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Determine the capacitor necessary to increase the power factor to 0.94 LEARNING EXTENSION P, =100KW,V, = 480(V)rms, pf =0.707| Rine = 0.12, f = 60HZ P = Re{S}=Scos(0, -0;)=S|xpf| 100 =141.44KVA |Qou l= VISolu P -P² =100.02(kVA). Sola = pf .707 - pf new pf new Enew →0. =0.363×P = 36.3KVA P cos Onew =0.94 tan 0, = 0.363 new :Qeapacitor = Qot -Qnew =100.02 – 36.3= 63.72KVA Ocapacitor =V. || Icapacitor Qcapacitor 0|VL (480)² × (27 x 60) 63.72x10 :C = = 0.000733(F) = 733µF| %3D

Determine the capacitor necessary to increase the power factor to 0.94 LEARNING EXTENSION P, =100KW,V, = 480(V)rms, pf =0.707| Rine = 0.12, f = 60HZ P = Re{S}=Scos(0, -0;)=S|xpf| 100 =141.44KVA |Qou l= VISolu P -P² =100.02(kVA). Sola = pf .707 - pf new pf new Enew →0. =0.363×P = 36.3KVA P cos Onew =0.94 tan 0, = 0.363 new :Qeapacitor = Qot -Qnew =100.02 – 36.3= 63.72KVA Ocapacitor =V. || Icapacitor Qcapacitor 0|VL (480)² × (27 x 60) 63.72x10 :C = = 0.000733(F) = 733µF| %3D

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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Problem 2 Reference slide #46 Load is changed to S= 100< 30° KVA, determine the supply voltage and overall pf of the source. If the power factor of the source is to be reduced by 20%, what is the size of the capacitor in microfarad?
Transcribed Image Text:Problem 2 Reference slide #46 Load is changed to S= 100< 30° KVA, determine the supply voltage and overall pf of the source. If the power factor of the source is to be reduced by 20%, what is the size of the capacitor in microfarad?
Determine the capacitor necessary to increase the power factor to 0.94 LEARNING EXTENSION P, =100k W,V, = 480(V )rms, pf =0.707 Rine = 0.12, f = 60HZ P = Re{S} =S|cos(0, – 0;) =S |×pf| 100 =141.44KVA||Qom l= VISolu P -P² =100.02(kVA). Sold = pf .707 pf new =Qnew =0.363×P = 36.3KVA Qnew P cos Onew =0.94= tan0, = 0.363 pf new new new ::Qeapacitor = Qot -Qnew =100.02 – 36.3 = 63.72KVA Ocapacitor =V1 || Icapacitor 63.72×10 Qcapacitor 0 |V, (480)²×(2x x 60) = 0.000733(F)=733µF| :C = %3D
Transcribed Image Text:Determine the capacitor necessary to increase the power factor to 0.94 LEARNING EXTENSION P, =100k W,V, = 480(V )rms, pf =0.707 Rine = 0.12, f = 60HZ P = Re{S} =S|cos(0, – 0;) =S |×pf| 100 =141.44KVA||Qom l= VISolu P -P² =100.02(kVA). Sold = pf .707 pf new =Qnew =0.363×P = 36.3KVA Qnew P cos Onew =0.94= tan0, = 0.363 pf new new new ::Qeapacitor = Qot -Qnew =100.02 – 36.3 = 63.72KVA Ocapacitor =V1 || Icapacitor 63.72×10 Qcapacitor 0 |V, (480)²×(2x x 60) = 0.000733(F)=733µF| :C = %3D
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