Problem Find Vo. 32 Ω ww 100/0° V -√24 02 j16 Ω 4:1 Ideal 3Ω www 2023 20/-20° V Answer Vo=24.95 (-62.65° )V
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- 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 1The sum of the following two e.m.fs will be e1 = 10 sin wt and e2 = 10 cos wt O 10 O 20 sin wt O 14.14 cos wt O 14.14 sin (wt + TT/4)DC-AC Circuits Express R1 using Rx, Ry, and Rz when performing wye to delta conversion R3 Rx Rz R2 R1 Ry O A. (Rx+Ry)(Ry+Rz)(Rx+Rz) Rx В. RyRz Rx+Ry+Rz O C. Ry+Rz RXRYRZ O D. RXRYRZ Ry+Rz OE. RxRy+RyRz+RXRZ Rx
- Two impedances of Z = 100 +j50 Q and Z2 = 20 -j20 Q are connected in series. what is the magnitude of the equivalent impedance in 2? %3D %3D Your Answer: AnswerP 269 W, Q = 150 VAR (capacitive). The power in the complex form is: %3D Select one: a. 150 + j269 VÀ b. None of these c. 150- j269 VA d. 269-j150 VÀPlease show your detailed solution Number 14 The current in a series inductive circuit is 7.5 A at 25 Hz. The circuit takes 425 W, and the power factor is 0.47. The resistance of the circuit is A. 7.32 ohms C. 7.98 ohms B. 7.86 ohms D. 7.56 ohms
- 2) 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 nohSolve thisthe topic is about : DELTA – WYE AND WYE –DELTA TRANSFORMATION4. 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 Vo
- From step1, how did you get that 50+(80*j60/80+j60) = 78.8+j38.4? also in step 2, how did you get that 5/(0.0125-0.0167) = 143.63 + j191.89?find the instantane ous expressions or the applied voltage V, voltage voltage figure. Als draw across resistos Vp and across aeross capaciter the phaser diengram. Vc in Ve R= 62 iz1o 5z sin (6250t)M 46 ll Asiacell O Mid electro... -> e: sketh tha oudiput veltage for t followling circut Vsi z0.7 kzo-8 aka ス0 -20 ç: Draw the oulput voltuge for tha following cirent Cand dertermine de Voltuse N: - 15