Experiment 6

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California State University, Northridge *

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240L

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Electrical Engineering

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Feb 20, 2024

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Fall 2022 California State University, Northridge Department of Electrical & Computer Engineering Experiment 6 Network Theorems ECE 240L Written By:

1. Purpose The intent of this experiment is to teach us how Thevenin’s Theorem, Norton’s Theorem, Maximum Power Theorem, and Superposition Theorem work in practice. In this experiment, we will be building circuits in order to demonstrate these theorems, as well as how to build them in PSPICE. 2. Equipment Used: a. DC Power Supply b. DMM c. Resistors 3. Parts Used a. Connecting cables b. Alligator clips c. Probes 4. Software Used a. PSPICE b. Microsoft Word 5. Theory: In Thevenin Circuits, the two main objects to evaluate are the R Th (Thevenin Resistor) and the V oc (open circuit voltage). Remove the portion of the circuit external to which the Thevenin's equivalent circuit is to be found. Compute the voltage across the open-loop terminals, this is V oc . Eliminate all the sources and compute the resistance across the open-loop terminal, this is R Th . This will be explained more during the procedure. In Norton’s Circuits, the main objects to evaluate are the I sc (short circuit current) and the R N (Norton Resistor). R N is the same as the R Th found during the Thevenin circuit analysis. Calculation will be explained more during the procedure. Maximum Power Theorem states that for the load resistor R L to dissipate the maximum power, its value must be equal to the Thevenin resistance.

Superposition Theorem states that in any linear resistive circuit containing two or more independent sources, the current through or voltage across any element is equal to the algebraic sum of the currents or voltages produced independently by each source. 6. Procedure In this experiment, we will be assembling 3 circuits, the first one is to be assembled as shown: This will be used to find the Thevenin and Norton equivalents. First, remove the portion of the circuit external to which the Thevenin's equivalent circuit is to be found. Then, Compute the voltage across the open-loop terminals to find Voc. In our case, the calculation is as follows. Vth = 1 k Ω 1 k Ω + 1 k Ω + 2.7 k Ω ( 10 V ) = 2.13 V Next, calculate R Th . Eliminate all the sources and compute the resistance across the open-loop terminal, this is R Th . In this case, the calculation is as follows RTh = ¿ The next value, I sc which can be calculated as follows Isc = Vth Rth = 2.13 V 1.93 k Ω = 1.10 x 10 3 A

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To measure each of these values, the digital multimeter and probes will be used. To measure V ab (V RL ), place probes at points A and B shown in the circuit above To measure V oc , remove the 3.3k resistor and probe points A and B again. To measure I sc , place a short between points A and B and probe points A and B again, this time testing the current. To measure R th , remove the 10-volt source and replace it with a short. Use an ohmmeter to measure R th to the left of point A and B. IL is calculated by taking the VRL over the 3.3k, the value of the far-right resistor. I L = V RL 3.3 k = 1.34 3.3 k = 406.36 x 10 − 6 The next circuit that will be tested is shown below This circuit will be used to showcase power transfer theorem. First, calculate VA and VB V A = 4.7 k 4.7 k + 2.7 k ( 10 V ) = 6.35 V V B = 1 k 1 k + 3.3 k ( 10 V ) = 2.32 V Calculate VTh using VA and VB

V Th = V A − V B = 4.03 V Rth is then calculated R Th = ¿ Now that VTh and RTh is known, power transfer can be calculated. P = 4.03 2 4 × 2.48 k = 1.63 × 10 − 3 W To further test this, take varying value resistors between 100 Ohms and 10,000 Ohms, and measure the voltage output. The Voltage VTh can be measured by using the probes on the resistor and measuring voltage. The Resistance RTh can be measured by using the probes on the resistor and measuring resistance. The last circuit to be tested is shown below. First set it up with a short instead of the 5V power source. V1 and V2 are calculated as shown below. V 1 = ( 1 k ∨ ¿ 1 k ) ¿¿ V 2 = 10 − 1.32 = 8.68 V

