Load ID L1 L2 L3 240/2 cas (1000+) V P 1 kW 0.98 KW 3.5 kW L1 Q O total P= 5.48 KW total Q 2.57 KVAR total S = 7.4 KVA L2 1 KVAR 3.57 KVAR L3 1 KVA 1.4 kVA 5 kVA pf 1.0 0.7 lead 0.7 lag overall power factor: 0.74 a. Using the same load details from table, how much Q must be compensated by the added capacitor to reach an overall power factor of 0.98 lagging? Use KVAR. b. What must be the value of the capacitor to achieve the new power factor? Use µF.

Load ID L1 L2 L3 240/2 cas (1000+) V P 1 kW 0.98 KW 3.5 kW L1 Q O total P= 5.48 KW total Q 2.57 KVAR total S = 7.4 KVA L2 1 KVAR 3.57 KVAR L3 1 KVA 1.4 kVA 5 kVA pf 1.0 0.7 lead 0.7 lag overall power factor: 0.74 a. Using the same load details from table, how much Q must be compensated by the added capacitor to reach an overall power factor of 0.98 lagging? Use KVAR. b. What must be the value of the capacitor to achieve the new power factor? Use µF.

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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