---
title: "1500 kVA alternator: excitation emf, voltage at leading pf"
subject: "Electrical Machines"
chapter: 6
key: "semester_4/electrical_machines:alt-1500kva-6600v-excitation-emf"
source: https://www.bctengineer.com/q/semester_4/electrical_machines/alt-1500kva-6600v-excitation-emf
site: BCT Engineering
---

# 1500 kVA alternator: excitation emf, voltage at leading pf

Electrical Machines · Chapter 6: Three Phase Synchronous Machines

- Asked in: 2079 Jestha, 8 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/electrical_machines/alt-1500kva-6600v-excitation-emf
- Whole chapter: https://www.bctengineer.com/md/semester_4/electrical_machines/chapter-6

## Question

A 1500 kVA, 6600 V, 3 phase star connected alternator with a resistance of $0.4\ \Omega$ and reactance of $6\ \Omega$ per phase delivers full load current at a power factor of 0.8 lagging and at normal rated voltage. Calculate the (i) excitation EMF (ii) terminal voltage for same excitation and the load current at 0.8 power factor leading.

## Short answer

For 0.8 lagging, per-phase Vph=3810.5 V, I=131.22 A, E_ph=4366 V (line 7562 V) from phasor sum. For 0.8 leading with same E and I, solving V²+2V·789.1·cos123.06°+789.1²=4366² gives terminal Vph=4746 V (line 8220 V).

## Answer

**Given:** 1500 kVA, 6600 V, star, $R_a = 0.4\ \Omega$, $X_s = 6\ \Omega$ per phase, full-load at 0.8 pf lagging, rated voltage.

$$
V_{ph} = \frac{6600}{\sqrt3} = 3810.5\ \text{V}, \qquad I = \frac{1500\times10^3}{\sqrt3\times6600} = 131.22\ \text{A}
$$

#### (i) Excitation emf (0.8 lagging)
Taking $V_{ph}$ as reference, $\vec I = 131.22\angle-36.87^\circ$, $Z_s = 0.4 + j6$:
$$
IR_a = 52.49\ \text{V}, \quad IX_s = 787.30\ \text{V}
$$
$$
\vec E = 3810.5 + 131.22\angle-36.87^\circ(0.4+j6)
$$
$$
E = \sqrt{(3810.5 + 52.49\times0.8 + 787.30\times0.6)^2 + (787.30\times0.8 - 52.49\times0.6)^2}
$$
$$
E_{ph} = 4366.07\ \text{V}, \qquad E_L = \sqrt3\times4366.07 = 7562.3\ \text{V}, \qquad \delta = 7.88^\circ
$$

#### (ii) Terminal voltage at 0.8 leading (same $E$ and $I$)
$\vec I = 131.22\angle+36.87^\circ$. The drop $\vec IZ_s$ has magnitude $131.22\times6.0133 = 789.1$ V at $36.87^\circ + 86.19^\circ = 123.06^\circ$ from $V$. With $|\vec V + \vec IZ_s| = E$:
$$
V^2 + 2V(789.1)\cos123.06^\circ + 789.1^2 = 4366.07^2
$$
Solving this quadratic gives (taking the positive root)
$$
V_{ph} = 4746.1\ \text{V}, \qquad V_L = \sqrt3\times4746.1 = 8220.5\ \text{V}
$$

**Answer:** (i) $E_{ph} = 4366$ V (line 7562 V); (ii) terminal voltage $\approx 8220$ V line (4746 V per phase). The voltage rises because the leading current's armature reaction is magnetising.

## Check yourself

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

1. For the 1500 kVA, 6600 V star alternator, what are Vph and full-load current?
   A. 6600 V and 131.22 A
   B. 3810.5 V and 131.22 A
   C. 3810.5 V and 227.3 A
   D. 6600 V and 227.3 A
   Answer: B. Vph = 6600/sqrt3 = 3810.5 V; I = 1500000/(sqrt3 x 6600) = 131.22 A. (Key point: Phase voltage and current)

2. How is E found for a lagging load?
   A. E = V - I(Ra + jXs)
   B. E = V + I(Ra + jXs), with I at -36.87 degrees to V
   C. E = V + I Xs only
   D. E = V + I(Ra + jXs), with I at +36.87 degrees
   Answer: B. For a lagging 0.8 pf generator the current lags V by 36.87 degrees and E = V + I Zs, giving about 4366 V per phase. (Key point: Excitation emf method)

3. What is the line excitation emf at full load, 0.8 lagging?
   A. About 4366 V
   B. About 7562 V
   C. About 6600 V
   D. About 8220 V
   Answer: B. E per phase = 4366.07 V, so the line value is sqrt3 x 4366 = 7562 V. (Key point: Result for (i))

4. Why is the terminal voltage higher (about 8220 V) with the same E and 0.8 leading current?
   A. Leading current armature reaction is demagnetising
   B. Leading current armature reaction is magnetising, so V rises for the same E
   C. The resistance becomes negative
   D. The power factor does not affect voltage
   Answer: B. With a leading current the IZs drop adds in a way that makes V larger than E would give for a lagging load; armature reaction is magnetising. (Key point: Leading load)

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