Power and efficiencyEdexcel International A Level Physics: Revision notes

Power is the rate at which energy is transferred, or the rate at which work is done:

$P = \frac{E}{t}$ and $P = \frac{W}{t}$

The unit is the watt (W): 1 W is a rate of 1 joule per second, 1 W = 1 J s⁻¹. Larger units are the kilowatt (kW, $10^3$ W) and megawatt (MW, $10^6$ W).

Power tells you how fast a transfer happens. Two machines can do the same work, but the one that does it in less time has the greater power.

When a force does work, the work is $W = Fs$ (or $Fs\cos\theta$ for a force at an angle), so the power is the work divided by the time taken. For a car that travels $750\ \text{m}$ in $30\ \text{s}$ against a constant $600\ \text{N}$ resistive force at constant speed:

$W = 600 \times 750 = 4.5 \times 10^5\ \text{J}$ and $P = \frac{4.5 \times 10^5}{30} = 1.5 \times 10^4\ \text{W}$.

For lifting a load, the useful energy is the gain in gravitational potential energy, $mg\Delta h$, so $P = \frac{mg\Delta h}{t}$.

No real device transfers all of its input energy usefully; some is dissipated, mostly as thermal energy. Efficiency compares the useful output with the total input:

$\text{efficiency} = \frac{\text{useful energy output}}{\text{total energy input}}$

$\text{efficiency} = \frac{\text{useful power output}}{\text{total power input}}$

Both give the same value because power is energy divided by the same time. Efficiency has no unit. It is a number between 0 and 1, or a percentage between 0 and 100%.

Energy is always conserved, so the energy input equals the useful output plus the energy wasted:

$\text{input} = \text{useful output} + \text{wasted energy}$

The wasted energy is dissipated to the surroundings, so say it is dissipated or transferred as thermal energy, never that it is 'lost' or 'used up'. Typical causes are friction, electrical resistance, air resistance and sound. Efficiency can be improved by reducing these, for example by lubrication or thicker wires.

A motor raises a $250\ \text{kg}$ load through $12\ \text{m}$ in $20\ \text{s}$ and takes electrical power of $2.4\ \text{kW}$.

Useful energy: $mg\Delta h = 250 \times 9.81 \times 12 = 2.94 \times 10^4\ \text{J}$.

Useful power: $\frac{2.94 \times 10^4}{20} = 1.47\ \text{kW}$.

Efficiency: $\frac{1.47}{2.4} = 0.61$, so 61%.

Wasted power: $2.4 - 1.47 = 0.93\ \text{kW}$.