---
title: "Back emf and new current of 5 kVA motor"
subject: "Electric Machines II"
chapter: 2
key: "semester_5/electric_machines_ii:motor-5kva-208v-power-angle"
source: https://www.bctengineer.com/q/semester_5/electric_machines_ii/motor-5kva-208v-power-angle
site: BCT Engineering
---

# Back emf and new current of 5 kVA motor

Electric Machines II · Chapter 2: Three Phase Synchronous Motor

- Asked in: 2080 Bhadra, 8 marks; 2072 Chaitra, 8 marks (asked 2 times)
- Answer checked by Claude Sonnet 5.5 on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/electric_machines_ii/motor-5kva-208v-power-angle
- Whole chapter: https://www.bctengineer.com/md/semester_5/electric_machines_ii/chapter-2

## Question

A 3-phase, 5 kVA, 208 V, 4-pole, 50 Hz star connected synchronous motor has negligible armature winding resistance and synchronous reactance of 8 ohm per phase. It is operated from the 3-phase, 208 V, 50 Hz power supply and field excitation is adjusted so that the power factor is unity and the motor draws a power of 3 kW from the supply.
(i) Find the back emf (or excitation) voltage and power angle.
(ii) Keeping the excitation voltage constant, the power angle is increased by 20% due to increase in load on the shaft. Calculate the new armature current and power factor.

## Short answer

With unity pf, per-phase V=120.09 V, I=8.327 A; back emf Eb=137.33 V/phase (237.9 V line), power angle δ=29.02°. Keeping excitation constant and increasing δ by 20% to 34.82° gives I=9.85 A and pf=0.996 lagging.

## Answer

**Given:** star, 5 kVA, 208 V, 50 Hz, $X_s = 8\ \Omega$, $R_a \approx 0$; unity pf, input 3 kW.

#### (i) Back emf and power angle
$$
\begin{aligned}
V &= \frac{208}{\sqrt3} = 120.09\ \text{V} \\
I &= \frac{3000}{\sqrt3\times208\times1} = 8.327\ \text{A}\quad(\text{in phase with } V) \\
\bar E_b &= \bar V - jX_s\bar I = 120.09 - j(8\times8.327) = 120.09 - j66.62 \\
E_b &= \sqrt{120.09^2 + 66.62^2} = 137.33\ \text{V per phase} \\
\delta &= \tan^{-1}\frac{66.62}{120.09} = 29.02^\circ\ (E_b \text{ lags } V)
\end{aligned}
$$
Line value of back emf $= \sqrt3\times137.33 = 237.9$ V.

```text
   O-------------------> V = 120.09
   O--------> I (upf)  |
     `-.               | jIXs = 66.62
        `-.            |
           `-. Eb = 137.33, delta = 29.02 deg
```

#### (ii) Power angle increased by 20%, excitation constant
$$
\begin{aligned}
\delta' &= 1.2\times29.02^\circ = 34.82^\circ \\
\bar E_b' &= 137.33\angle -34.82^\circ = 112.74 - j78.42 \\
\bar I' &= \frac{\bar V - \bar E_b'}{jX_s} = \frac{(120.09 - 112.74) + j78.42}{j8} = \frac{7.35 + j78.42}{j8} \\
&= 9.802 - j0.919\ \text{A} \\
I' &= 9.845\ \text{A},\quad \angle -5.36^\circ \\
\text{pf} &= \cos 5.36^\circ = 0.996\ \text{lagging}
\end{aligned}
$$
New input power $= 3\times120.09\times9.802 = 3531$ W.

**Answer: (i) E_b = 137.33 V/phase (237.9 V line), δ = 29.02°; (ii) I = 9.85 A, pf = 0.996 lagging (input ≈ 3.53 kW).**

## Check yourself

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

1. For the 208 V motor at unity pf drawing 3 kW, the armature current is found from:
   A. I = 3000/(sqrt3 x 208) = 8.33 A
   B. I = 3000/208 = 14.4 A
   C. I = 5000/(sqrt3 x 208) = 13.9 A
   D. I = 3000/(3 x 208) = 4.8 A
   Answer: A. P = sqrt3 VL I cos(phi) with cos(phi) = 1 gives I = 8.327 A. (Key point: Armature current)

2. Which equation gives the back emf of a motor with Ra neglected?
   A. Eb = V + jXs I
   B. Eb = V - jXs I
   C. Eb = jXs I
   D. Eb = V x I
   Answer: B. For a motor V = Eb + jXs I, so Eb = V - jXs I. (Key point: Motor emf equation)

3. What are Eb per phase and the power angle in part (i)?
   A. 120.09 V and 0 degrees
   B. 137.33 V and 29.02 degrees
   C. 66.62 V and 29.02 degrees
   D. 237.9 V and 29.02 degrees per phase
   Answer: B. Eb = 120.09 - j66.62 gives 137.33 V per phase and delta = atan(66.62/120.09) = 29.02 degrees; 237.9 V is the line value. (Key point: Result of (i))

4. In part (ii), with Eb constant and delta raised 20% to 34.82 degrees, how is the new current found?
   A. Multiply the old current by 1.2
   B. I' = (V - Eb')/(jXs) using Eb' = 137.33 at -34.82 degrees
   C. I' = Eb'/Xs
   D. Using Eb' = 120.09 at -34.82 degrees
   Answer: B. Eb magnitude stays 137.33 V but its angle changes; the current follows from the phasor difference, giving 9.85 A at 0.996 lagging. (Key point: New current method)

## Close variants set in other exams

- [Back emf and new current of 9 kVA motor](https://www.bctengineer.com/question.md?key=semester_5%2Felectric_machines_ii%3Amotor-9kva-320v-load-angle) — Electric Machines II, chapter 2, asked 3×

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