Transformers

Overview

A transformer is a device that uses electromagnetic induction to change alternating voltages.

It can:

• increase voltage in a step-up transformer
• decrease voltage in a step-down transformer

Transformers are essential in electrical power transmission because they allow efficient transfer of electrical energy over long distances.

This topic builds directly from:

• Electromagnetic Induction
• Alternating Current

Big Picture

A transformer does not transfer charge from the primary coil to the secondary coil. The two coils are not electrically connected.

Energy is transferred through changing magnetic flux in the iron core:

$$\text{AC in primary}\to \text{changing magnetic flux in core}\to \text{changing flux linkage in secondary}\to \text{induced emf}$$

Core Ideas

Transformer questions revolve around a compact set of linked ideas:

• AC in the primary coil produces changing magnetic flux in the core
• changing flux linkage induces emf in the secondary coil
• voltage ratio follows turns ratio
• current ratio changes inversely
• ideal transformers conserve power, but real transformers have losses
• high-voltage transmission reduces current and therefore reduces $I^{2}R$ cable loss

Minimum Exam Model

For an ideal transformer:

• the primary and secondary coils are not electrically connected
• the iron core links magnetic flux between the coils
• the supply must be AC so that magnetic flux changes continuously
• voltage ratio follows turns ratio
• current ratio is inverse to voltage ratio
• input power equals output power

For a real transformer:

• output power is less than input power
• energy is lost through copper loss, eddy currents, hysteresis, and flux leakage

Exam Relevance

Students are expected to:

• explain why transformers need AC
• apply voltage, current, and power ratios correctly
• distinguish step-up from step-down transformers
• explain transmission reasoning using $P=VI$ and $P_{\text{loss}}=I^{2}R$
• account for efficiency and practical losses

Prior Knowledge Links

Electromagnetic Induction

A changing magnetic flux linkage induces emf.

Alternating Current

Alternating current changes direction and magnitude continuously, so it creates changing magnetic flux in the transformer core.

Power Generation Source

AC supplied to transformers commonly comes from Alternating Current Generators.

Key Representations

Basic Construction

A simple transformer consists of:

• a primary coil with $N_p$ turns
• a secondary coil with $N_s$ turns
• a laminated soft iron core
• an AC supply connected to the primary coil
• a load connected to the secondary coil

Caption: A transformer uses mutual induction. Alternating current in the primary coil produces changing magnetic flux in the laminated soft iron core, inducing an emf in the secondary coil.

Roles of Components

Primary Coil

Receives AC input voltage $V_p$.

Secondary Coil

Delivers output voltage $V_s$.

Soft Iron Core

Provides a low-reluctance path for magnetic flux and improves magnetic coupling.

Laminated Core

Reduces eddy-current losses.

Principle of Operation

Step 1: Changing Current in Primary Coil

Alternating current in the primary coil changes continuously.

Step 2: Changing Magnetic Flux in Core

The changing current produces changing magnetic flux in the iron core.

Step 3: Induced emf in Secondary Coil

This changing flux links the secondary coil and induces emf by Faraday’s law.

This transfer of energy via changing magnetic flux is called mutual induction.

Why Transformers Need AC

Transformers require changing magnetic flux.

Supply to primary	Primary current	Magnetic flux	Secondary emf
AC	continuously changing	continuously changing	continuously induced
steady DC after switching	constant	constant	no continuous emf
DC switching on or off	briefly changing	briefly changing	brief transient emf

Exam Conclusion

A transformer does not operate properly with steady DC because steady DC does not produce a continuously changing magnetic flux.

Ideal Transformer Equations

For an ideal transformer:

Turns Ratio / Voltage Ratio

$$\frac{V_s}{V_p}=\frac{N_s}{N_p}$$

Voltage is proportional to number of turns.

Derivation from Faraday’s Law

Faraday’s law gives:

$$\epsilon =-N\frac{d\Phi }{dt}$$

In an ideal transformer:

• both coils are linked by the same changing magnetic flux $\Phi$ in the core
• each turn therefore experiences the same rate of change of flux

So for the primary and secondary coils:

$$V_p\approx {\epsilon }_p=-N_p\frac{d\Phi }{dt}$$ $$V_s\approx {\epsilon }_s=-N_s\frac{d\Phi }{dt}$$

Dividing gives:

$$\frac{V_s}{V_p}=\frac{N_s}{N_p}$$

This assumes negligible winding resistance and negligible flux leakage, so the induced emf magnitudes are approximately equal to the terminal p.d. magnitudes.

Current Ratio

$$\frac{I_p}{I_s}=\frac{N_s}{N_p}$$

The current ratio is not another direct turns-ratio rule to apply blindly. It follows from power conservation.

