IGCSE Physics Electrical Energy and Electrical Power Questions and Worked Solutions (4.2.5) | 0625

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IGCSE Physics 0625 · Topic 4.2.5

IGCSE Physics Electrical Energy and Electrical Power Questions and Worked Solutions

50 topic-wise practice questions on Electrical Energy and Electrical Power (Unit 4: Electricity and Magnetism), written to the Cambridge IGCSE Physics 0625 syllabus for 2026–2028, each with a step-by-step worked solution.

50 questions40 Core0 Extended10 Challenge129 marksTextbook §4.2 · pp. 184–208
Questions tagged Core are for all candidates (Papers 1 & 3). Questions tagged Extended are Supplement content (Papers 2 & 4). Questions 41–50 are harder Challenge questions. Try each question first, then tap Show solution to check your working.

Topic guide: Electrical Energy and Electrical Power

Electric circuits transfer energy from a source to components and the surroundings. Power is the rate of energy transfer. Electricity bills use the kilowatt-hour as the unit of energy.

Key formulas

P = I Vpower P (W), current I (A), p.d. V (V)Core
E = I V tenergy E (J), time t (s)Core
E (kW h) = P (kW) × t (h)household energyCore
cost = energy (kW h) × price per kW helectricity billCore

Prerequisite knowledge

  • Current and p.d. (4.2.2, 4.2.3)
  • Power (1.7.4)

Exam tips & common mistakes

  • Use seconds with E = IVt but kW and hours for kW h
  • Convert W to kW (÷ 1000) and minutes to hours (÷ 60)
  • Calculate current from P = IV when choosing fuses

Key facts from the syllabus (4.2.5)

  • Circuits transfer energy from a cell or mains supply to components and then to the surroundings
  • 1 kW h is the energy transferred by a 1 kW appliance in 1 hour (= 3.6 × 10⁶ J)
  • Higher-power appliances transfer more energy per second and cost more to run

Textbook: §4.2 Electrical quantities, pp. 184–208

1
Core[2 marks]

Describe the energy transfers in a circuit with a cell and a lamp.

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Method: Chemical → electrical → light and thermal.

Chemical energy in the cell → electrical work by the current → light and thermal energy in the surroundings
2
Core[1 mark]

State the equation for electrical power in terms of current and p.d.

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Method: P = IV.

P = IV
3
Core[2 marks]

A lamp draws 0.50 A from a 12 V supply. Calculate its power.

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Method: P = IV.

P = 0.50 × 12 = 6.00 W
4
Core[2 marks]

A 2300 W kettle runs on 230 V mains. Calculate the current.

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Method: I = P ÷ V.

I = 2300 ÷ 230 = 10.0 A
5
Core[2 marks]

A motor draws 3.0 A at 12 V for 5.0 minutes. Calculate the energy transferred.

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Method: E = IVt.

E = 3.0 × 12 × 300 = 10800 J
6
Core[2 marks]

A 60 W lamp is on for 4.0 hours. Calculate the energy transferred in J.

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Method: E = Pt.

E = 60 × 14400 = 864000 J
7
Core[1 mark]

Define the kilowatt-hour.

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Method: Energy of 1 kW for 1 h.

The energy transferred by a 1 kW appliance in 1 hour
8
Core[2 marks]

Calculate the energy in kW h used by a 3.0 kW heater for 2.5 h.

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Method: E = Pt.

E = 3.0 × 2.5 = 7.50 kW h
9
Core[3 marks]

A 2.0 kW iron is used for 45 minutes. Electricity costs 8 rupees per kW h. Calculate the cost.

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Method: E = 2.0 × 0.75.

E = 1.50 kW h; cost = 1.5 × 8 = ₹12.0
10
Core[3 marks]

A 100 W TV is on for 5 h a day for 30 days. Electricity costs 25 cents per kW h. Calculate the cost.

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Method: E = 0.1 × 150.

E = 15.0 kW h; cost = 15 × 25 = 375 cents
11
Core[2 marks]

Convert 1 kW h into joules.

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Method: 1000 W × 3600 s.

1 kW h = 1000 × 3600 = 3.6 × 10⁶ J
12
Core[3 marks]

An electricity meter changes from 12 450 kW h to 12 730 kW h in a month. At ₹7 per kW h, calculate the bill.

