---
title: "Back emf in dc motor"
subject: "Basic Electrical and Electronics Engineering"
chapter: 2
key: "semester_1/basic_electrical_and_electronics_engineering:back-emf-dc-motor"
source: https://www.bctengineer.com/q/semester_1/basic_electrical_and_electronics_engineering/back-emf-dc-motor
site: BCT Engineering
---

# Back emf in dc motor

Basic Electrical and Electronics Engineering · Chapter 2: Electrical Machines

- Asked in: 2081 Baishakh (new course), 1+2 marks
- Video: [What is Back EMF & what is its significance | DC Motor | TheElectricalGuy](https://www.youtube.com/watch?v=JgnPSOUoVJI) — Gaurav J - TheElectricalGuy
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_1/basic_electrical_and_electronics_engineering/back-emf-dc-motor
- Whole chapter: https://www.bctengineer.com/md/semester_1/basic_electrical_and_electronics_engineering/chapter-2

## Question

Describe back emf in a dc motor. Explain the significance of back emf in a dc motor.

## Short answer

Back emf Eb = (PφZN)/(60A) is the voltage induced in the armature opposing the applied voltage. It acts as a governor by automatically adjusting armature current with load, limits starting current, represents electrical power converted to mechnical power (Eb·Ia), and determines speed (N ∝ Eb/φ).

## Answer

**Back emf.** When the armature of a DC motor rotates, its conductors cut the main flux, and by Faraday's law an emf is induced in them. By Lenz's law this emf opposes the applied voltage (the cause producing it), so it is called the back (counter) emf:

$$
E_b=\frac{P\phi ZN}{60A}
$$

and the armature current is

$$
I_a=\frac{V-E_b}{R_a}
$$

**Significance:**
1. **Acts as a governor.** If load increases, speed falls, $E_b$ falls, $I_a$ rises, and torque increases to meet the load. If the load falls, the opposite occurs. The motor thus adjusts the input current automatically.
2. **Limits the armature current.** At standstill $E_b=0$, so the starting current $V/R_a$ is very large (a starter is therefore needed). At running speed, $E_b$ is close to $V$ and current is small.
3. **Indicates energy conversion.** The electrical power converted to mechanical power is $E_bI_a$; without back emf there would be no conversion.
4. **Determines speed**: $N\propto\frac{E_b}{\phi}$.

## Check yourself

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

1. Why does the emf induced in a motor armature oppose the applied voltage?
   A. Armature resistance reverses it
   B. Lenz's law: it opposes the cause producing it
   C. The brushes reverse polarity
   D. It is caused by the field winding's resistance
   Answer: B. Conductors cut flux and induce Eb, which by Lenz's law opposes the supply. (Key point: Origin of back emf)

2. When the load on a DC motor increases, what happens?
   A. Speed rises, Eb rises, Ia falls
   B. Speed stays fixed and Ia is unchanged
   C. Eb rises and torque falls
   D. Speed falls, Eb falls, Ia rises and torque increases
   Answer: D. Back emf acts as a governor: lower Eb means more Ia = (V - Eb)/Ra and more torque. (Key point: Governing action)

3. Why is a starter needed for a DC motor?
   A. At standstill Eb = 0, so Ia = V/Ra is very large
   B. At standstill Eb is maximum
   C. The field is zero at start
   D. Back emf is too large at start
   Answer: A. With no rotation there is no back emf, so only Ra limits the current. (Key point: Starting current)

4. What does Eb x Ia represent?
   A. Copper loss in the armature
   B. Input electrical power
   C. Electrical power converted to mechanical power
   D. Iron loss
   Answer: C. Without back emf there would be no conversion; Eb Ia is the mechanical power developed. (Key point: Energy conversion)

## Related in Basic Electrical and Electronics Engineering

- [DC shunt motor armature torque](https://www.bctengineer.com/question.md?key=semester_1%2Fbasic_electrical_and_electronics_engineering%3Adc-shunt-motor-torque) — Basic Electrical and Electronics Engineering, chapter 2

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