1010lab3
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University of Guelph *
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Course
1010
Subject
Electrical Engineering
Date
Dec 6, 2023
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5
Uploaded by JusticeFangIbis47
Lab 3 – Kirchhoff’s Rules
Experiment 1:
Kirchhoff’s rules, a simple circuit
Methods:
For experiment 1, I recreated the two-loop circuit given in the lab document. To do this, I first placed the
5 resistors in appropriate positions. I then connected the two 3.3V clips to the two 2000
Ω
resistors. I
then connected two ground clips to the two 1000
Ω resistors. Finally, I
connected a clip to the A7 port
which will be used to read potential across each resistor.
Upper Loop
Lower Loop
Results:
Conclusions:
Based on the small percent difference between the theoretical and actual values for current, the two
values agree with each other. The percent difference for the 1000
Ω
, 2000
Ω
and 4000
Ω resistors
are
0.4%, 1% and, 0.2% respectively. The small amount of error could be caused by the uncertainty in the
resistors or the uncertainty in the IOLab reading.
Resistance (Ω)
∆V
Current (A)
%difference
4700
2.507
0.0005334
0.226022994
2000
0.5378
0.0002689
1.046728972
1000
0.265
0.000265
0.414234607
Lower Loop
Resistance (Ω)
∆V
Current (A)
%difference
4700
2.507
0.0005334
0.226022994
2000
0.5378
0.0002689
1.046728972
1000
0.265
0.000265
0.414234607
Upper Loop
Experiment 2: A more complicated application of Kirchhoff’s Rules
Methods:
For experiment 2, I created the 3 loop circuit given in the lab document. To do this, I positioned each
resistor exactly how they appeared in the diagram. I then connected the 3.3V clip to the 4700
Ω resistor,
and
the DAC clip to the 1000Ω resistor on the right. I also connected the ground clip to the middle
1000Ω and 2000Ω resistors.
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Results:
I
1
I
1
I
2
I
2
I
3
I
3
I
4
I
5
I
1
I
2
I
3
I
4
I
5
I
1
I
2
I
3
I
4
I
5
Resistance (Ω)
∆V
Current (A)
%difference
4700
2.545
0.000541489
0.108463739
1000
0.743
0.000743
1.882374485
2000
1.192
0.000596
0.228471101
2000
0.442
0.000221
2.251118086
1000
0.813
0.000813
0.28080354
Both of Kirchhoff’s Rules are
practical examples of conservation laws in physics.
Which fundamental quantity is being conserved in the loop rule? Which other quantity is
conserved in the Junction rule?
The quantity being conserved in the loop rule is potential energy. As charges move through the circuit
they lose energy when passing through the resistors and gain energy when they mass through the
power source. Because of the total energy in the system remains constant.
The quantity being conserved in the junction rule is current. The total current flowing out of a junction
must be equal to total current flowing into the junction because current cannot be destroyed.
Conclusion:
The theoretical values agree with the experimental values based on the small % difference. The slight
difference can be caused by the uncertainty of the resistors, the internal resistance of the breadboard,
or the uncertainty of the IOLab measurements.
Related Documents
Related Questions
Shown in the figure below is an electrical circuit containing three resistors and two batteries.
I₁
0=
4
+
L
.
ww
R3
R₂
ww
1₂
R₁
ww
Write down the Kirchhoff Junction equation and solve it for I, in terms of I₂ and I3. Write the result here:
1₁ = 12-13
R₂=552
R₂ = 132
13
Write down the Kirchhoff Loop equation for a loop that starts at the lower left corner and follows the perimeter of the circuit diagram clockwise.
- IzR3 − LR₁ + 14
+ 10
Write down the Kirchhoff Loop equation for a loop that starts at the lower left corner and touches the components 4V, R₂, and R₁.
0 = 4-1₂R₂-11R₁
The resistors in the circuit have the following values:
R₁ = 12
Solve for all the following (some answers may be negative):
I₁ = 27.78
X Amperes
1₂ = 28.84
X Amperes
13 = 1.060
X Amperes
NOTE: For the equations, put in resistances and currents SYMBOLICALLY using variables like R₁,R₂,R3 and 11,12,13. Use numerical values of 10 and 4 for the voltages.
