W5 Law Report

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Grantham University *

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EN105

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Electrical Engineering

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Jan 9, 2024

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Kaila Gough G00215407 Week 5 Lab Grantham University October 23 rd , 2023

Introduction: Write one to two paragraphs about the Lab. Explain the following information for this lab:  What are the goals to achieve in the lab? The goals that this lab is set up to achieve are simply a good round about feel for the process and ways of properly working with DC circuits and using the various measuring tools to find the voltage, resistance, current, and power. Also, to learn how to apply Ohm’s Law to various situations within DC circuits as well when it comes to figuring out a missing value for a component.  What are the expectations of the lab? The expectation for this lab is to simply build self confidence in working with DC circuits, power sources, and building circuits properly. Anyone who is set out to become some type of engineer will need to know these things to succeed in the work field. Overall, the expectations being asked of whomever completes this lab is the values across the components. These will need to be strategically calculated using Ohm’s and Kirchhoff’s Law.  How will you be implementing this lab? This lab will be completed using the virtual Multisim Software to build and figure each of the component values either by using the tools within the software or by using Ohm’s Law.  What will you try to measure? All components within the circuit will be measured using the Multisim software virtual tools, such as the multimeter and the wattmeter. The values for current, voltage, resistance ,power, and the % Error will be measured and also calculated for later comparison and analysis. Equipment/Components: This lab requires a 20v DC Power source(V1), 3 grounds, and 6 different resistors. Each resistor has a different value. R1 = 1kohm 5%, R2 = 2.2kohm 5%, R3 = 3.3kohm 5%, R4 = 4.7kohm 5%, R5 = 5.6kohm 5%, R6 = 6.8kohm 5% Tolerance on each of the resistors as well. All have the same tolerance value of 5%. That completes the list of components needed. For the equipment needed to take the measurements, 1 or more multimeter(s) and a wattmeter is needed. Procedure: The lab can be worked out using the Ohm’s Law formula to find either the current, power, resistance, or voltage. Ohm’s law formula V = I * R, R = V * I, I = V * R; Also ohm’s law states that the current through a conductor between two points is directly proportional to the voltage across the two points. This statement is useful when figuring either voltage or current within a component. Voltage equals resistance multiplied by current. Resistance equals voltage divided by current. Current equals voltage divided by resistance. Since this is a parallel series circuit, the total resistance must be calculated separately. First the parallel resistors can be totaled up and the formula is different than the series circuit resistors. That formula will be listed under calculations. After you take that measurement, you will simply add it with the series

resistors total and find the sum of those two values. This total value will be the calculated total resistance value. The voltage will be the same across each resistor within a parallel circuit. The total current within the parallel branch of the circuit is going to be the sum of the resistances. The voltage of the parallel circuit has the same value as the voltage across each branch. The total current is the sum of the currents measured from all components. When measuring the current within the circuit you will set up the multimeter as if the meter were part of the circuit and run simulation so that the current travels through the positive lead and out the negative lead and then continues around the remaining part of the circuit. To measure the total power in the circuit, the wattmeter can be utilized and set up accordingly. The voltage half will be set up across the component and the current leads will be set up to measure the total circuits values. To do this, disconnect the source from the circuit and connect one lead to the beginning of the circuit and the other lead to the end of the circuit just after the ground. This will give a accurate measurement of the total power value. Calculations: To find the Rtotal (RT) of the circuit first the circuit must be simplified by combining the resistors in series and parallel. Second, apply Ohm’s law and Kirchhoff’s law to find the current and voltage across each resistor. Third, use the formula RT = V/I-total which means you divide the voltage by the total current to find the total resistance of the circuit. R 1+2 = 1000 x 2200/ 1000+2200 = 2200000/3200 = 687.51 Ω is the total resistance of the parallel resistors. Next, find the total for series resistors. Rt= R3+R4+R5+R6 Rt= 3300+4700+5600+6800= 20400Ω. Next find the equivalent resistance of R12 and R3456. Rt = 687.5 x 20400/687.5+20400 = 14025000/21087.5 = RT = 665.1Ω. The calculated Total Resistance of the circuit is equal to 665.1Ω. Total current is found using Ohms law, I=V/R = 20/665.1 = 0.0301A or simplified as 30.1mA. To find the voltage across VR1 you can use this formula: V1=I1xR1. the resistance value of R1 is 1000Ω. To find the current of R1, use this formula: I1=V/R1 VR1=V/R1 = 1.691/1000=0.001691A or simplified to 1.691v is the voltage through R1. To find the power across R1 the formula is P1=V1xR1 P= 1.691 x 0.001691 = 2.861mW Since R1 and R2 are both parallel this means that the voltages and current must be the same.

