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Edexcel International GCSE Physics · Topic Test
Electricity
FREE Online Physics GCSE Course45minutes
45marks
8questions
How it works
- Answer each question. Type into the boxes or pick an option, and use paper for working and drawings.
- Press Check my answers at the end of the question. Tick-box and number answers are marked for you.
- For written answers, compare yours with the mark scheme and tick each point you made. We highlight the key words we spot, but you decide, just like an examiner.
Which course are you taking?
Pick one and the page skips the questions that aren’t on your course. Double Award skips Questions 7 and 8 (static electricity), which are Physics only.
You will need: paper, a pen, a pencil, a ruler and a calculator. Your answers save automatically in this browser.
1
Question 1
2 marksCalculate the resistance of a filament lamp when the voltage across it is 4.5 V and the current in it is 51 mA.
Use the equation R = V ÷ I.
Ω
(2 marks)
Mark scheme
Resistance: 88 Ω (accept 88.2 Ω, or 0.088 kΩ)
💡 You used 51 for the current. 51 mA = 0.051 A: divide by 1000 first. (A power-of-ten slip like this still scores 1 mark: tick it below.)
💡 You multiplied. R = V ÷ I.
Final answer not right? You can still earn method marks. Tick each step you wrote down (max 1):
2
Question 2
4 marksA technician investigates different electrical devices that are used in a car.
The technician connects a device to the 12 V car battery and measures the current in the circuit and the potential difference (voltage) across the device.
Figure 1 shows the car battery and the device that is being tested.
Figure 1
(a)
Draw (on paper) how the circuit should be completed so that the current in the circuit and the voltage across the device can be measured.
(2 marks)
Mark scheme
Compare your drawing with the mark scheme. Which describes it best?
(b)(i)
The technician connects four devices to the car battery. Each device has its own switch and its own fuse. Figure 2 shows how they are connected. The current in each device is shown next to it.
Figure 2
Calculate the current in the wires to the battery when all the devices are switched on.
A
(1 mark)
Mark scheme
Current: 17.7 A
💡 You added 600 for the interior light. 600 mA = 0.6 A.
💡 Close: you left out the interior light (0.6 A).
(b)(ii)
State how the overall resistance of the circuit changes when any one of the devices is switched off.
(1 mark)
Mark scheme
The resistance: it increases
💡 Switching a branch off means less current from the same 12 V battery, so R = V ÷ I goes up.
3
Question 3
10 marksA student investigates resistors connected in parallel. Each resistor has the same resistance.
Figure 1 shows the circuit with one resistor, R.
Figure 1
(a)(i)
Copy the circuit on paper and add:
- a voltmeter to find the potential difference across resistor R
- another resistor in parallel with resistor R.
(2 marks)
Mark scheme
Compare your drawing with the answer above. Which describes it best?
(a)(ii)
State the measurements that the student must take to find the overall resistance of the resistors in parallel.
(2 marks)
Mark scheme
Tick each marking point your answer makes:
(a)(iii)
The student investigates how the overall resistance of the circuit changes when more identical resistors are added in parallel with R. Figure 2 shows the results.
Figure 2
State the resistance of a single resistor.
Ω
(1 mark)
Mark scheme
Resistance: 1 Ω
💡 That’s two resistors in parallel. Read the point at 1 resistor.
(a)(iv)
Comment on the relationship between the overall resistance of the circuit and the number of resistors in parallel. Use information from Figure 2 to support your answer.
(3 marks)
Mark scheme
Tick each marking point your answer makes:
(b)
A different circuit is then set up with two resistors, as shown in Figure 3.
Figure 3
Calculate the potential difference across the 15 Ω resistor.
V
(2 marks)
Mark scheme
Potential difference: 3 V
💡 You used both resistors (15 + 20). The question asks for the 15 Ω resistor only. (7 V scores 1 mark: tick it below.)
💡 You divided. V = I × R = 0.20 × 15.
Final answer not right? You can still earn method marks. Tick each step you wrote down (max 1):
4
Question 4
6 marks(a)
Which of these symbols is used to represent a thermistor in an electrical circuit?
(1 mark)
Mark scheme
Answer: D
(b)(i)
A student investigates how the current in a lamp changes with the potential difference across it, and uses the results to calculate the resistance of the lamp. Figure 1 shows the results.
| Potential difference in V | Current in A | Resistance in Ω |
|---|---|---|
| 1.0 | 0.09 | 11 |
| 2.0 | 0.14 | 14 |
| 3.0 | 0.18 | 17 |
| 4.0 | 0.22 | 18 |
| 5.0 | 0.26 | ? |
| 6.0 | 0.30 | 20 |
One value of resistance is missing from the table. Calculate the value of resistance that is missing.
Ω
(3 marks)
Mark scheme
Missing resistance: 19 Ω (accept 19.2)
💡 Upside down: R = V ÷ I = 5.0 ÷ 0.26.
💡 You multiplied. R = V ÷ I.
Final answer not right? You can still earn method marks. Tick each step you wrote down (max 2):
(b)(ii)
The student used a power supply that had fixed output voltage settings. Each of these outputs was a whole number of volts.
Describe how the student could add a component to the circuit that would give a continuously variable voltage across the lamp.
(2 marks)
Mark scheme
The second mark only counts if you got the first:
⚠️ An LDR, thermistor or fixed resistor doesn’t let the student choose the voltage. Use a variable resistor.
5
Question 5
7 marks(a)(i)
A resistor is connected to a power supply. The potential difference across the resistor is 6.0 V.
