cw BrainRelated papers
Problem 5
Electrical Circuit By Alexander Sadiku Solution Manual for Practice Problem
Muḥammad b. Khalaf Wakīʿ (m. 306/918), ou le désamour de Bagdad
al-ʿUṣūr al-wusṭā, 2024
This article provides an overview of the life and work of a historian largely overlooked by contemporary historiography, Muhammad b. Khalaf al-Ḍabbī, known as Wakīʿ (d. 306/918). This Baghdadian scholar and qadi authored several works in various fields, only one of which has come down to us in a single manuscript: the Akhbār al-quḍāt, one of the earliest sources on the history of Islamic judgeship. Wakīʿ received fierce criticism from his fellow scholars and their successors, before most of his work fell into oblivion. After tracing his training and career, I examine the author’s views about Baghdad. The marginal role he assigns to the Abbasid capital, in connection with the miḥna of the mid-third/ninth century, suggests that this period of inquisition traumatized historians who came from scholarly backgrounds, and had a profound impact on Islamic historiography.
CARTOGRAFÍAS VITALES La producción del territorio y el derecho a la ciudad en laderas de Medellín
Las localidades de las periferias o bordes urbano-rurales de la ciudad de Medellín han sido en la última década un escenario de profundas hibridaciones sociales, políticas y culturales que surgen como resultado de las dinámicas aceleradas de poblamiento y ocupación de las laderas como consecuencia de la migración de familias del campo a la ciudad en la mayoría de los casos a causa del conflicto armado. Los nuevos vecinos han traído consigo sus memorias, sus dolores, sus prácticas culturales, sus tradiciones orales, y principalmente su convicción plena en la reivindicación de condiciones de equidad y justicia, su derecho a habitar, apropiar y transformar ese lugar que eligieron para volver a comenzar. Estos procesos de re-significación del territorio promovidos por habitantes históricos y recientes de barrios periféricos han dado lugar a la construcción de metodologías participativas y de diálogo de saberes que tejen nuevos sentidos y proyectan dimensiones del territorio e interacciones sociales que aún no alcanzan a dimensionar las políticas institucionales de planeación del territorio. En estas experiencias de diálogo de saberes han sido apropiadas por comunidades y por actores académicos la metodología de la cartografía social crítica y el Documental Social Participativo, métodos que han permitido la construcción y difusión de nuevas miradas sobre el territorio que proyectan imaginarios, propuestas y escenarios de tensión que afronta la ciudad de Medellín en la perspectiva del derecho a la ciudad.
La rivoluzione collettivista spagnola (1936-1939)
Tesina da 6 cfu in Storia dell 'Europa Contemporanea (prof. Paul Anthony Ginsborg) Relazione di Lorenzo Alba e Lanfranco Rosso A.A.
Chapter 10, Solution 1.
We first determine the input impedance.
1H
1F
j L j1x10 j10
1
j C
1
j0.1
j10 x1
1
1
1
Zin 1
1.0101 j0.1 1.015 5.653o
j10 j0.1 1
1
I
2 0o
1.9704 5.653o
o
1.015 5.653
i(t) = 1.9704cos(10t+5.65˚) A
Chapter 10, Solution 2.
Using Fig. 10.51, design a problem to help other students better understand nodal
analysis.
Although there are many ways to work this problem, this is an example based on the
same kind of problem asked in the third edition.
Problem
Solve for V o in Fig. 10.51, using nodal analysis.
2
+
o
40 V
–j5
+
_
j4
Vo
–
Figure 10.51 For Prob. 10.2.
Solution
Consider the circuit shown below.
2
Vo
–j5
+
40 V- _
j4
o
At the main node,
V
V
4 Vo
o o
j5 j 4
2
40 Vo (10 j )
V o = 40/(10–j) = (40/10.05)5.71˚ = 3.985.71˚ V
Chapter 10, Solution 3.
4
2 cos(4t )
2 0
16 sin(4 t )
16 - 90 -j16
2H
jL j8
1
1
1 12 F
- j3
jC j (4)(1 12)
The circuit is shown below.