If a transformer steps up voltage, the secondary current must be smaller in the ideal model because the same power is transferred at a higher voltage.

Power Conservation

$$V_p{I}_p=V_s{I}_s$$

Input power equals output power in an ideal transformer. In AC transformer calculations, voltages and currents are normally rms values unless otherwise stated.

Exam Strategy: Ideal Transformer Ratios

Use the voltage ratio first when turns are given:

$$\frac{V_s}{V_p}=\frac{N_s}{N_p}$$

Then use ideal power conservation to find current:

$$V_p{I}_p=V_s{I}_s$$

This prevents the common error of making current increase in the same ratio as voltage. In an ideal transformer, if voltage is stepped up, current is stepped down.

Step-Up Transformer

A step-up transformer increases voltage.

Condition:

$$N_s>N_p$$

Therefore:

$$V_s>V_p$$

and current decreases:

$$I_s<I_p$$

Uses

• national-grid transmission
• X-ray equipment
• some industrial systems

Step-Down Transformer

A step-down transformer decreases voltage.

Condition:

$$N_s<N_p$$

Therefore:

$$V_s<V_p$$

and current increases:

$$I_s>I_p$$

Uses

• domestic appliances
• chargers
• electronics power supplies

Caption: A step-up transformer has more secondary turns than primary turns, so it increases voltage and decreases current. A step-down transformer does the opposite.

Step-up vs Step-down Summary

Transformer	Turns condition	Voltage effect	Current effect	Typical use
Step-up	$N_s>N_p$	$V_s>V_p$	$I_s<I_p$	long-distance transmission
Step-down	$N_s<N_p$	$V_s<V_p$	$I_s>I_p$	domestic supply / adaptors

An ideal transformer does not step up power. It changes the voltage-current combination while conserving total power.

Power Transmission

Why High Voltage Is Used

For AC transmission, use rms values:

$$P=V_{\text{rms}}{I}_{\text{rms}}$$

For fixed power transmitted:

• higher $V$
• lower $I$

Why Lower Current Helps

Power loss in cables:

$$P_{\text{loss}}=I^{2}R$$

So reducing current greatly reduces heating loss.

Caption: For a fixed transmitted power, increasing the transmission voltage reduces the current. Since cable heating loss is $I^{2}R$, a smaller current greatly reduces energy loss.

National Grid Context

1. Power stations generate AC.
2. Step-up transformers raise voltage for transmission.
3. Electricity travels through long-distance cables.
4. Step-down transformers lower voltage for consumers.

Exam Strategy: Transmission Questions

For transmission questions, keep two ideas separate:

• the useful transmitted power is usually treated as fixed
• the unwanted cable loss is calculated using $P_{\text{loss}}=I^{2}R$

The reason for stepping up voltage is not to create extra power. It is to deliver the same power with a smaller current, reducing heating in the cables.

Transmission Exam Workflow

1. Identify the useful transmitted power.
2. Use $P=VI$ to find the transmission current.
3. Use $P_{\text{loss}}=I^{2}R$ to calculate cable loss.
4. Compare losses at different transmission voltages.
5. Conclude that high voltage gives lower current, and lower current greatly reduces cable heating.

The current in $P_{\text{loss}}=I^{2}R$ is the current in the transmission cables, not the current in a domestic appliance after step-down.

Energy Losses in Practical Transformers

Real transformers are not perfect.

1. Copper Loss

Heating in coil wires due to resistance:

$$P=I^{2}R$$

2. Eddy Current Loss

Changing flux induces currents in the iron core, causing heating.

3. Hysteresis Loss

Repeated magnetisation and demagnetisation of the core wastes energy.

4. Flux Leakage

Not all magnetic flux from the primary links the secondary coil.

Caption: Practical transformer losses include copper loss, eddy-current loss, hysteresis loss, and flux leakage. Design choices reduce these losses but cannot remove them completely.

