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AP Physics 2 : Circuits

Study concepts, example questions & explanations for AP Physics 2

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Example Questions

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Example Question #1 : Resistivity

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 2.5 cm tall, with a radius of 1.5 cm .

You put electrodes on each face of the cylinder.

You determine the resistance to be 9.53 ohms .

What is the resistivity of "Type G?"

Possible Answers:

$1.9\times10^{-3}\Omega\cdot meters$

None of these

$1.3\times10^{3}\Omega\cdot meters$

$4.45\times10^{-3}\Omega\cdot meters$

$2.70\times10^{-3}\Omega\cdot meters$

Correct answer:

$2.70\times10^{-3}\Omega\cdot meters$

Explanation:

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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Example Question #2 : Resistivity

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 2.2 cm tall, with a radius of .2 cm .

You put electrodes on each face of the cylinder.

You determine the resistance to be .22 ohms .

What is the resistivity of "Type Z?"

Possible Answers:

$1.99\times 10^{-4} \Omega\cdot meters$

None of these

$1.26\times 10^{-4}\Omega\cdot meters$

$1.11\times 10^{-4}\Omega\cdot meters$

$1.75\times 10^{-4} \Omega\cdot meters$

Correct answer:

$1.26\times 10^{-4}\Omega\cdot meters$

Explanation:

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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Example Question #3 : Resistivity

A researcher is testing the electrical properties of a circuit that contains a resistor of unknown material. The resistor is a 3cm cylinder with a radius of 0.5cm . The researcher measures the resistance of the resistor to be $0.3\Omega$ . What is the resistivity of this unknown material?

Possible Answers:

$3.75\Omega\cdot m$

$1.6*10^{-2}\Omega\cdot m$

$7.85*10^{-4}\Omega\cdot m$

$8.75*10^{-3}\Omega\cdot m$

$9.1*10^{-4}\Omega\cdot m$

Correct answer:

$7.85*10^{-4}\Omega\cdot m$

Explanation:

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.

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

$\rho = 0.000785\Omega\cdot m$

$\rho = 7.85*10^{-4}\Omega\cdot m$

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Example Question #4 : Resistivity

A scientist carves out a cylinder of a new material she developed. It is 4.5 cm tall, with a radius of .60 cm . She put electrodes on each face of the cylinder. She determined the resistance to be $.75\Omega$ . What is the resistivity of this material?

Possible Answers:

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

$4.50 *10^{ -3} \Omega \cdot m$

$1.70 *10^{ -3} \Omega \cdot m$

None of these

$3.70 *10^{ -3} \Omega \cdot m$

Correct answer:

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

Explanation:

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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Example Question #5 : Resistivity

There is a sample of an unknown material under testing. It is 4.75 cm tall, with a radius of .65 cm . Electrodes are placed on each face of the cylinder. It is determined that the resistance to 10.5 ohms .

What is the resistivity?

Possible Answers:

$.0375 \Omega*meters$

$.0333 \Omega*meters$

$.0293 \Omega*meters$

None of these

$.0750 \Omega*meters$

Correct answer:

$.0293 \Omega*meters$

Explanation:

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

$A=\pi*r^2$

Combining equations

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

Plugging in values

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

$.0293 \Omega*meters=\rho$

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Example Question #871 : Ap Physics 2

A sample of a new material is being tested. It is 3.8 cm tall, with a radius of .38 cm . Electrodes are placec on each face of the cylinder. The resistance is determined to be 1.35 ohms .

What is the resistivity of this material?

Possible Answers:

$1.61*10^{-3}\Omega*meters$

$1.42*10^{-3}\Omega*meters$

$3.49*10^{-3}\Omega*meters$

None of these

$2.22*10^{-3}\Omega*meters$

Correct answer:

$1.61*10^{-3}\Omega*meters$

Explanation:

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

$A=\pi*r^2$

Combining equations

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

Plugging in values

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

$1.61*10^{-3}\Omega*meters=\rho$

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Example Question #52 : Circuits

A cylinder of an unknown material is being tested. It is 5.75 cm tall, with a radius of .95 cm . Electrodes are placed on each face of the cylinder. It is determined that the resistance is 59 ohms . What is the resistivity?

Possible Answers:

$.375 \Omega*meters$

$.762 \Omega*meters$

$.291\Omega*meters$

$.931\Omega*meters$

None of these

Correct answer:

$.291\Omega*meters$

Explanation:

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

$A=\pi*r^2$

Combining equations:

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

Plugging in values

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

$\rho=.291\Omega*meters$

Report an Error

Example Question #53 : Circuits

A cylinder of an unknown material is being tested for physical properties. It is 5.5 cm tall, with a radius of .65 cm . Electrodes are placed on each face of the cylinder. The resistance is $.35 \Omega$ . What is the resistivity?

Possible Answers:

None of these

$6.11*10^{-4} \Omega\cdot m$

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

$4.85*10^{-4}\Omega\cdot m$

$5.85*10^{-4} \Omega\cdot m$

Correct answer:

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

Explanation:

Use the relationship:

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

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:

$A=\pi*r^2$

Combine equations, convert to and plug in values:

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

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

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

Report an Error

Example Question #54 : Circuits

A cylinder of a new material is under testing. It is 2.0 cm tall, with a radius of .2 cm . Electrodes are placed on each face of the cylinder. The resistance is $75 \Omega$ . What is the resistivity?

Possible Answers:

None of these

$1.9*10^{-4}\Omega\cdot m$

$3.4*10^{-4} \Omega\cdot m$

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

$2.3*10^{-4}\Omega\cdot m$

Correct answer:

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

Explanation:

Use the relationship:

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

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:

$A=\pi*r^2$

Combine equations, convert to and plug in values.

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

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

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

Report an Error

Example Question #55 : Circuits

A sample of a new material is being tested for electrical resistivity. It is 1.5 cm tall, with a radius of .5 cm . Electrodes are placed on each face of the cylinder. Testing determines the resistance is $22 \Omega$ . What is the resistivity?

Possible Answers:

None of these

$1.15 \Omega\cdot m$

$.95 \Omega\cdot m$

$1.11 \Omega\cdot m$

$.67 \Omega\cdot m$

Correct answer:

None of these

Explanation:

Use the relationship:

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

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:

$A=\pi*r^2$

Combine these equations, convert to and plug in values:

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

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

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

Report an Error

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AP Physics 2 : Circuits

Study concepts, example questions & explanations for AP Physics 2

All AP Physics 2 Resources

Example Questions

Example Question #1 : Resistivity

Example Question #2 : Resistivity

Example Question #3 : Resistivity

Example Question #4 : Resistivity

Example Question #5 : Resistivity

Example Question #871 : Ap Physics 2

Example Question #52 : Circuits

Example Question #53 : Circuits

Example Question #54 : Circuits

Example Question #55 : Circuits

All AP Physics 2 Resources

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