---
title: "Internal emf and load angle of salient pole motor"
subject: "Electric Machines II"
chapter: 2
key: "semester_5/electric_machines_ii:salient-motor-200kw-load-angle"
source: https://www.bctengineer.com/q/semester_5/electric_machines_ii/salient-motor-200kw-load-angle
site: BCT Engineering
---

# Internal emf and load angle of salient pole motor

Electric Machines II · Chapter 2: Three Phase Synchronous Motor

- Asked in: 2078 Kartik, 7 marks
- Answer checked by Claude Sonnet 5.5 on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/electric_machines_ii/salient-motor-200kw-load-angle
- Whole chapter: https://www.bctengineer.com/md/semester_5/electric_machines_ii/chapter-2

## Question

A 3-phase, 11 kV, 50 Hz, 10 pole, 200 kW star connected salient pole synchronous motor has Xd = 1.2 p.u. and Xq = 0.8 p.u. It operates at 0.98 power factor leading. Determine the internal emf and load angle.

## Answer

**Assumptions:** the motor operates at rated voltage and rated current ($V = 1$ pu, $I_a = 1$ pu) with losses and $R_a$ neglected; bases are the machine ratings (11 kV line, 6.351 kV phase).

$\cos\phi = 0.98$ leading $\Rightarrow \phi = 11.48^\circ$, $\vec I_a = 1\angle 11.48^\circ = 0.98 + j0.199$ pu.

**Method (two-reaction theory, motor, $R_a = 0$):** $\vec V = \vec E_f + jX_d\vec I_d + jX_q\vec I_q$.

1. $\vec E' = \vec V - jX_q\vec I_a$ gives the direction of $E_f$ (the q-axis), so its angle is $-\delta$.
2. Angle between $I_a$ and the q-axis: $\psi = \delta - \phi$ ($\phi$ positive for lagging, negative for leading); $I_d = I_a\sin\psi$, $I_q = I_a\cos\psi$.
3. $E_f = E' + (X_d - X_q)I_d$.

#### Load angle

$$
\begin{aligned}
\vec E' &= 1 - j0.8(0.98 + j0.199) = 1 + 0.159 - j0.784 \\
&= 1.1592 - j0.784 = 1.3994\angle -34.07^\circ\ \text{pu}
\end{aligned}
$$

So the load angle is $\delta = 34.07^\circ$.

#### Internal emf

$$
\begin{aligned}
\psi &= \delta + 11.48^\circ = 45.55^\circ \ (\text{leading current}) \\
I_d &= 1 \times \sin45.55^\circ = 0.7139\ \text{pu},\quad I_q = \cos45.55^\circ = 0.7003\ \text{pu} \\
E_f &= 1.3994 + (1.2 - 0.8)(0.7139) = 1.3994 + 0.2856 = 1.685\ \text{pu}
\end{aligned}
$$

In volts: $E_f = 1.685 \times 6.351 = 10.70$ kV per phase $= 18.53$ kV line.

```text
 V = 1.0 ---------------------->
   \                  Ia (leads V by 11.5 deg)
    \ delta = 34.07
     \
      '--> Ef = 1.685 pu  (Ef > V: over-excited)
```

Check: $P = \dfrac{E_fV}{X_d}\sin\delta + \dfrac{V^2}{2}\left(\dfrac{1}{X_q}-\dfrac{1}{X_d}\right)\sin2\delta = 0.787 + 0.193 = 0.98$ pu $= V I\cos\phi$.

**Answer:** Internal emf $E_f = 1.685$ pu (10.70 kV/phase, 18.53 kV line); load angle $\delta = 34.07^\circ$.

## Check yourself

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

1. For 0.98 leading pf, how is the current phasor written with V as reference?
   A. 1 at -11.48 degrees
   B. 1 at +11.48 degrees = 0.98 + j0.199
   C. 1 at +90 degrees
   D. 0.98 at 0 degrees
   Answer: B. A leading current has a positive angle: cos(phi) = 0.98 gives phi = 11.48 degrees. (Key point: Current phasor)

2. What is the load angle and how is it found?
   A. 34.07 degrees, angle of E' = V - jXq Ia
   B. 11.48 degrees, from the pf
   C. 45.55 degrees, from psi
   D. 90 degrees
   Answer: A. E' = 1.1592 - j0.784 = 1.3994 at -34.07 degrees, so delta = 34.07 degrees. (Key point: Load angle)

3. What is Ef in pu?
   A. 1.685 pu
   B. 1.3994 pu
   C. 0.7139 pu
   D. 1.2 pu
   Answer: A. Ef = E' + (Xd - Xq) Id = 1.3994 + 0.4 x 0.7139 = 1.685 pu. (Key point: Internal emf)

4. Ef = 1.685 pu with leading pf indicates:
   A. Under-excitation
   B. Over-excitation (Ef > V)
   C. Zero excitation
   D. Unity pf operation
   Answer: B. Ef above V corresponds to an over-excited motor drawing leading current. (Key point: Excitation state)

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