R=0.15* 40-62 tor 4o kmメレニ(2スFL)* 40=2 (3.14) C6o) (l, 3263x10) CY0)XL 20n for 9.km(6+j 20Jn /per phastto find VR and IRPR=SROSORsos Oノ381MVASiapperint p(VA)|piactivep(w)reactinp(VAr)|PR = 3 8110* os 8 = 304.8xlo WatPR-J3 VR TR COS OR 304.8Xlo 3 X 220xla xIRxor8>IR loo0/-36.8 AVOR36.8LaggLagging'IRCLEARHow did he get thevalue inside the circle?

In AC every quantity is phasor which has magnitude as well as direction. Here after calculating…

R= 0.15 * 40 362 tor Yo km -3. XL=(27FL) * Yo = 2 (3.14) Cao) Clo 3263x10) (4) XL 20n for 90 km 2=R+jXL> 2:(6+j 20Jn /per phast to find VR and IR sos Q 381MVA Siapperint p(VA) piactive p(w) Qreactivnp(VAr) PR= 38116 *0.8 = 304.8x1o Wast PR J3 VR TR COS ORS304.8X10 3 X 220xla x IR xor8 3. > IR zlo00/-36.8 A OR36.8 Lagg Lagging

R= 0.15 * 40 362 tor Yo km -3. XL=(27FL) * Yo = 2 (3.14) Cao) Clo 3263x10) (4) XL 20n for 90 km 2=R+jXL> 2:(6+j 20Jn /per phast to find VR and IR sos Q 381MVA Siapperint p(VA) piactive p(w) Qreactivnp(VAr) PR= 38116 *0.8 = 304.8x1o Wast PR J3 VR TR COS ORS304.8X10 3 X 220xla x IR xor8 3. > IR zlo00/-36.8 A OR36.8 Lagg Lagging

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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