Lab #1 - Tech63

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San Jose State University *

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Electrical Engineering

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Jan 9, 2024

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Lab #1 Tech 63, Section 12, Analog and Digital Circuits Review of DC Circuit Analysis Fall Semester 2023 Group 6: Alfonso Reodica, Jezrael Flancia, Hailey Garibay 9/11/23

Objective: During this lab our objective is to understand the functions of a circuit using simulation software and hands-on circuit building. List of Equipment : Solderless Breadboard, this breadboard allows us to experiment and learn how a circuit physically works. A Keysight EDU36311A, Triple Output, Programmable DC Power Supply which allows us to input a measured amount of power into our circuit. And a Keysight EDU34450A 5½ Digit Digital Multimeter that lets us accurately measure the amount of power being outputted from the circuit. List of Materials : Multisim Software to simulate a digital breadboard circuit on PC. 1kΩ (quantity 2), 1kΩ(quantity 1), 18kΩ (quantity 2) resistors

Hand Calculations : R 3 ∨ ¿ 4 = R 3 R 4 R 3 + R 4 = 18 kΩ× 18 kΩ 18 kΩ + 18 kΩ = 9 kΩ R 2,3 ∨ ¿ 4 = R 2 + R 3 ∨ ¿ 4 = 1 kΩ + 9 kΩ = 10 kΩ R ( 2 , ( 3 ∨ ¿ 4 ))∨ ¿ 5 = R 2 , ( 3 ∨ ¿ 4 ) R 5 R 2 , ( 3 ∨ ¿ 4 ) + R 5 = 10 kΩ× 10 kΩ 10 kΩ + 10 kΩ = 5 kΩ R T = R 1 + R ( 2 , ( 3 ∨ ¿ 4 ))∨ ¿ 5 = 1 kΩ + 5 kΩ = 6 kΩ I T = V s R T = 6 V 6 kΩ = 1 mA I 1 = I T = V 1 R 1 = 1 V 1 kΩ = 1 mA I 2 = R 5 R 2,3 ∨ ¿ 4 + R 5 × I 1 = 10 kΩ 10 kΩ + 10 kΩ × 1 mA = 0.5 mA I 3 = R 4 R 3 + R 4 ×I 2 = 18 kΩ 18 kΩ + 18 kΩ × 0.5 mA = 0.25 mA I 4 = R 3 R 4 + R 3 ×I 2 = 18 kΩ 18 kΩ + 18 kΩ × 0.5 mA = 0.25 mA I 5 = R 2,3 ∨ ¿ 4 R 5 + R 2,3 ∨ ¿ 4 ×I 1 = 10 kΩ 10 kΩ + 10 kΩ × 1 mA = 0.5 mA V 1 = I T R 1 = 1 mA × 1 kΩ = 1 V V 2 = I 2 R 2 = 0.5 mA× 1 kΩ = 0.5 V V 3 = I 3 R 3 = 0.25 mA × 18 kΩ = 4.5 V V 4 = I 4 R 4 = 0.25 mA × 18 kΩ = 4.5 V V 5 = I 5 R 5 = 0.5 mA × 10 kΩ = 5 V

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Table 1 : RMS V 1 V 2 V 3 V 4 V 5 I 1 I 2 I 3 I 4 I 5 Calculated 1V 0.5V 4.5V 4.5V 5V 1mA 0.5mA 0.25mA 0.25mA 0.5mA Multisim 1V 0.5V 4.5V 4.5V 5V 1mA 0.5mA 0.25mA 0.25mA 0.5mA Measured 0.99V 0.49V 4.46V 4.46V 5.0V 1.01mA 0.51mA 0.25mA 0.25mA 0.51mA Percent Error 1% 2% 0.89% 0.89% 0% 1% 2% 0% 0% 2% − ¿ ❑ 100 = % PE ❑ ? = Expected − Measured NominalValue ( Expected ) 100 = ¿ Multisim

Breadboard

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Voltage Across Resistor 1 Voltage Across Resistor 2

Voltage Across Resistor 3 Voltage Across Resistor 4

Voltage Across Resistor 5 Current Across Resistor 1

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Current Across Resistor 2 Current Across Resistor 3

Current Across Resistor 4 Current Across Resistor 5

Question Answers : 1. What is the relation between current in resistor R2 and R5. Explain the reason why they are the same or why they are different. a. The current between R2 and R5 is the same because they are in a series circuit. The current is the same through each resistor because there is only one path to follow. 2. Compare the current in resistors R3 and R4. Explain the reason why they are the same or why they are different. a. The current in resistors R3 and R4 are also the same because they are in series. 3. Do the hand calculations, Multisim simulation, and breadboard measurements agree? Explain. a. Yes, the hand calculations, Multisim simulation and breadboard measurements agree. The breadboard measurements represent a live circuit that is affected by outside factors such as long wires so a percent error is needed to make up for that small difference. 4. Does percent error remain under 5%? If the percent error is greater than 5% then explain the reason for the error. a. Yes, the percent error is under 5%.

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