Methods to Reduce Losses

Loss	Where it occurs	Cause	Reduction method
Copper loss	coils	current heats wire resistance	thick, low-resistance copper wires
Eddy-current loss	iron core	changing flux induces currents in the core	laminated insulated core
Hysteresis loss	iron core	repeated magnetisation and demagnetisation	soft iron or low-hysteresis material
Flux leakage	around core and coils	not all primary flux links secondary	close winding and suitable core design

Efficiency

Efficiency:

$$\eta =\frac{\text{output power}}{\text{input power}}\times 100\%$$

or

$$\eta =\frac{P_s}{P_p}\times 100\%$$

Using rms electrical quantities:

$$\eta =\frac{V_s{I}_s}{V_p{I}_p}\times 100\%$$

where:

• $P_s$ = output power
• $P_p$ = input power
• $V_s{I}_s$ and $V_p{I}_p$ use rms values for AC circuits

Real transformers have:

$$\eta <100\%$$

Ideal vs Real Transformers

An ideal transformer is a model:

• no coil resistance
• no flux leakage
• no eddy-current loss
• no hysteresis loss

A real transformer has losses, so:

$$P_s<P_p$$

and:

$$\eta <100\%$$

Worked Examples

Example 1: Output Voltage

A transformer has:

• $N_p=200$
• $N_s=1000$
• $V_p=12\text{V}$

Use:

$$\frac{V_s}{V_p}=\frac{N_s}{N_p}$$ $$V_s=12\times \frac{1000}{200}=60\text{V}$$

Example 2: Current Ratio

If:

• $I_p=2.0\text{A}$
• $\frac{N_s}{N_p}=5$

Then:

$$\frac{I_p}{I_s}=5$$ $$I_s=\frac{2.0}{5}=0.40\text{A}$$

Example 3: Efficiency

Input power = $500\text{W}$
Output power = $450\text{W}$

$$\eta =\frac{450}{500}\times 100\%=90\%$$

Example 4: Efficiency from rms Voltage and Current

A real transformer has:

• $V_p=240\text{V}$
• $I_p=0.50\text{A}$
• $V_s=12\text{V}$
• $I_s=8.0\text{A}$

Input power:

$$P_p=V_p{I}_p=240(0.50)=120\text{W}$$

Output power:

$$P_s=V_s{I}_s=12(8.0)=96\text{W}$$

Efficiency:

$$\eta =\frac{96}{120}\times 100\%=80\%$$

Common Exam Traps

Trap 1

Thinking transformers work with steady DC.

Fix: changing flux is required, so AC is needed.

Trap 2

Thinking current passes directly from the primary coil to the secondary coil.

Fix: the coils are not electrically connected. Energy is transferred through changing magnetic flux in the core.

Trap 3

Reversing the turns ratio.

Use:

$$\frac{V_s}{V_p}=\frac{N_s}{N_p}$$

Trap 4

Thinking step-up increases power.

An ideal transformer changes the voltage-current ratio, not the total power.

Trap 5

Forgetting current decreases in a step-up transformer.

Fix: use power conservation. If $V_s>V_p$ and the transformer is ideal, then $I_s<I_p$.

Trap 6

Using $I^{2}R$ loss incorrectly.

Loss depends strongly on current.

Fix: first find the transmission current from the transmitted power and transmission voltage, then use $P_{\text{loss}}=I^{2}R$ for the cables.

Trap 7

Confusing useful transmitted power with cable loss.

Fix: useful transmitted power is the power being delivered along the grid. Cable loss is the unwanted heating power dissipated in the cable resistance.

Trap 8

Forgetting rms values in AC power calculations.

Fix: transformer power calculations normally use rms voltage and rms current unless the question states otherwise.

Concept Checkpoints

Operation

• Why are the primary and secondary coils not electrically connected?
• What role does the iron core play?
• Why must the magnetic flux change?

Ideal Ratios

• If $N_s>N_p$, what happens to $V_s$?
• What happens to $I_s$ in an ideal step-up transformer?
• Does the transformer create extra power?

Transmission

• Why does increasing transmission voltage reduce cable loss?
• Which current should be used in $P_{\text{loss}}=I^{2}R$?

Losses

• Which loss is reduced by lamination?
• Which loss is reduced by using thick copper wire?
• Why is efficiency less than $100\%$ in a real transformer?

Formula Sheet

Ideal Transformer Ratio

$$\frac{N_s}{N_p}=\frac{V_s}{V_p}=\frac{I_p}{I_s}$$

Voltage Ratio

$$\frac{V_s}{V_p}=\frac{N_s}{N_p}$$

Current Ratio

$$\frac{I_p}{I_s}=\frac{N_s}{N_p}$$

Ideal Power

$$V_p{I}_p=V_s{I}_s$$

Cable Loss

$$P_{\text{loss}}=I^{2}R$$

Efficiency

$$\eta =\frac{\text{output power}}{\text{input power}}\times 100\%$$

Summary

A transformer works by mutual induction:

1. AC in the primary coil
2. changing magnetic flux in the core
3. induced emf in the secondary

It allows voltage to be changed efficiently and is crucial for power transmission.

Links

• Electromagnetic Induction
• Alternating Current
• Alternating Current Generators
• RMS and AC Power