---
title: "Internal emf and armature reaction reactance"
subject: "Electric Machines II"
chapter: 1
key: "semester_5/electric_machines_ii:internal-emf-armature-reactance"
source: https://www.bctengineer.com/q/semester_5/electric_machines_ii/internal-emf-armature-reactance
site: BCT Engineering
---

# Internal emf and armature reaction reactance

Electric Machines II · Chapter 1: Three Phase Synchronous Generator

- Asked in: 2072 Kartik, 6 marks
- Answer checked by Claude Sonnet 5.5 on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/electric_machines_ii/internal-emf-armature-reactance
- Whole chapter: https://www.bctengineer.com/md/semester_5/electric_machines_ii/chapter-1

## Question

A 3-phase, star-connected, 60 kVA, 400 V, 50 Hz alternator has effective resistance of 0.25 Ω/ph. The synchronous reactance is 3.5 Ω/ph and leakage reactance is 0.75 Ω/ph. Determine at rated load and 0.8 lagging P.F (i) the internal emf Ei, (ii) the value of armature reactance which represents armature reaction.

## Short answer

Internal emf behind leakage impedance is 289.9 V per phase (502.1 V line). Armature reaction reactance Xa = Xs - Xl = 2.75 Ω per phase.

## Answer

**Given:** star, 60 kVA, 400 V, $R_a = 0.25\ \Omega$, $X_s = 3.5\ \Omega$, $X_l = 0.75\ \Omega$ per phase; rated load at 0.8 pf lagging.

#### Phase voltage and current
$$
\begin{aligned}
V_{ph} &= \frac{400}{\sqrt3} = 230.94\ \text{V} \\
I &= \frac{60\times10^3}{\sqrt3\times400} = 86.60\ \text{A}
\end{aligned}
$$

#### (i) Internal emf $E_i$
The internal (air-gap) emf is the voltage behind the leakage impedance $R_a + jX_l$:
$$
\begin{aligned}
V\cos\phi + IR_a &= 184.75 + 86.60\times0.25 = 206.40\ \text{V} \\
V\sin\phi + IX_l &= 138.56 + 86.60\times0.75 = 203.52\ \text{V} \\
E_i &= \sqrt{206.40^2 + 203.52^2} = 289.86\ \text{V per phase} \\
E_{i,L} &= \sqrt3\times289.86 = 502.1\ \text{V}
\end{aligned}
$$

#### (ii) Armature reactance (armature reaction)
Since $X_s = X_l + X_a$:
$$
X_a = X_s - X_l = 3.5 - 0.75 = 2.75\ \Omega\ \text{per phase}
$$

```text
              E0
             /|
            / |  j I Xa  (armature reaction)
           /  |
          /   + Ei
         /   /|  j I Xl
        O---+-+ V + I Ra
         `-.
            `-. I (0.8 lag)
```

(For reference, the no-load emf using $X_s$ would be $E_0 = \sqrt{206.40^2 + (138.56 + 303.11)^2} = 487.5$ V per phase.)

**Answer: (i) $E_i$ = 289.9 V per phase (502.1 V line); (ii) $X_a$ = 2.75 Ω per phase.**

## Check yourself

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

1. Which impedance is used to find Ei in this problem?
   A. Ra + jXs = 0.25 + j3.5
   B. Ra + jXl = 0.25 + j0.75
   C. jXa = j2.75
   D. Ra only
   Answer: B. The internal emf is the voltage behind the leakage impedance Ra + jXl. (Key point: Impedance for Ei)

2. What is the rated current of the 60 kVA, 400 V star alternator?
   A. 150 A
   B. 86.60 A
   C. 130 A
   D. 50 A
   Answer: B. I = 60000/(sqrt3 x 400) = 86.60 A. (Key point: Rated current)

3. What is Ei per phase?
   A. 487.5 V
   B. 230.9 V
   C. 289.9 V
   D. 502.1 V
   Answer: C. Ei = sqrt(206.40^2 + 203.52^2) = 289.86 V per phase; 502.1 V is the line value. (Key point: Value of Ei)

4. How is the armature reactance Xa found?
   A. Xa = Xs + Xl = 4.25 ohm
   B. Xa = Xs - Xl = 2.75 ohm
   C. Xa = Xs x Xl
   D. Xa = Xs / Xl
   Answer: B. Since Xs = Xl + Xa, Xa = 3.5 - 0.75 = 2.75 ohm per phase. (Key point: Armature reactance)

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