Problem
Current is the amount of charge that is flowing through a component per unit of time.
For a given voltage, Ohm's law tells us that if we increase the resistance, then the current must decrease.
But what's actually happening to decrease the current?
My reasoning so far – is it correct?
More resistance (if we're talking about something of the same size for simplicity) is more 'stuff' in the way (higher resistivity), so more collisions. More collisions means it takes longer for the charge to 'get through' the component. The charge is moving slower and so the current is lower.
Or, is it that the speed is always the same and that somehow if we have more resistance then it means there is just less charge able to flow, hence a lower current. If so, why?
Best Answer
The answer depends on the circumstances: how do you change the resistance? Both the drift velocity and the number of available charge carriers can be changed.
In a basic Drude model for electrical transport both, $n$, the charge carrier density and $\tau$, the time between collisions determine the resistance:
$$\mathbf{J} = \left( \frac{n q^2 \tau}{m} \right) \mathbf{E}.$$ Here, $\mathbf{J}$ is the current density and $\mathbf{E}$ the applied electrical field. The term in parentheses characterizes the material properties, i.e. its resistance.
By changing to a different material you can influence both, $n$ and $\tau$, so both cases are possible.