AQA GCSE PHYSICS · PAPER 1 · FOUNDATION & HIGHER
Exam Practice: Electrical power
Electrical power · Electricity · Lesson 6 of 10 · 14 marks · 15 minutes
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Instructions
• Answer all the questions.
• Write your answers in the spaces provided.
• The marks for each question are shown in brackets - use this as a guide to how much to write.
• The answers are on separate pages at the back. Attempt every question before you look at them.
• Answer all questions. Use the mark allocation as a guide to how much to write.
Question 1 CALCULATE (2 marks)
A kettle is connected to the 230 V mains and the current is 10 A. Calculate the power of the kettle. Use the equation: power = potential difference × current
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(Total for Question 1 = 2 marks)
Question 2 CALCULATE (3 marks)
A current of 6.0 A flows through a resistor of resistance 5.0 Ω. Calculate the power. Use the equation: power = (current)² × resistance
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(Total for Question 2 = 3 marks)
Question 3 CALCULATE (3 marks)
A lamp has a power of 60 W and works at 230 V. Calculate the current in the lamp.
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(Total for Question 3 = 3 marks)
Question 4 EXPLAIN (2 marks)
A student says that the power of a device only depends on the current. Explain why the student is not correct.
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(Total for Question 4 = 2 marks)
Question 5 CALCULATE (4 marks)
An electric heater has a resistance of 46 Ω and is connected to a 230 V supply. Calculate the power of the heater.
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(Total for Question 5 = 4 marks)
TOTAL FOR PAPER = 14 MARKS
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Answers and mark scheme
Check your answer only once you have written one.
Question 1 (2 marks)
P = 230 × 10 = 2300 W
• Correct substitution 1 mark
• 2300 W 1 mark
Question 2 (3 marks)
P = 6.0² × 5.0 = 180 W
• Squares the current 1 mark
• Correct substitution 1 mark
• 180 W 1 mark
Question 3 (3 marks)
I = P ÷ V = 60 ÷ 230 = 0.26 A
• Rearranges to I = P ÷ V 1 mark
• Correct substitution 1 mark
• 0.26 A 1 mark
Question 4 (2 marks)
Power depends on both the current and the potential difference (P = VI), or on the current and the resistance (P = I²R).
• Power = V × I 1 mark
• Depends on pd (or resistance) as well as current 1 mark
Question 5 (4 marks)
I = 230 ÷ 46 = 5.0 A; P = 230 × 5.0 = 1150 W
• Rearranges V = IR 1 mark
• 5.0 A 1 mark
• Correct substitution into P = VI 1 mark
• 1150 W 1 mark