The transformer

1. Overview

A transformer is an electrical device used to change the voltage of an alternating current (AC). By increasing or decreasing voltages, transformers allow electricity to be transmitted efficiently over long distances, minimizing energy loss in the form of heat.

Key Definitions

• Primary Coil: The input coil of a transformer, connected to the AC power supply.
• Secondary Coil: The output coil of a transformer, where the transformed voltage is delivered to the load.
• Soft-Iron Core: A ferromagnetic core that links the magnetic field of the primary coil to the secondary coil.
• Step-up Transformer: A transformer that increases the voltage (has more turns on the secondary coil than the primary).
• Step-down Transformer: A transformer that decreases the voltage (has fewer turns on the secondary coil than the primary).

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Core Content

Construction of a Simple Transformer

A simple transformer consists of two coils of insulated wire (the primary and secondary) wound around a laminated soft-iron core.

• The coils are not electrically connected to each other.
• The soft-iron core is used because it is easily magnetized and demagnetized, which helps concentrate the magnetic field.

The Transformer Equation

The ratio of the voltages is equal to the ratio of the number of turns on the coils: $$\frac{V_p}{V_s} = \frac{N_p}{N_s}$$

• Where $V$ is voltage and $N$ is the number of turns.
• Step-up: $N_s > N_p$, therefore $V_s > V_p$.
• Step-down: $N_s < N_p$, therefore $V_s < V_p$.

Worked Example: A transformer has 200 turns on the primary coil and 4000 turns on the secondary coil. If the input voltage is 12V, calculate the output voltage.

1. Identify: $V_p = 12$, $N_p = 200$, $N_s = 4000$.
2. Rearrange: $V_s = V_p \times (\frac{N_s}{N_p})$
3. Calculate: $V_s = 12 \times (\frac{4000}{200}) = 12 \times 20 = 240\text{V}$.

High-Voltage Transmission

Electricity is transported from power stations to homes via the National Grid.

1. Step-up transformers increase the voltage to very high levels (e.g., 400,000V) for long-distance transmission.
2. Step-down transformers decrease the voltage to safe levels (e.g., 230V) for use in homes.

Advantages of High-Voltage Transmission:

• By increasing the voltage, the current in the cables is reduced.
• Lower current results in significantly less energy being lost as heat in the transmission wires.
• This makes the process more efficient and cheaper.

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Extended Content (Extended Only)

Principle of Operation

1. An alternating current (AC) flows through the primary coil.
2. This creates a constantly changing magnetic field within the primary coil.
3. The soft-iron core directs this changing magnetic field through the secondary coil.
4. The changing magnetic field cuts through the secondary coil, inducing an alternating e.m.f. (voltage) in it via electromagnetic induction. Note: Transformers do NOT work with Direct Current (DC) because DC produces a steady magnetic field that does not induce a voltage.

100% Efficiency Equation

For an ideal transformer (100% efficient), the power input equals the power output: $$I_p V_p = I_s V_s$$ This means that if voltage is "stepped up," the current must be "stepped down" to conserve energy.

Worked Example: A transformer steps down 240V to 12V. If the output current is 2A, calculate the input current.

1. Identify: $V_p = 240$, $V_s = 12$, $I_s = 2$.
2. Equation: $I_p \times 240 = 2 \times 12$
3. Calculate: $I_p = \frac{24}{240} = 0.1\text{A}$.

Power Loss in Cables

The power lost as heat in a wire is given by: $$P = I^2 R$$

• Because the power loss is proportional to the square of the current, even a small reduction in current leads to a massive reduction in power loss.
• By using a transformer to increase voltage, the current $I$ decreases, which minimizes the $I^2 R$ "heating loss" in long-distance cables.

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Key Equations

Equation	Symbols	Units
$\frac{V_p}{V_s} = \frac{N_p}{N_s}$	$V$=Voltage, $N$=Number of turns	V (Volts), No units for $N$
$I_p V_p = I_s V_s$	$I$=Current, $V$=Voltage	A (Amps), V (Volts)
$P = I^2 R$	$P$=Power loss, $I$=Current, $R$=Resistance	W (Watts), A (Amps), $\Omega$ (Ohms)

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Common Mistakes to Avoid

• ❌ Wrong: Inverting the voltage ratio (e.g., calculating $12 \times \frac{200}{4000}$ when you meant to step-up).
  • ✅ Right: Always check if your answer makes sense—if it's a step-up transformer, your output voltage must be higher than the input.
• ❌ Wrong: Assuming current remains the same in both coils.
  • ✅ Right: If voltage increases, current must decrease ($I_p V_p = I_s V_s$).
• ❌ Wrong: Using the power equation $P=VI$ to calculate power loss in cables without considering resistance.
  • ✅ Right: Use $P = I^2 R$ specifically when explaining why high voltage (low current) reduces energy loss.
• ❌ Wrong: Forgetting that transformers only work with alternating current (AC).
  • ✅ Right: State clearly that a changing magnetic field is required for induction.

