AP Physics 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 .

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?

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.

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

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

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

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

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

Increasing the cross-sectional area of the cord

Decreasing the cross-sectional area of the cord

Increasing the length of the cord

Increasing the resistivity

Explanation

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.

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?

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

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

Student B's wire should be the same length 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 four times as long as 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:

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

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 .

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

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

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

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

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

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 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:

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

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

The resistivity will double

The resistivity will be quadrupled

The resistivity will be halved

The resistivity will not change

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.

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.

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:

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

Solve:

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

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 $10 \Omega$ ?

960m

38m

9.6m

10cm

2.6cm

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.

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

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

$0.2m\Omega\cdot m$

$0.8m\Omega\cdot m$

$80n\Omega\cdot m$

$20n\Omega\cdot m$

$80\mu\Omega\cdot m$

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 $\rho=(100\Omega )\frac{4mm^2(\frac{1m}{1000mm})^2}{2m}=.0002\Omega\cdot m=0.2m\Omega\cdot m$ .

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?

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

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

$\Omega$

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 $\frac{\pi}{4} d^{2}$ .

Resistivity

Help Questions

Explanation

Explanation

Explanation

Explanation

Explanation

Explanation

Explanation

Explanation

Explanation

Explanation