AP Physics 2 - 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.

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

None of the above

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

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

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

Explanation

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:

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

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

You have 4 resistors, , , , and , set up like this:

4resistorcircuit

Their resistance are as follows:

$\begin{align*} A&=1\Omega \ B&=3\Omega \ C&=4\Omega \ D&= 2\Omega \end{align*}$

If the battery has 8V, what is the total power dissipated through the resistors?

Explanation

The equation for power is

In order to get the power, we need the current. To find the current, we need to get the total resistance, and use Ohm's Law ().

To find the total resistance, remember the equations for adding resistors is this:

$(Series)\ R_{tot}&=R_1+R_2+...+R_n$

$(Parallel)\ \frac{1}{R_{tot}}&=\frac{1}{R_1}+\frac{1}{R_2}=...=\frac{1}{R_n}$

Resistors and are in series, resistors and are in parallel, and resistors and are in series.

$R_{AB}&=1+3$

$R_{AB}= 4$

$R_{ABC}&=(\frac{1}{R_{AB}}+\frac{1}{R_C})^{-1}$

$R_{ABC}= (\frac{1}{4}+\frac{1}{4})^{-1}$

$R_{ABC}= 2$

$R_{total}&=R_{ABC}+R_{D}$

$R_{total}= 2+2$

$R_{total}= 4$

Now, we can find the current.

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

$I=\frac{8}{4}$

Finally, we can find the power.

Therefore, the power is 16W (watts).

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

None of the above

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

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

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

Explanation

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:

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

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

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?

$.291\Omega*meters$

None of these

$.375 \Omega*meters$

$.931\Omega*meters$

$.762 \Omega*meters$

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

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

$\rho=.291\Omega*meters$

3 sets of parallel resistors

$R_1=R_4=1\Omega$

$R_2=R_5= 2\Omega$

$R_3=R_6 = 3\Omega$

Determine the total resistance of the given circuit.

$2.62\Omega$

$3\Omega$

$12\Omega$

$3.67\Omega$

Explanation

In order to find the total resistance of the circuit, we need to combine all of the parallel resistors first, then add them together as resistors in series.

Combine with :

$\frac{1}{R_1}+\frac{1}{R_2}=\frac{1}{R_{1,2}}$

$\frac{1}{1}+\frac{1}{2}=\frac{1}{R_{1,2}}$

$R_{1,2}=\frac{2}{3}\Omega$

Combine with :

$\frac{1}{R_3}+\frac{1}{R_4}=\frac{1}{R_{3,4}}$

$\frac{1}{3}+\frac{1}{1}=\frac{1}{R_{3,4}}$

$R_{3,4}=\frac{3}{4}\Omega$

Combine with :

$\frac{1}{R_5}+\frac{1}{R_6}=\frac{1}{R_{5,6}}$

$\frac{1}{2}+\frac{1}{3}=\frac{1}{R_{5,6}}$

$R_{5,6}=\frac{6}{5}\Omega$

Then, add the combined resistors, which are now all in series:

$R_{1,2}+R_{3,4}+R_{5,6}=R_{Total}$

$R_{Total}=\frac{2}{3}+\frac{3}{4}+\frac{6}{5}=2.62\Omega$

3 sets of parallel resistors

$R_1=R_4=1\Omega$

$R_2=R_5=2\Omega$

$R_3=R_6= 3\Omega$

Determine the voltage drop across

None of these

Explanation

The first step is to find the total resistance of the circuit.

In order to find the total resistance of the circuit, it is required to combine all of the parallel resistors first, then add them together as resistors in series.

Combining with , with , with .

$\frac{1}{R_1}+\frac{1}{R_2}=\frac{1}{R_{1,2}}$

$\frac{1}{R_3}+\frac{1}{R_4}=\frac{1}{R_{3,4}}$

$\frac{1}{R_5}+\frac{1}{R_6}=\frac{1}{R_{5,6}}$

Then, combining $R_{1,2}$ with $R_{3,4}$ and $R_{5,6}$ :

$R_{1,2}+R_{3,4}+R_{5,6}=R_{Total}$

$R_{1,2}=.667 \Omega$

$R_{3,4}=.75 \Omega$

$R_{5,6}=1.2 \Omega$

$R_{Total}= 2.62\Omega$

Ohms is used law to determine the total current of the circuit

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

Combing all voltage sources for the total voltage.

Plugging in given values,

$\frac{9V}{2.62\Omega}=I$

The voltage drop across parallel resistors must be the same, so:

$V_{R1}=V_{R2}$

Using ohms law:

It is also true that:

$I_1+I_2=I_{Total}$

Using Subsitution

$I_1(1+\frac{R_1}{R_2})=I_{Total}$

Solving for :

$\frac{I_{Total}}{(1+\frac{R_1}{R_2})}=I_1$

$\frac{3.43}{(1+\frac{1}{2})}=I_1$

Plugging back into ohms law in order to find the voltage drop.

