Electrical Transformer

Key Stage 4

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.

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.

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.


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. \(9 = 0.75 \times V_s\) \(V_s = 12V\)