Electrical energy and electrical power: past paper questions
Real questions on electrical energy and electrical power
The newest 20 of 71 written questions. Open one to see it on the real paper.
- 0625/31 · May/June 2026 · Q4a(ii)[3]
The current in the motor is 1.2 A. The potential difference (p.d.) across the motor is 24 V. Calculate the energy transferred to the motor in a time of 30 s. energy transferred = J
See it on the paper → - 0625/31 · May/June 2026 · Q4a(iii)[2]
Explain one way in which energy is wasted as the electric motor turns the wheel.
See it on the paper → - 0625/32 · May/June 2026 · Q8b[3]
Calculate the power of the heater. power = W
See it on the paper → - 0625/33 · May/June 2026 · Q8b[3]
Calculate the power of the heater. power = W
See it on the paper → - 0625/41 · May/June 2026 · Q7b(i)[2]
Calculate the work done by the battery. work done = J (i)
See it on the paper → - 0625/41 · May/June 2026 · Q7b(iii)[1]
Show that the current in the motor is 0.66 A.
See it on the paper → - 0625/41 · May/June 2026 · Q8b(i)[3]
Calculate the energy transferred to the lawnmower in 0.75 h. Give your answer in kW h. energy = kW h
See it on the paper → - 0625/41 · May/June 2026 · Q8b(ii)[1]
Calculate the cost of the energy calculated in 8(b)(i). cost = $
See it on the paper → - 0625/42 · May/June 2026 · Q2a(i)[1]
State the direction of the force acting on the satellite.
See it on the paper → - 0625/42 · May/June 2026 · Q2a(ii)[3]
A rocket is used to carry the satellite into space. The initial mass of the rocket and satellite is 1.5 × 10 4 kg. The initial force provided by the rocket is 1.9 × 10 5 N. Calculate the initial acceleration of the rocket. acceleration = m / s 2
See it on the paper → - 0625/42 · May/June 2026 · Q8a(i)[2]
Complete the definition of an ‘electric field’. An electric field is a region where 8
See it on the paper → - 0625/42 · May/June 2026 · Q8a(ii)[2]
Figure 8.1 shows two parallel, oppositely charged metal plates. – – – – – + + + + + Figure 8.1 On Figure 8.1, draw four field lines to show the pattern and the direction of the electric field between the two parallel plates.
See it on the paper → - 0625/42 · May/June 2026 · Q8b(i)[1]
Identify component X.
See it on the paper → - 0625/42 · May/June 2026 · Q8b(ii)[3]
The temperature increases. The lamp lights. Complete the sentences to explain, in terms of potential difference (p.d.) and resistance, why the lamp lights. When the temperature of the thermistor increases, the of the thermistor . This causes the p.d. across the coil . The relay coil attracts , to increase. The magnetic...
See it on the paper → - 0625/42 · May/June 2026 · Q8b(iii)[2]
The current in the lamp is 3.2 A. Calculate the charge that flows through the lamp in 120 s. charge = C
See it on the paper → - 0625/32 · February/March 2026 · Q8d[3]
The current in one lamp is 0.45 A and the potential difference across the lamp is 6.7 V. The lamp is switched on for a time of 50 s. Calculate the energy supplied to the lamp. energy supplied = J
See it on the paper → - 0625/42 · February/March 2026 · Q4c[2]
The heater has a power of 2.5 kW. The heater is switched on for 3 hours. Calculate the energy, in kWh, transferred electrically by the heater. energy = kWh
See it on the paper → - 0625/31 · October/November 2025 · Q8c[4]
The potential difference (voltage) across a resistor is 64 V. The current in the resistor is 2.2 mA. Calculate the power of the resistor. power = W
See it on the paper → - 0625/31 · October/November 2025 · Q8d[3]
Calculate the cost of using the television for 6.0 hours. cost for 6 hours = cents (d) The energy used by the television in one hour is 0.14 kWh. The cost of one kWh of energy is 36 cents.
See it on the paper → - 0625/32 · October/November 2025 · Q7b(ii)[3]
Calculate the energy transferred by the lamp in 140 s. energy transferred = J
See it on the paper →
What examiners said
Comprehensive responses • Most candidates were able to select the correct information from the question and diagram and then to substitute these values into a correct equation to calculate the energy supplied to the lamp as 150 J. Limited responses • Some candidates rearranged the equation for calculating energy transferred. A commonly seen approach was dividing I V by the time rather than multip...
— Examiner report, February/March 2026, 0625/32 Q8(d) (Cambridge)
Comprehensive responses • Most candidates knew how to calculate the electrical energy transferred by the heater. • Many candidates used the data in the question to calculate the energy in kWh. Limited responses • A few candidates attempted conversion of the data to give an answer in joules. • A few candidates were confused about the terms energy and power in their equation and performed the calcu...
— Examiner report, February/March 2026, 0625/42 Q4(c) (Cambridge)
Many candidates scored full credit, but the majority did not multiply the three terms. They often only multiplied 2 terms, and a common incorrect answer was 1200 from 50 24 or 0.045 from 0.0090 50.
— Examiner report, May/June 2025, 0625/31 Q9(b) (Cambridge)
Past papers, syllabus and examiner reports © UCLES (Cambridge University Press & Assessment). Questions are matched to topics from the official syllabus. AssessPro is independent and not endorsed by Cambridge.