FROM: TO: DATE: SUBJECT: Redha Al-Hamoud XLIX Management Team November 29, 2013 Electrical Circuit Lab Report The purpose of this project was to practice the engineering design process in designing a simple electrical circuit with limitations and requirements. This project objective included completing calculations using Ohm’s Law, Kirchhoff’s Voltage Law, and Kirchhoff’s Current Law. It also included building an electrical circuit and test it, and finally, documenting all of the work done in a formal laboratory report. The design process started with writing a problem statement that included all design requirements, constrains, and assumptions. The next step was to use Ohm’s Law, Kirchhoff’s Voltage Law, and Kirchhoff’s Current Law to solve and calculate what the resistors needed to complete this circuit successfully. After choosing the right resistors, the circuit was then constructed. The project was completed successfully as the experimental values of the nodes were about ± 5% of the theoretical values. The goals of this project were reached and improved the understanding of the design process and electrical circuits. Redha Al-Hamoud 2 ELECTRICAL CIRCUIT PROJECT Prepared for: ENGR 1201-013 Prepared by: Redha Al-Hamoud Date Submitted: 11/29/13 I have neither given nor received any unauthorized help on this assignment, nor witnessed any violation of the UNC Charlotte Code of Academic Integrity. Redha Al-Hamoud November 29, 2013 3 Summary One of the objectives of designing an electrical circuit was to practice the engineering design process. Practicing Ohm’s Law, Kirchhoff’s Voltage Law, and Kirchhoff’s Current Law was also a main propose of this project. Another goal was to test the electrical circuit designed to make sure that the product has met all the requirements. The project began by assigning each participant with a sheet that contained the voltages at each node, the current, and the source voltage value. Each participant was also provided with a breadboard and 25 resistors that had ± 5% tolerance. In this project, an electrical circuit must be designed and built. An 18 V DC power source was used to power this circuit with a current of 0.0011 amperes. One of the conditions of making this circuit was to use a minimum of one parallel resistor configuration in the design. Research was done regarding electrical circuits, voltage dividers, Ohm’s Law, and Krichhoff’s Voltage and Current Laws. It was found that Ohm’s law is a law that explains how current, resistance, and voltage are linked together as the electrical current value in the unit of amperes is proportional to the voltage, but inversely proportional to the resistance. Engineering calculations were made using Ohm’s law to find the resistance needed for each node. Then, multiple combinations were tested until a combination that was equivalent to the required resistance was found. The circuit was then constructed using the breadboard and the resistors provided. It was tested several days before the test day to insure that the nodes values in volts were equivalent or extremely close to the theoretical values obtained from calculations. On test day, the electrical circuit was tested. The actual values were very close to the theoretical values. Since the maximum percent difference was -0.80%, the product was considered successful. 