Physics 10.28 - 10.31 - Electrical power

Power as a rate

Imagine that two devices each transfer $600\,\mathrm{J}$ of energy. One takes $10\,\mathrm{s}$ and the other takes $60\,\mathrm{s}$. The first device is more powerful because it transfers the same energy in less time.

Power and energy are different quantities. Energy, measured in joules, is an amount transferred. Power, measured in watts, tells you how rapidly that transfer happens. A high-power device can transfer a large amount of energy in a short time.

This equation gives the average power over the stated time interval. Before substituting, energy must be in joules and time must be in seconds.

Worked example

A motor transfers $18\,\mathrm{kJ}$ of energy in $12\,\mathrm{s}$. Calculate its average power.

Solution

First convert the energy:

Select the equation and substitute:

So the average power is $1500\,\mathrm{W}$, or $1.5\,\mathrm{kW}$. This is sensible because $1500\,\mathrm{J}$ transferred in each of $12$ seconds gives $18\,000\,\mathrm{J}$ altogether.

Current, potential difference and power

For one circuit device, two quantities determine the rate of electrical energy transfer:

• the potential difference across the device, which is the energy transferred per unit charge;
• the current through the device, which is the rate of flow of charge.

This relationship can be explained from the meanings of current and potential difference:

The unit of potential difference is $\mathrm{J\,C^{-1}}$.

The unit of current is $\mathrm{C\,s^{-1}}$.

The units therefore combine as

The coulombs cancel, leaving energy transferred per second. This is why $P = IV$ describes power, rather than being an arbitrary rule.

If current increases while potential difference is held constant, power increases in direct proportion. If potential difference increases while current is held constant, power also increases in direct proportion. In a real circuit, changing one quantity may cause the other to change too, so the "held constant" condition matters.

Worked example

A small circuit device has a potential difference of $12\,\mathrm{V}$ across it and a current of $250\,\mathrm{mA}$ through it. Calculate its power.

Solution

Convert the current before substitution:

The result is sensible: $0.250\,\mathrm{C}$ passes each second and each coulomb transfers $12\,\mathrm{J}$, giving $3.0\,\mathrm{J}$ each second.