Do it asap A 500-kV 60-Hz three phase transmission line is 463 km long. The series impedance isz=0.06+j0.4 2 per phase per km and the shunt admittance is y=js*106 Siemens per phaseper km. If the T.L. compensated with a series compensation factor of 60%; The compensatedvalue of constant B in 2 is;O A 23.93492 j6141012O B. 25.25773-j113.17002O C23.93492+j63.05183O D. 25.25773+j114.43530

Given : z=0.6 +j0.4 per phase per kmy=j5×10-6 siemens per phase per kmTransmission line 463 Km Long.…

er km. If the T.L. compensated with a series compensation factor of 60%; The compensated alue of constant B in N is; PA 2393492 j61.41012 PE 25.25773-j113.17002 o C2393492+63.05183 OD. 2525773-j114.43530

er km. If the T.L. compensated with a series compensation factor of 60%; The compensated alue of constant B in N is; PA 2393492 j61.41012 PE 25.25773-j113.17002 o C2393492+63.05183 OD. 2525773-j114.43530

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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