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Electrical Power, Energy and the Cost of Electricity — Answers

Answer sheet with model answers and success criteria for the Year 10 energy, light, sound & electricity worksheet.

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Success TutoringWorksheetsAnswer sheetYear 10 · Science · Energy, Light, Sound & Electricity

Electrical Power, Energy and the Cost of Electricity — Answers

Part A: Power, voltage and current

Power is the rate at which energy is transferred, measured in watts (W); 1 W = 1 joule per second. For an electrical device, P = VI (power = voltage × current).

Also, P = E ÷ t, so E = P × t (energy = power × time).

1.A kettle has a power rating of 2000 W and runs on a 230 V supply. Calculate the current it draws, to one decimal place.

Answer: 8.7 A

  1. I = P ÷ V
  2. 2000 ÷ 230 = 8.695…
  3. ≈ 8.7 A
2.A microwave oven draws 5 A from a 230 V supply. What is its power?

Answer: 1150 W

  1. P = VI = 230 × 5 = 1150 W
3.A 12 V car headlight draws a current of 0.5 A. What is its power?

Answer: 6 W

  1. P = 12 × 0.5 = 6 W
4.What does a power rating of 1 watt mean?
  1. A1 joule of energy is transferred every second
  2. B1 coulomb of charge flows every second
  3. C1 volt pushes 1 amp
  4. D1 kilowatt-hour is used every hour

Answer: 1 joule of energy is transferred every second

Power is the rate of energy transfer: 1 W = 1 J/s.

5.A phone charger on a 230 V supply draws 0.05 A. Calculate its power, and the energy it transfers in one hour in joules.

Answer: 11.5 W; 41 400 J

  1. P = VI = 230 × 0.05 = 11.5 W
  2. E = P × t = 11.5 × 3600 = 41 400 J

Part B: Joules and kilowatt-hours

The joule is a very small unit, so electricity companies measure energy in kilowatt-hours (kWh). Energy (kWh) = power (kW) × time (h). 1 kWh = 1000 W × 3600 s = 3 600 000 J.

Cost = energy used (kWh) × price per kWh.

6.A 1500 W hair dryer is used for 10 minutes. How much energy does it transfer in joules?

Answer: 900 000 J

  1. 10 min = 600 s
  2. E = P × t = 1500 × 600 = 900 000 J
7.Convert your answer to kilowatt-hours.

Answer: 0.25 kWh

  1. 1 kWh = 3 600 000 J
  2. 900 000 ÷ 3 600 000 = 0.25 kWh
  3. Check: 1.5 kW × (10 ÷ 60) h = 0.25 kWh ✓
8.A 2 kW kettle is used for 6 minutes. Electricity costs 30 cents per kWh. How much does it cost to boil the kettle?

Answer: 6 cents

  1. 6 min = 0.1 h
  2. Energy = 2 × 0.1 = 0.2 kWh
  3. Cost = 0.2 × 30 = 6 cents
9.A kilowatt-hour is a unit of power.

Answer: False

A kilowatt-hour is a unit of energy: the energy transferred by a 1 kW device in 1 hour (3 600 000 J).

Part C: Household electricity costs

10.An electricity meter read 45 672 kWh at the start of a month and 45 912 kWh at the end. At $0.30 per kWh, what is the cost for the month?

Answer: $72.00

  1. Energy used = 45 912 − 45 672 = 240 kWh
  2. Cost = 240 × 0.30 = $72.00
11.A 2.4 kW heater is used for 1.5 hours each day for 30 days. At $0.30 per kWh, what does it cost to run?

Answer: $32.40

  1. Energy per day = 2.4 × 1.5 = 3.6 kWh
  2. 30 days: 3.6 × 30 = 108 kWh
  3. Cost = 108 × 0.30 = $32.40
12.A 60 W incandescent light bulb and an 8 W LED bulb give the same light. Each is used 5 hours a day for a year (365 days). Calculate the energy each uses in kWh and the yearly saving from using the LED, at $0.30 per kWh.

Answer: Incandescent 109.5 kWh; LED 14.6 kWh; saving $28.47

  1. Incandescent: 0.060 kW × 5 × 365 = 109.5 kWh
  2. LED: 0.008 kW × 5 × 365 = 14.6 kWh
  3. Saving = 109.5 − 14.6 = 94.9 kWh × $0.30 = $28.47
13.Explain why the LED bulb uses so much less energy for the same light output.

Answer: The LED is much more efficient: it converts a larger fraction of the electrical energy into light, while the incandescent bulb wastes most of its energy as heat.

14.Which appliances usually use the most electricity in a home, and what do they have in common?

Answer: Appliances that heat or cool, such as space heaters, air conditioners, water heaters, clothes dryers and ovens. They have high power ratings and are often used for long periods.

Part D: Evaluate

15.A family wants to cut its electricity bill. Evaluate three changes they could make. For each, explain how it reduces the energy used (using E = P × t) and comment on its cost or practicality.[5 marks]

Model answer: 1. Replace old light bulbs with LEDs: this lowers the power (P) of each light, for example from 60 W to 8 W, so less energy is used for the same time. LEDs cost a little more to buy but last much longer and pay for themselves quickly. 2. Turn appliances off at the wall instead of leaving them on standby, and switch off lights in empty rooms: this reduces the time (t) devices draw power. It costs nothing, but it relies on people remembering. 3. Improve insulation and set the heater or air conditioner a couple of degrees closer to the outside temperature: the heater or cooler then runs for less time, reducing E. Insulation has an upfront cost, but heating and cooling use the most energy in many homes, so the savings are large. Installing solar panels is another option: it does not reduce use, but it generates electricity so less is bought; it is expensive to install but saves money over many years.

Success criteria:

  • Three relevant changes
  • Each linked to reducing power or time in E = P × t
  • Comments on cost or practicality for each
  • Shows evaluation (comparing benefits and drawbacks)
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