Transformers and their operation

What you'll learn

Transformers are essential devices in the National Grid that allow electricity to be transmitted efficiently across long distances. This guide covers how transformers work using electromagnetic induction, the difference between step-up and step-down transformers, and how to perform calculations using the transformer equation.

Key terms and definitions

Transformer — a device that changes the voltage of an alternating current supply

Primary coil — the input coil of a transformer where alternating current enters

Secondary coil — the output coil of a transformer where the transformed voltage is produced

Step-up transformer — a transformer that increases voltage from primary to secondary coil

Step-down transformer — a transformer that decreases voltage from primary to secondary coil

Turns ratio — the ratio of the number of turns on the secondary coil to the number of turns on the primary coil

Iron core — a magnetically soft material that links the primary and secondary coils and concentrates the magnetic field

Electromagnetic induction — the process of generating a potential difference (voltage) in a conductor when it experiences a changing magnetic field

Core concepts

Structure of a transformer

A transformer consists of two coils of wire wound around an iron core:

• The primary coil connects to the input alternating voltage supply
• The secondary coil connects to the output circuit
• Both coils are wound around the same iron core but are not electrically connected to each other
• The iron core is made from magnetically soft iron that is easily magnetised and demagnetised

The coils are insulated from each other and from the core. Energy is transferred between the coils through the magnetic field in the core, not through direct electrical connection.

How transformers work

Transformers operate through electromagnetic induction:

1. An alternating current flows through the primary coil
2. This creates a changing magnetic field in the iron core
3. The iron core concentrates and links this magnetic field to the secondary coil
4. The changing magnetic field passes through the secondary coil
5. This induces an alternating voltage across the secondary coil by electromagnetic induction
6. If the secondary coil is part of a complete circuit, an alternating current flows

Critical requirement: Transformers only work with alternating current (AC). Direct current (DC) produces a constant magnetic field which does not induce a voltage in the secondary coil. The changing nature of AC is essential for the continuous changing magnetic field needed for electromagnetic induction.

Step-up and step-down transformers

The voltage across the secondary coil depends on the number of turns on each coil:

Step-up transformers:

• Have more turns on the secondary coil than the primary coil
• Increase the voltage from primary to secondary
• Decrease the current from primary to secondary (assuming 100% efficiency)
• Used at power stations to increase voltage to 400,000 V for transmission in the National Grid

Step-down transformers:

• Have fewer turns on the secondary coil than the primary coil
• Decrease the voltage from primary to secondary
• Increase the current from primary to secondary (assuming 100% efficiency)
• Used near homes and businesses to reduce voltage to safe, usable levels (230 V)
• Also used in mobile phone chargers and laptop power supplies

The transformer equation

For a transformer, the ratio of voltages is equal to the ratio of turns:

V_p / V_s = n_p / n_s

or rearranged:

V_s / V_p = n_s / n_p

Where:

• V_p = potential difference across primary coil (V)
• V_s = potential difference across secondary coil (V)
• n_p = number of turns on primary coil
• n_s = number of turns on secondary coil

This equation applies to both ideal (100% efficient) and real transformers.

Power and efficiency in transformers

For an ideal (100% efficient) transformer, electrical power is conserved:

Power in primary coil = Power in secondary coil

V_p × I_p = V_s × I_s

Where:

• I_p = current in primary coil (A)
• I_s = current in secondary coil (A)

This shows that:

• When voltage increases (step-up), current decreases proportionally
• When voltage decreases (step-down), current increases proportionally

Real transformers are not 100% efficient. Energy is wasted through:

• Resistance in the coils — causes heating in the copper wire
• Eddy currents — induced currents in the iron core that cause heating
• Magnetic losses — not all magnetic field lines link both coils

Modern transformers achieve approximately 98-99% efficiency. For real transformers:

Efficiency = (V_s × I_s) / (V_p × I_p) × 100%

Applications in the National Grid

Transformers are crucial for efficient electricity transmission:

At the power station:

• Step-up transformers increase voltage to 400,000 V or 275,000 V
• This reduces current for the same power transmitted
• Lower current means less energy wasted as heat in transmission cables (since energy wasted = I²R)

Near consumers:

• Step-down transformers at local substations reduce voltage to 33,000 V
• Further step-down transformers reduce voltage to 230 V for homes and businesses
• This provides safe, usable voltage levels

The use of high voltages for transmission significantly reduces energy losses. For the same power transmitted, doubling the voltage halves the current, which reduces heating losses to one quarter (since power loss is proportional to I²).

