---
title: "Ideal transformer principle and emf equation"
subject: "Electrical Machines"
chapter: 2
key: "semester_4/electrical_machines:ideal-transformer-principle-emf"
source: https://www.bctengineer.com/q/semester_4/electrical_machines/ideal-transformer-principle-emf
site: BCT Engineering
---

# Ideal transformer principle and emf equation

Electrical Machines · Chapter 2: Transformer

- Asked in: 2072 Asoj
- Video: [EMF Equation of Transformer & Transformation Ratio](https://www.youtube.com/watch?v=LoYaD8vnF5A) — Electrical Deck
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/electrical_machines/ideal-transformer-principle-emf
- Whole chapter: https://www.bctengineer.com/md/semester_4/electrical_machines/chapter-2

## Question

Explain the operating principle of an ideal transformer and derive the emf equation.

## Short answer

A transformer works on mutual induction: AC primary current sets up an alternating core flux that links the secondary and induces an emf by Faraday’s law. In an ideal transformer, the induced rms emf is E = 4.44 f N φm (φm = maximum flux), giving E1/E2 = N1/N2.

## Answer

#### Operating principle

A transformer works on **mutual induction**. Two windings (primary of $N_1$ turns, secondary of $N_2$ turns) are wound on a common laminated core. An alternating voltage on the primary sends a current that sets up an alternating flux $\phi$ in the core. This flux links the secondary and, by Faraday's law, induces an emf in it. Energy is transferred from primary to secondary at the same frequency, with voltage and current changed in the ratio of turns.

An ideal transformer has no winding resistance, no leakage flux, no core loss, and a core of infinite permeability.

```text
        core (laminated)
     +--------------------+
  ~  |  N1            N2  |  load
 V1  |  primary   secondary|  Z
     +--------------------+
        <----- flux ----->
```

#### emf equation

Let the flux be $\phi = \phi_m\sin\omega t$, where $\phi_m$ is the maximum flux (Wb).

$$
\begin{aligned}
e_1 &= -N_1\frac{d\phi}{dt} = -N_1\omega\phi_m\cos\omega t \\
E_{m1} &= N_1\omega\phi_m = 2\pi fN_1\phi_m
\end{aligned}
$$

The rms value is

$$
E_1 = \frac{E_{m1}}{\sqrt2} = \frac{2\pi}{\sqrt2}fN_1\phi_m = 4.44\,fN_1\phi_m
$$

Similarly,

$$
E_2 = 4.44\,fN_2\phi_m
$$

$$
\boxed{E_1 = 4.44fN_1\phi_m = 4.44fN_1B_mA\ \text{V},\qquad \frac{E_1}{E_2} = \frac{N_1}{N_2}}
$$

where $B_m$ is the maximum flux density and $A$ the core cross-section. The induced emf lags the flux by 90°. For the ideal case $V_1 = E_1$ and $V_2 = E_2$, hence $V_1/V_2 = N_1/N_2 = a$.

## Check yourself

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

1. A transformer works on the principle of:
   A. Self-induction only
   B. Electrostatic induction
   C. Mutual induction
   D. Motional emf
   Answer: C. Alternating flux from the primary links the secondary and induces an emf by Faraday's law. (Key point: Principle)

2. The maximum induced emf Em1 equals:
   A. f N1 phi_m
   B. 4.44 f N1 phi_m
   C. pi f N1 phi_m / 2
   D. 2 pi f N1 phi_m
   Answer: D. e1 = -N1 dphi/dt gives amplitude N1 omega phi_m. (Key point: Peak emf)

3. Dividing the peak by sqrt(2) gives the rms emf:
   A. E1 = 4.44 f N1 phi_m
   B. E1 = 2.22 f N1 phi_m
   C. E1 = 6.28 f N1 phi_m
   D. E1 = 1.414 f N1 phi_m
   Answer: A. 2 pi/sqrt(2) = 4.44. (Key point: RMS emf)

4. The induced emf relative to the flux:
   A. Is in phase with the flux
   B. Lags the flux by 90 degrees
   C. Leads the flux by 90 degrees
   D. Lags by 180 degrees
   Answer: B. e = -N dphi/dt is a cosine when flux is a sine. (Key point: Phase of emf)

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