Explain the principle of operation of a simple iron-cored transformer

Published by Patrick Mutisya · 14 days ago

Cambridge IGCSE Physics 0625 – The Transformer

4.5.6 The Transformer

Objective

Explain the principle of operation of a simple iron‑cored transformer.

1. What is a Transformer?

A transformer is a static electrical device that transfers electrical energy between two or more circuits through electromagnetic induction. It is used to change (step‑up or step‑down) the voltage level while keeping the frequency unchanged.

2. Basic Construction

  • Iron core – provides a low‑reluctance path for the magnetic flux.

  • Primary winding – coil of wire connected to the input (source) voltage.

  • Secondary winding – coil of wire where the transformed voltage is induced.

Suggested diagram: Simple iron‑cored transformer showing primary and secondary windings around a laminated iron core.

3. Principle of Operation

When an alternating current (AC) flows through the primary winding, it creates a time‑varying magnetic flux \$\Phi\$ in the iron core. According to Faraday’s law of electromagnetic induction, a changing magnetic flux linking a coil induces an emf in that coil.

For a coil of \$N\$ turns, the induced emf is

\$\mathcal{E} = -N \frac{d\Phi}{dt}\$

Because the same core flux \$\Phi\$ links both the primary and secondary windings, an emf is also induced in the secondary winding:

\$\mathcal{E}\text{s} = -N\text{s} \frac{d\Phi}{dt}\$

Dividing the secondary emf by the primary emf gives the turns ratio relationship:

\$\frac{V\text{s}}{V\text{p}} = \frac{N\text{s}}{N\text{p}}\$

where \$V\text{p}\$ and \$V\text{s}\$ are the rms voltages across the primary and secondary, and \$N\text{p}\$ and \$N\text{s}\$ are the numbers of turns.

4. Voltage and Current Relationships

Assuming an ideal transformer (no losses), the apparent power is the same on both sides:

\$V\text{p} I\text{p} = V\text{s} I\text{s}\$

Combining this with the turns ratio gives the current relationship:

\$\frac{I\text{s}}{I\text{p}} = \frac{N\text{p}}{N\text{s}}\$

5. Example Calculation

Given a transformer with \$N\text{p}=500\$ turns, \$N\text{s}=200\$ turns, and an applied primary voltage of \$V_\text{p}=240\ \text{V}\$ (rms):

  1. Find the secondary voltage:

    \$V\text{s}=V\text{p}\frac{N\text{s}}{N\text{p}}=240\ \text{V}\times\frac{200}{500}=96\ \text{V (rms)}\$

  2. If the secondary supplies a load drawing \$I_\text{s}=2\ \text{A}\$, the primary current is:

    \$I\text{p}=I\text{s}\frac{N\text{s}}{N\text{p}}=2\ \text{A}\times\frac{200}{500}=0.8\ \text{A}\$

6. Efficiency and Real‑World Considerations

Real transformers are not ideal. Losses arise from:

  • Core losses (hysteresis and eddy currents) – depend on the material and frequency.

  • Copper losses – \$I^{2}R\$ heating in the windings.

  • Leakage flux – magnetic flux that does not link both windings.

Typical efficiencies for power transformers are 95 % – 99 %.

7. Summary Table

ParameterIdeal TransformerReal Transformer
Voltage ratio\$\displaystyle \frac{V\text{s}}{V\text{p}}=\frac{N\text{s}}{N\text{p}}\$Same, but small deviation due to regulation
Current ratio\$\displaystyle \frac{I\text{s}}{I\text{p}}=\frac{N\text{p}}{N\text{s}}\$Same, with additional loss‑related current
Power\$V\text{p}I\text{p}=V\text{s}I\text{s}\$Input power \$>\$ output power (losses)
Efficiency100 %Typically 95 %–99 %

8. Key Points to Remember

  • The magnetic flux in the core is produced by the primary current.

  • Both windings experience the same changing flux, so the induced emf is proportional to the number of turns.

  • Voltage transformation follows the turns ratio; current transformation follows the inverse of the turns ratio.

  • Power is conserved (ignoring losses); real transformers have efficiency less than 100 %.