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Method: Units used = difference.

Units = 280 kW h; cost = 280 × 7 = ₹1960
13
Core[3 marks]

Three appliances run for 2 hours: 1.5 kW, 500 W and 100 W. Calculate the total energy in kW h.

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Method: Add powers in kW.

E = (1.5 + 0.5 + 0.1) × 2 = 4.20 kW h
14
Core[2 marks]

A heater transfers 72 kJ in 60 s from a 12 V supply. Calculate the current.

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Method: P = E/t, I = P/V.

P = 1200 W; I = 1200 ÷ 12 = 100 A
15
Core[2 marks]

A 9.0 W LED bulb replaces a 60 W filament bulb. Calculate the energy saved in 1000 hours in kW h.

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Method: ΔP × t.

Saved = 0.051 × 1000 = 51.0 kW h
16
Core[3 marks]

A 230 V hairdryer draws 4.0 A. Calculate its power and the energy used in 10 minutes.

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Method: P = IV; E = Pt.

P = 920 W; E = 920 × 600 = 552000 J
17
Core[2 marks]

A device uses 5400 J in 3.0 minutes. Calculate its power.

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Method: P = E/t.

P = 5400 ÷ 180 = 30.0 W
18
Core[3 marks]

A 12 V battery delivers 2.5 A. Calculate the time for it to deliver 54 kJ of energy.

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Method: t = E ÷ (IV).

t = 54000 ÷ 30 = 1800 s = 30 min
19
Core[2 marks]

State why kW h is used for household bills instead of joules.

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Method: J is too small.

The joule is very small; household energy in J would give inconveniently large numbers
20
Core[3 marks]

A 3.0 kW shower is used for 12 minutes each day for a week. Calculate the energy used in kW h.

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Method: t = 84 min = 1.4 h.

E = 3.0 × 1.4 = 4.20 kW h
21
Core[2 marks]

A 12 V car headlamp draws 4.5 A. Calculate its power.

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Method: P = IV.

P = 54.0 W
22
Core[2 marks]

A 230 V microwave has power 1150 W. Calculate the current.

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Method: I = P/V.

I = 5.00 A
23
Core[2 marks]

A 3.0 V torch bulb has power 0.75 W. Calculate the current.

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Method: I = P/V.

I = 0.250 A
24
Core[2 marks]

A 230 V drill draws 2.0 A for 3.0 minutes. Calculate the energy transferred.

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Method: E = IVt.

E = 82800 J
25
Core[2 marks]

A 6.0 V motor draws 0.50 A for 20 s. Calculate the energy transferred.

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Method: E = IVt.

E = 60.0 J
26
Core[2 marks]

Calculate the energy in kW h used by a 500 W TV for 6.0 h.

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Method: E = Pt.

E = 3.00 kW h
27
Core[3 marks]

A 2.5 kW washing machine runs for 1.5 h, three times a week. At ₹8 per kW h, calculate the weekly cost.

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Method: E = 2.5 × 4.5.

E = 11.25 kW h; cost = ₹90.0
28
Core[3 marks]

A family uses 450 kW h in a month. At 15 cents per kW h, calculate the cost in dollars.

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Method: Multiply.

Cost = 6750 cents = $67.5
29
Core[2 marks]

State the energy transfer in an electric heater.

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Method: Electrical → thermal.

Electrical work → internal (thermal) store of the heater and surroundings
30
Core[2 marks]

Explain why a 3 kW heater costs more to run per hour than a 1 kW heater.

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Method: More energy per hour.

It transfers three times as much energy each hour
31
Core[3 marks]

A 1.8 kW kettle takes 2.5 minutes to boil water. Calculate the energy used in J and in kW h.

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Method: E = Pt.

E = 1800 × 150 = 270000 J = 0.0750 kW h
32
Core[2 marks]

A device uses 0.60 kW h in 2.0 hours. Calculate its power.

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Method: P = E/t.

P = 0.300 kW = 300 W
33
Core[2 marks]

State two units of energy used in electricity calculations.

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Method: J and kW h.

joule (J) and kilowatt-hour (kW h)
34
Core[3 marks]

A 230 V shower draws 40 A. Calculate its power in kW and the energy used in a 10-minute shower in kW h.

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Method: P = IV.

