PHY 2053L Physics 2 lab - Miami Dade College, Miami. Factors Affecting Electrical Resistance Experim

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Jennifer Bahamonde Shakira Gonzalez Alvaro Lopez Jessica Lopez September 23, 2019

Factors Affecting Electrical Resistance INTRODUCTION: Metals are generally excellent conductors of heat and electricity. Because of this, we tend to neglect any resistance in wires when we examine the behavior of ordinary loads (bulbs, resistors, motors, etc.) in electrical circuits.

But metals are not perfect conductors, otherwise the wires in our toasters would not get hot. In this experiment, you will examine the factors that affect the electrical resistance of metals.

R = ρ (L/A) OBJECTIVE: The purpose of this experiment is to collect the potential difference and current data for various metal rods and determine the electrical resistance for each of the readings for the rods. Then determine how the length, crosssectional area, and type of material affect the resistance of the metal rods.

MATERIALS:        

Vernier data-collection interface Vernier Resistivity Rod Set Conventional power supply Logger Pro or LabQuest App Vernier Current Probe Ruler in centimeters Vernier Instrumentation Amplifier clip leads

a. Lab Set-Up

PRELIMANARY QUESTIONS: 1. The wires of a toaster carry electrical current between the electrical outlet and the toaster. The wires inside the toaster also carry electrical current.

During its operation, the wires inside the toaster get hot enough to glow, yet the wires to the outlet do not. List some factors that might determine how hot a wire will get when it conducts electricity. 

The metal used in the inside of toasters tends to slow down current flow by adding resistance. This increase in resistance is what cause and increase in temperature of the metal.

2. Which one(s) of the following factors affects Electrical resistance and resistivity? Explain  The wire length:  This will affect the electrical resistance. The electrical resistance is directly proportional to the wire length.  The wire cross-section:  This will decrease resistance, the larger the cross-section area of a wire the less interaction there will be between electrons.  The wire shapes:  N/A PROCEDURE: Setup using conventional power supply and Lab-Pro 1. Connect a Vernier Current Probe to the interface. Connect the current probe in series to the negative terminal on the power supply and to one end of the brass rod (see Figure 3).

2. Connect an Instrumentation Amplifier to the interface. Set the range to 0–20 mV. Use a lead to connect the black post on the instrumentation amplifier to ground (usually the black terminal on the power supply).

Start the data-collection program.

3. Connect clip leads to the red and black terminals of the instrumentation amplifier; these will act as the probe leads for your measurements of potential. Connect the black lead to a position just inside the lead carrying current to the rod. Note this as the zero position. Touch the other lead (hereafter referred to as the red lead) to the black lead on the rod and zero both sensors (see Figure 3).

4. Your instructor will provide specific instructions that will enable you to run a DC current of approximately 0.5 A through the rod.

Part 1 Effect of length 1. Change the data-collection mode to Events with Entry. Enter length as the Name and m as the Units.

2. Connect the red lead right next to the black lead at the zero position.

Click or tap Keep. Enter 0 for the length.

3. Move the red lead 5 cm from the black lead at the zero position. Click or tap Keep. Enter 5 for the length.

4. Move the red lead 5 cm farther from the zero position. Click or tap Keep and enter the new length.

5. Repeat Step 4 until you have five more readings. Store this run, then turn off the power to the apparatus.

Part 2 Effect of material 1. Replace the brass rod with one made of a different metal. It should have the same diameter as the rod you used in Part 1. (Why?) 

Keeping the diameter constant allows us to see how length, cross-sectional area, and type of material affect the resistance of the metal rods.

2. As you did for Part 1, set up a current of approximately 0.5 A in the rod.

Repeat Steps 2–4 from Part 1. Store this run, and then turn off the power to the apparatus.

3. Repeat Step 2 using additional metal rods. Store these runs.

Part 3 Effect of diameter 1. Obtain one of the other brass rods; record the diameter. Connect the apparatus as you did for Part 1.

2. Turn on the power and adjust the current to approximately 0.5 A. Connect the red lead at the 25-cm mark. Rather than open a new experiment file with diameter as the independent variable, you can continue to use your file and simply record the value of the current and potential from the sensors in your notebook.

