---
title: "Series motor: resistance to limit speed to 2400 rpm"
subject: "Electrical Machines"
chapter: 4
key: "semester_4/electrical_machines:dcm-back-emf-importance-series-resistance"
source: https://www.bctengineer.com/q/semester_4/electrical_machines/dcm-back-emf-importance-series-resistance
site: BCT Engineering
---

# Series motor: resistance to limit speed to 2400 rpm

Electrical Machines · Chapter 4: DC Motor

- Asked in: 2078 Poush, 3+5 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/electrical_machines/dcm-back-emf-importance-series-resistance
- Whole chapter: https://www.bctengineer.com/md/semester_4/electrical_machines/chapter-4

## Question

State the importance of back EMF in dc motor. A 240 V dc series motor has total resistance of $0.2\ \Omega$. When the speed is 1800 rpm, the motor draws a current of 40 A. Calculate the value of resistance to be connected in series with the armature so as to limit the speed to 2400 rpm when the line current is 10 A.

## Short answer

Back emf limits armature current, provides self-regulation, and converts electrical power to mechanical power. For the series motor, with unsaturated flux Eb∝NI. At 1800 rpm, 40 A, Eb=232 V. For 2400 rpm, 10 A, Eb=77.33 V. Required total resistance = 16.267 Ω, external resistance = 16.07 Ω.

## Answer

#### Importance of back emf

- The back emf $E_b = \dfrac{\Phi ZNP}{60A}$ opposes the supply voltage, so $I_a = (V - E_b)/R_a$. It **limits the armature current** to a safe value once the motor runs.
- It is **self-regulating**: when the load increases the speed and $E_b$ fall, more current flows and the torque rises to meet the load; the reverse happens for a lighter load.
- It is the means of energy conversion: the electrical power converted to mechanical power is $E_bI_a$.
- At starting $E_b = 0$, so a starter is needed to limit the current.
- Its value shows the speed of the motor, since $N\propto E_b/\Phi$.

#### Numerical

For a series motor (unsaturated) $\Phi\propto I$, so $E_b\propto N I$.

**Condition 1:** 1800 rpm, 40 A

$$
E_{b1} = V - IR = 240 - 40(0.2) = 232.0\ \text{V}
$$

**Condition 2:** 2400 rpm, 10 A, with an extra series resistance $R_x$

$$
\frac{E_{b2}}{E_{b1}} = \frac{N_2I_2}{N_1I_1} \;\Rightarrow\; E_{b2} = 232.0\times\frac{2400}{1800}\times\frac{10}{40} = 77.33\ \text{V}
$$

The total resistance in the circuit is found from the voltage equation:

$$
V = E_{b2} + I_2(R + R_x) \;\Rightarrow\; 240 = 77.33 + 10(0.2 + R_x)
$$

$$
0.2 + R_x = \frac{240 - 77.33}{10} = 16.267 \;\Rightarrow\; R_x = 16.07\ \Omega
$$

**Answer:** an external resistance of $16.07\ \Omega$ must be connected in series with the armature.

## Check yourself

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

1. What is a key importance of back emf in a DC motor?
   A. It increases the supply voltage
   B. It limits armature current and self-regulates the motor to the load
   C. It raises the flux
   D. It reduces torque to zero
   Answer: B. Ia = (V - Eb)/Ra; falling speed with more load lowers Eb and draws more current. (Key point: Role of back emf)

2. Electrical power converted to mechanical power in the armature is:
   A. V x Ia
   B. Ia^2 Ra
   C. Eb x Ia
   D. Eb / Ia
   Answer: C. Mechanical power developed equals Eb Ia. (Key point: Power conversion)

3. For an unsaturated series motor, Eb is proportional to:
   A. N only
   B. I only
   C. N / I
   D. N x I
   Answer: D. Phi is proportional to I, so Eb proportional to Phi N is proportional to N I. (Key point: Series motor emf scaling)

4. Eb1 = 232 V; at 2400 rpm and 10 A, Eb2 = 77.33 V. The extra series resistance is:
   A. 16.07 ohm
   B. 16.27 ohm
   C. 23.2 ohm
   D. 0.2 ohm
   Answer: A. 0.2 + Rx = (240 - 77.33)/10 = 16.267, so Rx = 16.07 ohm. (Key point: Required resistance)

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