Next, swap the short and power source so that the 5V power source is hooked up and the 10V power source is replaced by the short. V1 and V2 are found as follows. V 1 = ( 1 k ∨ ¿ 3.3 k ) ¿¿ V 2 = V 1 = 2.18 V The Voltages for both the 5V and the 10V can be measured using the probes. 7. Results The results for the first circuit are as follows Figure 6.4 Vab (V) Voc (V) Isc (A) Rth (Ohm) VRL (V) IL (A) Calculate d 1.34E+ 0 2.13E+ 0 1.10E- 3 1.94E+3 1.34E+ 0 406.36E- 6 Measured 1.34E+ 0 2.13E+ 0 1.10E- 3 1.91E+3 1.35E+ 0 408.18E- 6 % Error 0.07 0.09 0.13 1.39 0.45 0.45 The measured Vab for the first circuit. The measured Voc for the first circuit.

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The measured Isc for the first circuit. The measured Rth for the first circuit. The measured Vrl for the first circuit. The results for the second circuit are as follows. Figure 6.5 Vth (V) Rth (Ohm) P (W) Calculate d 4.03E+ 0 2.48E+3 1.63E- 3 Measured 4.00E+ 0 2.43E+3 1.65E- 3 % Error 0.62 2.10 0.88 The measured Vth for the second circuit.

The measured Rth for the second circuit. Power was calculated as shown above in the Procedures. Additionally, power across a range of resistors was tested. The results are as follows. Figure 6.5 R (Ohm) VR (V) P (V/4R) 100 0.16 64.0E-6 470 0.672 240.2E-6 1000 1.201 360.6E-6 2400 2.02 425.0E-6 3300 2.32 407.8E-6 4700 2.65 373.5E-6 5100 2.73 365.3E-6 6200 2.89 336.8E-6 7500 3.04 308.1E-6 9100 3.18 277.8E-6 10000 3.24 262.4E-6 We found that the maximum power occurs around 2.4k Ohms. The measured voltage using a 100 Ohm resistor for the second circuit. 100 470 1000 2400 3300 4700 5100 6200 7500 910010000 0.0E+0 50.0E-6 100.0E-6 150.0E-6 200.0E-6 250.0E-6 300.0E-6 350.0E-6 400.0E-6 450.0E-6 Resistance Vs. Power

The measured voltage using a 2.4k Ohm resistor for the second circuit. The measured voltage using a 10k Ohm resistor for the second circuit. The data for the third and final figure are as follows. Figure 6.6 V1 (V) V2 (V) V1 (V) [Total] V2 (V) [Total] Calc. 5V 2.18E+ 0 2.18E+ 0 3.49E+0 -6.51E+0 Calc. 10V 1.32E+ 0 8.68E+ 0 Meas. 5V 2.17E+ 0 2.17E+ 0 3.50E+0 -6.47E+0 Meas. 10V 1.33E+ 0 8.64E+ 0 % Error 5V 0.23 0.23 0.24 0.60 % Error 10V 1.01 0.51 The measured V1 and V2 for the third circuit at 5 volts . The measured V1 for the third circuit at 10 volts.

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The measured V2 for the third circuit at 10 volts 8. Computer Aided Analysis. Using PSPICE to construct each circuit yields these results. PSPICE diagram of Circuit 1 Finding the Vab from this diagram is as shown V AB = 7.510 − 6.169 = 1.341 V 1.341 V is very close to you calculated VAB, 1.34V.

Modified PSPICE diagram of Circuit 1 Finding the Voc from this diagram is as shown V OC = 7.872 − 5.745 = 2.127 V 2.127V is very close to our calculated VOC, 2.128V. Using the calculated resistance from before we can find ISC Isc = 2.127 V 1.93 k Ω = 1.10 x 10 3 A Circuit 2 built in PSPICE. VTH can be calculated as such

V TH = 6.351 − 2.326 = 4.025 V 4.025V is actually the exact calculated VTH value. Circuit 3 built in PSPICE with both voltages in. The V1 value 3.487V is similar to our values of V1 3.49V. 9. Conclusions This lab helped to introduce the concepts of Thevenin, Norton’s, Maximum Power Theorem and Superposition Theorem. While we did go over the theory and the steps on how to calculate each required value, there is still much to learn in each of these concepts. This lab served as a simple introduction if these techniques. We also learned how to use PSPICE to analyze Thevenin’s Theorem and how to account for the voltage changes by modifying the circuit.

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