---
title: "Generator on infinite bus with 15% excitation rise"
subject: "Electric Machines II"
chapter: 1
key: "semester_5/electric_machines_ii:generator-infinite-bus-excitation-15"
source: https://www.bctengineer.com/q/semester_5/electric_machines_ii/generator-infinite-bus-excitation-15
site: BCT Engineering
---

# Generator on infinite bus with 15% excitation rise

Electric Machines II · Chapter 1: Three Phase Synchronous Generator

- Asked in: 2071 Shrawan, 6 marks
- Video: [Effect of Change in Excitation | Operation of an Alternator with an Infinite Bus Bar | V-Curves](https://www.youtube.com/watch?v=xgmyt0JK3OA) — Dr. Afroz Alam
- Answer checked by Claude Sonnet 5.5 on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/electric_machines_ii/generator-infinite-bus-excitation-15
- Whole chapter: https://www.bctengineer.com/md/semester_5/electric_machines_ii/chapter-1

## Question

A 3-phase, 5 kVA, 208 V, 4-pole, 60 Hz, star connected synchronous generator has negligible armature winding resistance and synchronous reactance of 10 ohms per phase. The generator is first connected to an infinite bus of 208 V, 60 Hz.
i) Determine the excitation voltage and the power angle when the generator is delivering rated kVA at 0.8 pf lagging.
ii) If the field excitation is now increased by 15% (keeping turbine power constant), find the stator current, power factor, active and reactive power constant.

## Short answer

At rated kVA, 0.8 pf lagging, excitation is 231.7 V per phase, power angle 28.63°. After 15% field increase with constant turbine power, stator current is 16.51 A, pf 0.673 lagging, active power 4.0 kW, reactive power 4.40 kVAR.

## Answer

**Given:** star, 5 kVA, 208 V, 60 Hz, $X_s = 10\ \Omega$, $R_a \approx 0$, on a 208 V, 60 Hz infinite bus.

#### (i) Excitation voltage and power angle at rated kVA, 0.8 pf lagging
$$
\begin{aligned}
V &= \frac{208}{\sqrt3} = 120.09\ \text{V} \\
I &= \frac{5000}{\sqrt3\times208} = 13.88\ \text{A},\quad \bar I = 13.88\angle -36.87^\circ \\
\bar E &= \bar V + jX_s\bar I = 120.09 + j10\times13.88(0.8 - j0.6) \\
&= (120.09 + 83.27) + j111.03 = 203.36 + j111.03 \\
E &= 231.70\ \text{V per phase},\quad \delta = 28.63^\circ
\end{aligned}
$$
$P = 3VI\cos\phi = 4000$ W, $Q = 3VI\sin\phi = 3000$ VAR.

#### (ii) Excitation increased by 15%, turbine power constant
$$
\begin{aligned}
E' &= 1.15\times231.70 = 266.45\ \text{V} \\
\sin\delta' &= \frac{E\sin\delta}{E'} = \frac{111.03}{266.45} = 0.4167,\quad \delta' = 24.63^\circ \\
\bar E' &= 266.45\angle 24.63^\circ = 242.22 + j111.03 \\
\bar I' &= \frac{\bar E' - \bar V}{jX_s} = \frac{122.13 + j111.03}{j10} = 11.10 - j12.21 \\
I' &= 16.51\ \text{A},\quad \angle -47.73^\circ \\
\text{pf} &= \cos 47.73^\circ = 0.673\ \text{lagging}
\end{aligned}
$$

#### Active and reactive power
$$
\begin{aligned}
P &= 3\times120.09\times11.10 = 4000\ \text{W (unchanged)} \\
Q &= 3\times120.09\times12.21 = 4400\ \text{VAR (lagging, delivered)}
\end{aligned}
$$
Increasing the field only raises the reactive power supplied to the bus; active power is set by the turbine.

**Answer: (i) E = 231.7 V/phase (401.3 V line), δ = 28.63°; (ii) I = 16.51 A, pf = 0.673 lagging, P = 4.0 kW (constant), Q = 4.40 kVAR.**

## Check yourself

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

1. At rated kVA, 0.8 pf lagging with Xs = 10 ohm, the excitation voltage per phase is:
   A. 120.1 V
   B. 231.7 V
   C. 266.5 V
   D. 401.3 V
   Answer: B. E = 120.09 + j10 x 13.88(0.8 - j0.6) = 203.36 + j111.03, so E = 231.7 V per phase. (Key point: Excitation emf (i))

2. How is the new power angle after a 15% rise in excitation found?
   A. delta stays at 28.63 degrees
   B. From E' sin(delta') = E sin(delta) since P is constant
   C. From delta' = 1.15 x delta
   D. From cos(delta') = E/E'
   Answer: B. Constant turbine power means E sin(delta) is constant, so sin(delta') = 111.03/266.45 giving 24.63 degrees. (Key point: New load angle)

3. What happens to P and Q when excitation rises by 15%?
   A. P rises, Q unchanged
   B. Both unchanged
   C. P unchanged at 4 kW, Q rises to about 4.40 kVAR
   D. P falls, Q rises
   Answer: C. Active power is set by the turbine; extra field increases the lagging reactive power supplied to the bus. (Key point: Active and reactive power)

4. What are the new stator current and power factor?
   A. 16.51 A, 0.673 lagging
   B. 13.88 A, 0.8 lagging
   C. 16.51 A, 0.673 leading
   D. 11.10 A, unity
   Answer: A. I' = 11.10 - j12.21 A, magnitude 16.51 A, pf = cos 47.73 degrees = 0.673 lagging. (Key point: New current and pf)

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