---
title: "Internal emf, no-load emf and regulation"
subject: "Electric Machines II"
chapter: 1
key: "semester_5/electric_machines_ii:internal-noload-emf-50kva"
source: https://www.bctengineer.com/q/semester_5/electric_machines_ii/internal-noload-emf-50kva
site: BCT Engineering
---

# Internal emf, no-load emf and regulation

Electric Machines II · Chapter 1: Three Phase Synchronous Generator

- Asked in: 2074 Chaitra, 7 marks
- Answer checked by Claude Sonnet 5.5 on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/electric_machines_ii/internal-noload-emf-50kva
- Whole chapter: https://www.bctengineer.com/md/semester_5/electric_machines_ii/chapter-1

## Question

A star connected 50 kVA, 440 V, 50 Hz alternator has effective armature resistance of 0.25 Ω per phase, synchronous reactance is 3.2 Ω per phase and the leakage reactance is 0.5 Ω per phase. At rated load and unity power factor, Determine: (i) Internal emf (ii) No-load emf (iii) Percentage voltage regulation at full load

## Short answer

Using leakage impedance, internal emf is 272.4 V per phase (471.8 V line). With synchronous impedance, no-load emf is 342.4 V per phase (593.0 V line), giving 34.77% regulation.

## Answer

**Given:** star, 50 kVA, 440 V, $R_a = 0.25\ \Omega$, $X_s = 3.2\ \Omega$, leakage reactance $X_l = 0.5\ \Omega$ per phase; rated load at unity pf.

- The **internal (air-gap) emf** $E_i$ is found using only the leakage impedance $R_a + jX_l$.
- The **no-load emf** $E_0$ (excitation emf) uses the full synchronous impedance $R_a + jX_s$, since $X_s = X_l + X_a$ includes armature reaction.

$$
\begin{aligned}
V_{ph} &= \frac{440}{\sqrt3} = 254.03\ \text{V} \\
I &= \frac{50\times10^3}{\sqrt3\times440} = 65.61\ \text{A}\quad(\text{in phase with } V)
\end{aligned}
$$

#### (i) Internal emf
$$
\begin{aligned}
\bar E_i &= V + I(R_a + jX_l) = (254.03 + 65.61\times0.25) + j(65.61\times0.5) \\
&= 270.44 + j32.80 \\
E_i &= 272.42\ \text{V per phase}\quad(471.8\ \text{V line})
\end{aligned}
$$

#### (ii) No-load emf
$$
\begin{aligned}
\bar E_0 &= V + I(R_a + jX_s) = 270.44 + j(65.61\times3.2) \\
&= 270.44 + j209.95 \\
E_0 &= 342.36\ \text{V per phase}\quad(593.0\ \text{V line})
\end{aligned}
$$

#### (iii) Voltage regulation
$$
\%\text{Reg} = \frac{E_0 - V}{V}\times100 = \frac{342.36 - 254.03}{254.03}\times100 = 34.77\%
$$

```text
              E0
             /|
            / |  I Xa (armature reaction)
           /  |
          / Ei+
         / /  |  I Xl
        O-----+  V + I Ra
        O-------> I  (unity pf)
```

**Answer: (i) $E_i$ = 272.4 V/phase (471.8 V line); (ii) $E_0$ = 342.4 V/phase (593.0 V line); (iii) regulation = 34.77%.**

## Check yourself

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

1. Which impedance is used to find the internal (air-gap) emf Ei?
   A. Ra + jXs
   B. Only Ra
   C. Ra + jXl (leakage)
   D. jXa only
   Answer: C. The internal emf is the voltage behind the leakage impedance, Ei = V + I(Ra + jXl). (Key point: Internal emf uses leakage)

2. Why does the no-load emf E0 use Xs instead of Xl?
   A. Xs = Xl + Xa includes armature reaction
   B. Xs is smaller than Xl
   C. Xs excludes the resistance
   D. Xs is a mechanical quantity
   Answer: A. The synchronous reactance accounts for both leakage and armature reaction, so E0 = V + I(Ra + jXs). (Key point: No-load emf uses Xs)

3. What is the rated current of the 50 kVA, 440 V star alternator, and its phase relation to V at unity pf?
   A. 113.6 A, leading V
   B. 65.61 A, in phase with V
   C. 65.61 A, lagging V by 90 degrees
   D. 113.6 A, in phase with V
   Answer: B. I = 50000/(sqrt3 x 440) = 65.61 A, and at unity pf it is in phase with V. (Key point: Rated current)

4. What is the regulation at full load, unity pf?
   A. 7.2%
   B. 34.77%
   C. 56.2%
   D. 12.3%
   Answer: B. Reg = (342.36 - 254.03)/254.03 x 100 = 34.77%, using E0 not Ei. (Key point: Regulation)

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