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
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Question
![How did he get the value inside the circle?
Can you explain step by step and
clear line please?
= cUR + DIR
158.77
X103
%3D
+ 723.24/-30-79°
: 60 3-607/ -l0•269°
Is
%3D
A
torque 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]
Pi
373.105 MW
%3D
= output power =
J3 X 275 Xlo x 840 X 0- 85
340.088 MW
Loss
PLOSS
Pi -P. = 373.105-34o.088 =
33 MW
%3D
Po
Xlco =
Pi
340 1088 x lo0
etticiency -1 =
%3D
373.105
E 9 1· 15 •J.
14](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F77999ab1-4cad-4428-a98d-64e6574e5f4d%2Ff11fdedd-b3c6-4661-8bd1-a1daf91a9609%2Fpkoqfyn_processed.jpeg&w=3840&q=75)
Transcribed Image Text:How did he get the value inside the circle?
Can you explain step by step and
clear line please?
= cUR + DIR
158.77
X103
%3D
+ 723.24/-30-79°
: 60 3-607/ -l0•269°
Is
%3D
A
torque 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]
Pi
373.105 MW
%3D
= output power =
J3 X 275 Xlo x 840 X 0- 85
340.088 MW
Loss
PLOSS
Pi -P. = 373.105-34o.088 =
33 MW
%3D
Po
Xlco =
Pi
340 1088 x lo0
etticiency -1 =
%3D
373.105
E 9 1· 15 •J.
14
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