Z3 Z4 I3 L E2 I2 I5 Z2 I6 Z5 Z6 E1 Z1 Given: Z1= Z2 = 202-30° ohm %3D Z3 = Z6 = 30260° ohm Z4 = Z5 = 1040° ohm E1 = E2 = 100430° V The branch currents should be: h= 3.44450.153° A 12 = 3.44450.12° A I3 = 0.00096L-81.76° A I4 = 3.44450.13° A Is = 3.44450.14° A I6 = 0.001381Z-39.86 A %3D %3D

Introductory Circuit Analysis (13th Edition)
13th Edition
ISBN:9780133923605
Author:Robert L. Boylestad
Publisher:Robert L. Boylestad
Chapter1: Introduction
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solutions using nodal analysis, solve for branch currents

Z3
Z4
I3
E2
I2
I3
Z2
I6
Zs
E1
Z1
Given:
Z1= Z2 = 202-30° ohm
Z3 = Z6 = 30260° ohm
Z4 = Z5 = 1020° ohm
E1 = E2 = 100230° V
%3D
%3D
The branch currents should be:
1= 3.44450.153° A
%3D
I2 = 3.44450.12° A
13 = 0.00096Z-81.76° A
l4 = 3.44450.13° A
Is = 3.44450.14° A
%3D
%3D
I6 = 0.001381Z-39.86 A
%3D
Transcribed Image Text:Z3 Z4 I3 E2 I2 I3 Z2 I6 Zs E1 Z1 Given: Z1= Z2 = 202-30° ohm Z3 = Z6 = 30260° ohm Z4 = Z5 = 1020° ohm E1 = E2 = 100230° V %3D %3D The branch currents should be: 1= 3.44450.153° A %3D I2 = 3.44450.12° A 13 = 0.00096Z-81.76° A l4 = 3.44450.13° A Is = 3.44450.14° A %3D %3D I6 = 0.001381Z-39.86 A %3D
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