Transformer Calculations

The output potential difference (voltage) of a transformer depends on:
- The number of turns on the primary and secondary coils
- The input potential difference (voltage)

It can be calculated using the equation:

$\frac{P . D . across primary coil}{P . D . across secondary coil} = \frac{number of turns on primary coil}{number of turns on secondary coil}$

This equation can be written using symbols as follows:

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

Where
- V_p = potential difference (voltage) across the primary coil in volts (V)
- V_s = potential difference (voltage) across the secondary coil in volts (V)
- n_p = number of turns on primary coil
- n_s = number of turns on secondary coil

The equation above can be flipped upside down to give:

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

The equations above show that:
- The ratio of the potential differences across the primary and secondary coils of a transformer is equal to the ratio of the number of turns on each coil

Worked example

A transformer has 20 turns on the primary coil and 800 turns on the secondary coil. The input potential difference across the primary coil is 500 V.

a) Calculate the output potential difference

b) State what type of transformer this is

Answer

Part (a)

Step 1: List the known quantities

Number of turns in primary coil, $N_{P}$ = 20
Number of turns in secondary coil, $N_{S}$ = 800
Voltage in primary coil, $V_{P}$ = 500 V

Step 2: Write the equation linking the output potential difference ( $V_{S}$ ) to the known quantities

There will be less rearranging to do if $V_{S}$ is on the top of the fraction

$\frac{N_{S}}{N_{P}} = \frac{V_{S}}{V_{P}}$

Step 3: Rearrange the equation to make $V_{S}$ the subject

$V_{S} = \frac{N_{S} V_{P}}{N_{P}}$

Step 4: Substitute the known values into the equation

$V_{S} = \frac{800 \times 500}{20} = 20 000 V$

Part (b)

The secondary voltage is larger than the primary, therefore this is a step-up transformer

Examiner Tip

When you are using the transformer equation make sure you have used the same letter (p or s) in the numerators (top line) of the fraction and the same letter (p or s) in the denominators (bottom line) of the fraction.

There will be less rearranging to do in a calculation if the variable which you are trying to find is on the numerator (top line) of the fraction.

The individual loops of wire going around each side of the transformer should be referred to as turns and not coils.

High-Voltage Transmission

Transformers have a number of roles:
- They are used to increase the potential difference of electricity before it is transmitted across the national grid
- They are used to lower the high voltage electricity used in power lines to the lower voltages used in houses
- They are used in adapters to lower mains voltage to the lower voltages used by many electronic devices

Advantages of High Voltage Transmission

When electricity is transmitted over large distances, the current in the wires heats them, resulting in energy loss
To transmit the same amount of power as the input power the potential difference at which the electricity is transmitted should be increased
- This will result in a smaller current being transmitted through the power lines
- This is because P = IV, so if V increases, I must decrease to transmit the same power
A smaller current flowing through the power lines results in less heat being produced in the wire
- This will reduce the energy loss in the power lines

Model of High-Voltage Transmission

power-lines, IGCSE & GCSE Physics revision notes

Electricity is transmitted at high voltage, reducing the current and hence power loss in the cables

Transformer Calculations (Cambridge O Level Physics)

Revision Note