A Level Physics 9.2 Potential Difference and Power MCQs – 30 Practice Questions with Answers (Cambridge 9702)

Free practice MCQs for Cambridge International AS & A Level Physics (9702), Section 9.2 Potential Difference and Power. Test yourself on potential difference as energy transferred per unit charge, the three electrical power equations, and energy calculations in joules and kilowatt-hours, then open the Show Answer panels to check the worked solutions.

Formula recap: V = W / Q, P = V I = I2R = V2 / R, W = P t = V I t and 1 kWh = 3.6 × 106 J. See the full A Level Physics Formula Sheet for every equation.

Potential Difference & Energy Transfer

Q1. What is the definition of the potential difference between two points in a circuit?

  • (A) The energy transferred between the points per unit time
  • (B) The energy transferred per unit charge as charge passes between the points
  • (C) The charge passing between the points per unit time
  • (D) The force acting on a unit charge between the points
Show Answer

Answer: B – Potential difference is defined by V = W / Q: the work done, or energy transferred, per unit charge as the charge moves between the two points. Option (A) describes a rate of energy transfer (power) and option (C) describes current.

Q2. A charge of 5.0 C passes through a component across which the potential difference is 12 V. How much energy is transferred?

  • (A) 2.4 J
  • (B) 17 J
  • (C) 60 J
  • (D) 120 J
Show Answer

Answer: C – W = VQ = 12 × 5.0 = 60 J.

Q3. A battery transfers 240 J of energy as it drives charge around a circuit. The potential difference of the battery is 24 V. How much charge flows?

  • (A) 0.10 C
  • (B) 10 C
  • (C) 264 C
  • (D) 5760 C
Show Answer

Answer: B – From W = VQ, Q = W / V = 240 / 24 = 10 C.

Q4. One volt is equal to

  • (A) one coulomb per joule
  • (B) one joule per second
  • (C) one ampere per second
  • (D) one joule per coulomb
Show Answer

Answer: D – Since V = W / Q, the unit of potential difference is the joule per coulomb: 1 V = 1 J C−1.

Q5. An electron (charge 1.6 × 10−19 C) passes through a potential difference of 1.0 V. The energy it gains is

  • (A) 1.6 × 10−19 J
  • (B) 1.0 J
  • (C) 1.6 × 10−19 C
  • (D) 6.3 × 1018 J
Show Answer

Answer: A – W = QV = (1.6 × 10−19) × 1.0 = 1.6 × 10−19 J. This amount of energy is called one electronvolt: 1 eV = 1.6 × 10−19 J.

Q6. An electron is accelerated from rest through a potential difference of 500 V. Its gain in kinetic energy is

  • (A) 8.0 × 10−19 J
  • (B) 8.0 × 10−17 J
  • (C) 1.6 × 10−16 J
  • (D) 3.1 × 10−21 J
Show Answer

Answer: B – W = QV = (1.6 × 10−19) × 500 = 8.0 × 10−17 J, which is a gain of 500 eV.

Q7. A 12 V supply is connected across a resistor and a lamp in series. The potential difference across the resistor is 5.0 V. The potential difference across the lamp is

  • (A) 5.0 V
  • (B) 12 V
  • (C) 7.0 V
  • (D) 17 V
Show Answer

Answer: C – In a series circuit the potential differences across the components add up to the supply potential difference (energy per unit charge is conserved): 12 − 5.0 = 7.0 V.

Q8. A voltmeter is correctly connected in a circuit when it is placed

  • (A) in series with the component, with zero resistance
  • (B) in parallel with the component, with very low resistance
  • (C) in series with the component, with ideally infinite resistance
  • (D) in parallel with the component, with ideally infinite resistance
Show Answer

Answer: D – A voltmeter measures the potential difference between two points, so it is connected in parallel across the component. Its resistance is ideally infinite so that it draws no current and does not change the circuit it is measuring.

Q9. A battery does 45 J of work in moving 3.0 C of charge through a circuit. The potential difference of the battery is

  • (A) 0.067 V
  • (B) 5.0 V
  • (C) 15 V
  • (D) 135 V
Show Answer

Answer: C – V = W / Q = 45 / 3.0 = 15 V.

Q10. A charge of 2.0 C flows through a resistor and transfers 36 J of energy to it. The potential difference across the resistor is

  • (A) 18 V
  • (B) 72 V
  • (C) 0.056 V
  • (D) 38 V
Show Answer

Answer: A – V = W / Q = 36 / 2.0 = 18 V.

