Question

PLEASE SHOW ALL WORK AND TO FOLLOW DIRECTIONS. PLEASE USE PSPICE.
STUDENT ID= 7220849

3. Using a DC Sweep analysis to graph currents and voltages Let R1 through R4 equivalent to the digits 1 through 4 in your StudentID in k 2 (use 10 k12 for a digit of 0). • For example, if your StudentID is 9870654 then R1 =9 KS2, R2 =8 kN, R3 = 7k2 and R4 = 10k12 A. Analyze Circuit 3 by hand to determine The currents through R2 and R4 • The voltages across R1 and R3 B. Analyze Circuit 3 using PSPICE as follows: Use a DC Sweep analysis and let the voltage Vx vary from 0 to 100V in increments of 2V. Note that current and voltage printers are not required since the results will only be graphed. • Be sure to add text to the schematic as indicated in the Sample Report. • Do not use a Bias Point analysis or show any Bias Point values on the schematic. • Be sure to label all of the nodes. Create a graph showing the currents through R2 and R4 (if they are negative, you will need to rotate the resistors by 180 degrees). The voltage Vx should be on the x-axis. Turn on a cursor in and place it as close as possible to where Vx = 50V on I(R2). Mark this point. Move the cursor to the waveform for I(R4) and similarly mark the point where Vx = 50V. These points should match your results from part A. Be sure to add text to each graph (name, course, problem number, and purpose). Create a second graph showing the voltages across R1 and R3. Note that the voltage VI will be referred to as V(A,B) as PSPICE only uses node voltages. Similarly mark points on each curve where Vx = 50V. Again add text as required for all graphs. D. Print the schematic and both of the graphs. Include a table comparing all hand values with all PSPICE values marked on the graphs when the source voltage is 50V. Also include a brief discussion of the results. vi V3 R1 R3 12 14 50V Vx R2 R4 Ix 20mA Circuit 3

Answer 1

+ R R3 A VB {I2 AKS Vc In .Ru JOKDL SOY Vse 8K2 & R₂ 20mA Apply Nodal Analysis at Node v8: → + VB-50 Ri VB Rai + VB-VC 20 R3 VB (kit kat kha) - velkz) = sa NB (5 + 5 + ) - velt)- Apply Nodal Analysis 9 Analysis at Node ve : Ve 20 mA Vc - VB Rz = 25mA Vc Vc 7 Ry l Rz + Ru) - velks) 2 + to) - vB 1 ) = 20 mA By solving eqh 0 and eqh ③ , we get VB = 58.71 v Vc = 117, -SD-O ķE VB TV V = 50-VB = 50 - 58.71 VB - VG-Veso V3 = VB -vc = 58.71-117 H3 = -58.29 y -8.71v)

58.71 VB R. =7.33mA I 2 27.33mA I 2? 8 In ve u 117 11-7mA I4² 11.7mA 10 VB = 58.71 Vc = 117 V Vi = -8.717 Vz=-58.29V I q = 7.33mA I4 = 11:7mA.

PSPICE WORK

R1 R3 WW Sk 11 Ovde R2 BK R4 10k 20mAdo

16nn (5006 113590m) 1201 (50 000 7340 m) AnA Ann 100 DI(R?) 260 3 A DU 6 OU 1 ou 101 1(RN) UUT ADU (50 000 - 1290) OU - ADU (50,000 58 1721 - DU 21 3 ABU 1 DU U(B) - U(C) SOU 6 00 7 00 U(A) - U(B) BU 100

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