86 ma158 k ω 203 v e th 165 v 203 v 368 v 19 a r th

= (12.86 mA)(1.58 k Ω ) = 20.3 V
E
Th
= 16.5 V + 20.3 V = 36.8 V
19. a. R
Th
:
¨ R
Th
= 51 k Ω || 10 k Ω = 8.36 k Ω
E
Th
:
E
Th
=
10 k Ω (20 V)
10 k Ω + 51 k Ω
= 3.28 V
R
Th
5 mA
5 mA
1.58 k Ω
E
Th
+
−
+
−
1.2 k Ω 3.3 k Ω
+ −
I'
7.86 mA
7.86 mA

102 CHAPTER 9
b. I
E
R
E
+ V
CE
+ I
C
R
C
= 20 V
but I
C
= I
E
and I
E
( R
C
+ R
E
) + V
CE
= 20 V
or I
E
=
20 V - V
CE
R
C
+ R
E
-
20 V - 8 V
2.2 k Ω + 0.5 k Ω
=
12 V
2.7 k Ω
= 4.44 mA
c.
E
Th
- I
B
R
Th
- V
BE
- V
E
= 0
and I
B
=
E
Th
- V
BE
- V
E
R
Th
=
3.28 V - 0.7 V - (4.44 mA)(0.5 k Ω )
8.36 k Ω
= = 43.06 µ A
d. V
C
= 20 V - I
C
R
C
= 20 V - (4.44 mA)(2.2 k Ω )
= 20 V - 9.77 V
= 10.23 V
20. a. E
Th
= 20 V
I = 1.6 mA = = 12.5 k Ω
b. E
Th
= 60 mV , R
Th
= 2.72 k Ω
c. E
Th
= 16 V , R
Th
= 2.2 k Ω
21.
R
Th
= 4 Ω || (2 Ω + 2 Ω ) =
4 Ω
2
= 2 Ω
V
4 Ω
=
2 Ω (6 V)
2 Ω + 4 Ω + 2 Ω
=
12 V
8 Ω
= 1.5 V
V
2 Ω
= V
4 Ω
= 1.5 V
E
Th
= V
4 Ω
+ V
2 Ω
= 1.5 V + 1.5 V = 3 V

CHAPTER 9 103
22. a. From Problem 9, R
N
= R
Th
= 6 Ω
R
T
= 6 Ω + 3 Ω || 4 Ω
= 6 Ω + 1.714 Ω = 7.714 Ω
I
s
=
E
R
T
=
18 V
7.714 Ω
= 2.333 A
I
N
=
3 Ω (2.333 A)
3 Ω + 4 Ω
= 1 A
b. R
Th
= 6 Ω , E
Th
= I
N
R
N
= (1 A)(6 Ω ) = 6 V
c. same results
23. a. From Problem 10, R
N
= R
Th
= 4.1 k Ω
I ¢ =
2.4 k Ω (120 mA)
2.4 k Ω + (1.2 k Ω || 3.3 k Ω )
= 87.80 mA
I
N
=
1.2 k Ω (87.80 mA)
1.2 k Ω + 3.3 k Ω
= 23.41 mA
b. R
Th
= 4.1 k Ω , E
Th
= I
N
R
N
= (23.41 mA)(4.1 k Ω ) = 96 V
c. same results.
24. From Problem 12, R
N
= R
Th
= 2.18 Ω
I
N
= 6 A − 1.5 A = 4.5 A
25. From Problem 13, R
N
= R
Th
= 1.58 k Ω
I
N
= 8 mA − 7.27 mA = 0.73 mA
26. From Problem 14, R
N
= R
Th
= 7.56 Ω
3 Ω
10 Ω ⇒
8 Ω
12 V
18 V
I
N
+
−
+
−
6 A 8 Ω
I
N
3 Ω 1.5 A

104 CHAPTER 9
I ¢ =
4 Ω (2 A)
4 Ω + 3 Ω + 6 Ω || 2 Ω
=
4 Ω (2 A)
7 Ω + 1.5 Ω
= 0.941 A
I ¢¢ =
2 Ω
¢
I
2 Ω + 6 Ω
=
12 Ω (0.941 A)
8 Ω
= 0.235 A
I
N
= 2 A − I ¢¢
= 2 A − 0.235 A
= 1.765 A
27. From Problem 16, R
N
= R
Th
= 10 Ω
R
T
= 20 Ω + 5 Ω (12 Ω + 1.778 Ω ) = 23.67 Ω
I
s
=
E
T
R
T
=
20 V
23.67 Ω
= 844.95 mA
I
12 Ω
=
5 Ω (844.95 mA)
5 Ω + (12 Ω + 1.778 Ω )
=224. 98 mA
I
N
=
16 Ω (224.98 mA)
16 Ω + 2 Ω
= 200 mA
28. From Problem 18, R
N
= R
Th
= 6.08 k Ω
I
N
: Starting with figure from problem 18:
and coverting sources:
I ¢ =
12.42 V - 6 V
1.58 k Ω + 3.3 k Ω + 1.2 k Ω
= 1.06 mA
3 Ω
2 Ω
4 Ω
I ''
6 Ω
2 A
I '
2 A
1.58 k Ω
1.2 k Ω
3.3 k Ω
7.86 mA
I
N
5 mA
I
N
I'
1.58 k Ω 3.3 k Ω
12.42 V
+
−
16 V
+
−
1.2 k Ω
I'