R10 w ΧΙΩ 302 000 w V2 E 1002 150 V Z0° Where: R₁ = 52, and and XL = 25 2. Find the values of : Total impedance ZT Mag= ,ZT Angle V₁ Mag= V,V1 Angle= V2 Mag = ✓ V,V2 Angle =
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- P 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 VÀFor 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°2A certain impedance is given by 1012° N. Does the current lead or lag the voltage, and by how many degrees? Fill in the blanks. The current the voltage by
- 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 RxQUESTION 5 A resistor, a capacitor, an inductor, and a mystery component are connected in parallel. The resistor has a resistance of 300N, the capacitor has a reactance of 2500, and the inductor has a reactance of 6000. The total impedance of the circuit is 59.52-j119.640. What is the mystery component? O there is no mystery component O resistor O inductor O capacitor0 Consider the image below. Derive an expression for E as a function of the excitation voltage E¡ and the resistances R₁ and R₂. You may have to use Ohm's law and/or Kirchhoff's laws. Ei allt R1 Eo 0 R2
- 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 Vo2-A resultant current is made of two components: a 10 A D.C. component and a sinusoidal component of maximum value 14.14. The average value of the resultant current is amperes (a) 0 (b) 24.14 (c) 10 (d) 4.14 and r.m.s. value is - amperes (e) 10 (f) 14.14 (g) 24.14 (h) 100Two impedances Z1 = 2+j6 N and Z2 = 6-j12 Q are connected in series. What is the resultant impedance in the polar form? 10cis-36.9° O 10cis36.9° 10cis53.1° 10cis-53.1°
- The dioces are ideal. write the Fransfer Characteidie equati ons ( &, as a funchiar g &) Plot &o agaiwet &, imdicaling all intrcepts, slopes and Volt ope lerels. Sketch bo f bq =40 sin wt. Indicate al voltage lezels - %3D 1OK lok lovConsider a drcuk where an ideal inductor L=D.144H is connected in parallel to a resistor R=1.00k. This combination is connected in series with a capacitor C=3.67e-07. This combination in connected to a power supply source of frequency f-3.4kHz. To study this circuit you need to drawa phasor diagram. Note the sum cf the current phasors through the resistor and the inductor must be equal to the phasor representing the current through the capacitor fie. the total current). You should draw the phasor for the current through the capadtor on the horizontal axis, since this is also the total ourrent delivered by the supply. The angie between this phasor and that of the power supply represents the phase angle o. Note that the voltage across the inductor equals the voltage aoross the resistor since they conmected parallel to esch other. Furthermore the sum of the phasors representing the currents through the inductor and the resistor equals the phasor representing the current through the…For this question i get all the way to the point where i have a + bj [ 0.0166 + 0.1502j ] however, calculating the phase of this specific question just does not match up with the answer provided by my university. the formula that i am using is that the phase = 90-arctan(Imaginary/Real) which in this particular problem should add output a phase of -96,31 degrees. However it does only output 6,31 and the input voltage does not contain -90 so i dont get it.