Course: Combined Science + Separate Physics | Tier: Foundation + Higher | Equation sheet: given in the exam
The equation
efficiency = useful power output ÷ total power input
| Quantity | Unit |
|---|---|
| efficiency | no unit (a decimal, or × 100 for %) |
| useful power output | watts  |
| total power input | watts |
Rearranged: useful output = efficiency × total input | total input = useful output ÷ efficiency | wasted = input − useful
Use FIFA for every answer
- F – Formula: write the equation as it appears on the sheet.
- I – Insert: put in the numbers (use the same units top and bottom; turn % into a decimal).
- F – Fix: rearrange to make the unknown the subject.
- A – Answer: calculate and give the unit (efficiency has none).
Foundation: aim for Q1–8. Higher: aim for Q4–10. Most questions need rearranging.
Questions
Q1 (F) A drill has a total power input of 200 W and a useful power output of 150 W. Calculate its efficiency.
Show solution
F: efficiency = useful power output ÷ total power input
I: efficiency = 150 ÷ 200
F: efficiency is already the subject
A: 0.75 (75%)
Q2 (F) A 60 W filament bulb is 10% efficient. Calculate its useful light output.
Show solution
F: efficiency = useful output ÷ total input
I: 0.10 = useful ÷ 60
F: useful = 0.10 × 60
A: 6 W
Q3 (F) An electric motor is 90% efficient and gives a useful output of 1800 W. Calculate its total power input.
Show solution
F: efficiency = useful output ÷ total input
I: 0.90 = 1800 ÷ input
F: input = 1800 ÷ 0.90
A: 2000 W
Q4 (F) A 2.5 kW pond pump has an efficiency of 0.4. Calculate its useful power output in W.
Show solution
Convert first: 2.5 kW = 2500 W
F: efficiency = useful output ÷ total input
I: 0.4 = useful ÷ 2500
F: useful = 0.4 × 2500
A: 1000 W
Q5 (F) An LED bulb is 75% efficient and gives out 6 W of light. Calculate its power input.
Show solution
Convert first: 75% = 0.75
F: efficiency = useful output ÷ total input
I: 0.75 = 6 ÷ input
F: input = 6 ÷ 0.75
A: 8 W
Q6 (F/H) A 1.2 kW vacuum cleaner is 65% efficient. Calculate the power wasted.
Show solution
Convert first: 1.2 kW = 1200 W; 65% = 0.65
F: efficiency = useful output ÷ total input
I: 0.65 = useful ÷ 1200
F: useful = 0.65 × 1200 = 780 W
A: wasted = 1200 − 780 = 420 W
Q7 (F/H) A wind turbine is 40% efficient and generates 1.5 MW of electrical power. Calculate the power of the wind hitting it.
Show solution
F: efficiency = useful output ÷ total input
I: 0.40 = 1.5 ÷ input
F: input = 1.5 ÷ 0.40
A: 3.75 MW
Q8 (F/H) A power station is 36% efficient and delivers 1.8 × 109 W of electrical power. Calculate its total power input.
Show solution
F: efficiency = useful output ÷ total input
I: 0.36 = 1.8 × 109 ÷ input
F: input = 1.8 × 109 ÷ 0.36
A: 5.0 × 109 W
Q9 (H) A 500 W motor is 70% efficient. How long does it take to lift a 50 kg load by 7.0 m? (g = 9.8 N/kg)
Show solution
Step 1 – useful power
F: efficiency = useful output ÷ total input
I: 0.70 = useful ÷ 500
F: useful = 0.70 × 500 = 350 W
Step 2 – energy needed: Eₚ = m g h = 50 × 9.8 × 7.0 = 3430 J
Step 3 – time: t = E ÷ P = 3430 ÷ 350
A: 9.8 s
Q10 (H) An electric car motor is 92% efficient and delivers 40 kW of useful power. Calculate the power input, to 2 significant figures.
Show solution
F: efficiency = useful output ÷ total input
I: 0.92 = 40 ÷ input
F: input = 40 ÷ 0.92 = 43.47… kW
A: 43 kW (2 s.f.)
