---
title: "Emf and regulation of 1500 kVA alternator"
subject: "Electric Machines II"
chapter: 1
key: "semester_5/electric_machines_ii:regulation-1500kva-12000v"
source: https://www.bctengineer.com/q/semester_5/electric_machines_ii/regulation-1500kva-12000v
site: BCT Engineering
---

# Emf and regulation of 1500 kVA alternator

Electric Machines II · Chapter 1: Three Phase Synchronous Generator

- Asked in: 2080 Bhadra, 6 marks; 2072 Chaitra, 8 marks (asked 2 times)
- Answer checked by Claude Sonnet 5.5 on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/electric_machines_ii/regulation-1500kva-12000v
- Whole chapter: https://www.bctengineer.com/md/semester_5/electric_machines_ii/chapter-1

## Question

A 3-phase star-connected alternator is rated at 1500 kVA, 12000 V. The armature effective resistance and synchronous reactance are 2 Ω and 25 Ω respectively per phase. Calculate the emf generated and percentage voltage regulation for a load of 1250 kW at power factors of: (i) 0.75 leading (ii) 0.75 lagging.

## Short answer

Using per-phase voltage 6928 V and current 80.2 A. At 0.75 leading pf: generated emf = 5945 V/phase (10.30 kV line), regulation = –14.2%. At 0.75 lagging pf: emf = 8490 V/phase (14.71 kV line), regulation = +22.6%. Calculated via E = √[(Vcosφ+IR_a)² + (Vsinφ ∓ IX_s)²].

## Answer

**Given:** star connected, $V_L$ = 12000 V, $R_a = 2\ \Omega$, $X_s = 25\ \Omega$ per phase, load 1250 kW at 0.75 pf.

$$
\begin{aligned}
V_{ph} &= \frac{12000}{\sqrt3} = 6928.2\ \text{V} \\
I &= \frac{1250\times10^3}{\sqrt3\times12000\times0.75} = 80.19\ \text{A} \\
\cos\phi &= 0.75,\quad \sin\phi = 0.6614 \\
V\cos\phi + IR_a &= 5196.15 + 80.19\times2 = 5356.53\ \text{V} \\
V\sin\phi &= 4582.58\ \text{V},\quad IX_s = 80.19\times25 = 2004.69\ \text{V}
\end{aligned}
$$

(Note: this load is $1250/0.75 = 1667$ kVA, which is above the 1500 kVA rating. The calculation follows the data given in the question.)

#### (i) 0.75 pf leading
$$
\begin{aligned}
E &= \sqrt{(V\cos\phi + IR_a)^2 + (V\sin\phi - IX_s)^2} \\
&= \sqrt{5356.53^2 + (4582.58 - 2004.69)^2} \\
&= \sqrt{5356.53^2 + 2577.89^2} = 5944.57\ \text{V per phase} \\
E_L &= \sqrt3 \times 5944.57 = 10296\ \text{V} \\
\%\text{Reg} &= \frac{5944.57 - 6928.2}{6928.2}\times100 = -14.20\%
\end{aligned}
$$

#### (ii) 0.75 pf lagging
$$
\begin{aligned}
E &= \sqrt{5356.53^2 + (4582.58 + 2004.69)^2} \\
&= \sqrt{5356.53^2 + 6587.26^2} = 8490.26\ \text{V per phase} \\
E_L &= \sqrt3 \times 8490.26 = 14706\ \text{V} \\
\%\text{Reg} &= \frac{8490.26 - 6928.2}{6928.2}\times100 = 22.55\%
\end{aligned}
$$

**Answer: (i) 0.75 lead: E = 5944.6 V/phase (10.30 kV line), regulation = −14.20%; (ii) 0.75 lag: E = 8490.3 V/phase (14.71 kV line), regulation = +22.55%.**

## Check yourself

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

1. For the star alternator, the phase voltage and current at 1250 kW, 0.75 pf are
   A. 6928 V and 80.19 A
   B. 12000 V and 80.19 A
   C. 6928 V and 138.9 A
   D. 12000 V and 138.9 A
   Answer: A. Vph = 12000/sqrt3 = 6928 V; I = 1250e3/(sqrt3*12000*0.75) = 80.19 A. (Key point: Phase quantities)

2. E for the 0.75 leading case uses which reactive term?
   A. V sinφ + I Xs = 6587 V
   B. V sinφ - I Xs = 2577.9 V
   C. V cosφ - I Xs
   D. I Xs only
   Answer: B. For leading pf the reactance drop subtracts from V sinφ. (Key point: Leading formula)

3. The regulation at 0.75 pf leading is
   A. +22.55 %
   B. -22.55 %
   C. -14.20 % (E = 5944.6 V/phase)
   D. +14.20 %
   Answer: C. E = 5944.6 V is less than V = 6928 V, giving negative regulation. (Key point: Leading result)

4. For 0.75 lagging, E per phase and regulation are
   A. 5944 V and -14.20 %
   B. 6928 V and 0 %
   C. 8490 V and -22.55 %
   D. 8490 V and +22.55 %
   Answer: D. Reactive term is 4582.6 + 2004.7 = 6587.3 V, E = 8490 V > V, so regulation is positive. (Key point: Lagging result)

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