Transformer Calculations (Oxford AQA IGCSE Physics)

Revision Note

Written by: Dan Mitchell-Garnett

Reviewed by: Caroline Carroll

Transformer Calculations

The output potential difference across the secondary coil of a transformer depends on:
- The number of turns on the primary and secondary coils
- The input potential difference across the primary coil

The transformer equation

The transformer equation can be used to calculate the potential difference across each coil:

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

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

Step-up transformer

A step-up transformer increases the potential difference of a power source
A step-up transformer has more turns on the secondary coil than on the primary coil (n_s > n_p)
- The number of turns increases from primary to secondary
- So the potential difference increases from primary to secondary

Step-down transformer

A step-down transformer decreases the potential difference of a power source
A step-down transformer has fewer turns on the secondary coil than on the primary coil (n_s < n_p)
- The number of turns decreases from primary to secondary
- So the potential difference decreases from primary to secondary

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 the primary coil, n_p = 20
Number of turns in the secondary coil, n_s = 800
Potential difference across the primary coil, V_p = 500 V

Step 2: Write the transformer equation from the equation sheet

The transformer equation on the equation sheet is given as

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

Flipping this equation with the secondary coil's variables on the top will make rearranging this equation much easier

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

Step 3: Rearrange for V_s

The equation with secondary potential difference as the subject is:

$V_{s} = V_{p} \times \frac{n_{s}}{n_{p}}$

Step 4: Substitute the known quantities

$V_{s} = 500 \times \frac{800}{20}$

$V_{s} = 20 000 V$

Part b)

Step 1: Determine if the potential difference has increased or decreased

The potential difference across the primary coil is 500 V
The potential difference in the secondary coil is 20 000 V
Therefore, the potential difference has increased from primary to secondary

Step 2: State whether the transformer is step-up or step-down

The potential difference has increased
So this is a step-up transformer

Examiner Tips and Tricks

When you are using the transformer equation, 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 Ideal Transformer Equation

An ideal transformer would be 100% efficient
- Although transformers can increase the voltage of a power source, due to the law of conservation of energy, they cannot increase the power output

If a transformer is 100% efficient:

$input power = output power$

The equation to calculate electrical power is:

$P = I \times V$

Where:
- P = power in watts (W)
- V = potential difference in volts (V)
- I = current in amps (A)

Therefore, if a transformer is 100% efficient then:

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

Where:
- V_p = potential difference across primary coil in volts (V)
- I_p = current through primary coil in Amps (A)
- V_s = potential difference across secondary coil in volts (V)
- I_s = current through secondary coil in Amps (A)

The equation above could also be written as:

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

Where:
- P_s = output power (power produced in secondary coil) in watts (W)

Worked Example

A transformer in a travel adapter steps up a 115 V ac mains electricity supply to the 230 V needed for the hair dryer. A current of 5 A flows through the hairdryer.

Assuming that the transformer is 100% efficient, calculate the current drawn from the mains supply.

Answer:

Step 1: List the known quantities

Voltage in primary coil, V_p = 115 V
Voltage in secondary coil, V_s = 230 V
Current in secondary coil, I_s = 5 A

Step 2: Write the equation linking the known values to the current from the supply, I_p

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

Step 3: Rearrange the equation to make I_pthe subject

$I_{p} = \frac{I_{s} \times V_{s}}{V_{p}}$

Step 4: Substitute int he known values to calculate

$I_{p} = \frac{5 \times 230}{115}$

Step 5: Calculate a value for I_p and include the correct unit

$I_{p} = 10 A$

Turn Ratio

It is the ratio of turns on the primary coil to turns on the secondary coil that determine the change in potential difference
- This can be shown by the transformer equation

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

The right-hand side of the equation is the turn ratio
- The change in potential difference in the transformer can be determined without knowing the number of turns in each coil, as long as the ratio of turns is known
- If the ratio is increased, the change in potential difference is also increased by the same factor
The turns ratio is selected to produce the required output potential difference from a given input potential difference

Worked Example

A step-down transformer is connected to a supply of 0.16 kV. The primary coil has 32× more turns than the secondary coil.

Calculate the output potential difference.

Answer:

Step 1: List the known quantities

Potential difference across the primary coil, V_p = 0.16 kV
Number of turns on primary coil = n_p
Number of turns on secondary coil = n_s

Step 2: Convert kV into V

There are 1000 V in 1 kV

$V_{p} = 0.16 \times 1000 = 160 V$

Step 3: Write an expression relating the numbers of turns

There are 32 times more turns on the primary coil than on the secondary coil

$n_{p} = 32 n_{s}$

Step 4: Write down the transformer equation

From the equation sheet:

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

Flip the equation to find secondary potential difference

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

Step 5: Substitute in the expression relating number of turns

$\frac{n_{p}}{n_{s}} = 32$

$\frac{V_{p}}{V_{s}} = \frac{n_{p}}{n_{s}} = 32$

Step 6: Rearrange for potential difference across the secondary coil

$\frac{V_{p}}{V_{s}} = 32$

$V_{s} = \frac{V_{p}}{32}$

Step 7: Substitute in the potential difference across the primary coil

$V_{s} = \frac{160}{32}$

$V_{s} = 5.0 V$

Last updated: 4 April 2024

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Transformer Calculations (Oxford AQA IGCSE Physics)

Revision Note

Transformer Calculations

The transformer equation

Step-up transformer

Step-down transformer

Worked Example

Examiner Tips and Tricks

The Ideal Transformer Equation

Worked Example

Turn Ratio

Worked Example

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Author: Dan Mitchell-Garnett

Author: Caroline Carroll