Figure 4:Pgen,Pgen,Pload, O200 MWPload,550 MWLineLine Reactance (Per Unit)x12-0,1x13=0.125x23=0,21-21-32-3Pgen,Pload,100 MWIn the following system, Bus 1 is the phase reference and the base power is 100 MVA.If the power flows on the lines 1-2 and 1-3 are PF12=158.1776 MW and PF23= -57.2186 MW, then 02 is:Select one:O a.02= -0.1582 radO b. None of theseO c.02= -0.0437 radO d.02= +0.183 rad

Power delivered by the line having sending end voltage V1 and receiving end voltage V1 with del is…

Answered: Figure 4: Pgen, Pgen, Pload. 200 MW…

Introductory Circuit Analysis (13th Edition)

13th Edition

ISBN:9780133923605

Author:Robert L. Boylestad

Publisher:Robert L. Boylestad

Chapter1: Introduction

Section: Chapter Questions

Problem 1P: Visit your local library (at school or home) and describe the extent to which it provides literature...

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A three-phase alternator generating unbalanced voltages is connected to an unbalanced load through a 3-phase transmission line as shown in figure. The neutral of the alternator and the star point of the load are solidly grounded. The phase voltages of the alternator are Ea = 1020°, Eb = 102-90°, Ec = 10/120⁰ E E E The positive sequence component of the load current is (A) 1.3102-107⁰ A (B) 0.332Z –120° A (C) 0.9962-120º A (D) 3.5104-81⁰ A +) + j1.002 0000 j1.002 0000 j1.0Ω vovo j1.0Ω 0000 j2.002 voor j3.002 oooo
In a three-phase system the per-unit phase voltages are 100° , Vôn = 0.8Z180° Van 0.920° , Von = 0.92 - а What are the per-unit symmetrical components V0) and V)? Express in polar notation.
Question 1
2-What is the relationship between phase and line quantities in balanced star and delta connections in three phase circuit balanced ? Why?
Question 3. A three-phase impedance load balanced D a connected load in parallel with a balanced ? connected load made of bonding. D each arm for the load ?D = 6 + 6? and ? each arm for the connected load It is given as ?? = 2 + 2?. ? the connected load ?? = 1? is earthed through the neutral impedance. To this impedance load, the symmetrical component voltages ?0 = 10∠60°, ?1 = 100∠0° and ?2 = Unbalanced phase-to-earth voltages ??, ?? and ?? of 15∠200° volts were applied. a) Draw the zero, positive and negative sequence circuits. b) Find the complex powers imparted to each component circuit.
A three-phase alternator generating unbalanced voltages is connected to an unbalanced load through a 3-phase transmission line as shown in figure. The neutral of the alternator and the star point of the load are solidly grounded. The phase voltages of the alternator are Ea = 1020º, Eb = 102-90°, Ec = 10/120⁰ E E E The positive sequence component of the load current is (A) 1.3102-107⁰ A (B) 0.3322-120° A (C) 0.9962-120º A (D) 3.5104-81⁰ A +) + j1.002 0000 j1.002 0000 j1.0Ω vovo j1.0Ω 0000 j2.002 voor j3.002 oooo
Positive sequence components consist of 3 equal sized phasors. The phase difference between these phasors is 180 °. And the phasors are in the phase order of the true phasors. Is this statement true or false?
1- Your example deduced the total strength of the following three loads: the first load 1200VA at a power factor of 0.7 late, the second load 350VA at a power factor of 0.5 was late, and the third load 3275VA at a power factor of 1.0.
A. 3-8 50kw, 4-pole, soHZ inducffon motor haß a coinding Cacy desigmed for delta connection. the coinding has 24 conducforx por Mot mranged in Go slots, the mo value * the line current is 48A. Aind the fumdomental of the mmf coane f phase -A when the cunnent is passing throyph its maximum value. what ig the speed and peak volue the resultant mmf/ pule. folne il sume otheroise &leip. Al ame incorreet dorit copy fom Qne Anscwor Another Ansawer
What is the max load demand of the following feeders 4152 MW, 2.891 MW, 815.20 KW, 905.12 KWand 1.972 MW A) 10,735.32 KW B) 12,815.32 KW C) 10,571 MW D) 12.052 MW
Consider a 10 kVA, 200/400 V, 50 HZ: single-phase tranformer has the folloving test results:0.C. test: 200 V, 0.6A, 63 W (L.V.side)S.C. test : 20 V, 25A, 85 W (H.V. side)Deternine:i- The eficiengy at 75 % offiull-load at 0.9 leading power factor.ii -The secondary terminal voltage on fill-load at tunity power facton

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