The transformer
Cambridge IGCSE Physics (0625) · Unit 4: Electricity and magnetism · 16 flashcards
The transformer is topic 4.5.6 in the Cambridge IGCSE Physics (0625) syllabus , positioned in Unit 4 — Electricity and magnetism , alongside Simple phenomena of magnetism, Electric charge and Electric current.
This topic is examined in Paper 1 (multiple-choice) and Papers 3/4 (theory), plus Paper 5 or Paper 6 (practical / alternative to practical).
The deck below contains 16 flashcards — covering the precise wording mark schemes reward.
What the Cambridge 0625 syllabus says
Official 2026-2028 specThese are the exact learning objectives Cambridge sets for this topic. Match the command word (Describe, Explain, State, etc.) in your answer to score full marks.
- Describe Describe the construction of a simple transformer with a soft-iron core, as used for voltage transformations
- Use Use the terms primary, secondary, step-up and step-down
- Recall Recall and use the equation VN p VN p where p and s refer to primary and secondary
- Describe Describe the use of transformers in high-voltage transmission of electricity
- State State the advantages of high-voltage transmission
- Explain Explain the principle of operation of a simple iron-cored transformer Supplement
- Recall Recall and use the equation for 100% efficiency in a transformer IV = IV where p and s refer to primary and secondary Supplement
- Recall Recall and use the equation P = I2R to explain why power losses in cables are smaller when the voltage is greater Supplement
Describe the key components and construction of a simple transformer used for voltage transformation.
A simple transformer consists of two coils of insulated wire, the primary and secondary coils, wound around a shared soft iron core. The soft iron core concentrates the magnetic field, improving the efficiency of the transformer. The primary coil is connected to the input AC voltage, while the secondary coil provides the output AC voltage.
State two reasons why soft iron is used for the core of a transformer.
1. Soft iron is easily magnetised and demagnetised, reducing energy losses due to hysteresis.
2. Soft iron has a high permeability, meaning it concentrates the magnetic field lines effectively, improving the magnetic coupling between the primary and secondary coils.
A transformer has 500 turns on its primary coil and 50 turns on its secondary coil. If the input voltage to the primary coil is 240 V, calculate the output voltage from the secondary coil. State whether this is a step-up or step-down transformer.
Output Voltage = (Number of Secondary Turns / Number of Primary Turns) * Input Voltage
Output Voltage = (50 / 500) * 240 V = 24 V
This is a step-down transformer because the output voltage is lower than the input voltage.
Explain the difference between a step-up transformer and a step-down transformer, referring to the number of turns on the primary and secondary coils.
A step-up transformer has more turns on the secondary coil than on the primary coil. This increases the voltage. A step-down transformer has fewer turns on the secondary coil than on the primary coil. This decreases the voltage.
A transformer has 500 turns on its primary coil and 25 turns on its secondary coil. If the primary voltage is 240V, calculate the secondary voltage.
Formula: Vp/Vs = Np/Ns
Rearrange: Vs = Vp * (Ns/Np)
Vs = 240V * (25/500)
Vs = 12V
Answer: The secondary voltage is 12V. This shows the voltage is reduced because there are fewer turns on the secondary coil (step-down transformer).
A transformer has 2 turns on its secondary coil and 500 turns on the primary coil. If the primary voltage is 100V, state what type of transformer this is and explain how this affects the secondary voltage.
This is a step-down transformer because the number of turns on the secondary coil (2 turns) is less than the number of turns on the primary coil (500 turns).
Since Vp/Vs = Np/Ns, Vs = Vp * (Ns/Np). Therefore Vs = 100V * (2/500) = 0.4V.
The step-down transformer reduces the secondary voltage.
A transformer steps down the voltage for domestic use. The power input to the transformer is 2000 W at 250 V. The voltage output is 25 V. Assuming the transformer is 100% efficient, calculate the current in the secondary coil.
Step 1: Calculate the secondary power: Since efficiency is 100%, secondary power = primary power = 2000 W.
Step 2: Use the power equation: P = VI, so I = P/V
Step 3: Calculate the secondary current: I = 2000 W / 25 V = 80 A
Answer: The current in the secondary coil is 80 A.
Explain why high-voltage transmission of electricity is more efficient than low-voltage transmission.
High-voltage transmission reduces current for a given power output (P = VI). Lower current reduces the energy lost as heat in the transmission cables (P = I^2R, where R is the resistance of the cables). Reducing heat loss increases the overall efficiency of the transmission.
State two advantages of transmitting electrical power at high voltage.
1. Reduces current in the cables: (P = IV, higher V means lower I for same power)
2. Reduces power loss/heating in the cables: (Power loss ∝ I²R, lower I means less power loss)
Explain why electrical power is transmitted at high voltage over long distances.
Transmitting power at high voltage reduces the current in the wires (since P = IV). A lower current reduces the power lost as heat in the wires (since Power loss = I²R). Therefore, high-voltage transmission is more efficient over long distances.
Explain how a simple iron-cored transformer works to step down voltage. Include in your answer the role of the primary coil, secondary coil, and iron core.
1. Alternating current in the primary coil: An alternating current flowing through the primary coil creates a changing magnetic field.
2. Iron core: The iron core concentrates and directs this changing magnetic field towards the secondary coil. Iron is easily magnetised, improving efficiency.
3. Induced current in the secondary coil: The changing magnetic field lines cut through the secondary coil, inducing an alternating voltage (and thus current if a circuit is connected). The voltage is stepped down because the secondary coil has fewer turns than the primary coil. The ratio of turns dictates the voltage transformation.
A transformer has 500 turns on its primary coil and 100 turns on its secondary coil. If the primary voltage is 240V, what is the secondary voltage?
Formula: Vp / Vs = Np / Ns
Where:
Vp = primary voltage
Vs = secondary voltage
Np = number of turns in the primary coil
Ns = number of turns in the secondary coil
Working:
240 / Vs = 500 / 100
240 / Vs = 5
Vs = 240 / 5
Vs = 48V
Answer: The secondary voltage is 48V.
A transformer has a primary voltage of 240V and a primary current of 2.0A. Assuming the transformer is 100% efficient, calculate the secondary current if the secondary voltage is 12V.
Formula: VpIp = VsIs
Rearrange: Is = (VpIp) / Vs
Is = (240V * 2.0A) / 12V
Is = 40A
Answer: The secondary current is 40A. This calculation uses the principle of conservation of energy and assumes all power input to the primary coil is transferred to the secondary coil in an ideal transformer.
Explain why the product of voltage and current (V x I) is the same for both primary and secondary coils in a 100% efficient transformer. State the condition that makes this true.
In a 100% efficient transformer, there are no energy losses (
A power cable has a resistance of 2.0 Ω. It carries a current of 10.0 A. Calculate the power loss in the cable.
Formula: P = I²R
Calculation: P = (10.0 A)² * 2.0 Ω = 200 W
Answer: 200 W
Explanation: The power loss is due to the resistance of the cable converting electrical energy into heat.
Explain why transmitting electrical power at a higher voltage reduces power losses in the cables, given P=I²R. Assume the power transmitted remains constant.
If the power (P) transmitted is constant and voltage (V) is increased, the current (I) must decrease (since P=VI). According to P=I²R, if the current (I) decreases, the power loss due to heating in the cable (P) decreases significantly because the current is squared.
More topics in Unit 4 — Electricity and magnetism
The transformer sits alongside these Physics decks in the same syllabus unit. Each uses the same spaced-repetition system, so progress in one informs the next.
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