ohms law
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Florida Agricultural and Mechanical University *
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Course
2048L
Subject
Electrical Engineering
Date
Apr 3, 2024
Type
docx
Pages
4
Uploaded by ProfMusicEmu4
Ohm's Law
Written by: Sevaughn Clarington
Partners:
Jeniya Strobridge
Jayme Colbert-Williams
Due: February 19th, 2024
Abstract: This experiment served to calculate the Ohms by measuring the value of voltage and amperes throughout eight trials. The ohms were calculated using two equations.The first equation was voltage divided by amperes, and the second equation was the rise over run for the slope. The first
equation resulted in R= 2.7175, and the second equation resulted in R= 2.66.
Introduction:
In this lab, we're exploring the basic principles of electrical resistance discovered by Ohm. Ohm found that when voltage changes across a resistor, the current through it changes, too. Ohm's groundbreaking discovery revealed that in a resistor, the current (I) flowing through it is directly proportional to the voltage (V) applied across it, while inversely proportional to the resistance (R), expressed mathematically as I = V/R. By conducting experiments with different resistors, we will investigate the relationship between current and voltage in Ohmic and non-Ohmic materials.
Theory: In this experiment, the constant resistance (Ohms) was calculated by measuring the voltage and amperes throughout eight trials. Ohms was and calculated by this equation: (1) R= V/A. This equation translates to: Ohms is equal to voltage divided by amperes. The equation to find Ohms from slope is:
(2) R= y/x. This translates to Ohms is equal to voltage 1 minus voltage 2, divided by ampere
1 minus ampere 2.
Procedure: Using the diagram of the circuits, the assembly was made. The rheostat was adjusted
to get the current on the voltameter at a minimum and the voltage and current. The rheostat was then adjusted to 50 mA and the voltage along with the current was recorded. Step three was then repeated for a total of eight times to get eight readings. The power supply was then turned off and the circuit disassembled. R x
= V/I was then calculated for each trial (with error estimate) to find the average value of R x for all the trials. With the data gathered from the last step V vs. I was plotted and R x (with an error estimate) was found from the slope of the graph.
Results: Trial#
Current Voltage
R
1
0.1
0.15
1.5
2
0.1
0.3
3
3
0.2
0.5
2.5
4
0.25
0.7
2.8
5
0.3
0.9
3
6
0.35
1.1
3.14
7
0.4
1.2
3
8
0.5
1.4
2.8
Average R= 2.7175 Slope: y=3.1161x-0.0757
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Discussion:
As a result, we calculated 2.7175 as the average Ohms after eight trials. The result of the Ohms calculated through slope was 3.1161x-0.0757. These results are close, with small uncertainty. There is a possibility of a systematic error, such as misreading either the voltage or ampere measurement. However, it would not affect the results drastically. Conclusion:
In summary, this lab improved our comprehension of the fundamental differences between ohmic and non-ohmic materials as well as the connections between current, resistance, and voltage. We were able to obtain varied voltages and currents, observe our findings, and create graphs by applying a potential difference.
Questions : 1.
Compare the resistance, R
x , found as an average value to the R
x
from the slope of the graph.
-
The resistance, R
x was 2.7175 and the average value to the R
x
from the slope of the graph was 3.1161x-0.0757.
2.
Does the resistor have a constant resistance? Why or why not?
-
Since the average and slope aren’t equal, the resistor didn’t have a constant resistance.
Related Questions
1. Answer the subparts A&B with the step, based on the Image. This question Q&A please no reject and solve thank u
A) Based on the measurement data above, indicate that there is an error in the measurement
data. Give an explanation of the answers given.
B) Determine in which element there is an error in the measurement data. What should be
the correct measurement result. Give an explanation of the answers given.
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I need helpo with this. I need step by step solutions for this and I cant figure it out.
Wire to be used in buildings and homes in the United States is standardized according to the AWGsystem (American Wire Gauge). Wires are assigned a gauge number according to their diameter.The scale ranges from gauge 0000 to gauge 36 in 39 steps. The equation for the wire gauge interms of the wire diameter in millimeters is as follows.Answer the following questions with your group.(a) What is the equation for the diameter in millimeters in terms of the wire gauge N?(b) You have two wires, one that has five times the diameter of the other. By how much do thewire gauges differ for the two wires? (Give your answer to the nearest integer.)
arrow_forward
Given the circuit shown in the figure, use the words “increases,” “decreases,” or “stays the same” to complete the following statements.
