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Efficiency: useful output and total input

Key idea: Efficiency is the useful energy output divided by the total energy input. The rest is dissipated: it still exists, but it does not do the intended job.

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Real devices often transfer some input energy to outputs other than the job you want. A motor warms up, a lamp heats the room, and a car engine sends much of its fuel’s energy out through the exhaust and radiator. Efficiency measures what fraction of the input does the intended job, so you can compare devices and see where energy is wasted.

What efficiency measures

Efficiency is the useful energy output divided by the total energy input:

efficiency = (useful energy output)/(total energy input) × 100%

The job the device is meant to do decides which output is useful. For a lifting motor, the useful output is the gain in the load’s gravitational potential energy. For a lamp, it is the energy carried away by light.

Use input and output values for the same time interval. If the device runs steadily, the power ratio gives the same answer, because each power is an energy divided by the same time:

efficiency = (useful power output)/(total power input) × 100%

Efficiency is a ratio of two energies or two powers, so it has no unit. Never divide an energy by a power.

Account for the whole input

Energy is conserved, so every joule of input goes somewhere. If the device does not build up stored energy:

total energy input = useful energy output + energy dissipated

A Sankey diagram shows this split. The width of each arrow is proportional to the energy it carries.

Sankey diagram for a sixty-percent efficient deviceA 100 joule input arrow splits into a 60 joule useful output arrow and a narrower 40 joule arrow for energy dissipated to the surroundings. The labelled widths are proportional to the energy transferred.Energy flow through a deviceArrow width represents the amount of energy transferred.100 J inputelectrical input60 J usefuloutput40 J dissipatedEfficiency = 60 J ÷ 100 J × 100% = 60%
Sankey diagram for a sixty-percent efficient deviceA 100 joule input arrow splits into a 60 joule useful output arrow and a narrower 40 joule arrow for energy dissipated to the surroundings. The labelled widths are proportional to the energy transferred.Energy flow through a deviceArrow width represents the amount of energy transferred.100 J inputelectrical input60 J usefuloutput40 J dissipatedEfficiency = 60 J ÷ 100 J × 100% = 60%
The useful and dissipated arrow widths are proportional to their energy shares. Here, 60 J useful from 100 J input gives an efficiency of 60%.

If the device warms up, some input remains in its internal energy store. Include that gain separately: input = useful output + energy transferred to surroundings + gain in stored energy.

Useful output is a fraction of total input, so efficiency lies between 0% and 100%. An answer above 100% often means you reversed the ratio or compared values over different time intervals.

Worked example 1

The useful share of an input

Problem

In the diagram, a device receives 100 J and transfers 60 J as useful output during the same interval. It stores no additional energy. Find its efficiency and the fraction of input dissipated.

Show full solution

  1. Calculate the useful fraction

    Method

    Divide useful output by total input, then express the fraction as a percentage.

    Reason

    Efficiency measures how much of the whole input serves the intended task.

    Working

    Answer

    efficiency = (60 J)/(100 J) × 100% = 60%

  2. Account for the rest

    Method

    Subtract the useful output from the input.

    Reason

    With no net storage, all the input is accounted for by useful output and dissipation.

    Working

    Answer

    E_dissipated = 100-60 = 40 J; 40/100 = 0.40

  3. State both shares

    Working

    Answer

    The efficiency is 60% and 40% of the input is dissipated.

Check your understanding 1 Account for a warming motor

A motor receives 200 J. It transfers 140 J to a load and stores 20 J as it warms; the rest heats the surroundings. Find its efficiency and the energy transferred to the surroundings.

Unit: %
Unit: J
Show answer

Efficiency = 140/200 × 100% = 70%. Energy to surroundings = 200-140-20 = 40 J. The motor retains 20 J during this interval; the full 60 J of non-useful input has not all left it.

Exam-style question 1

Energy dissipated by a lifting motor

4 marks

Examination question

An electric motor lifts a 10 N load through 5.0 m at constant speed. Assume no net change in energy stored in the motor. The electrical energy input is 65 J. Calculate the energy dissipated by the motor and its efficiency. [4 marks]

Show useful energy, dissipation and efficiency

Show the steps Optional working
Unit: J
Unit: J
Unit: %
Show solution step by step

  1. Calculate useful output energy

    1 mark

    Method

    Find the work done lifting the load.

    Reason

    The load’s gain in GPE is the useful output.

    Working

    Answer

    E_useful = Fh = (10)(5.0) = 50 J

  2. Find dissipated energy

    1 mark

    Reason

    Input energy equals useful output plus dissipated energy.

    Working

    Answer

    E_dissipated = 65-50 = 15 J

  3. Form the efficiency ratio

    1 mark

    Method

    Divide useful output by total input.

