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Cambridge IGCSE Physics revision notes

The transformer Revision Notes

Cambridge IGCSE revision notes for The transformer.

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The transformer

CambridgeIGCSEPhysicsElectricity and magnetism

Complete revision guide 1 of 8

Basic Transformer Construction and the Role of the Soft-Iron Core

This note explains the construction of a basic transformer, detailing its components and the function of the soft-iron core in changing voltage levels efficiently in alternating current (AC) circuits.

Syllabus source

Section 3, Topic 4.5.6, page 33, objective 4.5.6.1

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What you should be able to do

  • ✓Identify the main components of a basic transformer.
  • ✓Explain how the primary and secondary coils interact via electromagnetic induction.
  • ✓Describe the role of the soft-iron core in enhancing the magnetic field.
  • ✓Understand how transformers change voltage levels in AC circuits.
1

1. Components of a Basic Transformer

A transformer consists primarily of two coils of insulated wire: the primary coil and the secondary coil. These coils are wound around a core made of soft iron. The primary coil is connected to an alternating current (AC) power supply, while the secondary coil is connected to the output circuit.

The soft-iron core is laminated to reduce energy losses due to eddy currents. The coils are insulated from each other and from the core to prevent electrical short circuits. The number of turns in each coil can vary depending on the desired voltage transformation.

  • Primary coil: connected to AC supply, creates magnetic field.
  • Secondary coil: voltage induced by changing magnetic field.
  • Soft-iron core: provides a path for magnetic flux, increases efficiency.
  • Coils are insulated and wound around the core.
2

2. Electromagnetic Induction in Transformers

When alternating current flows through the primary coil, it produces a time-varying magnetic field around the coil. This magnetic field passes through the soft-iron core, which guides and concentrates the magnetic flux.

According to Faraday's law of electromagnetic induction, a changing magnetic flux through the secondary coil induces an electromotive force (emf) or voltage in it. This induced voltage depends on the rate of change of magnetic flux and the number of turns in the secondary coil.

The transformer operates only with alternating current because a steady magnetic field (from direct current) does not induce a voltage in the secondary coil.

  • AC in primary coil creates changing magnetic field.
  • Soft-iron core channels magnetic flux efficiently.
  • Changing flux induces voltage in secondary coil (Faraday's law).
  • Voltage induced depends on number of turns and flux change rate.
3

3. Role of the Soft-Iron Core

The soft-iron core is made of a material with high magnetic permeability, which means it can easily become magnetised and demagnetised. This property allows the core to concentrate the magnetic flux generated by the primary coil and direct it through the secondary coil.

Without the soft-iron core, much of the magnetic field would spread into the air, reducing the efficiency of induction. The core increases the magnetic flux linkage between the coils, improving the transformer's efficiency and enabling effective voltage transformation.

The core is made of soft iron rather than hard iron to avoid residual magnetism, which would reduce efficiency and cause energy losses.

  • Soft-iron core has high magnetic permeability.
  • Concentrates and directs magnetic flux between coils.
  • Increases efficiency of voltage induction.
  • Soft iron avoids residual magnetism and energy loss.
4

4. Changing Voltage Levels Using a Transformer

The transformer changes voltage levels by varying the number of turns in the primary and secondary coils. If the secondary coil has more turns than the primary, the transformer increases voltage (step-up transformer). If it has fewer turns, it decreases voltage (step-down transformer).

The voltage induced in the secondary coil (Vs) is proportional to the ratio of the number of turns in the secondary coil (Ns) to the number of turns in the primary coil (Np), expressed as:

\[ \frac{V_p}{N_p} = \frac{V_s}{N_s} \] where Vp is the primary voltage. This relationship shows that voltage changes depend on coil turns, with the soft-iron core ensuring efficient magnetic coupling to enable this transformation.

  • Voltage ratio depends on turns ratio: Vp/Np = Vs/Ns.
  • Step-up transformer: Ns > Np, increases voltage.
  • Step-down transformer: Ns < Np, decreases voltage.
  • Soft-iron core ensures efficient energy transfer.

At-a-glance summary

Comparison table

StepPhysics DetailEvidenceExam Use
1AC current in primary coil produces changing magnetic fieldAlternating current symbol and coil diagramExplain why AC is necessary for transformer operation
2Soft-iron core concentrates magnetic fluxDiagram showing magnetic flux lines through coreDescribe role of core in transformer efficiency
3Changing magnetic flux induces voltage in secondary coilFaraday's law formula and coil turnsRelate coil turns to voltage change
4Voltage ratio depends on turns ratioEquation Vp/Np = Vs/NsCalculate voltage changes in transformers

Active recall

Quick check

1. What are the main components of a basic transformer?

Primary coil, secondary coil, and soft-iron core.

2. Why must the primary coil be connected to an alternating current supply?

Because only a changing magnetic field from AC induces voltage in the secondary coil.

3. What is the role of the soft-iron core in a transformer?

It concentrates and directs the magnetic flux between coils, increasing efficiency.

4. How does the number of turns in the coils affect the voltage?

Voltage is proportional to the number of turns; more turns in the secondary coil increase voltage, fewer turns decrease it.

