R=0.15*40=62 tor 4o km-3メレニ(2スFL)* 402 (3.14) Che) (l, 3263x10) CY0)XL 2on for 4.km+Z= R+j XL 2:(6+j 20Jn /par phastto find VR and IRP=SIRL0..1-36.8 AOR36.8LaggLaggingIRCLEARHow did he get thevalue inside the circle?

In the given question the receiving end power is calculated by the formula: P=Scosθ Once we get its…

R= 0.15 * 40 = 62 tor Yo km -3. XL=(27FL)* Yo = 2 (3,14) Cho) (l;3263x16) C40) XL 2on for 40 km 2: R+j XL > 2:(6+520Jn /per phast to Fnd VR and IR P: S cos PR=SREOS@R Siapperent p(VA) piactive p(w). Qreactivep(vAr). 381MVA %3D : PR = 3RI 1o *0.8= 304.8 xlo Wet *0.8=304.8xlo Wast 3. PR J 3 VR TR COS OR 304.8X10 3 X220X1a xIR xo.8 IR Lo00/-36.8A VOR36.8 Lagg IR Lagging CLEAR How did he get the

R= 0.15 * 40 = 62 tor Yo km -3. XL=(27FL)* Yo = 2 (3,14) Cho) (l;3263x16) C40) XL 2on for 40 km 2: R+j XL > 2:(6+520Jn /per phast to Fnd VR and IR P: S cos PR=SREOS@R Siapperent p(VA) piactive p(w). Qreactivep(vAr). 381MVA %3D : PR = 3RI 1o 0.8= 304.8 xlo Wet 0.8=304.8xlo Wast 3. PR J 3 VR TR COS OR 304.8X10 3 X220X1a xIR xo.8 IR Lo00/-36.8A VOR36.8 Lagg IR Lagging CLEAR How did he get the

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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