3 sets of parallel resistors

$R_1=R_4=1\Omega$

$R_2=R_5= 2\Omega$

$R_3=R_6 = 3\Omega$

Determine the total resistance of the given circuit.

$2.62\Omega$

$3\Omega$

$12\Omega$

$3.67\Omega$

Explanation

In order to find the total resistance of the circuit, we need to combine all of the parallel resistors first, then add them together as resistors in series.

Combine with :

$\frac{1}{R_1}+\frac{1}{R_2}=\frac{1}{R_{1,2}}$

$\frac{1}{1}+\frac{1}{2}=\frac{1}{R_{1,2}}$

$R_{1,2}=\frac{2}{3}\Omega$

Combine with :

$\frac{1}{R_3}+\frac{1}{R_4}=\frac{1}{R_{3,4}}$

$\frac{1}{3}+\frac{1}{1}=\frac{1}{R_{3,4}}$

$R_{3,4}=\frac{3}{4}\Omega$

Combine with :

$\frac{1}{R_5}+\frac{1}{R_6}=\frac{1}{R_{5,6}}$

$\frac{1}{2}+\frac{1}{3}=\frac{1}{R_{5,6}}$

$R_{5,6}=\frac{6}{5}\Omega$

Then, add the combined resistors, which are now all in series:

$R_{1,2}+R_{3,4}+R_{5,6}=R_{Total}$

$R_{Total}=\frac{2}{3}+\frac{3}{4}+\frac{6}{5}=2.62\Omega$

You have 4 resistors, , , , and , set up like this:

4resistorcircuit

Their resistance are as follows:

$\begin{align*} A&=1\Omega \ B&=3\Omega \ C&=4\Omega \ D&= 2\Omega \end{align*}$

If the battery has 8V, what is the total power dissipated through the resistors?

Explanation

The equation for power is

In order to get the power, we need the current. To find the current, we need to get the total resistance, and use Ohm's Law ().

To find the total resistance, remember the equations for adding resistors is this:

$(Series)\ R_{tot}&=R_1+R_2+...+R_n$

$(Parallel)\ \frac{1}{R_{tot}}&=\frac{1}{R_1}+\frac{1}{R_2}=...=\frac{1}{R_n}$

Resistors and are in series, resistors and are in parallel, and resistors and are in series.

$R_{AB}&=1+3$

$R_{AB}= 4$

$R_{ABC}&=(\frac{1}{R_{AB}}+\frac{1}{R_C})^{-1}$

$R_{ABC}= (\frac{1}{4}+\frac{1}{4})^{-1}$

$R_{ABC}= 2$

$R_{total}&=R_{ABC}+R_{D}$

$R_{total}= 2+2$

$R_{total}= 4$

Now, we can find the current.

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

$I=\frac{8}{4}$

Finally, we can find the power.

Therefore, the power is 16W (watts).

3 sets of parallel resistors

$R_1=R_4=1\Omega$

$R_2=R_5=2\Omega$

$R_3=R_6= 3\Omega$

Determine the voltage drop across

None of these

Explanation

The first step is to find the total resistance of the circuit.

In order to find the total resistance of the circuit, it is required to combine all of the parallel resistors first, then add them together as resistors in series.

Combining with , with , with .

$\frac{1}{R_1}+\frac{1}{R_2}=\frac{1}{R_{1,2}}$

$\frac{1}{R_3}+\frac{1}{R_4}=\frac{1}{R_{3,4}}$

$\frac{1}{R_5}+\frac{1}{R_6}=\frac{1}{R_{5,6}}$

Then, combining $R_{1,2}$ with $R_{3,4}$ and $R_{5,6}$ :

$R_{1,2}+R_{3,4}+R_{5,6}=R_{Total}$

$R_{1,2}=.667 \Omega$

$R_{3,4}=.75 \Omega$

$R_{5,6}=1.2 \Omega$

$R_{Total}= 2.62\Omega$

Ohms is used law to determine the total current of the circuit

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

Combing all voltage sources for the total voltage.

Plugging in given values,

$\frac{9V}{2.62\Omega}=I$

The voltage drop across parallel resistors must be the same, so:

$V_{R1}=V_{R2}$

Using ohms law:

It is also true that:

$I_1+I_2=I_{Total}$

Using Subsitution

$I_1(1+\frac{R_1}{R_2})=I_{Total}$

Solving for :

$\frac{I_{Total}}{(1+\frac{R_1}{R_2})}=I_1$

$\frac{3.43}{(1+\frac{1}{2})}=I_1$

Plugging back into ohms law in order to find the voltage drop.

$.291\Omega*meters$

None of these

$.375 \Omega*meters$

$.931\Omega*meters$

$.762 \Omega*meters$

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

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

$\rho=.291\Omega*meters$

Circuit Properties

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