Complete revision guide 2 of 8

Roles of Primary and Secondary Coils in Transformers and Step-up/Step-down Concepts

This note explains the function and definitions of the primary and secondary coils in transformers, and the concepts of step-up and step-down transformers, essential for understanding voltage conversion in electrical circuits.

Syllabus source

Section 3, Topic 4.5.6, page 33, objective 4.5.6.2

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What you should be able to do

  • ✓Define the primary coil and its role in a transformer.
  • ✓Define the secondary coil and its role in a transformer.
  • ✓Explain how the primary coil induces voltage in the secondary coil via electromagnetic induction.
  • ✓Distinguish between step-up and step-down transformers based on voltage changes.
  • ✓Understand the practical significance of step-up and step-down transformers in electrical systems.
1

Structure of a Transformer

A transformer is an electrical device used to change the voltage of alternating current (AC) in a circuit. It consists mainly of two coils of insulated wire, called the primary coil and the secondary coil, wound around a soft iron core. The soft iron core enhances the magnetic field produced by the coils, improving efficiency.

The primary coil is connected to the input AC voltage source. When AC flows through this coil, it produces a changing magnetic field in the iron core. This changing magnetic field passes through the secondary coil, which is not electrically connected to the primary coil but is magnetically linked via the core.

  • Primary coil: connected to the input voltage source.
  • Secondary coil: connected to the output circuit.
  • Soft iron core: provides a path for the magnetic flux, increasing efficiency.
2

Role of the Primary Coil

The primary coil receives the input alternating voltage. When an alternating current flows through it, the coil generates a time-varying magnetic field around and inside the coil. This magnetic field is concentrated and guided by the soft iron core to pass through the secondary coil.

The alternating magnetic flux in the core induces an electromotive force (emf) in the secondary coil by electromagnetic induction, according to Faraday's law. The magnitude of this induced voltage depends on the number of turns in the coils and the rate of change of magnetic flux.

  • Primary coil carries input AC current.
  • Creates a changing magnetic field in the iron core.
  • Induces voltage in the secondary coil via electromagnetic induction.
3

Role of the Secondary Coil

The secondary coil is the coil in which the output voltage is induced. It is magnetically linked to the primary coil through the iron core but is electrically isolated from it. The voltage induced in the secondary coil depends on the number of turns of wire it contains.

The secondary coil delivers the output voltage to the external circuit. The voltage can be higher or lower than the input voltage depending on the relative number of turns in the primary and secondary coils.

  • Secondary coil is magnetically linked but electrically isolated from primary.
  • Voltage induced depends on number of turns in secondary coil.
  • Delivers output voltage to the load or circuit.
4

Step-up and Step-down Transformers

Transformers are classified as step-up or step-down depending on whether they increase or decrease voltage from primary to secondary.

A step-up transformer has more turns on the secondary coil than on the primary coil. This means the output voltage is greater than the input voltage.

A step-down transformer has fewer turns on the secondary coil than on the primary coil, so the output voltage is less than the input voltage.

  • Step-up transformer: Ns > Np, Vs > Vp (increases voltage).
  • Step-down transformer: Ns < Np, Vs < Vp (decreases voltage).
  • Ns = number of turns on secondary coil; Np = number of turns on primary coil; Vs = secondary voltage; Vp = primary voltage.
5

Practical Importance of Transformers

Transformers are essential in electrical power distribution. High voltages are used to transmit electrical energy efficiently over long distances to reduce energy loss. Step-up transformers increase voltage for transmission, and step-down transformers reduce voltage to safe levels for domestic or industrial use.

Understanding the roles of primary and secondary coils and the step-up/step-down concepts is fundamental to grasping how transformers control voltage levels in AC circuits.

  • Step-up transformers increase voltage for efficient power transmission.
  • Step-down transformers reduce voltage for safe usage.
  • Primary coil receives input voltage; secondary coil delivers output voltage.

At-a-glance summary

Comparison table

AspectStep-up TransformerStep-down Transformer
Number of turns in secondary coilMore turns than primary coil (Ns > Np)Fewer turns than primary coil (Ns < Np)
Voltage relationshipSecondary voltage greater than primary voltage (Vs > Vp)Secondary voltage less than primary voltage (Vs < Vp)
PurposeIncrease voltage for transmissionDecrease voltage for safe use
Current relationshipSecondary current less than primary current (Is < Ip)Secondary current greater than primary current (Is > Ip)

Active recall

Quick check

1. What is the role of the primary coil in a transformer?

It receives the input alternating voltage and produces a changing magnetic field that induces voltage in the secondary coil.

2. How does a step-up transformer change voltage?

It increases the voltage because the secondary coil has more turns than the primary coil.

3. Are the primary and secondary coils electrically connected?

No, they are electrically isolated but magnetically linked through the iron core.

4. What happens to the voltage in a step-down transformer?

The voltage decreases because the secondary coil has fewer turns than the primary coil.

Complete revision guide 3 of 8

Transformer Voltage and Turns Ratio Relationship

This note explains the fundamental relationship between the voltages and number of turns in the primary and secondary coils of a transformer. It covers the transformer structure, the principle of electromagnetic induction, the derivation and use of the equation Vp/Vs = Np/Ns, and includes worked examples, common errors, and practical considerations.

