LAB Report Physics OHM\'S LAW Group 1 PDF

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UNIVERSITI TEKNOLOGI MARAPHYSICSLAB REPORT(Ohm’s Law)(Group 1)PREPARED FOR:PREPARED BY:NO NAME STUDENT ID12345DATE OF SUBMISSION23 February 2022PRE LAB QUESTIONPRE-LAB QUESTIONS Answer the following questions and submit your group answers to the instructor. What do you understand by the term “ohmic ...

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UNIVERSITI TEKNOLOGI MARA PHYSICS098 LAB REPORT

(Ohm’s Law) (Group 1) PREPARED FOR: PREPARED BY: NO

NAME

1 2 3 4 5

DATE OF SUBMISSION 23 February 2022

STUDENT ID

PRE LAB QUESTION PRE-LAB QUESTIONS Answer the following questions and submit your group answers to the instructor. 1. What do you understand by the term “ohmic material”? = Materials that obey the ohm’s law. Where current across the material is directly proportional to

the difference, thus calling it ohmic.

2. Sketch a graph of current, I versus voltage, V for an ohmic material. What does the

gradient of the graph indicate?

The gradient indicates the resistance

3. If you are given two different lengths of wires, a unit of voltmeter, a unit of ammeter, a

unit of power supply and a rheostat, how can the value of the wire resistance be determined? Show your experimental setup based on the stated apparatus.

=

ABSTRACT This experiment is conducted to determine the conductor’s resistance, R with different lengths 0.4m and 0.6m. It is conducted virtually by a website with a circuit and adjustable length of conductor. The resistance, R is determined through a formula of Ohm’s Law, with V is potential difference, I is current V=IR The graph is plotted for the two lengths to obtain the value of R is constant for each length throughout the experiment. The value of R is precise, the graph produced is a straight line and proves that the conductor obeyed the Ohm’s Law. The differences in R values for the different lengths shows that R is directly proportional to L. Area, A is kept constant throughout the experiment because R is inversely proportional to A

INTRODUCTION Ohm’s Law states that current through a conductor between two points is directly proportional to the potential difference or voltage across two points and inversely proportional to the resistance. The relationship can be described as

𝐼⍺

𝑉 𝑅

I represents the current in the unit of Ampere, V represents the potential difference measured across the resistance in the unit of Volt and R represents the resistance of the conductor in the unit of Ohm. The law was introduced by German physicist Georg Ohm and was published in 1827. He discovered that for metallic conductors there is a substantially constant ratio of the potential difference between the end conductor. From the relationship above, a plot of graph I versus V is linear, that is, resistance, R is independent of V. the resistance of the wire is the ratio of voltage to current as stated in equation below

𝑅 =

𝑉 𝐼

The resistance depends on the length of wire, l, cross-sectional area, A and resistivity of the wire, ⍴ as stated in the equation below

𝑅=

ρ𝑙 𝐴

A digital ammeter and voltmeter is a commonly used device in measuring the current and potential difference in a circuit. Rheostat is included in this experiment to provide variable resistance and as result it varies the potential difference and current in the circuit. On the other hand, the use of two different lengths of wire may vary the resistance in the circuit as the relationship between the resistance and length of wire is proportional. The objective of this experiment is to determine the resistance, R with two different lengths of conducting wires made of the same materials.

METHODOLOGY Click the 'Show circuit diagram' checkbox inside the simulator window to see the circuit diagram. The length of the resistance wire can be changed by using the slider which is 0.4m. By dragging the mouse from one connecting terminal to the other connecting terminal of the devices to be linked, connections can be created as shown in the circuit diagram. To connect the 0.4m wire, it is dragged and placed on the voltmeter. Finally, the key in the middle is inserted into the switch once all connections have been made. To alter the current flow, slowly move the rheostat contact between 0.98A, 0.50A, 0.33A, 0.25A, and 0.20A. The readings are shown both in Voltmeter and Ammeter. The resistance of the wire is calculated using the formula R=V/I . The experiment is repeated using different lengths of wire which is 0.6m.

Diagram of the procedure

RESULT AND ANALYSIS

Length of wire, L (m)

Experiment

Voltmeter reading, V (V)

Ammeter reading, I (A)

Resistance, R (Ω) R=V/I

0.4

1

0.199

0.980

0.203

2

0.100

0.495

0.202

3

0.067

0.331

0.202

4

0.050

0.249

0.201

5

0.040

0.199

0.201

Table 1

Length of wire, L (m)

Experiment

Voltmeter reading, V (V)

0.6

1

0.295

0.971

0.304

2

0.150

0.493

0.304

3

0.100

0.330

0.303

4

0.075

0.248

0.302

5

0.060

0.199

0.302

Table 2

Ammeter reading, I (A)

Resistance, R ( Ω) R=V/I

GRAPH ANALYSIS

Graph 1

Graph 2

Material that obeys ohm’s law is referred to as ohmic, because the current that passes through the material is directly proportional to the potential difference. The graph that will be expressed by ohmic material will have a linear relationship and produce a straight line. In this experiment, we have found out that different lengths of wire result in different resistance value outcomes. This is because length of wire is one of the factors that affect the resistance. The resistance of a ohmic conductor is directly proportional to its length, L and inversely proportional to its cross-sectional area, A. We can clearly see that as the potential difference increases, the current will also increase. Length of wire will also affect the gradient of the graph. Resistance is calculated by using formula

