Resistivity - AP Physics 1

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Question

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 $10 \Omega$ ?

Answer

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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Question

Which of the following factors would decrease the resistance through an electrical cord?

Answer

The equation for resistance is given by $R = \frac{\rho L}{A}$ .

From this equation, we can see the best way to decrease resistance is by increasing the cross-sectional area, , of the cord. Increasing the length, , of the cord or the resistivity, $\rho$ , will increase the resistance.

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Question

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.

Answer

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:

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

Solve:

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

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Question

Consider the circuit:

$R1=2\Omega,\ R2=4\Omega,\ R3=R4$

The current flowing through the entire circuit is . What is the value of R3?

Answer

We know the voltage and total current of the circuit, so we can calculate an equivalent resistance using Ohm's law:

$V = IR_{eq}$

$R_{eq}=\frac{V}{I} = \frac{12V}{40A}= 0.3 \Omega$

Now we can use the expression for condensing parallel resistors to calculate R3:

$\frac{1}{R_{eq}}= \sum\frac{1}{R}= \frac{1}{R1}+\frac{1}{R2}+\frac{1}{R3}+\frac{1}{R4}$

$\frac{1}{0.3\Omega} = \frac{1}{2\Omega}+\frac{1}{4\Omega} +\frac{1}{R3}+\frac{1}{R4}=\frac{1}{2\Omega}+\frac{1}{4\Omega} +\frac{2}{R3}$

$R3 = 0.77\Omega$

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Question

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

Answer

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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Question

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?

Answer

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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Question

$V_{in}=12V$

$Z_1=3\Omega$

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

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

Answer

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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Question

What is the resistance of a length of round copper wire with a radius of ?

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

Answer

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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Question

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

Answer

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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Question

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

Answer

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 was supplied, and drops at the station, that means that 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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Question

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

Answer

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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Question

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?

Answer

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

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Question

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

Answer

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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