XLIX
Engineering Design Firm
9201 University City Blvd
Charlotte, NC 28223
Transmittal
Date:
Section:
To:
Monday, September 16, 2019
1201 – 018
Don Blackmon
Worth Powell
From:
Subject:
Joseph Robinson
Electrical Circuit Design Package
We are submitting:
☐ Memorandum
☒ Design Package
☐ Problem Set
☐ Report - Draft
☐ Test Data Sheet
☐ Extra Credit
☐ Report - Final
☐ Other: Enter Other Item
Date
Description
9/16/2019
Electrical Circuit Design Package Draft
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These are transmitted as checked below:
☒ Individual Assignment
☒ For grading
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☐ Other: Enter Description
☐ Team Assignment -- Team No. ________-_
Problem Statement
The XLIX design firm is looking for an individual to create a simple circuit on a breadboard that was
made from network problem set 25 and test five nodes with an approximation of predicted voltages. An
18 V DC power source, 400-point breadboard, and 25 resistors is provided for the testing. To do this,
having the knowledge of basic mathematical skills and knowing how read and place resistors in a
breadboard is required.
In order to build the circuit, the first step is to find the needed voltage drop at each resistor combination
between the five nodes. All the resistors provided, has a gold band ± 5% tolerance. The resistance
between each node is equal to one resistor combination. Step two, with any formations set-up of series
and parallel, at least one resistor combination needs to be parallel. The third and final step is to use the
calculations and place it on the voltage divider to be tested.
The resistors and breadboard are the only supplies that are to be used during the project and its testing.
Calculations
To calculate the resistance needed at each node, manipulating Kirchhoff’s Law of voltage 𝑉 = 𝐼 ∗ 𝑅, to
solve for resistance, 𝑅 = 𝐼 ∗ 𝑉 at each node. By calculating the total resistance, series: 𝑅𝑇 = 𝑅1 + 𝑅2 +
𝑅3 and parallel: 𝑅𝑇 = (𝑅1 + 𝑅2 + 𝑅3 )−1 will be evaluated to find the resistor combinations. Network
problem set 25, has a total current of 𝐼 = 0.0012𝐴 and the values for each node was provided; with this
information, finding the resistance between nodes one through five is shown below:
Node 1: 14.04V
∆V = 18V − 14.04V = 3.96V
3.96𝑉
= 3300Ω
0.0012𝐴
Node 2: 10.884V
∆V = 14.04V − 10.884V = 3.156V
3.156𝑉
= 2630Ω
0.0012𝐴
Node 3: 7.98V
∆V = 10.884V − 7.98V = 2.904V
2.904𝑉
= 2420Ω
0.0012𝐴
Node 4: 5.868V
∆V = 7.98V − 5.868V = 2.112V
2.112𝑉
= 1760Ω
0.0012𝐴
Node: 0V
∆V = 2.112V − 0V = 2.112V
3.96𝑉
= 1760Ω
0.0012𝐴
With having 25 different resistors with five different values-220, 330, 2200, 3300, 10000; the calculations
needed to find the resistor combinations at each node are seen below:
Node one: 3300 Ω
Solution: 3300 Ω
1
1
Alternate solution: (2200 + 2200)−1 + 2200 = 3300 Ω
Node two: 2630 Ω
Solution: 2200 + 220 + 220 = 2640 Ω
1
1
1 −1
+
+
)
10000
10000
2200
Alternate solution:(
+ 220 + 220 + 330 + 330 = 2627 Ω
Node three: 2420 Ω
Solution: 2200 + 220 = 2420 Ω
1
1
1
Alternate solution:( 10000 + 10000 + 2200)−1 + 220 + 220 + 220 + 220 = 2420 Ω
Node four: 1760 Ω
1
1
+
)−1
2200Ω
2200Ω
Solution: (
+ 330Ω + 330Ω = 1760 Ω
1
1
1 −1
+
+
)
3300
3300
3300
Alternate solution: (
+ 330 + 330 = 1760 Ω
Node five: 1760 Ω
1
1
1 −1
+
+
)
3300
3300
3300
Solution: (
1
+ 330 + 330 = 1760 Ω
1
Alternate Solution: (2200Ω + 2200Ω)−1 + 330Ω + 330Ω = 1760 Ω
I have neither given nor received any unauthorized help on this assignment, nor witnessed any violation
of the UNC Charlotte Code of Academic Integrity.
September 16, 2019