Transformer Calculations (CIE IGCSE Physics: Co-ordinated Sciences (Double Award))

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Katie M

Last updated

28 August 2024

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Transformer calculations

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The voltages across the primary and secondary coils of a transformer can be calculated using the transformer equation, which states

The ratio of the voltages across the primary and secondary coils of a transformer is equal to the ratio of the number of turns on each coil

It can be expressed by the equation:

$\frac{primary voltage}{secondary voltage} = \frac{number of turns on primary}{number of turns on secondary}$

It can be expressed in symbols as follows:

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

Where
- $V_{p}$ = voltage across the primary coil, in volts (V)
- $V_{s}$ = voltage across the secondary coil, in volts (V)
- $N_{p}$ = number of turns on the primary coil
- $N_{s}$ = number of turns on the secondary coil

The transformer equation can be flipped upside down to give:

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

Rearranging for the secondary voltage:

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

This equation shows that the output (secondary) voltage of a transformer depends on:
- the number of turns on the primary and secondary coils
- the input (primary) voltage

In a step-up transformer, $V_{s} > V_{p}$ and $N_{s} > N_{p}$
In a step-down transformer, $V_{s} < V_{p}$ and $N_{s} < N_{p}$

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Worked example

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

Calculate the output voltage of the secondary coil.

State whether this is a step-up or step-down transformer.

Answer

Part (a)

Step 1: List the known quantities

Number of turns on the primary coil, $N_{p}$ = 20
Number of turns on the secondary coil, $N_{s}$ = 800
Voltage across the primary coil, $V_{P}$ = 500 V

Step 2: Write down the transformer equation

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

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

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

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

Step 4: Substitute the known values into the equation

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

Part (b)

The secondary voltage is larger than the primary, $V_{s} > V_{p}$
There are more turns on the secondary coil than on the primary, $N_{s} > N_{p}$
Therefore, this is a step-up transformer

Examiner Tip

When carrying out transformer calculations, 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.

Ideal transformer equation

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A transformer which is 100% efficient is called an ideal transformer
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, then the input power in the primary coil is equal to the output power of the secondary coil:

$power in primary = power in secondary$

The equation to calculate electrical power is:

$P = V \times I$

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

Therefore, the equation for an ideal transformer is:

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

Where:
- $I_{p}$ = primary current, in amps (A)
- $V_{p}$ = primary voltage, in volts (V)
- $I_{s}$ = secondary current, in amps (A)
- $V_{s}$ = secondary voltage, in volts (V)

The equation above could also be written as a ratio:

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

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Worked example

A transformer in a travel adapter steps up a 115 V a.c. mains electricity supply to the 230 V needed for a 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 down the equation for an ideal transformer

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

Step 3: Substitute in the known values

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

Step 4: Rearrange the equation to find the primary current

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

$I_{p} = 10 A$

High-voltage transmission

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Electricity is transmitted through power cables at a low current to prevent dissipation of energy
- When current flows in a wire, there is heating in the wire due to resistance
- Therefore, energy is dissipated to the surroundings, this energy is wasted
- The lower the current, the more efficient the energy transfer

Electrical power is equal to voltage × current, or $P = V I$

This means that a low current can be achieved by increasing the voltage, so electricity must be transmitted at a high voltage
- A smaller current flowing through the power lines results in less heat being produced in the wire
- This reduces the energy loss in the power lines

Power Loss National Grid, downloadable AS & A Level Physics revision notes

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

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Calculating power losses

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The power dissipated in the wire due to resistance is given by:

$P = I^{2} R$

Where:
- P = power, in watts (W)
- I = current, in amps (A)
- R = resistance, in ohms (Ω)

A step-up transformer is used to increase the voltage and decrease the current of electricity before transmission
- A high-voltage transmission ensures the same power transfer with a smaller current
- A smaller current means less thermal energy will be lost due to the resistance in the wire

A step-down transformer is used to decrease the voltage and increase the current of electricity after transmission
- High-voltage electricity is dangerous for use in homes, so it must be lowered before the current reaches consumers