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PROBLEM SOLUTIONS CHAPTER 3.
Solution 3.1.
Select the bottom node as the reference node, and write a node equation at the positive
terminal of the V
1
resistor:
V
1

V
0
3
R
+
V
1
6
R
+
V
1

4
V
0
6
R
=
0
⇒
2
V
1

2
V
0
+
V
1
+
V
1

4
V
0
=
0
⇒
4
V
1
=
6
V
0
⇒
V
1
=
1.5
V
0
Solution 3.2
Write a node equation at the top node:
0.6

V
x
100

2
V
x
100

V
x
50
=
0
⇒ 
V
x

2
V
x

2
V
x
= 
60
⇒ 
5
V
x
= 
60
⇒
V
x
=
12
V
Solution 3.3
0.6

V
x
100
+
25
V
x
100

V
x
50

V
x

0.2
V
x
40
=
0
⇒ 
3
V
x
100
+
25
V
x
100

8
V
x
400
= 
0.6
⇒
V
x
= 
0.6
×
400
80
⇒
V
x
= 
3
V
Solution 3.4 (a)
It is evident from the figure that V
c
= 20. We need to write two equations in V
a
and V
b
and put them in
matrix form. In this case, we can write the matrix equation by inspection. Note that the resistors are
identified by conductance values.
15
m

5
m

5
m
35
m
V
a
V
b
=
0.5
0.5
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Solve the matrix equation by inverting the leftmost matrix:
V
a
V
b
=
1
525

25
35
m
5
m
5
m
15
m
0.5
0.5
=
1
0.5
35
5
5
15
0.5
0.5
=
40
20
(c)
V
x
=
V
ab
=
V
a

V
b
=
20
V
, V
da
= V
a
= 40, V
db
= V
b
= 20.
(d)
P
i
= 0.5
×
40 = 20W, P
v
= 20
×
(2020) = 0. P
diss
= 40
×
40
×
10m + 20
×
20
×
5m + 20
×
20
×
5m
= 20 W.
Power delivered equals dissipated power.
Solution 3.6
Write two nodal equations:
V
s
1

V
1
3000
=
I
s
3
+
V
1

V
2
6000
V
2
30000
=
V
s
2

V
2
12000
+
V
1

V
2
6000
Rewrite equations as:
2
V
s
1

2
V
1
=
6000
I
s
3
+
V
1

V
2
2
V
2
=
5
V
s
2

5
V
2
+
10
V
1

10
V
2
Cast into a matrix equation

3
1

10
17
V
1
V
2
=
6000
I
s
3

2
V
s
1
5
V
s
2
Solving the matrix equation yields:
V
1
V
2
=
181.46
124.39
Power absorbed by the 6k resistor is (V
1
V
2
)
2
/R = 0.5429
W
.
Similarly, P
s1
= (V
s1
V
1
)/3000
×
V
s1
= 4.7W,
P
s
2
=
(
V
s
2

V
2
) /12000
×
V
s
2
= 
0.32
W
P
s
3
=
I
s
3
×
(
V
s
2

V
1
)
= 
1.21
W
Solution 3.7
(a)
Again, the matrix equation can be written by inspection:
G
1
+
G
2
+
G
4

G
4

G
4
G
3
+
G
4
+
G
s
V
B
V
C
=
50
G
1
50
G
3
(b)
Substituting the values of conductances and inverting the above matrix equation yields:
V
B
=
34.0132V
V
C
= 33.6842V
(c)
Power delivered is 80.7566
W
.
Using the Principle of Conservation of Power:
P
del
=
P
1
+
P
2
+
P
3
+
P
4
+
P
5
or,
P
del
=
50
×
V
A

V
B
20
+
V
A

V
C
20
=
80.7566
W
(d)
In this part, we take the above matrix equation and solve it for each value of G
s
. If we do this, we can
get a feel for the behavior of V
B
and V
C
w.r.t. changes in G
s
. The following plot is the voltage difference
between the two nodes as a function of G
s
, and hence as a function of temperature.
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