6.6 Calculating Resistance from Resistivity from Office of Academic Technologies on Vimeo.

**6.06 Calculating Resistance from Resistivity**

Since both of these quantities, resistance and resistivity, are in a way linked to the number of collisions that the charge carriers are making as they drift from high-potential region to low-potential region, then we can expect a relationship between these two quantities. To be able to show this relationship, let’s consider a piece of wire with a length of, say, *l*. And let’s connect the ends of this wire to a power supply, which generates *V* volts of potential difference between its terminals. Therefore, as soon as we turn the switch on we’re going to generate *V* volts of potential difference between these two ends of this wire. And, again, at the moment that we turn the switch on, we will set up an electric field pointing from positive end of this wire towards the negative end.

Let’s say that the cross-sectional area of the wire is *A*. So, *A* represents the cross-sectional area. The potential difference between the ends of this wire will be equal to, as you recall, integral of **E** dot *d***l**, integrated along the length of this wire. Well, if you do this, since the electric field magnitude is constant, and by choosing a path from one end to the other end, *d***l** represents the incremental displacement vector along this path. So the angle between these two vectors, vector-field vector times the incremental displacement field vector *d***l** times the cosine of the angle between these two vectors — in this case it’s going to be 0 degree — will give us the expanded form of **E** dot *d***l**.

Again, since **E** is constant, we can take it outside of the integral, and cosine of 0 is just 1, this quantity is going to be equal to **E** times integral of *d***l** along the length of this wire, and its length — let’s say we’ll put our origin at one end — then it’s going to go from 0 to l, so the integral of *d***l** is addition of these incremental distances, *d***l**‘s to one another along the length of the wire, will give us whatever the length of that wire is. So the potential difference is going to be equal to the electric field along the wire times its length, *l*.

From there we can solve for the electric field, which will be equal to *V* divided by the potential difference between the ends of the wire divided by its length. On the other hand, we know that the current density, **J**, is equal to current flowing through the wire, divided by the cross-sectional area of this wire.

Now, by recalling the definition of resistivity, which was the ratio of the electric field to the current density, we can express these quantities as *V* over *l* for the electric field divided by *i* over *A* for the current density. Going one further step, this is going to be equal to *V* over *i* times *A* over *l*. Well, *V* over *i* , by definition, that is the potential difference between the ends of this wire, *V*, divided by the amount of current flowing through this wire, and that is *i*, which is going to originate from the positive terminal and enter into the negative terminal.

This quantity is therefore nothing but, by definition, just the resistance, *R*, of this wire. So, we end up with *R* times *A* over *l*. If we solve this expression for the resistance, then that becomes equal to resistivity *ρ* times *l* over *a*. That’s the relationship between resistance and resistivity for a wire with length *l* and cross-sectional area *A*. From here we can easily see that the resistance is directly proportional to the length of the wire. In other words, the longer the wire means the greater the resistance is, which will eventually cause resistive losses that we will study in a moment.

So in order to avoid resistive losses — in other words, in order to avoid the waste of electrical potential energy eventually in the form of heat due to this resistance — as we design our circuits we’d like to keep the length of the wires as short as possible. On the other hand, we see that the resistance is inversely proportional to the cross-sectional area of the wire. So that tells us that the thicker the wires will cause less amount of resistive losses. But of course when we do the physical calculations that doesn’t mean that we just take our wires very, very thick so that we will decrease the resistive losses because such a process will even cause more in comparing to the amount of electrical potential energy wasted in the form of heat.

So the major part that we have to be careful is the length of the wires, and we try to keep them as short as possible in order to decrease the resistive losses in the electric circuits. And of course, as we expect, the resistance is directly proportional to the resistivity because both of these quantities are, in a way, measure of the number of collisions that the charge carriers are making as they drift from high-potential regions to low-potential regions.