Transformer

4.3 Transformer - Physics Form 5
PHYSICS • Form 5 • Chapter 4: Electromagnetism

4.3 Transformer

A transformer is an electrical device that changes an alternating input voltage to an output voltage of a different value. Operating on the principle of electromagnetic induction, transformers are essential in powering domestic appliances with appropriate voltages and minimising electrical energy loss during long-distance transmission from power stations to homes and industries.

Learning Objectives

  • Describe the working principle of a simple step-up and step-down transformer.
  • State and apply the transformer equation relating primary and secondary voltages to the number of turns.
  • Define an ideal transformer and calculate efficiency, power, voltage, and current using transformer equations.
  • Identify the four causes of energy loss in a transformer and explain ways to increase transformer efficiency.
  • Describe daily life applications of transformers and their role in the electricity transmission and distribution system.

Working Principle and Energy Loss Reductions in a Transformer

See how alternating flux induces voltage across coils and how four design modifications minimise transformer energy losses.

Structure of a Simple Transformer

Simple Transformer Primary coil (Np) Secondary coil (Ns) a.c. supply Vp Output Vs Changing magnetic flux Soft iron core (laminated)

Four Ways to Reduce Energy Loss

1. Resistance of coils

Use thicker copper wire to lower the electrical resistance of the coils.

2. Eddy currents

Use a laminated soft iron core made of thin iron sheets insulated from one another.

3. Hysteresis

Use a soft iron core, which requires a smaller amount of energy to magnetise and demagnetise.

4. Leakage of magnetic flux

Wind the secondary coil directly on top of the primary coil.

Short Explanation

What is a Transformer?

A transformer is an electrical device that changes an alternating input voltage to an output voltage of a different value. It operates on the principle of electromagnetic induction.

Why a.c. and Not d.c.?

A direct current produces a steady, non-changing magnetic field. Since there is no change in magnetic flux linking the secondary coil, no voltage is induced across the secondary coil.

Step-up Transformer

Increases voltage: \(V_s > V_p\) and \(N_s > N_p\).

Step-down Transformer

Decreases voltage: \(V_s < V_p\) and \(N_s < N_p\).

Transformer Equation

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

\(V_p\) = voltage across primary coil (input voltage), \(V_s\) = voltage across secondary coil (output voltage), \(N_p\) = number of turns of the primary coil, \(N_s\) = number of turns of the secondary coil.

Efficiency of a Transformer

\(\eta = \dfrac{\text{Output power}}{\text{Input power}} \times 100\% = \dfrac{V_s I_s}{V_p I_p} \times 100\%\)

An ideal transformer experiences zero energy loss (\(\eta = 100\%\)), so output power equals input power: \(V_p I_p = V_s I_s\).

Cause of Energy Loss Effect Way to Reduce
Resistance of coils Coils consist of long wires with resistance; current causes Joule heating, releasing heat energy to the surroundings. Use thicker copper wire to lower the electrical resistance of the coils.
Eddy currents The changing magnetic field induces swirling eddy currents in the iron core, heating up the core. Use a laminated soft iron core made of thin iron sheets insulated from one another with lacquer/glue.
Hysteresis Continuous magnetisation and demagnetisation of the core consumes energy that is not fully recovered, dissipating as heat. Use a soft iron core, which requires a smaller amount of energy to magnetise and demagnetise.
Leakage of magnetic flux Magnetic flux produced by the primary coil is not completely linked to the secondary coil. Wind the secondary coil directly on top of the primary coil.

Useful Application of Eddy Currents

In an induction cooker, a high-frequency alternating current in a coil beneath a ceramic plate creates a rapidly changing magnetic field. This induces large eddy currents at the metal base of a steel pot, heating the pot directly to cook food efficiently.

Uses of Transformers in Daily Life

Step-down transformers: mobile phone chargers, notebook computer chargers, photocopy machines, welding machines.

Step-up transformers: microwave ovens, defibrillators, X-ray machines.

Electricity Transmission and Distribution System

Electricity generated at the power station (e.g., \(20\text{ kV}\)) passes through a step-up transformer to be stepped up to high voltages (e.g., \(275\text{ kV}\)) for transmission. Stepping up the voltage reduces the transmission current (\(I\)). Since power loss in cables is given by \(P_{\text{loss}} = I^2 R\), a smaller current drastically reduces electrical energy lost as heat in the transmission cables.

During distribution, step-down transformers reduce voltage in stages: main receiving substation stepped down to \(132\text{ kV}\), main distribution substation to \(33\text{ kV}\) (heavy industry), further substations to \(11\text{ kV}\) (light industry), and finally to \(230\text{ V}\) for residential homes.

