AQA GCSE PHYSICS · PAPER 2 · HIGHER TIER ONLY
Transformers
Magnetism and electromagnetism · Lesson 8 of 8
Teacher copy - includes the notes for whoever is teaching from it.
Warm-up
Answer each one, then check.
1. What is the National Grid?
The network that transfers electrical power
2. Why is high voltage used in the grid?
To reduce energy lost as heat
3. What is power?
The rate of transfer of energy
4. What is a.c.?
Alternating current
5. What is a coil?
Loops of wire
Learning Objectives
1. Describe the structure and function of a transformer.
2. Explain how a step-up and a step-down transformer work.
3. Use V_p / V_s = n_p / n_s.
4. Use V_s I_s = V_p I_p for a 100% efficient transformer.
5. Explain why the National Grid uses transformers.
The Transformer Equations
V_p/(V_s) = n_p/(n_s) and V_s × I_s = V_p × I_p
V_p, I_p, n_p are for the primary coil; V_s, I_s, n_s for the secondary. The second equation assumes 100% efficiency (power in = power out).
A Transformer
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The iron core links the magnetic field between the coils. |
A transformer with primary and secondary coils wound on a soft iron core.
How a Transformer Works
Follow the steps.
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1 a.c. in the primary An alternating current flows in the primary coil. |
2 Changing field It produces a changing magnetic field in the iron core. |
3 Field through the secondary The changing field passes through the secondary coil. |
4 Generator effect An alternating p.d. is induced in the secondary coil. |
5 Turns ratio The ratio of turns sets the ratio of potential differences. |
Step-up or Step-down?
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STEP-UP TRANSFORMER |
STEP-DOWN TRANSFORMER |
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▸ More turns on the secondary coil. ▸ Increases the potential difference. ▸ Used at power stations before the grid. |
▸ Fewer turns on the secondary coil. ▸ Decreases the potential difference. ▸ Used near homes to bring the p.d. down to 230 V. |
Using the Turns Ratio
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A transformer has 200 turns on its primary coil and 1000 on its secondary. The input is 12 V. Calculate the output potential difference. |
1. Write the equation
V_p / V_s = n_p / n_s
2. Rearrange
V_s = V_p × n_s ÷ n_p
3. Substitute
V_s = 12 × 1000 ÷ 200
4. Answer
V_s = 60 V
Answer: 60 V
Power In Equals Power Out
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A 100% efficient transformer has a primary p.d. of 230 V and current of 2.0 A. The secondary p.d. is 11.5 V. Calculate the secondary current. |
1. Write the equation
V_s I_s = V_p I_p
2. Rearrange
I_s = V_p I_p ÷ V_s
3. Substitute
I_s = 230 × 2.0 ÷ 11.5
4. Answer
I_s = 40 A
Answer: 40 A
Why the National Grid Uses Them
High voltage, low current.
▸ Step-up at the power station. Raises the p.d. (up to 400 000 V) and lowers the current.
▸ Lower current. Less energy is wasted as heat in the cables.
▸ Step-down near homes. Reduces the p.d. to 230 V for safe use.
▸ Efficiency. The grid is efficient because transformers waste little energy.
Key Terms
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Transformer A device that changes the potential difference of an alternating supply. |
Primary coil The input coil of a transformer. |
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Secondary coil The output coil of a transformer. |
Step-up transformer A transformer that increases the potential difference. |
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Step-down transformer A transformer that decreases the potential difference. |
Soft iron core A core that easily becomes magnetised and demagnetised, linking the coils. |
Your Task: Transformer Sums
15 minutes
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A transformer has n_p = 50 and n_s = 500. Input 10 V a.c. Find V_s. If the input current is 2.0 A, find I_s assuming 100% efficiency. 1. Step-up: p.d. up, current down. |
A good answer shows: Vs = 10 × 500 ÷ 50 = 100 V. Power in = 20 W, so Is = 20 ÷ 100 = 0.20 A.
Note: Ask what happens to the current in a step-up transformer.
Can I...?
☐ Name the parts.
☐ Explain the working.
☐ State step-up and step-down.
☐ Use the turns ratio.
☐ Use power in = power out.
☐ Explain why a.c. is needed.
☐ Explain the National Grid.
☐ Give typical voltages.
Summary
✓ V_p / V_s = n_p / n_s.