P = 9.20 kW; E = 9.2 × 1/6 = 1.53 kW h
35
Core[2 marks]

A phone charger is 5.0 V, 2.0 A. Calculate its power.

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Method: P = IV.

P = 10.0 W
36
Core[3 marks]

A 60 W bulb is left on 8 hours a day for 30 days. Electricity costs ₹6 per kW h. Calculate the cost.

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Method: E = 0.06 × 240.

E = 14.4 kW h; cost = ₹86.4
37
Core[2 marks]

Calculate the energy transferred by a 12 V, 3.0 A heater in 1.0 hour, in J.

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Method: E = IVt.

E = 129600 J
38
Core[2 marks]

A heater uses 4.0 kW h in 2.0 hours. Calculate its power in W.

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Method: P = E/t.

P = 2000 W
39
Core[3 marks]

Compare the energy cost of running a 10 W LED and a 60 W filament bulb for 1000 h at ₹7 per kW h.

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Method: E = Pt.

LED: 10 kW h = ₹70; filament: 60 kW h = ₹420; saving ₹350
40
Core[2 marks]

A battery delivers 1.5 A at 9.0 V. Calculate the time for it to deliver 8100 J.

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Method: t = E/(IV).

t = 8100 ÷ 13.5 = 600 s

Challenge questions

Harder, multi-step questions at the top end of the exam. Attempt these once you are confident with the questions above.

41
ChallengeCore[4 marks]

A 230 V kettle has power 2.2 kW. Calculate the current and the energy used in 4.0 minutes in J and kW h.

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Method: I = P/V; E = Pt.

I = 9.57 A; E = 528000 J = 0.147 kW h
42
ChallengeCore[4 marks]

A household uses: 3.0 kW heater for 2 h, 100 W TV for 5 h, and four 10 W lamps for 6 h each day. At ₹8 per kW h, calculate the daily cost.

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Method: Sum kW h.

E = 6.0 + 0.5 + 0.24 = 6.74 kW h; cost = ₹53.92
43
ChallengeCore[4 marks]

A 12 V, 48 W lamp and a 12 V, 24 W lamp are both connected to 12 V. Calculate the current and resistance of each.

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Method: I = P/V; R = V/I.

48 W: 4.00 A, 3.00 Ω; 24 W: 2.00 A, 6.00 Ω
44
ChallengeCore[4 marks]

An electric car battery stores 60 kW h. It is charged from empty by a 7.2 kW charger. Calculate the time to charge and the cost at ₹10 per kW h.

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Method: t = E/P.

t = 8.33 h; cost = ₹600
45
ChallengeCore[3 marks]

Explain why replacing a 60 W filament bulb with a 9 W LED of the same brightness saves money.

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Method: Less energy per hour.

The LED transfers far less energy per hour for the same light, so fewer kW h are bought
46
ChallengeCore[4 marks]

A 230 V immersion heater is rated 3.0 kW. Calculate the current, then the energy it transfers in 45 minutes, and the cost at 18 cents per kW h.

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Method: Steps.

I = 13.0 A; E = 2.25 kW h; cost = 40.5 cents
47
ChallengeCore[4 marks]

A 6.0 V motor draws 0.80 A and lifts a 2.0 N weight 1.5 m in 5.0 s. Calculate the electrical energy supplied and the efficiency.

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Method: E = IVt.

E = 24.0 J; useful = 3.00 J; efficiency = 12.5%
48
ChallengeCore[4 marks]

Compare the energy transferred by a 2.0 kW heater in 30 minutes with a 500 W heater in 2.0 hours.

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Method: E = Pt.

Both transfer 1.0 kW h (3.6 MJ)
49
ChallengeCore[4 marks]

A meter reads 45 216 kW h on 1 March and 45 591 kW h on 1 April. Calculate the cost at ₹7.50 per kW h and the average power used.

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Method: Units used; P = E/t.

375 kW h; ₹2812.5; P = 375 ÷ 744 h = 0.504 kW
50
ChallengeCore[3 marks]

Explain why the kilowatt-hour is a unit of energy and not of power.

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Method: kW × h.

It is power (kW) multiplied by time (h), which gives energy (1 kW h = 3.6 MJ)
All IGCSE Physics topics← 4.2.4 Resistance4.3.1 Circuit Diagrams and Circuit Components →

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