Use the terms primary, secondary, step-up and step-down

4.5.6 The Transformer

Learning Objective

By the end of this lesson you should be able to:

  • Identify the primary and secondary windings of a transformer.
  • Distinguish between a step‑up and a step‑down transformer.
  • Apply the turn‑ratio and power equations to solve typical IGCSE problems.
  • Explain why a laminated core is used and name the two main types of loss in a real transformer.

1. Construction of a Simple Transformer

  • Core material: soft‑iron (usually laminated).
    • Lamination reduces **eddy‑current loss** by breaking up large circulating currents in the iron.
    • The iron also exhibits **hysteresis loss** due to the continual magnetisation‑demagnetisation cycle.
  • Windings: insulated copper wire wound around the core.
    • Primary winding – connected to the source (input).
    • Secondary winding – delivers the transformed voltage (output).
  • Both windings are electrically isolated from the core and from each other.

2. Principle of Operation (AC only)

When an alternating current flows through the primary winding it creates a time‑varying magnetic flux \( \Phi \) in the core. According to Faraday’s law an emf is induced in any coil that links this changing flux:

\( \displaystyle \mathcal{E} = -N\,\frac{d\Phi}{dt} \)

The same flux links the secondary winding, so an emf is induced there as well. Because the primary and secondary share the same core, the ratio of the induced emfs equals the ratio of the numbers of turns.

3. Turn‑Ratio Equations (Ideal Transformer)

For an ideal transformer (no losses):

\( \displaystyle \frac{V_{s}}{V_{p}} = \frac{N_{s}}{N_{p}} \qquad\text{and}\qquad \frac{I_{s}}{I_{p}} = \frac{N_{p}}{N_{s}} \)

  • \(V_{p}\) – primary (input) voltage
  • \(V_{s}\) – secondary (output) voltage
  • \(I_{p}\) – primary current
  • \(I_{s}\) – secondary current
  • \(N_{p}\) – number of turns in the primary winding
  • \(N_{s}\) – number of turns in the secondary winding

4. Power Relations

In an ideal transformer the input power equals the output power (100 % efficiency):

\( \displaystyle V_{p}I_{p}=V_{s}I_{s} \)

Real transformers have two principal losses:

  • Copper loss – \( P_{\text{copper}} = I^{2}R \) (heating of the windings).
  • Core loss – the sum of hysteresis loss and eddy‑current loss in the iron core.

The efficiency is therefore

\( \displaystyle \eta = \frac{P_{\text{out}}}{P_{\text{in}}}\times100\% \)

5. Step‑Up vs. Step‑Down Transformers

Feature Step‑Up Transformer Step‑Down Transformer
Turn ratio \((N_{s}/N_{p})\) \(>1\) \(<1\)
Secondary voltage \((V_{s})\) Higher than \(V_{p}\) Lower than \(V_{p}\)
Secondary current \((I_{s})\) Lower than \(I_{p}\) Higher than \(I_{p}\)
Typical uses Power‑transmission stations (raise voltage for long‑distance lines) Household adapters, chargers (reduce mains voltage to safe levels)

6. Real‑World Application – Power Grids

7. Advantages of High‑Voltage Transmission

  • **Reduced \(I^{2}R\) losses** – power loss varies with the square of the current.
  • **Smaller conductor cross‑section** – lower current allows thinner, cheaper cables.
  • **Improved overall efficiency** – less energy wasted as heat, lowering operating costs.

8. Worked Example

Problem: A transformer has 500 turns on the primary and 1500 turns on the secondary. The primary is connected to a 230 V AC supply.

  1. State whether the transformer is step‑up or step‑down.
  2. Calculate the secondary voltage.
  3. If the secondary supplies a lamp drawing 2 A, find the primary current (ignore losses).

Solution

  1. Type: \(N_{s}=1500 > N_{p}=500\) → step‑up transformer.
  2. Secondary voltage: \[ \frac{V_{s}}{V_{p}} = \frac{N_{s}}{N_{p}} = \frac{1500}{500}=3 \qquad\Rightarrow\qquad V_{s}=3\times230\;\text{V}=690\;\text{V} \]
  3. Primary current (ideal): \[ P_{s}=V_{s}I_{s}=690\;\text{V}\times2\;\text{A}=1380\;\text{W} \] Assuming 100 % efficiency, \(P_{p}=P_{s}\): \[ I_{p}= \frac{P_{p}}{V_{p}} = \frac{1380\;\text{W}}{230\;\text{V}} \approx 6.0\;\text{A} \]

9. Common Misconceptions

  • “A transformer changes the frequency.” – The output frequency is the same as the input frequency; only voltage and current are transformed.
  • “More turns always give a higher voltage.” – Voltage change depends on the **ratio** of secondary to primary turns, not on the absolute number of turns.
  • “Power is created or destroyed in a transformer.” – For an ideal transformer power is conserved; real devices have small losses (copper loss and core loss).

10. Suggested Diagram

Cross‑section of a transformer showing the laminated soft‑iron core, primary winding (left), secondary winding (right), and the direction of the induced emf.

Summary

Transformers are indispensable for changing AC voltages in power systems. The primary winding receives the input voltage, the secondary delivers the transformed voltage, and the turn ratio determines whether the device is a step‑up or step‑down transformer. Understanding the relationships between turns, voltage, current, and power, together with the reasons for a laminated core and the nature of real‑world losses, equips you to analyse electrical networks and to answer typical IGCSE exam questions confidently.

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