Elec Sys Lab 3 Report

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University of Texas, Rio Grande Valley *

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2317

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Electrical Engineering

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Apr 3, 2024

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Lab #3: Kirchhoff’s Laws & Superposition ELEE 2317 – 01 Ever Acosta Gerardo Zapata 02/16/2024 Page 1 of 4

A. OBJECTIVES • Further practice using voltmeters, ammeters, and ohmmeters for DC measurements. • Observe application of Kirchhoff’s Voltage and Current Laws to circuits. B. EQUIPMENT REQUIRED • Benchtop Digital multimeter • Breadboard • Power Supply (Dual or Triple Output) • Miscellaneous Cables • 1/4 Watt or 1/2 Watt Resistors; one each of following values: 1.2 kilohms, 1.5 kilohms, 2.0 kilohms, 2.4 kilohms, 3.6 kilohms • Hook-up wire (#20 or #22 solid conductor) • E. IN THE LAB To set up the circuit, first, identify the resistors from the parts list and measure their resistance using an ohmmeter, recording both the measurements and color codes. Next, configure the power supply to produce two outputs: one at 14 volts and the other at 5 volts. Verify the accuracy of these voltages using a voltmeter with a tolerance of +/- 0.1 volts. Afterward, construct the circuit according to Figure 3-4, aiming to arrange the components on the breadboard in a layout resembling the schematic diagram. This step ensures a systematic and accurate assembly of the circuit.  Red/Yellow/Red/Gold (2.4 kOhms) = 2.386 kOhms  Red/Black/Red/Gold (2.0 kOhms) = 1.980 kOhms  Brown/Green/Red/Gold (1.5 kOhms) = 1.494 kOhms  Orange/Blue/Red/Gold (3.6 kOhms) = 3.514 kOhms  Brown/Red/Red/Gold (1.2 kOhms) = 1.172 kOhms Page 2 of 4

Then we changed the multimeter to operate as an ammeter and moved the red test lead to a new jack on the front of the multimeter. We measured the currents shown in the image below and made sure to have the correct signs. We added the current algebraically for node N1 (3,4,-5) and node N2 (5,6,-7). After this, we added currents I3, I4, I6, and I7 and did the same for I5 and I8.  N1(3,4,-5) = (-4.68) + (2.43) - (2.25) = 0  N2(5,6,-7) = (-2.25) + (2.17) – (-0.08) = 0  I3 = -4.68 mA  I4 = 2.43 mA  I5 = -2.25 mA  I6 = 2.17 mA  I7 = -0.08 mA  I8 = 2.25 mA Figure 3-6 (a) and (b) Surfaces (a) "Kill" the 5V source in your circuit by disconnecting the 5V supply and replacing it with a short circuit. This is shown in Figure 3-7 both schematically and in a breadboard drawing. Measure the voltage VA shown in the figure, as well as the current IB. VA 1 = 9.62 V IB 1 = 1.39 mA (b) Restore the 5V source by removing the short circuit wire and reconnecting the 5V supply. Then "kill" the 14 Volt source, using a method similar to that used in the previous step. (No figure is shown.) Again, measure the voltage VA and the current IB. Page 3 of 4

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VA 2 = 0.8545 V IB 2 = 0.79 mA (c) Restore the 14V source, so that the circuit is back to its original configuration. Again, measure the voltage VA and the current IB. Note that the results in this case should closely match the results for V4 and I5 in Figure 3.5. If there is a significant difference, you have probably not properly restored the circuit to its original configuration. VA = 10.47 V IB = 2.20 mA (d) Finally, add the results for VA 1 and VA 2 from steps (4a) and (4b) and compare the result with that from step (4c). Does superposition seem to be working in this case? Do a similar comparison for the current IB 1 + IB 2 . Show your work: VA 1 +VA 2 = 10.4745 V IB 1 +IB 2 = 2.18 mA Does superposition work? Explain . Superposition DOES work and this was proven on part D when we added VA 1 +VA 2 and it was equal to VA in part C. And, when IB 1 +IB 2 was summed it also was equal to IB from part C. Figure 3-7 Circuit with 5V Source "Killed" Page 4 of 4