To compare two devices fairly, fix the task, then compare the energy each one needs and the time each one takes. Power is how quickly a device works, not how much energy a job costs.
This follows calculating power from energy and time and belongs to power and energy resources.
Which quantities should I compare?
For an equal task, three quantities tell the story:
- Energy input (J or kWh): the total supplied. Lower means a lower running cost.
- Time taken (s or h): shorter means a higher useful power.
- Efficiency = useful energy output ÷ total energy input. Higher means less wasted energy.
Useful power = useful energy output ÷ time. Input power = total energy input ÷ time. Keep these two separate.
Worked example
Two motors each lift a 500 N load through 12 m. (Invented example data.) Motor A has an input power of 400 W and takes 20 s. Motor B has an input power of 300 W and takes 30 s.
Useful energy output (same for both): work done = force × distance = 500 × 12 = 6000 J.
Energy input: A: 400 × 20 = 8000 J. B: 300 × 30 = 9000 J.
Efficiency: A: 6000 ÷ 8000 = 0.75 = 75%. B: 6000 ÷ 9000 = 0.667 = 67% (2 significant figures).
Useful power output: A: 6000 ÷ 20 = 300 W. B: 6000 ÷ 30 = 200 W.
Conclusion: Motor A has the higher input power, yet it uses less energy (8000 J against 9000 J), is more efficient and finishes sooner. Motor B’s lower power rating did not make it cheaper to run for this task.
Check: 8000 − 6000 = 2000 J is wasted by A, and 9000 − 6000 = 3000 J by B, which fits B being less efficient.
The mistake to watch for
A common slip is to choose the device with the lower power rating as the one that “uses less energy”.
Mistaken answer: “Motor B uses less energy because 300 W is less than 400 W.”
This compares power only. It ignores that B runs for 30 s instead of 20 s.
The correction is to multiply by time for each device, then compare the joules. Ask yourself: “Is the task equal, and have I compared energy, not just power?”
Check yourself
1. An LED lamp (8 W) and a filament lamp (40 W) give the same light for 10 hours. Find the energy each transfers in kWh, and the difference.
Show answer
LED: 8 W × 10 h = 80 Wh = 0.08 kWh. Filament: 40 W × 10 h = 400 Wh = 0.40 kWh. Difference = 0.40 − 0.08 = 0.32 kWh.
2. A machine takes 9000 J of input to do 6000 J of useful work. What is its efficiency?
Show answer
6000 ÷ 9000 = 0.667, so about 67%.
3. Two heaters of 1.5 kW and 2.0 kW must each transfer 3.0 MJ to a room. How long does each take?
Show answer
3.0 MJ = 3 000 000 J. 1.5 kW: 3 000 000 ÷ 1500 = 2000 s (about 33 min). 2.0 kW: 3 000 000 ÷ 2000 = 1500 s (25 min). The 2.0 kW heater is quicker, and both transfer the same 3.0 MJ.
Where this leads next
Next, read supplied generation data, where the same power × time idea is applied to tables from power stations. Try the practice set afterwards.
If you can do the arithmetic but struggle to decide what the question is comparing, that is a reasoning habit a teacher can shape in online one-to-one Physics tuition.