---
title: "Back emf and its role in DC motor"
subject: "Electrical Machines"
chapter: 4
key: "semester_4/electrical_machines:back-emf-role-dc-motor"
source: https://www.bctengineer.com/q/semester_4/electrical_machines/back-emf-role-dc-motor
site: BCT Engineering
---

# Back emf and its role in DC motor

Electrical Machines · Chapter 4: DC Motor

- Asked in: 2078 Baisakh, 8 marks; 2071 Magh, 2+4 marks (asked 2 times)
- 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_4/electrical_machines/back-emf-role-dc-motor
- Whole chapter: https://www.bctengineer.com/md/semester_4/electrical_machines/chapter-4

## Question

What is back emf in dc motor? How back emf helps to develop required torque according to load applied in the shaft?

## Short answer

Back emf E_b = (Φ Z N P)/(60 A) opposes supply voltage. When load increases, speed drops, E_b falls, I_a = (V − E_b)/R_a rises, increasing torque to match load. This self-regulation ensures the motor draws only the current needed by the load.

## Answer

#### Back emf

When the armature of a DC motor rotates, its conductors cut the magnetic flux, so an emf is induced in them (generator action). By Lenz's law this emf **opposes the supply voltage** which is the cause of the current. It is called the **back emf** or counter emf:

$$
E_b = \frac{\Phi ZNP}{60A}\ \text{V}
$$

The voltage equation of the motor is

$$
V = E_b + I_aR_a \quad\Rightarrow\quad I_a = \frac{V - E_b}{R_a}
$$

#### How back emf adjusts the torque to the load

The torque is $T = K_a\Phi I_a$, so the motor must draw the armature current that its load requires. Back emf adjusts this current automatically:

```text
 Load up -> speed N falls -> Eb falls -> (V - Eb) rises
        -> Ia rises -> torque rises -> speed stops falling
 Load down -> N rises -> Eb rises -> Ia falls -> torque falls
```

1. When the load on the shaft increases, the load torque exceeds the motor torque, so the speed falls.
2. $E_b\propto N$ falls, so $(V - E_b)$ increases and a larger armature current $I_a = (V - E_b)/R_a$ flows.
3. The torque $K_a\Phi I_a$ rises until it equals the new load torque, and the motor settles at a slightly lower speed.
4. When the load decreases, the reverse happens: speed rises, $E_b$ rises, $I_a$ falls and torque falls to match the lighter load.

So back emf acts as a **self-regulating governor** that makes the motor draw only the current needed by the load. At no load $E_b\approx V$ and the current is just enough to supply the losses. The power converted is $E_bI_a = T\omega$.

Back emf is also why the starting current is large: at start $E_b = 0$, so $I_a = V/R_a$ and a starter is required.

## Check yourself

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

1. Back emf in a dc motor:
   A. Is induced by the rotating armature and opposes the supply voltage
   B. Aids the supply
   C. Is zero at all speeds
   D. Is produced in the field winding
   Answer: A. By Lenz's law it opposes the cause. (Key point: Back emf)

2. The armature current is:
   A. (V + Eb)/Ra
   B. (V - Eb)/Ra
   C. V/Eb
   D. Eb/Ra
   Answer: B. V = Eb + IaRa. (Key point: Voltage equation)

3. When the load increases, what is the sequence?
   A. Speed rises, Eb rises, Ia falls
   B. Eb rises, Ia rises
   C. Speed falls, Eb falls, Ia rises, torque rises
   D. Ia stays constant
   Answer: C. Lower Eb increases V - Eb and so Ia and torque. (Key point: Load increase)

4. At no load the back emf is:
   A. Zero
   B. Twice V
   C. Equal to IaRa
   D. Approximately equal to V, with small current to meet losses
   Answer: D. Ia is just enough to supply losses. (Key point: No load)

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