    Reason

    Efficiency measures the useful fraction of the input.

    Working

    Answer

    efficiency = 50/65 × 100%

  4. State the efficiency

    1 mark

    Reason

    The result must be below 100% because some energy is dissipated.

    Working

    Answer

    efficiency = 76.9% ≈ 77%

Check your understanding 2 Match energy with energy

A motor takes 500 W for 20 s. It lifts a 400 N load through 6.0 m at constant speed. Find the input energy, useful output energy and efficiency.

Unit: J
Unit: J
Show the steps Optional working
Unit: J
Unit: J
Unit: %
Show answer

Input energy = 500 × 20 = 10 000 J. Useful energy = 400 × 6.0 = 2400 J. Efficiency = 2400/10 000 × 100% = 24%.

Alternatively, useful power = 2400/20 = 120 W, so efficiency = 120/500 × 100% = 24%. Dividing an energy directly by a power gives a time, rather than an efficiency.

Work backwards from an efficiency

Write a percentage as a decimal before you use it as a multiplier: 20% = 0.20. Then:

  • to find the useful output, multiply the input by the efficiency: E_useful = 0.20 Eᵢₙₚᵤₜ;
  • to find the input, divide the useful output by the efficiency: Eᵢₙₚᵤₜ = E_useful÷0.20.

Check the size of your answer. For the no-storage-change account, input is at least as large as useful output. At 100% efficiency they are equal; for a positive efficiency below 100%, dividing useful output by the efficiency gives a larger input. If your input comes out smaller than the useful output, you multiplied when you should have divided.

Spot the mistake 2

Useful energy from input power

About 6 min

Learner response

An electric motor has constant input power 1.0 kW. In half an hour, 60% of the input energy is dissipated. A student states that the efficiency is therefore 60%. Assume no net change in energy stored in the motor and that all remaining output is useful. Find the useful energy transferred and the actual efficiency.

Separate useful and dissipated fractions

Show the steps Optional working
Unit: W
Unit: s
Unit: J
Unit: J
Unit: %
Show solution step by step

  1. Find the useful fraction

    Method

    Subtract the dissipated share from 100%.

    Reason

    Every joule of input is either useful or dissipated, and efficiency is the useful share.

    Working

    Answer

    efficiency = 100%-60% = 40%

  2. Convert power and time

    Method

    Use watts and seconds.

    Reason

    E = Pt gives joules only when power is in watts and time in seconds.

    Working

    Answer

    1.0 kW = 1000 W, 0.5 h = 30 × 60 = 1800 s

  3. Find the total input energy

    Reason

    Power is the rate of input energy transfer.

    Working

    Answer

    Eᵢₙ = Pt = (1000)(1800) = 1.8 × 10⁶ J

  4. Find the useful energy

    Working

    Answer

    E_useful = 0.40(1.8 × 10⁶) = 7.2 × 10⁵ J

Try it yourself 3

Electrical power for a crane

Minimal support

Problem

A crane raises a weight of 200 N through a vertical height of 8.0 m in 4.0 s at constant speed. Its conversion efficiency is 20%, with no net change in energy stored in the crane. Find the useful output power and the electrical input power.

Work from useful output back to total input

Show the steps Optional working
Unit: J
Unit: W
Show the steps Optional working
Unit: W
Unit: W

Hints

Hint 1

find useful output power

How can you find the useful work and its rate of transfer?
Hint 2

reverse the efficiency ratio

Write 0.20 = Pₒᵤₜ/Pᵢₙ.
Show solution step by step

  1. Find useful output energy

    Method

    Calculate the work done on the load.

    Reason

    The vertical lift is the crane’s useful transfer.

    Working

    Answer

    E_useful = Fh = (200)(8.0) = 1600 J

  2. Find useful output power

    Reason

    Power is energy transferred per unit time.

    Working

    Answer

    Pₒᵤₜ = 1600/4.0 = 400 W

  3. Infer the input power

    Method

    Rearrange the efficiency equation for input power.

    Reason

    The 400 W useful output is only 20% of the electrical input.

    Working

    Answer

    Pᵢₙ = 400/0.20 = 2000 W = 2.0 kW

Common mistakes

  • Giving the dissipated share as the efficiency.
  • Dividing an energy by a power, or comparing an input and an output measured over different times.
  • Multiplying by the efficiency when the input is the unknown. For the no-storage-change account, input must be at least as large as useful output.
  • Multiplying by 20 instead of 0.20 when the efficiency is 20%.
  • Reversing the ratio and obtaining an efficiency above 100%.
  • Confusing power with efficiency. A powerful machine transfers energy quickly, but it can still waste a large fraction of it.

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