Efficiency and Power

Efficiency = useful output / total input; P = E/t; reducing energy waste

# Efficiency and Power — GCSE Physics

No energy transfer is 100% efficient. Every device wastes some energy, usually as heat. Understanding efficiency tells us how much useful energy we actually get, while power tells us how quickly energy is transferred.


1. Efficiency

Efficiency measures how much of the input energy (or power) is usefully transferred.

Efficiency=useful output energytotal input energy\boxed{\text{Efficiency} = \frac{\text{useful output energy}}{\text{total input energy}}}

Or equivalently:

Efficiency=useful output powertotal input power\text{Efficiency} = \frac{\text{useful output power}}{\text{total input power}}

Efficiency can be expressed as a decimal (e.g., 0.75) or a percentage (e.g., 75%).

To convert: multiply the decimal by 100.

Key Points

  • Efficiency is always less than 1 (or less than 100%) — you can never get more useful energy out than you put in
  • Wasted energy = input energy − useful output energy
  • The wasted energy usually goes to thermal energy stores (heating surroundings)

Typical Efficiencies

Device Approximate Efficiency
LED light bulb 30–50%
Filament light bulb 5%
Electric motor 80–95%
Petrol car engine 25–30%
Electric heater ~100% (all energy becomes heat)
Solar cell 15–25%
Gas boiler 90%
Human body 25% (for mechanical work)

2. Reducing Energy Waste

To improve efficiency:

  • Lubrication — reduces friction between moving parts, less energy wasted as heat
  • Streamlining — reduces air resistance/drag
  • Insulation — reduces heat loss (e.g., in homes, engines)
  • Better design — e.g., LED bulbs instead of filament bulbs
  • Regenerative braking — converts kinetic energy back to electrical energy (electric cars)

3. Power

Power is the rate at which energy is transferred (or work is done).

P=Et\boxed{P = \frac{E}{t}}

Where:

  • PP = power (in watts, W)
  • EE = energy transferred (in joules, J)
  • tt = time (in seconds, s)

1 watt = 1 joule per second (1 W = 1 J/s)

Larger Units

  • 1 kilowatt (kW) = 1000 W
  • 1 megawatt (MW) = 1,000,000 W

Also: P=F×vP = F \times v (power = force × velocity) — useful for moving objects.

Rearranging

E=P×tt=EPE = P \times t \qquad t = \frac{E}{P}


Worked Example: Efficiency

Problem

Question: A motor uses 500 J of electrical energy and converts 375 J to kinetic energy. Calculate the efficiency.

Efficiency=375500=0.75=75%\text{Efficiency} = \frac{375}{500} = 0.75 = 75\%

Solution

Worked Example: Finding Useful Output

Problem

Question: A 60% efficient engine uses 10,000 J of fuel energy. Calculate the useful energy output.

Useful output=0.60×10,000=6000 J\text{Useful output} = 0.60 \times 10{,}000 = 6000 \text{ J}

Solution

Worked Example: Power

Problem

Question: A crane lifts a 200 kg load to a height of 15 m in 30 seconds. Calculate the power. (g=9.8g = 9.8 N/kg)

E=mgh=200×9.8×15=29,400 JE = mgh = 200 \times 9.8 \times 15 = 29{,}400 \text{ J} P=Et=29,40030=980 WP = \frac{E}{t} = \frac{29{,}400}{30} = 980 \text{ W}

Solution

Worked Example: Efficiency with Power

Problem

Question: A light bulb has an input power of 60 W and a useful light output of 9 W. Calculate the efficiency and the power wasted as heat.

Efficiency=960=0.15=15%\text{Efficiency} = \frac{9}{60} = 0.15 = 15\% Wasted power=609=51 W\text{Wasted power} = 60 - 9 = 51 \text{ W}

Solution

Worked Example: Energy Cost

Problem

Question: A 2 kW heater is used for 3 hours. Calculate the energy transferred in kWh and the cost at 30p per kWh.

E=P×t=2×3=6 kWhE = P \times t = 2 \times 3 = 6 \text{ kWh} Cost=6×30=180p=£1.80\text{Cost} = 6 \times 30 = 180\text{p} = £1.80


Solution

5. The Kilowatt-Hour (kWh)

Electricity bills use kilowatt-hours instead of joules:

Energy (kWh)=Power (kW)×Time (hours)\text{Energy (kWh)} = \text{Power (kW)} \times \text{Time (hours)}

Converting: 1 kWh = 1000×3600=3,600,0001000 \times 3600 = 3{,}600{,}000 J = 3.6 MJ

Cost=Energy (kWh)×Price per kWh\text{Cost} = \text{Energy (kWh)} \times \text{Price per kWh}


6. Practice Questions

    1. A motor has an input energy of 800 J and does 520 J of useful work. Calculate the efficiency. (2 marks)
    1. An LED bulb is 40% efficient and has an input power of 10 W. Calculate the useful light output. (2 marks)
    1. Calculate the power of an athlete who transfers 9000 J of energy in 12 seconds. (2 marks)
    1. A 3 kW oven is used for 2.5 hours at 28p/kWh. Calculate the cost. (2 marks)
    1. A coal power station has an efficiency of 35%. If it burns fuel containing 1000 MJ of energy, how much useful electrical energy is produced? How much is wasted? (3 marks)

    Answers

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Frequently Asked Questions

Can efficiency be 100%?

In theory, an electric heater is close to 100% efficient since all electrical energy becomes thermal energy. However, for devices that do mechanical work or produce light, efficiency is always less than 100%.

Is a higher wattage always better?

Not necessarily. Higher wattage means more energy per second, but if the device is less efficient, much of that energy is wasted. A 10W LED can produce as much light as a 60W filament bulb.


Summary

  • Efficiency = useful output ÷ total input (as decimal or percentage)
  • Efficiency is always < 1 (< 100%); wasted energy usually becomes heat
  • Power = energy ÷ time: P=E/tP = E/t (watts)
  • 1 kWh = energy from 1 kW for 1 hour = 3.6 MJ
  • Cost = energy (kWh) × price per kWh
  • Reduce waste: lubrication, insulation, streamlining, better technology

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