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Lab #2: Introduction to Circuit Simulation ECOR 1043: circuits Lab Section: L 11 A-2 Group member 1: 101154328 Thana Taher (Question 2) Group member 2: 1011457765 Hamza Adan (Question 1&3) Performed on: 29/09/2020 Submitted on : 06/10/2020

Part 5.5 - (6)Take a clear screenshot of the graph after ensuring that it is a horizontal line at 2.0V [ /1] - Figure 1: screenshot of the horizontal line at 2.0V

- (9) Take a screenshot of the resulting graph. [ /1] Figure 2: screenshot of the two series resistors breadboard and schematic - (10) What value do you now see across the same resistor (R1) and why it is different this time? (Note that you can see the exact value on the graph by clicking on the Toggle cursor button . This will show you the exact value of the line in the bottom right window). Answer: VR1 1.0 = [ /1]

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- (11,12) Now move the V+ probe to the left of R2 and V− probe to the right of R2 so that you can measure the voltage drop across R2. Run the simulation and take the screenshot.[ /1] Figure 3: screenshot of the V+ probe and V- probe across R2 - (13)What is the voltage drop across R2? Answer: VR2 = 1.0 [ /1] - (16)Calculate the voltage drop across R1 by showing your steps Answer: VR1 = 2.000V [ /1]

- (18)Take a screenshot with all voltages and components clearly visible. [ /1] Figure 4: screenshot of the voltages and components across the circuit

Part 5.6 - (5)Take screenshot of the resulting graph. [ /1] Figure5: screenshot of the resulting graph of the currents over the circuit - (6)Record the values of the following three currents: Answer: IR1 = 754.6uA Answer: IR2 = 340.6uA Answer: IR3 =414.6uA [ /3]

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- (10)Take a screenshot with all currents and components clearly visible [ /1] Figure 6: screenshot of the current and components over the circuit Part 5.7 - (5) This will show power value labels on various components. Record their values [ /4] Answer: PR1 = 1.253 mW Answer: PR2 = 115.6uW Answer: PR3 =140.9uW Answer: PV 1 = -1.509mW

- Take a screenshot with all the powers and components clearly visible. [ /1] Figure7: screenshot of the power of the curit over R1,R2, R3.

Part 5.8 - (4)Take three separate screenshots (move the labels if needed for clarity) which should show the values of the following parameters (on the circuit diagram and not on the graph) for all the five components of the circuit [ /3] (a) Voltages Figure8: the voltage value across the circuit.

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(b) Currents Figure 9: the current value across the circuit. (c) Powers Figure 10: the power value across the circuit.

Table 1: Circuit Parameters, the current and the voltage and power over R1,R2,R3,V1, and V2 R1 R2 R3 V1 V2 i(uA) 928.6 142.9 785.7 928.6 142.9 v(V) 25.00 15.71 15.00 0 0 P(mW) 8.622 102.0 12.35 -23.21 2.143 Discussion Question: 1) In my pre lab using the voltage divider rule to calculate the voltage drop across the two resistors (R1 & R2 ) in figure 2 the two voltages I got were for VR1 = 1.375V, and VR2: 0.645. Furthermore in the lab we calculated the voltage drop using the software available and we got a value of 1.0 for Vr1 and 1.0 for Vr2. I think we made a mistake during the lab this is the only possible answer for having no voltage drop across both resistors, I think this could have happened when we made circuit we might have built an open circuit, which means there is no current flowing in the circuit thereby no voltage drop because voltage drop is equal to IR.

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This were the results for the prelab V_V1 N10 25V R_R1 N1 N2 10k Ω R_R2 N2 N3 5kΩ R_R3 0N2 20 k Ω V_V2 N30 15V From figure 3 and the net results of the prelab, when comparing the net list there is a major similarity between these two.