---
title: "New current and pf when back emf rises 25%"
subject: "Electric Machines II"
chapter: 2
key: "semester_5/electric_machines_ii:motor-back-emf-increase-25"
source: https://www.bctengineer.com/q/semester_5/electric_machines_ii/motor-back-emf-increase-25
site: BCT Engineering
---

# New current and pf when back emf rises 25%

Electric Machines II · Chapter 2: Three Phase Synchronous Motor

- Asked in: 2073 Shrawan, 7 marks
- Answer checked by Claude Sonnet 5.5 on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/electric_machines_ii/motor-back-emf-increase-25
- Whole chapter: https://www.bctengineer.com/md/semester_5/electric_machines_ii/chapter-2

## Question

A 660 V, 3-phase, star-connected synchronous motor draws 50 kW at power factor of 0.8 lagging. Find the new current and power factor when the back e.m.f increases by 25%. The machine has synchronous reactance of 3 Ω and effective resistance is negligible.

## Short answer

Original Ef = 311.61 V per phase; increased by 25% to 389.52 V. Power unchanged, so using P = 3VEf sinδ/Xs gives δ' = 19.69°. New armature current = 44.00 A, power factor = 0.994 lagging.

## Answer

**Assumptions:** star connection; the shaft load (input power 50 kW) stays the same when the excitation is raised; $R_a = 0$.

#### Original operating point

$$
\begin{aligned}
V &= \frac{660}{\sqrt3} = 381.05\ \text{V},\quad I_a = \frac{50000}{\sqrt3 \times 660 \times 0.8} = 54.67\ \text{A} \\
\vec I_a &= 54.67\angle -36.87^\circ = 43.74 - j32.80\ \text{A} \\
jX_s\vec I_a &= j3(43.74 - j32.80) = 98.41 + j131.22 \\
\vec E_f &= \vec V - jX_s\vec I_a = 282.64 - j131.22 = 311.61\angle -24.90^\circ\ \text{V}
\end{aligned}
$$

#### New back emf (+25 %)

$$
E_f' = 1.25 \times 311.61 = 389.52\ \text{V per phase}
$$

Power is unchanged, so $P = \dfrac{3VE_f'}{X_s}\sin\delta'$:

$$
\begin{aligned}
\sin\delta' &= \frac{P X_s}{3VE_f'} = \frac{50000 \times 3}{3 \times 381.05 \times 389.52} = 0.3369 \\
\delta' &= 19.69^\circ
\end{aligned}
$$

#### New current and power factor

$$
\begin{aligned}
\vec E_f' &= 389.52\angle -19.69^\circ = 366.75 - j131.22 \\
\vec E_r &= \vec V - \vec E_f' = 14.30 + j131.22 \\
\vec I_a' &= \frac{\vec E_r}{jX_s} = \frac{14.30 + j131.22}{j3} = 43.74 - j4.77 \\
&= 44.00\angle -6.22^\circ\ \text{A}
\end{aligned}
$$

Power factor $= \cos6.22^\circ = 0.9941$ lagging.

Check: $\sqrt3 \times 660 \times 44.00 \times 0.9941 = 50.0$ kW.

Raising the excitation reduced the reactive (lagging) part of the current, so the current fell and the pf moved close to unity.

**Answer:** New current $= 44.00$ A at a power factor of **0.994 lagging** (originally 54.67 A at 0.8 lagging).

## Check yourself

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

1. How is the original Ef found from 50 kW at 0.8 lagging?
   A. Ef = V - jXs Ia
   B. Ef = V + jXs Ia
   C. Ef = V x 0.8
   D. Ef = Xs Ia
   Answer: A. For a motor, V = Ef + jXs Ia, so Ef = V - jXs Ia = 311.6 V at -24.9 degrees. (Key point: Original Ef)

2. With the load unchanged and Ef raised by 25 percent, how is the new load angle found?
   A. The angle stays the same
   B. From P = 3 V Ef' sin(delta')/Xs, with P constant, giving delta' = 19.69 degrees
   C. From Ia' = Ia
   D. From delta' = 1.25 x delta
   Answer: B. Constant power means Ef sin(delta) is constant, so a larger Ef gives a smaller delta. (Key point: New load angle)

3. What are the new current and power factor?
   A. 54.67 A at 0.8 lagging
   B. 44.0 A at 0.994 lagging
   C. 44.0 A at 0.994 leading
   D. 65 A at 0.9 leading
   Answer: B. Ia' = 44.0 A at -6.22 degrees, so the pf is 0.994 lagging; extra excitation cut the lagging reactive current. (Key point: Result)

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