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AP Physics 1 : Circuits

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Example Questions

Example Question #3 : Resistivity

An electrician wishes to cut a copper wire $(\rho = 1.724 * 10^{-8}\Omega m)$ that has no more than $10 \Omega$ of resistance. The wire has a radius of 0.725mm. Approximately what length of wire has a resistance equal to the maximum $10 \Omega$ ?

Possible Answers:

2.6cm

960m

38m

9.6m

10cm

Correct answer:

960m

Explanation:

To relate resistance R, resistivity $\rho$ , area A, and length L we use the equation $R = \frac{\rho L}{A}$ .

Rearranging to isolate the quantity we wish to solve for, L, gives the equation $L = \frac{RA}{\rho }$ . We must first solve for A using the radius, 0.725mm.

$A = \pi r^{2} = \pi (0.000725m)^{2}=1.65 * 10^{-6}m^{2}$

Plugging in our numbers gives the answer, 960m.

$L = \frac{10\Omega * 1.65 * 10^{-6}m^{2}}{1.724*10^{-8}\Omega m} = 957m \approx960m$

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Example Question #1 : Resistivity

What is the current in a circuit with a $2\Omega$ resistor followed by a $3\Omega$ resistor that are both in parallel with a $5\Omega$ resistor? The voltage supplied to the circuit is 5V.

Possible Answers:

1.5A

0.25A

Correct answer:

Explanation:

Resistors in serries add according to the formula:

$R_{f}=R_{1}+R_{2}$

Resistors in parallel add according to the formula:

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

We can find the total equivalent resistance:

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

$R_f=2.5\Omega$

Now we can use Ohm's law to find the current:

V=IR

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

Solve:

$I = \frac{5 V}{2.5 \Omega }=2 A$

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Example Question #1 : Resistivity

Basic circuit

$V_{in}=12V$

$Z_1=3\Omega$

In the circuit above, your aim is to limit the current to , so you must design a resistor to serve in the place of Z_2 .

To make this resistor, you have a spool of mystery metal, which has a cross sectional area of 1mm^2 and a resistivity of, $\rho = 3\cdot 10^{-3}\Omega m$ . What length of wire should you cut?

Possible Answers:

15mm

150mm

3mm

1mm

3000mm

Correct answer:

3mm

Explanation:

First, find out how much total resistance should be in the circuit in order to get the desired current:

$R_{tot}=\frac{V}{I}=\frac{12V}{1A}=12\Omega$

Determine what the second resistance should be.

$Z_2=12\Omega-Z_1$

$Z_2=9\Omega$

Find the necessary length of wire.

$Z_2=\frac{\rho L }{A}$

$L=\frac{Z_2A}{\rho}$

$L=\frac{(9\Omega)(1mm^2)}{3\cdot 10^{-3}\Omega m}=\frac{(9\Omega)(10^{-6}m^2)}{3\cdot 10^{-3}\Omega m}$

$L=3\cdot 10^{-3}m=3mm$

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Example Question #1 : Resistivity

What is the resistance of a 100m length of round copper wire with a radius of 0.3mm ?

$\rho_{copper}=1.68\cdot 10^{-8}\Omega m$

Possible Answers:

$118.8\Omega$

$37.7\Omega$

$3.77\Omega$

$5.94\Omega$

$59.4\Omega$

Correct answer:

$5.94\Omega$

Explanation:

Resistance and resistivity are related as follows:

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

$A={\pi}(0.0003m)^2$

$A\approx 2.83\cdot 10^{-7}$

$R\approx \frac{1.68\cdot10^{-8}\Omega m(100m)}{2.83\cdot10^{-7}m^2}$

$R\approx 5.94\Omega$

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Example Question #2 : Resistivity

Two students are performing a lab using lengths of wire as resistors. The two students have wires made of the exact same material, but Student B has a wire the has twice the radius of Student A's wire. If Student B wants his wire to have the same resistance as Student A's wire, how should Student B's wire length compare to Student A's wire?

Possible Answers:

Student B's wire should be $\frac{1}{2}$ the length of Student A's wire

Student B's wire should be four times as long as Student A's wire

Student B's wire should be twice as long as Student A's wire

Student B's wire should be $\frac{1}{4}$ the length of Student A's wire

Student B's wire should be the same length as Student A's wire

Correct answer:

Student B's wire should be $\frac{1}{4}$ the length of Student A's wire

Explanation:

Resistance is proportional to length, and inversely proportional to cross-sectional area. Area depends on the square of the radius: $A= \pi r^{2}$ , so Student B's wire has $2^{2}=4$ times the cross-sectional area of Student A's wire. In order to compensate for the increased area, Student B must make his wire $\frac{1}{4}$ the length of Student A's wire. This can be shown mathematically using the equation for resistance:

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

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Example Question #2 : Resistivity

By how much will resistivity change if resistance and length are constant, and cross sectional area is doubled?

