Transformers: IGCSE Physics questions with answers
Free Transformers questions for IGCSE Physics (0625), part of Electricity & magnetism. Each question has the answer and the marking points an examiner would look for.
A step-down transformer is 100 % efficient. The output current is larger than the input current. Why?
A The input power equals the output power, so a lower voltage needs a larger current.
B The primary loses current as heat.
C The secondary has more turns.
D Current is created in the core.
Show answer
Answer: A
Question 2Multiple choice · 1 mark
A transformer has 200 turns on the primary coil and 50 turns on the secondary coil. The input voltage is 6 V. What is the output voltage?
A 1.5 V
B 6 V
C 24 V
D 300 V
Show answer
Answer: A
Question 3Structured · 4 marksA* challenge
Electricity is transmitted over long distances at very high voltages.
(a)
Explain why a high voltage reduces energy losses in the cables.
[3]
(b)
State the type of transformer used at a power station.
[1]
Show answers and mark scheme
Part
Answer and marking points
Marks
(a)
smaller current, less I²R heating
for the same power, a higher voltage means a smaller current B1 power lost in cables = I²R B1 so less energy is wasted as heat B1
3
(b)
step-up
step-up transformer B1
1
Question 4Structured · 8 marks
A generator supplies 50 kW of electrical power to a village through cables with a total resistance of 2.0 Ω. Transformers are used to change the voltage at which the power is transmitted.
(a)
Calculate the current in the cables and the power lost in them if the power is transmitted at 2500 V.
[3]
(b)
The generator output is 2500 V. Calculate the turns ratio of the transformer needed to step it up to 11000 V, assuming it is 100% efficient.
[1]
(c)
Explain how a transformer produces an alternating output voltage in the secondary coil.
[3]
(d)
Explain why a transformer does not work with a steady direct current.
[1]
Show answers and mark scheme
Part
Answer and marking points
Marks
(a)
20.0 A; 800 W
I = P/V = 20.0 A C1 P = I²R C1 800 W A1
3
(b)
1 : 4.40
Np : Ns = 2500 : 11000 = 1 : 4.40 B1
1
(c)
alternating current in primary; changing magnetic field in iron core; induces e.m.f. in secondary
an alternating current in the primary coil produces a changing magnetic field B1 the soft-iron core carries the changing field to the secondary coil B1 a changing field linking the secondary induces an (alternating) e.m.f. B1
3
(d)
no changing magnetic field, so no induced e.m.f.
a d.c. gives a constant magnetic field, so no e.m.f. is induced B1
1
Question 5Structured · 5 marks
Describe the structure of a simple transformer.
(a)
Name the three main parts.
[2]
(b)
Explain how a transformer produces an output voltage.
[3]
Show answers and mark scheme
Part
Answer and marking points
Marks
(a)
primary coil, secondary coil, iron core
Any 2 of: primary coil; secondary coil; soft-iron core (all three = 2) B2
2
(b)
a.c. → changing field in core → induced e.m.f.
alternating current in the primary produces a changing magnetic field B1 the core carries the field to the secondary B1 the changing field induces an e.m.f. in the secondary B1
3
Question 6Structured · 10 marks
Fig. 6.1 shows a transformer.
Fig. 6.1
(a)
Explain why a transformer only works with alternating current.
[2]
(b)
Use the information in Fig. 6.1 to calculate the output voltage.
[2]
(c)
State why the core is made of soft iron.
[1]
(d)
A power station transmits 2.0 × 106 W at 132 kV through cables of total resistance 10 Ω. Calculate the power lost in the cables.
[3]
(e)
Explain why electrical energy is transmitted at very high voltage.
[2]
Show answers and mark scheme
Part
Answer and marking points
Marks
(a)
a.c. produces a changing magnetic field in the core; changing field induces e.m.f. in secondary
a.c. in the primary produces a changing magnetic field in the iron core B1 the changing field induces an e.m.f. in the secondary coil (d.c. gives a constant field) B1
2
(b)
24.0 V
Vs/Vp = Ns/Np C1 24.0 V A1
2
(c)
easily magnetised and demagnetised
iron is easily magnetised and demagnetised / concentrates the magnetic field B1
1
(d)
2300 W
I = P/V = 15.2 A C1 P = I²R C1 2300 W A1
3
(e)
smaller current for same power; less energy lost as heat in cables (I²R)
for the same power a higher voltage means a smaller current B1 power loss is proportional to I² so less energy is wasted heating the cables B1
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Original Cambridge-style practice questions written for mentorriyaz.in. These are not Cambridge past paper questions, and this site is not affiliated with or endorsed by Cambridge University Press & Assessment. Updated 2026-11-04. Paper generator · mentorriyaz.in