Efficiency ($\eta$) is the ratio of useful energy output to total energy input, expressed as a percentage. It quantifies how much of the energy put into a system is converted into the desired output, and how much is wasted — typically as heat.
KEY TAKEAWAY: Efficiency is always between 0% and 100% for real systems. The “missing” energy is not destroyed — it is converted to an unwanted form (usually heat) through friction, electrical resistance, or other loss mechanisms.
$$\eta = \frac{E_{output}}{E_{input}} \times 100\% = \frac{P_{output}}{P_{input}} \times 100\%$$
| Symbol | Quantity | Unit |
|---|---|---|
| $\eta$ | Efficiency | % |
| $E_{output}$ | Useful output energy | Joule (J) |
| $E_{input}$ | Total input energy | Joule (J) |
| $P_{output}$ | Useful output power | Watt (W) |
| $P_{input}$ | Total input power | Watt (W) |
Three algebraic forms (rearranging for different unknowns):
$$\eta = \frac{E_{out}}{E_{in}} \times 100\% \qquad E_{out} = \frac{\eta}{100} \times E_{in} \qquad E_{in} = \frac{E_{out}}{\eta / 100}$$
Mechanical efficiency is the ratio of useful work output to work input. Losses arise from friction at bearings, gear teeth, pulley surfaces, and pivot joints.
Efficiency from MA and VR:
$$\eta = \frac{MA}{VR} \times 100\%$$
This form was developed in the mechanical systems strand and applies to all simple machines (levers, pulleys, gear trains, inclined planes).
Worked example 1 — Gear train:
A gear train has VR = 5 (5:1 reduction). Under load, the effort force is 40 N and the load force is 160 N.
$$MA = \frac{F_{load}}{F_{effort}} = \frac{160}{40} = 4$$
$$\eta = \frac{4}{5} \times 100\% = 80\%$$
Worked example 2 — Pulley system:
A 4-rope block-and-tackle (VR = 4) is used to lift a 500 N load. The effort required is 145 N.
$$MA = \frac{500}{145} = 3.45$$
$$\eta = \frac{3.45}{4} \times 100\% = 86.2\%$$
In electrical circuits, energy losses are due to resistance in conductors and components. Useful output may be light, mechanical work, sound, or chemical energy.
Worked example 3 — LED circuit:
A 12 V supply drives a circuit. The LED produces 0.5 W of light output and 0.1 W of heat. The current-limiting resistor dissipates 0.3 W.
$$E_{input} = V \times I$$
Total power consumed = 0.5 + 0.1 + 0.3 = 0.9 W from the supply
$$\eta_{LED} = \frac{P_{light}}{P_{LED total}} = \frac{0.5}{0.6} \times 100\% = 83.3\%$$
$$\eta_{circuit} = \frac{P_{light}}{P_{supply}} = \frac{0.5}{0.9} \times 100\% = 55.6\%$$
Note: circuit efficiency includes resistor losses; LED efficiency considers only the LED itself.
VCAA FOCUS: VCAA questions often specify “system efficiency” (including all losses) vs. “component efficiency” (losses in one component). Read the question carefully to determine which boundary to use.
An electric motor converts electrical input power to mechanical output power:
$$\eta_{motor} = \frac{P_{mechanical}}{P_{electrical}} \times 100\% = \frac{\tau \omega}{VI} \times 100\%$$
Worked example 4 — DC motor:
A motor takes 3 A from a 24 V supply. It produces 0.8 N·m at 800 rpm.
$$P_{electrical} = 24 \times 3 = 72 \text{ W}$$
$$\omega = \frac{2\pi \times 800}{60} = 83.8 \text{ rad/s}$$
$$P_{mechanical} = 0.8 \times 83.8 = 67.0 \text{ W}$$
$$\eta = \frac{67.0}{72} \times 100\% = 93.1\%$$
Losses = \$72 - 67 = 5$ W (heat in motor windings and friction in bearings).
When multiple components operate in series, the overall efficiency is the product of individual efficiencies:
$$\eta_{system} = \eta_1 \times \eta_2 \times \eta_3 \times \ldots$$
Worked example 5 — Drive system:
Motor efficiency: 90%, gearbox efficiency: 85%, belt drive efficiency: 95%.
$$\eta_{system} = 0.90 \times 0.85 \times 0.95 = 0.726 = 72.6\%$$
If the motor input is 500 W:
$$P_{output} = 0.726 \times 500 = 363 \text{ W}$$
$$P_{losses} = 500 - 363 = 137 \text{ W}$$
COMMON MISTAKE: Efficiency values should be expressed as decimals (0.85) when used in calculations, and as percentages (85%) in written descriptions. Multiplying percentages directly (e.g. 90 × 85 × 95) gives a meaningless result.
| Application | Efficiency formula |
|---|---|
| General | $\eta = (E_{out}/E_{in}) \times 100\%$ |
| Mechanical (MA/VR) | $\eta = (MA/VR) \times 100\%$ |
| Motor | $\eta = (\tau\omega / VI) \times 100\%$ |
| Multi-stage | $\eta_{sys} = \eta_1 \times \eta_2 \times \ldots$ |
STUDY HINT: Every efficiency problem can be structured as: (1) identify input power/energy, (2) identify useful output power/energy, (3) apply the formula. The hardest part is correctly identifying what counts as “useful output” — read the question context carefully.
