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AP Physics 1 : Resistivity

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Example Question #1 : Resistivity

Consider the circuit:

Circuit_4

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

The current flowing through the entire circuit is 40A . What is the value of R3?

Possible Answers:

$0.77\Omega$

$1.5\Omega$

$0.29\Omega$

$2.6\Omega$

$0.54\Omega$

Correct answer:

$0.77\Omega$

Explanation:

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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Example Question #1 : Resistivity

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

Possible Answers:

Decreasing the cross-sectional area of the cord

Increasing the length of the cord

Increasing the cross-sectional area of the cord

Increasing the resistivity

Correct answer:

Increasing the cross-sectional area of the cord

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.

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Example Question #2 : 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 #2 : 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:

3mm

1mm

150mm

15mm

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 #3 : 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 the same length 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 $\frac{1}{2}$ the length of 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

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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AP Physics 1 : Resistivity

Study concepts, example questions & explanations for AP Physics 1

All AP Physics 1 Resources

Example Questions

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 #3 : Resistivity

Example Question #2 : Resistivity

Example Question #1 : Resistivity

Example Question #1 : Resistivity

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