---
title: "Series resistance to halve series motor speed"
subject: "Electrical Machines I"
chapter: 4
key: "semester_4/electrical_machines_i:series-motor-600rpm-250v-resistance-300rpm"
source: https://www.bctengineer.com/q/semester_4/electrical_machines_i/series-motor-600rpm-250v-resistance-300rpm
site: BCT Engineering
---

# Series resistance to halve series motor speed

Electrical Machines I · Chapter 4: DC Motor

- Asked in: 2079 Chaitra, 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_i/series-motor-600rpm-250v-resistance-300rpm
- Whole chapter: https://www.bctengineer.com/md/semester_4/electrical_machines_i/chapter-4

## Question

A dc series motor running at 600 rpm on 250 V mains draws a current of 40 A. The total resistance of armature and series field of the machine is 0.20 ohm. Calculate the value of resistance connected in series with the machine which will reduce the speed to 300 rpm, assume the load torque being half of the previous value.

## Short answer

Torque is proportional to I², so halving torque gives I2=40/√2=28.28 A. With Eb1=242 V and N2/N1=0.5, Eb2=85.56 V. Required series resistance R=(250−85.56)/28.28−0.2≈5.61 Ω.

## Answer

For a series motor (unsaturated), $\phi \propto I$, so $T \propto I^2$ and $N \propto E_b / I$.

**Given:** $V = 250$ V, $N_1 = 600$ rpm, $I_1 = 40$ A, $R_a + R_{se} = 0.2\ \Omega$, $N_2 = 300$ rpm, $T_2 = 0.5\,T_1$.

#### Step 1: New current
$$
\frac{T_2}{T_1} = \left(\frac{I_2}{I_1}\right)^2 = 0.5 \;\Rightarrow\; I_2 = \frac{40}{\sqrt 2} = 28.28\ \text{A}
$$

#### Step 2: Back emfs
$$
E_{b1} = 250 - 40 \times 0.2 = 242\ \text{V}
$$

$$
\begin{aligned}
\frac{N_2}{N_1} &= \frac{E_{b2}}{E_{b1}} \times \frac{I_1}{I_2} \\
E_{b2} &= E_{b1} \times \frac{N_2}{N_1} \times \frac{I_2}{I_1} \\
&= 242 \times \frac{300}{600} \times \frac{28.28}{40} = 85.56\ \text{V}
\end{aligned}
$$

#### Step 3: Series resistance
$$
\begin{aligned}
E_{b2} &= V - I_2 (0.2 + R) \\
0.2 + R &= \frac{250 - 85.56}{28.28} = 5.814\ \Omega \\
R &= 5.814 - 0.2 = 5.61\ \Omega
\end{aligned}
$$

**Answer:** Extra series resistance $R \approx 5.61\ \Omega$ (current falls to 28.28 A).

## Check yourself

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

1. Load torque is halved on a series motor. How does current change?
   A. I2 = I1/sqrt(2)
   B. I2 = I1/2
   C. I2 = I1
   D. I2 = I1/4
   Answer: A. T ∝ I^2 so I2 = 40/1.414 = 28.28 A. (Key point: Torque and current)

2. Which relation gives Eb2?
   A. Eb2 = Eb1·(N2/N1)·(I1/I2)
   B. Eb2 = Eb1·(N2/N1)·(I2/I1)
   C. Eb2 = Eb1·(N1/N2)
   D. Eb2 = V - I2·Ra only
   Answer: B. N ∝ Eb/I, so Eb ∝ N·I. (Key point: Speed relation)

3. What is Eb1?
   A. 250 V
   B. 258 V
   C. 242 V
   D. 210 V
   Answer: C. Eb1 = 250 - 40×0.2. (Key point: Initial back emf)

4. How is the extra series resistance found?
   A. R = (V - Eb2)/I1
   B. R = V/I2
   C. R = Eb2/I2 - 0.2
   D. R = (V - Eb2)/I2 - 0.2
   Answer: D. Gives about 5.61 Ω. (Key point: Resistance)

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