Electrical Power

Q11. The equation P = V I for electrical power follows from the definitions of potential difference and current because

  • (A) V is the energy transferred per unit charge and I is the charge passing per unit time, so V I is the energy transferred per unit time
  • (B) V is the energy transferred per unit time and I is the total charge, so V I is the energy per unit charge
  • (C) V is the charge passing per unit time and I is the energy per unit charge, so V I is the total charge
  • (D) V and I are both rates, so V I is the energy stored in the component
Show Answer

Answer: A – Power is energy transferred per unit time. P = W / t = (V Q) / t = V (Q / t) = V I, because V = W / Q and I = Q / t.

Q12. A component has a potential difference of 12 V across it and a current of 3.0 A through it. The power dissipated in the component is

  • (A) 0.25 W
  • (B) 36 W
  • (C) 4.0 W
  • (D) 15 W
Show Answer

Answer: B – P = V I = 12 × 3.0 = 36 W.

Q13. A current of 2.0 A flows through a resistor of resistance 10 Ω. The power dissipated in the resistor is

  • (A) 20 W
  • (B) 200 W
  • (C) 40 W
  • (D) 5.0 W
Show Answer

Answer: C – P = I2R = (2.0)2 × 10 = 4.0 × 10 = 40 W.

Q14. A potential difference of 24 V is applied across a resistor of resistance 12 Ω. The power dissipated in the resistor is

  • (A) 48 W
  • (B) 2.0 W
  • (C) 288 W
  • (D) 12 W
Show Answer

Answer: A – P = V2 / R = (24)2 / 12 = 576 / 12 = 48 W.

Q15. The potential difference across a resistor and the resistance of the resistor are known, but the current through it has not been measured. Which expression gives the power dissipated in the resistor?

  • (A) P = V I
  • (B) P = I2R
  • (C) P = V / R
  • (D) P = V2 / R
Show Answer

Answer: D – P = V2 / R uses only V and R. The forms P = V I and P = I2R both need the current, and V / R gives the current, not the power. Choosing the form built from the quantities you know is the key skill.

Q16. Two identical resistors, each of resistance 6.0 Ω, are connected in parallel across a 12 V supply. The total power dissipated in the combination is

  • (A) 12 W
  • (B) 24 W
  • (C) 48 W
  • (D) 96 W
Show Answer

Answer: C – In parallel, each resistor has the full 12 V across it, so each dissipates P = V2 / R = 144 / 6.0 = 24 W. Total power = 2 × 24 = 48 W. (In series the same two resistors would dissipate only 12 W in total, so a parallel combination always dissipates more power from the same supply.)

Q17. A resistor of resistance 25 Ω dissipates a power of 100 W. The current through the resistor is

  • (A) 2.0 A
  • (B) 0.50 A
  • (C) 4.0 A
  • (D) 20 A
Show Answer

Answer: A – From P = I2R, I2 = P / R = 100 / 25 = 4.0, so I = 2.0 A.

Q18. An electric kettle operates at 230 V and has a power rating of 2300 W. Which fuse is the most suitable for the kettle?

  • (A) 3 A
  • (B) 5 A
  • (C) 10 A
  • (D) 13 A
Show Answer

Answer: D – The normal working current is I = P / V = 2300 / 230 = 10 A. The fuse rating must be just above the normal current so that it does not melt in normal use but still protects the cable if a fault occurs, so a 13 A fuse is chosen. A 10 A fuse would be right at its limit during normal operation.

Q19. Resistors of resistance 2.0 Ω and 4.0 Ω are connected in series and carry a current of 3.0 A. The power dissipated in the 4.0 Ω resistor is

  • (A) 18 W
  • (B) 36 W
  • (C) 12 W
  • (D) 24 W
Show Answer

Answer: B – P = I2R = (3.0)2 × 4.0 = 9.0 × 4.0 = 36 W. In series the current is the same in both resistors, so the larger resistance dissipates the larger power.

Q20. Resistors of resistance 3.0 Ω and 6.0 Ω are connected in parallel across a 12 V supply. The total power dissipated in the combination is

  • (A) 24 W
  • (B) 48 W
  • (C) 72 W
  • (D) 108 W
Show Answer

Answer: C – Both resistors have 12 V across them. P(3.0 Ω) = V2 / R = 144 / 3.0 = 48 W and P(6.0 Ω) = 144 / 6.0 = 24 W, giving a total of 48 + 24 = 72 W. In parallel, the smaller resistance dissipates the larger power.


