AQA GCSE PHYSICS · PAPER 1 · FOUNDATION & HIGHER
Exam Practice: Series and parallel circuits
Series and parallel circuits · Electricity · Lesson 4 of 10 · 18 marks · 19 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 (3 marks)
The diagram shows a 12 V battery connected to two resistors in series. Calculate (a) the total resistance and (b) the current in the circuit.
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(Total for Question 1 = 3 marks)
Question 2 CALCULATE (3 marks)
Use your answer to the previous question to calculate the potential difference across the 8.0 Ω resistor. The current is 1.0 A.
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(Total for Question 2 = 3 marks)
Question 3 USE THE DIAGRAM (3 marks)
The diagram shows two lamps connected in parallel to a 6.0 V supply. Ammeter A1 reads 1.2 A and A2 reads 0.80 A. (a) Calculate the reading on A3. (b) State the pd across each lamp.
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(Total for Question 3 = 3 marks)
Question 4 EXPLAIN (2 marks)
Explain why the lights in a house are connected in parallel rather than in series.
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(Total for Question 4 = 2 marks)
Question 5 EXPLAIN (3 marks)
Explain, in terms of the flow of current, why adding a second resistor in series increases the total resistance but adding a second resistor in parallel decreases the total resistance.
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(Total for Question 5 = 3 marks)
Question 6 CALCULATE (4 marks)
A 9.0 V cell is connected to a 6.0 Ω resistor and a 12 Ω resistor in series. Calculate the potential difference across the 12 Ω resistor.
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(Total for Question 6 = 4 marks)
TOTAL FOR PAPER = 18 MARKS
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Answers and mark scheme
Check your answer only once you have written one.
Question 1 (3 marks)
(a) R = 4.0 + 8.0 = 12 Ω. (b) I = 12 ÷ 12 = 1.0 A
• Adds resistances 1 mark
• 12 Ω 1 mark
• 1.0 A 1 mark
Question 2 (3 marks)
V = IR = 1.0 × 8.0 = 8.0 V
• Correct equation 1 mark
• Correct substitution 1 mark
• 8.0 V 1 mark
Question 3 (3 marks)
(a) 1.2 + 0.80 = 2.0 A. (b) 6.0 V across each lamp.
• Adds the branch currents 1 mark
• 2.0 A 1 mark
• 6.0 V 1 mark
Question 4 (2 marks)
In parallel each lamp has the full supply potential difference across it, and each lamp can be switched on and off independently of the others.
• Each lamp gets the full supply pd 1 mark
• Lamps can be switched independently or one failing does not stop the others 1 mark
Question 5 (3 marks)
In series the current must pass through both resistors so the opposition is greater. In parallel there are two paths, so more current can flow for the same potential difference, so the total resistance is less.
• Series: current passes through both, so resistance is greater 1 mark
• Parallel: two paths 1 mark
• More current for the same pd, so total resistance is less 1 mark
Question 6 (4 marks)
R = 18 Ω; I = 9.0 ÷ 18 = 0.50 A; V = 0.50 × 12 = 6.0 V
• Total resistance 18 Ω 1 mark
• Current 0.50 A 1 mark
• Correct substitution 1 mark
• 6.0 V 1 mark