---
title: "Generator on bus with 20% excitation increase"
subject: "Electric Machines II"
chapter: 1
key: "semester_5/electric_machines_ii:generator-infinite-bus-excitation-20"
source: https://www.bctengineer.com/q/semester_5/electric_machines_ii/generator-infinite-bus-excitation-20
site: BCT Engineering
---

# Generator on bus with 20% excitation increase

Electric Machines II · Chapter 1: Three Phase Synchronous Generator

- Asked in: 2071 Chaitra, 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-20
- Whole chapter: https://www.bctengineer.com/md/semester_5/electric_machines_ii/chapter-1

## Question

A 3-phase, 5 kVA, 208 V, 4-pole, 50 Hz star connected synchronous machine has negligible stator winding resistance and synchronous reactance of 8 Ω/phase. The machine is operated as generator in parallel with 3-phase, 208 V and 50 Hz supply. Then,
i) Determine the excitation voltage and power angle when machine is delivering rated kVA at 0.8 pf lagging.
ii) If the excitation is increased by 20% without changing prime mover power, find the stator current and power factor.

## Short answer

At rated kVA, 0.8 pf lagging, excitation voltage is 206.8 V per phase, power angle 25.44°. With 20% higher excitation and constant prime mover power, stator current becomes 17.83 A at 0.623 lagging pf.

## Answer

**Given:** star, 5 kVA, 208 V, 50 Hz, $X_s = 8\ \Omega$, $R_a \approx 0$, in parallel with a 208 V bus (infinite bus).

#### (i) Excitation voltage and power angle at rated kVA, 0.8 pf lagging
$$
\begin{aligned}
V &= \frac{208}{\sqrt3} = 120.09\ \text{V} \\
I &= \frac{5000}{\sqrt3\times208} = 13.88\ \text{A},\quad \bar I = 13.88\angle -36.87^\circ \\
\bar E &= \bar V + jX_s\bar I = 120.09 + j8\times13.88(0.8 - j0.6) \\
&= (120.09 + 66.62) + j88.82 = 186.71 + j88.82 \\
E &= 206.76\ \text{V per phase},\quad \delta = 25.44^\circ
\end{aligned}
$$
Check: $P = \dfrac{3EV}{X_s}\sin\delta = \dfrac{3\times206.76\times120.09}{8}\sin 25.44^\circ = 4000$ W $= 5\times0.8$ kW.

#### (ii) Excitation increased by 20%, same input power
$$
\begin{aligned}
E' &= 1.2\times206.76 = 248.11\ \text{V} \\
E'\sin\delta' &= E\sin\delta = 88.82 \quad (P \text{ constant}) \\
\sin\delta' &= \frac{88.82}{248.11} = 0.3580,\quad \delta' = 20.98^\circ \\
\bar E' &= 248.11\angle 20.98^\circ = 231.66 + j88.82 \\
\bar I' &= \frac{\bar E' - \bar V}{jX_s} = \frac{111.58 + j88.82}{j8} = 11.10 - j13.95 \\
I' &= 17.83\ \text{A},\quad \angle -51.48^\circ \\
\text{pf} &= \cos 51.48^\circ = 0.623\ \text{lagging}
\end{aligned}
$$
The active current component stays 11.10 A (power unchanged); extra excitation only increases the lagging reactive current, so the generator supplies more reactive power ($Q$ rises from 3 kVAR to 5.02 kVAR).

**Answer: (i) E = 206.8 V/phase (358.1 V line), δ = 25.44°; (ii) I = 17.83 A, pf = 0.623 lagging.**

## Check yourself

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

1. How is E found for a generator with negligible Ra?
   A. E = V - jXs I
   B. E = V + jXs I
   C. E = V + Xs I
   D. E = V x Xs
   Answer: B. For a generator E = V + jXs I, with I at 36.87 degrees lagging giving E = 186.71 + j88.82. (Key point: Excitation phasor equation)

2. What are E and delta at rated kVA, 0.8 pf lagging?
   A. 206.76 V per phase, 25.44 degrees
   B. 120.09 V per phase, 0 degrees
   C. 186.7 V per phase, 36.87 degrees
   D. 358.1 V per phase, 25.44 degrees
   Answer: A. E = sqrt(186.71^2 + 88.82^2) = 206.76 V per phase (358.1 V is the line value), delta = 25.44 degrees. (Key point: Result for part (i))

3. When excitation is raised 20% with prime mover power unchanged, what stays constant?
   A. Load angle delta
   B. E sin(delta), so the active current component
   C. Stator current
   D. Power factor
   Answer: B. P = 3EV sin(delta)/Xs is fixed, so E sin(delta) = 88.82 stays the same. (Key point: Constant power condition)

4. What happens to the stator current and pf after raising excitation by 20%?
   A. Current falls and pf improves to unity
   B. Current rises to 17.83 A and pf drops to 0.623 lagging as reactive power rises
   C. Current stays 13.88 A with same pf
   D. Active power rises
   Answer: B. The active current stays 11.10 A while the lagging reactive current rises to 13.95 A, so I = 17.83 A at 0.623 lagging. (Key point: Effect of more excitation)

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