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AP Physics 1 : Resistivity

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Example Questions

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 #11 : Resistivity

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

Possible Answers:

Tripling the cross-sectional area of the wire

Halving the cross-sectional area of the wire

Doubling the cross-sectional area of the wire

Tripling the length of the wire

Halving the length 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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Example Question #1381 : Ap Physics 1

Find the resistivity of a cylindrical wire with resistance $100\Omega$ , length , and cross sectional area of 4 mm^2 .

Possible Answers:

$20n\Omega\cdot m$

$0.8m\Omega\cdot m$

$0.2m\Omega\cdot m$

$80\mu\Omega\cdot m$

$80n\Omega\cdot m$

Correct answer:

$0.2m\Omega\cdot m$

Explanation:

There exist a formula that directly relates resistance and resistivity. The formula is $R=\rho\frac{L}{A}$ .

is resistance, $\rho$ is resistivity, is length, and is cross sectional area. Solving for $\rho$ , we get $\rho=R\frac{A}{L}$ . Plugging in our givens, we get $\rho=(100\Omega )\frac{4mm^2(\frac{1m}{1000mm})^2}{2m}=.0002\Omega\cdot m=0.2m\Omega\cdot m$ .

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AP Physics 1 : Resistivity

Study concepts, example questions & explanations for AP Physics 1

All AP Physics 1 Resources

Example Questions

Example Question #11 : Resistivity

Example Question #11 : Resistivity

Example Question #1381 : Ap Physics 1

All AP Physics 1 Resources

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