Syllabus source

Section 3, Topic 4.5.6, page 33, objective 4.5.6.3

Open source document ↗

What you should be able to do

  • ✓Understand the construction and function of a transformer.
  • ✓Recall and use the equation relating primary and secondary voltages to the number of turns in the coils: Vp/Vs = Np/Ns.
  • ✓Apply the transformer equation to calculate unknown voltages or turns in problems.
  • ✓Recognise the conditions under which the transformer equation is valid (ideal transformer assumptions).
  • ✓Identify common errors in using the transformer equation and how to avoid them.
1

1. Structure and Function of a Transformer

A transformer is an electrical device used to change the voltage of an alternating current (AC) supply. It consists of two coils of insulated wire, called the primary coil and the secondary coil, wound around a common soft iron core. The primary coil is connected to the input voltage source, and the secondary coil provides the output voltage.

The soft iron core enhances the magnetic flux linking the two coils, allowing efficient transfer of energy by electromagnetic induction. When an alternating current flows in the primary coil, it produces a changing magnetic field in the core, which induces an alternating voltage in the secondary coil.

  • Primary coil: coil connected to the input voltage source.
  • Secondary coil: coil connected to the output circuit.
  • Soft iron core: provides a path for magnetic flux, increasing efficiency.
  • Transformer changes voltage levels without changing frequency.
2

2. Electromagnetic Induction and Transformer Equation

The operation of a transformer is based on Faraday's law of electromagnetic induction, which states that a changing magnetic flux through a coil induces an electromotive force (emf) in the coil. The magnitude of the induced emf is proportional to the rate of change of magnetic flux and the number of turns in the coil.

For an ideal transformer (no energy losses), the magnetic flux linking both coils is the same. Therefore, the induced emf per turn is the same in both coils. This leads to the fundamental relationship between the voltages and the number of turns:

\[ \frac{V_p}{V_s} = \frac{N_p}{N_s} \] where: Vp = primary voltage (volts, V) Vs = secondary voltage (volts, V) Np = number of turns on the primary coil Ns = number of turns on the secondary coil

  • Voltage ratio equals turns ratio: Vp/Vs = Np/Ns.
  • Assumes ideal transformer with 100% efficiency.
  • Voltages are root mean square (rms) values for AC.
3

3. Using the Transformer Equation: Worked Example

Example: A transformer has 500 turns on the primary coil and 100 turns on the secondary coil. The primary voltage is 240 V. Calculate the secondary voltage.

Step 1: Write down the formula: \( \frac{V_p}{V_s} = \frac{N_p}{N_s} \)

Step 2: Rearrange to find \( V_s \): \[ V_s = V_p \times \frac{N_s}{N_p} \] Step 3: Substitute values: \[ V_s = 240 \times \frac{100}{500} = 240 \times 0.2 = 48 \, \text{V} \] Step 4: State the answer with units: The secondary voltage is 48 volts (V).

  • Always keep units consistent (volts for voltage, turns as a count).
  • Rearrange the formula carefully to isolate the unknown.
4

4. Common Errors and How to Avoid Them

A frequent mistake is to confuse the ratio of voltages with the ratio of currents or to invert the transformer equation incorrectly. Remember that the voltage ratio equals the turns ratio exactly as \( V_p / V_s = N_p / N_s \).

Another error is to apply the equation to a non-ideal transformer without considering losses, which can cause discrepancies in real measurements.

Also, ensure that the coils are correctly identified as primary and secondary, and that the number of turns corresponds to the correct coil.

  • Do not invert the ratio: \( V_p / V_s \neq N_s / N_p \).
  • Use the equation only for AC transformers, not DC.
5

5. Practical Considerations and Diagram Description

In practice, transformers are used to step up (increase) or step down (decrease) voltages in power distribution. Step-up transformers have more turns on the secondary coil than the primary, increasing voltage. Step-down transformers have fewer turns on the secondary coil, decreasing voltage.

To draw a transformer diagram: Draw two coils side by side, label the left coil as primary with Np turns and voltage Vp, and the right coil as secondary with Ns turns and voltage Vs. Between them, draw a rectangular soft iron core linking both coils. Indicate the direction of alternating current in the primary coil and the induced emf in the secondary coil.

  • Step-up transformer: Ns > Np, Vs > Vp.
  • Step-down transformer: Ns < Np, Vs < Vp.

At-a-glance summary

Comparison table

StepPhysics DetailEvidenceExam Use
Identify coilsPrimary coil connected to input, secondary to outputCircuit diagrams, coil labellingMarks for correct coil identification
Write formulaUse \( V_p / V_s = N_p / N_s \)Given in syllabus and formula sheetsRecall and apply formula correctly
Rearrange formulaIsolate unknown variable, e.g. \( V_s = V_p \times N_s / N_p \)Algebraic manipulationMarks for correct rearrangement
Substitute valuesInsert numerical values with unitsGiven problem dataMarks for correct substitution
Calculate answerPerform arithmetic and include unitsCalculator or manual calculationMarks for correct final answer

Active recall

Quick check

1. What is the relationship between the primary voltage (Vp), secondary voltage (Vs), and the number of turns on the primary (Np) and secondary (Ns) coils of a transformer?

