IGCSE Physics Transformer Questions and Worked Solutions (4.5.6) | 0625

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IGCSE Physics 0625 · Topic 4.5.6

IGCSE Physics Transformer Questions and Worked Solutions

50 topic-wise practice questions on The Transformer (Unit 4: Electricity and Magnetism), written to the Cambridge IGCSE Physics 0625 syllabus for 2026–2028, each with a step-by-step worked solution.

50 questions21 Core19 Extended10 Challenge136 marksTextbook §4.5 · pp. 229–250
Questions tagged Core are for all candidates (Papers 1 & 3). Questions tagged Extended are Supplement content (Papers 2 & 4). Questions 41–50 are harder Challenge questions. Try each question first, then tap Show solution to check your working.

Topic guide: The Transformer

A transformer changes the size of an alternating p.d. using two coils on a soft-iron core. Step-up transformers increase p.d.; step-down transformers decrease it. High-voltage transmission reduces energy losses in cables.

Key formulas

Vp ÷ Vs = Np ÷ Nsp.d.s and number of turns on primary (p) and secondary (s)Core
Ip Vp = Is Vs100% efficient transformer (power in = power out)Extended
P = I² Rpower lost in transmission cablesExtended

Prerequisite knowledge

  • Electromagnetic induction (4.5.1)
  • Power P = IV (4.2.5)
  • Ratios

Exam tips & common mistakes

  • Check: step-up means Vs > Vp
  • If voltage goes up by a factor, current goes down by the same factor
  • Halving current reduces losses to a quarter

Key facts from the syllabus (4.5.6)

  • Construction: primary coil, secondary coil, laminated soft-iron core
  • Step-up: more secondary turns; step-down: fewer secondary turns
  • Transformers only work with a.c.
  • High-voltage transmission: step up at the power station, step down near homes
  • Advantages: smaller current, less energy lost as heat, thinner cables
  • Extended a.c. in the primary → changing magnetic field in the core → induced alternating e.m.f. in the secondary
  • Extended for the same power, higher voltage → lower current → much smaller I²R loss

Textbook: §4.5 Electromagnetic effects, pp. 229–250

1
Core[2 marks]

Describe the construction of a simple transformer.

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Method: Two coils, soft-iron core.

A primary coil and a secondary coil wound on a laminated soft-iron core
2
Core[1 mark]

State the difference between a step-up and a step-down transformer.

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Method: Ns vs Np.

Step-up: more turns on the secondary (increases p.d.); step-down: fewer turns on the secondary (decreases p.d.)
3
Core[2 marks]

A transformer has 200 primary turns and 1000 secondary turns. The primary p.d. is 12 V. Calculate the secondary p.d.

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Method: Vp/Vs = Np/Ns.

Vs = 12 × 1000 ÷ 200 = 60.0 V
4
Core[2 marks]

A transformer steps 230 V down to 11.5 V. The primary has 4000 turns. Calculate the secondary turns.

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Method: Ns = Np × Vs/Vp.

Ns = 4000 × 11.5 ÷ 230 = 200
5
Core[2 marks]

A transformer has 500 primary and 25 secondary turns. The primary p.d. is 230 V. Calculate the secondary p.d.

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Method: Vs = Vp × Ns/Np.

Vs = 230 × 25 ÷ 500 = 11.5 V
6
Core[2 marks]

State whether the transformer in the previous question is step-up or step-down.

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Method: Ns < Np.

Step-down
7
Core[2 marks]

Describe the use of transformers in high-voltage transmission of electricity.

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Method: Step-up at station, step-down near homes.

Step-up transformers raise the p.d. at the power station for transmission; step-down transformers reduce it for homes
8
Core[2 marks]

State two advantages of high-voltage transmission.

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Method: Lower current, less loss.

Lower current, so less energy lost as heat in cables; thinner, cheaper cables can be used
9
Core[2 marks]

Explain why transformers do not work with a steady d.c. supply.

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Method: No changing field.

A steady current produces a constant magnetic field, so no e.m.f. is induced in the secondary
10
Core[2 marks]

Explain why the core of a transformer is made of soft iron.

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Method: Easily magnetised and demagnetised.

Soft iron is easily magnetised and demagnetised as the current alternates, linking the field to the secondary
11
Core[3 marks]

A phone charger transformer has 2300 primary turns and reduces 230 V to 5.0 V. Calculate the number of secondary turns.

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Method: Ns = Np × Vs/Vp.

