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AP Physics 2 : Circuits

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Example Questions

Example Question #871 : Ap Physics 2

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

Possible Answers:

$.0714\Omega\cdot m$ $\displaystyle .0714\Omega\cdot m$

$.805\Omega*m$ $\displaystyle .805\Omega*m$

$.567\Omega \cdot m$ $\displaystyle .567\Omega \cdot m$

$.0626\Omega\cdot m$ $\displaystyle .0626\Omega\cdot m$

$.0209\Omega\cdot m$ $\displaystyle .0209\Omega\cdot m$

Correct answer:

$.0209\Omega\cdot m$ $\displaystyle .0209\Omega\cdot m$

Explanation:

Use the following relationship:

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

Where $\rho$ $\displaystyle \rho$ is the resistivity

$\displaystyle R$ is the resistance

$\displaystyle l$ is the length

$\displaystyle A$ is the cross sectional area

Convert $\displaystyle cm$ and $\displaystyle mm$ to $\displaystyle m$ and plug in values:

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

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

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Example Question #872 : Ap Physics 2

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

Possible Answers:

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

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

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

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

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

Correct answer:

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

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:

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

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, $\rho$ $\displaystyle \rho$.

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

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

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

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

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Example Question #873 : Ap Physics 2

A single resistor of radius 1mm $\displaystyle 1mm$ and length 1.5cm $\displaystyle 1.5cm$ is placed into a circuit with a single voltage source of $\displaystyle 3V$. If the current is 2.3A $\displaystyle 2.3A$, determine the resistivity of the resistor.

Possible Answers:

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

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

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

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

None of these

Correct answer:

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

Explanation:

Using

V=IR $\displaystyle V=IR$

and

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

Combining equations:

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

Solving for $\rho$ $\displaystyle \rho$:

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

Converting $\displaystyle mm$ and $\displaystyle cm$ to $\displaystyle m$ and plugging in values:

$\rho=\frac{3*.001^2*\pi}{2.3*.015}$ $\displaystyle \rho=\frac{3*.001^2*\pi}{2.3*.015}$

$\rho=2.73\Omega\cdot m$ $\displaystyle \rho=2.73\Omega\cdot m$

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Example Question #231 : Electricity And Magnetism

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

Possible Answers:

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

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

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

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

None of these

Correct answer:

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

Explanation:

Using Ohm's law:

V=IR $\displaystyle V=IR$

Converting $\displaystyle mA$ to $\displaystyle A$ and plugging in values

$\displaystyle 3=.340*R$

Solving for resistance:

$R=8.82\Omega$ $\displaystyle R=8.82\Omega$

Using the equation for resistivity:

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

Converting mm^2 $\displaystyle mm^2$ to $\displaystyle m$ and $\displaystyle cm$ to $\displaystyle m$ and plugging in values:

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

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

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Example Question #232 : Electricity And Magnetism

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

Possible Answers:

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

None of the above

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

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

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

Correct answer:

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

Explanation:

Using Ohm's law:

V=IR $\displaystyle V=IR$

Converting $\displaystyle mA$ to $\displaystyle A$ and plugging in values

$\displaystyle 3=.340*R$

Solving for resistance:

$R=8.82\Omega$ $\displaystyle R=8.82\Omega$

Using the equation for resistivity:

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

Converting mm^2 $\displaystyle mm^2$ to $\displaystyle m$ and $\displaystyle cm$ to $\displaystyle m$ and plugging in values:

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

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

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Example Question #233 : Electricity And Magnetism

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

Possible Answers:

$7.77\Omega$ $\displaystyle 7.77\Omega$

None of these

$4.05\Omega$ $\displaystyle 4.05\Omega$

$3.95\Omega$ $\displaystyle 3.95\Omega$

$6.36\Omega$ $\displaystyle 6.36\Omega$

Correct answer:

$6.36\Omega$ $\displaystyle 6.36\Omega$

Explanation:

