Question

Vw Vout V2 SR, The circuit above has R1 = 1.5 k22, R2 = 2 k22, R3 = 1 k2, R4 = 2 ks, R6 = 2 k2, R2 = 750 12, Rg = 4 k22 and Rg = 1 k12. If V2 = 3 V + 3 sin(2000nt) V, determine R5 and the dc voltage v1 so that the output is 5.4 sin(2000nt) V. O v1 = -0.5 Vdc and R5 = 1 k12 Ov1 = 0.7 Vdc and R5 = 2 ks2 Ov1 = -2 Vdc and R5 = 1 k12 O v1 = -1.2 Vdc and R5 = 2 k12

Answer 1

View vor RG RC Roll- VO & Ra Ri= 1.5kr, R2 = 2k , R3 = 1k , Ry-2ks, R6214 Ry=1502-0.25kr R8 = uker & Rg=lku U2 = 3 V+3 sin (2000 ) vout= 5.4 sin (2000) v a) = -0.5 dc & Rs=1kr. b) v=0.3 Udc & R5 = 2kr. c) 0,= –2vdc & Rs=ke d) Vis -7-2 Oda & R$ 2k find up up I 2 Up 21.1K Kaun up - 1 h & I Up 2

ground concept [A-] due to virtual upaon ปก 6 find Vol R6=2kr uring nodal analysie RS uno & Un Vol Ud VO R8=4 fullt R5 vole {Razku ( +2) Una vol= (1 + 2 har un ground concept find vor due to virtual (A=00 due to a R2=21s Up-un valente a i uportuno = Up Uring nodal Analysis o-v o -voz =0 + 2 A 15 0 10.2 =

Ruzak u ponu de vout using nodal Analysy volvo Voz-vout to 2 (1+1) Volvo – bout it 1823 [142 JUD- (+40) Siy 5.4 cin(roudTt) u = 2:7 sinhrott) 3 [142] [3+3.4in (2010)] +0, Since here we have 2 Unknowns so we connot find R5 & W, directly. So with the help of geven options we will find &, & R5. Tirst check option (a) y os UHI & RS alku Substitude &, & Rs in equation 6, u (1+2)/34 85TN (2007) 7 + x - 22.7 sincronit) -1.8 + 2.7 sin luont) + 29 sin (Zvolt lv to U -0.5 Ude & Rroku are not Dequired values

so we check for option (b) in equation 0 URO.nunc & Break substitute these values 2.7 sincrovont) 2.734 118 sin (wollt) these are not required valle So another option (C) I so will we Ud & RS = 1ks substitute these values in equation 0 2:7 sin ( 2000lt) = 2-7 sin ( 2dit) These values are gete satisfied so we need not check other option.. so requered values are 12 Ude & R5=1ka