Simple iron-core ideal transformers

Key idea: H2 Physics lessons on magnetic flux, induction laws, applications and ideal transformers.

  • GCE A-Level H2 Physics 2027

Learn the idea

Big question: How can changing flux transfer power between two circuits?

An alternating primary current creates changing flux in an iron core, linking a secondary coil. For an ideal transformer Vs/Vp = Ns/Np and input power equals output power, so current changes inversely with voltage. A steady direct current cannot maintain the required changing flux.

Use common changing core flux

An alternating primary current creates changing magnetic flux in a shared core. This links both coils and induces e.m.f.s. For an ideal transformer, V_s/V_p = N_s/N_p because both coils share the same rate of flux change per turn.

A transformer requires changing flux, so a steady d.c. input gives no sustained secondary e.m.f. after the switching transient. Step-up means N_s > N_p and V_s > V_p; it does not mean energy is created.

Check your understanding: A transformer has 200 primary and 50 secondary turns at 240 V. Find ideal output voltage.

V_s = 240(50/200) = 60 V.

Use the ideal power balance

For an ideal transformer, input and output powers are equal: V_pI_p = V_sI_s. Combining this with the turns ratio gives N_s/N_p = V_s/V_p = I_p/I_s, so stepping voltage up steps current down by the reciprocal factor.

The ideal model assumes the two coils share the changing core flux and that no energy is lost in the windings or core. Keep primary quantities together on one side of each ratio before substituting.

Check your understanding: An ideal transformer doubles the voltage. What happens to the current?

The current halves, so input and output power remain equal.

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.

Key ideas to keep

  • A transformer requires changing flux, normally from a.c.
  • Step-up voltage means step-down current in the ideal model.
  • The turns ratio compares corresponding primary and secondary quantities.

Worked example

Explain energy transfer through a transformer

Question: Explain how an iron-core transformer transfers energy without conducting current between windings.

  1. Step 1: Create changing core flux

    Why: Only changing flux can induce a sustained e.m.f.

    Working: Alternating primary current produces alternating magnetic flux in the iron core.

  2. Step 2: Link the secondary

    Why: The common core guides nearly the same changing flux through both windings.

    Working: Faraday's law induces an alternating secondary e.m.f.; Vₛ/Vₚ = Nₛ/Nₚ ideally.

  3. Step 3: Account for power

    Why: The ideal model neglects winding and core losses.

    Working: VₚIₚ = VₛIₛ, so stepping voltage up steps current down.

Answer: Alternating primary current produces changing core flux. The iron core links this flux through both windings; Faraday's law induces alternating secondary e.m.f. Turns ratio sets voltage ratio. In the ideal model, power is conserved and current ratio is inverse to voltage ratio.

Check: A steady d.c. primary produces no sustained secondary e.m.f. after switching.

Practise with support

Try this

An ideal transformer steps voltage up by factor 5. State the current factor.

Hint: Ideal power is conserved.

Check your answer

Current steps down by factor 5 because VpIp = VsIs and Ip/Is = Ns/Np.

Practise independently

Your turn

State the ideal-transformer assumptions behind the voltage and current ratios and explain what a step-up device does.

Check your answer

All changing core flux links both windings, winding resistance and core losses are negligible, and input power equals output power. Ns/Np = Vs/Vp = Ip/Is. A step-up transformer raises voltage and lowers current by the same ratio; it does not create power.

Common mistakes

Common mistake

A step-up transformer creates power.

What is wrong with this reasoning?

Show better thinking

An ideal transformer conserves power and trades increased voltage for decreased current.

Common mistake

A steady d.c. primary produces a continuous secondary e.m.f.

What is wrong with this reasoning?

Show better thinking

After the switching transient, steady d.c. gives constant flux and no induced secondary e.m.f.; transformer operation requires changing flux.

Exam guidance

Write the turns ratio and ideal power equation separately before combining them.

Exam-style practice [7 marks]

An ideal 240 V transformer with 1500 primary and 75 secondary turns supplies 4.0 A. Find secondary voltage, primary current and output power.

Plan before you answer

  • Use the turns ratio for voltage.
  • Use ideal power for current.
  • Check both powers agree.
Mark your answer and compare the model

Marking points

Tick each point only if your answer states it clearly.

Model answer

Vs = 240(75/1500) = 12 V. Ip = Is(Ns/Np) = 0.20 A. Output power is 12(4.0) = 48 W, equal to input power.

Check what stayed with you

Recall question

An ideal transformer has Ns/Np = 0.10. State Vs/Vp and Ip/Is.

Check the answer

Vs/Vp = 0.10 and Ip/Is = 0.10; secondary current is ten times primary current.

Try this next

Use the longer mixed questions to connect the ideas, calculations and diagrams from this topic.

Open Electromagnetic Induction structured practice

Syllabus and review details

This lesson covers the listed H2 Physics 9478 outcomes. Flux uses area perpendicular to B; flux linkage is NΦ for N linked turns. Faraday's law uses the rate of change of linkage and Lenz's law fixes polarity from the change being opposed. The simple Blv motional-e.m.f. form requires mutually perpendicular conductor length, velocity and uniform field. Ideal transformer ratios assume common linked flux, alternating operation and no winding or core losses. More advanced induction applications are not required here.

  • GCE A-Level H2 PhysicsTopic 18(g) · 2027Checked against the syllabus · partial topic coverageOfficial 9478 syllabus
Course and syllabus information
Course
GCE A-Level H2 Physics
Edition
GCE A-Level H2 Physics 2027