Using

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

Plugging in values

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

$\rho=3.5\Omega*mm$ $\displaystyle \rho=3.5\Omega*mm$

Using

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

Plugging in values

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

$R=6.36\Omega$ $\displaystyle R=6.36\Omega$

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Example Question #61 : Circuit Properties

$\displaystyle 3$ identical resistors are placed in parallel. They are placed in a circuit with a $\displaystyle 4V$ battery. If the current through the battery is $\displaystyle 3A$, determine the resistivity of each resistor. Each resistor is 3cm $\displaystyle 3cm$ long and has a diameter of 1cm $\displaystyle 1cm$.

Possible Answers:

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

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

None of these

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

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

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

Correct answer:

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

Explanation:

Since each resistor is in parallel, the voltage drop across each will be $\displaystyle 4V$.

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

Using

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

and

$\displaystyle I_1=I_2=I_3$

It is determined that

$\displaystyle I_1=I_2=I_3=1A$

Using

V=IR $\displaystyle V=IR$

3=1*R $\displaystyle 3=1*R$

$R=3\Omega$ $\displaystyle R=3\Omega$ for all three resistors

Using definition of resistivity:

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

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

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

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Example Question #235 : Electricity And Magnetism

Possible Answers:

$6\Omega\cdot mm$ $\displaystyle 6\Omega\cdot mm$

$2\Omega\cdot mm$ $\displaystyle 2\Omega\cdot mm$

$4\Omega\cdot mm$ $\displaystyle 4\Omega\cdot mm$

None of these

$4\Omega\cdot mm$ $\displaystyle 4\Omega\cdot mm$

Correct answer:

$4\Omega\cdot mm$ $\displaystyle 4\Omega\cdot mm$

Explanation:

Use the equation for resistivity:

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

Plugging in values

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

$\rho=4\Omega\cdot mm$ $\displaystyle \rho=4\Omega\cdot mm$

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Example Question #61 : Circuit Properties

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

Possible Answers:

$4.17\Omega$ $\displaystyle 4.17\Omega$

$6.37\Omega$ $\displaystyle 6.37\Omega$

None of these

$10.7\Omega$ $\displaystyle 10.7\Omega$

$4\Omega$ $\displaystyle 4\Omega$

Correct answer:

$10.7\Omega$ $\displaystyle 10.7\Omega$

Explanation:

Use the equation for resistivity:

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

Plugging in values

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

$\rho=4\Omega\cdot mm$ $\displaystyle \rho=4\Omega\cdot mm$

Using

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

Plugging in values

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

$R=10.7\Omega$ $\displaystyle R=10.7\Omega$

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Example Question #237 : Electricity And Magnetism

Possible Answers:

.11A $\displaystyle .11A$

.89A $\displaystyle .89A$

.23A $\displaystyle .23A$

None of these

.56A $\displaystyle .56A$

Correct answer:

.56A $\displaystyle .56A$

Explanation:

Use the equation for resistivity:

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

Plugging in values

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

$\rho=4\Omega\cdot mm$ $\displaystyle \rho=4\Omega\cdot mm$

Rearrange the same equation and solve for resistance:

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

Plugging in values

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

$R=10.7\Omega$ $\displaystyle R=10.7\Omega$

Use Ohm's law to find current:

V=IR $\displaystyle V=IR$

$\displaystyle 6=I*10.7$

I=.56A $\displaystyle I=.56A$

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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 #871 : Ap Physics 2

Example Question #872 : Ap Physics 2

Example Question #873 : Ap Physics 2

Example Question #231 : Electricity And Magnetism

Example Question #232 : Electricity And Magnetism

Example Question #233 : Electricity And Magnetism

Example Question #61 : Circuit Properties

Example Question #235 : Electricity And Magnetism

Example Question #61 : Circuit Properties

Example Question #237 : Electricity And Magnetism

All AP Physics 2 Resources

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