Question

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

2. Using a DC Sweep analysis to generate a table of values in the .OUT file: See Circuit 2 below: Let R1 through R6 equivalent to the digits 1 through 6 in your StudentID in k12 (use 10 k22 for a digit of 0). • For example, if your StudentID is 9870654 then R1 =9 kN, R2 =8 kN, R3 = 7kN, R4 = 10KN, R5 6 kQ and R6 = 4k.22 A. Analyze Circuit 2 by hand to determine • The current through R5 • The voltage across R2 B. Analyze Circuit 2 using PSPICE as follows: Use a DC Sweep analysis to find the quantities listed in part A above. Let the voltage source vary from 0 to 50V in increments of 5V. • The OUT file should contain two tables of values. One table will show the voltage across R2 as the source voltage varies. The line in the table where the source voltage is 25V should match your voltage calculation from part A (highlight this line when you print it.) Similarly, the second table will show the current through R6 as the source voltage varies. The line in the table where the source voltage is 25V should match your current calculation from part A (highlight this line when you print • Do not use a Bias Point analysis or show any Bias Point values on the schematic. • Be sure to add text to the schematic as indicated in the Sample Report. C. Print the schematic and the .OUT file. Include a table comparing hand values and PSPICE values from the OUT file when the source voltage is 25V. Also include a brief discussion of the results. it.) R6 + R1 V2 R2 R5 25V Is R3 R4 W Circuit 2

Answer 1

Ang 4K A + Zqk ak V2 Is + 25 6k 4k} LOK В This curauit can be drawn as Чk M A t * This is unbalanced qk .8k V₂ + wheat stone bridge. 25 in C 'D 6k Slok * This circuit can be qk solve by making its B Thevenins equivalent circuit A circuit. so thevenin's equivalent will be look like ; {across across resistor 6K} REL here we assume 6k Ск th as Road Resistor D when Now Rth = equivalent Resistance of ciraut all independent sources are remove & lead resistance will be open. for Rth circuit A will be A 9k 8k Yk p B lok 7k

A ak 88k mm 44 Ik Lok B A Rth = 16k {uk {4.4446 В Rth = 1.060kr Now th= open circuit volt, when load resistor is remove. Yk www A 9k 8k + 25v tar 7k lok B Apply ke at Node A. V-25 v t t V = 0 126 166 ЧЕ 25 V- xk Чk 0 368 V16 981 volt

Νοω Vak Il 16.981 x x 7K 16K Vak = 7.429 volt Vok = 16.981 % lok 1884 NIOK = 9.483 volts Now 4k M 250 qk ak at the 7.429² 9.433 ot KvL in that loop. Nth = 7.429 -9.433 = 2.004 volk. Now circuit will be 1.gbok Is = -2.004 6k 1-86kt6k 2.0046 Y IS Is = -0.2549 mA -ve means, actual direction of current will be from node 'o ato mode C. V2 Now 니냐 8K {qk V2 장 Vq Vio I GK D k 0.254mA 3 lok

kal at c V -V2 + (-0.25५) V + १ Va Vr k = * ०.254 १ + 1254 ०.२७५ V - V2 +। 7 (1) 2.285 V= = 0.437 V2 +1 Kch at A V2-25 + 2-V + Vz-Vio -0 ५ १६ 26 2x0.५२८ २१२८ 25 4 + 0.048512 to + VLO 25 V२ = + 0.।।। Vio + ५०.435 0.435 80.435 (२). VIO V2 14.28 + 0.2537 + 820.4375. keh at Node D Vio VIO-VZ 8 + 0.254 mA + 30 lok Vo - V२ + 0.254 7 (3) 80.225 0.225 from en (2) & (3) VO : 8.0+ + Vox0.15873. Vo - १.592 volti