Define power, apply power equals work over time, and calculate efficiency as useful output over input

What this dot point is asking

SEAB wants you to define power as the rate of doing work (or transferring energy), to use $P = W/t$, and to calculate efficiency as the fraction of input energy that is usefully transferred. The big idea is that power tells you how fast energy is used, while efficiency tells you how much of the energy supplied does the job you wanted.

The answer

What power is

Power is the rate of doing work, or the rate of transferring energy:

Energy is in joules and time in seconds, so the unit of power is the watt ($\text{W}$), where $1\ \text{W} = 1\ \text{J s}^{-1}$. A 60-watt bulb transfers $60\ \text{J}$ of energy every second.

Why power matters

Two motors might do the same total work, but the more powerful one does it faster. A powerful crane lifts the same load in less time, and a powerful engine accelerates a car more quickly, because each transfers energy at a higher rate.

Efficiency

No real device transfers all its input energy usefully; some is always wasted as heat. Efficiency measures the useful fraction:

Efficiency is a percentage with no unit. It can never be more than $100\%$, because that would mean creating energy, which conservation of energy forbids.

Efficiency in terms of power

Because power is energy per second, efficiency can also be written using powers:

This is handy when a question gives power ratings rather than energy values.

Examples in context

Example 1. Energy-saving bulbs. An old filament bulb turns only a small fraction of its electrical input into light, wasting most as heat, so it is inefficient. An LED bulb of the same brightness uses far less power because a much larger fraction of its input becomes light, so it is more efficient and cheaper to run.

Example 2. Car engines. A petrol engine is only about a quarter efficient: most of the chemical energy in the fuel becomes heat in the exhaust and engine rather than useful motion. Engineers improve efficiency to get more useful power from each litre of fuel, but conservation of energy means it can never reach $100\%$.

Try this

Q1. A machine transfers $600\ \text{J}$ of energy in $3.0\ \text{s}$. Calculate its power. [2 marks]

• Cue. $P = \dfrac{W}{t} = \dfrac{600}{3.0} = 200\ \text{W}$.

Q2. A device takes in $400\ \text{J}$ and usefully outputs $240\ \text{J}$. Calculate its efficiency. [2 marks]

• Cue. Efficiency $= \dfrac{240}{400} \times 100\% = 60\%$.

Q3. Explain why no device can be more than $100\%$ efficient. [2 marks]

• Cue. Useful output cannot exceed the energy supplied; a value over $100\%$ would mean creating energy, which conservation of energy forbids.