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AP Physics 1 : Resistivity

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Example Questions

Example Question #1 : Resistivity

Consider the circuit:

Circuit_4

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

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

Possible Answers:

$0.77\Omega$ $\displaystyle 0.77\Omega$

$2.6\Omega$ $\displaystyle 2.6\Omega$

$0.29\Omega$ $\displaystyle 0.29\Omega$

$1.5\Omega$ $\displaystyle 1.5\Omega$

$0.54\Omega$ $\displaystyle 0.54\Omega$

Correct answer:

$0.77\Omega$ $\displaystyle 0.77\Omega$

Explanation:

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

$V = IR_{eq}$ $\displaystyle V = IR_{eq}$

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

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

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

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

$R3 = 0.77\Omega$ $\displaystyle R3 = 0.77\Omega$

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Example Question #1 : Resistivity

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

Possible Answers:

Increasing the resistivity

Increasing the length of the cord

Increasing the cross-sectional area of the cord

Decreasing the cross-sectional area of the cord

Correct answer:

Increasing the cross-sectional area of the cord

Explanation:

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

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

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Example Question #2 : Resistivity

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

Possible Answers:

960m

9.6m

2.6cm

38m

10cm

Correct answer:

960m

Explanation:

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

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

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

Plugging in our numbers gives the answer, 960m.

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

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Example Question #1 : Resistivity

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

Possible Answers:

$\displaystyle 2A$

$\displaystyle 1A$

0.25A $\displaystyle 0.25A$

1.5A $\displaystyle 1.5A$

$\displaystyle 5A$

Correct answer:

$\displaystyle 2A$

Explanation:

Resistors in serries add according to the formula:

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

Resistors in parallel add according to the formula:

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

We can find the total equivalent resistance:

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

$R_f=2.5\Omega$ $\displaystyle R_f=2.5\Omega$

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

V=IR $\displaystyle V=IR$

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

Solve:

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

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Example Question #3 : Resistivity

Basic circuit

$V_{in}=12V$ $\displaystyle V_{in}=12V$

$Z_1=3\Omega$ $\displaystyle Z_1=3\Omega$

In the circuit above, your aim is to limit the current to $\displaystyle 1A$, so you must design a resistor to serve in the place of Z_2 $\displaystyle Z_2$.

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

Possible Answers:

150mm $\displaystyle 150mm$

15mm $\displaystyle 15mm$

3mm $\displaystyle 3mm$

3000mm $\displaystyle 3000mm$

1mm $\displaystyle 1mm$

Correct answer:

3mm $\displaystyle 3mm$

Explanation:

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

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

Determine what the second resistance should be.

$Z_2=12\Omega-Z_1$ $\displaystyle Z_2=12\Omega-Z_1$

$Z_2=9\Omega$ $\displaystyle Z_2=9\Omega$

Find the necessary length of wire.

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

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

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

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

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Example Question #1 : Resistivity

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

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

Possible Answers:

$118.8\Omega$ $\displaystyle 118.8\Omega$

$37.7\Omega$ $\displaystyle 37.7\Omega$

$3.77\Omega$ $\displaystyle 3.77\Omega$

$5.94\Omega$ $\displaystyle 5.94\Omega$

$59.4\Omega$ $\displaystyle 59.4\Omega$

Correct answer:

$5.94\Omega$ $\displaystyle 5.94\Omega$

Explanation:

Resistance and resistivity are related as follows:

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

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

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

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

$R\approx 5.94\Omega$ $\displaystyle R\approx 5.94\Omega$

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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 twice as long as Student A's wire

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

Student B's wire should be $\frac{1}{4}$ $\displaystyle \frac{1}{4}$ 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 $\frac{1}{2}$ $\displaystyle \frac{1}{2}$ the length of Student A's wire

Correct answer:

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

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

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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 be quadrupled

The resistivity will double

The resistivity will be halved

The resistivity will not change

Correct answer:

The resistivity will double

Explanation:

Recall the formula for resistance $\displaystyle R$ is given by

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

Solve for resistivity:

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

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

$A \propto \rho$ $\displaystyle A \propto \rho$

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

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Example Question #2 : 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 $\displaystyle 1:2$

1:1 $\displaystyle 1:1$

4:1 $\displaystyle 4:1$

2:1 $\displaystyle 2:1$

Correct answer:

1:2 $\displaystyle 1:2$

Explanation:

Resistivity $\rho$ $\displaystyle \rho$ is given by:

$\rho=\frac{RA}{L}$ $\displaystyle \rho=\frac{RA}{L}$, where $\displaystyle R$ is the resistance, $\displaystyle A$ is the area, and $\displaystyle L$ 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 $\displaystyle 1:2$

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Example Question #1 : Resistivity

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

Possible Answers:

$\displaystyle .05$ $\Omega$ $\displaystyle \Omega$

6.9 $\displaystyle 6.9$ $\Omega$ $\displaystyle \Omega$

$\displaystyle 13$ $\Omega$ $\displaystyle \Omega$

$\displaystyle .23$ $\Omega$ $\displaystyle \Omega$

Correct answer:

$\displaystyle .23$ $\Omega$ $\displaystyle \Omega$

Explanation:

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

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AP Physics 1 : Resistivity

Study concepts, example questions & explanations for AP Physics 1

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