Try to Answer First

Answer in your mind, then press “Check Answer”.

1

Why is an alternating current power supply used instead of a direct current power supply in a transformer?

Check Answer
Answer: An alternating current produces a magnetic field that changes continuously in magnitude and direction, which is necessary to induce a voltage in the secondary coil by electromagnetic induction. A direct current produces a constant magnetic field, so no voltage is induced.
2

An ideal transformer connected to a \(240\text{ V}\) supply has \(800\) turns in its primary coil and \(60\) turns in its secondary coil. If it supplies \(27\text{ W}\) of power to an appliance, what is the output voltage and secondary current?

Check Answer
Answer: Using \(\dfrac{V_s}{V_p} = \dfrac{N_s}{N_p}\), \(V_s = \dfrac{60}{800} \times 240\text{ V} = 18\text{ V}\). Since output power is \(27\text{ W}\), \(I_s = \dfrac{\text{Power}}{V_s} = \dfrac{27}{18} = 1.5\text{ A}\).
3

How does a laminated iron core help improve the efficiency of a transformer?

Check Answer
Answer: It consists of thin iron sheets separated by insulating glue, which breaks up the paths of circulating currents, thereby reducing eddy currents and minimising heat loss in the core.

Common Mistakes

  • !Forgetting that direct current cannot operate a transformer; students often assume any voltage input produces an output.
  • !Inverting the transformer equation ratio, such as incorrectly writing \(\dfrac{V_s}{V_p} = \dfrac{N_p}{N_s}\).
  • !Confusing current and voltage relationships in ideal transformers: when voltage is stepped up, current is stepped down (\(V_p I_p = V_s I_s\)).
  • !Misidentifying the role of soft iron: soft iron is used to reduce hysteresis loss because it is easily magnetised and demagnetised, whereas lamination is used to reduce eddy currents.

Concept Misunderstandings

Misunderstanding

A transformer creates extra energy when it steps up voltage.

Correct Concept

Transformers obey the principle of conservation of energy. For an ideal transformer, input power equals output power (\(V_p I_p = V_s I_s\)). Stepping up voltage results in a proportional reduction in current, so power output is never greater than power input.

Misunderstanding

Eddy currents are always harmful and have no useful applications.

Correct Concept

While eddy currents cause unwanted heat loss in transformers, they are deliberately generated in devices such as induction cookers to heat cooking pots directly and efficiently.

Summary

  • A transformer transfers electrical energy between circuits via electromagnetic induction across a soft iron core using an alternating current.
  • Turn ratio determines function: \(N_s > N_p\) steps up voltage, while \(N_s < N_p\) steps down voltage, obeying \(\dfrac{V_s}{V_p} = \dfrac{N_s}{N_p}\).
  • An ideal transformer has \(100\%\) efficiency with \(V_p I_p = V_s I_s\).
  • Four causes of energy loss are coil resistance, eddy currents, hysteresis, and flux leakage, which are reduced by thicker copper wire, laminating the core, using soft iron, and winding coils together.
  • High-voltage transmission using step-up transformers lowers transmission current, drastically minimising energy loss along power cables.

Short Activity

Objective: Test your understanding of transformers, energy loss and transmission.

A. Objective Quiz

1 Why are step-up transformers used at power stations before transmitting electricity through long-distance cables?

2 Which modification specifically reduces energy loss due to hysteresis in a transformer?

B. Fill in the Blanks

3 For an ideal transformer, the output power is equal to the input power, which can be expressed by the formula \(V_p I_p =\) .

4 In a step-down transformer, the number of turns in the secondary coil is than the number of turns in the primary coil.

C. Matching / Drag and Drop

Match each cause of energy loss in a transformer to its corresponding reduction method. If using a phone, tap the answer first, then tap the matching box.

Causes of Energy Loss
Resistance of coils
Eddy currents
Hysteresis
Leakage of magnetic flux
Methods to Reduce Loss
Use thicker copper wire so resistance is smallerCorrect ✓Try againEmpty
Use a laminated iron core made of thin insulated sheetsCorrect ✓Try againEmpty
Use soft iron as the core materialCorrect ✓Try againEmpty
Wind the secondary coil directly on the primary coilCorrect ✓Try againEmpty
 

Keywords

Transformer Step-up Transformer Step-down Transformer Ideal Transformer Eddy Currents Hysteresis Laminated Core National Grid Transmission

Transformer

4.3 Transformer - Physics Form 5
PHYSICS • Form 5 • Chapter 4: Electromagnetism

4.3 Transformer

A transformer is an electrical device that changes an alternating input voltage to an output voltage of a different value. Operating on the principle of electromagnetic induction, transformers are essential in powering domestic appliances with appropriate voltages and minimising electrical energy loss during long-distance transmission from power stations to homes and industries.

