IR lc R C V ICA IR V

Introductory Circuit Analysis (13th Edition)
13th Edition
ISBN:9780133923605
Author:Robert L. Boylestad
Publisher:Robert L. Boylestad
Chapter1: Introduction
Section: Chapter Questions
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3)) please choose the correct answer with complete solution
A 25 µF (25E-06 F) capacitor is
connected in parallel with an 90
resistor across a Vrms = 230<0° V, f =
60 Hz supply. Calculate (a) the
current in each branch, (b) the supply
current, (c) the circuit phase
angle, (d) the circuit impedance.
Xc = (-j/(2nf*C)
IR
R
Ich
C
44
HH
Ic
V
IR = 2.56<0° Q, Ic= 2.168<90° Q
IR = 2.56<0° Q, Ic= 2.168<-90° Q
power factor angle = 40.26°, power
factor= cos(40.26°) = 0.763 leading
I-total = 3.35<40.26° A
IR
power factor angle = 90°, power
factor = cos(90°) = 0
I-total = 3.35<-40.26° A
power factor angle = 0°, power factor
= cos(0°) = 1, unity
I-total = -3.35<40.26° A
IR = 2.56<-90° Q, Ic = 2.168<0° Q
None from the above
V
Transcribed Image Text:A 25 µF (25E-06 F) capacitor is connected in parallel with an 90 resistor across a Vrms = 230<0° V, f = 60 Hz supply. Calculate (a) the current in each branch, (b) the supply current, (c) the circuit phase angle, (d) the circuit impedance. Xc = (-j/(2nf*C) IR R Ich C 44 HH Ic V IR = 2.56<0° Q, Ic= 2.168<90° Q IR = 2.56<0° Q, Ic= 2.168<-90° Q power factor angle = 40.26°, power factor= cos(40.26°) = 0.763 leading I-total = 3.35<40.26° A IR power factor angle = 90°, power factor = cos(90°) = 0 I-total = 3.35<-40.26° A power factor angle = 0°, power factor = cos(0°) = 1, unity I-total = -3.35<40.26° A IR = 2.56<-90° Q, Ic = 2.168<0° Q None from the above V
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