(a) If R7 increases, the potential difference between A and E _____. Assume no resistance in the ammeter and the power supply
(b) If R7 increases, the potential difference between A and E _____. Assume nonzero resistance in the ammeter and the power supply
(c) If R7 increases, the voltage drop across R4 _____.
(d) If R2 decreases, the current through R1 _____.
(e) If R2 decreases, the current through R6 _____.
(f) If R2 decreases, the current through R3 _____.
(g) If R5 increases, the voltage drop across R2 _____.
(h) If R5 increases, the voltage drop across R4 _____.
(i) If R2, R5, and R7 increase, ℇ (r = 0) _____.
arrow_forward
I need answers with clear hand writing or using Microsoft word . ASAP
measurements , describe how to read the values given on the measurement tools and to write about measurement tools
note: i want a lot of information
arrow_forward
Please help me solve questions 1-5 using the given values found.
arrow_forward
Suppose the three branch currents in this circuit are I₁ = -3 A, I₂ = -18 A, and I3 = -15 A. The voltage drop across each circuit element is as given in the table below. From this information, determine, for each of these circuit elements,
(i) whether an active or passive sign convention is being used for that element,
(ii) whether that element is absorbing or producing a net (positive) amount of electrical power.
In each answer box within the table below, type the correct choice from among the bold-faced words above.
V₂ B
A
B
A
C
D
1₁
A
B
1₂
1₂
+
Circuit element Voltage drop Sign convention? Absorbing or producing net electrical power?
-9 V
-2 V
C
9 V
-11 V
D
arrow_forward
Make three plots using Excel:
1. Voltage vs current for the linear resistor data (voltage on vertical axis, current on
horizontal). Add a linear trendline with the equation and R² value shown.
2. Voltage vs current for the lightbulb data (voltage on vertical axis, current on
horizontal).
3. A plot showing both curves on the same set of axes (voltage on vertical axis,
current on horizontal).
arrow_forward
QUESTION 7
Match Collumn A and Collumn B? Properties of Metal
v Britleness
A. the property of a metal which allows little bending or deformation without shattering
Conductivity
B. the property of metal to resist deformation and stress without breaking
the property of a metal which can be hammered, rolled. or pressed into various shape without
C.
v Hardness
cracking, breaking or leaving some other detrimental effect
v Strength
D. the property of a metal to resist cutting, penetration and abbrasion
v Elasticity
E. the weight of a unit volume of a material
F. the property of a metal to become liquid by the application of heat
v Malleability
the property of metal which enables a material to return to its original state when the force
G.
applied is removed
v Fusibility
v Density
H. the property of a metal to conduct electricity or heat
Click Save and Submit to save and submit. Click Save All Answers to save all answers.
Beno
arrow_forward
201503131
3. Two batteries with E, = 3V and E₂ = 5 V are connected with three resisters R₁ =
10 ohms, R₂ = 20 ohms and R3 = 30 ohms as shown in the Figure. (8-points)
a) Find the currents I₁, I2 and 13 passing through the resistors R₁, R₂ and R3
consecutively and indicate their directions
in the figure. (4-points)
I
Iz
V.P
23 13 = I₁+I₂
R₂
10
3 &
3-101₁-3013 =0
5-20 I 2-10 I₁ = 0
I
R₁
ww
R3 30
20
25
arrow_forward
8.) Suppose a resistor has a voltage drop of 0.000034 volts across it. How many µ (micro) volts is dropped across the resistor?
a. 34 µV
b. 0.34 µV
c. None of These
d. 3.4 µV
10.) How much current flows in a short circuit?
a. About 2 A
b. A very large amount
c. A very small amount
d. About 20 A
11.) How much resistance is in the following circuit: (picture inserted below)
a. A very small amount
b. A very large amount
arrow_forward
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- Delmar's Standard Textbook Of ElectricityElectrical EngineeringISBN:9781337900348Author:Stephen L. HermanPublisher:Cengage Learning

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