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Features of an A.C. Transformer

Key idea: Identify transformer parts and use the ideal transformer ratios Vs/Vp = Ns/Np and VpIp = VsIs to solve basic transformer questions (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. Transformer

A transformer changes an a.c. voltage from one value to another using electromagnetic induction.

B. Key parts

  • Primary coil: input coil with turns Nₚ.
  • Secondary coil: output coil with turns Nₛ.
  • Core (laminated soft iron): guides magnetic flux to improve flux linkage and reduce eddy current losses.
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.

2. Key Ideas (What Earns Marks)

  • Transformers require a changing magnetic flux, so they work with a.c., not steady d.c.
  • Ideal transformer equations (syllabus):
    • Vₛ/Vₚ = Nₛ/Nₚ
    • Vₚ Iₚ = Vₛ Iₛ (ideal: 100% efficient)
    • so Iₛ/Iₚ = Nₚ/Nₛ
  • Step-up vs step-down:
    • step-up: Nₛ > Nₚ ⇒ Vₛ > Vₚ and Iₛ < Iₚ
    • step-down: Nₛ < Nₚ ⇒ Vₛ < Vₚ and Iₛ > Iₚ
Syllabus scope (9478)

Transformer principle + ideal transformer ratio is a core induction outcome (18g).

3. Detailed Explanations

A. Why laminations are used

The core is laminated (thin insulated sheets) to reduce eddy current loops and therefore reduce heating losses.

B. What “ideal transformer” assumes

In an ideal transformer:

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

4. Common Mistakes

  • Treating a transformer as “working for d.c.” (steady d.c. gives no changing flux, so no induced e.m.f.).
  • Swapping the ratio (using Vₛ/Vₚ = Nₚ/Nₛ).
  • Forgetting the current changes inversely with turns ratio (for ideal power conservation).

5. Exam Tips

  • If you only need Vₛ, use Vₛ/Vₚ = Nₛ/Nₚ.
  • If you also need current, use power: VₚIₚ = VₛIₛ.
  • State “ideal transformer” before using power conservation.

6. Worked Examples

Modelled example 1

Turns ratio

Core

Problem

An ideal transformer has Nₚ = 500 and Nₛ = 100. The primary is connected to Vₚ = 240 V a.c. Find Vₛ.
Study the worked solution

  1. Write a consistently ordered ratio

    Method

    Vₛ/Vₚ = Nₛ/Nₚ.

    Reason

    Secondary quantities must appear in the numerator on both sides.

    Working

    Vₛ/240 = 100/500

  2. Evaluate

    Method

    Vₛ = 48 V.

    Reason

    The secondary has one fifth as many turns, so it has one fifth of the primary voltage.

    Working

    Vₛ = 240(100/500) = 48 V

Guided practice 2

Current ratio (ideal)

About 4 min

Problem

In Example 1, the secondary supplies Iₛ = 2.0 A at 48 V. Find the primary current at 240 V.

Try this before viewing the solution

Unit: A

Hints

Hint 1: choose the power relation
Use VₚIₚ = VₛIₛ and isolate Iₚ.
View solution step by step

  1. Conserve ideal power

    Method

    VₚIₚ = VₛIₛ.

    Reason

    The stated ideal model excludes energy losses.

    Working

    240Iₚ = (48)(2.0)

  2. Calculate

    Method

    Iₚ = 0.40 A.

    Reason

    The fivefold voltage decrease is accompanied by a fivefold current increase on the secondary side.

    Working

    Iₚ = (48)(2.0)/240 = 0.40 A

Common misconception 3

Turns needed for a target voltage

Find and correct the mistake

Learner claim

An ideal transformer must reduce 240 V a.c. to 12 V. With Nₚ = 1200, a learner inverts the ratio and obtains Nₛ = 24,000. Diagnose the error and find the required Nₛ.

Try this before viewing the solution

Unit: turns

View solution step by step

  1. Use the physical direction

    Method

    The required secondary must have fewer, not more, turns.

    Reason

    A lower secondary voltage requires Nₛ < Nₚ.

    Working

    12/240 = 1/20

  2. Repair the ratio

    Method

    Vₛ/Vₚ = Nₛ/Nₚ.

    Reason

    The learner paired secondary voltage with primary turns, which inverted the factor.

    Working

    12/240 = Nₛ/1200

  3. Evaluate

    Method

    Nₛ = 60 turns.

    Reason

    The secondary needs one twentieth of the primary turns.

    Working

    Nₛ = 1200(12/240) = 60

Examiner practice 4

Load power and input current (ideal)

4 marks

Examination question

An ideal transformer steps down 240 V to 24 V. The secondary supplies a Pₒᵤₜ = 120 W load. Find (i) Iₛ and (ii) Iₚ. [4 marks]

Try this before viewing the solution

View solution step by step

  1. Relate output power and current

    1 mark

    Method

    Pₒᵤₜ = VₛIₛ.

    Reason

    The load power and secondary voltage are given.

    Working

    120 = 24Iₛ

  2. Find secondary current

    1 mark

    Method

    Iₛ = 5.0 A.

    Reason

    Current is power divided by voltage.

    Working

    Iₛ = 120/24 = 5.0 A

  3. Use ideal input power

    1 mark

    Method

    Pᵢₙ = Pₒᵤₜ = VₚIₚ.

    Reason

    Only an ideal transformer permits equality of input and output power.

    Working

    120 = 240Iₚ

  4. Find primary current

    1 mark

    Method

    Iₚ = 0.50 A.

    Reason

    The primary voltage is ten times the secondary voltage, so its current is one tenth as large.

    Working

    Iₚ = 120/240 = 0.50 A

Challenge 5

Identify step-up vs step-down from turns

Minimal support

Independent transfer

An ideal transformer has Nₚ = 400, Nₛ = 2000, Vₚ = 12 V and Iₚ = 3.0 A. Classify it, then find Vₛ and Iₛ.

Try this before viewing the solution

Hints

Hint 1: track both tradeoffs
Voltage follows the turns factor; ideal current changes by its inverse.
View solution step by step

  1. Classify from turns

    Method

    It is a step-up transformer.

    Reason

    Nₛ > Nₚ, so the secondary voltage exceeds the primary voltage.

    Working

    Nₛ/Nₚ = 2000/400 = 5

  2. Find secondary voltage

    Method

    Vₛ = 60 V.

    Reason

    Voltage increases by the turns factor of five.

    Working

    Vₛ = 12(2000/400) = 60 V

  3. Find secondary current

    Method

    Iₛ = 0.60 A.

    Reason

    Ideal power is conserved, so the fivefold voltage increase requires a fivefold current decrease.

    Working

    Iₛ = IₚNₚ/Nₛ = 3.0(400/2000) = 0.60 A

7. Mind Stretchers

Mind stretcher 1: Why current decreases in a step-up transformerExtension

Explain why a step-up transformer reduces output current (ideal case).

Show Answer

For an ideal transformer, power is conserved: VₚIₚ = VₛIₛ.

If Vₛ is increased (step-up), then Iₛ must decrease so that the product VI stays the same.

Mind stretcher 2: Why must the input be a.c.?Extension

Explain why a transformer does not work with a steady d.c. supply.

Show Answer

A transformer needs a changing magnetic flux in the core so that an e.m.f. is induced in the secondary (Faraday’s law).

A steady d.c. current produces a steady flux (after a brief transient), so dΦ/dt = 0 and the induced secondary e.m.f. becomes zero.

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