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Energy Losses In Transformer

Key idea: Identify the main transformer losses (I²R, eddy currents, hysteresis, flux leakage) and the design features used to reduce them (A Level Physics).

  • A-level H2 Physics topic extensions
On this page

Learning objectives

  • Explain energy losses in practical transformers and the design features that reduce them beyond the H2 syllabus.

1. Definitions (Must Know)

A. Efficiency idea

A real transformer is efficient but not perfect: some input power becomes unwanted heating or is lost because not all flux links the secondary.

2. Key Ideas (What Earns Marks)

Main loss mechanisms:

LossWhat causes itTypical reduction method
Copper lossresistance of windings → I²R heatinglow-resistance copper wire, thicker wire
Eddy current lossinduced currents in the corelaminated core (thin insulated sheets)
Hysteresis lossrepeated magnetisation of the coresoft iron / low-hysteresis material
Flux leakagenot all primary flux links secondarygood core design, close coupling
Scope note

Simple iron-core transformer operation and ideal ratios are core outcome 18g. Detailed loss mechanisms are useful extension material rather than a separately named 9478 requirement.

3. Detailed Explanations

A. Eddy current loss and lamination

Changing flux in the core induces currents in the metal core. Those eddy currents dissipate energy as heat.

Lamination reduces this by breaking up large current loops into small loops with higher resistance.

Cross-sections of a transformer core with changing flux into the page: a solid core carries one large eddy-current loop, while a laminated core confines eddy currents to small loops inside thin insulated sheets.
Lamination reduces eddy currents by preventing large current loops in the core.

B. Hysteresis loss (qualitative)

In a.c. operation, the core is repeatedly magnetised and demagnetised. This process uses energy each cycle, which appears as heat. Using soft iron (low hysteresis) reduces this loss.

C. Flux leakage

If some magnetic flux produced by the primary does not link the secondary, less energy is transferred to the load. A well-designed core improves coupling.

4. Common Mistakes

  • Saying “lamination reduces hysteresis loss” (it mainly reduces eddy currents).
  • Confusing “flux leakage” (field lines missing the secondary) with “resistance loss” (I²R).

5. Exam Tips

  • If asked “how to reduce losses”, give a matched pair: loss → design feature.
  • In explanations, keep it causal: “changing flux induces eddy currents → heating”.

6. Worked Examples

Modelled example 1

Match the loss to the fix

Core

Problem

State one design feature that reduces each transformer loss: eddy-current loss, hysteresis loss and copper loss.
Study the worked solution

  1. Reduce eddy currents

    Method

    Use a core made from thin insulated laminations.

    Reason

    Lamination interrupts large conducting loops and raises the resistance of induced current paths.

    Working

    laminated core ⇒ I_eddy↓

  2. Reduce hysteresis

    Method

    Use soft iron or another low-hysteresis core material.

    Reason

    It requires less energy for each magnetisation reversal.

    Working

    low-hysteresis material ⇒ E_cycle↓

  3. Reduce copper loss

    Method

    Use thick, low-resistance copper windings.

    Reason

    Lower winding resistance reduces I²R heating at a given current.

    Working

    R↓ ⇒ P_copper = I²R↓

Guided practice 2

Copper loss calculation

About 4 min

Problem

A transformer winding of resistance 0.80 Ω carries 2.5 A. Find its copper-loss power.

Try this before viewing the solution

Unit: W

Hints

Hint 1: square the current
Copper loss uses I²R, not IR.
View solution step by step

  1. Identify the loss model

    Method

    P_copper = I²R.

    Reason

    The winding behaves as a resistor carrying current.

    Working

    P = I²R

  2. Calculate the loss

    Method

    P_copper = 5.0 W.

    Reason

    Substitute the current and winding resistance stated.

    Working

    P_copper = (2.5)²(0.80) = 5.0 W

Common misconception 3

Efficiency from losses

Find and correct the mistake

Learner claim

A transformer receives 500 W, with 12 W copper loss and 8 W core loss. A learner reports efficiency as (12 + 8)/500 = 4%. Diagnose the ratio, then find output power and efficiency.

Try this before viewing the solution

Unit: W
Unit: %

View solution step by step

  1. Add the loss pathways

    Method

    Total loss is 20 W.

    Reason

    Copper and core losses both remove power from the useful output.

    Working

    Pₗₒₛₛ = 12 + 8 = 20 W

  2. Find output power

    Method

    Pₒᵤₜ = 480 W.

    Reason

    Input power divides between useful output and losses.

    Working

    Pₒᵤₜ = 500-20 = 480 W

  3. Use the efficiency ratio

    Method

    η = 96%.

    Reason

    Efficiency compares useful output with total input.

    Working

    η = 480/500 = 0.96 = 96%

Examiner practice 4

Identify the dominant loss from a symptom

3 marks

Examination question

A transformer becomes warm even when no load is connected. Identify the dominant loss category and explain why copper loss is relatively small. [3 marks]

Try this before viewing the solution

Dominant loss category

View solution step by step

  1. Identify core losses

    1 mark

    Method

    Eddy-current and hysteresis losses dominate the observed no-load heating.

    Reason

    They occur in the magnetic core whenever it experiences alternating flux.

    Working

    P_core = P_eddy + P_hysteresis

  2. Explain why they persist

    1 mark

    Method

    The primary still produces changing core flux with no secondary load.

    Reason

    The core continues to be magnetised and demagnetised, and eddy currents are still induced.

    Working

    Φ(t) alternates at no load

  3. Compare copper loss

    1 mark

    Method

    Copper loss is relatively small because no-load current is small.

    Reason

    Winding heating scales as I²R.

    Working

    I small ⇒ I²R small

Challenge 5

Flux leakage effect (qualitative)

Minimal support

Independent transfer

A real transformer has significant flux leakage. State and explain one effect on its secondary output compared with an ideal transformer.

Try this before viewing the solution

Secondary output effect

Hints

Hint 1: focus on linked flux
Faraday’s law depends on the flux linkage through the secondary, not all flux produced by the primary.
View solution step by step

  1. Identify the linkage change

    Method

    Less of the primary’s changing flux links the secondary winding.

    Reason

    Leaked field lines do not pass through all secondary turns.

    Working

    NₛΦ_linked < NₛΦ_ideal

  2. Infer the induced e.m.f.

    Method

    The secondary e.m.f. is reduced.

    Reason

    Its magnitude depends on the rate of change of secondary flux linkage.

    Working

    |Eₛ| = |d(NₛΦ_linked)/dt|

  3. State the observable output

    Method

    Output voltage under load is lower than the ideal prediction.

    Reason

    The induced secondary e.m.f. supplies that output.

    Working

    V_(out,real) < V_(out,ideal)

7. Mind Stretchers

Mind stretcher 1: Why do laminated sheets need insulation?Extension

Explain why the laminated sheets are insulated from each other.

Show Answer

Insulation stops charge carriers from crossing between sheets, which breaks up large current loops and increases the resistance of any induced currents.

That reduces eddy current magnitude and therefore reduces I²R heating.

Mind stretcher 2: Trade-offs in reducing copper lossExtension

Why can’t we always eliminate copper loss by using extremely thick wire for the windings?

Show Answer

Thicker wire reduces resistance but increases size, mass and cost, and may not fit in the available space.

It can also reduce the number of turns you can pack onto the core, affecting design requirements.

Continue with the next resource in this course.

Course and syllabus information
Course
A-level H2 Physics topic extensions
Syllabus scope
Beyond the syllabus
Edition
A-level H2 Physics topic extensions