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CAMBRIDGE INTERNATIONAL IGCSE™ PHYSICS 0625 · EXAMINATIONS 2026–2028

Efficiency

Find out where the energy goes, compare useful shares fairly, and practise calculations with clear reasons for every step.

Core: qualitative efficiency, useful and unwanted transfers. Supplement / Extended: efficiency equations, percentage calculations and Sankey interpretation. This is Cambridge IGCSE, not Egyptian Baccalaureate / Thanaweya Amma or Year 3.

Core 1.7.3.3 · Supplement 1.7.3.7 · Recap: energy conservation and Sankey diagrams (1.7.1) · PDF pp.126–131 / printed pp.118–123.

STEP 1 OF 6 · Core

First ask: what job is the device doing?

Useful energy is the part of the input that achieves the intended purpose. For a fan, moving air is useful; unwanted sound and motor heating are not. For a room heater, heating the room is the purpose. Thermal energy is not automatically wasted.

Worked example

A reading lamp is intended to illuminate a page. Light reaching the page is useful for that purpose. Heating the lamp and air is generally unwanted. The same warming of air can be useful when the intended task is heating a room.

Think first: Does a warm device always have poor efficiency?

No. You must define its job and identify all inputs and outputs. Warmth alone is not a measurement of the useful share.

SUPPLEMENT / EXTENDED

A reliable efficiency method

  1. Name the useful job and identify every input. Match the same interval and system.
  2. Choose energy/energy or power/power. Put numerator and denominator in matching units.
  3. Useful output ÷ total input × 100% gives percentage efficiency.
  4. For useful output, multiply input by efficiency/100. For input, divide useful output by that fraction.
  5. Check conservation and reasonableness. If stored energy changes, account for it rather than calling every remainder an output.
Worked example: lifting work to efficiency

A hoist lifts 400 N through 3.0 m steadily in 8.0 s while receiving 250 W electrically. Useful work = 400 × 3.0 = 1200 J. Input energy = 250 × 8.0 = 2000 J. Efficiency = 1200/2000 × 100% = 60%. Equivalently, useful power = 1200/8.0 = 150 W, so 150/250 × 100% = 60%. With no net stored-energy change, unwanted energy = 800 J. Do not divide 1200 J directly by 250 W to calculate efficiency: that gives seconds.

CORE VISUAL COMPARISON · SUPPLEMENT NUMERICAL PREDICTION

Change the input and useful share

Predict the useful share

Model motor: choose input energy and the useful share in each 100 parts. The diagram uses equal-length bars: widths show the shares, not joules across different input settings. No net energy remains stored in the motor. The 0% and 100% settings are limiting cases, not claims about a real motor.

All input: 200 JPurple: useful 70 partsGold: unwanted 30 parts
A share bar, not a Sankey diagram. Numerical useful-output prediction is Supplement / Extended.

Investigate: keep the useful share fixed and double input. Both useful and unwanted joules double, but efficiency stays the same. Next keep input fixed and raise the useful share: useful output rises and waste falls. Changing a control clears your previous prediction.

12 ORIGINAL QUESTIONS · CORE + SUPPLEMENT · MIXED FORMATS

Practise, then explain

Four multiple-choice questions, four numerical responses and four written tasks. Each question is labelled by tier. Percentage efficiency calculations are Supplement / Extended; Core includes qualitative comparisons and simple energy conservation. Numerical answers are checked automatically; written answers use a self-assessment guide. The suggested marks are for this original practice, not an official Cambridge mark scheme.

Practice stays only while this page is open. Refreshing or leaving clears these responses. Copy useful written work into your notes before leaving.

Auto-checked correct: 0 / 8

Question 1 · Useful depends on purpose · Core · 1 mark · Multiple choice

Which transfer is useful for an electric fan whose purpose is to move air?

Optional hint

Start with the job the fan is meant to do.

Source clarifications

The original PDF is unchanged. Its statement that every real device is always below 100% is too absolute without defining the useful purpose: electrical room heating can be approximately 100% efficient on a room-heating basis. For real motors and lamps, unwanted transfers keep useful efficiency below 100%. Fully accounted useful output cannot exceed total input; check omitted inputs, stored-energy changes or measurement errors if it appears to do so.

The source also says waste energy “cannot be recovered”. Some waste heat can be recovered for another job; dispersed energy is generally less useful, not destroyed. Our lamp comparison uses the source’s given light powers, not a claim that all lamps have those efficiencies. Figure 18.1 is interpreted by arrow width, not arrow length. The simple model used here explicitly assumes no net stored-energy increase.

Source alignment: supplied Physics-0625 Volume A, PDF pp.126–131 / printed pp.118–123, Figure18.1; official Cambridge 2026–2028 syllabus, section1.7.3. Original explanations, activity and questions. Original source-download access is unchanged.

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