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Ideal Transformer Operation and Ratios

Key idea: Explain how an a.c. transformer works via changing flux and Faraday’s law, and use Vs/Vp = Ns/Np with power conservation in exam problems (A Level Physics).

  • GCE A-Level H2 Physics 2027
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Learning objectives

  • Explain simple iron-core transformer operation and apply ideal transformer ratios.

1. Definitions (Must Know)

A. Ideal transformer assumptions

An ideal transformer is a model where:

  • all flux produced by the primary links the secondary (no flux leakage),
  • coil resistances are negligible (no I²R loss),
  • energy losses are negligible, so power in = power out.

B. Faraday’s law for each coil

If the flux through the core is Φ, then the induced e.m.f. magnitudes are:

Vₚ = Nₚ|dΦ/dt|, Vₛ = Nₛ|dΦ/dt|

2. Key Ideas (What Earns Marks)

  • A transformer works because a.c. produces changing flux, so it induces an e.m.f. (Faraday’s law).
  • Voltage ratio (ideal): Vₛ/Vₚ = Nₛ/Nₚ
  • Power conservation (ideal): VₚIₚ = VₛIₛ so current ratio: Iₛ/Iₚ = Nₚ/Nₛ
A one-line memory aid

Voltage follows turns: V ∝ N. Current goes opposite: I ∝ 1/N (ideal case).

3. Detailed Explanations

A. What actually happens in an a.c. transformer

Simple iron-core transformerPrimary and secondary coils with different numbers of turns are wound on opposite sides of a closed iron core. Arrows show alternating input, changing core flux and output to a load.primary Nₚsecondary Nₛloadchanging flux ΦVₛ/Vₚ = Nₛ/Nₚ = Iₚ/Iₛ
Scroll diagram horizontally to read all labels.
The alternating primary current produces changing core flux linking both coils. For the ideal model, voltage follows turns while current changes inversely.
  1. An alternating voltage applied to the primary drives an a.c. current.
  2. That current produces a changing magnetic flux Φ in the iron core.
  3. The changing flux links both coils, so an e.m.f. is induced in the primary and in the secondary.
  4. If a load is connected, the secondary e.m.f. drives a current and transfers energy to the load.

B. Deriving the voltage ratio

For each coil, Faraday’s law gives:

Vₚ = Nₚ|dΦ/dt|, Vₛ = Nₛ|dΦ/dt|

Divide:

Vₛ/Vₚ = Nₛ/Nₚ

C. Deriving the current ratio (ideal power)

For an ideal transformer, power in equals power out:

VₚIₚ = VₛIₛ

Rearrange using Vₛ/Vₚ = Nₛ/Nₚ:

Iₛ/Iₚ = Vₚ/Vₛ = Nₚ/Nₛ

4. Common Mistakes

  • Saying “transformers work with d.c.” (steady d.c. gives no changing flux, so no induced e.m.f.).
  • Swapping the turns ratio (writing Vₛ/Vₚ = Nₚ/Nₛ).
  • Using VₚIₚ = VₛIₛ without stating “ideal transformer”.

5. Exam Tips

  • If you see “ideal transformer”, you can use:
    • Vₛ/Vₚ = Nₛ/Nₚ
    • VₚIₚ = VₛIₛ
  • If asked “why must it be a.c.”, answer: “need changing flux for induction”.

6. Worked Examples

Modelled example 1

Step-down transformer

Core

Problem

An ideal transformer has Nₚ = 1200, Nₛ = 300 and Vₚ = 240 V. Find Vₛ and classify the transformer.
Study the worked solution

  1. Use the voltage ratio

    Method

    Vₛ/Vₚ = Nₛ/Nₚ.

    Reason

    Both windings link the same changing core flux in the ideal model, so induced e.m.f. is proportional to turns.

    Working

    Vₛ/Vₚ = Nₛ/Nₚ

  2. Calculate secondary voltage

    Method

    Vₛ = 60 V.

    Reason

    The secondary has one quarter as many turns as the primary.

    Working

    Vₛ = 240(300/1200) = 60 V

  3. Classify

    Method

    This is a step-down transformer.

    Reason

    Secondary voltage and turns are lower than the primary values.

    Working

    Vₛ < Vₚ and Nₛ < Nₚ.

Guided practice 2

Finding current (ideal)

About 4 min

Problem

The transformer above supplies Iₛ = 4.0 A at 60 V. Find the ideal primary current at 240 V.

Try this before viewing the solution

Unit: A

Hints

Hint 1: conserve power
Set VₚIₚ = VₛIₛ and isolate Iₚ.
View solution step by step

  1. Apply ideal power conservation

    Method

    VₚIₚ = VₛIₛ.

