Power is a measure of how much potential energy is dissipated (i.e. converted into heat, light and other forms of energy) by a component or circuit in one second. This is due to a drop in the potential energy, and so the voltage, of charge. Power is measured in Watts (commonly abbreviated W), where 1 W is 1 Js-1. It can be calculated by finding the product of the current flowing through a component / circuit and the potential difference across the component / circuit. This gives us the equation:

$P = \frac{E}{t} = IV$

where P is the power dissipated (in W), E is the drop in potential energy (in Joules, J), t is the time taken (in s), I is the current (in A) and V is either potential difference or electromotive force (in V), depending on the component being measured.

Since power is the amount of energy changing form per. second, the amount of energy being given out each second will equal the power of the component giving out energy.

You should be able to substitute in values for I and V from other formulae (V=IR, Q=It) in order to relate power to resistance, conductance, charge and time, giving formulae like these:

$P = I^2R$

$P = \frac{V^2}{R}$

$P = \frac{QV}{t}$

## Questions

1. The potential difference across a 9W light bulb is 240V. How much current is flowing through the light bulb?

2. How much energy is dissipated by a 10W component in 1 hour?

3. The potential difference across a top-notch kettle, which can hold up to 1 litre of water, is 240V, and the current is 12.5 A. 4.2 kJ of energy is required to raise the temperature of 1kg of water by 1°C. Assuming 100% efficiency and that the temperature has to be raised 80°C (20°C to 100°C), how long does it take to boil 1 litre of water?

4. How much energy is dissipated by a 100Ω resistor in 10 seconds if 2A of current are flowing?

5. The charge on an electron is -1.6 x 10-19 C. How long does it take for a mole (6 x 1023 particles) of electrons to flow through a 40W light bulb on a 240V ring main?

Worked Solutions