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EXAMPLE: V = 250 20⁰ V Req 40 L PL QL Z = 25020°V z + 30j V Key I, R 1Ω w IL 353.55 cos j4Ω wt MW 39Ω j26Ω Vr Ir cos (Ov- 0₁) = (234.36) (5) wor (-3.18⁰ + 36.87" Vr Iv sin (Ov - 0₁) = (234-36) (5) sin (-3.18° +36-87) 250 L 0° =5L-36-87° 40 + 30 j A (52-36.87°) (39 + 26j) = 234 - 13 i =234.36 L-3.18' V . = Load current? 52-36.87° A Load voltage? 234. 362-3.1798° V Load P? 975 W (consumed) Load Q? 650 VARS (consumed) = Line P? 25 W (consumed) Line Q? 100 VARS (consumed) 1975 W/ 650 VARS please answer Item 5 & 6

EXAMPLE: V = 250 20⁰ V Req 40 L PL QL Z = 25020°V z + 30j V Key I, R 1Ω w IL 353.55 cos j4Ω wt MW 39Ω j26Ω Vr Ir cos (Ov- 0₁) = (234.36) (5) wor (-3.18⁰ + 36.87" Vr Iv sin (Ov - 0₁) = (234-36) (5) sin (-3.18° +36-87) 250 L 0° =5L-36-87° 40 + 30 j A (52-36.87°) (39 + 26j) = 234 - 13 i =234.36 L-3.18' V . = Load current? 52-36.87° A Load voltage? 234. 362-3.1798° V Load P? 975 W (consumed) Load Q? 650 VARS (consumed) = Line P? 25 W (consumed) Line Q? 100 VARS (consumed) 1975 W/ 650 VARS please answer Item 5 & 6

Introductory Circuit Analysis (13th Edition)
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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EXAMPLE: V = 250 20⁰ V Req 40 L PL QL Z = 25020°V z + 30j V Kev I, R = 1Ω M IL 353.55 cos j4Ω wt M Vr Ir cos (Ov- 0₁) = (234.36) (5) cor Vr Iv sin (Ov - 0₁) = (234-36) (5) sin 39Ω j26Ω 250 L 0° =5L-36-87° 40 + 30 j A (52-36.87°) (39 + 26j) = 234 - 13 i =234.36 L-3.18' V (-3.18⁰ + 36.87" (-3.18° +36-87) . Load current? 52-36.87° A Load voltage? 234. 362-3.1798° V Load P? 975 W (consumed) Load Q? 650 VARS (consumed) = Line P? 25 W (consumed) Line Q? 100 VARS (consumed) 1975 W/ 650 VARS please answer Item 5&6
Transcribed Image Text:EXAMPLE: V = 250 20⁰ V Req 40 L PL QL Z = 25020°V z + 30j V Kev I, R = 1Ω M IL 353.55 cos j4Ω wt M Vr Ir cos (Ov- 0₁) = (234.36) (5) cor Vr Iv sin (Ov - 0₁) = (234-36) (5) sin 39Ω j26Ω 250 L 0° =5L-36-87° 40 + 30 j A (52-36.87°) (39 + 26j) = 234 - 13 i =234.36 L-3.18' V (-3.18⁰ + 36.87" (-3.18° +36-87) . Load current? 52-36.87° A Load voltage? 234. 362-3.1798° V Load P? 975 W (consumed) Load Q? 650 VARS (consumed) = Line P? 25 W (consumed) Line Q? 100 VARS (consumed) 1975 W/ 650 VARS please answer Item 5&6
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