EE300: Basic Electrical Engineering I
LABORATORY REPORT
Basic Circuit Analysis
Experiment No. 2
Written by: Jo Dahle
Instructor: Dr. Jason Sternhagen
Lab Section 01
Date Performed: 9 September 2020
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Introduction
Series and parallel circuits are the foundation for electronics. As a function in every electronic, it
is vital to understand how resistance can change based off of connection and value. This lab seeks to
deepen that understanding by having circuits in series, parallel, and a combination, and comparing
resistance and voltage based off of differing resistors in the circuit. This will allow more experience with
the different values of resistors and how they relate to one another, and how that affects the resulting
current. By measuring current and voltage at different points, this lab will also illustrate the changes in
voltage and current throughout the circuit as a whole.
Theory
When two resistors are in series, the total resistance found by adding the resistance of the
individual pieces. Since the current should be constant throughout the circuit, the current can be found
by dividing the source voltage by the total resistance. The voltage across a single resistor can then be
found by taking the current across that resistance multiplied by the current found before. Using that
information, power can be found at a specific resistor by multiplying the voltage at the resistor of choice
multiplied by the current.
For example, for a current with an R1 of 1 k Ω and a 9 V source, for a varying R2, the results will
be as follows.
R2= 10 Ω
R2= 100 Ω
R2= 1 kΩ
R2= 10 kΩ
R2= 100 kΩ
Total Resistance 1010 Ω
1100 Ω
2k Ω
11k Ω
101 k Ω
I
8.9 mA
8.2 mA
4.5mA
.8 18mA
.089 mA
V1
8.91 V
8.81 V
4.5 V
.82 V
.089 V
V2
.09 V
.82 V
4.5 V
8.2 V
8.9 V
Power @ R2
.81 mW
6.72 mW
20.25 mW
6.69 mW
.7921 mW
This changes when resistors are instead placed in parallel. The total resistance of resistors placed
in parallel can be found by using the following equation:
1
R equivalent
=
1 1
1
+ + …+
R1 R 2
Rn
Once the total resistance is found, current can be found by taking the voltage from the source
divided by the total resistance. This will be the current going into the parallel circuit as a whole. In order
to find the individual currents, take the voltage source divided by the resistance chosen, because in
parallel, the voltage will be constant, meaning that each resistor will have the same voltage but a
different current. With the previous information, power for a resistor than can be found by multiplying
the voltage from the source by the current across that resistor.
For example, for a circuit with a 12 V source and two resistors in parallel, where R 1=1 kΩ, the
results are as follows.
R2= 100 Ω
R2= 1 kΩ
R2= 10 kΩ
R2= 100 kΩ
R2= 1 MΩ
Total Resistance 90.9 Ω
50 kΩ
.91k Ω
.99 k Ω
.999 k Ω
Vin
12 V
12 V
12 V
12 V
12 V
Iin
132 mA
24 mA
13.2 mA
12.12 mA
12 mA
I1
12 mA
12 mA
12 mA
12 mA
12 mA
I2
120 mA
12 mA
1.2 mA
.12 mA
.012 mA
Power @ R2
1.44 W
144mW
14.4 mW
1.44 mW
.144 mW
When there is a combination of series and parallel circuits, then node voltage analysis can be
used. This is done by choosing a point and calculating the voltage, by taking the unknown voltage,
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subtracting the voltage on the other side of the resistor, and diving by the resistor. The sum of all of
those into the node is then set equal to zero to solve. The two voltages found were V1= 3.724 V and V2 =
4.055 V.
Example for a circuit with 5 resistors:
Resistor:
5V Supply 100 Ω
220Ω, lef
220Ω, right
510Ω, lef
510Ω, right
Current (mW) 15.25
9.45
5.8
1.8
7.3
7.95
Power (mW)
76.25
8430
7400
628
27100
3220
Procedures
Resistors in Series: Set up a breadboard to have 9 V running through it, and have two resistors,
connected in series (end to end), one resistor with a constant 1 kΩ, and the second being variable. Using
R2= 10 Ω, 100 Ω, 1k Ω, 10k Ω, and 100k Ω, to measure I, V 1, and V2. Add the calculated V values and use
those to calculate the power at the variable resistor. Record data.
