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Electrical Transformer

Key Stage 4

Meaning

A picture of a transformer in the National Grid.

A transformer is a device used to take the alternating current in a circuit and transmit it to another circuit to produce electricity with a different potential difference and current.

About Transformers

Transformers use a coil of wire in one circuit (primary coil) wrapped around a soft iron core and second coil of wire in another circuit (secondary coil). When an alternating current passes through the primary coil this creates a changing magnetic field which induces an alternating potential difference across the secondary coil. If that secondary coil is connected in a circuit it will produce an alternating current.
There are two types of transformer you should know:
Step up transformers are used in the National Grid to increase the potential difference and decrease the current for long distance transmission.
TransformerDiagram.png
This diagram of a transformer shows that an alternating current in the primary coil induces an alternating potential difference in the secondary coil.

Equation

NB: You must be able to use this equation, but you do not need to remember it.

(Primary Potential Difference)/(Secondary Potential Difference)=(Primary Coils)/(Secondary Coils)

\(\frac{V_p}{V_s} = \frac{n_p}{n_s}\)

\(V_p\) = The potential difference across the primary coil.

\(V_s\) = The potential difference across the secondary coil.

\(n_p\) = The number of loops on the primary coil.

\(n_s\) = The number of loops on the secondary coil.

Example Calculations

TransformerQ1.png
TransformerQ2.png
Calculate the potential difference across the secondary coil. Calculate the potential difference across the secondary coil.
1. State the known quantities in SI Units.

\(V_p\) = 3V

\(n_p\) = 4 coils

\(n_s\) = 8 coils

1. State the known quantities in SI Units.

\(V_p\) = 9V

\(n_p\) = 6 coils

\(n_s\) = 8 coils

2. Substitute the numbers and evaluate.

\(\frac{V_p}{V_s} = \frac{n_p}{n_s}\)

\(\frac{3}{V_s} = \frac{4}{8}\)

\(\frac{3}{V_s} = 0.5\)

2. Substitute the numbers and evaluate.

\(\frac{V_p}{V_s} = \frac{n_p}{n_s}\)

\(\frac{9}{V_s} = \frac{6}{8}\)

\(\frac{9}{V_s} = 0.75\)

3. Rearrange the equation and solve.

\(3 = 0.5 \times V_s\)

\(V_s = 6V\)

3. Rearrange the equation and solve.

\(9 = 0.75 \times V_s\)

\(V_s = 12V\)

References

AQA

Transformer, page 130, GCSE Chemistry; Student Book, Collins, AQA
Transformer, pages 68-9, 243, 264-5, 267, GCSE Physics; Student Book, Collins, AQA
Transformer; step-down, pages 243, 264-5, GCSE Physics; Student Book, Collins, AQA
Transformer; step-up, pages 243, 264-5, GCSE Physics; Student Book, Collins, AQA
Transformers, page 191, GCSE Combined Science; The Revision Guide, CGP, AQA
Transformers, page 309, GCSE Combined Science Trilogy 1, Hodder, AQA
Transformers, pages 308-312, GCSE Physics; The Complete 9-1 Course for AQA, CGP, AQA
Transformers, pages 34, 98, GCSE Physics; The Revision Guide, CGP, AQA
Transformers, pages 55, 236-7, GCSE Physics, Hodder, AQA
Transformers, pages 64, 226-229, GCSE Physics; Third Edition, Oxford University Press, AQA
Transformers, pages 92, 93, GCSE Combined Science Trilogy; Physics, CGP, AQA
Transformers; in the national grid, pages 95, 310, GCSE Physics; The Complete 9-1 Course for AQA, CGP, AQA