The Transformer

CIE IGCSE Physics· 4.5.6· 25 min read

1. Core: Transformer Structure & Operating Principle★★☆☆☆⏱ 5 min

A basic transformer consists of two separate insulated copper coils (primary and secondary) wrapped around a shared laminated soft iron core. There is no direct electrical connection between the two coils.

📘 Definition

Transformer

A device that uses electromagnetic induction to change the voltage of an alternating current (AC) supply, with no change to the frequency of the supply.

📐 Worked Example

Name the two core components of a transformer and state the material used for the shared core, including one property that makes this material suitable.

1
Identify the two coil components: primary coil (connected to input AC supply) and secondary coil (connected to output circuit).
2
State the core material is laminated soft iron, which is easily magnetised and demagnetised to maximise magnetic field transfer between coils.

2. Core: Turns Ratio & Voltage Calculations★★★☆☆⏱ 7 min

The ratio of the output voltage across the secondary coil to the input voltage across the primary coil is equal to the ratio of the number of turns on the secondary and primary coils, called the turns ratio.

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

If $N_{s} > N_{p}$ , the transformer is step-up, so $V_{s} > V_{p}$ . If $N_{s} < N_{p}$ , it is step-down, so $V_{s} < V_{p}$ .

📐 Worked Example

A step-down transformer has 1200 turns on the primary coil and 60 turns on the secondary coil. The input primary voltage is 240 V AC. Calculate the output secondary voltage.

1
List known values: $N_{p} = 1200$ , $N_{s} = 60$ , $V_{p} = 240$ V.
2
Rearrange the turns ratio equation to solve for $V_{s}$ : $V_{s} = V_{p} \times \frac{N _{s}}{N _{p}}$ .
3
$V_s = 240 \times \frac{60}{1200} = 240 \times 0.05 = 12\ \text{V}$
4
Final output voltage is 12 V, consistent with a step-down transformer.

✓ Quick check

A step-up transformer has 50 primary turns and an input voltage of 10 V. If the output voltage is 60 V, how many secondary turns does it have?
- 300
- 8.3
- 110
Reveal answer
300 —
Rearrange $N_{s} = N_{p} \times \frac{V _{s}}{V _{p}} = 50 \times \frac{60}{10} = 300$ turns.

3. Core: Transformers in Power Transmission★★☆☆☆⏱ 5 min

Transformers are critical for national power grids to reduce energy waste during long-distance electricity transmission. For a fixed power value, voltage and current are inversely proportional ( $P = V I$ ), so transmitting power at high voltage reduces the current flowing through cables.

📐 Worked Example

Explain why step-up transformers are installed at power stations before electricity is sent through overhead transmission lines.

1
Step-up transformers increase the voltage of the electricity generated at the power station to very high values (often 400 kV for long-distance transmission).
2
Higher voltage results in much lower current for the same power output, so less energy is wasted as heat from cable resistance.
3
Step-down transformers are used near homes and businesses to reduce the voltage to safe, usable levels (230 V in most countries).

4. Extended Only: Transformer Power & Efficiency★★★★☆Extended only⏱ 10 min

For an ideal (100% efficient) transformer, there are no energy losses, so the total power input to the primary coil equals the total power output from the secondary coil.

V_p I_p = V_s I_s

Real transformers are not 100% efficient due to energy losses from coil resistance, eddy currents in the core, and energy used to repeatedly magnetise and demagnetise the core. Efficiency is calculated as the percentage of input power converted to useful output power.

\text{Efficiency} = \frac{\text{Power Output}}{\text{Power Input}} \times 100\% = \frac{V_s I_s}{V_p I_p} \times 100\%

📐 Worked Example

An ideal transformer has an input voltage of 230 V and input current of 3 A. The output voltage is 115 V. Calculate the output current.