Ns = 2300 × 5.0 ÷ 230 = 50
12
Core[2 marks]

State the purpose of a transformer.

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Method: Change a.c. p.d.

To change the size of an alternating p.d.
13
Core[2 marks]

A transformer has 100 primary turns and 2500 secondary turns. State whether it is step-up or step-down.

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Method: Ns > Np.

Step-up
14
Core[2 marks]

A transformer steps 11 kV up to 132 kV. The primary has 500 turns. Calculate the secondary turns.

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Method: Ns = Np × Vs/Vp.

Ns = 500 × 132 ÷ 11 = 6000
15
Core[2 marks]

A step-down transformer has a turns ratio 20:1. The input is 240 V. Calculate the output.

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Method: Divide.

Vs = 12.0 V
16
Core[2 marks]

A transformer has 1200 primary turns and 60 secondary turns. The secondary p.d. is 11.5 V. Calculate the primary p.d.

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Method: Vp = Vs × Np/Ns.

Vp = 230 V
17
Core[2 marks]

State where step-up transformers are used in the National Grid.

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Method: Power station.

At the power station, to raise the p.d. for transmission
18
Core[2 marks]

State where step-down transformers are used in the National Grid.

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Method: Substations.

At local substations, to reduce the p.d. for homes and businesses
19
Core[2 marks]

State why electricity is transmitted at high voltage.

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Method: Reduce losses.

To reduce the current and so the energy lost as heat in the cables
20
Core[2 marks]

Explain why the primary and secondary coils are not connected electrically.

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Method: Linked by field.

Energy is transferred by the changing magnetic field in the core, not by current flowing between them
21
Core[2 marks]

State the material used for the core of a transformer.

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Method: Soft iron.

Soft iron (laminated)
22
Extended[3 marks]

Explain the principle of operation of a transformer.

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Method: Changing field induces e.m.f.

Alternating current in the primary produces a changing magnetic field in the iron core; this changing field links the secondary and induces an alternating e.m.f.
23
Extended[1 mark]

State the equation for a 100% efficient transformer.

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Method: Power in = power out.

IpVp = IsVs
24
Extended[3 marks]

A 100% efficient transformer has Vp = 230 V, Vs = 12 V and Is = 4.6 A. Calculate Ip.

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Method: Ip = IsVs/Vp.

Ip = 4.6 × 12 ÷ 230 = 0.240 A
25
Extended[3 marks]

A transformer steps up 25 kV to 400 kV and is 100% efficient. The primary current is 800 A. Calculate the secondary current.

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Method: Is = IpVp/Vs.

Is = 800 × 25 ÷ 400 = 50.0 A
26
Extended[2 marks]

Calculate the power lost in a cable of resistance 5.0 Ω carrying 50 A.

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Method: P = I²R.

P = 50² × 5.0 = 12500 W
27
Extended[4 marks]

1.0 MW is transmitted through cables of resistance 4.0 Ω at (a) 10 kV and (b) 100 kV. Calculate the power loss in each case.

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Method: I = P/V, loss = I²R.

(a) I = 100 A, loss = 40000 W; (b) I = 10 A, loss = 400 W
28
Extended[2 marks]

Using P = I²R, explain why power losses in cables are smaller at higher voltage.

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Method: Higher V → lower I.

For the same power, higher voltage means a smaller current; loss depends on I², so it is much smaller
29
Extended[2 marks]

The current in a transmission cable is halved. State the factor by which the power loss changes.

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Method: P ∝ I².

It becomes one quarter
30
Extended[3 marks]

A transformer with 100% efficiency gives 6.0 V and 2.0 A from a 240 V supply. Calculate the primary current and the output power.

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Method: P = IV.

P = 12.0 W; Ip = 12 ÷ 240 = 0.0500 A
31
Extended[3 marks]

Explain why an alternating current is needed in the primary coil.

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Method: Changing field.

Only a changing current produces a changing magnetic field, which induces an e.m.f. in the secondary
32
Extended[3 marks]

An ideal transformer has Vp = 240 V, Ip = 0.50 A and Vs = 12 V. Calculate Is.

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Method: IpVp = IsVs.

Is = 120 ÷ 12 = 10.0 A
33
Extended[3 marks]

An ideal transformer delivers 60 W at 24 V from a 230 V supply. Calculate the primary and secondary currents.

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Method: P = IV.

Is = 2.50 A; Ip = 0.261 A
34
Extended[3 marks]

A transmission line of resistance 10 Ω carries 200 A. Calculate the power lost.