✓ V_s I_s = V_p I_p for 100% efficiency.
✓ Step-up increases the p.d. and decreases the current.
✓ a.c. is needed for a changing field.
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EXAM FOCUS A transformer has 200 primary turns and 1000 secondary turns. The input is 12 V. Calculate the output. (2 marks) Vs = Vp × ns ÷ np. |
Exam Practice: Transformers
Answer all questions. Use the mark allocation as a guide to how much to write. · 17 minutes
▸ Question 1 · 2 marks · Calculate. A transformer has 200 turns on its primary coil and 1000 turns on its secondary coil. The input potential difference is 12 V. Calculate the…
▸ Question 2 · 4 marks · Explain. Explain how a transformer changes the potential difference of an alternating supply.
▸ Question 3 · 3 marks · Calculate. A transformer is 100% efficient. The primary p.d. is 230 V and the primary current is 2.0 A. The secondary p.d. is 11.5 V. Calculate the…
▸ Question 4 · 3 marks · Explain. Explain why a transformer does not work with a direct current.
▸ Question 5 · 4 marks · Explain. Explain why electricity is transmitted in the National Grid at a very high potential difference.
Question 1 · 2 marks · Calculate
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“A transformer has 200 turns on its primary coil and 1000 turns on its secondary coil. The input potential difference is 12 V. Calculate the output potential difference. Use the equation: Vp ÷ Vs = np ÷ ns” |
HOW TO ANSWER IT Command word: Calculate. Worth 2 marks, so plan before writing.
Question 1 · mark scheme
2 marks available. Award a mark for each point made.
▸ Correct rearrangement and substitution. 1 mark
▸ 60 V. 1 mark
▸ Model answer. V_s = 12 × 1000 ÷ 200 = 60 V
Question 2 · 4 marks · Explain
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“Explain how a transformer changes the potential difference of an alternating supply.” |
HOW TO ANSWER IT Command word: Explain. Worth 4 marks, so plan before writing.
Question 2 · mark scheme
4 marks available. Award a mark for each point made.
▸ a.c. in the primary makes a changing magnetic field. 1 mark
▸ Field carried by the iron core to the secondary. 1 mark
▸ Induced p.d. in the secondary (generator effect). 1 mark
▸ Turns ratio sets the p.d. ratio. 1 mark
▸ Model answer. An alternating current in the primary coil produces a changing magnetic field in the iron core. This changing field passes through the secondary coil and induces an alternating potential difference in it. The ratio of the turns determines the ratio of the potential differences.
Question 3 · 3 marks · Calculate
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“A transformer is 100% efficient. The primary p.d. is 230 V and the primary current is 2.0 A. The secondary p.d. is 11.5 V. Calculate the secondary current.” |
HOW TO ANSWER IT Command word: Calculate. Worth 3 marks, so plan before writing.
Question 3 · mark scheme
3 marks available. Award a mark for each point made.
▸ Vs Is = Vp Ip. 1 mark
▸ Correct substitution. 1 mark
▸ 40 A. 1 mark
▸ Model answer. I_s = 230 × 2.0 ÷ 11.5 = 40 A
Question 4 · 3 marks · Explain
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“Explain why a transformer does not work with a direct current.” |
HOW TO ANSWER IT Command word: Explain. Worth 3 marks, so plan before writing.
Question 4 · mark scheme
3 marks available. Award a mark for each point made.
▸ d.c. produces a constant field. 1 mark
▸ No change in field in the secondary. 1 mark
▸ No p.d. induced. 1 mark
▸ Model answer. A direct current gives a steady magnetic field in the core. No change in field passes through the secondary coil, so no potential difference is induced.
Question 5 · 4 marks · Explain
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“Explain why electricity is transmitted in the National Grid at a very high potential difference.” |
HOW TO ANSWER IT Command word: Explain. Worth 4 marks, so plan before writing.
Question 5 · mark scheme
4 marks available. Award a mark for each point made.
▸ Step-up transformer raises p.d.. 1 mark
▸ Current is lower for the same power. 1 mark
▸ Less energy wasted as heat. 1 mark
▸ Step-down near homes for safety. 1 mark
▸ Model answer. A step-up transformer increases the potential difference and so, for the same power, the current is lower. A lower current means less energy is wasted as heat in the cables, so the transmission is more efficient. Step-down transformers then reduce the p.d. for use in homes.