---
title: "Shunt motor loaded speed at 50 A"
subject: "Electrical Machines"
chapter: 4
key: "semester_4/electrical_machines:dcm-shunt-1-25kw-loaded-speed"
source: https://www.bctengineer.com/q/semester_4/electrical_machines/dcm-shunt-1-25kw-loaded-speed
site: BCT Engineering
---

# Shunt motor loaded speed at 50 A

Electrical Machines · Chapter 4: DC Motor

- Asked in: 2076 Baisakh, 8 marks
- Video: [DC Shunt Motor](https://www.youtube.com/watch?v=T1yt3ZUHTpU) — Lee Thian Pau
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/electrical_machines/dcm-shunt-1-25kw-loaded-speed
- Whole chapter: https://www.bctengineer.com/md/semester_4/electrical_machines/chapter-4

## Question

A 1.25 kW, 250 V dc shunt motor on no load runs at 1000 rpm. The armature and field circuit resistance are $0.2\ \Omega$ and $250\ \Omega$ respectively. Calculate the speed of motor when it is loaded and draws current of 50 A.

## Short answer

The no-load input of 1.25 kW gives a line current of 5 A, armature current of 4 A, and back emf 249.2 V at 1000 rpm. When loaded with 50 A line current, armature current is 49 A, back emf drops to 240.2 V, and speed becomes 1000 × (240.2/249.2) ≈ 963.9 rpm.

## Answer

The flux is constant for a shunt motor, so $N\propto E_b$.

The 1.25 kW figure is the motor's input on no load (a 1.25 kW rated motor could not draw 50 A on load), so the no-load line current is $I_0 = 1250/250 = 5$ A.

**Field current:** $I_{sh} = \dfrac{250}{250} = 1$ A.

**No load (1000 rpm):**

$$
I_{a0} = 5 - 1 = 4\ \text{A}, \qquad E_{b0} = V - I_{a0}R_a = 250 - 4(0.2) = 249.2\ \text{V}
$$

**Loaded (line current 50 A):**

$$
I_a = 50 - 1 = 49\ \text{A}
$$

$$
E_b = V - I_aR_a = 250 - 49(0.2) = 240.2\ \text{V}
$$

$$
N = 1000\times\frac{E_b}{E_{b0}} = 1000\times\frac{240.2}{249.2} = 963.9\ \text{rpm}
$$

**Answer:** loaded speed $\approx 963.9\ \text{rpm}$. (If the no-load armature drop is neglected, taking $E_{b0}\approx 250$ V, the result is $960.8$ rpm.)

## Check yourself

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

1. At no load, Eb0 is taken as:
   A. 0
   B. Approximately V = 250 V, since no-load Ia drop is negligible
   C. 200 V
   D. 240.2 V
   Answer: B. The tiny no-load armature current gives negligible drop. (Key point: No-load back emf)

2. With line current 50 A and Ish = 1 A, Ia is:
   A. 50 A
   B. 51 A
   C. 1 A
   D. 49 A
   Answer: D. Ia = 50 - 1. (Key point: Armature current)

3. Eb when loaded is:
   A. 240.2 V
   B. 250 V
   C. 245 V
   D. 239 V
   Answer: A. 250 - 49 x 0.2 = 240.2 V. (Key point: Loaded Eb)

4. Loaded speed is about:
   A. 1040 rpm
   B. 1000 rpm
   C. 960.8 rpm
   D. 900 rpm
   Answer: C. 1000 x 240.2/250 = 960.8 rpm. (Key point: Speed)

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