Learning Objectives

  • Describe the working principle of a simple step-up and step-down transformer.
  • State and apply the transformer equation relating primary and secondary voltages to the number of turns.
  • Define an ideal transformer and calculate efficiency, power, voltage, and current using transformer equations.
  • Identify the four causes of energy loss in a transformer and explain ways to increase transformer efficiency.
  • Describe daily life applications of transformers and their role in the electricity transmission and distribution system.

Working Principle and Energy Loss Reductions in a Transformer

See how alternating flux induces voltage across coils and how four design modifications minimise transformer energy losses.

Structure of a Simple Transformer

Simple Transformer Primary coil (Np) Secondary coil (Ns) a.c. supply Vp Output Vs Changing magnetic flux Soft iron core (laminated)

Four Ways to Reduce Energy Loss

1. Resistance of coils

Use thicker copper wire to lower the electrical resistance of the coils.

2. Eddy currents

Use a laminated soft iron core made of thin iron sheets insulated from one another.

3. Hysteresis

Use a soft iron core, which requires a smaller amount of energy to magnetise and demagnetise.

4. Leakage of magnetic flux

Wind the secondary coil directly on top of the primary coil.

Short Explanation

What is a Transformer?

A transformer is an electrical device that changes an alternating input voltage to an output voltage of a different value. It operates on the principle of electromagnetic induction.

Why a.c. and Not d.c.?

A direct current produces a steady, non-changing magnetic field. Since there is no change in magnetic flux linking the secondary coil, no voltage is induced across the secondary coil.

Step-up Transformer

Increases voltage: \(V_s > V_p\) and \(N_s > N_p\).

Step-down Transformer

Decreases voltage: \(V_s < V_p\) and \(N_s < N_p\).

Transformer Equation

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

\(V_p\) = voltage across primary coil (input voltage), \(V_s\) = voltage across secondary coil (output voltage), \(N_p\) = number of turns of the primary coil, \(N_s\) = number of turns of the secondary coil.

Efficiency of a Transformer

\(\eta = \dfrac{\text{Output power}}{\text{Input power}} \times 100\% = \dfrac{V_s I_s}{V_p I_p} \times 100\%\)

An ideal transformer experiences zero energy loss (\(\eta = 100\%\)), so output power equals input power: \(V_p I_p = V_s I_s\).

Cause of Energy Loss Effect Way to Reduce
Resistance of coils Coils consist of long wires with resistance; current causes Joule heating, releasing heat energy to the surroundings. Use thicker copper wire to lower the electrical resistance of the coils.
Eddy currents The changing magnetic field induces swirling eddy currents in the iron core, heating up the core. Use a laminated soft iron core made of thin iron sheets insulated from one another with lacquer/glue.
Hysteresis Continuous magnetisation and demagnetisation of the core consumes energy that is not fully recovered, dissipating as heat. Use a soft iron core, which requires a smaller amount of energy to magnetise and demagnetise.
Leakage of magnetic flux Magnetic flux produced by the primary coil is not completely linked to the secondary coil. Wind the secondary coil directly on top of the primary coil.

Useful Application of Eddy Currents

In an induction cooker, a high-frequency alternating current in a coil beneath a ceramic plate creates a rapidly changing magnetic field. This induces large eddy currents at the metal base of a steel pot, heating the pot directly to cook food efficiently.

Uses of Transformers in Daily Life

Step-down transformers: mobile phone chargers, notebook computer chargers, photocopy machines, welding machines.

Step-up transformers: microwave ovens, defibrillators, X-ray machines.

Electricity Transmission and Distribution System

Electricity generated at the power station (e.g., \(20\text{ kV}\)) passes through a step-up transformer to be stepped up to high voltages (e.g., \(275\text{ kV}\)) for transmission. Stepping up the voltage reduces the transmission current (\(I\)). Since power loss in cables is given by \(P_{\text{loss}} = I^2 R\), a smaller current drastically reduces electrical energy lost as heat in the transmission cables.

