
V\cos\phi + IR_a &= 6373.95 + 110.45 = 6484.40\ \text{V} \\
V\sin\phi &= 4780.46\ \text{V},\quad IX_s = 2416.09\ \text{V}
\end{aligned}
$$
The power angle $\delta$ is the angle between $E$ and $V$. If $\theta$ is the angle of $E$ from the current phasor, $\theta = \tan^{-1}\dfrac{V\sin\phi \pm IX_s}{V\cos\phi + IR_a}$, then $\delta = \theta - \phi$ for lagging pf and $\delta = \phi - \theta$ for leading pf.

#### (a) 0.8 pf lagging
$$
\begin{aligned}
E &= \sqrt{6484.40^2 + (4780.46 + 2416.09)^2} = \sqrt{6484.40^2 + 7196.55^2} = 9686.98\ \text{V} \\
\theta &= \tan^{-1}\frac{7196.55}{6484.40} = 47.98^\circ \\
\delta &= 47.98^\circ - 36.87^\circ = 11.11^\circ \\
\%\text{Reg} &= \frac{9686.98 - 7967.43}{7967.43}\times100 = 21.58\%
\end{aligned}
$$

#### (b) 0.8 pf leading
$$
\begin{aligned}
E &= \sqrt{6484.40^2 + (4780.46 - 2416.09)^2} = \sqrt{6484.40^2 + 2364.37^2} = 6902.00\ \text{V} \\
\theta &= \tan^{-1}\frac{2364.37}{6484.40} = 20.03^\circ \\
\delta &= 36.87^\circ - 20.03^\circ = 16.84^\circ \\
\%\text{Reg} &= \frac{6902.00 - 7967.43}{7967.43}\times100 = -13.37\%
\end{aligned}
$$
(In both cases $E$ leads $V$, as it must for a generator.)

**Answer: (a) regulation = 21.58%, power angle = 11.11°; (b) regulation = −13.37%, power angle = 16.84°.**

## Check yourself

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

1. I for 1320 kW at 0.8 pf, 13.8 kV star is
   A. 119.6 A
   B. 69.03 A
   C. 95.6 A
   D. 57.5 A
   Answer: B. I = 1320e3/(sqrt3 x 13800 x 0.8) = 69.03 A. (Key point: Load current)

2. The power angle δ is
   A. the angle between E and I
   B. the angle between I and V
   C. the angle between E and V
   D. the angle of Zs
   Answer: C. δ = θ - φ for lagging and φ - θ for leading, where θ is the angle of E from I. (Key point: Power angle)

3. For 0.8 lagging, regulation and δ are
   A. -13.37 % and 16.84°
   B. 21.58 % and 16.84°
   C. 11.11 % and 21.58°
   D. 21.58 % and 11.11°
   Answer: D. E = 9687 V; θ = 47.98°, φ = 36.87°. (Key point: Lagging)

4. For 0.8 leading, regulation and δ are
   A. -13.37 % and 16.84°
   B. 21.58 % and 11.11°
   C. -13.37 % and 11.11°
   D. 13.37 % and 16.84°
   Answer: A. E = 6902 V; δ = φ - θ = 36.87° - 20.03°. (Key point: Leading)

## Related in Electric Machines II

- [Regulation of 10 kVA round-rotor generator](https://www.bctengineer.com/question.md?key=semester_5%2Felectric_machines_ii%3Aregulation-10kva-zero-pf) — Electric Machines II, chapter 1
- [Power angle and regulation of 10 kVA alternator](https://www.bctengineer.com/question.md?key=semester_5%2Felectric_machines_ii%3Apower-angle-regulation-10kva) — Electric Machines II, chapter 1
- [Synchronous reactance and regulation from test data](https://www.bctengineer.com/question.md?key=semester_5%2Felectric_machines_ii%3Aregulation-1500kva-2300v-test) — Electric Machines II, chapter 1
- [Regulation of delta connected 1600 kVA alternator](https://www.bctengineer.com/question.md?key=semester_5%2Felectric_machines_ii%3Aregulation-delta-1600kva) — Electric Machines II, chapter 1
- [Emf and regulation of 1500 kVA alternator](https://www.bctengineer.com/question.md?key=semester_5%2Felectric_machines_ii%3Aregulation-1500kva-12000v) — Electric Machines II, chapter 1, asked 2×
- [Emf, regulation and loading with given load impedance](https://www.bctengineer.com/question.md?key=semester_5%2Felectric_machines_ii%3Aemf-regulation-load-impedance) — Electric Machines II, chapter 1

## Related in other subjects

- [1600 kVA alternator: emf, regulation, power angle](https://www.bctengineer.com/question.md?key=semester_4%2Felectrical_machines%3Aalt-1600kva-regulation-power-angle) — Electrical Machines, chapter 6
- [Reactance diagram: 20 MVA generator, 15 kV/25 MVA base](https://www.bctengineer.com/question.md?key=semester_4%2Fpower_system_analysis_i%3Areactance-diagram-20mva-15kv-25mva) — Power System Analysis I, chapter 3
- [Efficiency of 5 kVA transformer at 2 kW load](https://www.bctengineer.com/question.md?key=semester_4%2Felectrical_machines_i%3A5kva-1100-110-max-efficiency-40a) — Electrical Machines I, chapter 2
- [Phase order and star load (6+j8)](https://www.bctengineer.com/question.md?key=semester_1%2Fbasic_electrical_engineering%3Aphase-order-star-load-6j8) — Basic Electrical Engineering, chapter 8