Video: Resistance and Resistivity of Conductors

In this video, we will learn how to relate the dimensions of and the motion of free electrons through an object to its resistance.

13:10

Video Transcript

In this lesson, we will learn how to relate the dimensions of and the motion of free electrons through an object to its resistance. We’ll also look at how resistance is affected by the dimensions of the material, the length, and area as well as by a property called resistivity.

The resistance of a material 𝑅 is dependent on the resistivity of the material 𝜌, the length of the material 𝐿, and the cross-sectional area of the material 𝐴. Let’s look at each individual variable to determine its effect on a resistance, starting with the resistivity, 𝜌.

The resistivity of material is what opposes the charge flow based on its atomic makeup. Materials with low resistivity are called conductors, where materials with high resistivity are considered insulators. Conductors allow charge to flow more easily, where insulators do not. Let’s compare two conductors with differing resistivities to determine how this will affect the resistance.

Let’s look at the diagram below, which represents a lattice of copper ions, as shown by the red dots. Metals have a lot of free electrons as almost every ion in the grid has an associated free electron. However, we will diagram just one electron’s path. The electron goes through relatively unobstructed. There’s minimal disturbance in the lattice for the electron to scatter, which is what gives the copper a low resistivity.

Let’s compare the electron movement through copper to the electron movement through brass, where brass is a copper–zinc alloy. We have chosen brass that is made up of 10 percent zinc ions, as represented by the larger red dots in our grid. Zinc disturbances provide more opportunities for the electrons to scatter as compared to the more uniform copper lattice. The increase in irregularity in the brass causes the increase in scattering of the electrons. As we can see, in the brass, the electron takes a more zigzagging path, which is longer than the more direct path in copper. Traveling such a path takes more time, which results in a higher resistivity. This tells us that the resistivity of brass is greater than the resistivity of copper.

Let’s look at what this means for the resistance of both materials. Resistivity is directly related to resistance, which means that the greater the resistivity, the greater the resistance. Copper, with its low resistivity, is used in electronics or when we want to keep the resistance low.

Let’s look at our next variable length and the effect that it has on resistance. The length, just like the resistivity, is directly related to the resistance. So the longer the wire, the greater the resistance. Let’s take a closer look at two wires, one short and one long.

Let’s compare two wires that are made of the same material and have the same cross-sectional area but different lengths, with one being a short wire and one being a long wire. By diagramming the journey of a single electron through each of these wires, we could see the difference in resistance.

Let’s start by diagramming the journey for the short wire. The electron experiences relatively few collisions and travels a relatively short distance as the length of the wire is small. And so the flow of charge experiences a small resistance. Now, let’s compare that to the journey of an electron through a long wire. The electron experienced a relatively larger number of collisions and traveled a relatively longer path as it moved along the larger length of wire. This shows that the flow of charge experiences a larger resistance, which brings us back to the fact that when we have a longer length of wire, we have a larger resistance.

Now, we’ll analyze the effect that cross-sectional area has on resistance. The cross-sectional area of the wire is inversely related to the resistance, which means as the area increases, the resistance decreases. To understand this relationship better, we can compare the diagram for a thin wire to the diagram for a thick wire, assuming that the wires have the same resistivity and same length.

The thin wire has a small cross-sectional area, which means if we were to slice the wire and look at the end of it, we would see a circle with a small surface area, whereas the thick wire would have a large cross-sectional area. If we once again were to cut our wire, we’d have a circle with a large surface area.

Let’s take a look at how the electrons drift across the wire. Specifically, we’re going to want to get eight charge carriers to the other side of the wire. We could see that it took a relatively long time for all eight charge carries to get to the other side of the wire, meaning that our wire has a large resistance. Thin wires that have small areas have large resistances.

Now, let’s look how long it takes eight charge carriers to move through the thick wire. We don’t have to wait for all the charge carries to drift through the wire, only the eight that were closest to the end. The eight charge carriers in the thick wire experience relatively fewer collisions than those in the thin wire as they don’t travel as far, thereby saying that the charge carriers experience a small resistance. Wires that are thick have large cross-sectional areas and small resistance.

Now that we know how the different variables affect the resistance, let’s plug in the values so we can practice calculating resistance. Let’s determine the resistance of a copper wire that has a cross-sectional area of 0.02 meter squared and a length of 0.80 meters. Starting with the equation 𝑅 equals 𝜌𝐿 over 𝐴, we can plug in our variables. When we look up the resistivity of copper in a table, we see that it is 1.68 times 10 to the negative eighth ohms times meters. For the length, we use 0.80 meters. And for the cross-sectional area, we use 0.02 meters squared. When we multiply 1.68 times 10 to the negative eighth ohms times meters by 0.80 meters and divide that by 0.02 meter squared, we get a value of the resistance of 6.72 times 10 to the negative seventh ohms.

There is one more variable that does affect the resistance of materials but doesn’t show up in our equation. And that is temperature. The temperature will affect the resistance of metals differently than it will the resistance of nonmetals and semiconductors. For metals, the higher the temperature, the higher the resistance. For nonmetals and semiconductors, as we increase the temperature, the resistance gets lower. Let’s explore the effect of temperature on resistance a little bit more with some diagrams.