Power & Energy Calculations

Q21. A 60 W lamp is switched on for 5.0 minutes. The electrical energy transferred by the lamp is

  • (A) 300 J
  • (B) 18000 J
  • (C) 12000 J
  • (D) 300000 J
Show Answer

Answer: B – W = P t = 60 × (5.0 × 60) = 60 × 300 = 18000 J. The time must be converted from minutes to seconds.

Q22. A heater operates at 230 V with a current of 2.0 A for 10 minutes. The energy transferred by the heater is

  • (A) 276 kJ
  • (B) 4.6 kJ
  • (C) 27.6 kJ
  • (D) 460 J
Show Answer

Answer: A – W = V I t = 230 × 2.0 × (10 × 60) = 230 × 2.0 × 600 = 276000 J = 276 kJ.

Q23. A 2.0 kW heater is used for 3.0 hours. How many kilowatt-hours (units) of electrical energy does it use?

  • (A) 1.5 kWh
  • (B) 5.0 kWh
  • (C) 6000 kWh
  • (D) 6.0 kWh
Show Answer

Answer: D – Energy in kWh = power in kW × time in hours = 2.0 × 3.0 = 6.0 kWh.

Q24. An appliance uses 5.0 kWh of electrical energy. This energy, expressed in joules, is

  • (A) 1.8 × 106 J
  • (B) 3.6 × 106 J
  • (C) 1.8 × 107 J
  • (D) 5.0 × 106 J
Show Answer

Answer: C – One kilowatt-hour is the energy transferred by a 1 kW device in 1 hour: 1 kWh = 1000 × 3600 = 3.6 × 106 J. So 5.0 kWh = 5.0 × 3.6 × 106 = 1.8 × 107 J.

Q25. An electric kettle has a power rating of 2760 W when connected to a 230 V supply. The current it draws from the supply is

  • (A) 12 A
  • (B) 0.083 A
  • (C) 120 A
  • (D) 2530 A
Show Answer

Answer: A – I = P / V = 2760 / 230 = 12 A.

Q26. An electric motor takes an input power of 500 W and produces a useful mechanical output power of 350 W. The efficiency of the motor is

  • (A) 30%
  • (B) 65%
  • (C) 70%
  • (D) 143%
Show Answer

Answer: C – Efficiency = useful power out / total power in = 350 / 500 = 0.70, so the efficiency is 70%.

Q27. A filament lamp takes 60 W of electrical power, of which only 9.0 W is transferred as useful light energy. The efficiency of the lamp is

  • (A) 15%
  • (B) 85%
  • (C) 6.7%
  • (D) 51%
Show Answer

Answer: A – Efficiency = 9.0 / 60 = 0.15, so the efficiency is 15%. The remaining 85% of the input power is transferred to the surroundings as thermal energy.

Q28. A lamp is rated “12 V, 24 W”. It is instead connected to a 6.0 V supply. Assuming the resistance of the lamp stays constant, the power it dissipates is

  • (A) 12 W
  • (B) 6.0 W
  • (C) 24 W
  • (D) 48 W
Show Answer

Answer: B – This is a two-step problem: the rating only applies at 12 V. First find the resistance from the rating: R = V2 / P = (12)2 / 24 = 144 / 24 = 6.0 Ω. Then at 6.0 V: P = V2 / R = (6.0)2 / 6.0 = 36 / 6.0 = 6.0 W. Halving the voltage quarters the power.

Q29. An immersion heater connected to a 230 V supply carries a current of 8.0 A. It is switched on for 20 minutes. The energy transferred is approximately

  • (A) 36.8 kJ
  • (B) 110 kJ
  • (C) 1.84 MJ
  • (D) 2.2 MJ
Show Answer

Answer: D – W = V I t = 230 × 8.0 × (20 × 60) = 230 × 8.0 × 1200 = 2208000 J, which is approximately 2.2 MJ.

Q30. A 3.0 kW storage heater is switched on for 4.0 hours. Electricity costs $0.30 per kWh. The cost of the energy used is

  • (A) $0.90
  • (B) $36.00
  • (C) $1.20
  • (D) $3.60
Show Answer

Answer: D – Energy used = 3.0 × 4.0 = 12 kWh. Cost = 12 × 0.30 = $3.60.

Keep practising: work through more topic-wise questions on the Physics MCQs hub, and keep the A Level Physics Formula Sheet open while you revise – knowing V = W / Q and all three forms of the power equation by heart makes every one of these questions faster.

These are original practice questions written for revision. They are not Cambridge past-paper questions, and this page is not affiliated with Cambridge Assessment International Education.