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AP Physics 2 : Circuit Properties

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Example Questions

Example Question #61 : Circuits

A cylindrical resistor is 3cm long with a diameter of 4mm . Determine the resistivity of the material if the resistance is $50\Omega$ .

Possible Answers:

$.805\Omega*m$

$.0714\Omega\cdot m$

$.0209\Omega\cdot m$

$.0626\Omega\cdot m$

$.567\Omega \cdot m$

Correct answer:

$.0209\Omega\cdot m$

Explanation:

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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Example Question #222 : Electricity And Magnetism

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?

Possible Answers:

$9.68\cdot 10^{-5}\: \Omega\cdot m$

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

$3.14\cdot 10^{-5}\: \Omega\cdot m$

$4.38\cdot 10^{-5}\: \Omega\cdot m$

$6.20\cdot 10^{-5}\: \Omega\cdot m$

Correct answer:

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

Explanation:

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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Example Question #62 : Circuit Properties

A single resistor of radius 1mm and length 1.5cm is placed into a circuit with a single voltage source of . If the current is 2.3A , determine the resistivity of the resistor.

Possible Answers:

$9.09*10^{-4}\Omega\cdot m$

None of these

$6.26*10^{-4}\Omega\cdot m$

$7.14*10^{-4}\Omega\cdot m$

$2.73*10^{-4}\Omega\cdot m$

Correct answer:

$2.73*10^{-4}\Omega\cdot m$

Explanation:

Using

V=IR

and

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

Combining equations:

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

Solving for $\rho$ :

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

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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Example Question #63 : Circuit Properties

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.

Possible Answers:

$8.221*10^{-3}\ \Omega\cdot \textup{m}$

None of these

$9.996*10^{-3}\ \Omega\cdot \textup{m}$

$1.764*10^{-3}\ \Omega\cdot \textup{m}$

$3.445*10^{-3}\ \Omega\cdot \textup{m}$

Correct answer:

$1.764*10^{-3}\ \Omega\cdot \textup{m}$

Explanation:

Using Ohm's law:

V=IR

Converting to and plugging in values

3=.340*R

Solving for resistance:

$R=8.82\Omega$

Using the equation for resistivity:

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

Converting mm^2 to and to and plugging in values:

$\rho=8.82\frac{2*10^{-6}}{.01}$

$\rho=1.764*10^{-3}\Omega\cdot m$

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Example Question #64 : Circuit Properties

A resistor has cross sectional area $16\textup{ mm}^2$ and length $2\textup{ cm}$ . When placed in series with a $3\textup{ V}$ battery, a current of $56.0\textup{ mA}$ is produced. Determine the resistivity.

Possible Answers:

None of the above

$\rho=1.213*10^{-3}\ \Omega\cdot \textup{m}$

$\rho=3.751*10^{-3}\ \Omega\cdot \textup{m}$

$\rho=1.764*10^{-3}\ \Omega\cdot \textup{m}$

$\rho=2.028*10^{-3}\ \Omega\cdot \textup{m}$

Correct answer:

$\rho=1.764*10^{-3}\ \Omega\cdot \textup{m}$

Explanation:

Using Ohm's law:

V=IR

Converting to and plugging in values

3=.340*R

Solving for resistance:

$R=8.82\Omega$

Using the equation for resistivity:

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

Converting mm^2 to and to and plugging in values:

$\rho=8.82\frac{2*10^{-6}}{.01}$

$\rho=1.764*10^{-3}\Omega*m$

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Example Question #65 : Circuit Properties

A resistor is made out of a material. The resistor has a cross-sectional area of 4mm^2 and a length of 8mm . It is found to have a resistance of $7\Omega$ . A new resistor is built that has a length of 6mm and a cross sectional area of 3.3mm^2 . Determine the resistance of the new resistor.

