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

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

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

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.

With AC Supply

Current changes continuously, so magnetic flux changes continuously.

Therefore emf is continuously induced in the secondary coil.

With Steady DC Supply

Current becomes constant after switching.

So magnetic flux becomes constant.

Hence:

• no continuous induced emf in the secondary
• only a brief transient emf when switching on or off

Exam Conclusion

A transformer does not operate properly with steady DC.

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.

Current Ratio

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

If voltage increases, current decreases correspondingly.

Power Conservation

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

Input power equals output power in an ideal transformer. In AC contexts, these are typically interpreted using rms values.

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

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.

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.

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.

Methods to Reduce Losses

Copper Loss

Use thick low-resistance copper wires.

Eddy Current Loss

Use a laminated core with insulated layers.

Hysteresis Loss

Use soft iron or suitable magnetic materials.

Flux Leakage

Wind coils closely on the same core for tight magnetic coupling.

Efficiency

Efficiency:

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

or

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

where:

• $P_s$ = output power
• $P_p$ = input power

Real transformers have:

$$\eta <100\%$$

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 = $450\text{W}$

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

Common Exam Traps

Trap 1

Thinking transformers work with steady DC.

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

Trap 2

Reversing the turns ratio.

Use:

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

Trap 3

Thinking step-up increases power.

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

Trap 4

Forgetting current decreases in a step-up transformer.

Trap 5

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

Loss depends strongly on current.

Formula Sheet

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