# The Transformer

> CIE IGCSE Physics · 0625 (2026-2028)
> Source: https://www.owlsprep.com/study/cie-0625-u4-the-transformer/

This guide covers transformer structure, operating principle, turns ratio calculations, and power transmission applications for CIE IGCSE Physics 0625, including both Core and Extended tier content.

**Prerequisites:** [Understanding of electromagnetic induction](https://www.owlsprep.com/study/cie-0625-u4-electromagnetic-induction/); [Knowledge of AC vs DC current properties](https://www.owlsprep.com/study/cie-0625-u4-alternating-current/)

## Learning objectives

- Describe transformer structure and operating principle for Core tier
- Apply the turns ratio equation to calculate voltage and turn counts
- Explain why transformers only operate with alternating current
- Explain the use of transformers in grid power transmission
- Apply ideal power and efficiency calculations for Extended tier
- Identify sources of energy loss in real transformers

## Core: Transformer Structure & Operating Principle

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.

**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.

> **info**
>
> When AC flows through the primary coil, it produces a continuously changing magnetic field in the soft iron core. This changing field cuts across the secondary coil, inducing an AC voltage across the secondary coil ends.

**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.

## Core: Turns Ratio & Voltage Calculations

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.

**Check your understanding**

1. 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

   *Why:* Rearrange $N_s = N_p \times \frac{V_s}{V_p} = 50 \times \frac{60}{10} = 300$ turns.

## Core: Transformers in Power Transmission

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 ($P = VI$), so transmitting power at high voltage reduces the current flowing through cables.

> **tip**
>
> Lower current in transmission cables reduces energy lost as heat from the resistance of the cables, making the entire transmission system far more efficient.

**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).

## Extended Only: Transformer Power & Efficiency

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: $\text{Efficiency} = \frac{\text{Output Power}}{\text{Input Power}} \times 100\%$.
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.

## Common pitfalls

- **Wrong:** Applying transformer equations to DC supply inputs
  - Why it fails: DC produces a constant magnetic field, so no induction occurs in the secondary coil, and there is no output voltage
  - Correct: Only use transformer calculations for alternating current (AC) supplies, and state AC requirement in explanation questions
- **Wrong:** Swapping primary and secondary values in the turns ratio equation
  - Why it fails: This reverses step-up/step-down results, leading to incorrect voltage and turn count answers
  - Correct: Always label $N_p$, $V_p$ (primary input) and $N_s$, $V_s$ (secondary output) clearly before substituting values
- **Wrong:** Using the ideal power equation for non-ideal transformers without efficiency adjustments
  - Why it fails: Real transformers lose energy, so output power is always lower than input power
  - Correct: 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:** Claiming transformers increase total power in transmission systems
  - Why it fails: Power is conserved (minus losses), so higher voltage corresponds to lower current, not higher power
  - Correct: State transformers change voltage and current values, but do not increase total power output
- **Wrong:** Stating the transformer core is made of steel or hard iron
  - Why it fails: Hard iron retains magnetism, leading to much higher energy losses and lower efficiency
  - Correct: Always specify the core is made of laminated soft iron for maximum efficiency

## 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 | $\frac{\text{Output Power}}{\text{Input Power}} \times 100\%$ |

## 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](https://www.owlsprep.com/study/cie-0625-u4-electromagnetic-induction/)

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