Applying the KCL at top node we get: e e e R + X, + xis S e e e + 3k j4k =5Z – 20° mA - j2k O e( -i+)=52 – 20' 1 3 4 e(0.4164Z36. 897°)=5Z – 20° - e=12Z – 56. 897° V

Introductory Circuit Analysis (13th Edition)
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ISBN:9780133923605
Author:Robert L. Boylestad
Publisher:Robert L. Boylestad
Chapter1: Introduction
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How did We get the e=12∠-56.897° V Please explain to me.
Applying the KCL at top node we get:
e
e
+Ý + x=is
R
XL
Xc
S.
e
e
=5Z – 20° mA
e
+
3k
+
· j2k
j4k
1
÷+3)=5Z – 20°
-
4
e(0. 4164Z36. 897°)=5Z – 20°
e=12Z – 56. 897° V
Transcribed Image Text:Applying the KCL at top node we get: e e +Ý + x=is R XL Xc S. e e =5Z – 20° mA e + 3k + · j2k j4k 1 ÷+3)=5Z – 20° - 4 e(0. 4164Z36. 897°)=5Z – 20° e=12Z – 56. 897° V
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