\( \frac{V_p}{V_s} = \frac{N_p}{N_s} \)

2. A transformer has 200 turns on the primary coil and 50 turns on the secondary coil. If the primary voltage is 240 V, what is the secondary voltage?

\( V_s = 240 \times \frac{50}{200} = 60 \, \text{V} \)

3. Why does the transformer equation only apply to AC and not DC?

Because transformers rely on changing magnetic flux to induce voltage, which only occurs with alternating current (AC), not direct current (DC).

Complete revision guide 4 of 8

Transformers in High-Voltage Transmission of Electrical Energy

Transformers are essential devices used in the transmission of electrical energy over long distances. They adjust voltage levels to reduce energy loss during transmission and ensure safe voltage levels for end users.

Syllabus source

Section 3, Topic 4.5.6, page 33, objective 4.5.6.4

Open source document ↗

What you should be able to do

  • ✓Understand the role of transformers in electrical power transmission.
  • ✓Explain why high voltage is used for transmitting electrical energy over long distances.
  • ✓Describe how transformers step up and step down voltage in transmission systems.
  • ✓Recognise the relationship between voltage, current, and power loss in transmission cables.
1

Introduction to Transformers in Power Transmission

Transformers are electrical devices that change the voltage of alternating current (AC) electricity without changing its frequency. They consist of two coils of wire, called the primary and secondary coils, wound around a soft iron core. The primary coil receives the input voltage, and the secondary coil delivers the output voltage.

In electrical power transmission, transformers are used to adjust voltage levels to improve efficiency and safety. This directly affects the result because electrical energy is generated at power stations and must be transmitted over long distances to consumers.

The use of transformers allows the voltage to be increased (stepped up) for transmission and then decreased (stepped down) for safe use in homes and businesses.

  • Transformers operate only with alternating current (AC).
  • They change voltage levels by electromagnetic induction.
  • They consist of primary and secondary coils wound on a soft iron core.
2

Why High Voltage is Used for Transmission

When electrical energy is transmitted through cables, some energy is lost as heat due to the resistance of the cables. This loss is given by the power loss formula:

P_loss = I²R, where P_loss is power loss in watts (W), I is current in amperes (A), and R is resistance in ohms (Ω).

To reduce power loss, the current in the transmission cables must be minimised. Since power (P) transmitted is the product of voltage (V) and current (I), P = VI, increasing the voltage allows the current to decrease for the same power transmitted. This reduces I²R losses significantly, making high-voltage transmission more efficient.

  • Power loss in cables is proportional to the square of the current.
  • Increasing voltage reduces current for the same power transmitted.
  • High voltage reduces energy lost as heat in transmission cables.
3

Stepping Up Voltage for Transmission

At the power station, a step-up transformer increases the voltage from the generator output to a much higher voltage suitable for transmission. For example, voltage may be increased from a few kilovolts (kV) to hundreds of kilovolts.

The step-up transformer has more turns on the secondary coil than on the primary coil, increasing voltage according to the transformer equation:

Vp / Vs = Np / Ns, where V is voltage, N is number of turns, and subscripts p and s refer to primary and secondary coils respectively.

  • Step-up transformers increase voltage for efficient transmission.
  • They have more turns on the secondary coil than the primary coil.
  • Voltage increase reduces current and power loss in cables.
4

Stepping Down Voltage for Safe Use

Near the point of use, step-down transformers reduce the high transmission voltage to safer, usable levels for homes and businesses, typically to 230 V in many countries.

Step-down transformers have fewer turns on the secondary coil than on the primary coil, decreasing voltage according to the same transformer equation.

This ensures electrical devices receive the correct voltage and improves safety for consumers.

  • Step-down transformers reduce voltage to safe levels.
  • They have fewer turns on the secondary coil than the primary coil.
  • They protect electrical devices and users from high voltage hazards.
5

Summary of Transformer Use in Transmission

The combination of step-up and step-down transformers enables efficient and safe electrical energy distribution. High voltage reduces current and power loss during transmission, while low voltage ensures safety and compatibility with electrical appliances.

Transformers rely on electromagnetic induction and require alternating current to function. Their design with a soft iron core enhances magnetic flux linkage between coils, improving efficiency.

  • Transformers enable efficient long-distance power transmission.
  • High voltage reduces energy loss in cables.
  • Low voltage ensures safety and appliance compatibility.
  • Transformers operate only with AC and use electromagnetic induction.

At-a-glance summary

Comparison table

StepPhysics DetailEvidenceExam Use
Step-up TransformerIncreases voltage by having more turns on secondary coil than primary coilVoltage ratio Vp/Vs = Np/Ns; reduces current for same powerExplain why voltage is increased for transmission to reduce losses
Transmission at High VoltageHigh voltage reduces current, lowering I²R power loss in cablesPower loss P = I²R; lower current means less heat lossDescribe why high voltage is used in transmission lines
Step-down TransformerDecreases voltage by having fewer turns on secondary coil than primary coilVoltage ratio Vp/Vs = Np/Ns; provides safe voltage for consumersDescribe how voltage is reduced for safe domestic use
Transformer CoreSoft iron core increases magnetic flux linkage between coilsImproves efficiency of voltage change by electromagnetic inductionIdentify role of iron core in transformer operation

Active recall

Quick check

1. Why are transformers used in high-voltage transmission?

To increase voltage for transmission, reducing current and power loss in cables.