3. Repeat Steps 2 and 3 with the rest of the brass rods in the set of materials for this lab. Save the file.

Resistivity of the material used in this lab Material

Copper

Resistivity( Ω ∙m)

1.72×10

−8

Stainless Steel

9 .78×10

−8

Length Current Potential (m) (A) (mV) 0 0.2064 Rod 1 Diameter: 3.18mm0.002 mV A Material: Brass 5 0.2042 0.613 mV A Length Current Potential 10(A) 0.2065 (m) (mV) 1.233 mV A 0 0.224 A 0.005 mV 15 0.2065 1.906 mV 5 0.224 A 0.090 mV A 10 0.224 A 0.158 mV 20 0.2064 2.484 mV 15 0.224 A 0.227 mV A 20 0.224 A 0.300 mV 25 0.2071 3.097 mV 25 0.224 A 0.372 mV A 30 0.224 A 0.2079 0.447 mV 30 3.771 mV 35 0.224 A 0.519 mV A 35 3.18mm 0.2081 4.387 mV Rod 3 Diameter: Material: A Cooper DATA

Length (m) 0

Current (A) 0.014 A

5

0.027 A

10

0.041 A

15

0.061 A

20

0.078 A

25

0.094 A

30

0.112 A

35

0.128v

Potential (mV) 0.2268 mV 0.2270 mV 0.2271 mV 0.2273 mV 0.2273 mV 0.2274 mV 0.2273 mV 0.2273 mV

Aluminum −8

2.75×10

Brass −8

7 .50×10

TABLES: Rod 2 Diameter: 3.18mm Material: Stainless Steel

Rod 4 Diameter: 3.18mm Material: Aluminum Length Current Potential (m) (A) (mV) 0 0.2200 0.001 mV A 5 0.2201 0.183 mV A 10 0.2197 0.368 mV A 15 0.2190 0.553 mV A 20 0.2202 0.731 mV A 25 0.2186 0.919 mV A 30 0.2185 1.096 mV A 35 0.2197 1.280 mV A

b. Run 1- Brass Rod

c. Run 2- Stainless Steel Rod

Current (A)

e. Run 3- Cooper Rod

f. Potential (mV) vs. Lenght (cm) Graph

R (Ω)

dRun 4- Aluminum Rod

g. Current (A) vs. Lenght (cm) Graph

h. Resistivity Graph

EVALUATION OF DATA:

Part 1 1. Disconnect your sensors, then view your first run.

2. Choose New Calculated Column from the Data menu in Logger Pro. In LabQuest App, create this calculated column by choosing New Calculated Column from the Table menu on the Table tab. Enter resistance as the Column Name and ohms as the Units. Using Ohm’s law, determine the equation to use in the equation field. Keep in mind that the potential was measured in mV. 3. Choose to view only one graph. Select resistance as the label of your vertical axis. If the relationship between resistance and length appears to be linear, fit a straight line to your data.

Part 2 1. Repeat the evaluation you did in Part 1 for the runs you performed with different types of metal.

2. What does the slope of the best-fit line for a metal tell you about its relative ability to conduct electricity? 

The slope indicates the relationship between resistance and length. The greater the slope the more resistance there will be there is an increase in length; hence this reduces conductivity as length increase.

Part 3 1. Calculate the resistance from your recorded values of current and potential difference.

2. Choose New from the File menu.

3. Manually enter data for diameter and resistance for the brass rods you examined.

4. Create a new calculated column showing the cross-sectional area of the rods.

5. Change the horizontal axis of your graph to cross-sectional area. What relationship appears to exist between resistance and the cross-sectional area of the rods?



The relationship between cross-sectional area and resistance is an inverse relationship. As the cross-sectional area of the wire decreases, the resistance increases.

6. Test your conclusion by creating another calculated column and modifying the cross-sectional area so as to produce a linear plot. If successful, write a statement describing the relationship you have found.

Part 4 Consolidation of your findings 1. At this point, you have two proportional relationships. In your class discussion you will discuss how these can be combined to form a single proportionality relating resistance, length, and the cross-sectional area of the rod.

2. The name given to the constant of proportionality for the relationship you have determined is resistivity, . What are the units of this constant? 

Measured in Ω∙ m

Resistance (Ω) 2.23214E05 0.00040178 6 0.00070535 7 0.00101339 3 0.00133928 6 0.00166071 4 0.00199553 6 0.00231696 4 Avg: 0.00118192 4. Using known values for resistance, length, and area, determine the value of  for your particular brass. How does this value compare to the table value for this metal?

