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Workings of a Transformer

Key idea: O Level electromagnetic induction: how an iron-cored transformer works, step-up/step-down equations, and why high voltage transmission reduces cable losses.

  • SEC G3 Physics 2027
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Learning objectives

  • State the properties of magnets
  • Describe induced magnetism
  • Distinguish temporary and permanent magnets
  • Determine magnetic-field direction with a compass or bar magnet
  • Interpret bar-magnet field patterns
  • Draw the magnetic field pattern around a bar magnet and between the poles of two bar magnets
  • Interpret the field pattern around a straight current-carrying wire
  • Draw the magnetic field pattern around a straight current-carrying wire
  • Interpret the field pattern around a current-carrying solenoid
  • Draw the magnetic field pattern around a current-carrying solenoid
  • Relate current magnitude and direction to magnetic field
  • Describe electromagnet applications
  • Describe experiments showing the force on a current-carrying conductor in a magnetic field
  • Describe magnetic force on a charged-particle beam
  • Predict force reversal when current or field reverses
  • Use Fleming’s left-hand rule
  • Explain the turning effect on a current-carrying coil
  • Explain how current and turns increase the turning effect
  • Describe split-ring commutator action
  • Describe the effect of winding a motor coil on a soft-iron cylinder
  • Deduce that a changing magnetic field can induce an e.m.f.
  • Deduce that induced e.m.f. opposes the change producing it
  • Deduce factors affecting induced e.m.f. magnitude
  • Describe a simple a.c. generator and slip rings
  • Sketch a simple a.c. generator voltage–time graph
  • Describe a simple iron-cored transformer
  • Apply ideal-transformer equations
  • Explain cable loss and high-voltage transmission

1. Definition

A. Transformer

A transformer changes an a.c. voltage using electromagnetic induction.

B. Step-up and step-down

  • Step-up: higher output voltage.
  • Step-down: lower output voltage.
Transformer principle and high-voltage transmissionAn iron-cored transformer labels primary and secondary turns, voltages and currents. A reasoning chain shows that higher transmission voltage gives lower current and lower resistive cable loss for fixed power and resistance.How the secondary e.m.f. is producedPrimaryVP, IP, NPalternating inputSecondaryVS, IS, NSinduced outputchanging magnetic fluxlaminated soft-iron coreWhy transmit at high voltage?Compare at fixed transmitted power P and fixed cable resistance R:higher voltage Vstep-up transformerlower current Ibecause P = VIlower cable lossPloss = I²R
Scroll diagram horizontally to read all labels.
An alternating primary current produces changing flux in the iron core and an induced secondary e.m.f. For fixed transmitted power, stepping voltage up reduces current and therefore reduces I²R cable loss.
What you need for this course

You should be able to describe a simple iron-cored transformer, apply the ideal transformer equations and explain why high-voltage transmission reduces cable losses.

2. Key Ideas

  • Transformers need a.c. (changing current → changing magnetic field). A steady d.c. does not produce continuous induction.
  • Primary coil: input V_P, turns N_P. Secondary coil: output V_S, turns N_S.
  • Ideal transformer equations:

V_P/V_S = N_P/N_S For an ideal transformer only:

V_P I_P = V_S I_S

  • Step-up: N_S > N_P so V_S > V_P.
    Step-down: N_S < N_P so V_S < V_P.

3. Detailed Explanations

A. How a transformer works (principle)

  1. An a.c. current in the primary coil produces a changing magnetic field in the iron core.
  2. The changing magnetic field links the secondary coil.
  3. This changing flux induces an e.m.f. (voltage) in the secondary coil.

B. Turns ratio controls voltage ratio

More turns on the secondary coil gives a larger induced voltage.

C. Energy loss in cables and high-voltage transmission

Power transmitted is P = VI. Compare transmission systems carrying the same power through cables with the same resistance R.

For fixed transmitted power:

  • higher voltage means smaller current, since I = P/V
  • cable heating power is Pₗₒₛₛ = I²R
  • therefore smaller current gives much less heating loss in the same cables

So power is transmitted at high voltage and then stepped down near homes for safety and use.

Ideal versus real

V_P I_P = V_S I_S assumes no energy loss. A real transformer has V_S I_S < V_P I_P because some energy is dissipated, for example by heating in the coils and core.

Explore the transformer

Use the Motor, Generator & Transformer Lab to compare primary and secondary turns, then predict whether each setup is step-up or step-down.

4. Common Mistakes

  • Using transformer equations for d.c. (needs changing flux, so use a.c.).
  • Mixing up N_P and N_S in the turns ratio.
  • Saying a step-up transformer “increases power”. In an ideal transformer, power in equals power out; in a real transformer, output power is lower.
  • Saying high voltage alone reduces loss. The comparison must keep transmitted power and cable resistance fixed: high V gives low I, so I²R is smaller.