4
-j16 V
-j3
Vo
1
+
j8
20 A
Applying nodal analysis,
Vo
Vo
- j16 Vo
2
4 j3
1 6 j8
1
1
- j16
V
2 1
4 j3
4 j3 6 j8 o
Vo
Therefore,
3.92 j2.56 4.682 - 33.15
3.835 - 35.02
1.22 j0.04
1.2207 1.88
v o ( t ) 3.835cos(4t – 35.02) V
6
Chapter 10, Solution 4.
Step 1.
Convert the circuit into the frequency domain and solve for the node
voltage, V 1 , using analysis. The find the current I C = V 1 /[1+(1/(j4x0.25)] which
then produces V o = 1xI C . Finally, convert the capacitor voltage back into the time
domain.
Ix
j1
–j1 Ω
V1
160º V
0.5I x
Vo
Note that we represented 16sin(4t–10º) volts by 160º V. That will make our
calculations easier and all we have to do is to offset our answer by a –10º.
Our node equation is [(V 1 –16)/j] – (0.5I x ) + [(V 1 –0)/(1–j)] = 0. We have two
unknowns, therefore we need a constraint equation. I x = [(16–V 1 )/j] = j(V 1 –16).
Once we have V 1 , we can find I o = V 1 /(1–j) and V o = 1xI o .
Step 2.
Now all we need to do is to solve our equations.
[(V 1 –16)/j] – [0.5j(V 1 –16] + [(V 1 –0)/(1–j)] = [–j–j0.5+0.5+j0.5]V 1 +j16+j8 = 0
or
[0.5–j]V 1 = –j24 or V 1 = j24/(–0.5+j) = (2490º)/(1.118116.57º)
= 21.47–26.57º V.
Finally, I x = V 1 /(1–j) = (21.47–26.57º) (0.707145º) = 15.18118.43º A and
V o = 1xI o = 15.18118.43º V or
v o (t) = 15.181sin(4t–10º+18.43º) = 15.181sin(4t–8.43º) volts.
Chapter 10, Solution 5.
0.25H
2 F
j L j0.25 x 4 x10 j1000
3
1
j C
1
j125
6
j4 x10 x 2 x10
3
Consider the circuit as shown below.
Io
2000
Vo
250o V +
_
-j125
j1000
+
–
At node V o ,
Vo 25 Vo 0 Vo 10I o
0
j125
2000
j1000
Vo 25 j2Vo j16Vo j160I o 0
(1 j14)Vo j160I o 25
But I o = (25–V o )/2000
(1 j14)Vo j2 j0.08Vo 25
Vo
25 j2
25.084.57
1.7768 81.37
1 j14.08 14.11558.94
Now to solve for i o ,
25 Vo 25 0.2666 j1.7567
12.367 j0.8784 mA
2000
2000
12.3984.06
Io
i o = 12.398cos(4x103t + 4.06˚) mA.
10I o
Chapter 10, Solution 6.
Let V o be the voltage across the current source. Using nodal analysis we get:
Vo
Vo 4Vx
20
Vo
3
0 where Vx
20 j10
20 j10
20
Combining these we get:
Vo
4Vo
Vo
3
0 (1 j0.5 3)Vo 60 j30
20 20 j10
20 j10
Vo
60 j30
20(3)
or Vx
2 j0.5
2 j0.5
29.11–166˚ V.
Chapter 10, Solution 7.
At the main node,
V
V
120 15 o V
630 o
40 j20
j30 50
1
j
1
V
40 j20 30 50
V
115.91 j31.058
5.196 j3
40 j20
3.1885 j4.7805
124.08 154 o V
0.04 j0.0233
Chapter 10, Solution 8.
200,
100mH
50F
jL j200x 0.1 j20
1
1
j100
jC j200x 50x10 6
The frequency-domain version of the circuit is shown below.
0.1 V o
40
V1
20
615 o
+
Vo
-
Io
V2
-j100
j20
At node 1,
or
V
V1
V V2
1
615 o 0.1V1 1
20 j100
40
5.7955 j1.5529 (0.025 j 0.01)V1 0.025V2
(1)
At node 2,
V1 V2
V
0.1V1 2
40
j20
From (1) and (2),
0 3V1 (1 j2)V2
(0.025 j0.01) 0.025 V1 (5.7955 j1.5529)
3
(1 j2) V2
0
Using MATLAB,
or
AV B
(2)
V = inv(A)*B
leads to V1 70.63 j127.23,
V2 110.3 j161.09
V V2
Io 1
7.276 82.17 o
40
Thus,
i o ( t ) 7.276 cos( 200t 82.17 o ) A
Chapter 10, Solution 9.