𝑅=

𝑉 𝐼

CALCULATIONS Diameter of the wire=0.2mm=0.0002m 2

Cross sectional area=πr2= 3.142(0. 0001) = 3. 142𝑥10

−8

2

𝑚

0.4 meter silver wire gradient=

1 𝑦2−𝑦1 = 𝑅 𝑥2−𝑥1

Thus, R=

𝑥2−𝑥1 𝑦2−𝑦1

(0.199−0.067)𝑉

R= (0.980−0.331)𝐴 = 0.203Ω (0.203+0.202+0.202+0.201+0.201)Ω = 0.202Ω 5

Average resistance=

Resistivity coefficient calculation R=ρ

𝐿 𝐴

0.203Ω=ρ

0.4𝑚 −8

3.142𝑥10

𝑚

2

−8

ρ= 1. 595 𝑥10 Ω. 𝑚

- Silver actual resistivity= 1. 59𝑥10−8Ω. 𝑚 𝐸𝑥𝑝𝑒𝑟𝑖𝑚𝑒𝑛𝑡𝑎𝑙 𝑣𝑎𝑙𝑢𝑒− 𝑇ℎ𝑒𝑜𝑟𝑒𝑡𝑖𝑐𝑎𝑙 𝑣𝑎𝑙𝑢𝑒 x 100% 𝑇ℎ𝑒𝑜𝑟𝑒𝑡𝑖𝑐𝑎𝑙 𝑣𝑎𝑙𝑢𝑒

Percentage of error=

−8

−8

(1.595𝑥10

=

− 1.590𝑥10 −8

1.59𝑥10

)Ω.𝑚

x 100% = 0.31%

Ω.𝑚

0.6 meter silver wire R=

𝑥2−𝑥1 𝑦2−𝑦1 (0.295−0.100)𝑉

R= (0.971−0.330)𝐴 = 0.304Ω (0.304+0.304+0.302+0.302+0.303)Ω = 0.303Ω 5

Average resistance=

Resistivity coefficient calculation 0.304Ω =ρ

0.6𝑚 −8

3.142𝑥10

𝑚

2

−8

ρ=1.592x10 Ω.m −8

Percentage of error=

(1.592𝑥10

−8

− 1.590𝑥10 )Ω.𝑚 −8

1.59𝑥10 Ω.𝑚

x 100% = 0.13%

DISCUSSION In this experiment , we manipulated the length of conducting wire which is 0.4 m and 0.6 m of silver wire where we want to observe the value resistance , R. The experiment was repeated five times for each length of wire to get more accurate results for the resistance value. Based on the result obtained from 0.4 m of wire, we plotted a graph and calculated the slope to get the value of resistance. As result , the value of resistance gain is 0.203 Ω and by using 0.6 m of silver wire, we obtain that the value of resistance is increasing which is 0.304 Ω. This result shows that the resistance of the wire is directly proportional to the length of the wire as the resistance increases when the length of wire increases. In addition ,we also calculated the resistivity from the value obtained and compare it with the actual resistivity of silver wire where in this experiment , the value of resistivity calculated is −8

−8

1. 595 𝑥10 Ω. 𝑚 for 0.4 m wire and 1.592x10 Ω.m for 0.6 m wire. The percentage error of both resistivity value for 0.4 m wire and 0.6 m wire from the actual value is 0.31% and 0.13% −8

respectively where the actual value is 1. 59𝑥10 Ω. 𝑚. Other than that , when plotting the graph of current, I versus voltage, V for both 0.4 and 0.6 m of wire , we gain that the current is directly proportional to voltage and the value resistance calculated from the slope shows that resistance is inversely proportional with current and directly proportional with voltage. This is proven when the value of voltage increases, the value of resistance also increases while when the value of current increases the resistance decreases. As from this result , we can conclude that silver is one of the ohmic materials as it obeys ohm’s law.

CONCLUSION As a conclusion, this experiment was carried out to determine the resistance of wire that is made up of the same material but different in lengths by using Ohm’s Law. From the experiment, we can conclude that resistance is directly proportional to the potential difference and inversely proportional to current. To prove this, we had plotted a graph of current, I against potential difference, V and got a linear relationship which means as the voltage increased, the current also increased, verifying Ohm’s Law. Based on the graph plotted, the slopes were calculated to get the resistance value by using the formula V = IR. The resistance value that we got for the shorter wire is lower than the resistance value for the longer wire. There are also differences in percentage error of the resistivity between both of the wires and from the actual value. These results support the theory that increasing the length of a wire will increase the resistance, since resistance is directly proportional to resistivity.

POST LAB QUESTION 1. Suppose that the conducting wire in this experiment is replaced with a semiconductor wire. Does the Ohm’s Law can be observed in this situation? Explain your answer. = No, we cannot see the linear relationship in conductors while using semiconductors. Because semiconductors have the properties of conductors and insulators. Thus it does not have a constant resistance and shows an exponential characteristic.

2. Besides the wire length, how does the cross-sectional area affect the resistance of the wire? = In equation R=ρ

𝐿 , we know that resistance of the wire is inversely proportional to the cross 𝐴

sectional area. Larger cross sectional area of the wire will allow charge inside the wire to move more easily with less resistance. The rate of charge flow in a wider wire is higher than in a narrow one.

REFERENCES 1. the Physics Classroom. (2000). Resistance. Physicsclassroom.com.

https://www.physicsclassroom.com/class/circuits/Lesson-3/Resistance 2. Ohm’s law and resistance (Simulator) : Class 12 : Physics : Amrita Online Lab. (n.d.). Amrita.olabs.edu.in. http://amrita.olabs.edu.in/?sub=1&brch=6&sim=22&cnt=4...