Physics 9702 · Paper 2

Efficiency

No real machine turns all of its input energy into useful output. Some always leaks away, usually as heat.
measures the useful fraction.

The definition

\text{efficiency} = \dfrac{\text{useful output energy}}{\text{total input energy}} \times 100\%

Symbols

$\text{useful output}$ = the energy (or power) doing the wanted job (J or W)
$\text{total input}$ = all the energy (or power) supplied (J or W)

The same ratio works with powers instead of energies, over the same time interval.
Efficiency is a ratio of two energies, so it has no unit (9702/11/M/J/24 Q18). It is written as a percentage or a decimal, and for a real device it is always below 100%.
The wrong options to avoid: useful output over wasted energy, or the ratio upside down (9702/12/M/J/25 Q20).

A Sankey diagram for a car engine. The widths are drawn to scale: 100% in, 80% lost as heat, 20% useful. The total width never changes, because energy is conserved.

Worked example

Smallest case: one machine

A machine takes in 500 J and gives 180 J of useful output. Find its efficiency.

$\text{efficiency} = \dfrac{180}{500} \times 100\% = 36\%$ .

Answer

36\%

Worked example

Exam version: the motor from lesson 5.07

The lifting motor has input power 900 W and useful output power 600 W. Find its efficiency. (9702/21/O/N/24 Q3(c)(iii))

$\text{efficiency} = \dfrac{600}{900} = 0.67 = 67\%$ .

Answer

67\%

Using it in both directions

Given the input, multiply by the efficiency to get the useful output: a 1.7 kW motor at 53% gives $0.53 \times 1.7 = 0.90 \text{ kW}$ (9702/13/M/J/25 Q17).
Given the useful output, divide by the efficiency to get the input, which must come out larger.

Worked example

Chain: efficiency joins the electrical circuit

An electric car motor delivers a useful output of 6.1 × 10⁴ W. The p.d. across it is 600 V and its efficiency is 85%. Find the current it draws. (9702/21/M/J/25 Q3(b)(iii))

Electrical input power: $P_{\text{in}} = VI = 600I$ (chapter 9 formula, first used here).
$0.85 = \dfrac{6.1 \times 10^{4}}{600I}$ , so $I = \dfrac{6.1 \times 10^{4}}{0.85 \times 600} = 120 \text{ A}$ .

Answer

I = 120 \text{ A}

Your turn— tap to reveal the worked answer (9702/22/M/J/24 Q1(b)(iii)-(iv))

Sunlight delivers 750 W to a solar panel, which produces 160 W of useful electrical output. Find the percentage efficiency. Then: a smaller panel gives the same output in the same sunlight. Is its efficiency greater, less, or the same? (9702/22/M/J/24 Q1(b)(iii)-(iv))

$\dfrac{160}{750} \times 100\% = 21\%$ .
A smaller area collects less input power. Same useful output ÷ smaller input = greater efficiency.

Answer: 21%; greater.

Common mistake

When the useful output and the efficiency are given, students multiply and get an input smaller than the output. That is impossible for a real machine: divide by the efficiency instead, and check the input comes out larger.

PreviousPower & Speed: P = Fv NextHooke’s Law & the Spring Constant

Efficiency

No real machine turns all of its input energy into useful output. Some always leaks away, usually as heat.
measures the useful fraction.

The definition

\text{efficiency} = \dfrac{\text{useful output energy}}{\text{total input energy}} \times 100\%

Symbols

$\text{useful output}$ = the energy (or power) doing the wanted job (J or W)
$\text{total input}$ = all the energy (or power) supplied (J or W)

The same ratio works with powers instead of energies, over the same time interval.
Efficiency is a ratio of two energies, so it has no unit (9702/11/M/J/24 Q18). It is written as a percentage or a decimal, and for a real device it is always below 100%.
The wrong options to avoid: useful output over wasted energy, or the ratio upside down (9702/12/M/J/25 Q20).

A Sankey diagram for a car engine. The widths are drawn to scale: 100% in, 80% lost as heat, 20% useful. The total width never changes, because energy is conserved.

Worked example

Smallest case: one machine

A machine takes in 500 J and gives 180 J of useful output. Find its efficiency.

$\text{efficiency} = \dfrac{180}{500} \times 100\% = 36\%$ .

Answer

36\%

Worked example

Exam version: the motor from lesson 5.07

The lifting motor has input power 900 W and useful output power 600 W. Find its efficiency. (9702/21/O/N/24 Q3(c)(iii))

$\text{efficiency} = \dfrac{600}{900} = 0.67 = 67\%$ .

Answer

67\%

Using it in both directions

Given the input, multiply by the efficiency to get the useful output: a 1.7 kW motor at 53% gives $0.53 \times 1.7 = 0.90 \text{ kW}$ (9702/13/M/J/25 Q17).
Given the useful output, divide by the efficiency to get the input, which must come out larger.

Worked example

Chain: efficiency joins the electrical circuit

An electric car motor delivers a useful output of 6.1 × 10⁴ W. The p.d. across it is 600 V and its efficiency is 85%. Find the current it draws. (9702/21/M/J/25 Q3(b)(iii))

Electrical input power: $P_{\text{in}} = VI = 600I$ (chapter 9 formula, first used here).
$0.85 = \dfrac{6.1 \times 10^{4}}{600I}$ , so $I = \dfrac{6.1 \times 10^{4}}{0.85 \times 600} = 120 \text{ A}$ .

Answer

I = 120 \text{ A}

Your turn— tap to reveal the worked answer (9702/22/M/J/24 Q1(b)(iii)-(iv))

$\dfrac{160}{750} \times 100\% = 21\%$ .
A smaller area collects less input power. Same useful output ÷ smaller input = greater efficiency.

Answer: 21%; greater.

Common mistake