Lab 4 Report

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COMMUNITY COLLEGE OF PHILADELPHIA PHILADELPHIA, PENNSYLVANIA Experiment Number: 4 Title: Network Theorems and Devices Performed by: Partners: N/A Engineering Course: Engineering 102 Lab Section No: 901 Lab Day/Hour: Tuesday 1:00 – 3:00 PM Lab Instructor: Dr.W.A.Gontar Date Performed: 10/05/2021 Date Due: 10/12/2021

Purpose The purpose of this experiment is to understand the two important network theorems that are Maximum Power Transfer Theorem and Thevenin’s Theorem. Procedures In part 1, Maximum Power Transfer Theorem would be verified in practice. The four given resistors was measured by using the Data Acquisition Unit (DaQ) and were recorded the values. Next, the circuit was set up as shown in Figure 1, and R L was replaced by the decade box. The circuit was applied 10 volts by using the Triple Output Power Supply. To measure the voltage across R L , adjust the decade box from 500Ω to 15.5 kΩ in 1 kΩ increments, and then the results were displayed on the screen of the DaQ; their values were recorded. Power consumed was computed by the formula P = V 2 R L . A graph of power consumed versus load resistance values would be drawn. After that, the peak of the graph would be verified. In part 2, to measure the Thevenin equivalent voltage (V TH ), remove the decade box from the circuit of Part 1, and then record its value. Next, power supply was disengaged creating a new circuit as shown in Figure 2. The Thevenin equivalent resistance (R TH ) value was measured by using the DaQ and it was recorded. R L value for the peak from Part 1 was compared with the measured R TH value.

The Thevenin Theorem would be proved by the following. Percent error between the measured and calculated R TH by the formula R TH = R 4 + R 3 ( R 1 + R 2 ) R 1 + R 2 + R 3 would be computed. Percent error between the measured and calculated V TH by the formula V TH = ( R 3 R 1 + R 2 + R 3 ¿ V S would be computed. The values of the voltage measured in Part 1 should be equal to that in a Thevenin equivalent circuit. To check them, link the power supply (where to read the V TH value) with the decade box (R L ) in series, and apply this equivalent circuit to a second decade box representing a load; then adjust the load resistance to a few of the values used in Part 1 to obtain the voltage values. A schematic of the circuit would be drawn. The detailed procedure that was followed can be found in Physics Laboratory Instructions, Volume I, Community College of Philadelphia, John Wiley & Sons, 2005.

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Data sheets Part I – Maximum Power Transfer Measurements ► Resistors Resistor Color Code Nominal Value ± Tolerance Measured Value 1 Brown-Black-Red-Gold 1000 ± 50 Ω 0.983 kΩ = 983 Ω 2 Brown-Black-Red-Gold 1000 ± 50 Ω 0.983 kΩ = 983 Ω 3 Brown-Green-Red-Gold 1500 ± 75 Ω 1.468 kΩ = 1468 Ω 4 N/A N/A 0.498 kΩ = 498 Ω ► Power consumed Voltage [V] R L [kΩ] P [W] 1.165 0.5 2.71 2.261 1.5 3.41 2.784 2.5 3.10 3.092 3.5 2.73 3.294 4.5 2.41 3.437 5.5 2.15 3.543 6.5 1.93 3.626 7.5 1.75 3.691 8.5 1.60 3.745 9.5 1.48 3.790 10.5 1.37 3.828 11.5 1.27 3.860 12.5 1.19 3.888 13.5 1.12 3.912 14.5 1.06 3.933 15.5 1 Part II – Thevenin Equivalent Measurements Thevenin equivalent voltage 4.272 [V] Thevenin equivalent resistance 1.339 [kΩ]

Graphs and Diagrams Part I – Graph of Power Consumed vs Load Resistor 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 0 0.5 1 1.5 2 2.5 3 3.5 4 Power Consumed vs Load Resistor Load Resistance Values [kΩ] Power Consumed [W] Part II – Schematic of the circuit

