Convert this electrical network to block diagram, simplify and derive the transfer function T(s)=Vo(s)/lo(s). Show your block reduction process. 6H www 20 lo(t) Vi(t) 40 thi #5F 70
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- for the circuit shown below If v_in=5 sin2m10t then corner frequency will be ......Hz 1 UF 2 k Ohm 1 k Ohim 050 060 070 080 vinApplication: Construct the circuit in figure. R= 2,2KN, L = 1 mH, and C= 10 nF. L Necesssary Formulae: 1 Xc 2.π.f.C X.-2. π.f. L By using the formulea perform the calculations of Xc and XL then fill the table. f (Hz) Xc XL 50000 60000 70000 80000 90000 100000 110000 120000 130000 140000For the circuit diagram in figure Q22: which of the following is the total impedance? 12 Ri 10R R2||5R Figure Q22 V 220V40 50HZ L1 315mH C1=350uF O a. 3.33/0° O b. 4.71290°2 O c. 7.3304-19.66° O d. 10.37261.19°2
- Draw a seven-segment display, show the number highlighted in figure 1 (d3 -d0 are the input, a -g are the outputs). III. Simulation Wavetorm Editor - Cers/Ahmad Fainuz/Desanop/DLD Lab Practice/bedseg - beaeg - bcaseg 2018o03121936sim.vwwt Ee ep Ea w smutation search altera.com Haster Time Bar: ops Pointer: s29 us Start End iterval 529 us o ps 0 ps २ 2.56 us S12 us 75 us 10.24 us 12 s 15.36 us 172 us value at Name Ops do Figure 12) IL a cherecterio kc eq. of a conto/ oystem iJ ginen as s456+25+ 4K = 0,?u+ this eg. ia a form 1+ KGG) = 0 QS %3D draw roe t lous cw.r.t. k . cen nohCalibri (Body) -11 by AaBbCcDx AaBbCcD AaBbC AaBbCc AaB AaBbCc. mat Painter BIU 1Normal 1No Spaci. Heading 1 Heading 2 Title Subtitle Chang Styles d Font Paragraph Styles IMPEDANCE QUIZ # 2 1. What capacitance when connected in series with a 500Q resistor will limit the current drawn from a 48-mV 465-kHz source to 20µA? * a. 144pF b. 145pF c. 146pF d. 147pF MY ANSWER: c.146pF 2. What is the total impedance at 20kHz of a series circuit consisting of a 1.5mH inductance, a 100Q resistance, and a 0.08uF capacitance? * a. 1340 41.7 b. 1930 L-37.2" C. 920 290 d. 530 2-12.6°
- sohw circuit diagram and step by step solution pls4. A capacitor now is connected in parallel with R1 to reduce the ripple. Find the value of the capacitance needed to make the following ripple index less than 5% through PSpice simulation study. V -×100% -V o,min 0,max rpp RFP = -x100% = V dc dc D1 Dbreak R1 V1 1k VAMPL = 169.71 FREQ = 50 VoA series RL with R = 15-02 and L = 73-H connected across a 24 VDC source for a LONG TIME, what is the voltage acroos the inductor. Write the values only (round to four decimal places).
- Consider the circuit below: Simplify the circuit and write the super position circuits for I1, 12, and 13 2 AZ45° 1220 v 4 AZ0° O 20 3 VZ-90°TE is given as RI-102, Rz=7L, Xes= 10R, Xez=52. XM=6R, Xc=71 in the circuit below. Calculate I. curment and Ve voltage in polar form and mork then the following ilems. in in Angles are given in degr ees. I, JXL1 R2 100/60° jXM -jXc +Vc a) Vo 29.696/17.552 b) Ve 25.870/25.402 Ve 14.702/10.440" d) Vo 27.661/177.887° 1.908,/62 552" 1.805/27,130 4.058/17.003" 21.808/22.713" h) Ve 3.352,/17.887" 2.446/9.456in the figure (UBEI O.7U,R1= 6.41 khonm, in the circuit given ,vcC= 1o. 00 v, R2 = 27. 68 khom, ec =1.06 khom, RE = 3. 1g khom Beta = IC curren t by full canaly sis. when making your 127,00 calculate the the Poin t. tronsaction s, 2 steps will be +akın a fter Cz Re RE VCC Ri 122