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

# Ideal transformer principle and emf equation

Electrical Machine · 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_machine/ideal-transformer-emf-equation
- Whole chapter: https://www.bctengineer.com/md/semester_4/electrical_machine/chapter-2

## Question

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

## Short answer

Alternating flux Φ = Φₘ sin ωt induces emf e = −N dΦ/dt. RMS value E = (2π/√2) f N Φₘ = 4.44 f N Φₘ. For primary E₁ = 4.44 f N₁ Φₘ, for secondary E₂ = 4.44 f N₂ Φₘ.

## Answer

A transformer is a static device that transfers electrical energy from one circuit to another at the same frequency by **mutual induction**. An ideal transformer has no winding resistance, no leakage flux and no core loss.

#### Operating principle
- When an alternating voltage V1 is applied to the primary winding (N1 turns), a small magnetising current flows and sets up an alternating flux Φ in the laminated core.
- This flux links the secondary winding (N2 turns) as well. By Faraday's law, an emf is induced in each winding:  e = −N dΦ/dt.
- The emf in the primary (E1) is the counter-emf opposing V1, so E1 = V1. The secondary emf E2 appears across the load terminals.
- So the voltage is changed in the ratio of turns, V2/V1 = N2/N1 = K, while the frequency stays the same.

#### Emf equation
Let the flux vary sinusoidally: Φ = Φm sin ωt, ω = 2πf.
```text
e1 = −N1 dΦ/dt = −N1 Φm ω cos ωt
Maximum value        Em1 = N1 Φm ω = 2π f N1 Φm
RMS value            E1  = Em1/√2 = (2π/√2) f N1 Φm
                         = 4.44 f N1 Φm
```
Similarly for the secondary:

**E1 = 4.44 f N1 Φm   and   E2 = 4.44 f N2 Φm** (volts)

where f is in Hz, Φm is the maximum flux in Wb. If the maximum flux density is Bm and the core area is *a*, then Φm = Bm a, so E = 4.44 f N Bm a.

**Example:** N1 = 500, f = 50 Hz, Φm = 2 mWb gives E1 = 4.44 × 50 × 500 × 0.002 = 222 V.

From the two equations, E1/N1 = E2/N2 = 4.44 f Φm (the emf per turn is the same in both windings).

## Check yourself

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

1. On what principle does a transformer work?
   A. Mutual induction
   B. Motional emf
   C. Self induction alone
   D. Electrostatic induction
   Answer: A. Flux set up by one winding links the other, inducing emf by mutual induction at the same frequency. (Key point: Principle)

2. What factor gives E = 4.44 f N Phi_m?
   A. 2 pi f N Phi_m times sqrt(2)
   B. f N Phi_m divided by 2
   C. pi f N Phi_m
   D. 2 pi f N Phi_m divided by sqrt(2)
   Answer: D. Em = 2 pi f N Phi_m and E = Em/sqrt(2), so 2pi/sqrt(2) = 4.44. (Key point: RMS emf)

3. What follows from the two emf equations?
   A. N1 = N2
   B. E1/N1 = E2/N2 = 4.44 f Phi_m
   C. E depends on load
   D. E1 = E2 always
   Answer: B. Both windings link the same flux, so emf per turn is equal. (Key point: Emf per turn)

4. N1 = 500, f = 50 Hz, Phi_m = 2 mWb. Find E1.
   A. 444 V
   B. 22.2 V
   C. 111 V
   D. 222 V
   Answer: D. E1 = 4.44 x 50 x 500 x 0.002 = 222 V. (Key point: Numerical check)

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