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How would I approach this question? Do I need to use Kirchoff's junction rules or can I just subtract the voltage drop?
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Given the circuit below and that R1=1.7ohm,R2=3.5ohm,Emf1=2.1V and Emf2=6.3V. What is the current through the section ac?
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Return the simplified circuit to its original form and get currents i1 to i9 (flowing through each resistor R1 to R9) as well as the voltage v1 to v9 across each resistor.
It=13A
Rt= 3.08
Vt=40.04V
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E,=?
R₁=552
V₁ = 2V
in
R3=1532
For this DC series circuit, what is the value of Etotal?
12V
30V
0.4V
2V
R₂=1052
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4. Use mesh analysis to determine the current supplied by each DC source in the circuit shown in the figure
below:
5V
50
U8
B4
) 30
| 20
iz B5
B1-
-20v( i
iz
- 30V
B2-
5V
5V
B3
arrow_forward
An ammeter is a tool that measures the current (a better name would be
V Show Current
O Electrons
ampmeter) that passes through it. In the circuit simulation you can get
Conventional
ammeters from the upper right toolbar.
Labels
Values
When current is flowing in through the meter a number value will replace
the question mark.
Current
Votmeter
Ammeters
Wire Resistivity
First make a simple series circuit with a battery, switch, single light bulb
and ammeter as shown at right.
It's actually not necessary to include the switch but if I left it out then my
snapshots would all give away the answers.
With the switch closed/current flowing in the circuit, record the current that
Current
is flowing out of the light bulb and back to the battery.
0.00 A
The current is
A
Now switch the light bulb and the battery so that the ammeter comes
before the bulb.
Current
0.00 A
With the switch closed/current flowing in the circuit, record the current that
is flowing out of the battery and into the light…
arrow_forward
How do I go about solving this problem?
To solve thus problem, I know there needs to be 5 equations prior and the currents should be negative, but the three loops keep confusing me. I'm not sure how to progress with this and would appreciate some guidance. Thanks.
arrow_forward
Using these diagrams of circuits, what are the steps and materis to create these kind of circuits?
arrow_forward
Explain this diagram
arrow_forward
Q5
On a common set of axes, plot the v-i characteristics
for resistors
R₁ = 4.7k²2, R₂=47k52, R₂ = 47052
arrow_forward
SEE MORE QUESTIONS
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Delmar's Standard Textbook Of Electricity
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ISBN:9781337900348
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Related Questions
- Shown in the figure below is an electrical circuit containing three resistors and two batteries. I₁ 0= 4 + L . ww R3 R₂ ww 1₂ R₁ ww Write down the Kirchhoff Junction equation and solve it for I, in terms of I₂ and I3. Write the result here: 1₁ = 12-13 R₂=552 R₂ = 132 13 Write down the Kirchhoff Loop equation for a loop that starts at the lower left corner and follows the perimeter of the circuit diagram clockwise. - IzR3 − LR₁ + 14 + 10 Write down the Kirchhoff Loop equation for a loop that starts at the lower left corner and touches the components 4V, R₂, and R₁. 0 = 4-1₂R₂-11R₁ The resistors in the circuit have the following values: R₁ = 12 Solve for all the following (some answers may be negative): I₁ = 27.78 X Amperes 1₂ = 28.84 X Amperes 13 = 1.060 X Amperes NOTE: For the equations, put in resistances and currents SYMBOLICALLY using variables like R₁,R₂,R3 and 11,12,13. Use numerical values of 10 and 4 for the voltages.arrow_forwardHow would I approach this question? Do I need to use Kirchoff's junction rules or can I just subtract the voltage drop?arrow_forwardGiven the circuit below and that R1=1.7ohm,R2=3.5ohm,Emf1=2.1V and Emf2=6.3V. What is the current through the section ac?arrow_forward
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Recommended textbooks for you
- Delmar's Standard Textbook Of ElectricityElectrical EngineeringISBN:9781337900348Author:Stephen L. HermanPublisher:Cengage Learning

Delmar's Standard Textbook Of Electricity
Electrical Engineering
ISBN:9781337900348
Author:Stephen L. Herman
Publisher:Cengage Learning