10 cos(10 3 t )
10 0, 10 3
10 mH
50 F
jL j10
1
1
- j20
3
jC j (10 )(50 10 -6 )
Consider the circuit shown below.
20
100 V
At node 1,
At node 2,
V1
-j20
V2
j10
Io
+
20
4 Io
30
10 V1 V1 V1 V2
20
20
- j20
10 (2 j) V1 jV2
Vo
(1)
V1 V2
V
V2
V1
( 4) 1
, where I o
has been substituted.
20
- j20
20 30 j10
(-4 j) V1 (0.6 j0.8) V2
0.6 j0.8
(2)
V1
V2
-4 j
Substituting (2) into (1)
(2 j)(0.6 j0.8)
10
V2 jV2
-4 j
170
or
V2
0.6 j26.2
Vo
Therefore,
+
30
3
170
V2
6.154 70.26
30 j10
3 j 0.6 j26.2
v o ( t ) 6.154 cos(103 t + 70.26) V
Chapter 10, Solution 10.
50 mH
2F
jL j2000x50 x10 3 j100,
1
1
j250
jC j2000 x 2x10 6
2000
Consider the frequency-domain equivalent circuit below.
V1
360o
2k
-j250
j100
V2
0.1V 1
4k
At node 1,
36
V1
V
V V2
1 1
2000 j100 j250
36 (0.0005 j0.006)V1 j0.004V2
(1)
0 (0.1 j0.004)V1 (0.00025 j0.004)V2
(2)
At node 2,
V1 V2
V
0.1V1 2
j250
4000
Solving (1) and (2) gives
Vo V2 535.6 j893.5 8951.193.43o
v o (t) = 8.951 sin(2000t +93.43o) kV
Chapter 10, Solution 11.
Consider the circuit as shown below.
–j5
Io
2
2
V1
40o V
+
_
V2
j8
2I o
At node 1,
V V2
V1 4
2I o 1
0
2
2
V1 0.5V2 2I o 2
But, I o = (4–V 2 )/(–j5) = –j0.2V 2 + j0.8
Now the first node equation becomes,
V 1 – 0.5V 2 + j0.4V 2 – j1.6 = 2 or
V 1 + (–0.5+j0.4)V 2 = 2 + j1.6
At node 2,
V2 V1 V2 4 V2 0
0
j5
2
j8
–0.5V 1 + (0.5 + j0.075)V 2 = j0.8
Using MATLAB to solve this, we get,
>> Y=[1,(-0.5+0.4i);-0.5,(0.5+0.075i)]
Y=
1.0000
-0.5000
-0.5000 + 0.4000i
0.5000 + 0.0750i
>> I=[(2+1.6i);0.8i]
I=
2.0000 + 1.6000i
0 + 0.8000i
>> V=inv(Y)*I
V=
4.8597 + 0.0543i
4.9955 + 0.9050i
I o = –j0.2V 2 + j0.8 = –j0.9992 + 0.01086 + j0.8 = 0.01086 – j0.1992
= 199.586.89˚ mA.
Chapter 10, Solution 12.
Using Fig. 10.61, design a problem to help other students to better understand Nodal analysis.
Although there are many ways to work this problem, this is an example based on the same kind
of problem asked in the third edition.
Problem
By nodal analysis, find i o in the circuit in Fig. 10.61.
Figure 10.61
Solution
20 sin(1000t )
20 0, 1000
10 mH
50 F
jL j10
1
1
- j20
3
jC j (10 )(50 10 -6 )
The frequency-domain equivalent circuit is shown below.
2 Io
V1
200 A
At node 1,
10
20
V2
-j20
Io
j10
20 2 I o
Io
V1 V1 V2
,
20
10
V2
j10
2V2 V1 V1 V2
20
j10 20
10
400 3V1 (2 j4) V2
(1)
At node 2,
or
2V2 V1 V2
V
V
2 2
j10
10
- j20 j10
j2 V1 (-3 j2) V2
V1 (1 j1.5) V2
(2)
Substituting (2) into (1),
400 (3 j4.5) V2 (2 j4) V2 (1 j0.5) V2
V2
Io
Therefore,
400
1 j0.5
V2
40
35.74 - 116.6
j10 j (1 j0.5)
i o ( t ) 35.74 sin(1000t – 116.6) A
where
Chapter 10, Solution 13.
