Transformers

In a nutshell

The potential difference of an alternating current can be changed using a transformer. The two types of transformers are step-up and step-down transformers. Transformers help to increase the efficiency of the national grid.

Equations

Variable definitions

Note: For the AQA examination, this lesson is only required for high level single science.

Transformers

A transformer is used to change the size of the potential difference of an alternating current.

A. Primary coil B. Iron core C. Secondary coil

There are two coils of wire on a transformer: the primary coil and the secondary coil. They are wrapped around either side of an iron core. Iron is a magnetic material which means its magnetism can be induced.

When an alternating current flows through the primary coil it produces a changing magnetic field. The iron core magnetises and demagnetises quickly. The changing magnetic field induces an alternating potential difference in the secondary coil.

Note: If the secondary coil is part of a complete circuit, a current will be induced.

Transformer equation

The potential difference induced depends on the number of turns (loops) used in the primary and secondary coils. This can be written as a ratio.

​$$\frac{input \space potential \space difference}{output \space potential \space difference} = \frac{number \space of \space turns \space in \space primary \space coil}{number \space of \space turns \space in \space secondary\space coil} \\ \ \\ \ \\ \frac{V_{p}}{V_{s}}= \frac{N_{p}}{N_{s}}$$

Tip: The ratio also works the other way around ( $$\frac{V_{s}}{V_{p}} = \frac{N_{s}}{N_{p}}$$ ) which makes rearranging easier!

Example

A transformer has an alternating potential difference input of $$5\,V$$​. The number of turns in the secondary coil is $$100$$​ and the number of turns in the primary coil is $$40$$​. What is the output potential difference?

First, write out the quantities given and check they are in the correct form:

$$V_{p} = 5\,V \\ \ \\ N_{p} = 40\,turns \\ \ \\ N_{s} = 100\,turns$$​​

Next, write down the equation you need to use:

$$V_{s} = \frac{N_{s}}{N_{p}} \times V_{p}$$​​

Then, substitute the quantities into the equation:

​$$V_{s} = \frac{40}{100} \times 5= 0.4 \times 5 = 2$$​​

Make sure to include the correct units:

$$\text{output potential difference, }V_{s} = 2\,V$$​​

The output potential difference is $$\underline{2\,V}$$.​

Efficiency and power

Transformers have an efficiency that's close to $$100\%$$​. This means that the input power can be considered equal to the output power.

The input and output power can be equated using an equation for electrical power.

$$electrical \space power = current \times potential\space difference\\ \ \\ P = I\times V$$

​$$P_{p} = P_{s} \\ \ \\ I_{p}\times V_{p} = I_{s} \times V_{s}$$​​

​​​​​

Transformers in the national grid

The national grid transports electricity from power stations to places that need it, such as homes, hospitals and shops. It is made up of wires and transformers.

1. Power station 2. Step up transformer 3. Transmission lines 4. Step down transformer 5. Homes, hospitals and shops

Transmission lines are used to carry the electricity from place to place. They waste a large amount of energy by heating. When the current is made smaller, there is less energy wasted. The higher the voltage carried by these, lines the lower the energy wasted.

Transformers are used to increase the voltage so that the current is reduced. This increases the efficiency of the national grid.

Step-up transformers increase the voltage by having more turns in the secondary coil than primary. They are used to step-up the potential difference before the electricity is carried by transmission lines.

Step-down transformers decrease the voltage by having more turns in the primary coil than secondary. They are used to step-down the potential difference after the electricity has been carried by transmission lines.

Power wasted by heating can be worked out using the following equation.

​$$electrical \space power = current^{2} \times resistance \\ \ \\ P = I^{2} \times R$$​​

Energy wasted by heating over a certain time can be worked out using the equation

$$energy \space transferred = power \times time \\ \ \\ E = P\times t$$​​

Example

The national grid uses transmission lines to transport electricity. It sends electricity to a town with a power requirement of $$100\,MW$$. The resistance of these lines is $$2\,\Omega$$ and they transmit voltages of $$400\,kV$$​. How much energy would a transmission line lose in $$1\, hour$$?

First, write out the quantities given and check they are in the correct form:

$$P = 100 \,MW = 100 \times 10^{6} = 1\times 10^{8}\,W\\ \ \\ R = 2\,\Omega \\ \ \\ V = 400\,kV = 400\: 000\,V \\ \ \\ t = 1 \,hr = 1 \times60\times60 = 3600\,s$$​​

Next, write down the equations needed:

$$I = \frac{P}{V}\\ \ \\ P_{wasted} = I^2\times R\\ \ \\ E = P\times t$$​​

Then, substitute the values into the equations:

$$I = \frac{10 \times 10^{8}}{400\,000} = 250\,A\\ \ \\ P_{wasted} = 200^2 \times 2 = 80 \,000\,W\\ \ \\ E = 80\,000 \times 3600 = 2.88 \times 10^{8}$$​​

Make sure to include units and round to two significant figures:

$$\text{wasted energy, }E = 2.9 \times 10^{8}\,J$$​​

The energy lost in an hour is $$\underline{2.9 \times 10^{8}\,J}$$.

​​

Tip: Conversion for megawatts to watts: $$1\,MW = 1\times10^{6}\,W$$.