Length: 45cm ≈ 0.45m Area: πD2/4= π(0.00318m)2/4= 7.9 x 10-6 m2

Resistance: 1.18 X10-3Ω R= ρ⋅

L A

R⋅ A =ρ L (1.18 ⋅10−3 Ω)(7.9⋅10−6 m2 ) =2.1⋅10−8 Ω⋅m 0.45 m

This compares to the expected resistivity value of 7.50 x 10-8 Ω∙m.

The discrepancies between the resistivity values may be due to cross-sectional area and length of the experimental brass rod versus the ideal brass rod

CONCLUSION Definitions: 





Electric Current: When a circuit is formed, charge can move (or flow) through the wires of the circuit, from one terminal of the battery to the other, as long as the conducting path is continuous. Any flow of charge such as this is called an electric current. (Chapter 18, Section 2, Page 504) Resistance: the proportionality factor between the voltage V (between the ends of the wire) and the current I (passing through the wire). (Chapter 18, Section 3, Page 506) Ohm’s Law: Law that was found experimentally that states that in metal conductors R is a constant independent of V. Essentially, that for any given resistance the current is directly proportional to voltage.

(Chapter 18, Section 3, Page 506)



Ampere: Electric current is measured in coulombs per second, and the unit for this is called ampere or Amps (Chapter 18, Section 3, Page 506)



Resistors: These are used to control the amount of current. Resistors have resistances ranging from less than an ohm to millions of ohms (Chapter 18, Section 3, Page 507)



Electric potential: (or simply the potential when “electric” is understood) as the electric potential energy per unit charge. (Chapter 17, Section 1, Page 474)



Volt: The unit of electric potential, and of potential difference, is joules/coulomb and is given a special name, volt. (Chapter

Meaning of the slope and y intercepts: For the first graph (Potential (mV) vs. Length (cm)), the slopes for each of the lines indicate just how the potential increases as the length increases. In turn, it means that the further you go from the zero point, the higher the potential. Going in order, Stainless steel is the highest within the graph, showing that the potential increases greatly (at 0.1254 mV/cm) the further it is away from 0cm. Second is Aluminum, sitting with a slope of 0.03654 mV/cm, it’s not as strong as stainless steel, but it’s still very conductive. Then comes brass, which has a slope of 0.01452 mV/cm, which is less resistive. Finally, aluminum has a slope of 0.003338 mV/cm, being the least resistive of the others.

The y-intercept for each of these lines is 0, which indicates that the closer you are to the zero position on the rod, there is no electric potential going through it.

For the second graph (Current(A) vs Length (cm)), there is no real slope (apart from it being 0) and the y-intercept stays the same throughout runs. This is because throughout the entire experiment and through all of the runs, the current stays the complete same even as length increases.

Lastly, for the third graph (Resistance(Ω) vs Length (cm)), the graph and slopes are similar to that of the first graph, Potential vs Length. As such, the slopes indicate that the further you get from the zero point on the rod, the higher the resistance. Going in order: Stainless steel is the highest within the graph, showing that its resistance has a slope of 0.6022 Ω /cm the further it is away from 0cm. Second is Aluminum, sitting with a slope of 0.1668 Ω /cm, it’s not as strong as stainless steel, but it’s still very conductive. Then comes brass, which has a slope of 0.06480 Ω /cm, which is less resistive. Finally, aluminum has a slope of 0.01468 Ω /cm, being the least resistive of the others.

The y-intercept for each of these lines is 0, meaning that the closer you are to the zero position on the rod, there is no resistance occurring.

Equations: Electric Current: I = ∆Q/∆t Where ∆Q is the amount of charge that passes through the conductor at any location during the time interval ∆t.

Ohm’s Law: V = IR Electric Potential: Va = PEa/q Difference in Potential: Vba = Vb - Va = (PEb - PEa)/q = -Wba/q Resistance: R = ρ (L/A) Divergent results When it comes to error, error could have occurred through a few means.

Failure to keep the distance between each recording even could have caused error in the slopes. Failure to calculate the resistance in the third graph could have also led to errors in that graph as well. There may also be error within the analysis questions involving calculations, as failure to understand a formula or inputting the incorrect numbers or failure to convert values may lead to erroneous results.