Resistivity - AP Physics 2

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Question

You have a sample that is a cube of volume . You attach electrodes to opposite faces and find the resistance to be $3\Omega$ .

What is the resistivity of the material?

Answer

Since we have a cube of volume , the length of the material must be . Use the equation for resistivity.

$R=\rho \frac{L}{A}$

Rearrange the equation and plug in known values.

$\rho = \frac{RA}{L}$

$\rho=\frac{3\Omega\cdot 1cm^2}{1cm}=3\Omega\cdot cm$

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Question

You create a cylinder of an unknown material that has a diameter of , and a height of . You attach the electrodes to the faces of the cylinder and find the sample has a resistance of $25\Omega$ .

What is the resistivity of the material?

Answer

Use the equation for resistivity:

$R= \rho \frac{L}A{}$

First, find the cross sectional area.

$A=\pi r^{2}$

Plug in known values.

$R= \rho \frac{L}A{}$

$\rho = \frac{25\Omega\cdot 3.14cm^2}{1cm}$

$\rho =78.5\Omega \cdot cm$

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Question

You have a 4cm long copper wire with a radius of 0.5mm. You have experimentally determined the resistance of the wire to be $0.8709m\Omega$ . What is the resistivity of copper?

Answer

The resistivity of a material is how much the material resists the flow of charge through it. Metals have low resistivities (which makes them good conductors), while things like glass or plastic have high resistivities.

The equation for resistivity is as follows:

$R=\frac{\rho L}{A}$

is the length of the wire, is the cross-sectional area, is the resistance, and $\rho$ is the resistivity. We have values for and , and we're given the radius of the wire so we can find , so we're trying to solve for $\rho$ . If we rearrange the equation, we end up with this:

$\frac{RA}{L}=\rho$

We're given the radius of the wire, so to find the area, we square the radius and multiply it by $\pi$ .

$A=\pi r^2$

$A=\pi (0.0005)^2$

$A=7.854\cdot 10^{-7}$

Now, we can plug in our values.

$\frac{RA}{L}=\rho$

$\frac{(0.8709\cdot 10^{-3}) \cdot (7.854\cdot 10^{-7})}{0.04}=\rho$

$1.71\cdot 10^{-8}=\rho$

Therefore, the answer is $1.71\cdot10^{-8}\Omega\cdot m$ .

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Question

You have a material with resistivity $3500\Omega\cdot m$ . You build a component of a fuel cell out of this material with a cross sectional area of , and a thickness of .

What will be the resistance of this component?

Answer

Use the resistivity equation:

$R= \rho \frac{L}{A}$

$R= 3500\Omega \cdot m \cdot \frac{0.01m}{0.01m^2}$

$R=3500\Omega$

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Question

You are researching new materials for usage in spacecraft electronics.

You have a new material, known as "Type F."

You carve out a cylinder of the material. It is tall, with a radius of .

You put electrodes on each face of the cylinder.

You determine the resistance to be $935\Omega$ .

What is the resistivity of "Type F?"

Answer

We will use the relationship

$\rho=R\frac{A}{l}$

Where $\rho$ is the resistivity, is the resistance, is the surface area of the face the current is coming in or out of, and is the length from one face to the other.

Remember that the area of a circle is

$A=\pi\cdot r^2$

Combining our equations we get

$\rho=R\frac{\pi\cdot r^2}{l}$

We then need to plug in our values

$\rho=935\frac{\pi\cdot{.0265}^2}{.045}$

$45.8\Omega\cdot meters=\rho$

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Question

You are researching new materials for usage in spacecraft electronics.

You have a new material, known as "Type G."

You carve out a cylinder of the material. It is tall, with a radius of .

You put electrodes on each face of the cylinder.

You determine the resistance to be .

What is the resistivity of "Type G?"

Answer

We will use the relationship

$\rho=R\frac{A}{l}$

Where $\rho$ is the resistivity, is the resistance, is the surface area of the face the current is coming in or out of, and is the length from one face to the other.

Remember that the area of a circle is

$A=\pi\cdot r^2$

Combining our equations we get

$\rho=R\frac{\pi\cdot r^2}{l}$

We then need to plug in our values

$\rho=9.53\frac{\pi\cdot .015^2}{.025}$

$2.70\times 10^{-3}\Omega\cdot meters=\rho$

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Question

You are researching new materials for usage in spacecraft electronics.

You have a new material, known as "Type Z."

You carve out a cylinder of the material. It is tall, with a radius of .

You put electrodes on each face of the cylinder.

You determine the resistance to be .

What is the resistivity of "Type Z?"

