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Let vo = vo1 + vo2 + vo3, where vo1, vo2 , and vo3, are due to the 20-V, 1-A, and 2-A sources respectively. For vo1, consider the circuit below.
Let i = i1 + i2 + i3, where i1 , i2 , and i3 are due to the 20-V, 2-A, and 16-V sources. For i1, consider the circuit below.
Transform the voltage sources to current sources. The result is shown in Fig. (a),
Soln0454_76.pdf
This problem does not have a solution as it was originally stated. The reason for this is that the load resistor is in series with a current source which means that the only equivalent circuit that will work will be a Norton circuit where the value of R
RTh = (10 + 20)||(50 + 40) 30||90 = 22.5 ohms
We need the Thevenin equivalent across the resistor R. To find RTh, consider the circuit below.
We need to first find RTh and VTh.
Nội dung
Chapter 4, Solution 1.
i i
o
5
Ω
8
Ω
1
Ω
+
−
1 V
3 Ω
Ω=+ 4)35(8 ,
5
1
41
1
i =
+
=
===
10
1
i
2
1
i
o
0.1A
Chapter 4, Solution 2.
,3)24(6 Ω=+ A
2
1
i
21
==i
,
4
1
i
2
1
i
1o
== ==
oo
i2v 0.5V
5
Ω
4
Ω
i
2
8
Ω
i
1
i
o
6
Ω
1 A
2 Ω
If i
s
= 1µA, then v
o
= 0.5µV
Chapter 4, Solution 3.
R
+
v
o
−
3R
i
o
3R
3R
R
+
−
3R
+
−
1 V
1.5R
V
s
(b)
(a)
(a)
We transform the Y sub-circuit to the equivalent
∆
.
,R
4
3
R
4
R3
R3R
2
== R
2
3
R
4
3
R
4
3
=+
2
v
v
s
o
= independent of R
i
o
= v
o
/(R)
When v
s
= 1V, v
o
= 0.5V, i
o
= 0.5A
(b)
When v
s
= 10V, v
o
= 5V, i
o
= 5A
(c)
When v
s
= 10V and R = 10Ω,
v
o
= 5V, i
o
= 10/(10) = 500mA
Chapter 4, Solution 4.
If I
o
= 1, the voltage across the 6Ω resistor is 6V so that the current through the 3Ω
resistor is 2A.
+
v
1
−
3A
I
s
2
Ω
4 Ω
i
1
3A
1A
I
s
2A
6
Ω
4
Ω
3 Ω
2 Ω 2
Ω
(a)
(b)
Ω= 263
, v
o
= 3(4) = 12V, .A3
4
v
o
1
==
i
Hence I
s
= 3 + 3 = 6A
If I
s
= 6A I
o
= 1
I
s
= 9A I
o
= 6/(9) = 0.6667A
Chapter 4, Solution 5.
If v
o
= 1V, V21
3
1
V
1
=+
=
3
10
v
3
2
2V
1s
=+
=
If v
s
=
3
10
v
o
= 1
Then v
s
= 15 v
o
= =15x
10
3
4.5V
v
o
3
Ω
2 Ω
+
−
6
Ω
6
Ω
6 Ω
v
1
V
s
Chapter 4, Solution 6
Let
s
T
T
oT
V
RR
R
V
RR
RR
RRR
132
32
32
then ,//
+
=
+
==
133221
32
1
32
32
32
32
1
RRRRRR
RR
R
RR
RR
RR
RR
RR
R
V
V
k
T
T
s
o
++
=
+
+
+
=
+
==
Chapter 4, Solution 7
We find the Thevenin equivalent across the 10-ohm resistor. To find V
Th
, consider the
circuit below.
3V
x
5Ω 5
Ω
+
+
4V 15
Ω
V
Th
- 6
Ω
-
+ V
x -
From the figure,
V3)4(
515
15
,0 =
+
==
Thx
VV
To find R
Th,
consider the circuit below:
3V
x
5
Ω 5
Ω
V
1
V
2
+
4V 15
Ω
1A
- 6
Ω
+ V
x
-
At node 1,
12
2111
73258616,
515
3
5
4
VVxV
VVV
V
V
xx
−=→==
−
++=
−
(1)
At node 2,
950
5
31
21
21
−=→=
−
++ VV
VV
V
x
(2)
Solving (1) and (2) leads to V
2
= 101.75 V
mW 11.22
75.1014
9
4
,75.101
1
2
max
2
===Ω==
xR
V
p
V
R
Th
Th
Th
Chapter 4, Solution 8.
Let i = i
1
+ i
2
,
where i
1
and i
L
are due to current and voltage sources respectively.
6
Ω
i
1
+
−
20V
i
2
5 A
4 Ω 6 Ω
4 Ω
(a)
(b)
i
1
= ,A3)5(
46
6
=
+
A2
46
20
2
=
+
=i
Thus i = i
1
+ i
2
= 3 + 2 = 5A
Chapter 4, Solution 9.
Let
i
2
x
1
xx
ii +=
where
i is due to 15V source and i is due to 4A source,
1
x
2
x
12
Ω
-4A
40Ω 10
Ω
i
x2
40
Ω
i
10 Ω
i
x1
12 Ω
+
−
15V
(a)
(b)
For i
x1
, consider Fig. (a).
10||40 = 400/50 = 8 ohms, i = 15/(12 + 8) = 0.75
i
x1
= [40/(40 + 10)]i = (4/5)0.75 = 0.6
For i
x2
, consider Fig. (b).
