---
title: "Back emf and new current of 9 kVA motor"
subject: "Electric Machines II"
chapter: 2
key: "semester_5/electric_machines_ii:motor-9kva-320v-load-angle"
source: https://www.bctengineer.com/q/semester_5/electric_machines_ii/motor-9kva-320v-load-angle
site: BCT Engineering
---

# Back emf and new current of 9 kVA motor

Electric Machines II · Chapter 2: Three Phase Synchronous Motor

- Asked in: 2079 Bhadra, 6 marks
- Answer checked by Claude Sonnet 5.5 on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/electric_machines_ii/motor-9kva-320v-load-angle
- Whole chapter: https://www.bctengineer.com/md/semester_5/electric_machines_ii/chapter-2

## Question

A 3-phase 9 kVA, 320 V, 4-pole, 50 Hz synchronous motor has negligible armature winding resistance and synchronous reactance of 7.2 ohm per phase is operated from a 3-phase 320 V, 50 Hz power supply and field supply is adjusted so that power factor is unity and motor draws power of 4.6 kW from the supply.
(i) Find back emf and power angle.
(ii) Keeping excitation voltage constant, load on the shaft of motor increased so that load angle increases by 25%. Calculate new value of current drawn by the motor and power factor.

## Short answer

Per-phase V=184.75 V, unity-pf Ia=8.299 A. Eb = V – jXs Ia = 194.18 V/phase (336.3 V line), δ=17.92°. With excitation constant and δ increased 25% to 22.40°, new current I'= (V–Eb∠-22.4°)/jXs = 10.30 A, pf=0.9975 lagging.

## Answer

**Assumption:** star connection (not stated); $R_a$ neglected.

**Given:** $V_L = 320$ V, $X_s = 7.2\ \Omega$, $P = 4.6$ kW at unity pf.

#### (i) Back emf and power angle

$$
\begin{aligned}
V &= \frac{320}{\sqrt3} = 184.75\ \text{V per phase} \\
I_a &= \frac{4600}{\sqrt3 \times 320 \times 1} = 8.299\ \text{A} \\
I_aX_s &= 8.299 \times 7.2 = 59.76\ \text{V}
\end{aligned}
$$

Motor: $\vec E_f = \vec V - j X_s \vec I_a$, with $I_a$ in phase with $V$:

$$
\begin{aligned}
\vec E_f &= 184.75 - j59.76 \\
E_f &= \sqrt{184.75^2 + 59.76^2} = 194.18\ \text{V per phase} \\
\delta &= \tan^{-1}\frac{59.76}{184.75} = 17.92^\circ
\end{aligned}
$$

Line value of $E_f = \sqrt3 \times 194.18 = 336.3$ V.

```text
  V = 184.75 -------------->
   \ delta = 17.92 deg      |
    \                       | jXs Ia = 59.76
     \--------------------->|
          Ef = 194.18
  Ia in phase with V
```

#### (ii) Load angle increased by 25 %

$$
\delta' = 1.25 \times 17.92^\circ = 22.40^\circ, \qquad \vec E_f = 194.18\angle -22.40^\circ = 179.52 - j74.01\ \text{V}
$$

$$
\begin{aligned}
\vec I_a' &= \frac{\vec V - \vec E_f}{jX_s} = \frac{(184.75 - 179.52) + j74.01}{j7.2} \\
&= \frac{5.23 + j74.01}{j7.2} = 10.279 - j0.727 \\
&= 10.30\angle -4.04^\circ\ \text{A}
\end{aligned}
$$

Power factor $= \cos 4.04^\circ = 0.9975$ lagging.

Check: new input $= 3 \times 184.75 \times 10.30 \times 0.9975 = 5.70$ kW, equal to $\dfrac{3VE_f}{X_s}\sin22.40^\circ$.

**Answer:** (i) $E_f = 194.2$ V/phase (336.3 V line), $\delta = 17.92^\circ$. (ii) $I_a = 10.30$ A at **0.9975 lagging** pf.

## Check yourself

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

1. What connection was assumed because it was not stated, and what is the per-phase voltage?
   A. Delta; 320 V
   B. Star; 184.75 V
   C. Star; 320 V
   D. Delta; 184.75 V
   Answer: B. Star was assumed, so V = 320/sqrt3 = 184.75 V. (Key point: Assumption)

2. At unity pf with Ia = 8.299 A and Xs = 7.2 ohm, what are Ef and delta?
   A. 194.18 V per phase and 17.92 degrees
   B. 184.75 V and 0 degrees
   C. 59.76 V and 17.92 degrees
   D. 336.3 V per phase and 17.92 degrees
   Answer: A. Ef = 184.75 - j59.76 gives magnitude 194.18 V and delta = atan(59.76/184.75) = 17.92 degrees; 336.3 V is the line value. (Key point: Back emf)

3. After the load angle increases by 25% with Ef constant, what is the new angle?
   A. 42.92 degrees
   B. 22.40 degrees
   C. 17.92 degrees x 25
   D. 30 degrees
   Answer: B. 1.25 x 17.92 = 22.40 degrees. (Key point: New load angle)

4. What is the resulting current and pf?
   A. 10.30 A at 0.9975 lagging
   B. 8.30 A at unity
   C. 10.30 A at 0.9975 leading
   D. 5.7 A at 0.5 lagging
   Answer: A. I = (V - Ef)/(jXs) = 10.279 - j0.727 A, i.e. 10.30 A at -4.04 degrees, which is lagging. (Key point: New current and pf)

## Close variants set in other exams

- [Back emf and new current of 5 kVA motor](https://www.bctengineer.com/question.md?key=semester_5%2Felectric_machines_ii%3Amotor-5kva-208v-power-angle) — Electric Machines II, chapter 2, asked 3×

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