Problem 6.7 Obtain the nodal equations for the circuit shown and find the node voltages. Ans V₁ = 2.07 -26.6 V V₁ 16/-70° A 20/30° A4S 8 S www V₂ -j16 S j12 S HE 8 S 18/35° A 18 S V3 -j20 S 72/30° A
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- Solve the power triangle at Z6 in the circuit diagram below.V1, V2, and V3 = 220∠0°, 50 HzZ1 = 7+j8 ohmsZ2 = 6+j3 ohmsZ3 = 5+j5 ohmsZ4 = 2+j3 ohmsZ5 = 8+j4 ohmsZ6 = 1+j9 ohmsfor the R-th part. isn't R3 and R5 in series not parrallel? If not, explain why. If so, fix the solution. THank you.Application: 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 140000
- 4. 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 VoPlease 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 ohmsDC-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
- Solve the power triangle at Z6 using the following network law below.1.) Kirchorff's LawGiven:V1, V2, and V3 = 220∠0° V, 50 HzZ1 = 7+j8 ohmsZ2 = 6+j3 ohmsZ3 = 5+j5 ohmsZ4 = 2+j3 ohmsZ5 = 8+j4 ohmsZ6 = 1+j9 ohmsThree impedance Za= (3+j4) ohms, Zc= (4-j4) ohms and Zr=0+j3 ohms are connected in parallel. Solve for the pf of the combination.Find the current in the resistor (R1) using the principle of superposition in the figure below. jón R. 62 -j82 V,=10260° V 1 JL=220 A