How did he get the value inside the circle?Can you explain step by step andclear line please?= cUR + DIR158.77X103%3D+ 723.24/-30-79°: 60 3-607/ -l0•269°Is%3DAtorque angle (S) = - 2 3-31'† 1= - 22-31"%3D= in put power = V3 x 428.358 X lo°x603.607 x cos 23-31+10.269]Pi373.105 MW%3D= output power =J3 X 275 Xlo x 840 X 0- 85340.088 MWLossPLOSSPi -P. = 373.105-34o.088 =33 MW%3DPoXlco =Pi340 1088 x lo0etticiency -1 =%3D373.105E 9 1· 15 •J.14

In this question we will write about how we will find sending current with given data...

How did he get the value inside the circle? Can you explain step by step and clear line please? i. I, = cUR +DIR = cVR Is 948° 158.77 t X103 %3D + 723.24/-30-79° I : 60 3-60+/ -10.269° A %3D Horque angle (S) = - 2 3·31't 1 = -22-31 %3D %3D = in put power : 3 x 428-358 Y10³x co3.607 x Cos (23-31'+10.269) Pi 373.105 MW output powerr = J3 X 275 Xlo x 840 X 0- 85 %3D %3D 340.088 MW YLoss = P: -P. = 373.105-340.088 = 33 MW Po Xlo = Pi %3D 340 088 etticiency =1 = xl00 %3D 373.105

How did he get the value inside the circle? Can you explain step by step and clear line please? i. I, = cUR +DIR = cVR Is 948° 158.77 t X103 %3D + 723.24/-30-79° I : 60 3-60+/ -10.269° A %3D Horque angle (S) = - 2 3·31't 1 = -22-31 %3D %3D = in put power : 3 x 428-358 Y10³x co3.607 x Cos (23-31'+10.269) Pi 373.105 MW output powerr = J3 X 275 Xlo x 840 X 0- 85 %3D %3D 340.088 MW YLoss = P: -P. = 373.105-340.088 = 33 MW Po Xlo = Pi %3D 340 088 etticiency =1 = xl00 %3D 373.105

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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