(c) Current of the capacitor (d) Value of the capacitance reactance (-jxc= (e) Draw circuit (shown below) (f) Equations used and units (g) Capacitor size Found in class: Resistor current-69.44 A Inductor current=-j166.7 A Given: V=72/0KV R= 103.682 jx=43.22 P-500kW Pf= 0.95
(c) Current of the capacitor (d) Value of the capacitance reactance (-jxc= (e) Draw circuit (shown below) (f) Equations used and units (g) Capacitor size Found in class: Resistor current-69.44 A Inductor current=-j166.7 A Given: V=72/0KV R= 103.682 jx=43.22 P-500kW Pf= 0.95
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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Question
please do a,b,and c
![Show:
(a) Power factor angle
(b) Total current
(c) Current of the capacitor
(d) Value of the capacitance reactance (-jXc=_
(e) Draw circuit (shown below)
(f) Equations used and units
(g) Capacitor size
Found in class:
Resistor current=69.44 A
Inductor current--j166.7 A
Given: V=72/0°KV
R= 103.682
jx=43.252
{R
JX
P-500kW
Pf= 0.95
-jxc](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fd53b5d21-36fa-4ee2-932f-fd40dc0982c5%2F1b51478f-7775-40d9-805a-f130ee7aa2fb%2F7onjyhd_processed.jpeg&w=3840&q=75)
Transcribed Image Text:Show:
(a) Power factor angle
(b) Total current
(c) Current of the capacitor
(d) Value of the capacitance reactance (-jXc=_
(e) Draw circuit (shown below)
(f) Equations used and units
(g) Capacitor size
Found in class:
Resistor current=69.44 A
Inductor current--j166.7 A
Given: V=72/0°KV
R= 103.682
jx=43.252
{R
JX
P-500kW
Pf= 0.95
-jxc
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