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Efficiency: efficiency = useful output power ÷ input power

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Specification: AQA A-level Physics (7408) 3.4.1.7 | Paper: 1 (and 2 as assumed knowledge) | Data and formulae booklet: given in the exam

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📌 KEY DEFINITIONS – learn them word for word

Efficiency: the ratio of the useful output power to the total input power (or useful energy output to total energy input). It has no unit and can never be greater than 1 (100%).

Useful output power: the rate at which energy is transferred to the store we want, e.g. the rate of gain of gravitational potential energy when a motor lifts a load.

Wasted power: input power − useful output power. It is usually dissipated to the surroundings as thermal energy, e.g. by friction or resistive heating.

The equation

efficiency = useful output power ÷ input power efficiency = P_out ÷ P_in (also = useful energy output ÷ total energy input)

Quantity	Symbol	Unit
efficiency	–	no unit (decimal or %)
useful output power	P_out	watts (W)
input power	P_in	watts (W)

Rearranged: P_out = efficiency × P_in | P_in = P_out ÷ efficiency | Change percentages to decimals (38% = 0.38) before you calculate.

Use FIFA for every answer

F – Formula: write the equation as it appears on the sheet.
I – Insert: put in the numbers (convert to standard units first).
F – Fix: rearrange to make the unknown the subject.
A – Answer: calculate and always give the unit.

Core: Q1–4. Exam standard: Q5–8. Stretch: Q9–10.

Questions

Q1 (Core) An electric motor has an input power of 250 W and a useful output power of 180 W. Calculate its efficiency as a decimal and as a percentage.

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F: efficiency = P_out ÷ P_in
I: efficiency = 180 ÷ 250
F: efficiency is already the subject
A: 0.72 (no unit), which is 72%

Q2 (Core) A filament lamp is supplied with 3600 J of electrical energy and emits 180 J of energy as visible light. Calculate its efficiency as a percentage.

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F: efficiency = useful energy output ÷ total energy input (the same ratio as the powers, because both energies are for the same time)
I: efficiency = 180 ÷ 3600
F: efficiency is already the subject
A: 0.050 = 5.0%

Q3 (Core) A gas-fired power station has an input power of 1.8 GW and an efficiency of 38%. Calculate its useful electrical output power in MW, to 2 s.f.

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Convert first: 1.8 GW = 1.8 × 10⁹ W; 38% = 0.38
F: efficiency = P_out ÷ P_in
I: 0.38 = P_out ÷ (1.8 × 10⁹)
F: P_out = 0.38 × 1.8 × 10⁹
A: 6.84 × 10⁸ W = 680 MW (2 s.f.)
Watch out: using 38 instead of 0.38 gives an output 38 times bigger than the input. An efficiency can never be more than 1.

Q4 (Core) A solar panel has an efficiency of 0.18. In 10 minutes it delivers 27 kJ of useful electrical energy. Calculate the power of the sunlight falling on the panel.

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Convert first: 27 kJ = 27 000 J; 10 minutes = 600 s
Step 1 – useful output power: P = E ÷ t = 27 000 ÷ 600 = 45 W
Step 2 – input power
F: efficiency = P_out ÷ P_in
I: 0.18 = 45 ÷ P_in
F: P_in = 45 ÷ 0.18
A: 250 W

Q5 (Exam) An electric motor lifts a 120 kg crate vertically at a constant speed of 0.35 m s⁻¹. The input power to the motor is 750 W. Calculate the efficiency of the motor, to 2 s.f.

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Step 1 – useful output power: at constant speed the lifting force equals the weight, so P_out = Fv = mgv = 120 × 9.81 × 0.35 = 412.02 W
Step 2 – efficiency
F: efficiency = P_out ÷ P_in
I: efficiency = 412.02 ÷ 750
F: efficiency is already the subject
A: 0.5493… = 0.55 (2 s.f.), or 55%
Watch out: the force in P = Fv is the weight mg = 1177.2 N, not the mass of 120 kg.

