A 500v, 50kva. 1-p altemator has an effective resistance of 0.2w. A field current of 10a produces an armature current of 200a on short-circuit and On emf. Of 450v on apen-Circuit Calulate the ful- load regulation at pf. 0.8 lag

Introductory Circuit Analysis (13th Edition)
13th Edition
ISBN:9780133923605
Author:Robert L. Boylestad
Publisher:Robert L. Boylestad
Chapter1: Introduction
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DEGREE: ELECTRICAL ENGINEERING 
SUBJECT/COURSE: AC MACHINES
 
NOTE: Please solve in this way. 
1. Please have a good handwriting, some of the answers are not readable. Thank you!
2. GIVEN. (Include symbols and units)
3. REQUIRED/FIND/MISSING (with symbol/s and units)
4. ILLUSTRATIONS such as diagrams and graphs are Required
5. Step-by-step SOLUTION with Formulas and Symbols. No Shortcut, no skipping, and detailed as possible
6. FINAL ANSWERS must be rounded up to three decimal places with the corresponding units and must be enclosed in a rectangle.

A 500v, 50kva. 1-w altemator haG an effective
vesistance of 0.2w. A field curvent of 10a produces
an Armature Current of 200a on short-circuit and
an emf. Of 450v on open-circuit Coaloulate the full-
load regulation at pf. 0.8 lag
Transcribed Image Text:A 500v, 50kva. 1-w altemator haG an effective vesistance of 0.2w. A field curvent of 10a produces an Armature Current of 200a on short-circuit and an emf. Of 450v on open-circuit Coaloulate the full- load regulation at pf. 0.8 lag
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