---
title: "Emf, regulation and loading with given load impedance"
subject: "Electric Machines II"
chapter: 1
key: "semester_5/electric_machines_ii:emf-regulation-load-impedance"
source: https://www.bctengineer.com/q/semester_5/electric_machines_ii/emf-regulation-load-impedance
site: BCT Engineering
---

# Emf, regulation and loading with given load impedance

Electric Machines II · Chapter 1: Three Phase Synchronous Generator

- Asked in: 2073 Shrawan, 7 marks
- Video: [Synchronous Impedance Method or EMF Method to find Voltage Regulation of 3-phase Alternator](https://www.youtube.com/watch?v=j6EiKcY3vTk) — We Learn by Sharun
- Answer checked by Claude Sonnet 5.5 on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/electric_machines_ii/emf-regulation-load-impedance
- Whole chapter: https://www.bctengineer.com/md/semester_5/electric_machines_ii/chapter-1

## Question

A 3-phase, star-connected 3-phase synchronous generator is rated as 1500 kVA, 11 kV. The armature winding resistance is 0.8 ohm per phase and synchronous reactance is 4 ohms per phase. If the generator is supplying power to a three phase balanced load of (80+j60) ohm per phase at rated terminal voltage, calculate emf generated and voltage regulation. Is the generator overloaded OR under-loaded? Calculate the percentage by which it is overloaded OR under-loaded.

## Short answer

With load impedance 80+j60 Ω, generated emf is 6546.2 V per phase (11.34 kV line), regulation 3.08%. The generator supplies 1210 kVA, so it is under-loaded by 19.33%.

## Answer

**Given:** star, 1500 kVA, 11 kV, $R_a = 0.8\ \Omega$, $X_s = 4\ \Omega$; load $Z_L = 80 + j60\ \Omega$ per phase at rated voltage.

#### Load current
$$
\begin{aligned}
V_{ph} &= \frac{11000}{\sqrt3} = 6350.85\ \text{V} \\
Z_L &= 80 + j60 = 100\angle 36.87^\circ\ \Omega \\
I &= \frac{6350.85}{100} = 63.51\ \text{A},\quad \text{pf} = \frac{80}{100} = 0.8\ \text{lagging}
\end{aligned}
$$

#### EMF generated
$$
\begin{aligned}
V\cos\phi + IR_a &= 5080.68 + 63.51\times0.8 = 5131.49\ \text{V} \\
V\sin\phi + IX_s &= 3810.51 + 63.51\times4 = 4064.55\ \text{V} \\
E &= \sqrt{5131.49^2 + 4064.55^2} = 6546.20\ \text{V per phase} \\
E_L &= \sqrt3\times6546.20 = 11338\ \text{V}
\end{aligned}
$$

#### Voltage regulation
$$
\%\text{Reg} = \frac{6546.20 - 6350.85}{6350.85}\times100 = 3.08\%
$$

#### Overloaded or under-loaded?
$$
\begin{aligned}
I_{rated} &= \frac{1500\times10^3}{\sqrt3\times11000} = 78.73\ \text{A} \\
\text{Load kVA} &= 3\times6350.85\times63.51\times10^{-3} = 1210\ \text{kVA} \\
\text{Under-load} &= \frac{78.73 - 63.51}{78.73}\times100 = \frac{1500 - 1210}{1500}\times100 = 19.33\%
\end{aligned}
$$

**Answer: E = 6546.2 V/phase (11.34 kV line); regulation = 3.08%; the generator is under-loaded by 19.33% (supplies 1210 kVA, 968 kW).**

## Check yourself

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

1. A load of 80 + j60 ohm per phase at 6350.85 V phase voltage gives:
   A. I = 63.51 A at 0.8 lagging
   B. I = 63.51 A at 0.6 lagging
   C. I = 79.4 A at 0.8 lagging
   D. I = 63.51 A at 0.8 leading
   Answer: A. |Z| = 100 ohm so I = 63.51 A, and pf = 80/100 = 0.8 lagging because the load is inductive. (Key point: Load current and pf)

2. What is the generated emf per phase?
   A. 5131.5 V
   B. 6350.9 V
   C. 7500 V
   D. 6546.2 V
   Answer: D. E = sqrt(5131.49^2 + 4064.55^2) = 6546.2 V per phase. (Key point: EMF calculation)

3. What is the voltage regulation?
   A. 19.33%
   B. 3.08%
   C. 30.8%
   D. -3.08%
   Answer: B. (6546.2 - 6350.85)/6350.85 x 100 = 3.08%. (Key point: Regulation)

4. How is the generator found to be under-loaded by 19.33%?
   A. Load current 63.51 A is below rated 78.73 A, (78.73 - 63.51)/78.73
   B. Load voltage is below rated
   C. Regulation is below 20%
   D. Power factor is below 1
   Answer: A. Rated current is 1500 kVA/(sqrt3 x 11 kV) = 78.73 A, and the load takes 1210 kVA, 19.33% below rating. (Key point: Under-load percentage)

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