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AP Physics 2 : Electricity and Magnetism

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6 Diagnostic Tests 149 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

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:

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

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

None of these

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

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

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 #64 : Circuit Properties

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:

None of the above

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

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

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

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

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 #61 : Circuit Properties

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:

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

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

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

None of these

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

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 #66 : 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:

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

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

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

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

None of these

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

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 #231 : Electricity And Magnetism

Possible Answers:

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

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

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

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

None of these

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 #231 : Electricity And Magnetism

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:

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

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

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

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

None of these

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 #881 : Ap Physics 2

Possible Answers:

.11A $\displaystyle .11A$

.23A $\displaystyle .23A$

.89A $\displaystyle .89A$

.56A $\displaystyle .56A$

None of these

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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Example Question #71 : Circuits

A scientist is testing resistors made of the same material. Resistor $\displaystyle A$ has a cross sectional area of 3mm^2 $\displaystyle 3mm^2$ and a length of 3mm^2 $\displaystyle 3mm^2$. Resistor $\displaystyle B$ has a cross sectional area of 6mm^2 $\displaystyle 6mm^2$ and a length of 3mm^2 $\displaystyle 3mm^2$. How will the resistance of $\displaystyle B$ compare to that of $\displaystyle A$?

Possible Answers:

The resistance will quadruple

None of these

The resistance will double

The resistance will be cut in half

The resistance will stay the same

Correct answer:

The resistance will be cut in half

Explanation:

Use the equation for resistivity to visualize the proportions:

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

$\rho$ $\displaystyle \rho$ will stay constant as the material is the same. $\displaystyle A$ doubles and $\displaystyle l$ stays the same. Thus, the resistance, $\displaystyle R$, will get cut in half.

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Example Question #71 : Circuit Properties

Possible Answers:

$4.7\Omega\cdot mm$ $\displaystyle 4.7\Omega\cdot mm$

$6.5\Omega\cdot mm$ $\displaystyle 6.5\Omega\cdot mm$

$7.0\Omega\cdot mm$ $\displaystyle 7.0\Omega\cdot mm$

$10.3\Omega\cdot mm$ $\displaystyle 10.3\Omega\cdot mm$

$3.5\Omega\cdot mm$ $\displaystyle 3.5\Omega\cdot mm$

Correct answer:

$3.5\Omega\cdot mm$ $\displaystyle 3.5\Omega\cdot mm$

Explanation:

Using

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

Plugging in values

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

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

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Example Question #71 : Circuits

Possible Answers:

.94A $\displaystyle .94A$

None of these

3.25A $\displaystyle 3.25A$

1.94A $\displaystyle 1.94A$

1.25A $\displaystyle 1.25A$

Correct answer:

.94A $\displaystyle .94A$

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

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

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

Using

V=IR $\displaystyle V=IR$

Solving for $\displaystyle I$

$I=\frac{V}{R}$ $\displaystyle I=\frac{V}{R}$

$I=\frac{6}{6.36}$ $\displaystyle I=\frac{6}{6.36}$

I=.94A $\displaystyle I=.94A$

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AP Physics 2 : Electricity and Magnetism

Study concepts, example questions & explanations for AP Physics 2

All AP Physics 2 Resources

Example Questions

Example Question #231 : Electricity And Magnetism

Example Question #64 : Circuit Properties

Example Question #61 : Circuit Properties

Example Question #66 : Circuit Properties

Example Question #231 : Electricity And Magnetism

Example Question #231 : Electricity And Magnetism

Example Question #881 : Ap Physics 2

Example Question #71 : Circuits

Example Question #71 : Circuit Properties

Example Question #71 : Circuits

All AP Physics 2 Resources

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