4 Introduction A particular electrical circuit must be designed and created with some limitations. To power this circuit, an 18 V DC source will be used with a current of 0.0011 amperes. The circuit will include five dissimilar resistors; an equal resistance must be developed by placing resistors in series, parallel, or both. However, a minimum of one parallel resistor configuration must exist in the design. In this project, resistors with ± 5% tolerance will be used. The voltage must be 16.427 V at Node 1, 8.741 V at Node 2, 5.111 V at Node 3, 0.271 V at Node 4, and 0 V at Node 5. Each participant will be provided with a bread board, four wires, and five each of 220 Ω resistors, 330 Ω resistors, 2200 Ω resistors, 3300 Ω, and 10,000 Ω resistors. Figure 1. The Required Electrical Circuit Design. 5 Background Information • An electrical circuit is a closed a circle which is activated by a power source. To power an object, the electricity travels through the circuit and return to the power source, which keeps the current flowing from the positive terminal to the negative terminal. The load is the object which consumes the moving energy and transform it to work. For example, a light bulb would convert the electricity to heat and light (Partington). • There are three forms of circuits that could be used: the series circuit, the parallel circuit, and the series-parallel circuit. In a series circuit, there would only be one way for the energy to flow, whereas the parallel series includes more than one path for the travelling electricity. This is an important factor because when a series circuit is broken, all of the load devices will not function. However, when using a parallel circuit, if one path could not be used, the other paths will still be operating (Partington). • Voltage dividers are one of the widely used basic circuits, and they could be defined as the circuits that turn a high voltage into a low one. An output voltage that is smaller than the input can be built using two series resistors and an input voltage ("Voltage Dividers"). • George Simon Ohm was a professor in Germany in the early 1800s. He documented his findings in a book, which was rejected at first, but recognized and accepted later, which led to the creation of the Ohm’s law. Ohm’s law is a law that demonstrates the way a simple electricity circuit function. This law proves how current, resistance, and voltage are related when used in circuits. The law says that the electrical current value in the unit of amperes is proportional to the voltage, but inversely proportional to the resistance. The mathematical formula for this law is: 𝐼 = 𝑉𝑅, where I is the current in amperes, V is the voltage measured in volts, and R is the resistance measured in ohms (Uy). • The resistor is one of the essential parts of an electrical circuit; it is mainly used to make and keep safe currents in electrical components. Resistors work to resist the current flowing inside the circuit, and they are designed to drop the voltage as the current flows through the circuit. The resistance value could differ between resistors depending on the material used in creating them. For