The graph shows how the resistivity of copper varies all the way down to absolute zero, the lowest possible temperature. It gives us two clues about why metals have electrical resistance.

	What the graph shows Explanation Resistance decreases as copper (or any other metal) is cooled. Its atoms vibrate less, and so they impede the flow of electrons less. Pure copper has less resistance than copper containing impurities. Impurity atoms are a different size to copper atoms, so they get in the way of moving electrons.

Here's how we picture the way electrons move through copper.

Now lets concentrate on a few electrons. Each time an electron collides with an ion, it loses energy. This energy is transferred to the ion (which vibrates more); eventually the energy from billions of collisions is shared amongst all the atoms, and the material gets hot.

This is often called joule heating, and it is something electrical engineers try to reduce – unless they are designing a heater!

Once the ions are vibrating, they hinder the flow of the electrons. This is because the electrons will be scattered off the vibrating ions.

As the wire gets hotter, the ions vibrate more vigorously. This makes it more difficult for electrons to pass along the wire. Hence the resistance increases.

In fact:

electrical power = voltage × current

The voltage is the potential difference - i.e. the difference in potential energy per unit charge. It tells us how much energy is transferred per coulomb of charge that flows. The bigger the voltage, the more energy each coulomb will transfer. So power will increase with voltage.		A good way to remember this is through the more basic units of the volt. One volt is equivalent to one joule per coulomb. [voltage] = V = J/C [coulomb is the unit of charge]
The current is how quickly charge is flowing. The bigger the current the more charge flows through the wire in a given time and the more energy it can transfer. So the bigger the current, the bigger the power.		One amp is equivalent to one coulomb per second. [current] = A = C/s

We can put the units into the equation for power:

[power] = [voltage] x [current] = J/C × C/s = J/s = W

This is the unit for power. So the equation works!

[You can get this equation by substituting V = I R into the first equation for power.]