---
title: "Series motor speed at half torque"
subject: "Electrical Machines"
chapter: 4
key: "semester_4/electrical_machines:dcm-series-220v-800rpm-half-torque"
source: https://www.bctengineer.com/q/semester_4/electrical_machines/dcm-series-220v-800rpm-half-torque
site: BCT Engineering
---

# Series motor speed at half torque

Electrical Machines · Chapter 4: DC Motor

- Asked in: 2065 Chaitra (old course), 8 marks
- Video: [(B)EM(5th ed) || End Chapter  Q 8.13  || Series DC Motors || Speed || Efficiency || Torque](https://www.youtube.com/watch?v=QgQw4coRH0k) — Electrical Engineering Academy
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/electrical_machines/dcm-series-220v-800rpm-half-torque
- Whole chapter: https://www.bctengineer.com/md/semester_4/electrical_machines/chapter-4

## Question

A 220 V series motor is running at a speed of 800 rpm and draws 100 A. Calculate at what speed the motor will run developing half the torque. Total resistance of the armature and field is $0.1\ \Omega$. Assume that the magnetic circuit is unsaturated.

## Short answer

For an unsaturated series motor, torque is proportional to the square of armature current, so halving torque reduces current to 70.71 A. The new back emf is 212.93 V, and speed becomes 800 × (212.93/210) × (100/70.71) ≈ 1147 rpm.

## Answer

For an unsaturated series motor $\Phi\propto I$, so the torque $T\propto\Phi I\propto I^2$, and the speed $N\propto E_b/\Phi\propto E_b/I$.

**Case 1:** 800 rpm, $I_1 = 100$ A, $R = 0.1\ \Omega$.

$$
E_{b1} = 220 - 100(0.1) = 210\ \text{V}
$$

**Case 2:** torque $T_2 = T_1/2$.

$$
\frac{T_2}{T_1} = \left(\frac{I_2}{I_1}\right)^2 = \frac12 \;\Rightarrow\; I_2 = \frac{100}{\sqrt{2}} = 70.71\ \text{A}
$$

$$
E_{b2} = 220 - I_2R = 220 - (70.71)(0.1) = 212.93\ \text{V}
$$

$$
\frac{N_2}{N_1} = \frac{E_{b2}}{E_{b1}}\times\frac{I_1}{I_2} \;\Rightarrow\; N_2 = 800\times\frac{212.93}{210}\times\frac{100}{70.71} = 1147\ \text{rpm}
$$

**Answer:** the motor runs at about $1147\ \text{rpm}$ when developing half the torque.

## Check yourself

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

1. For an unsaturated series motor, torque is proportional to:
   A. I
   B. I^2
   C. 1/I
   D. N
   Answer: B. Phi proportional to I, so T proportional to Phi I proportional to I^2. (Key point: Torque relation)

2. For half torque, I2 is:
   A. 50 A
   B. 75 A
   C. 70.71 A
   D. 25 A
   Answer: C. (I2/I1)^2 = 1/2 gives I2 = 100/sqrt(2). (Key point: New current)

3. The speed ratio uses:
   A. N2/N1 = Eb2/Eb1
   B. N2/N1 = (Eb2/Eb1)(I1/I2)
   C. N2/N1 = I2/I1
   D. N2/N1 = V2/V1
   Answer: B. N is proportional to Eb/Phi proportional to Eb/I. (Key point: Speed relation)

4. The speed at half torque is about:
   A. 566 rpm
   B. 800 rpm
   C. 1147 rpm
   D. 1600 rpm
   Answer: C. 800 x (212.93/210) x (100/70.71) = 1147 rpm. (Key point: Final speed)

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