example, to receive higher power levels, a wirewound resistor would be used, which is made out of metal wire and plastic or fiberglass tubing. To create a resistor that is useful in a high temperature environment, cermet, or tantalum would be the correct choices because they are known to absorb heat (Dellaporta). Formulas Used: • 𝑉𝑜𝑙𝑡𝑎𝑔𝑒 (𝑉) = 𝐶𝑢𝑟𝑟𝑒𝑛𝑡 (𝐴) × 𝑅𝑒𝑠𝑖𝑠𝑡𝑎𝑛𝑐𝑒 (Ω) • When resistors are in series: 𝑅𝑒𝑞 = 𝑅1 + 𝑅2 + 𝑅3 + ⋯ • When resistors are in parallel: 𝑅 • % Difference = 1 𝑒𝑞 𝐴𝑐𝑡𝑢𝑎𝑙−𝑇ℎ𝑒𝑟𝑜𝑡𝑖𝑐𝑎𝑙 𝑇ℎ𝑒𝑟𝑜𝑡𝑖𝑐𝑎𝑙 1 1 1 = 𝑅 +𝑅 +𝑅 +⋯ 1 2 × 100% 2 6 Methods and Procedures 1. Each participant was assigned with a specific required circuit to design. The current and voltage were included as a constant. 2. Each participant was provided with a breadboard four wires, and five each of 220 Ω resistors, 330 Ω resistors, 2200 Ω resistors, 3300 Ω, and 10,000 Ω resistors. 3. Design Package process began. a) Problem statement was identified and written. b) Research for related theories and background information was completed. 4. Engineering calculations began. a) ΔV values between the nodes were calculated. b) Req was calculated for each combination. c) Resistance needed was obtained. d) Selecting the best resistance combination between the offered resistors was done. 5. A sketch of the final circuit was made. 6. Construction of the circuit was done following the sketched design. 7. The circuit was tested prior to the test day to insure it was working under the requirements. 8. The circuit was tested on the test day and passed successfully. 9. A formal laboratory report was prepared which documented the details of the project entirely. Figure 2. The Final Circuit Design. 7 Sample Calculations 𝛥𝑉 (𝑉) 𝐶𝑢𝑟𝑟𝑒𝑛𝑡 (𝐴) 1.573 𝑉 0.0011 𝐴 • 𝑅(𝛺) = • Resistors in parallel: • Resistors in series: 𝑅𝑒𝑞1 + 𝑅𝑒𝑞2 = 1320 𝛺 + 110 𝛺 = 1430 𝛺 • % Difference = = 1 𝑅𝑒𝑞1 = = 1430 𝛺 1 2200 𝛺 16.40 𝑉−16.427 𝑉 16.427 𝑉 + 1 3300 𝛺 = 1 1320 𝛺 ⟹ 𝑅𝑒𝑞1 = 1320 𝛺 × 100% = −0.16% Observation and Results After testing the electrical circuit, the experimental values were found to be very close to the theoretical values. Since the values were within ±5% of the theoretical values, the project was considered successfully completed. Table 1 below shows all the values with the calculated percent differences. Table 1: The theoretical and experimental values of the nodes: Node 1 Node 2 Node 3 Node 4 Node 5 Theoretical 16.427 V 8.741 V 5.111 V 0.271 V 0V Experimental 16.40 V 8.75 V 5.07 V 0.27 V 0V Figure 3. The Percent Differences at Each Node. % Difference -0.16 % 0.10 % -0.80 % -0.37 % 0% 8 Discussion The experimental values obtained after the test were very near to the theoretical values found from the engineering calculations. The percent differences were quite low and reasonable. The differences in the values were most probably due to the material used in constructing the electrical circuit. One factor might be the resistors used. Resistors may sometimes not function properly due to manufacturing errors. Also being limited to using specific resistors was a main factor. The results obtained means that designing an electrical circuit needs to be done by following the engineering design process. This is important because once these steps are followed, percentage