2. What happens to voltage and current in a step-up transformer?

Voltage increases and current decreases.

3. Write the transformer equation relating voltage and number of turns.

Vp / Np = Vs / Ns, where V is voltage and N is number of turns.

4. Why is high voltage used to transmit electrical energy over long distances?

Because it reduces current, which reduces energy lost as heat due to resistance in cables.

5. What is the role of the soft iron core in a transformer?

It increases magnetic flux linkage between coils, improving efficiency of voltage change.

Complete revision guide 5 of 8

Benefits of Transmitting Electricity at High Voltages

Transmitting electrical energy at high voltages significantly improves efficiency and reduces costs by minimising energy losses, enabling long-distance transmission, and allowing the use of smaller conductors.

Syllabus source

Section 3, Topic 4.5.6, page 33, objective 4.5.6.5

Open source document ↗

What you should be able to do

  • ✓Understand why electricity is transmitted at high voltages.
  • ✓Explain how high voltage transmission reduces energy losses.
  • ✓Identify the advantages of using smaller conductors in power transmission.
  • ✓Recognise the importance of high voltage transmission for long-distance power delivery.
  • ✓Describe how high voltage transmission increases the capacity of power systems.
1

Energy Losses in Electrical Transmission

When electrical power is transmitted through cables, some energy is lost as heat due to the resistance of the wires. This loss is given by the formula:

\[ P_{loss} = I^2 R \] where \(I\) is the current in amperes (A) and \(R\) is the resistance of the cable in ohms (Ω).

For a given power \(P\) transmitted, the current \(I\) is related to the voltage \(V\) by: \[ P = V I \] Rearranging gives \( I = \frac{P}{V} \). Increasing the voltage \(V\) reduces the current \(I\), which in turn reduces the power loss \(P_{loss}\) because it depends on the square of the current.

  • Power loss in cables is proportional to the square of the current (\(I^2 R\)).
  • Higher voltage means lower current for the same power transmitted.
  • Lower current reduces resistive heating and energy loss.
2

Use of Smaller Conductors

Because higher voltage reduces the current, the cables do not need to be as thick to carry the same power safely. Thinner cables have less metal, making them lighter and cheaper to manufacture and install. This reduces the overall cost of the transmission infrastructure.

  • Smaller cross-sectional area of cables is sufficient at high voltage.
  • Reduced material cost and easier installation.
  • Lighter cables reduce mechanical support requirements.
3

Long-Distance Transmission

High voltage transmission allows electricity to be sent over long distances with minimal energy loss. This is important because power stations are often located far from where electricity is used, such as in remote hydroelectric or thermal power plants. Efficient long-distance transmission ensures that electricity reaches consumers with less wastage.

  • Enables connection between remote power stations and cities.
  • Reduces energy loss over long cables.
  • Supports national and regional power grids.
4

Increased Transmission Capacity

High voltage systems can carry more power through the same cables compared to low voltage systems. This is essential to meet the growing demand for electricity in homes, industries, and businesses. Increasing voltage is a practical way to increase capacity without building new cables.

  • Higher voltage increases power capacity (\(P = V I\)).
  • Supports growing electricity demand.
  • Improves efficiency of existing infrastructure.

At-a-glance summary

Comparison table

AspectPhysics DetailPractical EvidenceExam Use
Energy LossesPower loss \(P_{loss} = I^2 R\) decreases with lower current at high voltageMeasurement of temperature rise in cables at different voltagesExplain why high voltage reduces energy loss
Conductor SizeCurrent reduction allows thinner cables with less resistanceCost and weight comparison of cablesState why smaller conductors are used
Transmission DistanceVoltage drop over distance is less at high voltageVoltage measurements at various distancesDescribe benefits for long-distance transmission
CapacityPower \(P = V I\) increases with voltage for same currentPower delivered through cables at different voltagesState how capacity is increased

Active recall

Quick check

1. Why does transmitting electricity at high voltage reduce energy losses?

Because higher voltage means lower current for the same power, and energy loss depends on the square of current (\(I^2 R\)), so losses decrease.

2. What is the formula relating power loss to current and resistance?

\( P_{loss} = I^2 R \), where \(I\) is current and \(R\) is resistance.

3. How does high voltage transmission affect the size of conductors?

It allows the use of smaller, thinner conductors because the current is reduced.

4. Why is high voltage transmission important for long distances?

It reduces energy loss over long cables, enabling efficient delivery from remote power stations.

Complete revision guide 6 of 8

Operation of a Simple Iron-Cored Transformer

A transformer is an electrical device that changes the voltage of alternating current (AC) using electromagnetic induction. A simple iron-cored transformer consists of two coils wound around a soft iron core. The alternating current in the primary coil produces a changing magnetic flux in the iron core, which induces a voltage in the secondary coil. The voltage change depends on the ratio of turns in the coils, allowing the transformer to step up or step down voltage efficiently.