During distribution, step-down transformers reduce voltage in stages: main receiving substation stepped down to \(132\text{ kV}\), main distribution substation to \(33\text{ kV}\) (heavy industry), further substations to \(11\text{ kV}\) (light industry), and finally to \(230\text{ V}\) for residential homes.

Try to Answer First

Answer in your mind, then press “Check Answer”.

1

Why is an alternating current power supply used instead of a direct current power supply in a transformer?

Check Answer
Answer: An alternating current produces a magnetic field that changes continuously in magnitude and direction, which is necessary to induce a voltage in the secondary coil by electromagnetic induction. A direct current produces a constant magnetic field, so no voltage is induced.
2

An ideal transformer connected to a \(240\text{ V}\) supply has \(800\) turns in its primary coil and \(60\) turns in its secondary coil. If it supplies \(27\text{ W}\) of power to an appliance, what is the output voltage and secondary current?

Check Answer
Answer: Using \(\dfrac{V_s}{V_p} = \dfrac{N_s}{N_p}\), \(V_s = \dfrac{60}{800} \times 240\text{ V} = 18\text{ V}\). Since output power is \(27\text{ W}\), \(I_s = \dfrac{\text{Power}}{V_s} = \dfrac{27}{18} = 1.5\text{ A}\).
3

How does a laminated iron core help improve the efficiency of a transformer?

Check Answer
Answer: It consists of thin iron sheets separated by insulating glue, which breaks up the paths of circulating currents, thereby reducing eddy currents and minimising heat loss in the core.

Common Mistakes

  • !Forgetting that direct current cannot operate a transformer; students often assume any voltage input produces an output.
  • !Inverting the transformer equation ratio, such as incorrectly writing \(\dfrac{V_s}{V_p} = \dfrac{N_p}{N_s}\).
  • !Confusing current and voltage relationships in ideal transformers: when voltage is stepped up, current is stepped down (\(V_p I_p = V_s I_s\)).
  • !Misidentifying the role of soft iron: soft iron is used to reduce hysteresis loss because it is easily magnetised and demagnetised, whereas lamination is used to reduce eddy currents.

Concept Misunderstandings

Misunderstanding

A transformer creates extra energy when it steps up voltage.

Correct Concept

Transformers obey the principle of conservation of energy. For an ideal transformer, input power equals output power (\(V_p I_p = V_s I_s\)). Stepping up voltage results in a proportional reduction in current, so power output is never greater than power input.

Misunderstanding

Eddy currents are always harmful and have no useful applications.

Correct Concept

While eddy currents cause unwanted heat loss in transformers, they are deliberately generated in devices such as induction cookers to heat cooking pots directly and efficiently.

Summary

  • A transformer transfers electrical energy between circuits via electromagnetic induction across a soft iron core using an alternating current.
  • Turn ratio determines function: \(N_s > N_p\) steps up voltage, while \(N_s < N_p\) steps down voltage, obeying \(\dfrac{V_s}{V_p} = \dfrac{N_s}{N_p}\).
  • An ideal transformer has \(100\%\) efficiency with \(V_p I_p = V_s I_s\).
  • Four causes of energy loss are coil resistance, eddy currents, hysteresis, and flux leakage, which are reduced by thicker copper wire, laminating the core, using soft iron, and winding coils together.
  • High-voltage transmission using step-up transformers lowers transmission current, drastically minimising energy loss along power cables.

Short Activity

Objective: Test your understanding of transformers, energy loss and transmission.

A. Objective Quiz

1 Why are step-up transformers used at power stations before transmitting electricity through long-distance cables?

2 Which modification specifically reduces energy loss due to hysteresis in a transformer?

B. Fill in the Blanks

3 For an ideal transformer, the output power is equal to the input power, which can be expressed by the formula \(V_p I_p =\) .

4 In a step-down transformer, the number of turns in the secondary coil is than the number of turns in the primary coil.

C. Matching / Drag and Drop

Match each cause of energy loss in a transformer to its corresponding reduction method. If using a phone, tap the answer first, then tap the matching box.

Causes of Energy Loss
Resistance of coils
Eddy currents
Hysteresis
Leakage of magnetic flux
Methods to Reduce Loss
Use thicker copper wire so resistance is smallerCorrect ✓Try againEmpty
Use a laminated iron core made of thin insulated sheetsCorrect ✓Try againEmpty
Use soft iron as the core materialCorrect ✓Try againEmpty
Wind the secondary coil directly on the primary coilCorrect ✓Try againEmpty
 

Keywords

Transformer Step-up Transformer Step-down Transformer Ideal Transformer Eddy Currents Hysteresis Laminated Core National Grid Transmission