---
title: "3.3 kV synchronous motor back emf"
subject: "Electrical Machines"
chapter: 6
key: "semester_4/electrical_machines:sync-motor-3-3kv-back-emf"
source: https://www.bctengineer.com/q/semester_4/electrical_machines/sync-motor-3-3kv-back-emf
site: BCT Engineering
---

# 3.3 kV synchronous motor back emf

Electrical Machines · Chapter 6: Three Phase Synchronous Machines

- Asked in: 2078 Baisakh, 8 marks; 2071 Bhadra, 8 marks (asked 2 times)
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/electrical_machines/sync-motor-3-3kv-back-emf
- Whole chapter: https://www.bctengineer.com/md/semester_4/electrical_machines/chapter-6

## Question

A 3.3 kV, 3-phase star connected synchronous motor has impedance of $0.2 + j2.2\ \Omega$/phase of the armature winding. The motor is operated at 0.5 pf leading with line current of 100 A. Determine the back emf per phase and also draw phasor diagram.

## Short answer

Phase voltage 1905.3 V, line cur rent 100 A at 0.5 p f lead ing. Back emf per phase = Vph – I Zs = 2089.7 V, lagging the terminal voltage by 3.49°. The mot or is over-excited (E b > V).

## Answer

**Given:** 3.3 kV, 3-phase star motor, $Z_s = 0.2 + j2.2\ \Omega$ per phase, 0.5 pf leading, $I_L = 100$ A.

$$
V_{ph} = \frac{3300}{\sqrt3} = 1905.26\ \text{V}, \qquad I = 100\ \text{A}
$$
$\cos\phi = 0.5$ leading, so $\phi = 60^\circ$ and $\vec I = 100\angle +60^\circ$ A with $V_{ph}$ as reference.

For a motor, the back emf is
$$
\vec E_b = \vec V - \vec I Z_s
$$
Impedance drop:
$$
\vec I Z_s = 100\angle60^\circ\times(0.2 + j2.2) = 100\angle 60^\circ\times 2.209\angle 84.8^\circ = 220.9\angle144.8^\circ = -180.53 + j127.32\ \text{V}
$$
$$
\vec E_b = 1905.26 - (-180.53 + j127.32) = 2085.79 - j127.32\ \text{V}
$$
$$
E_b = \sqrt{2085.79^2 + 127.32^2} = 2089.66\ \text{V}, \qquad \delta = \tan^{-1}\!\left(\frac{-127.32}{2085.79}\right) = -3.49^\circ
$$
The line value is $\sqrt3\times 2089.66 = 3619.4$ V.

#### Phasor diagram
```text
         / I (leads V by 60 deg)
       /
     /
   /
 O------------------------> V
  `-._                     \
      `-._    Eb            \ I Zs = 220.9 V at 144.8 deg
          `-._               \ (from tip of Eb to tip of V)
              `-._            \
                  `-._         \
                      `-._      \
                          `-._   \
                              `-._*  tip of Eb
```
(Not to scale. $\vec V = \vec E_b + \vec I Z_s$; $E_b$ lags $V$ by $3.49^\circ$ and $I$ leads $V$ by $60^\circ$. The motor is over-excited because $E_b > V$.)

**Answer:** back emf per phase $= 2089.7$ V (line 3619 V), load angle $3.49^\circ$ ($E_b$ lagging $V$).

## Check yourself

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

1. Which equation gives the back emf of a motor?
   A. Eb = V + I Zs
   B. Eb = V - I Zs
   C. Eb = I Zs - V
   D. Eb = V/I Zs
   Answer: B. For a motor the supply voltage equals back emf plus drop, so Eb = V - I Zs. (Key point: Motor back emf equation)

2. What is the phase voltage of the 3.3 kV star motor?
   A. 3300 V
   B. 1905.26 V
   C. 5716 V
   D. 1100 V
   Answer: B. V_ph = 3300/sqrt3 = 1905.26 V. (Key point: Phase voltage)

3. At 0.5 pf leading, what is the current phasor with V as reference?
   A. 100 angle -60 deg A
   B. 100 angle +60 deg A
   C. 100 angle +30 deg A
   D. 100 angle 0 A
   Answer: B. cos(phi) = 0.5 gives 60 deg; leading means the angle is positive. (Key point: Current angle)

4. Why is the motor said to be over-excited?
   A. Eb is larger than V and the current leads
   B. Current lags V
   C. Eb is zero
   D. Load angle is 90 deg
   Answer: A. Eb (2089.7 V) exceeds V (1905.3 V) and the current leads V. (Key point: Over-excitation)

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