Syllabus source

Section 3, Topic 4.5.6, page 33, objective 4.5.6.6

Open source document ↗

What you should be able to do

  • ✓Understand the construction of a simple iron-cored transformer.
  • ✓Explain the role of the primary and secondary coils and the iron core.
  • ✓Describe how alternating current in the primary coil induces voltage in the secondary coil.
  • ✓Explain the relationship between the number of turns in coils and the voltage induced.
  • ✓Understand the principle of electromagnetic induction as applied to transformers.
1

Construction of a Simple Transformer

A simple transformer consists of two insulated coils of wire called the primary coil and the secondary coil. Both coils are wound around a core made of soft iron. The iron core is laminated to reduce energy losses due to eddy currents. The primary coil is connected to an alternating current (AC) supply, while the secondary coil is connected to the output circuit.

The soft iron core serves to concentrate and guide the magnetic flux generated by the primary coil, increasing the efficiency of magnetic coupling between the coils. The coils are insulated from each other and from the core to prevent electrical conduction between them.

  • Primary coil: connected to AC supply, creates magnetic field.
  • Secondary coil: receives induced voltage, supplies output circuit.
  • Soft iron core: enhances magnetic flux linkage, laminated to reduce eddy currents.
2

Principle of Operation: Electromagnetic Induction

When an alternating current flows through the primary coil, it produces a time-varying magnetic field around the coil. This changing magnetic flux passes through the iron core and links to the secondary coil. According to Faraday's law of electromagnetic induction, a changing magnetic flux through a coil induces an electromotive force (emf) or voltage in that coil.

The induced voltage in the secondary coil depends on the rate of change of magnetic flux and the number of turns in the coil. Since the magnetic flux is common to both coils, the voltage induced in each coil is proportional to its number of turns.

  • AC in primary coil produces changing magnetic flux.
  • Changing flux induces voltage in secondary coil (Faraday's law).
  • Voltage induced ∝ number of turns in the coil.
3

Voltage and Turns Ratio Relationship

The voltage induced in the secondary coil (Vs) and the voltage applied to the primary coil (Vp) are related to the number of turns in the secondary coil (Ns) and primary coil (Np) by the equation:

\[ \frac{V_p}{N_p} = \frac{V_s}{N_s} \] This means the voltage ratio equals the turns ratio.

If the secondary coil has more turns than the primary coil (Ns > Np), the transformer increases voltage and is called a step-up transformer. Conversely, if the secondary coil has fewer turns (Ns < Np), the transformer decreases voltage and is called a step-down transformer.

  • Voltage ratio equals turns ratio: Vp/Np = Vs/Ns.
  • Step-up transformer: Ns > Np, Vs > Vp.
  • Step-down transformer: Ns < Np, Vs < Vp.
4

Role of the Iron Core

The iron core provides a low reluctance path for the magnetic flux, which increases the magnetic coupling between the primary and secondary coils. This means more of the magnetic flux produced by the primary coil passes through the secondary coil, improving the efficiency of the transformer.

The core is made of soft iron because it is easily magnetised and demagnetised, which is essential for following the alternating current changes without significant energy loss. The core is laminated to reduce eddy currents, which are loops of induced current in the core that cause heating and energy loss.

  • Iron core concentrates magnetic flux linking both coils.
  • Soft iron chosen for easy magnetisation and demagnetisation.
5

Summary of Transformer Operation

1. An alternating current in the primary coil produces a changing magnetic flux in the iron core.

2. The changing magnetic flux induces a voltage in the secondary coil by electromagnetic induction.

3. The voltage induced in the secondary coil depends on the turns ratio between the coils, allowing voltage to be increased or decreased as needed.

  • AC in primary coil → changing magnetic flux in iron core.
  • Changing flux induces voltage in secondary coil.

At-a-glance summary

Comparison table

StepPhysics DetailEvidenceExam Use
1AC current in primary coil creates changing magnetic fluxAlternating current symbol and magnetic field lines around coilExplain how AC causes changing flux in transformer
2Magnetic flux passes through iron core linking primary and secondary coilsDiagram showing iron core with coils and magnetic flux linesDescribe role of iron core in flux linkage
3Changing magnetic flux induces emf in secondary coil (Faraday's law)Equation emf = -N dΦ/dt and induced voltage in secondary coilApply Faraday's law to transformer operation
4Voltage induced proportional to number of turns in coilFormula Vp/Np = Vs/NsCalculate voltage changes using turns ratio
5Transformer steps up or steps down voltage depending on turns ratioExamples of step-up and step-down transformersIdentify transformer type from coil turns and voltage

Active recall

Quick check

1. What causes the voltage to be induced in the secondary coil of a transformer?

A changing magnetic flux in the iron core induces voltage in the secondary coil by electromagnetic induction.

2. Why must the current in the primary coil be alternating?

Because only a changing current produces a changing magnetic flux necessary to induce voltage in the secondary coil.

3. How does the number of turns in the coils affect the voltage output?

The voltage induced in a coil is proportional to its number of turns; more turns produce higher voltage.

4. What is the purpose of the soft iron core in a transformer?

To concentrate and guide the magnetic flux between the coils, increasing efficiency.