of failing the project would be relatively low. Overall, the project was considered successful because the differences in values did not exceed ± 5%. Conclusion Due to following the design process steps, the project was effectively accomplished. Defining the problem was done by writing a problem statement, which included the design requirements, constrains, and assumptions. Information was gathered by researching and reading about related theories. After that, multiple combinations of resistors were selected depending on the resistance needed for each node. Finally, the circuit was tested prior to and on test day, and it met all the requirements. 9 References 1. 2. 3. 4. Dellaporta, J. "What is a Resistor?." wiseGEEK. . http://www.wisegeek.com/what-is-a-resistor.htm (accessed November 17, 2013). Partington, Rebecca. "What Is an Electrical Circuit?" wiseGEEK. . http://www.wisegeek.org/whatis-an-electrical-circuit.htm (accessed November 17, 2013). Uy, Karize. "What is Ohm's Law?." wiseGEEK. . http://www.wisegeek.com/what-is-ohmslaw.htm (accessed November 17, 2013). "Voltage Dividers." Sparkfun. . https://learn.sparkfun.com/tutorials/voltage-dividers/introduction (accessed November 17, 2013). 10 Appendix Formulas Used: • 𝑉𝑜𝑙𝑡𝑎𝑔𝑒 (𝑉) = 𝐶𝑢𝑟𝑟𝑒𝑛𝑡 (𝐴) × 𝑅𝑒𝑠𝑖𝑠𝑡𝑎𝑛𝑐𝑒 (Ω) • When resistors are in series: 𝑅𝑒𝑞 = 𝑅1 + 𝑅2 + 𝑅3 + ⋯ • When resistors are in parallel: 𝑅 • % Difference = 1 𝑒𝑞 𝐴𝑐𝑡𝑢𝑎𝑙−𝑇ℎ𝑒𝑟𝑜𝑡𝑖𝑐𝑎𝑙 𝑇ℎ𝑒𝑟𝑜𝑡𝑖𝑐𝑎𝑙 1 1 1 = 𝑅 +𝑅 +𝑅 +⋯ 1 2 2 × 100% Resistor Combination 1: 𝛥𝑉 = 𝑉𝑜𝑙𝑡𝑎𝑔𝑒 𝑆𝑜𝑟𝑐𝑒 − 𝑁𝑜𝑑𝑒 1 𝑉𝑜𝑙𝑡𝑎𝑔𝑒 = 18 𝑉 − 16.427 𝑉 = 1.573 𝑉 𝑅𝑒𝑞 = 1 𝑅𝑒𝑞1 1 𝑅𝑒𝑞2 𝛥𝑉 (𝑉) 1.573 𝑉 = = 1430 𝛺 𝐶𝑢𝑟𝑟𝑒𝑛𝑡 (𝐴) 0.0011 𝐴 = 1 1 1 + = ⟹ 𝑅𝑒𝑞1 = 1320 𝛺 2200 𝛺 3300 𝛺 1320 𝛺 = 1 1 1 + = ⟹ 𝑅𝑒𝑞2 = 110 𝛺 220 𝛺 220 𝛺 110 𝛺 𝑅𝑒𝑞1 + 𝑅𝑒𝑞2 = 1320 𝛺 + 110 𝛺 = 1430 𝛺 = 𝑅𝑒𝑞 Resistors used: 1 × 𝑂𝑂𝑅 (3300 𝛺) 1 × 𝑅𝑅𝑅 (2200 𝛺) 2 × 𝑅𝑅𝐵 (220 𝛺) Node 1 % Difference = 16.40 𝑉 − 16.427 𝑉 × 100% = −0.16% 16.427 𝑉 Resistor Combination 2: 𝛥𝑉 = 𝑁𝑜𝑑𝑒 1 𝑉𝑜𝑙𝑡𝑎𝑔𝑒 − 𝑁𝑜𝑑𝑒 2 𝑉𝑜𝑙𝑡𝑎𝑔𝑒 = 16.427 𝑉 − 8.741 𝑉 = 7.686 𝑉 𝑅𝑒𝑞 = 𝛥𝑉 (𝑉) 7.686 𝑉 = = 6987.27 𝛺 𝐶𝑢𝑟𝑟𝑒𝑛𝑡 (𝐴) 0.0011 𝐴 𝑅𝑒𝑞1 = 3300 𝛺 + 3300 𝛺 = 6600 𝛺 1 𝑅𝑒𝑞2 = 1 1 23 1 + = ⟹ 𝑅𝑒𝑞2 = 23 = 286.96 𝛺 2200 𝛺 330 𝛺 6600 𝛺 6600 𝛺 11 𝑅𝑒𝑞1 + 𝑅𝑒𝑞2 = 6600 𝛺 + 286.96 𝛺 = 6886.96 𝛺 ≈ 𝑅𝑒𝑞 Resistors used: 2 × 𝑂𝑂𝑅 (3300 𝛺) 1 × 𝑅𝑅𝑅 (2200 𝛺) 1 × 𝑂𝑂𝐵 (330 𝛺) Node 2 % Difference = 8.741 𝑉 − 8.75 𝑉 × 100% = 0.10% 8.741 𝑉 Resistor Combination 3: 𝛥𝑉 = 𝑁𝑜𝑑𝑒 2 𝑉𝑜𝑙𝑡𝑎𝑔𝑒 − 𝑁𝑜𝑑𝑒 3 𝑉𝑜𝑙𝑡𝑎𝑔𝑒 = 8.741 𝑉 − 5.111 𝑉 = 3.63 𝑉 𝑅𝑒𝑞 = 𝛥𝑉 (𝑉) 3.63 𝑉 = = 3300 𝛺 𝐶𝑢𝑟𝑟𝑒𝑛𝑡 (𝐴) 0.0011 𝐴 Resistors used: 1 × 𝑂𝑂𝑅 (3300 𝛺) Node 3 % Difference = 5.07 𝑉 − 5.111 𝑉 × 100% = −0.80% 5.111 𝑉 Resistor Combination 4: 𝛥𝑉 = 𝑁𝑜𝑑𝑒 3 𝑉𝑜𝑙𝑡𝑎𝑔𝑒 − 𝑁𝑜𝑑𝑒 4 𝑉𝑜𝑙𝑡𝑎𝑔𝑒 = 5.111 𝑉 − 0.271 𝑉 = 4.84 𝑉 𝑅𝑒𝑞 = 𝛥𝑉 (𝑉) 4.84 𝑉 = = 4400 𝛺 𝐶𝑢𝑟𝑟𝑒𝑛𝑡 (𝐴) 0.0011 𝐴 𝑅𝑒𝑞1 = 2200 𝛺 + 2200 𝛺 = 4400 𝛺 = 𝑅𝑒𝑞 Resistors used: 2 × 𝑅𝑅𝑅 (2200 𝛺) Node 4 % Difference = 0.27 𝑉 − 0.271 𝑉 × 100% = −0.37% 0.271 𝑉 Resistor Combination 5: 𝛥𝑉 = 𝑁𝑜𝑑𝑒 4 𝑉𝑜𝑙𝑡𝑎𝑔𝑒 − 𝑁𝑜𝑑𝑒 5 𝑉𝑜𝑙𝑡𝑎𝑔𝑒 = 0.271 𝑉 − 0 𝑉 = 0.271 𝑉 𝑅𝑒𝑞 = 1 𝑅𝑒𝑞1 𝛥𝑉 (𝑉) 0.271 𝑉 = = 246.36 𝛺 𝐶𝑢𝑟𝑟𝑒𝑛𝑡 (𝐴) 0.0011 𝐴 = 1 1 1 + = ⟹ 𝑅𝑒𝑞1 = 132 𝛺 220 𝛺 330 𝛺 132 𝛺 12 1 𝑅𝑒𝑞2 = 1 1 1 + = ⟹ 𝑅𝑒𝑞2 = 110 𝛺 220 𝛺 220 𝛺 110 𝛺 𝑅𝑒𝑞1 + 𝑅𝑒𝑞2 = 132 𝛺 + 110 𝛺 = 242 𝛺 ≈ 𝑅𝑒𝑞 Resistors used: 3 × 𝑅𝑅𝐵 (220 𝛺) 1 × 𝑂𝑂𝐵 (330 𝛺) Node 5 % Difference = 0 𝑉 × 100% = 0%
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