SIS TECHN tions for I, and I, we obtain 4 = 2 mA 0matr 1, = 1 mA V= IV 12 V. S1k2 1k0 2V IA 1 k2 4V 13 1 kN er Example 3.13. In this case, we will examine the network ere are four sources, two of which are dependent, only rom the outset we expect that a loop analysis will be me arly, the circuit contains six loops. Thus ve for ll the unknown curu in One of difh 2/K +1

Introductory Circuit Analysis (13th Edition)
13th Edition
ISBN:9780133923605
Author:Robert L. Boylestad
Publisher:Robert L. Boylestad
Chapter1: Introduction
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YSIS TECHNI
r Example 3.13. In this case, we will examine the network
arly, the circuit contains six loops. Thus
lions for / and L. we obtain
%3D2 mA
0 Amatr
1, = 1 mA
V = 1V
31k2
Vo
12
1k0
2Vx
1 k2
+.
2.
V.
4V
I3
1kQ
er Example 3.13. In this case, we will examine the netwot
ere are four sources, two of which are dependent, only o at
rom the outset we expect that a loop analysis will be moe
Ive for ll
difi
the unknown sur
in
Transcribed Image Text:YSIS TECHNI r Example 3.13. In this case, we will examine the network arly, the circuit contains six loops. Thus lions for / and L. we obtain %3D2 mA 0 Amatr 1, = 1 mA V = 1V 31k2 Vo 12 1k0 2Vx 1 k2 +. 2. V. 4V I3 1kQ er Example 3.13. In this case, we will examine the netwot ere are four sources, two of which are dependent, only o at rom the outset we expect that a loop analysis will be moe Ive for ll difi the unknown sur in
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