Question

Consider the circuit shown in Fig. 1: 1. Given the values of the resistors are R;=1.2k12 , Rz=3.3kN and R3=5.6k22, calculate values of the voltages V1, V2 and V3, and the currents I1, I2 and 13 using any other method except Superposition theorem. 2- Verify the results using QUCS. + V 1 + V3 Iz RI R E-15 V E-10 V Fig. 1 3- Consider the circuit shown in Fig. 2: + V I' I'3 + V's www I'2 R3 RI + E-15 V R2 Fig. 2 4- Calculate the values of the voltages V, V, and V, and the currents I, 1", and I'), 5- Verify the results using QUCS 6- Consider the circuit shown in Fig. 3: + V" I". + V"; - 1"} w 192 RI E-10 V R R2 Fig. 3 7- Calculate the values of the voltages V", V", and V)", and the currents 1", 1", and I". 8- Verify the results using QUCS 9- Sum respectively the voltages and the currents found from (4) and (7) and compare with the results found from (1). Comment on the results.

please ignore Qucs questions .

Answer 1

Answere 1 Given the values of the resistors are R,=1-2 ur R₂ = 3.3 kr R3² = 5.6 nr calculate the values of cunnents Tu, I2, I3 Voltages Vi, Ves V3 and Circuit + Is 3 + I V3 mums R3 mum Iz RI + + Vr Ez = 101 R, E, -15V at Node А Using Node analysis T + I3 I, V E, - V₂ V₂ - E > t R2 Ri R3

Putting values. 15 - V₂ V2 V2 - 10 + 102x103 3.3 X103 5.6 un 15 - > V₂ t + 10 + 5.68103 3.3 x103 5.68 103 1.2 X103 1,2x103 Nolts 10. 864 V2 = -3 Vz 10.864 A IL - 3.29 X 10 R₂ 33% 103 I 3.29 mA 15 - 10.864 E, - V2 -3 3,44 X10 A I, RU 1-2 x103 I = 3.44 mA - 15- 10.864 E-V₂ 11 st 4.136

Vi = 4.136 Volts Iz 11 Vo Ez 10.864 - 10 -6 =154.28X10 R3 5.6 x 103 Ig 154.28 M A ws 11 V2 - EL = 10.864 - 10 = 0,864 v v Va 0.864 Results we found - Using Hoad analysis 4.136 Volts I = 3.44 mA VE I · 3. 29 mA 2 V2 - 10. 864 Volts V3 0.864 Volts - 154.28 UA wy =

3 Find values N',v,', ve and currents T,', [' and I' + " Vs It' + I, 2 R3 t Ei 15. Rz 2 1 + V2 equivalent न Ro+CRIIR) Req Req - Rit R R₂ 3.3 x 5.6 + 11 = 1.2 R₂+R₂ 3.3 +5.6 Req= 3.27 kr I,' E, 15 s - = 4.57 M A Req 3-27 Купи 4.57 ma 며

11 I,' x R, = 4.57 x 1.2 = 5,484 v V? 5.484 Volts 5.6 TET X 4.57 X R3 Rat Rg 11 5.6 +3.3 = 2.87 m A 9.87 m A It 2.87 X 3,3 - 9.48 Volts = RO V2 N = 9.48 Is I,' - [' 2.87 4.57 Il - 1.70 mA cott - 1.70 1.70 mA V3 ķ' 9.48 v

Results of Figure 2. I,'. 4.57 m A Vi' - 5.484 v V₂ = 9.48 V Il = 2.87 mA Vs' = 9.48 v I'- 1.70 mA 6 In figure 3 calculate : 7 vi", vz" V" and I.", I,"15" Circuit u/ // + V 11 I, R₃ 13 muma RI um t I VS Vz + CV zlov 1

equivalent resistance of cricult Rea Rgt (Ri 11R₂) 3.3 X 1.2 5, 6 to 6.48 Kar 3.3 +1:2 10 11 10 I 6-48 kan Req / I 1. 543 mA 1 -1.543 x 5o to Ver 11 Iz" XR = 8.64 1 Volts ve" 8, 641 Volts 11 11 Rz 3. 3 I," T X -1.543 X RA TR₂ 3.3 + 1.2 - 1.191 mA 1," 11 -1.131 MA

V" TI" ORI -). 1314 1.2 - -1.35 Volts V,11 -1.35 V 11 11 w) I" - I" Al -1.131 -(-1.543) IN 11 I' O. 412 A N2" 11 I" XR₂ 0.412X 3.3 1.359V 1.359 Volts V₂ " 3 Results of figune vi" - 1.35 V I,''ccl.131 m A 11 V" = 1.359 = 0.412 MA V3" Ne -8.6410 I" - -1.543 M A

a Sum of Noltage and canned in 4 anda. Il VS vi' 1,35 t vi :5.484 s M 4.134 s 10-839 5 Vz 5 NatV2" & 4.48 + 1.359 V₂ = 10.839 v 1 " = 0.839 as V3 tv3 = 9.48 - 8.641 S V₂ = O. 839 V 1,5 It' + I," 4.57 -1.131 = 3.439 IS = 3.439 m A 11 IS = In th 2.87 +0.412 3.282 m IS c 3.282 MA

S TS 1 Iz to" 11 1.54 3 I 1:70 = 0.157 S „Н - 0.157 mA Finally comparison of sum of results with initial results Sum of Results Initial Results us= 4.134 V Vi - 4.1360 V2 = 10.86 4 U V₂S 10.839 v z = 0. 86.4 V VES 11 0.839 v I= 3.44 MA IS 3,439 m A 11 s I = 3.29 MA 1, 11 3.282 MA s Iz= 154.28 UA I 11 0.157 mA 157 MA 11 ( are approximately see we can two result's same. (upto one decimal values.]. calculation nicely done, so our w as