Electrical Transformer
Contents
Key Stage 4
Meaning
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 Transformer - A transformer which increase the potential difference.
- Step Down Transformer - A transformer which decrease the potential difference.
- Step up transformers are used in the National Grid to increase the potential difference and decrease the current for long distance transmission.
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
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