Efficiency and useful output

By UK GCSE Physics · independent study · myAudience

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Efficiency and useful output

Efficiency compares useful output with total input. It can be calculated as useful energy/total energy, or useful power/total power when measured on a consistent basis. Multiply the fraction by 100 for a percentage. Define what useful means for the actual task before choosing the numerator.

An efficiency of 0.80 means 80% of input reaches the chosen useful output. The remaining 20% is not destroyed; it is transferred in other ways. A measured efficiency above 100% usually indicates inconsistent accounting, units or measurements, not a device creating energy.

Higher-tier explanations of improvement connect a change with the intended transfer: suitable lubrication reduces frictional heating in machinery, while insulation reduces unwanted heating of surroundings. A change can have costs or trade-offs, so more efficient does not automatically mean more economical or more suitable for every job.

Worked example: A device receives 500 J and provides 350 J useful output. Calculate efficiency and other transfers.

Efficiency=350/500=0.70.

Percentage=0.70×100=70%.

Other transfers=500−350=150 J.

Answer: 70% efficient; 150 J transferred in other ways

Check your understanding: Can efficiency be calculated with 350 J divided by 500 W?

Answer: No.

Why: Input and output must be comparable quantities; mixing energy and power gives the wrong dimensions.