LIVE Course for free

Rated by 1 million+ students
Get app now
JEE MAIN 2023
JEE MAIN 2023 TEST SERIES
NEET 2023 TEST SERIES
NEET 2023
CLASS 12 FOUNDATION COURSE
CLASS 10 FOUNDATION COURSE
CLASS 9 FOUNDATION COURSE
CLASS 8 FOUNDATION COURSE
0 votes
55 views
in Electronics by (15 points)
edited by

A three-phase balanced delta connected load of (4+j8) Ω is connected across a 400V, 3 – Ø balanced supply. Determine the phase current IR. Assume the phase sequence to be RYB.

Please log in or register to answer this question.

1 Answer

0 votes
by (56.4k points)

Taking the line voltage VRY = V

∠0° as a reference

VRY = 400∠0°V,

VYB = 400∠-120°V 

and VBR = 400∠-240°V. 

Impedance per phase = (4+j8)

Ω = 8.94∠63.4°Ω.

Phase current IR = (400∠0°)/(8.94∠63.4°)

= 44.74∠-63.4°A.

Welcome to Sarthaks eConnect: A unique platform where students can interact with teachers/experts/students to get solutions to their queries. Students (upto class 10+2) preparing for All Government Exams, CBSE Board Exam, ICSE Board Exam, State Board Exam, JEE (Mains+Advance) and NEET can ask questions from any subject and get quick answers by subject teachers/ experts/mentors/students.

Categories

...