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
Linear Resistor data
Lightbulb data
i(mA)
v(V)
i(mA)
v(V)
0.0
0.00
0.0
0.000
19.0
0.50
11.7
0.588
25.4
1.00
19.9
0.998
31.2
1.50
29.8
1.495
36.3
2.00
39.8
1.997
41.0
2.50
49.7
2.490
45.5
3.00
59.8
3.000
49.6
3.50
69.7
3.500
53.5
4.00
79.7
4.000
57.1
4.50
89.8
4.500
60.8
5.00
99.7
5.000
64.2
5.50
109.6
5.500
67.4
6.00
119.5
6.000
70.6
6.50
129.6
6.500
73.7
7.00
139.5
7.000
76.6
7.50
149.5
7.500
79.5
8.00
159.5
8.000
82.3
8.50
169.6
8.500
85.1
9.00
179.5
9.000
87.8
9.50
189.4
9.500
90.4
10.00
199.4
10.000
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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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only 11
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A fuse wire of circular cross-section has a radius of 0.8 mm. The wire blows off at a current of 8A. Calculate the radius of the wire that will blow off at a current of 1 A. Also calculate the Value of blow off current when radius is 0.5 mm
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Please help with a, d, f, h. I don't know why my values are coming out incorrect even though the units are correct.
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D7.36 Design the circuit of Fig. 7.54 to obtain a de emitter current of 1 mA, maximum gain, and a ±2-V
signal swing at the collector, that is, design for Vc=+2.3 V. Let Vcc= 10 V and B = 100.
RB
Vec
Rc
Ra
VBE
Vcc
A
Rc
lc +18=lg
le
HE
Vc = VBE + IBRB
(a)
(b)
Figure 7.54 (a) A common-emitter transistor amplifier biased by a feedback resistor R₁. (b) Analysis of the
circuit in (a).
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
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How about solving this using KCL equations? Thank you
arrow_forward
Please don't copy answer i need step by step answer for understanding else I will give dislikes
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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
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3. Fast help needed to answer this Engineeing question. Check image.
Postive feedback for right answer :)
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Consider the circuit shown in Figure P1.68. a. Which elements are in series?b. Which elements are in parallel? c. Apply Ohm’s and Kirchhoff’s laws to solve for R x.
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I need the answer quickly
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In the figure the ideal batteries have emfs ɛ1 = 20.4 V, ɛ2 = 9.05 V, and ɛ3 = 5.40 V, and the resistances are each 2.40 Q. What are the (a)
size and (b) direction (left or right) of current i,? (c) Does battery 1 supply or absorb energy, and (d) what is its power? (e) Does battery
2 supply or absorb energy, and (f) what is its power? (g) Does battery 3 supply or absorb energy, and (h) what is its power?
(a) Number
i
Units
(b)
(c)
(d) Number
i
Units
(e)
(f) Number
i
Units
(g)
(h) Number
i
Units
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Related Questions
- Linear Resistor data Lightbulb data i(mA) v(V) i(mA) v(V) 0.0 0.00 0.0 0.000 19.0 0.50 11.7 0.588 25.4 1.00 19.9 0.998 31.2 1.50 29.8 1.495 36.3 2.00 39.8 1.997 41.0 2.50 49.7 2.490 45.5 3.00 59.8 3.000 49.6 3.50 69.7 3.500 53.5 4.00 79.7 4.000 57.1 4.50 89.8 4.500 60.8 5.00 99.7 5.000 64.2 5.50 109.6 5.500 67.4 6.00 119.5 6.000 70.6 6.50 129.6 6.500 73.7 7.00 139.5 7.000 76.6 7.50 149.5 7.500 79.5 8.00 159.5 8.000 82.3 8.50 169.6 8.500 85.1 9.00 179.5 9.000 87.8 9.50 189.4 9.500 90.4 10.00 199.4 10.000arrow_forward1. 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.arrow_forwardonly 11arrow_forward
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Recommended textbooks for you
- Delmar's Standard Textbook Of ElectricityElectrical EngineeringISBN:9781337900348Author:Stephen L. HermanPublisher:Cengage Learning
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