Possible Answers:

$3.95\Omega$

$6.36\Omega$

$7.77\Omega$

None of these

$4.05\Omega$

Correct answer:

$6.36\Omega$

Explanation:

Using

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

Plugging in values

$\rho=7*\frac{4}{8}$

$\rho=3.5\Omega*mm$

Using

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

Plugging in values

$R=3.5*\frac{6}{3.3}$

$R=6.36\Omega$

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Example Question #66 : Circuit Properties

identical resistors are placed in parallel. They are placed in a circuit with a battery. If the current through the battery is , determine the resistivity of each resistor. Each resistor is 3cm long and has a diameter of 1cm .

Possible Answers:

$\rho_1=\rho_2=\rho_3=.79\Omega\cdot cm$

$\rho_1=.79\Omega\cdot cm$

$\rho_2=\rho_3=1.50\Omega\cdot cm$

$\rho_1=\rho_2=\rho_3=2.27\Omega\cdot cm$

None of these

$\rho_1=\rho_2=\rho_3=1.50\Omega\cdot cm$

Correct answer:

$\rho_1=\rho_2=\rho_3=.79\Omega\cdot cm$

Explanation:

Since each resistor is in parallel, the voltage drop across each will be .

Since each resistor is identical, they all have the same resistance.

Using

$I_1+I_2+I_3=I_{total}=3A$

and

I_1=I_2=I_3

It is determined that

I_1=I_2=I_3=1A

Using

V=IR

3=1*R

$R=3\Omega$ for all three resistors

Using definition of resistivity:

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

$\rho=3*\frac{\pi*.5^2}{3}$

$\rho=.79\Omega\cdot cm$

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Example Question #21 : Resistivity

A resistor is made out of a material. The resistor has a cross-sectional area of 2mm^2 and a length of 3mm . It is found to have a resistance of $6\Omega$ . Determine the resistivity of the material.

Possible Answers:

$6\Omega\cdot mm$

$4\Omega\cdot mm$

None of these

$2\Omega\cdot mm$

$4\Omega\cdot mm$

Correct answer:

$4\Omega\cdot mm$

Explanation:

Use the equation for resistivity:

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

Plugging in values

$\rho=6*\frac{2}{3}$

$\rho=4\Omega\cdot mm$

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Example Question #68 : Circuit Properties

A resistor is made out of a material. The resistor has a cross-sectional area of 2mm^2 and a length of 3mm . It is found to have a resistance of $6\Omega$ . A new resistor is built that has a length of 4mm and a cross sectional area of 1.5mm^2 . Determine the resistance of the new resistor.

Possible Answers:

$4.17\Omega$

$4\Omega$

$10.7\Omega$

$6.37\Omega$

None of these

Correct answer:

$10.7\Omega$

Explanation:

Use the equation for resistivity:

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

Plugging in values

$\rho=6*\frac{2}{3}$

$\rho=4\Omega\cdot mm$

Using

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

Plugging in values

$R=4*\frac{4}{1.5}$

$R=10.7\Omega$

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Example Question #231 : Electricity And Magnetism

Possible Answers:

.89A

.23A

.11A

.56A

None of these

Correct answer:

.56A

Explanation:

Use the equation for resistivity:

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

Plugging in values

$\rho=6*\frac{2}{3}$

$\rho=4\Omega\cdot mm$

Rearrange the same equation and solve for resistance:

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

Plugging in values

$R=4*\frac{4}{1.5}$

$R=10.7\Omega$

Use Ohm's law to find current:

V=IR

6=I*10.7

I=.56A

Report an Error

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AP Physics 2 : Circuit Properties

Study concepts, example questions & explanations for AP Physics 2

All AP Physics 2 Resources

Example Questions

Example Question #61 : Circuits

Example Question #222 : Electricity And Magnetism

Example Question #62 : Circuit Properties

Example Question #63 : Circuit Properties

Example Question #64 : Circuit Properties

Example Question #65 : Circuit Properties

Example Question #66 : Circuit Properties

Example Question #21 : Resistivity

Example Question #68 : Circuit Properties

Example Question #231 : Electricity And Magnetism

All AP Physics 2 Resources

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