IGCSE Physics a.c. Generator Questions and Worked Solutions (4.5.2) | 0625

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IGCSE Physics 0625 · Topic 4.5.2

IGCSE Physics a.c. Generator Questions and Worked Solutions

50 topic-wise practice questions on The a.c. Generator (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 questions0 Core40 Extended10 Challenge129 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 a.c. Generator

(Extended) An a.c. generator rotates a coil in a magnetic field (or a magnet near a coil) to produce an alternating e.m.f. Slip rings and brushes connect the rotating coil to the circuit.

Key formulas

f = rotations per secondfrequency of the outputExtended
T = 1 ÷ fperiod of one cycleExtended

Prerequisite knowledge

  • Electromagnetic induction (4.5.1)
  • a.c. (4.2.2)

Exam tips & common mistakes

  • Slip rings (generator) ≠ split-ring commutator (motor)
  • Doubling speed doubles BOTH peak e.m.f. and frequency
  • Relate peaks and zeros to coil positions

Key facts from the syllabus (4.5.2)

  • A simple a.c. generator: coil rotating between magnet poles, slip rings and brushes
  • Rotating-magnet generators need no slip rings
  • The e.m.f.–time graph is sinusoidal
  • e.m.f. is maximum when the plane of the coil is parallel to the field (sides cutting field lines fastest)
  • e.m.f. is zero when the plane of the coil is perpendicular to the field
  • Faster rotation increases both the peak e.m.f. and the frequency

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

1
Extended[3 marks]

Describe a simple a.c. generator with a rotating coil.

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Method: Coil, magnets, slip rings, brushes.

A coil rotates between the poles of a magnet; an e.m.f. is induced as it cuts the field; the coil is connected to the circuit through slip rings and carbon brushes
2
Extended[2 marks]

State the purpose of slip rings and brushes in an a.c. generator.

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Method: Continuous connection.

To connect the rotating coil to the external circuit without the wires twisting
3
Extended[2 marks]

Explain why the output of a rotating-coil generator is alternating.

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Method: Sides move up then down.

Each side of the coil moves up through the field then down, so the induced e.m.f. reverses every half turn
4
Extended[2 marks]

State the position of the coil when the induced e.m.f. is maximum.

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Method: Plane parallel to field.

When the plane of the coil is parallel to the magnetic field lines (sides cutting field fastest)
5
Extended[2 marks]

State the position of the coil when the induced e.m.f. is zero.

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Method: Plane perpendicular to field.

When the plane of the coil is perpendicular to the field lines (sides moving along the field)
6
Extended[2 marks]

Sketch and describe the graph of e.m.f. against time for a simple a.c. generator.

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Method: Sine curve.

A sine-shaped curve alternating between positive and negative peaks
7
Extended[3 marks]

A generator coil rotates 50 times per second. State the frequency of the a.c. and the time for one cycle.

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Method: f = rotations per second.

f = 50 Hz; T = 1 ÷ 50 = 0.0200 s
8
Extended[3 marks]

State and explain two changes to a generator that would double the peak e.m.f.

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Method: Double speed or turns.

Double the speed of rotation, or double the number of turns (or use a stronger field)
9
Extended[2 marks]

The rotation speed of a generator is doubled. State the effect on the peak e.m.f. and the frequency.

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Method: Both double.

Both the peak e.m.f. and the frequency double
10
Extended[2 marks]

Describe a generator in which a magnet rotates inside a fixed coil.

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Method: Rotating magnet.

A magnet rotates inside or near a stationary coil, so the field through the coil changes, inducing an alternating e.m.f.; no slip rings are needed
11
Extended[2 marks]

State one advantage of a rotating-magnet generator over a rotating-coil generator.

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Method: No slip rings.

No slip rings or brushes needed, so less wear and sparking
12
Extended[2 marks]

The peaks of an e.m.f.–time graph occur every 0.010 s, alternately positive and negative. Calculate the frequency.

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Method: Positive peaks every 0.020 s.

T = 0.020 s; f = 50.0 Hz
13
Extended[2 marks]

Explain why the e.m.f. changes sign every half revolution.

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Method: Each side's direction reverses.