1
For ideal transformers, power input equals power output: $V_{p} I_{p} = V_{s} I_{s}$ .
2
Rearrange to solve for $I_{s}$ : $I_{s} = \frac{V _{p} I _{p}}{V _{s}}$ .
3
$I_s = \frac{230 \times 3}{115} = \frac{690}{115} = 6\ \text{A}$
4
Final output current is 6 A, twice the input current as expected for a step-down transformer.

📐 Worked Example

A real transformer has an input power of 800 W and an output power of 736 W. Calculate its percentage efficiency.

1
Use the efficiency formula: $Efficiency = \frac{Output Power}{Input Power} \times 100%$ .
2
$\text{Efficiency} = \frac{736}{800} \times 100 = 92\%$
3
The transformer is 92% efficient, with 8% of input energy lost as heat and sound.

5. Common Pitfalls

Wrong move:

Applying transformer equations to DC supply inputs

Why:

DC produces a constant magnetic field, so no induction occurs in the secondary coil, and there is no output voltage

Correct move:

Only use transformer calculations for alternating current (AC) supplies, and state AC requirement in explanation questions

Wrong move:

Swapping primary and secondary values in the turns ratio equation

Why:

This reverses step-up/step-down results, leading to incorrect voltage and turn count answers

Correct move:

Always label $N_{p}$ , $V_{p}$ (primary input) and $N_{s}$ , $V_{s}$ (secondary output) clearly before substituting values

Wrong move:

Using the ideal power equation for non-ideal transformers without efficiency adjustments

Why:

Real transformers lose energy, so output power is always lower than input power

Correct move:

Only use $V_{p} I_{p} = V_{s} I_{s}$ if the question explicitly states the transformer is ideal, multiply input power by efficiency to get output power for real transformers

Wrong move:

Claiming transformers increase total power in transmission systems

Why:

Power is conserved (minus losses), so higher voltage corresponds to lower current, not higher power

Correct move:

State transformers change voltage and current values, but do not increase total power output

Wrong move:

Stating the transformer core is made of steel or hard iron

Why:

Hard iron retains magnetism, leading to much higher energy losses and lower efficiency

Correct move:

Always specify the core is made of laminated soft iron for maximum efficiency

6. Quick Reference Cheatsheet

Concept	Core Tier Formula/Rule	Extended Tier Add-on
Turns-Voltage relation	$\frac{V _{s}}{V _{p}} = \frac{N _{s}}{N _{p}}$	Same as Core
Step-up transformer	$N_{s} > N_{p}$ , $V_{s} > V_{p}$	$I_{s} < I_{p}$
Step-down transformer	$N_{s} < N_{p}$ , $V_{s} < V_{p}$	$I_{s} > I_{p}$
Ideal Power	Not required	$V_{p} I_{p} = V_{s} I_{s}$
Efficiency	Not required	$\frac{Output Power}{Input Power} \times 100%$

7. Frequently Asked

Why do transformers not work with DC supply?

Direct current produces a constant magnetic field in the primary coil, so there is no change in magnetic flux linkage with the secondary coil. No electromotive force (emf) is induced in the secondary coil, so no output voltage is generated.

Why are laminated soft iron cores used in transformers?

Soft iron is easily magnetised and demagnetised to maximise magnetic field transfer. Lamination (thin insulated iron sheets) reduces energy loss from eddy currents induced in the core by the changing magnetic field.

Going deeper

syllabusCIE 0625 2026-2028 Syllabus
revision_guideElectromagnetic Induction Guide

What's Next

Now you have mastered transformer principles and calculations, you can apply this knowledge to broader electromagnetism topics for your CIE IGCSE Physics 0625 exam. Transformers are a key application of electromagnetic induction, so revisiting induction fundamentals will strengthen your understanding of how transformers operate. Be sure to practice linking transformer concepts to power transmission systems, which are frequently tested in extended response questions. For Extended tier students, practice efficiency calculations alongside other energy transfer topics to prepare for multi-step exam questions. Next, you can move on to revising other electromagnetism applications such as electric motors and generators, which are also common exam topics.

→Electromagnetic Induction Revision