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.