12||40 = 480/52 = 120/13
i
x2
= [(120/13)/((120/13) + 10)](-4) = -1.92
i
x
= 0.6 – 1.92 = -1.32 A
p = vi
x
= i
x
2
R = (-1.32)
2
10 = 17.43 watts
Chapter 4, Solution 10.
Let v
ab
= v
ab1
+ v
ab2
where v
ab1
and v
ab2
are due to the 4-V and the 2-A sources
respectively.
+
v
ab2
−
10
Ω
+
−
3v
ab2
2 A
10 Ω
+
−
+
v
ab1
−
+ −
3v
ab1
4V
(a) (b)
For v
ab1
, consider Fig. (a). Applying KVL gives,
- v
ab1
– 3 v
ab1
+ 10x0 + 4 = 0, which leads to v
ab1
= 1 V
For v
ab2
, consider Fig. (b). Applying KVL gives,
-
v
ab2
– 3v
ab2
+ 10x2 = 0, which leads to v
ab2
= 5
v
ab
= 1 + 5 = 6 V
Chapter 4, Solution 11.
Let i = i
1
+ i
2
, where i
1
is due to the 12-V source and i
2
is due to the 4-A source.
12V
4A
2Ω 2
Ω
i
x2
6Ω
4A
3
Ω
2Ω
i
2
3
Ω
i
o
(a)
2
Ω
i
1
6
Ω
+
−
(b)
For i
1
, consider Fig. (a).
2||3 = 2x3/5 = 6/5, i
o
= 12/(6 + 6/5) = 10/6
i
1
= [3/(2 + 3)]i
o
= (3/5)x(10/6) = 1 A
For i
2
, consider Fig. (b), 6||3 = 2 ohm, i
2
= 4/2 = 2 A
i = 1 + 2 =
3 A
Chapter 4, Solution 12.
Let v
o
= v
o1
+ v
o2
+ v
o3
, where v
o1
, v
o2
, and v
o3
are due to the 2-A, 12-V, and 19-V
sources respectively. For v
o1
, consider the circuit below.
5
Ω
5
Ω
+ v
o1
−
i
o
2A
2A
3Ω
4
Ω
6Ω 12
Ω
5 Ω
+ v
o1
−
6||3 = 2 ohms, 4||12 = 3 ohms. Hence,
i
o
= 2/2 = 1, v
o1
= 5io = 5 V
For v
o2
, consider the circuit below.
6 Ω 5 Ω 4
Ω
6
Ω
5 Ω
+ v
o2
−
3
Ω
3 Ω
+
v
1
−
+
−
12V
+ v
o2
−
12
Ω
+
−
3 Ω
12V
3||8 = 24/11, v
1
= [(24/11)/(6 + 24/11)]12 = 16/5
v
o2
= (5/8)v
1
= (5/8)(16/5) = 2 V
For v
o3
, consider the circuit shown below.
4 Ω 5 Ω 5
Ω
4 Ω
2
Ω
+ v
o3
−
12
Ω
+
v
2
−
+
−
19V
+
−
19V
6Ω
+ v
o3
−
12 Ω
3 Ω
7||12 = (84/19) ohms, v
2
= [(84/19)/(4 + 84/19)]19 = 9.975
v = (-5/7)v2 = -7.125
v
o
= 5 + 2 – 7.125 = -125 mV
,
Chapter 4, Solution 13
Let
iiii
321o
+
+
=
where i
1
, i
2
, and i
3
are the contributions to i
o
due to 30-V, 15-V, and 6-mA sources
respectively. For i
1
, consider the circuit below.
1 kΩ 2 k
Ω
3 k
Ω
+ i
1
30V
- 4 k
Ω
5 k
Ω
3//5 = 15/8 = 1.875 kohm, 2 + 3//5 = 3.875 kohm, 1//3.875 = 3.875/4.875 = 0.7949
kohm. After combining the resistors except the 4-kohm resistor and transforming the
voltage source, we obtain the circuit below.
i
1
30 mA
4 k
Ω
0.7949 k
Ω
Using current division,
mA 4.973mA)30(
7949.4
7949.0
1
==i
For i
2
, consider the circuit below.
1 k
Ω 2 k
Ω
3 k
Ω
i
2 -
15V
4 k
Ω
5 k
Ω
+
After successive source transformation and resistance combinations, we obtain the circuit
below:
2.42mA
i
2
4 k
Ω
0.7949 k
Ω
Using current division,
mA 4012.0mA)42.2(
7949.4
7949.0
2
−=−=i
For i
3
, consider the circuit below.
6mA
1 k
Ω 2 k
Ω
3 k
Ω
i
3
4 k
Ω
5 k
Ω
After successive source transformation and resistance combinations, we obtain the circuit
below:
3.097mA
i
3
4 k
Ω
0.7949 k
Ω
mA 5134.0mA)097.3(
7949.4
7949.0
3
−=−=i
Thus,
mA 058.4
321
=++= iiii
o
Chapter 4, Solution 14.
Let v
o
= v
o1
+ v
o2
+ v
o3
, where v
o1
, v
o2
, and v
o3
, are due to the 20-V, 1-A, and 2-A
sources respectively. For v
o1
, consider the circuit below.
6 Ω
20V
+
−
+
v
o1
−
4 Ω
2 Ω
3
Ω
6||(4 + 2) = 3 ohms, v
o1
= (½)20 = 10 V