Each side of the coil reverses its direction of motion through the field
14
Extended[3 marks]

An a.c. generator gives a peak e.m.f. of 12 V at 25 Hz. It is turned twice as fast. State the new peak e.m.f. and frequency.

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Method: Both double.

24 V and 50 Hz
15
Extended[2 marks]

State the energy transfer in an a.c. generator.

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Method: Kinetic → electrical.

Kinetic energy → electrical energy
16
Extended[2 marks]

A coil is at rest between the poles of a magnet. Explain why no e.m.f. is induced.

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Method: No change.

The field linking the coil is not changing
17
Extended[2 marks]

Explain why the brushes in a generator are made of carbon.

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Method: Conducts, low friction.

Carbon conducts electricity and slides easily over the slip rings with little wear
18
Extended[2 marks]

State how many times per rotation the e.m.f. of a simple a.c. generator is zero.

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Method: Twice.

Twice
19
Extended[3 marks]

An e.m.f.–time graph shows one complete cycle in 40 ms. Calculate the frequency and the rotation rate.

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Method: f = 1/T.

f = 1 ÷ 0.040 = 25.0 Hz; 25 rotations per second
20
Extended[2 marks]

Explain how a bicycle dynamo is a type of a.c. generator.

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Method: Rotating magnet.

A magnet turned by the wheel rotates near a coil, inducing an alternating e.m.f.
21
Extended[2 marks]

Name the parts of a simple a.c. generator.

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Method: Coil, magnet, slip rings, brushes.

Coil, magnets, slip rings, brushes (and axle)
22
Extended[2 marks]

Explain why the e.m.f. is zero when the coil's plane is perpendicular to the field.

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Method: Not cutting.

The sides are moving parallel to the field lines, not cutting them
23
Extended[2 marks]

Explain why the e.m.f. is maximum when the coil's plane is parallel to the field.

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Method: Cutting fastest.

The sides cut the field lines at the fastest rate
24
Extended[3 marks]

A coil starts with its plane perpendicular to the field. Describe the e.m.f. over one full turn.

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Method: Sine from zero.

Zero, rises to a positive peak at 90°, zero at 180°, negative peak at 270°, zero at 360°
25
Extended[2 marks]

State the effect of using a stronger magnet in a generator.

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Method: Larger peak.

The peak e.m.f. increases (frequency unchanged)
26
Extended[2 marks]

State the effect of adding more turns to a generator coil.

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Method: Larger peak.

The peak e.m.f. increases
27
Extended[3 marks]

A generator turns 3000 times per minute. Calculate the frequency of the a.c.

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Method: Per second.

f = 3000 ÷ 60 = 50.0 Hz
28
Extended[2 marks]

The time period of a generator output is 0.040 s. Calculate the frequency.

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Method: f = 1/T.

f = 25.0 Hz
29
Extended[3 marks]

A generator gives a peak e.m.f. of 6.0 V at 10 Hz. Its speed is halved. State the new peak e.m.f. and frequency.

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Method: Both halve.

3.0 V and 5.0 Hz
30
Extended[2 marks]

Explain the difference between a generator and a motor in terms of energy transfer.

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Method: Reverse transfers.

A generator transfers kinetic to electrical energy; a motor transfers electrical to kinetic
31
Extended[2 marks]

State why power station generators usually rotate electromagnets inside fixed coils.

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Method: Avoid large currents through brushes.

The large output current does not need to pass through sliding contacts
32
Extended[2 marks]

State the energy source that turns the turbine and generator in a wind farm.

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Method: Wind.

Kinetic energy of the wind
33
Extended[3 marks]

Sketch the e.m.f.–time graph for a generator, then describe the effect of doubling both the speed and the number of turns.

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Method: Peak ×4, frequency ×2.

The peak e.m.f. becomes four times larger and the period halves (frequency doubles)
34
Extended[2 marks]

Explain why a generator connected to nothing (open circuit) is easier to turn than one connected to a lamp.

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Method: No current, no opposing force.

No current flows, so there is no opposing force from induced current
35
Extended[2 marks]

State what the brushes in an a.c. generator press against.

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Method: Slip rings.