Possible Answers:

The resistivity will not change

The resistivity will double

The resistivity will be halved

The resistivity will be quadrupled

Correct answer:

The resistivity will double

Explanation:

Recall the formula for resistance is given by

$R=\frac{\rho L}{A}$ , where is the cross sectional area, $\rho$ is resistivity, and is length.

Solve for resistivity:

$\frac{AR}{L}=\rho$

From this, we can tell that resistivity is proportional to cross sectional area by:

$A \propto \rho$

Since is doubled, and resistance and length are constant, resistivity $\rho$ will also be doubled.

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Example Question #1 : Resistivity

Ratio is given by:

Resistivity of first resistor: Resistivity of second resistor

What is the ratio of resistivity of 2 resistors with identical resistances and area, where the first resistor is twice the length of the second resistor?

Possible Answers:

1:2

2:1

1:1

4:1

Correct answer:

1:2

Explanation:

Resistivity $\rho$ is given by:

$\rho=\frac{RA}{L}$ , where is the resistance, is the area, and is the length.

Since both have identical resistances and area, the first resistor will have half the resistivity since it has twice the length. Therefore the resistivity relation is

1:2

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Example Question #1 : Resistivity

What is the resistance of a copper rod with resistivity of $1.78\cdot 10^{-8} \Omega m$ , diameter of 10mm , and length of 1000m ?

Possible Answers:

$\Omega$

6.9 $\Omega$

$\Omega$

Correct answer:

$\Omega$

Explanation:

The equation for resistance is as follows: $R=\frac{\rho l}{A}$ . Where $\rho$ is resistivity, is the length of the wire, and is the cross section of the wire which can be found using $\frac{\pi}{4} d^{2}$ .

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Example Question #11 : Resistivity

You have a very long wire connected to an electric station. Even though you are suppling 120V from the source, by the time it reaches the station, there is a loss of voltage. The wire is 100 meters long.

If 100V reaches the power station, what is the resistivity of the wire? Assume a current of .

Possible Answers:

$1.45\frac{\Omega}{m}$

$0.10\frac{\Omega}{m}$

$2.2\frac{\Omega}{m}$

$0.55\frac{\Omega}{m}$

$0.15\frac{\Omega}{m}$

Correct answer:

$0.10\frac{\Omega}{m}$

Explanation:

The voltage drop from the source to the station (the "load") indicates that there is an internal resistance in the wire. According to the voltage law, the total amount of voltage drop is equal to the total amount of voltage supplied. Since 100V was supplied, and 100V drops at the station, that means that 20V drops along the wire.

$V_{source}=V_{wire}+V_{station}$

$V_{wire}=V_{source}-V_{station}$

$V_{wire}=120V-100V=20V$

Now that the voltage drop across the wire is known, Ohm's law will give the resistance of the wire:

$V=IR\rightarrow R=\frac{V}{I}=\frac{20V}{2A}=10\Omega$

$R=10\Omega$

The resistivity of the wire is equal to the resistance per unit length, therefore, in order to find resistivity you divide the total resistance by the length:

$\rho=\frac{R}{L}=\frac{10\Omega}{100m}=0.10\frac{ \Omega}{m}$

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Example Question #111 : Circuits

Which of the following actions would decrease the resistance of a wire by a factor of ?

Possible Answers:

Halving the cross-sectional area of the wire

Tripling the length of the wire

Halving the length of the wire

Doubling the cross-sectional area of the wire

Tripling the cross-sectional area of the wire

Correct answer:

Doubling the cross-sectional area of the wire

Explanation:

The equation for resistance is as follows:

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

Where is the resistance, $\rho$ is the resistivity of the material, is the length of the material and is the cross-sectional area of the material. Looking at this equation, by doubling the area we effectively reduce the resistance by a factor of two.

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AP Physics 1 : Circuits

Study concepts, example questions & explanations for AP Physics 1

All AP Physics 1 Resources

Example Questions

Example Question #3 : Resistivity

Example Question #1 : Resistivity

Example Question #1 : Resistivity

Example Question #1 : Resistivity

Example Question #2 : Resistivity

Example Question #2 : Resistivity

Example Question #1 : Resistivity

Example Question #1 : Resistivity

Example Question #11 : Resistivity

Example Question #111 : Circuits

All AP Physics 1 Resources

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