Complete revision guide 7 of 8

Ideal Transformer Power Relationship: IpVp = IsVs

This note explains the fundamental power relationship in an ideal transformer, where the product of primary current and voltage equals the product of secondary current and voltage. It covers the components of a transformer, the principle of operation, the derivation and use of the equation IpVp = IsVs, and includes worked examples, common errors, and practical considerations.

Syllabus source

Section 3, Topic 4.5.6, page 33, objective 4.5.6.7

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What you should be able to do

  • ✓Define the primary and secondary currents and voltages in a transformer.
  • ✓Understand the concept of an ideal transformer and its assumptions.
  • ✓Recall and use the equation IpVp = IsVs for an ideal transformer.
  • ✓Apply the equation to solve problems involving current and voltage in transformers.
  • ✓Recognise the implications of the power relationship for transformer efficiency.
1

1. Components and Operation of a Transformer

A transformer is an electrical device used to change the voltage of an alternating current (AC) supply. It consists of two coils of insulated wire: the primary coil and the secondary coil, wound around a common soft iron core. The primary coil is connected to the input voltage source, and the secondary coil delivers the output voltage to a load.

When an alternating voltage is applied to the primary coil, it creates a changing magnetic flux in the iron core. This changing flux induces an alternating voltage in the secondary coil by electromagnetic induction. The iron core serves to concentrate the magnetic flux and improve efficiency.

  • Primary coil: connected to input voltage source, carries primary current (Ip) and voltage (Vp).
  • Secondary coil: connected to output load, carries secondary current (Is) and voltage (Vs).
  • Soft iron core: provides a low reluctance path for magnetic flux, increasing efficiency.
  • Alternating current in primary produces changing magnetic flux, inducing voltage in secondary.
2

2. Ideal Transformer and Power Relationship

An ideal transformer is a theoretical model where there are no energy losses due to resistance, hysteresis, eddy currents, or leakage flux. Under these ideal conditions, all the electrical power input to the primary coil is transferred to the secondary coil.

Electrical power (P) is the product of current (I) and voltage (V), measured in watts (W). For the primary coil, power input is Pp = Ip × Vp. For the secondary coil, power output is Ps = Is × Vs.

Because power is conserved in an ideal transformer, the input power equals the output power: IpVp = IsVs. This equation is fundamental for analysing transformers and solving related problems.

  • Ideal transformer: no energy losses, 100% efficiency.
  • Power input (primary) = Power output (secondary).
  • Power formula: P = I × V, units in watts (W).
  • Key equation: IpVp = IsVs.
3

3. Using the Equation IpVp = IsVs

This equation relates the currents and voltages in the primary and secondary coils of an ideal transformer. It can be rearranged to find any one of the four variables if the other three are known.

For example, to find the secondary current Is: Is = (Ip × Vp) / Vs.

To find the primary current Ip: Ip = (Is × Vs) / Vp.

  • Rearrangement examples:
  • Is = (Ip × Vp) / Vs
  • Ip = (Is × Vs) / Vp
  • Use consistent units: current in amperes (A), voltage in volts (V), power in watts (W).
4

4. Worked Example

A transformer has a primary voltage of 240 V and a primary current of 2.0 A. The secondary voltage is 120 V. Calculate the secondary current, assuming the transformer is ideal.

Step 1: Write down the known values: Vp = 240 V, Ip = 2.0 A, Vs = 120 V.

Step 2: Use the equation IpVp = IsVs and rearrange to find Is: Is = (Ip × Vp) / Vs = (2.0 A × 240 V) / 120 V = 4.0 A.

  • Answer: Secondary current Is = 4.0 A.
  • Units: current in amperes (A).
  • Common error: forgetting to rearrange the formula correctly or mixing up primary and secondary values.
5

5. Practical Considerations and Efficiency

In real transformers, some energy is lost due to resistance in the coils, eddy currents in the core, and magnetic hysteresis. Therefore, the output power is slightly less than the input power, and the equation IpVp = IsVs applies only approximately.

Efficiency (η) of a transformer is defined as the ratio of output power to input power, usually expressed as a percentage: η = (Ps / Pp) × 100%.

Despite losses, transformers are highly efficient, often above 95%, making them essential for voltage conversion in power distribution.

  • Real transformers have losses; ideal transformer equation is an approximation.
  • Efficiency η = (output power / input power) × 100%.
  • High efficiency means output power is close to input power.
  • IpVp = IsVs is used for ideal transformer calculations and approximations.

At-a-glance summary

Comparison table

StepPhysics DetailEvidenceExam Use
Identify known quantitiesList given currents and voltages with unitsProblem statement or diagramExtract data correctly for calculations
Write down formulaUse IpVp = IsVs for ideal transformer powerSyllabus formula and theoryRecall and apply formula accurately
Rearrange formulaMake Is or Ip the subject as neededAlgebraic manipulationShow correct rearrangement for marks
Substitute valuesInsert numerical values with unitsGiven data from questionUse correct substitution and units
Calculate answerPerform arithmetic to find unknownCalculator or manual calculationProvide final answer with units
Check units and reasonablenessConfirm units are amperes or voltsDimensional analysisAvoid unit errors and check plausibility

Active recall

Quick check

1. What is the power input to the primary coil if Ip = 3 A and Vp = 120 V?

Power input Pp = Ip × Vp = 3 A × 120 V = 360 W.