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Method: I²R.

P = 400000 W
35
Extended[4 marks]

500 kW is transmitted through cables of total resistance 2.0 Ω at 25 kV. Calculate the current, the power lost and the percentage lost.

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Method: I = P/V.

I = 20.0 A; loss = 800 W; 0.160%
36
Extended[4 marks]

The same 500 kW is sent at 5.0 kV through the same 2.0 Ω cables. Calculate the power lost and compare.

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Method: I = 100 A.

Loss = 20000 W, 25 times more than at 25 kV
37
Extended[2 marks]

Explain why the transformer core is laminated.

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Method: Reduce eddy currents.

Thin insulated layers reduce eddy currents, so less energy is wasted as heat
38
Extended[2 marks]

State the relationship between voltage and current for an ideal step-up transformer.

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Method: Voltage up, current down.

The voltage increases by the same factor that the current decreases
39
Extended[3 marks]

A transformer has 400 primary and 1600 secondary turns. Vp = 50 V and Ip = 2.0 A (100% efficient). Calculate Vs and Is.

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Method: Ratios.

Vs = 200 V; Is = 0.500 A
40
Extended[2 marks]

Explain why transformers become warm in use.

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Method: Losses.

Current in the coils' resistance and eddy currents in the core produce heat (they are not 100% efficient)

Challenge questions

Harder, multi-step questions at the top end of the exam. Attempt these once you are confident with the questions above.

41
ChallengeCore[4 marks]

A transformer must convert 230 V mains to 6.0 V for a doorbell. The primary has 2300 turns. Calculate the secondary turns and state the type of transformer.

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Method: Ns = Np × Vs/Vp.

Ns = 60; step-down
42
ChallengeCore[4 marks]

Describe the construction of a transformer and explain why it will not work with a steady d.c. supply.

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Method: Coils, core.

Primary and secondary coils wound on a laminated soft-iron core; d.c. produces a constant field so no e.m.f. is induced in the secondary
43
ChallengeCore[4 marks]

Explain how transformers are used in transmitting electricity from a power station to homes, stating why the voltage is changed at each stage.

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Method: Step-up, step-down.

A step-up transformer raises the p.d. (to e.g. 400 kV) for transmission so the current and energy losses are small; step-down transformers reduce it in stages to 230 V for safe use in homes
44
ChallengeExtended[5 marks]

An ideal transformer has 800 primary turns and 40 secondary turns. The primary is connected to 240 V and the secondary supplies a 6.0 Ω heater. Calculate Vs, Is, the power and Ip.

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Method: Ratios and P = IV.

Vs = 12.0 V; Is = 2.00 A; P = 24.0 W; Ip = 0.100 A
45
ChallengeExtended[5 marks]

A power station supplies 20 MW. Compare the power lost in cables of total resistance 5.0 Ω if transmitted at 25 kV and at 400 kV.

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Method: I = P/V; P = I²R.

25 kV: I = 800 A, loss = 3.2 × 10⁶ W; 400 kV: I = 50 A, loss = 12500 W
46
ChallengeExtended[4 marks]

Calculate the percentage of the 20 MW lost in each case in the previous question and explain why high voltage is used.

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Method: Percentages.

16.0% and 0.062%; high voltage reduces the current, and losses depend on I²
47
ChallengeExtended[4 marks]

A transformer has an input of 230 V, 0.50 A and an output of 12 V, 8.5 A. Calculate its efficiency and suggest two causes of energy loss.

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Method: Pout/Pin.

Efficiency = 102 ÷ 115 × 100 = 88.7%; heating of the coils (resistance) and eddy currents in the core
48
ChallengeExtended[4 marks]

Explain the principle of operation of a transformer step by step.

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Method: Changing field.

An alternating current in the primary produces an alternating magnetic field in the soft-iron core; the changing field links the secondary coil and induces an alternating e.m.f. in it; the ratio of turns sets the ratio of voltages
49
ChallengeExtended[4 marks]

A step-up transformer increases the voltage by a factor of 50. If 100% efficient, explain what happens to the current and why this does not violate conservation of energy.

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Method: IpVp = IsVs.

The current decreases by a factor of 50; power in equals power out, so no energy is created
50
ChallengeExtended[4 marks]

The current in a transmission line is reduced from 400 A to 100 A while the cable resistance stays 8.0 Ω. Calculate the power loss before and after.

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Method: P = I²R.

Before: 1.3 × 10⁶ W; after: 8.0 × 10⁴ W (16 times smaller)
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