    Reason

    The ideal model excludes winding and core energy losses.

    Working

    240Iₚ = (60)(4.0)

  2. Evaluate

    Method

    Iₚ = 1.0 A.

    Reason

    The fourfold voltage step-down corresponds to a fourfold current step-up.

    Working

    Iₚ = (60)(4.0)/240 = 1.0 A

Common misconception 3

Show that current ratio is inverse turns ratio

Find and correct the mistake

Learner claim

A learner says current follows the turns ratio just like voltage, so Iₛ/Iₚ = Nₛ/Nₚ. Diagnose the claim for an ideal transformer and derive the correct ratio.

Try this before viewing the solution

Correct current ratio

View solution step by step

  1. Use voltage ratio

    Method

    Vₛ/Vₚ = Nₛ/Nₚ.

    Reason

    Voltage follows turns in the ideal common-flux model.

    Working

    Vₛ/Vₚ = Nₛ/Nₚ

  2. Use ideal power

    Method

    Iₛ/Iₚ = Vₚ/Vₛ.

    Reason

    VₚIₚ = VₛIₛ requires current to change inversely with voltage.

    Working

    Iₛ/Iₚ = Vₚ/Vₛ

  3. Substitute turns ratio

    Method

    Iₛ/Iₚ = Nₚ/Nₛ.

    Reason

    The voltage ratio is inverted when expressed as Vₚ/Vₛ.

    Working

    Iₛ/Iₚ = Nₚ/Nₛ

Examiner practice 4

Step-up example (voltage and current)

4 marks

Examination question

An ideal transformer has Nₚ = 400, Nₛ = 2000, Vₚ = 12 V and supplies Iₛ = 0.80 A. Find Vₛ, output power and Iₚ. [4 marks]

Try this before viewing the solution

View solution step by step

  1. Find turns factor

    1 mark

    Method

    Nₛ/Nₚ = 5.

    Reason

    The secondary has five times as many turns.

    Working

    2000/400 = 5

  2. Find secondary voltage

    1 mark

    Method

    Vₛ = 60 V.

    Reason

    Voltage follows the turns ratio.

    Working

    Vₛ = 12(5) = 60 V

  3. Find output power

    1 mark

    Method

    Pₛ = 48 W.

    Reason

    Use Pₛ = VₛIₛ.

    Working

    Pₛ = (60)(0.80) = 48 W

  4. Find primary current

    1 mark

    Method

    Iₚ = 4.0 A.

    Reason

    Ideal input power equals output power.

    Working

    Iₚ = 48/12 = 4.0 A

Challenge 5

Efficiency check phrasing (ideal vs real)

Minimal support

Independent transfer

Evaluate: “A transformer must reduce current because energy is lost.” Correct the statement for ideal step-up and step-down transformers, then identify what changes for a real transformer.

Try this before viewing the solution

Hints

Hint 1: separate ratio from efficiency
First apply ideal power conservation; then discuss losses only for the real device.
View solution step by step

  1. Reject the loss explanation

    Method

    The claim is false for an ideal transformer.

    Reason

    Ideal input and output powers are equal.

    Working

    VₚIₚ = VₛIₛ

  2. State the ideal tradeoff

    Method

    A step-up transformer lowers current; a step-down transformer raises current.

    Reason

    Current changes inversely with voltage and turns.

    Working

    Iₛ/Iₚ = Nₚ/Nₛ.

  3. Add the real-device limit

    Method

    A real transformer has output power below input power because of winding, core and leakage losses.

    Reason

    Energy is transferred to internal energy and other unwanted stores, but loss is not why ideal current ratios invert.

    Working

    Pₒᵤₜ < Pᵢₙ for a real transformer.

7. Mind Stretchers

Mind stretcher 1: Why step-up helps power transmissionExtension

Explain why stepping up voltage reduces power loss in transmission cables.

Show Answer

For a given transmitted power P, current is I = P/V.

Cable loss is Pₗₒₛₛ = I²R, so: Pₗₒₛₛ = (P/V)²R

Increasing V reduces I and therefore reduces I²R losses.

Mind stretcher 2: What sets the limit of an “ideal” model?Extension

Give two real-world reasons why a transformer is never perfectly ideal.

Show Answer

Examples:

  • Windings have resistance, so there is copper loss (I²R heating).
  • Not all flux links the secondary (flux leakage).
  • Eddy currents and hysteresis in the core cause heating losses.

8. Optional (Enrichment)

A. Real transformer losses (preview)

Real transformers have copper loss (I²R), eddy current loss, hysteresis loss, and flux leakage. See:

Continue with the next resource in this course.

Course and syllabus information
Course
GCE A-Level H2 Physics
Edition
GCE A-Level H2 Physics 2027