---
title: "Power angle and regulation of 10 kVA alternator"
subject: "Electric Machines II"
chapter: 1
key: "semester_5/electric_machines_ii:power-angle-regulation-10kva"
source: https://www.bctengineer.com/q/semester_5/electric_machines_ii/power-angle-regulation-10kva
site: BCT Engineering
---

# Power angle and regulation of 10 kVA alternator

Electric Machines II · Chapter 1: Three Phase Synchronous Generator

- Asked in: 2075 Chaitra, 8 marks
- Answer checked by Claude Sonnet 5.5 on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/electric_machines_ii/power-angle-regulation-10kva
- Whole chapter: https://www.bctengineer.com/md/semester_5/electric_machines_ii/chapter-1

## Question

A 3-phase, 10 kVA, 400 V, 50 Hz star-connected synchronous alternator supplies the rated load at 0.8 power factor lagging. If the armature resistance is 0.5 Ω and synchronous reactance is 10 Ω, find the power angle and voltage regulation.

## Short answer

Vph=230.94 V, I=14.43 A. Using E=√((Vcosφ+IRa)²+(Vsinφ+IXs)²) at 0.8 pf lagging gives E=341.9 V/phase. Power angle δ=θ−φ=18.97°. Voltage regulation = (341.9−230.94)/230.94×100 = 48.04%.

## Answer

**Given:** star, 10 kVA, 400 V, 50 Hz, $R_a = 0.5\ \Omega$, $X_s = 10\ \Omega$; rated load at 0.8 pf lagging.

#### Phase voltage and current
$$
\begin{aligned}
V_{ph} &= \frac{400}{\sqrt3} = 230.94\ \text{V} \\
I &= \frac{10\times10^3}{\sqrt3\times400} = 14.43\ \text{A} \\
\cos\phi &= 0.8,\quad \sin\phi = 0.6,\quad \phi = 36.87^\circ
\end{aligned}
$$

#### Excitation emf
$$
\begin{aligned}
V\cos\phi + IR_a &= 184.75 + 14.43\times0.5 = 191.97\ \text{V} \\
V\sin\phi + IX_s &= 138.56 + 14.43\times10 = 282.90\ \text{V} \\
E &= \sqrt{191.97^2 + 282.90^2} = 341.89\ \text{V per phase}\quad(592.2\ \text{V line})
\end{aligned}
$$

#### Power angle
The angle of $E$ from the current phasor is
$$
\begin{aligned}
\theta &= \tan^{-1}\frac{282.90}{191.97} = 55.84^\circ \\
\delta &= \theta - \phi = 55.84^\circ - 36.87^\circ = 18.97^\circ
\end{aligned}
$$

```text
                       E = 341.9 V
                      /
                     /  delta = 18.97 deg
                    O-------------- V = 230.9 V
                     \  phi = 36.87 deg
                      \ I = 14.43 A
```

#### Voltage regulation
$$
\%\text{Reg} = \frac{E - V}{V}\times100 = \frac{341.89 - 230.94}{230.94}\times100 = 48.04\%
$$
(The regulation is high because $X_s = 10\ \Omega$ is large compared with the rated impedance $V/I = 16\ \Omega$, i.e. $X_s \approx 0.625$ pu.)

**Answer: power angle δ = 18.97°, voltage regulation = 48.04% (E = 341.9 V per phase).**

## Check yourself

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

1. For the 10 kVA, 400 V star alternator, what are the phase voltage and rated current?
   A. 400 V and 25 A
   B. 230.94 V and 14.43 A
   C. 230.94 V and 25 A
   D. 400 V and 14.43 A
   Answer: B. V_ph = 400/sqrt3 = 230.94 V and I = 10000/(sqrt3 x 400) = 14.43 A. (Key point: Phase quantities)

2. Which expression gives E per phase at 0.8 pf lagging?
   A. sqrt((V cos(phi) + I Ra)^2 + (V sin(phi) + I Xs)^2)
   B. sqrt((V cos(phi) + I Ra)^2 + (V sin(phi) - I Xs)^2)
   C. V + I(Ra + Xs)
   D. V cos(phi) + I Xs
   Answer: A. For lagging pf, the reactive terms add: E = sqrt(191.97^2 + 282.90^2) = 341.89 V. (Key point: Excitation emf formula)

3. How is the power angle delta obtained from the angle theta = arctan(282.90/191.97) = 55.84 degrees?
   A. delta = theta + phi
   B. delta = theta
   C. delta = 90 - theta
   D. delta = theta - phi = 18.97 degrees
   Answer: D. theta is measured from the current phasor, so subtracting the power factor angle phi (36.87 degrees) gives the angle between E and V. (Key point: Power angle)

4. What is the percentage voltage regulation of this alternator?
   A. About 48.04%
   B. About 18.97%
   C. About 33.7%
   D. About 14.8%
   Answer: A. Reg = (341.89 - 230.94)/230.94 x 100 = 48.04%. (Key point: Voltage regulation)

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