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5.5 Appliances and Efficiency

GCSE Physics · AQA · Chapter 5: Electricity in the Home

A free GCSE Physics lesson for AQA (8463) from OBA School: energy supplied to appliances, efficiency, output power, wasted energy and the efficiency of the National Grid. Watch the video, read the explanation and worked examples, then try the revision questions. You can work through the full lesson, with every video, practice quizzes and printable worksheets with mark schemes, in our free online course.

What you will learn

  • Calculate the energy supplied to an appliance using E = P t, with P = V I
  • Calculate the efficiency of an appliance from its input and output power (or energy), as a decimal or a percentage
  • Use the efficiency to calculate the useful output power and the wasted power
  • Explain why appliances waste energy and use data to compare appliances that do the same job
  • Explain why transmitting electricity at a high pd reduces energy losses in the National Grid

Key words

  • Input power: The total power supplied to an appliance from the mains or a battery.
  • Output power: The useful energy transferred by the appliance each second.
  • Efficiency: The fraction of the energy (or power) supplied that is usefully transferred. It has no unit.
  • Wasted energy: Energy that is not transferred usefully, usually because it heats the surroundings.
  • Dissipated: Spread out to the surroundings, so it is less useful.
  • Sankey diagram: A flow diagram in which the width of each arrow is proportional to the energy or power it shows.

Energy supplied to an appliance

The energy supplied to an appliance depends on its power and how long it is on. A 2 kW heater used for 1 hour is supplied with the same energy as a 1 kW heater used for 2 hours, or a 500 W heater used for 4 hours.

energy transferred, E (J) = power, P (W) × time, t (s) rearranged: P = E ÷ t

power, P (W) = potential difference, V (V) × current, I (A)

Remember: 1 kW = 1000 W and 1 minute = 60 s.

Worked example: energy used by a heater

A 1.5 kW fan heater is used for 20 minutes.
P = 1500 W; t = 20 × 60 = 1200 s
E = P t = 1500 × 1200 = 1 800 000 J

Answer: 1.8 × 10⁶ J (1.8 MJ)

Efficiency of an appliance

No appliance transfers all the energy it is supplied with into the useful store. The efficiency tells you what fraction is useful:

efficiency = useful power output ÷ total power input (× 100% for a percentage)

or efficiency = useful energy output ÷ total energy input

Rearranged: useful output = efficiency × input wasted power = input − useful output

For the drill: efficiency = 420 W ÷ 600 W = 0.70, or 70%. The other 30% is wasted.

Worked example: output power of a food processor

A food processor takes 6.0 A from the 230 V mains. Its efficiency is 55%.
input power = V I = 230 × 6.0 = 1380 W
efficiency as a decimal = 55 ÷ 100 = 0.55
useful output power = 0.55 × 1380 = 759 W
wasted power = 1380 − 759 = 621 W

Answer: useful output 759 W; 621 W wasted

Exam tip: Efficiency can never be greater than 1 (100%). If your answer is bigger, you have probably divided the wrong way round. Change a percentage to a decimal (divide by 100) before multiplying.

Where does the wasted energy go?

  • Resistance heating: the current heats the wires and components inside the appliance because they have resistance. This energy is dissipated to the surroundings.
  • Friction: appliances with moving parts, such as motors, fans and drills, waste energy because friction between moving parts heats them up.
  • Sound: vibrating parts make noise, which also spreads energy into the surroundings.

Ways to reduce waste include using wires with lower resistance, lubricating moving parts, and choosing more efficient designs such as LEDs instead of filament lamps.

Comparing appliances

To compare appliances that do the same job, compare how much input power each needs for the same useful output. The chart compares four lamps that each give out 3 W of light.

The LED lamp needs only 7.5 W for the same light as the 60 W filament lamp, so it wastes far less energy and costs much less to run. New appliances such as fridges and washing machines carry an energy rating label to help buyers compare them. When choosing, also think about the purchase price, how long the appliance lasts and how often it is used.

Worked example: energy wasted in one hour

filament lamp: wasted power = 60 − 3 = 57 W, so wasted energy = 57 × 3600 = 205 200 J
LED lamp: wasted power = 7.5 − 3 = 4.5 W, so wasted energy = 4.5 × 3600 = 16 200 J

Answer: the filament lamp wastes about 13 times as much energy

Efficiency of the National Grid

The cables of the National Grid have resistance, so some power is always wasted heating them: Plost = I² R. For a given power, sending it at a higher pd needs a smaller current (I = P ÷ V). Because the loss depends on I², cutting the current makes a very big difference.

Worked example: power lost in the cables

At 400 kV: I = 100 000 000 ÷ 400 000 = 250 A
power lost = I² R = 250² × 2.0 = 125 000 W = 125 kW (only 0.125% of the power sent)
At 25 kV: I = 4000 A, so power lost = 4000² × 2.0 = 32 000 000 W = 32 MW (32%)

Answer: a high pd makes the grid far more efficient

(Higher tier, physics only) For a transformer that is 100% efficient, the power in equals the power out: Vp × Ip = Vs × Is. Stepping the pd up from 25 kV to 400 kV (16 times bigger) makes the current 16 times smaller: 4000 A in the primary gives 250 A in the secondary.

Common mistakes

  • Giving an efficiency greater than 1 or 100%. Check you have divided useful output by total input.
  • Multiplying by the percentage itself (e.g. × 60) instead of the decimal (× 0.60).
  • Forgetting to change kW to W or minutes to s in E = P t.
  • Saying a step-up transformer increases the power sent. The power stays (almost) the same; the pd goes up and the current goes down.
  • Saying wasted energy is “destroyed”. It is dissipated to the surroundings, mostly by heating.

Revision questions

Try these, then check your answers below each one.

Question 1. A 1.8 kW hairdryer is used for 5.0 minutes. Calculate the energy supplied to it, in joules.

Answer: 540000 J

Working: E = P t = 1800 W × 300 s = 540 000 J.

Question 2. A motor has an input power of 500 W and a useful output power of 350 W. Calculate its efficiency as a percentage.

Answer: 70 %

Working: 350 ÷ 500 = 0.70 = 70%.

Question 3. Four lamps all give the same light output of 4 W. Which one is the most efficient? (Choose from: Lamp needing 80 W input, Lamp needing 40 W input, Lamp needing 16 W input, Lamp needing 5 W input.)

Answer: Lamp needing 5 W input

Working: Efficiency = 4 ÷ input. The 5 W lamp: 4 ÷ 5 = 0.80 (80%), the highest of the four.

Related lessons

Continue in the full course

This lesson is section 5.5 (AGCPH0505) of our free GCSE Physics course for AQA. In the full course you get every video for this section, a self-marking practice quiz, a printable worksheet with a mark scheme, chapter tests, and your progress saved. Go to the full GCSE Physics course.

 

 

 

 

 

 

 

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