We have drawn the same grid that we drew earlier in the video to represent the lattice structure of copper. We need to once again remember that even though metals have lots of free electrons, we are going to look at the journey of just one electron. As the temperature of the copper increases, disturbances are being added to the lattice that will scatter the electrons. We have represented those disturbances by having the ions in some areas be closer together and farther apart.

The zigzag path that the electron takes is similar to how the electrons were scattered in the brass when the zinc that was being added to the copper created disturbances. In a semiconductor at cooler temperatures, there are very few electrons. However, as the temperature increases, a lot of covalent bonds are broken yielding more free electrons. With so many extra charge carriers, we don’t have to get this one electron over to the other side; one of the ones that are closer can move over. Just as we saw with length of a wire, the shorter the distance, the lower the resistance. As the temperature increases and the semiconductor has more free electrons, the path that an electron has to take in order to get one to the other side is much shorter, thereby decreasing the resistance.

Now that we have analyzed how temperature affects the resistance of materials, let’s do a few example problems using our equation from earlier.

A wire made of an unknown substance has a resistance of 125 milliohms. The wire has a length of 1.8 meters and a cross-sectional area of 2.35 times 10 to the negative fifth meter squared. What is the resistivity of the substance from which the wire’s made?

Let’s begin by drawing a picture of our wire. In the diagram, we have labeled the length of the wire 𝐿 as 1.8 meters, the cross-sectional area of the wire 𝐴 as 2.35 times 10 to the negative fifth meter squared, the resistance of the wire 𝑅 as 125 milliohms, and we are looking for the resistivity of the wire 𝜌. In order to solve for resistivity, we need an equation that relates the resistance, resistivity, length, and cross-sectional area of our wire.

The equation that relates these four variables together is 𝑅, the resistance of the wire, equals 𝜌, the resistivity, times 𝐿, the length, divided by 𝐴, the cross-sectional area. We are solving for the resistivity. Therefore, we must rearrange our formula so that it solves for 𝜌. To do this, we multiply both sides of the equation by 𝐴 over 𝐿. This will cancel out both the 𝐴 and the 𝐿 on the right side of the equation, leaving us with the relationship 𝐴𝑅 over 𝐿 is equal to 𝜌. We can now substitute in our values for our variables.

For the cross-sectional area, we use 2.35 times 10 to the negative fifth meter squared. For the resistance, we use 125 milliohms, and 1.8 meters for the length. We need to be careful with our units. Right now our resistance is in milliohms, but it needs to be converted into ohms. Recall that the prefix milli- means 10 to the negative third. 125 milliohms is the same thing as 125 times 10 to the negative third ohms. After we multiply out our numerator and divide by our denominator, we get 1.63 times 10 to the negative six ohm meters for our resistivity.

The length in the problem was given to us to two significant figures. Therefore, we must report our resistivity to two significant figures. 1.63 times 10 to the negative sixth ohm meters rounds to 1.6 times 10 to the negative sixth ohm meters. The resistivity of the substance from which the wire is made is 1.6 times 10 to the negative sixth ohm meters.

A copper wire with a resistance of 12.8 milliohms has a cross-sectional area of 1.15 times 10 to the negative fifth meter squared. Find the length of the wire. Use 1.7 times 10 to the negative eighth ohms times meters for the resistivity of copper.

Let’s begin by drawing a diagram of our copper wire. In our diagram, we’ve labeled our copper wire with a resistivity 𝜌 of 1.7 times 10 to the negative eighth ohm meters, a cross-sectional area 𝐴 of 1.15 times 10 to the negative fifth meter squared, a resistance 𝑅 of 12.8 milliohms, and we are solving for the length of the wire 𝐿.

Before we do any calculations, we need to recall an equation that relates the resistivity, cross-sectional area, length, and resistance of our wire to each other. That equation is 𝑅, the resistance of the wire, is equal to 𝜌, the resistivity of the wire, times 𝐿, the length of the wire, divided by 𝐴, the cross-sectional area of the wire. The problem asked us to solve for the length of the wire. Therefore, we must rearrange our formula so that we are solving for 𝐿.

To do this, we have to multiply both sides of the equation by 𝐴 over 𝜌. This will cancel out both the 𝐴 and the 𝜌 on the right side of the equation, leaving us with the relationship 𝐴 times 𝑅 divided by 𝜌 is equal to 𝐿. We can now substitute in our values for our variables. For our area, we use 1.15 times 10 to the negative fifth meter squared. The resistance is 12.8 milliohms. And the resistivity is 1.7 times 10 to the negative eighth ohms meters.

We need to be careful with our units. Our resistance is given to us in milliohms, but it needs to be converted into ohms. We need to remember that milli- is the prefix for 10 to the negative third. Therefore, we can replace 12.8 milliohms with 12.8 times 10 to the negative third ohms. When we multiply out our fraction, we get a length of the wire as 8.66 meters. The resistivity in our problem was given to two significant figures. Therefore, we must report back our length to two significant figures. 8.66 meters rounds to 8.7 meters. The length of the wire is 8.7 meters.

Key Points

The formula 𝑅 equals 𝜌𝐿 over 𝐴 is used to relate the resistivity, resistance, and dimensions of a wire. The motion of free electrons will be affected by varying the length, cross-sectional area, and resistivity of a resistor as well as its resistance. The motion of free electrons will be affected by varying the temperature of resistors as well as their resistivities.

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