---
title: "Speed of 220 V series motor at reduced load current"
subject: "Electrical Machines I"
chapter: 4
key: "semester_4/electrical_machines_i:series-motor-220v-38a-speed"
source: https://www.bctengineer.com/q/semester_4/electrical_machines_i/series-motor-220v-38a-speed
site: BCT Engineering
---

# Speed of 220 V series motor at reduced load current

Electrical Machines I · Chapter 4: DC Motor

- Asked in: 2081 Chaitra (new course), 3+3 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/electrical_machines_i/series-motor-220v-38a-speed
- Whole chapter: https://www.bctengineer.com/md/semester_4/electrical_machines_i/chapter-4

## Question

A 220 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. The brush voltage drop irrespective of load is 3.0 volts, find: a) The speed of the motor when the load current drops to 19 A. b) The speed on removal of load when the motor takes only 1 A from the supply.

## Short answer

Using N ∝ Eb/I and Eb=V−I(Ra+Rse)−3 V, with I1=38 A and Eb1=194.2 V: at 19 A, Eb2=205.6 V and N2≈1270 rpm; at 1 A, Eb3=216.4 V and N3≈25,400 rpm. This shows a series motor must never run without load.

## Answer

In a series motor flux is proportional to the series field (line) current, below saturation. So

$$
N \propto \frac{E_b}{\phi} \propto \frac{E_b}{I}, \qquad E_b = V - I(R_a + R_{se}) - V_{brush}
$$

**Given:** $V = 220$ V, $I_1 = 38$ A, $N_1 = 600$ rpm, $R_a + R_{se} = 0.4 + 0.2 = 0.6\ \Omega$, brush drop 3 V.

$$
E_{b1} = 220 - 38 \times 0.6 - 3 = 194.2\ \text{V}
$$

#### a) Speed at 19 A
$$
\begin{aligned}
E_{b2} &= 220 - 19 \times 0.6 - 3 = 205.6\ \text{V} \\
N_2 &= N_1 \times \frac{E_{b2}}{E_{b1}} \times \frac{I_1}{I_2} \\
&= 600 \times \frac{205.6}{194.2} \times \frac{38}{19} = 1270.4\ \text{rpm}
\end{aligned}
$$

#### b) Speed at 1 A (load removed)
$$
\begin{aligned}
E_{b3} &= 220 - 1 \times 0.6 - 3 = 216.4\ \text{V} \\
N_3 &= 600 \times \frac{216.4}{194.2} \times \frac{38}{1} = 25\,406\ \text{rpm}
\end{aligned}
$$

**Answer:** a) $N \approx 1270$ rpm; b) $N \approx 25\,400$ rpm.

This dangerously high value (about 42 times rated speed) shows why a **series motor must never be started or run without load**: as current falls, flux falls and speed rises without limit. (In practice residual flux and windage keep it a little lower, but it would still be destructive.)

## Check yourself

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

1. For a series motor below saturation, to which does speed relate?
   A. N ∝ Eb/I
   B. N ∝ Eb·I
   C. N ∝ Eb
   D. N ∝ I/Eb
   Answer: A. Flux ∝ I, so N ∝ Eb/φ ∝ Eb/I. (Key point: Series speed relation)

2. How is Eb computed here?
   A. Eb = V - I·Ra only
   B. Eb = V - I(Ra+Rse) - brush drop
   C. Eb = V + I(Ra+Rse) - brush drop
   D. Eb = V - brush drop
   Answer: B. Series field resistance and brush drop must be subtracted; Eb1 = 194.2 V. (Key point: Back emf)

3. What is the speed at 19 A?
   A. About 635 rpm
   B. About 600 rpm
   C. About 1270 rpm
   D. About 2540 rpm
   Answer: C. N2 = 600×(205.6/194.2)×(38/19) = 1270 rpm. (Key point: Speed at 19 A)

4. Why must a series motor never run without load?
   A. Current rises without limit
   B. Back emf becomes zero
   C. The field winding overheats
   D. Flux falls with current, so speed rises dangerously high
   Answer: D. At 1 A the speed comes to about 25,400 rpm. (Key point: No-load danger)

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