2. If an ideal transformer has a primary voltage of 240 V and secondary voltage of 120 V, and the primary current is 2 A, what is the secondary current?

Is = (Ip × Vp) / Vs = (2 A × 240 V) / 120 V = 4 A.

3. State the equation relating primary and secondary currents and voltages in an ideal transformer.

IpVp = IsVs.

Complete revision guide 8 of 8

Cable Power Loss and the Formula P = I²R

This note explains the concept of power loss in electrical cables due to their resistance when current flows through them. It introduces the formula P = I²R, where power loss depends on the square of the current and the resistance of the cable. The note covers the physical cause of power loss, the meaning of each term in the formula, units, practical implications for electrical transmission, and a worked example calculation.

Syllabus source

Section 3, Topic 4.5.6, page 33, objective 4.5.6.8

Open source document ↗

What you should be able to do

  • ✓Understand the physical cause of power loss in cables due to resistance.
  • ✓Recall and use the formula P = I²R to calculate power loss in cables.
  • ✓Interpret the units and meaning of each variable in the formula.
  • ✓Explain why power loss increases with the square of the current.
  • ✓Apply the formula to practical problems involving electrical transmission and cable design.
1

Physical Cause of Power Loss in Cables

When an electric current flows through a cable, the cable's resistance opposes the flow of charge. This opposition causes electrical energy to be converted into heat energy within the cable, a process known as Joule heating or resistive heating.

The heat produced represents energy lost from the electrical supply, reducing the efficiency of power transmission. This loss is called power loss in the cable.

The amount of power lost depends on how much current flows and the resistance of the cable material and dimensions.

  • Resistance (R) is a property of the cable measured in ohms (Ω).
  • Current (I) is the rate of flow of electric charge, measured in amperes (A).
  • Power loss manifests as heat, which can cause cables to warm up.
2

The Formula for Cable Power Loss: P = I²R

The power loss in a cable due to its resistance is given by the formula:

P = I²R, where: P is the power loss in watts (W), I is the current in amperes (A), and R is the resistance of the cable in ohms (Ω).

This formula shows that power loss increases with the square of the current, meaning that doubling the current causes the power loss to increase by a factor of four.

  • Power (P) is the rate at which energy is lost as heat.
  • Current (I) squared means I × I, emphasising the strong effect of current on power loss.
  • Resistance (R) depends on the cable's material, length, and cross-sectional area.
3

Units and Interpretation of the Formula

Each quantity in the formula has specific units: current (I) in amperes (A), resistance (R) in ohms (Ω), and power (P) in watts (W).

Since 1 watt = 1 ampere² × 1 ohm, the units are consistent:

W = A² × Ω, confirming the formula's dimensional correctness.

  • Power loss is energy lost per second as heat.
  • Higher resistance or higher current increases power loss.
  • Reducing current or resistance reduces power loss.
4

Worked Example Calculation

Calculate the power loss in a cable with resistance 5 Ω when a current of 3 A flows through it.

Using the formula: P = I²R = (3 A)² × 5 Ω = 9 × 5 = 45 W.

Therefore, the cable loses 45 watts of power as heat during operation.

  • Step 1: Write down the formula P = I²R.
  • Step 2: Substitute I = 3 A and R = 5 Ω.
  • Step 3: Calculate I² = 3² = 9.
  • Step 4: Multiply 9 × 5 Ω = 45 W.
  • Step 5: State the answer with correct units: 45 watts.
5

Practical Implications and Reducing Power Loss

Power loss in cables wastes energy and can cause cables to heat up, potentially damaging insulation or causing safety hazards.

To reduce power loss, electrical power is often transmitted at high voltages and low currents, since power loss depends on current squared.

Using cables with lower resistance (thicker cables or better conducting materials) also reduces power loss.

  • High voltage transmission reduces current for the same power, lowering I²R losses.
  • Choosing cables with low resistance minimises heating and energy waste.
  • Understanding P = I²R helps engineers design efficient electrical systems.

At-a-glance summary

Comparison table

StepPhysics DetailEvidenceExam Use
Identify variablesCurrent (I), Resistance (R), Power loss (P)Given in problem or known cable propertiesRecall definitions and units
Apply formulaP = I²RStandard formula for power loss in resistive cablesUse formula to calculate power loss
Substitute valuesInsert numerical values for I and RFrom problem statement or measurementsPerform calculation accurately
CalculateSquare current, multiply by resistanceMathematical operationShow working clearly for marks
State answerInclude units (watts)Final numerical resultWrite final answer with correct units

Active recall

Quick check

1. What physical quantity does P represent in the formula P = I²R?

P represents the power loss in the cable, measured in watts (W).

2. If the current doubles, by what factor does the power loss increase?

The power loss increases by a factor of four because power loss is proportional to the square of the current (I²).

3. Calculate the power loss in a cable with resistance 2 Ω carrying a current of 5 A.

P = I²R = (5 A)² × 2 Ω = 25 × 2 = 50 W.

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