Answer

We will use the relationship

$\rho=R\frac{A}{l}$

Where $\rho$ is the resistivity, is the resistance, is the surface area of the face the current is coming in or out of, and is the length from one face to the other.

Remember that the area of a circle is

$A=\pi\cdot r^2$

Combining our equations we get

$\rho=R\frac{\pi\cdot r^2}{l}$

We then need to plug in our values

$\rho=.22\frac{\pi\cdot.002^2}{.022}$

$\rho=1.26\times10^{-4}\Omega\cdot meters$

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Question

A researcher is testing the electrical properties of a circuit that contains a resistor of unknown material. The resistor is a cylinder with a radius of . The researcher measures the resistance of the resistor to be $0.3\Omega$ . What is the resistivity of this unknown material?

Answer

Use the equation for resistivity:

$\rho=R\frac{A}{L}$

Here, $\rho$ is the resistivity, is the resistance, is the cross-sectional area, and is the length of the resistor.

Begin by finding the cross-sectional area of the material, which is circular:

$A=\pi r^2$

$A=\pi (0.005m)^2$

$A\approx 0.0000785m^2$

Now plug in all known values and solve for resistivity.

$\rho=0.3\Omega\frac{0.0000785m^2}{0.03m}$

$\rho = 0.000785\Omega\cdot m$

$\rho = 7.8510^{-4}\Omega\cdot m$

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Question

A scientist carves out a cylinder of a new material she developed. It is tall, with a radius of . She puts electrodes on each face of the cylinder. She determines the resistance to be $.75\Omega$ . What is the resistivity of this new material?

Answer

Using the relationship:

$\rho =R\frac{A}{l}$

Here, $\rho$ is the resistivity, is the resistance, is cross sectional area, and is the length.

The area of a circle is:

$A=\pi r^2$

Substitute:

$\rho =R\frac{\pi r^2}{l}$

Plug in given values and solve.

$\rho =.75\Omega \frac{\pi (.008m)^2}{.05m}$

$\rho=3.0 *10^{ -3} \Omega \cdot m$

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Question

A scientist carves out a cylinder of a new material she developed. It is tall, with a radius of . She put electrodes on each face of the cylinder. She determined the resistance to be $.75\Omega$ . What is the resistivity of this material?

Answer

Use the relationship:

$\rho =R\frac{A}{l}$

Here, $\rho$ is the resistivity, is the resistance, is cross sectional area, and is the length.

The area of a circle is:

$A=\pi r^2$

Substitute:

$\rho =R\frac{\pi r^2}{l}$

Plug in given values and solve

$\rho =.75\Omega \frac{\pi (.006m)^2}{.045m}$

$\rho = 1.88 *10^{-3} \Omega\cdot m$

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Question

A cylinder of an unknown material is being tested. It is tall, with a radius of . Electrodes are placed on each face of the cylinder. It is determined that the resistance is . What is the resistivity?

Answer

Using the relationship

$\rho=R\frac{A}{l}$

Where $\rho$ is the resistivity, is the resistance, is the cross sectional area, and is the length from one face to the other.

The area of a circle is

$A=\pir^2$

Combining equations:

$\rho=R\frac{\pir^2}{l}$

Plugging in values

$\rho=59\frac{\pi.0095^2}{.0575}$

$\rho=.291\Omegameters$

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Question

A sample of a new material is being tested. It is tall, with a radius of . Electrodes are placec on each face of the cylinder. The resistance is determined to be .

What is the resistivity of this material?

Answer

Using the relationship

$\rho=R\frac{A}{l}$

Where $\rho$ is the resistivity, is the resistance, is the cross sectional area, and is the length from one face to the other.

The area of a circle is

$A=\pir^2$

Combining equations

$\rho=R\frac{\pir^2}{l}$

Plugging in values

$\rho=1.35\frac{\pi.0038^2}{.038}$

$1.6110^{-3}\Omegameters=\rho$

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Question

There is a sample of an unknown material under testing. It is tall, with a radius of . Electrodes are placed on each face of the cylinder. It is determined that the resistance to .

What is the resistivity?

Answer

Using the relationship

$\rho=R\frac{A}{l}$

Where $\rho$ is the resistivity, is the resistance, is the cross sectional area, and is the length from one face to the other.

The area of a circle is

$A=\pir^2$

Combining equations

$\rho=R\frac{\pir^2}{l}$

Plugging in values

$\rho=10.5\frac{\pi.0065^2}{.0475}$

$.0293 \Omegameters=\rho$

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Question

A sample of a new material is being tested for electrical resistivity. It is tall, with a radius of . Electrodes are placed on each face of the cylinder. Testing determines the resistance is $22 \Omega$ . What is the resistivity?