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Exam Tips

1. Check the turns: If $N_s > N_p$, it is a step-up transformer. If $N_p > N_s$, it is a step-down transformer. Always state which one it is if asked.
2. Show your steps: In the Extended paper, you often need to calculate current after finding the voltage. Show the $V_p/V_s$ step first, then the $IV$ step.
3. The "Why" of High Voltage: If asked why we use high voltage for transmission, use the phrase: "Low current results in less energy lost as heat due to the resistance of the cables ($P=I^2R$)."

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Exam-Style Questions

Practice these original exam-style questions to test your understanding. Each question mirrors the style, structure, and mark allocation of real Cambridge 0625 Theory papers.

Exam-Style Question 1 — Short Answer [5 marks]

Question:

A small model railway transformer is used to step down the mains voltage of 230 V to 12 V for the tracks.

(a) State what is meant by the term "step-down" transformer. [1]

(b) Describe the key features of the transformer's construction that allow it to function as a step-down transformer. [3]

(c) State one advantage of using a transformer in the transmission of electricity. [1]

Worked Solution:

(a)

1. A step-down transformer is one that reduces the voltage from the primary coil to the secondary coil. This is the definition of a step-down transformer.

How to earn full marks:

• State the voltage is reduced from primary to secondary.

(b)

1. The transformer has a soft iron core. Soft iron is easily magnetised and demagnetised, improving efficiency.
2. The primary coil has more turns than the secondary coil. This ensures the voltage is reduced.
3. Both coils are insulated from each other. This prevents a short circuit.

How to earn full marks:

• Mention the soft iron core.
• Mention more turns on the primary coil.
• Mention insulation of the coils.

(c)

1. A transformer allows for efficient transmission of electrical power over long distances. High voltages reduce current, reducing $I^2R$ power losses.

How to earn full marks:

• State that it allows efficient transmission of electricity, or enables high voltage transmission.

Common Pitfall: Make sure you understand the difference between step-up and step-down transformers. The number of turns in the primary and secondary coils determines whether the voltage is increased or decreased. Also, remember the soft iron core improves efficiency, and the coils must be insulated.

Exam-Style Question 2 — Short Answer [6 marks]

Question:

A student investigates a transformer with a primary coil connected to a 240 V a.c. supply. The secondary coil is connected to a lamp. The number of turns on the primary coil is 1200.

(a) The student measures the voltage across the lamp to be 12 V. Calculate the number of turns on the secondary coil. [3]

(b) The current in the primary coil is 0.1 A. Assuming the transformer is 100% efficient, calculate the current in the secondary coil. [3]

Worked Solution:

(a)

1. State the transformer equation: $\frac{V_p}{V_s} = \frac{N_p}{N_s}$ This is the correct equation to use.
2. Rearrange to find $N_s$: $N_s = \frac{V_s \times N_p}{V_p}$ Rearranging to make the subject $N_s$
3. Substitute the values and calculate: $N_s = \frac{12 \text{ V} \times 1200}{240 \text{ V}} = 60$ Substituting the values into the equation

How to earn full marks:

• State the correct transformer equation.
• Correct rearrangement of the equation.
• Correct calculation and final answer with the correct units $\boxed{60 \text{ turns}}$.

(b)

1. State the power equation (100% efficiency): $V_p I_p = V_s I_s$ This is the correct equation.
2. Rearrange to find $I_s$: $I_s = \frac{V_p I_p}{V_s}$ Rearranging to make the subject $I_s$
3. Substitute the values and calculate: $I_s = \frac{240 \text{ V} \times 0.1 \text{ A}}{12 \text{ V}} = 2 \text{ A}$ Substituting the values into the equation

How to earn full marks:

• State the correct power equation.
• Correct rearrangement of the equation.
• Correct calculation and final answer with the correct units $\boxed{2 \text{ A}}$.

Common Pitfall: Be careful when rearranging the transformer equation. It's easy to accidentally invert the voltage or turns ratio. Also, remember that the unit for number of turns is "turns," but it's often omitted in the final answer.

Exam-Style Question 3 — Extended Response [8 marks]

Question:

A power company uses transformers to transmit electrical power from a power station to a town. The power station generates electricity at 25 kV. The transmission lines have a total resistance of 10 Ω. The town requires a power of 5 MW.

