In a three-phase wye-connected source, the three phase currents are \(12 \angle 15^\circ \, \text{A}, 12 \angle -105^\circ \, \text{A},\) and \(12 \angle 135^\circ \, \text{A}.\) What must the three line currents be?- \(12 \angle 45^\circ \, \text{A}, 12 \angle -75^\circ \, \text{A},\) and \(12 \angle 165^\circ \, \text{A}\)- \(12 \angle 15^\circ \, \text{A}, 12 \angle -105^\circ \, \text{A},\) and \(12 \angle 135^\circ \, \text{A}\)- \(6.93 \angle 45^\circ \, \text{A}, 6.93 \angle -75^\circ \, \text{A},\) and \(6.93 \angle 165^\circ \, \text{A}\)- \(6.93 \angle 15^\circ \, \text{A}, 6.93 \angle -105^\circ \, \text{A},\) and \(6.93 \angle 135^\circ \, \text{A}\)

This question belongs to circuit theory . It is based on the concept of three phase circuit . It is…

In a three-phase wye-connected source, the three phase currents are 12/15° A, 12Z- 105° A, and 12/135° A. What must the three line currents be? O 12/45° A, 12Z – 75° A, and 12/165° A - O 12/15° A, 12Z 105° A. and 12/135° A - O 6.93/45° A, 6.93Z – 75° A, and 6.93Z165° A O 6.93/15° A. 6.93Z – 105° A. and 6.93Z135° A -

In a three-phase wye-connected source, the three phase currents are 12/15° A, 12Z- 105° A, and 12/135° A. What must the three line currents be? O 12/45° A, 12Z – 75° A, and 12/165° A - O 12/15° A, 12Z 105° A. and 12/135° A - O 6.93/45° A, 6.93Z – 75° A, and 6.93Z165° A O 6.93/15° A. 6.93Z – 105° A. and 6.93Z135° A -

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...

See similar textbooks

Similar questions

QUESTION 6 Find the total real power dissipation in a balanced wye-delta system. Load impedance is 5kN270° and phase voltage is 320V RMS O 63W 99W 150W 36W
Circuit theory
A balanced load is connected in star across a 3 phasec415V 50Hz supply. The load consists of a coil of resistance 25 ohms and inducrance 0.5 henry in paeallel with a purw resistance of 28 ohms and a pure capaciror of 60microfarads capacitance ,in each branch. Calculate a. The line currents The answer has to be 9.3A
Vs=240V 6042 Sousce current ? Total impedance ? R SAKA L 142mH с 1604 F Determine the total impedance of the circuit and sousce cursent Is. Give answer in Polas form
Use Kirchhoff's Laws to find the branch currents indicated in the following networks: Show complete solution and how you solve the equation using cramer's rule.
Find out the zero-phase sequence components of the following set of three unbalanced voltage vectors: Va =10cis30° ,Vb= 30cis-60°, Vc=15cis145°" (THOS PROBLEM IS MULTIPLE CHOICE PLEASEE CHOOSE THE LETTER WITH THE CORRECT ANSWER AND PLEASE PROVIDE YOUR CLEAR SOLUTION! A 16.809cis-72.579° B 5.603cis72.579° C 16.809cis-47.421° D 5.603cis-47.421°
Page No. Date The sesistance between the terrainal AB is B0-2, find Volue of R9 12 2020/12/31 10:50
For the series-parallel arrangement shown in Figure, determine the equivalent series circuit impedance, the supply current I, the circuit phase angle, the values of voltages V1 and V2, and the values of currents IA and IB
Solve for equivalent wye connected load. Za, Zb, & Zc in ohms.
34 Source Zine 30 Load Source: Zan = Zbn = Zen = 0.1 + j1.5 Q Ean = 12020° v %3D %3D %3D Eon = 1502 - 143.13° v Ecn = 90490° v Line: ZLineA= 1 + j2 Q; ZLines= 1 - j2 Q; ZLinec= 0 + j2.2361 Q Load: Zan= 12 + j9 Q Zon= 12 - j9 N Zen= 15 + j0 Q *Fault located on the side of the 3-phase load Find the load line currents la , Ib , lc under unbalanced conditions 1. Without the fault 2. With fault
In figure below, convert each delta network to a wye network. 1.0 MO RC Z RA RB. R = 1.0 N Rg = 2.7 N 560 kN `1.5 MN R = 2.2 N (a) (b)
!