Which of these corresponds to a potential difference of 6.0 V?
(1 mark)
Mark scheme
Answer: C 6.0 joules per coulomb
(a)(ii)
The resistor stays connected for a period of time. The current in the resistor is 200 mA. A total charge of 42 C flows through the resistor.
Calculate, in minutes, the time taken for this amount of charge to flow through the resistor.
minutes
(3 marks)
Mark scheme
Time: 3.5 minutes (3 minutes 30 seconds)
💡 210 is the time in seconds. The question asks for minutes: divide by 60. (Scores 2 marks: tick below.)
💡 You used 200 for the current: 200 mA = 0.200 A. (A power-of-ten slip like this scores 2 marks.)
💡 You multiplied. Q = I × t, so t = Q ÷ I = 42 ÷ 0.200.
Final answer not right? You can still earn method marks. Tick each step you wrote down (max 2):
(b)
Figure 1 shows a cardboard tube with a wire coming out of each end.
Figure 1
There are two 10 Ω resistors inside the cardboard tube. A potential difference of 6.0 V is connected between P and Q. There is a current of 1.2 A in the wires.
Deduce how the resistors have been arranged inside the cardboard tube.
(3 marks)
Mark scheme
Tick each marking point your answer makes:
6
Question 6
7 marks(a)
Figure 1 shows eight identical filament lamps connected in series to a 12 V power supply.
Figure 1
Calculate the potential difference across each lamp.
V
(1 mark)
Mark scheme
Potential difference: 1.5 V
💡 In series the 12 V is shared between the eight lamps.
💡 In series the p.d. is shared: divide by 8, don’t multiply.
(b)
Explain, with the aid of a circuit diagram, the method a student could use to investigate how the resistance of a single lamp changes with the potential difference across the lamp.
Draw your circuit diagram on paper.
(6 marks)
Mark scheme
Circuit diagram: tick each point you made
Method: tick each point you made
| 0 | No rewardable material. | 0 |
|---|---|---|
| Level 1 | Some relevant physics, but some of it is inaccurate, and the method lacks detail. Not in a logical order, with big gaps. | 1–2 |
| Level 2 | Mostly relevant physics, with some inaccuracies. The method is not fully detailed. A mostly clear, logical structure with minor steps missing. | 3–4 |
| Level 3 | Accurate, relevant physics throughout. The method is detailed and fully developed. Clear, coherent and logical. | 5–6 |
Your mark:
Show a model answer
- Connect the filament lamp in series with an ammeter, a variable resistor and a power supply. Connect a voltmeter in parallel across the lamp.
- Switch on and record the current from the ammeter and the potential difference from the voltmeter.
- Adjust the variable resistor to change the potential difference, and record at least six pairs of readings over a wide range.
- Calculate the resistance for each pair of readings using R = V ÷ I.
- Repeat the readings and calculate a mean for each potential difference.
- Plot a graph of resistance against potential difference and compare: the resistance increases as the potential difference increases, because the filament gets hotter.
7
Question 7
Physics only4 marksSkip this one. Static electricity is only on the Physics course, so it isn’t in your Double Award score.
Physics only. Double Award students: pick “Double Award” at the top and this question folds away.
A student gives a plastic strip an overall electric charge.
(a)(i)
Describe one way that the student can give the plastic strip an overall electric charge.
(1 mark)
Mark scheme
Tick each marking point your answer makes:
(a)(ii)
Figure 1 shows a gold leaf electroscope that can be used to investigate static electricity.
Figure 1
The electroscope has no overall charge. The gold leaf has a very small mass and can bend very easily.
The student brings a negatively charged plastic strip near to the cap of the electroscope. The gold leaf bends away from the metal rod.
Which diagram shows how the electric charge is now distributed?
(1 mark)
Mark scheme
Answer: C
(b)
Figure 2a shows another gold leaf electroscope that has been given an overall negative charge. A student connects its metal cap to earth with a piece of wire, as shown in Figure 2b.
Figure 2
Explain why the gold leaf has moved.
(2 marks)
Mark scheme
Tick each marking point your answer makes:
⚠️ Only electrons move, and they are negative. Protons stay put in the nucleus.
8
Question 8
Physics only5 marksSkip this one. Static electricity is only on the Physics course, so it isn’t in your Double Award score.
Physics only. Double Award students: pick “Double Award” at the top and this question folds away.
Figure 1 shows a paint sprayer. Some paint sprayers give the paint drops a positive charge as they leave the sprayer.
Figure 1
(a)
The paint drops have a positive charge because the sprayer
(1 mark)
Mark scheme
Answer: A removes electrons from the paint drops
(b)
Figure 2 shows the spray pattern from two different paint sprayers. Sprayer X does not charge the paint drops. Sprayer Y gives the paint drops a positive charge.
Figure 2
Explain how charging the paint drops changes the shape of the spray pattern.
(2 marks)
Mark scheme
Tick each marking point your answer makes:
⚠️ Drops with the same charge repel, they don’t attract each other.
(c)
Sprayer Y is used in a factory to paint a metal object. The object hangs by a metal wire that is connected to earth.
Explain why a metal wire is used to connect the object to earth.
(2 marks)
Mark scheme
Tick each marking point your answer makes:
Your results
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| Question | Q1 | Q2 | Q3 | Q4 | Q5 | Q6 | Q7 | Q8 | Total |
|---|---|---|---|---|---|---|---|---|---|
| Out of | 2 | 4 | 10 | 6 | 7 | 7 | 4 | 5 | 45 |
| Your mark | – | – | – | – | – | – | – | – | 0 |
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