(a) Explain why the power company uses high-voltage transmission lines. [3]

(b) The power is stepped up to 400 kV for transmission. Calculate the current in the transmission lines. [2]

(c) Calculate the power lost in the transmission lines due to resistance. [3]

Worked Solution:

(a)

1. Increasing the voltage, decreases the current for the same power. Explanation of the relationship between voltage and current.
2. Lower current reduces the power loss in the transmission lines. Explanation of the effect of lower current.
3. Power loss is proportional to the square of the current ($P = I^2R$). Power loss is proportional to the square of the current, so even small reductions in current result in significant power savings.

How to earn full marks:

• State that higher voltage reduces current.
• State that lower current reduces power loss.
• State power loss is proportional to the square of the current.

(b)

1. State the power equation: $P = VI$. This is the correct equation.
2. Rearrange to find $I$: $I = \frac{P}{V}$ Rearranging the equation.
3. Substitute the values and calculate: $I = \frac{5 \times 10^6 \text{ W}}{400 \times 10^3 \text{ V}} = 12.5 \text{ A}$ Substituting the values into the equation.

How to earn full marks:

• State the correct power equation.
• Correct rearrangement.
• Correct calculation and final answer with the correct units $\boxed{12.5 \text{ A}}$.

(c)

1. State the power loss equation: $P = I^2R$ This is the correct equation to use.
2. Substitute the values and calculate: $P = (12.5 \text{ A})^2 \times 10 \text{ } \Omega = 1562.5 \text{ W}$ Substituting the values into the equation.

How to earn full marks:

• State the correct power loss equation.
• Correct substitution.
• Correct calculation and final answer with the correct units $\boxed{1562.5 \text{ W}}$.

Common Pitfall: Remember that power loss in transmission lines is proportional to the square of the current. A small increase in current can lead to a significant increase in power loss. Also, pay close attention to units, especially when dealing with large numbers (e.g., kW, MW, kV).

Exam-Style Question 4 — Extended Response [9 marks]

Question:

A student sets up an experiment to investigate the relationship between the number of turns in the secondary coil of a transformer and the output voltage. The primary coil has a fixed number of turns and is connected to a 6.0 V a.c. supply. The student measures the output voltage for different numbers of turns in the secondary coil. The core of the transformer is made of laminated soft iron.

(a) Explain why the core of the transformer is made of soft iron. [2]

(b) Explain why the core of the transformer is laminated. [2]

(c) The student obtains the following data:

Number of turns in secondary coil	Output voltage (V)
100	2.0
200	4.0
300	6.0
400	8.0
500	10.0

Calculate the number of turns in the primary coil of the transformer. Show your working clearly. [3]

(d) Suggest one way the student could improve the accuracy of the experiment. [2]

Worked Solution:

(a)

1. Soft iron is easily magnetised and demagnetised. Explanation of the property of soft iron.
2. This allows for efficient transfer of magnetic flux between the primary and secondary coils. Explanation of why that property is desirable.

How to earn full marks:

• State that soft iron is easily magnetised and demagnetised.
• State that this allows efficient transfer of magnetic flux.

(b)

1. Lamination reduces eddy currents in the core. Explanation of what laminations do.
2. Eddy currents cause heating and energy loss. Explanation of why eddy currents are undesirable.

How to earn full marks:

• State that lamination reduces eddy currents.
• State that eddy currents cause energy loss.

(c)

1. State the transformer equation: $\frac{V_p}{V_s} = \frac{N_p}{N_s}$. This is the correct equation.
2. Choose a data point from the table (e.g., $N_s = 100$, $V_s = 2.0 \text{ V}$) and rearrange the equation to find $N_p$: $N_p = \frac{V_p \times N_s}{V_s}$ Rearranging the equation.
3. Substitute the values and calculate: $N_p = \frac{6.0 \text{ V} \times 100}{2.0 \text{ V}} = 300$ Substituting the values into the equation.

How to earn full marks:

• State the correct transformer equation.
• Show rearrangement and substitution.
• Correct calculation and final answer with the correct units $\boxed{300 \text{ turns}}$.

(d)

1. Use a more precise voltmeter to measure the output voltage. Increases resolution of measured values.
2. Repeat the experiment multiple times and calculate the average output voltage for each number of turns. Reduces the impact of random errors.

How to earn full marks:

• Suggest using more precise measuring instruments (voltmeter).
• Suggest repeating the experiment and calculating an average.

Common Pitfall: When calculating the number of turns, make sure you use corresponding values of voltage and number of turns from the same row of the table. Using values from different rows will give you an incorrect result. Also, remember to explain why a suggestion improves accuracy, not just state the suggestion.