Q6 (Exam) A crane motor with an efficiency of 0.62 lifts a 450 kg load vertically through 18 m in 25 s. Calculate the input power to the motor, in kW to 2 s.f.

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Step 1 – useful energy: ΔEₚ = mgΔh = 450 × 9.81 × 18 = 79 461 J
Step 2 – useful power: P_out = 79 461 ÷ 25 = 3178.44 W
Step 3 – input power
F: efficiency = P_out ÷ P_in
I: 0.62 = 3178.44 ÷ P_in
F: P_in = 3178.44 ÷ 0.62
A: 5126.5… W = 5.1 kW (2 s.f.)

Q7 (Exam) An electric car motor has an input power of 45 kW and an efficiency of 91%. Calculate the energy wasted in 15 minutes of steady driving, to 2 s.f.

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Convert first: 45 kW = 45 000 W; 91% = 0.91; 15 minutes = 900 s
Step 1 – useful power: P_out = 0.91 × 45 000 = 40 950 W
Step 2 – wasted power: 45 000 − 40 950 = 4050 W
Step 3 – wasted energy
F: E = Pt
I: E = 4050 × 900
F: E is already the subject
A: 3 645 000 J = 3.6 × 10⁶ J (2 s.f.), mostly dissipated as thermal energy in the motor and its surroundings

Q8 (Exam) A car travels at a constant 31 m s⁻¹ along a level road. The total resistive force on it is 650 N. The engine has an efficiency of 0.28. Calculate the rate at which energy is supplied by the fuel, to 2 s.f.

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Step 1 – useful output power: at constant velocity the resultant force is zero, so driving force = resistive force = 650 N and P_out = Fv = 650 × 31 = 20 150 W
Step 2 – input power
F: efficiency = P_out ÷ P_in
I: 0.28 = 20 150 ÷ P_in
F: P_in = 20 150 ÷ 0.28
A: 71 964… W = 7.2 × 10⁴ W (2 s.f.), or 72 kW
Watch out: constant velocity means zero resultant force, not zero driving force.

Q9 (Stretch) In a hydroelectric power station, water falls through a height of 85 m at a rate of 1200 kg s⁻¹. The turbine has an efficiency of 0.90, the generator 0.95 and the transformer 0.98. Calculate the overall efficiency to 3 s.f. and the electrical output power to 2 s.f. (Assume all the gravitational potential energy lost by the water is the input to the turbine.)

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Step 1 – overall efficiency: multiply the efficiencies: 0.90 × 0.95 × 0.98 = 0.8379 = 0.838 (3 s.f.)
Step 2 – input power: rate of loss of GPE = (m ÷ t) × g × h = 1200 × 9.81 × 85 = 1 000 620 W
Step 3 – output power
F: efficiency = P_out ÷ P_in
I: 0.8379 = P_out ÷ 1 000 620
F: P_out = 0.8379 × 1 000 620
A: 838 419… W = 8.4 × 10⁵ W (2 s.f.)
Watch out: for devices in a chain, multiply the efficiencies. Never add or average them.

Q10 (Stretch) An electric winch has a motor of efficiency 0.80 driving a gearbox of efficiency 0.75. The electrical input power is 2.0 kW. The winch raises a 150 kg load vertically at constant speed. Calculate the speed of the load and the time taken to raise it through 12 m, both to 2 s.f.

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Convert first: 2.0 kW = 2000 W
Step 1 – overall efficiency: 0.80 × 0.75 = 0.60
Step 2 – useful power: P_out = 0.60 × 2000 = 1200 W
Step 3 – speed
F: P = Fv = mgv
I: 1200 = 150 × 9.81 × v
F: v = 1200 ÷ 1471.5
A: 0.8154… m s⁻¹ = 0.82 m s⁻¹ (2 s.f.)
Step 4 – time: t = Δh ÷ v = 12 ÷ 0.8154… = 14.7… s = 15 s (2 s.f.)
Watch out: the useful output is the power after the gearbox. Using the motor’s output alone (1600 W) overestimates the speed.

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