The slip rings
36
Extended[3 marks]

An e.m.f.–time graph has peaks of +4.0 V and −4.0 V repeating every 0.10 s. State the peak e.m.f. and frequency.

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Method: Read graph.

Peak = 4.0 V; f = 10.0 Hz
37
Extended[2 marks]

Explain why the e.m.f. alternates even though the coil rotates in one direction.

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Method: Sides swap.

Each side moves up through the field then down, so the e.m.f. reverses each half turn
38
Extended[2 marks]

Explain why turning a generator faster makes it harder to turn when connected to a load.

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Method: Larger current.

A larger induced current produces a larger opposing force
39
Extended[2 marks]

State one similarity between an a.c. generator and a d.c. motor in their construction.

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Method: Coil and magnet.

Both have a coil rotating in a magnetic field with brushes
40
Extended[2 marks]

State one difference between an a.c. generator and a d.c. motor in their construction.

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Method: Slip rings vs commutator.

A generator uses slip rings; a d.c. motor uses a split-ring commutator

Challenge questions

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

41
ChallengeExtended[4 marks]

Describe the structure of a simple a.c. generator and explain the function of each part.

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Method: Parts.

Coil (cuts field, e.m.f. induced); magnets (field); slip rings (rotate with coil, continuous contact); brushes (connect to external circuit); axle (turned by an energy source)
42
ChallengeExtended[4 marks]

Sketch (describe) the e.m.f.–time graph for one rotation starting with the coil plane perpendicular to the field, and label the coil positions at zeros and peaks.

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Method: Sine starting at zero.

Zero at 0° and 180° (plane perpendicular to field), peaks at 90° (positive) and 270° (negative) where the plane is parallel to the field
43
ChallengeExtended[4 marks]

A generator gives a peak e.m.f. of 15 V at 50 Hz. It is changed to have 3 times as many turns and rotate at 2/3 of the speed. Calculate the new peak e.m.f. and frequency.

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Method: Peak ∝ turns × speed.

Peak = 15 × 3 × 2/3 = 30.0 V; f = 33.3 Hz
44
ChallengeExtended[4 marks]

The output of a generator has a period of 0.025 s. Calculate the frequency and the rotations per minute.

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Method: f = 1/T.

f = 40.0 Hz; 2400 rotations per minute
45
ChallengeExtended[4 marks]

Explain why a generator becomes harder to turn when a lamp is connected and switched on.

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Method: Lenz's law.

The induced current in the coil creates a force opposing the rotation, so more work must be done
46
ChallengeExtended[4 marks]

Explain why the e.m.f. is largest when the coil's plane is parallel to the field and zero when perpendicular.

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Method: Rate of cutting.

Parallel: the sides move at right angles to the field lines, cutting them fastest. Perpendicular: the sides move along the field lines, not cutting them
47
ChallengeExtended[3 marks]

Compare a rotating-coil generator with a rotating-magnet generator, giving one advantage of the second type.

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Method: No slip rings.

In the rotating-magnet type the coil is fixed, so no slip rings and brushes are needed — less wear and sparking and large currents are easier to take off
48
ChallengeExtended[4 marks]

A bicycle dynamo lights a lamp. Explain why the lamp flickers at very low speed but is steady and brighter at high speed.

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Method: Frequency and e.m.f.

At low speed the e.m.f. is small and the frequency low, so the lamp visibly dims between peaks; at high speed the e.m.f. is larger and alternates too quickly to see
49
ChallengeExtended[4 marks]

An e.m.f.–time graph peaks at +6.0 V and −6.0 V with 5 complete cycles in 0.20 s. State the peak e.m.f., period and frequency.

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Method: Read graph.

6.0 V; T = 0.0400 s; f = 25.0 Hz
50
ChallengeExtended[4 marks]

Explain the energy transfers in a power station generator and why it is not 100% efficient.

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Method: Kinetic → electrical; losses.

Kinetic energy of the turbine → electrical energy; some energy is wasted as heat in the coil resistance, friction and as sound
All IGCSE Physics topics← 4.5.1 Electromagnetic Induction4.5.3 Magnetic Effect of a Current →

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