Answer

Use the relationship:

$\rho=\frac{RA}{l}$

Where $\rho$ is the resistivity, is the resistance, is the cross sectional area, and is the length from one face to the other.

The area of a circle is:

$A=\pir^2$

Combine these equations, convert to and plug in values:

$\rho=\frac{R\pir^2}{l}$

$\rho=\frac{22\pi.005^2}{.015}$

$\rho=.115 \Omega\cdot m$

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Question

A cylinder of an unknown material is being tested for physical properties. It is tall, with a radius of . Electrodes are placed on each face of the cylinder. The resistance is $.35 \Omega$ . What is the resistivity?

Answer

Use the relationship:

$\rho=\frac{RA}{l}$

Where $\rho$ is the resistivity, is the resistance, is the cross sectional area, and is the length from one face to the other.

The area of a circle is:

$A=\pir^2$

Combine equations, convert to and plug in values:

$\rho=\frac{R\pir^2}{l}$

$\rho=\frac{.35\pi.0065^2}{.055}$

$\rho=8.45*10^{-4} \Omega\cdot m$

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Question

A cylinder of a new material is under testing. It is tall, with a radius of . Electrodes are placed on each face of the cylinder. The resistance is $75 \Omega$ . What is the resistivity?

Answer

Use the relationship:

$\rho=\frac{RA}{l}$

Where $\rho$ is the resistivity, is the resistance, is the cross sectional area, and is the length from one face to the other.

The area of a circle is:

$A=\pir^2$

Combine equations, convert to and plug in values.

$\rho=\frac{R\pir^2}{l}$

$\rho=\frac{75\pi.002^2}{.02}$

$\rho=4.7*10^{-4}\Omega\cdot m$

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Question

A cylindrical resistor is long with a diameter of . Determine the resistivity of the material if the resistance is $50\Omega$ .

Answer

Use the following relationship:

$\rho=R\frac{A}{l}$

Where $\rho$ is the resistivity

is the resistance

is the length

is the cross sectional area

Convert and to and plug in values:

$\rho=50\frac{.002^2\pi}{.03}$

$.0209\Omega\cdot m=\rho$

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Question

Suppose that the resistance of an unknown conductive material is found to be $3: \Omega$ . Considering that the cross-sectional area of this material is $7.85\cdot10^{-7}:m^{2}$ and its length is $4.5\cdot 10^{-2}:m$ , what is the resistivity of this unknown material?

Answer

In this question, we're presented with the length and area of a given material, as well as the resistance of the material. We're being asked to find the resistivity of this material.

First, it's important to distinguish between resistance and resistivity. Resistance represents the impediment to the flow of charge that is caused by such factors as the length and cross-sectional area of the material. Resistivity is a value that is intrinsic to each material; changing the cross-sectional area or the length will not affect the resistivity.

To solve this problem, we'll need to utilize the equation that relates resistance and resistivity:

$R=\rho \frac{L}{A}$

From this equation, we can see that as the length of the conductive material increases, so too does the resistance of that material. However, as the area increases, the resistance decreases.

We can go ahead and rearrange this equation in order to isolate the resistivity term, $\rho$ .

$\rho=R\frac{A}{L}$

Next, we can plug in the values that we have in order to solve for our answer:

$\rho=3: \Omega(\frac{7.85\cdot10^{-7}: m^{2}}{4.5\cdot 10^{-2}: m})$

$\rho=5.23\cdot 10^{-5}: \Omega\cdot m$

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Question

A single resistor of radius and length is placed into a circuit with a single voltage source of . If the current is , determine the resistivity of the resistor.

Answer

Using

and

$R=\rho \frac{l}{A}$

Combining equations:

$V=I\rho \frac{l}{A}$

Solving for $\rho$ :

$\rho=\frac{VA}{Il}$

Converting and to and plugging in values:

$\rho=\frac{3.001^2\pi}{2.3.015}$

$\rho=2.73\Omega\cdot m$

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Question

A resistor has cross sectional area $2\textup{ mm}^2$ and length $1\textup{ cm}$ . When placed in series with a $3\textup{ V}$ battery, a current of $340 \textup{ mA}$ is produced. Determine the resistivity.

Answer

Using Ohm's law:

Converting to and plugging in values

Solving for resistance:

$R=8.82\Omega$

Using the equation for resistivity:

$\rho=R\frac{A}{l}$

Converting to and to and plugging in values:

$\rho=8.82\frac{210^{-6}}{.01}$

$\rho=1.76410^{-3}\Omega\cdot m$

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