# GCSE Science/Transformers

Transformers are devices that use induction to transform a high a.c. voltage to a low one or vice versa. They are incredibly important devices, and (even more importantly) they are examiners' favourites. On this page you will be looking at how a transformer works and how they are used. You will also be given some practice in using some simple equations that let you work out what the output voltage and current must be.

## How a transformer worksEdit

There is nothing new to learn in this section. You've already covered GCSE Science/Magnetic effects of a current and GCSE Science/Induction. This section used ideas from those two pages to understand a transformer.

The simplest transformer consists of a soft iron core with two coils of wire independently wrapped around it. The first coil called the *primary coil* is connected to an a.c. power supply. The other coil, called the *secondary coil* is connected to the output.

The current in the primary coil causes a magnetic field. It's just an electromagnet but because the current is a.c. the magnetic field is constantly changing. If we consider one end of the soft iron core, it's first a north pole, then the field gets weaker and weaker until it becomes zero, then it starts to build up and the end becomes a south pole. The process repeats and the end becomes a north pole again.

Now let's think about what is going on in the secondary coil. The magnetic field of the first coil passes right through the secondary coil, what's more the field is constantly changing. This changing field induces a current in the secondary coil. So even though the coils are not physically connected, the current in one, causes the current through the other.

If you imagine the lines of field they appear to pulse as the field grows stronger and weaker. When the field grows weak they sweep inwards cutting the secondary coil.

**Q1)** What would the current in the secondary coil be like if you used d.c. in the primary coil? - d.c. in the primary coil will not create any current in the secondary coil.

Transformers are used to change the voltage. If the number of turns in the secondary coil is greater than the number of turns in the primary coil, then the voltage in the secondary coil will be higher than that in the primary coil. This is called a *step-up transformer*. Conversely, if the secondary coil has fewer turns than the primary coil, then the voltage will be lower, and it is called a *step-down transformer*.

## A simple demonstration transformerEdit

Your teacher may show you a demonstration of a transformer. (see here for details on how to set up a simple transformer). Look at the diagram below.

The second coil is on a separate soft iron core. This core can be brought up to the primary coil. At large distances the bulb is off, but as the coils are brought closer together they suddenly attract one another (the primary coil is a powerful electromagnet). Once the two cores are stuck together the bulb in the secondary coil starts to glow. The brightness of the bulb depends on the number of turns on the secondary coil. The more windings there are, the brighter the bulb.

## Important equationsEdit

There are two things you need to remember about transformers.

- The
**voltage**in the secondary coil depends on the voltage in the primary coil*and*the ratio of the number of turns in the secondary and primary coils.

- If we call:

- the voltage in and number of turns in the primary V
_{p}and N_{p} - the voltage in and number of turns in the secondary V
_{s}and N_{s}

- the voltage in and number of turns in the primary V

- then
- V
_{s}/V_{p}= N_{s}/N_{p}

- V

- or, rearranging:
- V
_{s}= V_{p}* N_{s}/N_{p}

- V

- Similarly, the
**current**in the secondary coil depends on the current in the primary coil*and*the ratio of the number of turns in the secondary and primary coils. However, the current is inversely related to the number of turns.

- If we call:

- the current in and number of turns in the primary I
_{p}and N_{p} - the current in and number of turns in the secondary I
_{s}and N_{s}

- the current in and number of turns in the primary I

- then
- I
_{s}/I_{p}= N_{p}/N_{s}

- I

- or, rearranging:
- I
_{s}= I_{p}* N_{p}/N_{s}

- I

- The power in the primary is equal to the power in the secondary (assuming a
**perfect**transformer). This is because energy is never created or destroyed.

- I
_{p}V_{p}= I_{s}V_{s}

- I

## Step up and Step down transformersEdit

Because the voltage in the secondary coil depends on the number of turns, we can use a transformer to *transform* one voltage to another. For example let's assume that the primary coil has 300 turns of wire and the secondary coil has 900 turns. This transformer will step up the voltage by a factor of 3. So if the primary coil has 2V applied to it, the secondary coil will have 6V across it.

A step down transformer is the opposite of a step up transformer. Let's say the primary has 900 turns and the secondary has 450 turns. This time the voltage will be half on the secondary as it is on the primary. So if the primary has 2V applied, the secondary will have only 1V across it.

### Strategy for answering questionsEdit

Questions on transformers will often be of the form "A coil has 3000 turns of wire on a soft iron. Another coil of 12000 turns is brought into contact with the first coil. This second coil has a 12V bulb attached to it completing a circuit.What voltage should applied to the first coil?"

to answer a question like this:

**Decide which is the primary coil and which is the secondary**-in this case the 3000 turn coil is the primary and the 12000 turn coil is the secondary**Decide if it is a step up or a step down transformer**in this case it is a step up transformer because the secondary coil has more turns on it than the primary.**Work out a multiplication factor**in this case 12000/3000 = 4**Use this multiplication factor to work out the required voltage**- the secondary coil has to supply 12V so the primary has to be 12V/4 = 3V

## Uses of transformersEdit

The most important use of transformers is in the routing of mains electricity. The power dissipated by a cable = I^{2}R, where I is the current and R is the resistance. There is little that can be done to lower the resistance, so in order to lose as little energy as possible in heating up the cable, the current needs to be kept as low as possible. This is done using a step up transformer which increases the voltage to many thousands of volts. Because power is voltage times current, a high voltage means a low current for a given power value.(Notice that transormers do not obey Ohm's law). At the other end of the distribution network, the voltage is stepped down using another transformer so that the voltage going into homes and factories is not excessively high.