ECE 205 Section A, B, and C. Homework #7 Solutions
3.39
For mesh 1,
0
6
10
2
10
2
1
=
−
+
−
−
I
I
I
x
But
2
1
I
I
I
x
−
=
.
Hence,
2
1
2
1
2
1
I
2
I
4
5
I
6
I
10
I
2
I
2
10
−
=
⎯→
⎯
−
+
+
−
=
(1)
For mesh 2,
2
1
1
2
4
3
6
0
6
8
12
I
I
I
I
−
=
⎯→
⎯
=
−
+
(2)
Solving (1) and (2) leads to
-0.9A
A,
8
.
0
2
1
=
=
I
I
3.40
Assume all currents are in mA and apply mesh analysis for mesh 1.
30 = 12i
1
– 6i
2
– 4i
3
15 = 6i
1
– 3i
2
– 2i
3
(1)
for mesh 2,
0 = - 6i
1
+ 14i
2
– 2i
3
0 = -3i
1
+ 7i
2
– i
3
(2)
for mesh 3,
0 = -4i
1
– 2i
2
+ 10i
3
0 = -2i
1
– i
2
+ 5i
3
(3)
Solving (1), (2), and (3), we obtain,
i
o
= i
1
=
4.286 mA
.
4 k
Ω
–
+
30V
i
1
i
3
6 k
Ω
2 k
Ω
4 k
Ω
6 k
Ω
i
2
2 k
Ω

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