---
title: "Excitation emf of motor drawing 160 A"
subject: "Electric Machines II"
chapter: 2
key: "semester_5/electric_machines_ii:motor-excitation-emf-160a"
source: https://www.bctengineer.com/q/semester_5/electric_machines_ii/motor-excitation-emf-160a
site: BCT Engineering
---

# Excitation emf of motor drawing 160 A

Electric Machines II · Chapter 2: Three Phase Synchronous Motor

- Asked in: 2069 Chaitra, 4 marks
- Answer checked by Claude Sonnet 5.5 on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/electric_machines_ii/motor-excitation-emf-160a
- Whole chapter: https://www.bctengineer.com/md/semester_5/electric_machines_ii/chapter-2

## Question

The full load current of 3.3 kVA, star-connected synchronous motor is 160 A at 0.8 pf lagging. The resistance and synchronous reactance of the motor are 0.8 Ω and 5.5 Ω per phase respectively. Calculate the excitation emf. Assume mechanical stray load loss to be 30 kW.

## Short answer

Using motor equation: V_ph = 1905.26 V, Ia = 160∠-36.87° A. Voltage drop Ia Zs = 630.4 + j627.2 V. Ef = V - Ia Zs = 1420.8 V per phase (2.46 kV line) lagging V by 26.20°.

## Answer

**Reading of data:** "3.3 kVA" is a misprint for **3.3 kV** (a 3.3 kVA motor cannot take 160 A). Star connection. The stray loss does not affect the excitation emf, since $E_f$ depends only on $V$, $I_a$ and $Z_s$.

$$
V = \frac{3300}{\sqrt3} = 1905.26\ \text{V per phase}
$$

Motor equation: $\vec E_f = \vec V - \vec I_a(R_a + jX_s)$, with $V$ as reference.

$$
\begin{aligned}
\vec I_a &= 160\angle -36.87^\circ = 128 - j96\ \text{A} \\
\vec I_aZ_s &= (128 - j96)(0.8 + j5.5) \\
&= 102.4 + 528 + j(704 - 76.8) = 630.4 + j627.2\ \text{V} \\
\vec E_f &= 1905.26 - (630.4 + j627.2) = 1274.86 - j627.2 \\
&= 1420.8\angle -26.20^\circ\ \text{V}
\end{aligned}
$$

Line value $= \sqrt3 \times 1420.8 = 2461$ V.

Since $E_f < V$ the motor is under-excited, which agrees with its lagging power factor.

(For reference: input $= \sqrt3 \times 3300 \times 160 \times 0.8 = 731.6$ kW, copper loss $= 3 \times 160^2 \times 0.8 = 61.44$ kW; the 30 kW stray loss would be used only for output or efficiency.)

**Answer:** Excitation emf $E_f = 1421$ V per phase (2.46 kV line), lagging $V$ by $26.20^\circ$.

## Check yourself

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

1. What is the interpretation of the rating '3.3 kVA'?
   A. It is correct as printed
   B. It is a misprint for 3.3 kV, since a 3.3 kVA motor cannot take 160 A
   C. It means 3.3 kW
   D. It means 3.3 kV delta
   Answer: B. A 3.3 kVA machine cannot draw 160 A, so the voltage is 3.3 kV. (Key point: Data correction)

2. Which equation gives the excitation emf of the motor?
   A. Ef = V + Ia(Ra + jXs)
   B. Ef = V - Ia(Ra + jXs)
   C. Ef = V - Ia Xs only
   D. Ef = Ia Zs
   Answer: B. For a motor the applied V = Ef + Ia Zs, so Ef = V - Ia Zs. (Key point: Motor equation)

3. Does the 30 kW stray loss affect Ef, and what is the result?
   A. Yes; Ef = 1275 V
   B. No; Ef depends only on V, Ia and Zs and equals 1421 V per phase, lagging V by 26.2 degrees
   C. Yes; Ef = 1905 V
   D. No; Ef = 2461 V per phase
   Answer: B. Ef = 1905.26 - (630.4 + j627.2) = 1421 V per phase; 2461 V is the line value. (Key point: Stray loss and result)

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