5. Exam Tips

  1. Write the ratio equation first: V_P/V_S = N_P/N_S.
  2. If current is asked, use V_P I_P = V_S I_S.
  3. For transmission: “for the same transmitted power, high V → low I; for the same cable resistance, Pₗₒₛₛ = I²R is smaller”.

6. Worked Examples

Modelled example 1

Voltage ratio

Core

Problem

An ideal transformer has N_P = 200, N_S = 50 and V_P = 240 V. Find V_S.
Study the worked solution

  1. Pair corresponding quantities

    Method

    Write V_P/V_S = N_P/N_S.

    Reason

    Primary voltage must align with primary turns, and secondary with secondary.

    Working

    240/V_S = 200/50 = 4

  2. Solve for secondary voltage

    Method

    Divide the primary voltage by 4.

    Reason

    The secondary has one quarter as many turns.

    Working

    V_S = 240/4 = 60 V

Guided practice 2

Current ratio

About 5 min

Problem

For the ideal transformer in Example 1, V_P = 240 V, V_S = 60 V and I_S = 2.0 A. Find I_P.

Use ideal input–output power equality

Unit: A

Hints

Hint 1: ideal condition
V_PI_P = V_SI_S.
Hint 2: rearrange
I_P = V_SI_S/V_P.
View solution step by step

  1. Apply power balance

    Method

    Equate ideal input and output power.

    Reason

    An ideal transformer has no energy loss.

    Working

    V_PI_P = V_SI_S

  2. Calculate primary current

    Method

    Substitute and divide by V_P.

    Reason

    The known secondary power is 60 × 2.0 W.

    Working

    I_P = (60)(2.0)/240 = 0.50 A

Common misconception 3

Step-up or step-down?

Find and correct the mistake

Learner response

A transformer has N_P = 400 and N_S = 1200. A learner calls it step-up and concludes that an ideal transformer produces three times as much output power. Diagnose the response.

Separate voltage ratio from power balance

View solution step by step

  1. Classify the voltage change

    Method

    Confirm that it is step-up and V_S = 3V_P.

    Reason

    The secondary has three times as many turns.

    Working

    V_S/V_P = 1200/400 = 3

  2. Correct the power claim

    Method

    State that ideal output power equals input power, not three times it.

    Reason

    The current changes inversely so V_PI_P = V_SI_S.

    Working

    V_S = 3V_P ⇒ I_S = I_P/3

Examiner practice 4

Finding turns

3 marks

Examination question

An ideal transformer steps 12 V up to 240 V. Its primary has 50 turns. Find the secondary turns. [3 marks]

Keep primary quantities on the same side

View solution step by step

  1. Set up the ratio

    1 mark

    Method

    Pair voltage and turns for each coil.

    Reason

    Corresponding primary and secondary ratios are equal.

    Working

    12/240 = 50/N_S

  2. Rearrange and calculate

    2 marks

    Method

    Multiply 50 by 240/12.

    Reason

    The voltage is stepped up by factor 20, so turns must be too.

    Working

    N_S = 50(240/12) = 1000 turns

Challenge 5

Transmission current comparison

Minimal support

Transmission transfer

20 kW is transmitted through the same cables at (i) 200 V and (ii) 20 kV. Compare current and cable-heating loss.

Hold transmitted power and cable resistance fixed

Hints

Hint 1: current
Convert 20 kW to 20000 W and use I = P/V.
Hint 2: loss
For the same cable resistance, compare I²R using the current ratio.
View solution step by step

  1. Calculate both currents

    Method

    Divide fixed power by each voltage.

    Reason

    P = VI for the transmitted power.

    Working

    I₂₀₀ = 20000/200 = 100 A; I_(20 000) = 20000/20000 = 1 A

  2. Compare the current

    Method

    State that high voltage reduces current by factor 100.

    Reason

    The voltage is 100 times larger for the same power.

    Working

    (I_(high V))/(I_(low V)) = 1/100

  3. Compare cable heating

    Method

    State that loss becomes 1/10000 as large.

    Reason

    Pₗₒₛₛ = I²R and the same cables have the same R.

    Working

    (P_(loss,high V))/(P_(loss,low V)) = (1/100)² = 1/10000

7. Mind Stretchers

Mind stretcher 1: Why step up then step down?Extension

Why are transmission lines typically stepped up and then stepped down again near homes?

Show Answer

Stepping up increases voltage and reduces current for the same power, so cable heating losses are smaller during transmission. Near homes, the voltage is stepped down to safer, usable levels.

Mind stretcher 2: Why d.c. does not workExtension

Why does a transformer not work properly with a steady d.c. supply?

Show Answer

Steady d.c. produces a (nearly) constant magnetic field in the core, so there is no changing flux to induce a voltage in the secondary coil.

8. Practice and next step

Solve a turns-ratio and transmission-loss comparison in the Motor–Generator–Transformer Lab, then complete the Magnetism Structured Practice.

Continue with the next resource in this course.

Course and syllabus information
Course
SEC G3 Physics
Edition
SEC G3 Physics 2027