---
title: "Series motor speed at reduced load"
subject: "Electrical Machine"
chapter: 4
key: "semester_4/electrical_machine:series-motor-38a-600rpm-speed"
source: https://www.bctengineer.com/q/semester_4/electrical_machine/series-motor-38a-600rpm-speed
site: BCT Engineering
---

# Series motor speed at reduced load

Electrical Machine · Chapter 4: DC Motor

- Asked in: 2072 Magh
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/electrical_machine/series-motor-38a-600rpm-speed
- Whole chapter: https://www.bctengineer.com/md/semester_4/electrical_machine/chapter-4

## Question

A 200 V DC series motor draws full-load line current of 38 A at the rated speed of 600 rpm. The motor has armature resistance of 0.4 Ω and the series field resistance is 0.2 Ω find:
i) The speed of the motor when the load current drops to 19 A
ii) The speed on removal of load when the motor takes only 1 A from supply

## Short answer

At 19 A, speed is about 1277 rpm. At 1 A, theoretical speed is about 25,660 rpm, a dangerously high runaway speed.

## Answer

For a series motor, Φ ∝ I (assumed unsaturated), and N ∝ E_b / Φ ∝ E_b / I.

#### Equivalent circuit
```text
 +--[Ra=0.4]--[Rse=0.2]--(Eb)--+
 |            I                  |
 +------------ V = 200 V --------+
```
Total resistance R = 0.4 + 0.2 = 0.6 Ω.

#### Full load (38 A, 600 rpm)
E_b1 = 200 − 38 × 0.6 = 200 − 22.8 = 177.2 V

#### i) Load current = 19 A
E_b2 = 200 − 19 × 0.6 = 188.6 V

N₂ = N₁ × (E_b2/E_b1) × (I₁/I₂) = 600 × (188.6/177.2) × (38/19)

= 600 × 1.0643 × 2 = **1277 rpm**

#### ii) Current = 1 A (no load)
E_b3 = 200 − 1 × 0.6 = 199.4 V

N₃ = 600 × (199.4/177.2) × (38/1) = 600 × 1.1253 × 38 = **≈ 25 660 rpm**

| Current (A) | E_b (V) | Speed (rpm) |
|---|---|---|
| 38 | 177.2 | 600 |
| 19 | 188.6 | 1277 |
| 1 | 199.4 | about 25 660 |

The second result is theoretical: with flux ∝ I, the speed becomes extremely high at light load. In practice, residual flux, saturation, friction and windage limit it, but the speed is still dangerously high. This is why a series motor must never run on no load.

**Answer: (i) about 1277 rpm; (ii) about 25 660 rpm (a runaway speed).**

## Check yourself

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

1. With Φ ∝ I, how is speed expressed?
   A. N ∝ E_b/I
   B. N ∝ E_b × I
   C. N ∝ I
   D. N ∝ 1/E_b
   Answer: A. N ∝ E_b/Φ and Φ ∝ I. (Key point: Series motor speed relation)

2. What is E_b at 38 A?
   A. 22.8 V
   B. 177.2 V
   C. 200 V
   D. 188.6 V
   Answer: B. E_b1 = 200 - 38 × 0.6 = 177.2 V. (Key point: Full-load back emf)

3. What is the speed at 19 A?
   A. About 639 rpm
   B. About 300 rpm
   C. About 1277 rpm
   D. About 600 rpm
   Answer: C. N2 = 600 × (188.6/177.2) × 2 = 1277 rpm. (Key point: Speed at 19 A)

4. Why is the 25 660 rpm result only theoretical?
   A. Because the motor stops
   B. Because R is 0
   C. Because E_b is zero
   D. Flux is not really proportional to I at light load; residual flux, saturation and friction limit it, but speed is still dangerously high
   Answer: D. The proportional-flux assumption fails at tiny currents, yet runaway speed is the lesson. (Key point: No-load result)

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