---
title: "Regulation of 1.5 MVA generator and zero-regulation pf"
subject: "Electric Machines II"
chapter: 1
key: "semester_5/electric_machines_ii:regulation-1-5mva-zero-pf"
source: https://www.bctengineer.com/q/semester_5/electric_machines_ii/regulation-1-5mva-zero-pf
site: BCT Engineering
---

# Regulation of 1.5 MVA generator and zero-regulation pf

Electric Machines II · Chapter 1: Three Phase Synchronous Generator

- Asked in: 2075 Asoj, 6 marks
- Answer checked by Claude Sonnet 5.5 on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/electric_machines_ii/regulation-1-5mva-zero-pf
- Whole chapter: https://www.bctengineer.com/md/semester_5/electric_machines_ii/chapter-1

## Question

A 3-phase, star connected synchronous generator is rated at 1.5 MVA, 11 kV. The armature effective resistance and synchronous reactance are 1.2 Ω and 25 Ω respectively per phase. Calculate the percentage voltage regulation for a load of 1.4375 MVA at i) 0.8 p.f. lagging and ii) 0.8 p.f. leading. Also find out the p.f. at which regulation is zero.

## Short answer

For 1.4375 MVA load, voltage regulation is 21.15% at 0.8 pf lagging and -13.12% at 0.8 pf leading. Zero regulation occurs at 0.981 leading power factor.

## Answer

**Given:** star, 1.5 MVA, 11 kV, $R_a = 1.2\ \Omega$, $X_s = 25\ \Omega$; load 1.4375 MVA at 0.8 pf.
$$
\begin{aligned}
V_{ph} &= \frac{11000}{\sqrt3} = 6350.85\ \text{V} \\
I &= \frac{1.4375\times10^6}{\sqrt3\times11000} = 75.45\ \text{A} \\
V\cos\phi + IR_a &= 5080.68 + 90.54 = 5171.22\ \text{V} \\
V\sin\phi &= 3810.51\ \text{V},\quad IX_s = 1886.23\ \text{V}
\end{aligned}
$$

#### (i) 0.8 pf lagging
$$
\begin{aligned}
E &= \sqrt{5171.22^2 + (3810.51 + 1886.23)^2} = 7693.79\ \text{V} \\
\%\text{Reg} &= \frac{7693.79 - 6350.85}{6350.85}\times100 = 21.15\%
\end{aligned}
$$

#### (ii) 0.8 pf leading
$$
\begin{aligned}
E &= \sqrt{5171.22^2 + (3810.51 - 1886.23)^2} = 5517.64\ \text{V} \\
\%\text{Reg} &= \frac{5517.64 - 6350.85}{6350.85}\times100 = -13.12\%
\end{aligned}
$$

#### Power factor for zero regulation
For $E = V$, with $Z_s = 1.2 + j25 = 25.029\angle 87.25^\circ\ \Omega$ and current angle $\phi$ (leading positive):
$$
\begin{aligned}
\cos(\theta_z + \phi) &= -\frac{IZ_s}{2V} = -\frac{75.45\times25.029}{2\times6350.85} = -0.1487 \\
\theta_z + \phi &= 98.55^\circ \\
\phi &= 98.55^\circ - 87.25^\circ = 11.30^\circ \\
\text{pf} &= \cos 11.30^\circ = 0.981\ \text{leading}
\end{aligned}
$$

**Answer: (i) 21.15%; (ii) −13.12%; zero regulation at pf = 0.981 leading.**

## Check yourself

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

1. What is the load current for 1.4375 MVA at 11 kV (three-phase)?
   A. 130.7 A
   B. 75.45 A
   C. 43.6 A
   D. 94.3 A
   Answer: B. I = 1.4375e6 / (sqrt3 x 11000) = 75.45 A. (Key point: Load current)

2. At 0.8 pf lagging, the emf and regulation are:
   A. 5517.6 V and -13.12%
   B. 7693.8 V and 21.15%
   C. 6350.9 V and 0%
   D. 7693.8 V and -21.15%
   Answer: B. E = sqrt(5171.22^2 + (3810.51 + 1886.23)^2) = 7693.79 V, giving +21.15%. (Key point: Lagging regulation)

3. Why is regulation negative at 0.8 pf leading?
   A. Because Ra is negative
   B. Because the reactive drop IXs subtracts from V sin(phi), so E < V
   C. Because the current is lower
   D. Because the speed rises
   Answer: B. E = sqrt(5171.22^2 + (3810.51 - 1886.23)^2) = 5517.64 V which is below V, giving -13.12%. (Key point: Leading regulation)

4. At what power factor is regulation zero for this machine?
   A. About 0.8 lagging
   B. Unity
   C. About 0.981 leading
   D. About 0.981 lagging
   Answer: C. Setting E = V gives phi = 11.30 degrees leading, i.e. pf = cos 11.30 degrees = 0.981 leading. (Key point: Zero regulation pf)

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