According to the supply given w=4 rad /sec 64° X₁₂ хс 8 In phasour domain, = WL : $85 j4 1 WC by Supernode $8.5 Vx m Vy - Vx Apply KCL at node Y i j4 = 4X1 reller 55 Supernode analysis. Vy -6 20° 8 Vy = 3.77 /-45 V Current i will Vy j4 8 4 Here node X and Y is super node because there is no resistance reactive element are present in อง the line joining X and Y. + = 2 10° = & 210° = = 2.75/-76 V 422 = 22 Vx وز Vx -ja Xy ( + + + + + + ) + ²x ( = √2 + + + 4) = √5 + ²/3 Vx 6 6 j4 8 (0.25-jo.25) Vy + (0.125 + j0.25) Vx = (0.75 -1.5) Salve eq and Vx 8 ww 8.2 0.9425 /-135° A Vx-6 40° ju = 0
According to the supply given w=4 rad /sec 64° X₁₂ хс 8 In phasour domain, = WL : $85 j4 1 WC by Supernode $8.5 Vx m Vy - Vx Apply KCL at node Y i j4 = 4X1 reller 55 Supernode analysis. Vy -6 20° 8 Vy = 3.77 /-45 V Current i will Vy j4 8 4 Here node X and Y is super node because there is no resistance reactive element are present in อง the line joining X and Y. + = 2 10° = & 210° = = 2.75/-76 V 422 = 22 Vx وز Vx -ja Xy ( + + + + + + ) + ²x ( = √2 + + + 4) = √5 + ²/3 Vx 6 6 j4 8 (0.25-jo.25) Vy + (0.125 + j0.25) Vx = (0.75 -1.5) Salve eq and Vx 8 ww 8.2 0.9425 /-135° A Vx-6 40° ju = 0
Introductory Circuit Analysis (13th Edition)
13th Edition
ISBN:9780133923605
Author:Robert L. Boylestad
Publisher:Robert L. Boylestad
Chapter1: Introduction
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Problem 1P: Visit your local library (at school or home) and describe the extent to which it provides literature...
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Transcribed Image Text:According to the supply
X₁ WL 4X1
640⁰
Here
хс
8
In phasour domain,
=
Vy
j4
=
node X and
$85
६82
Vx
WC
i
=
m
=
J4
no resistance
Or
the line joining X and Y.
relle
by supernode analysis.
Vy - Vx
= 2 10°
Apply KCL at node Y
Vy +
j4
Current i will
Vy
j4
8
Vy - 6 20⁰
8
Vy = 3.77 /-45 V
015
=
240°
given
Y is super node because there is
reactive element are present in
= 2.75/-76 V
4.2
= 2л
Vx
ja
-j2
D
6
Xy (1 + √x + + 7) + Xx ( = 2 + + + 4) = 6 + 4
Yy
vx
j4
**/00
(0.25-j0.25) Vy + (0.125 + j0.25)VX = (0.75 −j 1·5)
Salve eq and Q
Vx
w = 4 rad /sec
вл
0.94 25-135 A
Vx-6 40°
J4
= 0
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