A.3.13—Efficiency

Syllabus
First assessment 2025
Objective
Level
SL

Calculate Efficiency

Useful fraction

Efficiency is the ratio of useful output to total input:

η=EusefulEinput=PusefulPinput\eta=\frac{E_{useful}}{E_{input}}=\frac{P_{useful}}{P_{input}}

Choose matching quantities

Use energy ratios for the same process and time interval, or power ratios when input and output are rates. Efficiency is dimensionless and is often reported as a percentage.

Worked example from local Question Bank row 29709

Solar intensity is 240Wm2240\,\mathrm{W\,m^{-2}} over 2.50×104m22.50\times10^4\,\mathrm{m^2}, so input power is

Pin=(240)(2.50×104)=6.0×106W=6.0MWP_{in}=(240)(2.50\times10^4)=6.0\times10^6\,\mathrm{W}=6.0\,\mathrm{MW}

For a useful output of 1.6MW1.6\,\mathrm{MW},

η=1.66.0=0.27=27%\eta=\frac{1.6}{6.0}=0.27=27\%

The remaining input is transferred through non-useful pathways.

Common trap

Do not invert the ratio or use the total output, including unwanted transfers, as the useful output.

A.3.13 Exam Analysis

Assessment in practice

1–3 marks
How it is assessed

The evidence asks for motor input power from output power and efficiency, and for fuel mass or energy from a vehicle’s kinetic-energy gain and efficiency.

Command terms

Calculate / Identify

What earns marks

Use the useful-output/input ratio and convert the final fraction to a percentage when required. For a motor, calculate useful mechanical output first, then divide by electrical input power.

Watch for

Using the loss power as useful output or reporting 75 rather than 0.75 when using the ratio.

Representative question

Question 1

[Maximum number: 1]

An electric motor of efficiency 75 % raises a mass of 120 kg at a constant speed of 0.50 ms10.50 \mathrm{~ms}^{-1}. What is the power input to the motor?

A

20 W

B

450 W

C

600 W

D

800 W