--- title: "Ideal transformer principle and emf equation" subject: "Electrical Machines I" chapter: 2 key: "semester_4/electrical_machines_i:ideal-transformer-principle-emf" source: https://www.bctengineer.com/q/semester_4/electrical_machines_i/ideal-transformer-principle-emf site: BCT Engineering --- # Ideal transformer principle and emf equation Electrical Machines I · Chapter 2: Transformer - Asked in: 2077 Chaitra, 8 marks; 2072 Asoj, 8 marks; 2071 Bhadra, 8 marks; 2070 Bhadra, 8 marks (asked 4 times) - 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_i/ideal-transformer-principle-emf - Whole chapter: https://www.bctengineer.com/md/semester_4/electrical_machines_i/chapter-2 ## Question What do you mean by an ideal transformer? Explain the operating principle of an ideal single phase transformer and derive the emf equation. ## Short answer An ideal transformer has zero losses, infinite permeability, and all flux links both windings. Operating principle: ac voltage on primary sets up core flux that induces emfs in both coils per Faraday's law, voltage ratio equals turns ratio. Emf equation: E1 = 4.44 f N1 φm, E2 = 4.44 f N2 φm, derived from e = -N dφ/dt for sinusoidal flux. ## Answer An **ideal transformer** is one with zero winding resistance, no leakage flux (all flux links both windings), no hysteresis or eddy-current loss, and a core of infinite permeability, so it needs no magnetising current. Its efficiency is 100% and its voltage regulation is zero. #### Operating principle A transformer works on **mutual induction** (Faraday's law). ```text phi (alternating) +---------------------+ o---|] [|---o V1 |] N1 core N2 [| V2 -> load o---|] [|---o +---------------------+ primary secondary ``` 1. An ac voltage $V_1$ is applied to the primary winding of $N_1$ turns. 2. The current sets up an alternating flux φ in the laminated iron core. 3. The flux links the secondary winding of $N_2$ turns and induces in it an emf $E_2$ of the same frequency. The same flux induces a back emf $E_1$ in the primary that balances $V_1$. 4. When a load is connected, $I_2$ flows; the primary automatically draws a matching current so that $N_1I_1 = N_2I_2$ and $V_1I_1 = V_2I_2$. 5. Since emf per turn is the same in both windings, $V_1/V_2 = N_1/N_2$: the voltage is stepped up or down without changing frequency. #### EMF equation Let the flux be sinusoidal: $\phi = \phi_m\sin\omega t$, $\omega = 2\pi f$. $$ \begin{aligned} e_1 &= -N_1\frac{d\phi}{dt} = -N_1\omega\phi_m\cos\omega t \\ &= N_1\omega\phi_m\sin(\omega t - 90^\circ) \end{aligned} $$ So $e_1$ lags φ by 90°. Its maximum value is $$ E_{1m} = N_1\omega\phi_m = 2\pi f N_1\phi_m $$ RMS value: $$ \begin{aligned} E_1 &= \frac{E_{1m}}{\sqrt2} = \frac{2\pi}{\sqrt2} f N_1 \phi_m \\ E_1 &= 4.44\, f N_1 \phi_m = 4.44\, f N_1 B_m A \end{aligned} $$ Similarly for the secondary: $$ E_2 = 4.44\, f N_2 \phi_m $$ #### Transformation ratio $$ \frac{E_2}{E_1} = \frac{N_2}{N_1} = K, \qquad \frac{E_1}{N_1} = \frac{E_2}{N_2} = 4.44 f\phi_m \ (\text{emf per turn}) $$ For an ideal transformer $V_1 = E_1$, $V_2 = E_2$ and $I_1/I_2 = N_2/N_1 = K$. **Example:** a 50 Hz transformer with $N_1 = 500$ and $\phi_m = 9$ mWb gives $E_1 = 4.44\times50\times500\times0.009 = 999$ V. ## 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. Self induction of a single coil only B. Mutual induction (Faraday's law) by an alternating flux linking both windings C. The Hall effect in the core D. Electrostatic coupling between windings Answer: B. The alternating flux set up by the primary links the secondary and induces an emf there. (Key point: Working principle) 2. How does the induced emf e1 relate to a sinusoidal flux? A. It is in phase with the flux B. It lags the flux by 90°, since e1 = −N1 dφ/dt C. It leads the flux by 90° D. It lags the flux by 180° only Answer: B. Differentiating φm sin ωt gives a cosine, i.e. sin(ωt − 90°). (Key point: Emf lag) 3. How is the emf equation obtained from the maximum value? A. E1 = 2π f N1 φm × √2 = 8.89 f N1 φm B. E1 = π f N1 φm = 3.14 f N1 φm C. E1 = (2π f N1 φm)/√2 = 4.44 f N1 φm D. E1 = f N1 φm / 4.44 Answer: C. The maximum 2π f N φm divided by √2 gives the rms value 4.44 f N φm. (Key point: RMS emf) 4. What do emf per turn and the voltage ratio imply? A. V1/V2 = N2/N1 B. Voltage ratio depends on frequency C. Secondary emf per turn is smaller for step-up D. Emf per turn is the same in both windings, so V1/V2 = N1/N2 Answer: D. E/N = 4.44 f φm is common to both windings, giving E2/E1 = N2/N1. (Key point: Turns ratio) ## Close variants set in other exams - [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× ## Related in Electrical Machines I - [Operating principle of 3-phase induction motor](https://www.bctengineer.com/question.md?key=semester_4%2Felectrical_machines_i%3Ainduction-motor-operating-principle) — Electrical Machines I, chapter 5 - [Transformer no-load operation and ideal transformer properties](https://www.bctengineer.com/question.md?key=semester_4%2Felectrical_machines_i%3Atransformer-no-load-ideal-properties) — Electrical Machines I, chapter 2 ## Related in other subjects - [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 - [Transformer definition, principle, emf equation](https://www.bctengineer.com/question.md?key=semester_1%2Fbasic_electrical_and_electronics_engineering%3Atransformer-emf-equation) — Basic Electrical and Electronics Engineering, chapter 2 - [Output (kVA) equation of single phase transformer](https://www.bctengineer.com/question.md?key=semester_5%2Felectric_machine_design%3Aoutput-equation-single-phase-transformer) — Electric Machine Design, chapter 3, asked 3×