Recall and use the equation V_p / V_s = N_p / N_s where p and s refer to primary and secondary

4.5.6 The Transformer

Learning Objective

Recall and use the fundamental transformer equations

where the subscripts p and s denote the primary and secondary windings respectively.

1. Construction of a Simple (Soft‑Iron Core) Transformer

• Core material: laminated soft‑iron (or silicon‑steel)

  • Very high magnetic permeability – concentrates the alternating magnetic flux.
  • Lamination reduces eddy‑current losses.
  • Soft‑iron has low hysteresis loss because it can be magnetised and demagnetised easily.

• Windings: insulated copper wire wound in two separate coils

  • Primary coil (N_p) – connected to the source.
  • Secondary coil (N_s) – delivers the transformed voltage to the load.

• Magnetic circuit: the core provides a low‑reluctance path so that the same alternating flux Φ links both windings.

Diagram suggestion: cross‑section showing the laminated soft‑iron core, primary winding on one limb, secondary winding on the other, and the direction of the alternating flux.

2. How a Transformer Works

1. An alternating current in the primary creates a time‑varying (alternating) magnetic flux Φ in the core (Faraday’s law).
2. The same flux passes through the secondary winding, inducing an emf (voltage) in it.
3. Because the flux is common to both coils, the induced voltages are directly proportional to the number of turns in each coil.

3. Derivation of the Turns‑Ratio Equations (Ideal Transformer)

1. Faraday’s law for each winding

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

   where N is the number of turns and Φ the magnetic flux.

2. Since the same Φ links both windings, the ratio of the induced emfs equals the ratio of turns

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

3. For an ideal transformer (100 % efficiency) input power equals output power

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

   Combining this with the voltage‑ratio gives the current‑ratio (inverse of the voltage ratio)

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

4. Ideal vs. Real Transformers

When a question does not mention efficiency or losses, you may assume an ideal transformer.

5. Step‑Up and Step‑Down Transformers

6. Advantage of Using High‑Voltage Transmission

• For a given power, raising the voltage reduces the current (I = P/V).
• Lower current means much smaller I²R losses in the transmission lines, allowing thinner, cheaper cables and more efficient long‑distance power delivery.

7. Worked Example (Ideal Transformer)

Problem: A transformer has 500 turns on the primary and 200 turns on the secondary. The primary voltage is 240 V and the primary current is 0.5 A. Find the secondary voltage and secondary current (assume an ideal transformer).

1. Turns ratio:

   \$\frac{N{p}}{N{s}}=\frac{500}{200}=2.5\$

2. Voltage ratio:

   \$\$\frac{V{p}}{V{s}}=\frac{N{p}}{N{s}}\;\Rightarrow\;V{s}=V{p}\frac{N{s}}{N{p}}=

   240\;\frac{200}{500}=96\text{ V}\$\$

3. Power conservation (100 % efficiency):

   \$\$V{p}I{p}=V{s}I{s}\;\Rightarrow\;

   I{s}= \frac{V{p}I{p}}{V{s}}=

   \frac{240\times0.5}{96}=1.25\text{ A}\$\$

Answer: \(V{s}=96\text{ V}\), \(I{s}=1.25\text{ A}\).

8. Common Mistakes & How to Avoid Them

• Direction of the ratio: Always write \(\displaystyle\frac{V{p}}{V{s}}=\frac{N{p}}{N{s}}\). Reversing the fraction gives the wrong secondary voltage.
• Current ratio: Remember it is the inverse of the voltage ratio. Using the same ratio for current leads to an error.
• Assuming ideal behaviour: The equations are valid only for an ideal transformer. If the question gives an efficiency \(\eta\) or mentions losses, first calculate the output power as \(P{\text{out}}=\eta P{\text{in}}\).
• Core material: The syllabus expects you to know that a laminated soft‑iron core is used to concentrate flux and minimise core loss.

9. Quick Revision Checklist

1. Identify primary (source) and secondary (load) windings.
2. Record the number of turns: \(N{p}\) and \(N{s}\).
3. Write the voltage‑ratio equation \(\displaystyle\frac{V{p}}{V{s}}=\frac{N{p}}{N{s}}\).
4. If currents are required, use the power‑conservation equation for an ideal transformer: \(V{p}I{p}=V{s}I{s}\), or include efficiency: \(P{\text{out}}=\eta P{\text{in}}\).
5. Classify the transformer as step‑up or step‑down based on the turns ratio.
6. Check whether the question mentions a soft‑iron core, efficiency, or losses – adjust the answer accordingly.

10. Summary

The fundamental relationships for an ideal transformer are

These equations show how voltage changes with the turns ratio, while current changes in the opposite sense because power is conserved (100 % efficiency). Real transformers use a laminated soft‑iron core to keep core, copper and leakage losses low, giving typical efficiencies of 95–99 % at IGCSE level. Always state any assumptions (ideal vs. real) when answering exam questions.