Calculations Part I 1. The nominal value and tolerance of resistors a. Resistor 1 R = (1x10 + 0) x 100 = 1000 Ω The tolerance band is gold which indicates a 5% tolerance, so 950Ω ≤ R ≤ 1050Ω % error = ¿ 983 − 1000 1000 ∨ ¿ x 100% = 1.7 % b. Resistor 2 R = (1x10 + 0) x 100 = 1000 Ω The tolerance band is gold which indicates a 5% tolerance, so 950Ω ≤ R ≤ 1050Ω % error = ¿ 983 − 1000 1000 ∨ ¿ x 100% = 1.7 % c. Resistor 3 R = (1x10 + 5) x 100 = 1500 Ω The tolerance band is gold which indicates a 5% tolerance, so 1425Ω ≤ R ≤ 1575Ω % error = ¿ 1468 − 1500 1500 ∨ ¿ x 100% = 2.13 % 2. Power consumed P = V 2 R L P 1 = 1.165 2 0.5 = 2.71 [W] P 2 = 2.261 2 1.5 = 3.41 [W] P 3 = 2.784 2 2.5 = 3.10 [W] P 4 = 3.092 2 3.5 = 2.73 [W] P 5 = 3.294 2 4.5 = 2.41 [W]

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P 6 = 3.437 2 5.5 = 2.15 [W] P 7 = 3.543 2 6.5 = 1.93 [W] P 8 = 3.626 2 7.5 = 1.75 [W] P 9 = 3.691 2 8.5 = 1.60 [W] P 10 = 3.745 2 9.5 = 1.48 [W] P 11 = 3.790 2 10.5 = 1.37 [W] P 12 = 3.828 2 11.5 = 1. 27 [W] P 13 = 3.860 2 12.5 = 1.19 [W] P 14 = 3.888 2 13.5 = 1.12 [W] P 15 = 3.912 2 14.5 = 1.06 [W] P 16 = 3.933 2 15.5 = 1 [W]  The maximum value of power consumed is 3.41 [W] and it occurs at the point (1.5, 3.41). Data points around the peak is from 0.5 kΩ to 3.5 kΩ, so from the graph, we get: P at 1 kΩ = 3.1 [W] P at 2 kΩ = 3.25 [W] P at 3 kΩ = 2.9 [W] Part II 1. Thevenin equivalent resistance R TH = R 4 + R 3 ( R 1 + R 2 ) R 1 + R 2 + R 3 = 0.498 + 1.468 x ( 0.983 + 0.983 ) 0.983 + 0.983 + 1.468 = 1.338 kΩ

2. Percent error between the measured and calculated Thevenin resistance % Error = | 1.399 − 1.338 1.338 | x 100% = 4.56 % 3. Thevenin equivalent voltage V TH = ( R 3 R 1 + R 2 + R 3 ) V s = ( 1.468 0.983 + 0.983 + 1.468 ) x 10 = 4.275 [V] 4. Percent error between the measured and calculated Thevenin voltage % Error = | 4.272 − 4.275 4.275 | x 100% = 0.07 %

Answers to Questions Quiz A 1. For the circuit below, calculate the Thevenin’s equivalent voltage V TH across R 3 R 1 = 5 kΩ, R 2 = 2 kΩ, V 1 = 20 V, V 2 = 10 V To calculate V TH across R 3 , we remove R 3 from the circuit, so we get a new circuit:  V TH = V 1 + V 2 = 20 + 10 = 30 V 2. For the circuit above, calculate the Thevenin equivalent resistance (R TH ). To calculate R TH , we remove R 3 , V 1 and V 2 from the circuit, so we get a new circuit:  R TH = R 1 x R 2 R 1 + R 2 = 5 x 2 5 + 2 = 10 7 kΩ 3. b. one half 4. It is presented in the “Graphs and Diagrams” section.

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5. True

Analysis and Discussion Percent error between the measured and calculated R TH values is acceptable (4.56%), and percent error between the measured and calculated V TH values is very low (0.07%). Source of errors might be from an unstable electric current flowing through the circuit, or old equipment/wire also affects the results. In Maximum Power Transfer Theorem, a graph of power consumed versus load resistance values that has a peak where a maximum occurs was obtained. In Thevenin’s Theorem, a schematic of the circuit was obtained. The maximum power consumed (P peak ) in Maximum Power Transfer Measurements occurs when the load resistance (R L ) is equal to the Thevenin’s equivalent resistance (R TH ) in Thevenin Equivalent Measurements. Power is inversely proportional to resistance, which means when the value of power is low, the value of resistance will be high and vice versa. Therefore, the theory of the following formula P = V 2 R L is verified by the experiment. The purpose of these experiment is achieved.