---
title: "Transformer definition, principle, emf equation"
subject: "Basic Electrical and Electronics Engineering"
chapter: 2
key: "semester_1/basic_electrical_and_electronics_engineering:transformer-emf-equation"
source: https://www.bctengineer.com/q/semester_1/basic_electrical_and_electronics_engineering/transformer-emf-equation
site: BCT Engineering
---

# Transformer definition, principle, emf equation

Basic Electrical and Electronics Engineering · Chapter 2: Electrical Machines

- Asked in: 2081 Chaitra (new course), 6 marks
- Video: [How does a Transformer work  - Working Principle electrical engineering](https://www.youtube.com/watch?v=UchitHGF4n8) — The Engineering Mindset
- 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/transformer-emf-equation
- Whole chapter: https://www.bctengineer.com/md/semester_1/basic_electrical_and_electronics_engineering/chapter-2

## Question

Define transformer, describe operating principle of transformer and write emf equation for it.

## Short answer

A transformer operates on mutual induction, transferring power at constant frequency. RMS emf per winding is E =4.44 f N Φm, so E2/E1 = N2/N1. The core flux Φm links both windings.

## Answer

**Definition.** A transformer is a static electrical device that transfers electric power from one circuit to another at the same frequency through electromagnetic induction, normally raising or lowering the voltage.

**Operating principle.** It works on mutual induction. An alternating voltage applied to the primary winding drives an alternating flux in the laminated iron core. This flux links the secondary winding and induces an emf in it (Faraday's law). The emf in each winding is proportional to its number of turns, so $\frac{E_2}{E_1}=\frac{N_2}{N_1}$.

**emf equation.** Let $N_1$ = primary turns, $N_2$ = secondary turns, $\Phi_m$ = maximum flux in the core (Wb), $f$ = frequency (Hz). The flux varies sinusoidally:

$$
\Phi=\Phi_m\sin\omega t
$$

The average rate of change of flux in a quarter cycle is

$$
\frac{d\Phi}{dt}\Big|_{av}=\frac{\Phi_m}{T/4}=4f\Phi_m
$$

So the average emf per turn is $4f\Phi_m$ volts. For a sine wave, the form factor is 1.11, so rms emf per turn $=1.11\times4f\Phi_m=4.44f\Phi_m$. Hence:

$$
E_1=4.44\,f\,N_1\,\Phi_m \qquad E_2=4.44\,f\,N_2\,\Phi_m
$$

Dividing, the transformation ratio is

$$
\frac{E_2}{E_1}=\frac{N_2}{N_1}=K
$$

## Check yourself

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

1. Which describes a transformer?
   A. A rotating machine that changes frequency
   B. A static device transferring power between circuits at the same frequency by electromagnetic induction
   C. A device that converts AC to DC
   D. A device that stores charge
   Answer: B. It is a static device with no moving parts, working at constant frequency. (Key point: Definition)

2. What is the average emf per turn?
   A. 2 f Phi_m
   B. f Phi_m
   C. 4 f Phi_m (flux changes by Phi_m in a quarter cycle T/4)
   D. 8 f Phi_m
   Answer: C. Average rate of change in a quarter cycle is Phi_m/(T/4) = 4 f Phi_m. (Key point: Average emf per turn)

3. How does 4.44 arise?
   A. 4 x 1.11, form factor of a sine wave times 4 f Phi_m
   B. 4 x 1.57
   C. 4 x 0.707
   D. 4 x 0.637
   Answer: A. RMS = form factor x average = 1.11 x 4 f Phi_m = 4.44 f Phi_m per turn. (Key point: RMS factor)

4. Using E1 = 4.44 f N1 Phi_m and E2 = 4.44 f N2 Phi_m, what is E2/E1?
   A. N1/N2
   B. N2/N1 = K
   C. N1 N2
   D. Phi_m/f
   Answer: B. The same flux and frequency cancel, leaving the turns ratio. (Key point: Transformation ratio)

## Related in Basic Electrical and Electronics Engineering

- [Operating principle of MCB](https://www.bctengineer.com/question.md?key=semester_1%2Fbasic_electrical_and_electronics_engineering%3Amcb-operating-principle) — Basic Electrical and Electronics Engineering, chapter 4
- [Transformer principle and maximum efficiency](https://www.bctengineer.com/question.md?key=semester_1%2Fbasic_electrical_and_electronics_engineering%3Atransformer-principle-efficiency) — Basic Electrical and Electronics Engineering, chapter 2
- [25 kVA ideal transformer currents, flux](https://www.bctengineer.com/question.md?key=semester_1%2Fbasic_electrical_and_electronics_engineering%3Aideal-25kva-transformer) — Basic Electrical and Electronics Engineering, chapter 2

## Related in other subjects

- [Transformer operating principle and emf equation](https://www.bctengineer.com/question.md?key=semester_4%2Felectrical_machines_i%3Atransformer-principle-emf-equation) — Electrical Machines I, chapter 2, asked 6×
- [Ideal transformer principle and emf equation](https://www.bctengineer.com/question.md?key=semester_4%2Felectrical_machines_i%3Aideal-transformer-principle-emf) — Electrical Machines I, chapter 2, asked 6×
- [Ideal transformer principle and emf equation](https://www.bctengineer.com/question.md?key=semester_4%2Felectrical_machine%3Aideal-transformer-emf-equation) — Electrical Machine, chapter 2
- [Ideal transformer principle and emf equation](https://www.bctengineer.com/question.md?key=semester_4%2Felectrical_machines%3Aideal-transformer-principle-emf) — Electrical Machines, chapter 2