Capacitances C1 = 265.26 µF C2 = 1.33 mF Inductances Resistance R1 = 30 LI SRI R2 = 80 230v 60HZ L1 = 26.53 mH ir = 5.572 + j19.156° A ir = 19.95473.78° A For C1, ic, = 19.95473.78° A vc, =199.52 – 16.22°V For R1, iRi = 18. 82489.1° A %3D Vr1 = 56.47489°V For C2, ic, = 18. 82489. 1° A vc, = 37.642 – 0.9°V For R2, İR2 = 5.34.06° A VR2 = 42.424. 06°V For L1, İL, = 5.344.06° A Vi, = 5.32 – 94.06°V

Power System Analysis and Design (MindTap Course List)
6th Edition
ISBN:9781305632134
Author:J. Duncan Glover, Thomas Overbye, Mulukutla S. Sarma
Publisher:J. Duncan Glover, Thomas Overbye, Mulukutla S. Sarma
Chapter2: Fundamentals
Section: Chapter Questions
Problem 2.13P
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Question

Find the voltage and current for the AC circuit

Capacitances
C1 = 265.26 µF
C2 = 1.33 mF
LI
Resistance
R1 = 30
ERI
R2 = 80
230v
60HZ
R2
Inductances
L1 = 26.53 mH
ir = 5.572 + j19. 156° A
ir = 19.95473.78° A
For C1,
ic, = 19.95473.78° A
vc, =199.52 – 16.22°V
For R1,
İRi = 18. 82289.1° A
VR1 = 56.47489°V
For C2,
ic, = 18. 82289.1° A
= 37.642 – 0.9°V
For R2,
İR2 = 5.344.06° A
VR2 =
42.444.06°V
For L1,
= 5.344.06° A
VL1
= 5.32 – 94.06°V
一
Transcribed Image Text:Capacitances C1 = 265.26 µF C2 = 1.33 mF LI Resistance R1 = 30 ERI R2 = 80 230v 60HZ R2 Inductances L1 = 26.53 mH ir = 5.572 + j19. 156° A ir = 19.95473.78° A For C1, ic, = 19.95473.78° A vc, =199.52 – 16.22°V For R1, İRi = 18. 82289.1° A VR1 = 56.47489°V For C2, ic, = 18. 82289.1° A = 37.642 – 0.9°V For R2, İR2 = 5.344.06° A VR2 = 42.444.06°V For L1, = 5.344.06° A VL1 = 5.32 – 94.06°V 一
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