2.11 Electrical power and energy in the home
9 exam questions worth 21 marks. Write your answer, open the mark scheme, then give yourself the marks you earned. Your answers are saved on this device.
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An electric fire works from the 230 V mains and takes a current of 5 A. Calculate its power.
Mark scheme
P = V I = 230 × 5 (1); 1150 W (1)
Model answer
P = V I = 230 × 5 = 1150 W.
A current of 3 A passes through a resistor of resistance 5 Ω. Calculate the power transferred.
Mark scheme
P = I² R = 3² × 5 (1); 45 W (1)
Model answer
P = I² R = 9 × 5 = 45 W.
A kettle has a power of 2.2 kW. Calculate the energy it transfers in 3 minutes. Give your answer in joules.
Mark scheme
power 2200 W and time 180 s (1); E = P t = 2200 × 180 (1); 396 000 J (1)
Model answer
P = 2200 W and t = 3 × 60 = 180 s. E = P t = 2200 × 180 = 396 000 J.
A 3 kW heater is switched on for 4 hours. Electricity costs 30p per kWh. Calculate the cost.
Mark scheme
energy = 3 × 4 = 12 kWh (1); cost = 12 × 30 (1); 360p / £3.60 (1)
Model answer
Energy = 3 kW × 4 h = 12 kWh. Cost = 12 × 30p = 360p = £3.60.
Explain why the wire inside an electric kettle gets hot when a current passes through it.
Mark scheme
electrons collide with the ions in the wire / the wire has resistance (1); transferring energy to the thermal store so the ions vibrate more / the wire's temperature rises (1)
Model answer
As the electrons move through the wire they collide with the metal ions. This transfers energy to the ions, so they vibrate more and the temperature of the wire rises.
A kettle is rated at 2.2 kW and works from the 230 V mains. Which fuse should be fitted in its plug?
Choose one answer.
Correct Not quite: the answer is 13 A
Mark scheme
C: 13 A (I = P / V = 2200 / 230 = 9.6 A, so a 3 A or 5 A fuse would blow)
Model answer
13 A (C). The current is I = P / V = 2200 / 230 = 9.6 A, which is more than a 3 A or 5 A fuse allows.
A 100 W television is switched on for 5.0 hours each day for 30 days. One kilowatt-hour of electricity costs 30p. Calculate the cost.
Mark scheme
energy = 0.100 kW × 150 h = 15 kWh (1); cost = 15 × 30 (1); 450p (£4.50) (1)
Model answer
Energy = 0.100 × 150 = 15 kWh. Cost = 15 × 30p = 450p, or £4.50.
A heating element has a resistance of 46 Ω and is connected to the 230 V mains. Calculate its power.
Mark scheme
P = V² ÷ R (1); 230² ÷ 46 = 52 900 ÷ 46 (1); 1150 W (1)
Model answer
P = V² / R = 230² / 46 = 52 900 / 46 = 1150 W.
Explain why electricity bills use kilowatt-hours rather than joules.
Mark scheme
the joule is a very small unit (1); a household uses millions of joules, so the numbers would be huge (1 kWh = 3 600 000 J) (1)
Model answer
A joule is a very small amount of energy, and a household uses millions of joules, so the kilowatt-hour is a more convenient unit (1 kWh = 3.6 million J).
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