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So, VR2 = 1.691v IR1 = 1.691A The calculated power dissipation is equaled to 1.286mW. Next, we can calculate R3 voltage value by using Kirchhoff’s voltage law. First, we must find the current using this formula: I = V/R, then we can use this formula to find the power: P 3 =V 3 I, and finding the voltage can be done by this formula: V 3 =IR 3 The resistance of R3 is equal to 3.3kohms. The current is equal to the total current in the circuit which is 665.1 Ω . In order to calculate the percentage error of the measurements taken from the components within the circuit of this lap. The following formula can be followed: % Error = Actual measurement – expected value/ expected value * 100% So each component will have the % error calculation listed at the end of the analysis and also organized in a chart. Circuit design: This screenshot shows the circuit design of the circuit before any simulation has taken place. It has a 20V DC power supply beginning with a ground then to DC supply then on to R1 which is a 1kohm resistor and it is running in parallel with another resistor that is 2.2k ohm. Next it leads to another resistor in series that is 3.3kohm, then three more resistors that are also in series. R4 is equal to 4.7kohm, R5 is equal to 5.6k ohms and R6 is equal to 6.8k ohms. All the resistors have a 5% tolerance rating.

i. This screenshot shows the total Resistance within the circuit. Resistance is equal to 8.004k ohms ii. This screenshot shows the measured total current value of the circuit. IT = 2.488mA

iii. This screenshot shows the total measured value for power within the circuit. PT = 47.964mW

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iv. This screenshot shows the voltage values being measured across each of the components within the circuit. The voltage values for the components are as follows: V R1 = 1.704v VR2 = 1.704v VR3 = 8.179v VR4 = 4.663v VR5 = 5.454v VR6 = 10.117v

v.This screenshot shows the current through every component within the circuit. The current values are as follows: IR1 = 1.712mA IR2 = 2.472mA IR3 = 2.472mA IR4 = 999.893uA IR5 = 999.893uA IR6 = 1.473mA Analysis: The following table displays all the measurements that were taken within the lab. It contains the experimental and the theoretical measurements along with the percentage error between the two. Category Calculated Measured % Error Rt 6.651k 8.004k 16.90% It 3.01mA 2.488mA 20.90% VR1 17.12v 1.704v 904.60% VR2 19.8v 1.704v 1025.50% VR3 19.8v 8.179v 142.08% VR4 19.975v 4.663v 328.37% VR5 19.992v 5.454v 266.50% VR6 19.992v 10.117v 97.60% Pt 6.645mW 47.964mW 86.10% PR1 4mW 2.837mW 40.90% PR2 3.96mW 1.358mW 191.60% PR3 1.188mW 20.211mW 94.12% PR4 8.509mW 4.826mW 76.31% PR5 7.14mW 5.432mW 31.40% PR6 5.88mW 15.460mW 61.90%

Conclusion After double checking my work and measurements across all the components and within the entire circuit, I am confident that my values are correct. It was discovered that all the measurements were not going to be the same. There is always going to be some level of difference between the calculations and the measured values that were taken. I assume my calculations are correct as I followed Ohm’s Law to calculate the measurements of the current, voltage, power, and resistance of the total circuit as well as each component. Some of the calculations that were measured using multisim were relatively close with the calculated measurements. Although some were not, I am not exactly sure if I may have made a mistake or not. Although I looked over my work and I am feeling confident that all values are current. To troubleshoot this circuit, the equivalent resistance must be calculated along with the total current and the voltage drops using the series and parallel rules. In a real-world situation, normally when I am just fixing electronics that burn out and such, I would just look over the components on the greenboard and look for any kind of corrosion or visibly burnt-out components. 80% of the time I run across corrosion that just needs to be cleaned up and the current is then allowed to continue through the component that was corroded and shorting out the needed power to properly operate the device. Part 2 1. 2.876 + 17.124 = 20v Pt = 47.964mW

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“The circuit currents vary from branch to branch, and the component voltage drops depend on the branch currents and on the component resistances. The supply current depends on the supply voltage and on the circuit, resistance offered to the voltage source. Kirchhoff’s voltage and current laws are applied for analyzing series-parallel circuits.” 2. After analyzing the calculations that I took according to the resistor values. I determined that the R6 resistor is the fault. I concluded this because the answer that I figured compared to the multimeter value that is shown is not a match. The answer I calculated is 2.94v and the multimeter shows a much higher voltage of 17.124v. 3. When the resistor is delivering a much higher than normal voltage than it should it is due to it having an open fault. Higher resistance will drop a higher voltage. R6 is the resistor within the circuit that has a open fault present which is throwing a much higher voltage than it should. This fault can be due to bad solder, broken wires, or poor contacts. (Electrical Academia, 2022) electricalacademia.com/basic-electrical/series-parallel-circuit-series-parallel- circ…