2 N ZI E=150 LY Z3 Z2 Z4 26 Z 12 Z13 15 Z8 N Z7 Z16 Z14 Z9 215 Z5 ZIG ZI GIVEN Z₁ = Z2 = Z Z3 = 5 51° 2. 4440°52 Z4= 4 s-jazz ++js ˚. 25 = 10 ₤15°. NG= Z7 Z 8 = 3 = 7 L=44°2 3+j42. 4 53° = 10 -25°2 Z10 = 10 -972 Z11 = 10 -45°2 212 = 730° 213= 2 -6°2. 214 = 12 L-35 Z 15 = 25 L70° 216 = 20 25°.

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
Ise the long method STEP BY STEP process: 3. Solve for I2 4. Solve for I3
2
N
ZI
E=150 LY
Z3
Z2
Z4
26
Z 12
Z13
15
Z8
N
Z7
Z16
Z14
Z9
215
Z5
ZIG
ZI
GIVEN
Z₁ =
Z2 =
Z
Z3
=
5 51° 2.
4440°52
Z4= 4
s-jazz
++js ˚.
25 = 10 ₤15°.
NG=
Z7
Z 8
= 3
=
7 L=44°2
3+j42.
4 53°
= 10 -25°2
Z10 = 10
-972
Z11
= 10 -45°2
212 = 730°
213= 2 -6°2.
214 = 12 L-35
Z 15 = 25 L70°
216 =
20 25°.
Transcribed Image Text:2 N ZI E=150 LY Z3 Z2 Z4 26 Z 12 Z13 15 Z8 N Z7 Z16 Z14 Z9 215 Z5 ZIG ZI GIVEN Z₁ = Z2 = Z Z3 = 5 51° 2. 4440°52 Z4= 4 s-jazz ++js ˚. 25 = 10 ₤15°. NG= Z7 Z 8 = 3 = 7 L=44°2 3+j42. 4 53° = 10 -25°2 Z10 = 10 -972 Z11 = 10 -45°2 212 = 730° 213= 2 -6°2. 214 = 12 L-35 Z 15 = 25 L70° 216 = 20 25°.
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