Resistors in Parallel: Set up a breadboard to have 12 V running through it, and have two resistors,
connected in parallel, one resistor with a constant 1 kΩ, and the second being variable. Using R 2= 100 Ω,
1k Ω, 10k Ω, 100k Ω, and 1 M Ω to measure Vin, Iin, I1, and I2. Use Vin and I2 to calculate the power at the
variable resistor. Record data.
For the combination, connect the 5V source to a circuit using 1 100Ω resistor, 2 220Ω resistors,
and 2 510 Ω resistors as shown (Figure 10 from Lab 2, Sternhagen)
Measure voltage between the 220Ω and one 100Ω resistor and the voltage at the node
connecting the 100Ω, 220Ω, and right 510Ω resistors. Then measure the current at the 5V supply and
each of the resistors, using the found current and voltage to calculate power.
Results and Analysis
The theoretical results are shown above, under the “Theory” section. For the results found
experimentally, they are as follows:
In Series:
R2= 10 Ω
R2= 100 Ω
R2= 1 kΩ
R2= 10 kΩ
R2= 100 kΩ
Total Resistance 1010 Ω
1100 Ω
2k Ω
11k Ω
101 k Ω
I
.865 mA
8.6 mA
4.3 mA
.86 mA
.087 mA
V1
8.5 V
8.0 V
4.3 V
.78 V
.085 V
V2
.09 V
.767 V
4.3 V
7.8 V
8.5 V
V1 + V 2
8.59 V
8.767 V
8.6V
8.58 V
8.585 V
Power @ R2
7.7785 mW
6.596 mW
37.2 mW
6.708 mW
.7395 mW
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In Parallel:
Total Resistance
Vin
Iin
I1
I2
I1 + I2
Power @ R2
R2= 100 Ω
90.9 Ω
11.3 V
130 mA
11.4 mA
114 mA
125.4 mA
1288.2 mW
R2= 1 kΩ
50 kΩ
11.4 V
22.89 mA
11.6 mA
11.5 mA
23.5 mA
131.1 mW
With the Combination:
Resistor:
5V Supply 100 Ω
Current (mW) 15.3
9.45
Power (mW)
76.5
8930
R2= 10 kΩ
.91k Ω
11.4 V
12.6 mA
11.5 mA
1.2 mA
12.7 mA
13.68 mW
220Ω, lef
5.8
7400
R2= 100 kΩ
.99 k Ω
11.4 V
11.7 mA
11.6 mA
.12 mA
11.72 mA
1.368 mW
220Ω, right
1.5
495
510Ω, lef
7.4
27900
R2= 1 MΩ
.999 k Ω
11.4 V
11.6mA
11.6 mA
.017 mA
11.611 mA
.1254 mW
510Ω, right
8.0
3200
When comparing the theoretical calculations to the experimental results, there is a consistent
<5% error, due in part to resistance error, added resistance in the physical pieces, and error in the
measurement tools.
In series, the power is maximized when the 2 nd resistor’s value matches the 1st resistor. For
parallel, the power is maximized when the resistance is smallest, and when combined, the largest power
measurement occurs for the lef 550 Ω resistor.
When comparing the voltage across the resistors to the resistor values, they should match
closely to the current, but are off by a significant amount. When multiplying the current vs the resistor
values, the voltage should be able to be found, but there is a consistently higher theoretical this way
then there is measured.
Conclusions
By comparing theoretical with practical results, we can conclude that the material world holds
error and resistance to electricity. This limits our ability to build any electronic with extraordinarily
specific needs based solely on the theoretical values. This opens up a greater acceptance for Thevenin
resistance, because rather than calculating what should be based on every individual part, it looks at the
pieces as a whole. By using and measuring using this method, the accuracy and reliability of various
circuits increases, because it doesn’t take into consideration each part individually, but rather the sum of
the parts. And just as power is maximized when the variable resistor is equal to the other, this can lend
to understanding the importance and efficiency of a Thevenin resistance that is not only non-zero, but
also equal to that dissipated in the desired load. By using the Thevenin method, it would be easier to
measure in the real world, electronics such as car batteries, as it would limit the pieces being measured
and allow for the quickest and cheapest understanding. While this method is limited based off of how
the circuit was designed, what it was designed for, and the usual limits of size, resistor error, and
measuring error, the Thevenin method is still a good engineering compromise.
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