Worked examples

Example 1: Basic transformer calculation

Question: A transformer has 200 turns on its primary coil and 50 turns on its secondary coil. The input voltage is 240 V. Calculate the output voltage. [3 marks]

Solution:

Step 1: Write down the known values

• n_p = 200 turns
• n_s = 50 turns
• V_p = 240 V
• V_s = ?

Step 2: Select and write the equation [1 mark] V_s / V_p = n_s / n_p

Step 3: Substitute values [1 mark] V_s / 240 = 50 / 200

Step 4: Calculate the answer [1 mark] V_s = 240 × (50 / 200) V_s = 240 × 0.25 V_s = 60 V

Example 2: Step-up transformer with current

Question: A step-up transformer in the National Grid increases voltage from 25,000 V to 400,000 V. The primary coil carries a current of 800 A. Assuming the transformer is 100% efficient, calculate the current in the secondary coil. [4 marks]

Solution:

Step 1: Write down known values

• V_p = 25,000 V
• V_s = 400,000 V
• I_p = 800 A
• I_s = ?

Step 2: For 100% efficiency, state the principle [1 mark] Power in primary = Power in secondary

Step 3: Write the equation [1 mark] V_p × I_p = V_s × I_s

Step 4: Rearrange [1 mark] I_s = (V_p × I_p) / V_s

Step 5: Substitute and calculate [1 mark] I_s = (25,000 × 800) / 400,000 I_s = 20,000,000 / 400,000 I_s = 50 A

Example 3: Efficiency calculation

Question: A transformer has an input voltage of 230 V and an input current of 5 A. The output voltage is 12 V and the output current is 92 A. Calculate the efficiency of the transformer. [4 marks]

Solution:

Step 1: Calculate input power [1 mark] Power input = V_p × I_p = 230 × 5 = 1150 W

Step 2: Calculate output power [1 mark] Power output = V_s × I_s = 12 × 92 = 1104 W

Step 3: Write efficiency equation [1 mark] Efficiency = (Power output / Power input) × 100%

Step 4: Calculate efficiency [1 mark] Efficiency = (1104 / 1150) × 100% Efficiency = 96%

Common mistakes and how to avoid them

• Using the transformer equation with direct current — Remember that transformers only work with AC. If a question mentions DC supply, the transformer will not function (zero output voltage).

• Confusing primary and secondary coils — Always label your values with subscripts (p for primary, s for secondary) and check which coil is connected to the input and which to the output.

• Incorrectly inverting the turns ratio — The equation V_s / V_p = n_s / n_p shows that the voltage ratio equals the turns ratio. Check your ratios are the right way up (secondary over primary on both sides).

• Assuming all transformers are 100% efficient — Only use V_pI_p = V_sI_s when the question explicitly states the transformer is ideal or 100% efficient. Otherwise, calculate efficiency from the power values.

• Forgetting that current changes inversely to voltage — In a step-up transformer, voltage increases so current decreases. In a step-down transformer, voltage decreases so current increases (assuming 100% efficiency).

• Not using standard form for large voltages — National Grid voltages like 400,000 V can be written as 4 × 10⁵ V to reduce calculation errors. Ensure your calculator can handle large numbers or use standard form throughout.

Exam technique for "Transformers and their operation"

• For "explain how" questions about transformer operation, use the correct sequence: alternating current → changing magnetic field → iron core links the field → electromagnetic induction in secondary coil. Include the phrase "electromagnetic induction" for the mark.

• For calculations, always show your working — Write the equation, substitute values with units, then calculate. This allows partial credit even if your final answer is incorrect. A 3-mark calculation typically awards 1 mark for equation, 1 for substitution, 1 for correct answer.

• Command word "describe" requires you to state facts (e.g., "Step-up transformers increase voltage"). Command word "explain" requires reasons (e.g., "Step-up transformers increase voltage because there are more turns on the secondary coil than the primary coil").

• For questions on the National Grid, link transformers to efficiency: high voltage → low current → reduced heating in cables → less energy wasted. This shows understanding beyond simple recall.

Quick revision summary

Transformers change AC voltage using two coils around an iron core. Alternating current in the primary coil creates a changing magnetic field that induces voltage in the secondary coil through electromagnetic induction. Step-up transformers (more secondary turns) increase voltage; step-down transformers (fewer secondary turns) decrease voltage. The transformer equation V_s/V_p = n_s/n_p links voltage and turns. For 100% efficiency, V_pI_p = V_sI_s. Transformers in the National Grid reduce transmission energy losses by increasing voltage and decreasing current.