Nodal analysis is the best approach to use on this problem. We can make our work easier
by doing a source transformation on the right hand side of the circuit.
–j2
18
+
4030º V
+
Vx
3
Vx 4030 Vx Vx 50
0
3
18 j6
j2
which leads to V x = 29.3662.88˚ A.
j6
500º V
+
Chapter 10, Solution 14.
At node 1,
At node 2,
0 V1 0 V1 V2 V1
2030
- j2
10
j4
- (1 j2.5) V1 j2.5 V2 173.2 j100
(1)
V2 V2 V2 V1
2030
j2 - j5
j4
- j5.5 V2 j2.5 V1 173.2 j100
(2)
Equations (1) and (2) can be cast into matrix form as
1 j2.5 j2.5 V1 - 200 30
j2.5
- j5.5 V2 200 30
1 j2.5
j2.5
j2.5
- j5.5
1
2
20 j5.5 20.74 - 15.38
- 200 30 j2.5
j3 (20030) 600120
200 30 - j5.5
1 j2.5 - 20030
(200 30)(1 j5) 1020108.7
j2.5
20030
1
28.93135.38 V
2
49.18124.08 V
V2
V1
Chapter 10, Solution 15.
We apply nodal analysis to the circuit shown below.
5A
2
-j20 V
At node 1,
At node 2,
+
j
V1
-j2
V2
I
4
2I
- j20 V1
V
V V2
5 1 1
2
- j2
j
- 5 j10 (0.5 j0.5) V1 j V2
(1)
V1 V2 V2
,
j
4
V
where I 1
- j2
5
V2
V1
0.25 j
5 2I
(2)
Substituting (2) into (1),
- 5 j10
j5
0.5 (1 j) V1
0.25 j
j40
(1 j) V1 -10 j20
1 j4
160 j40
( 2 - 45) V1 -10 j20
17 17
V1 15.81313.5
I
V1
(0.590)(15.81313.5)
- j2
I 7.90643.49 A
Chapter 10, Solution 16.
Consider the circuit as shown in the figure below.
j4
V1
V2
+ Vx –
20o A
5
–j3
345o A
At node 1,
V 0 V1 V2
2 1
0
5
j4
(0.2 j0.25)V1 j0.25V2 2
(1)
At node 2,
V2 V1 V2 0
345 0
j4
j3
j0.25V1 j0.08333V2 2.121 j2.121
In matrix form, (1) and (2) become
0.2 j0.25
j0.25
j0.25 V1
2
j0.08333 V2 2.121 j2.121
Solving this using MATLAB, we get,
>> Y=[(0.2-0.25i),0.25i;0.25i,0.08333i]
Y=
0.2000 - 0.2500i
0 + 0.2500i
0 + 0.2500i
0 + 0.0833i
>> I=[2;(2.121+2.121i)]
I=
(2)
2.0000
2.1210 + 2.1210i
>> V=inv(Y)*I
V=
5.2793 - 5.4190i
9.6145 - 9.1955i
V s = V 1 – V 2 = –4.335 + j3.776 = 5.749138.94˚ V.
Chapter 10, Solution 17.
Consider the circuit below.
j4
10020 V
+
Io
1
2
V1
V2
3
-j2
At node 1,
10020 V1 V1 V1 V2
j4
3
2
V1
100 20
(3 j10) j2 V2
3
(1)
At node 2,
10020 V2 V1 V2 V2
1
2
- j2
100 20 -0.5 V1 (1.5 j0.5) V2
(2)
From (1) and (2),
10020 - 0.5
0.5 (3 j) V1
10020 1 j10 3
- j2 V2
- 0.5
1 j10 3
1.5 j0.5
- j2
0.1667 j4.5
1
10020 1.5 j0.5
-55.45 j286.2
10020
- j2
2
- 0.5
10020
-26.95 j364.5
1 j10 3 10020
1
64.74 - 13.08
2
V2
81.17 - 6.35
V V2 1 2 - 28.5 j78.31
Io 1
2
2
0.3333 j 9
V1
I o 9.25-162.12 A
Chapter 10, Solution 18.
Consider the circuit shown below.
8
V1
2
445 A
j6 V
2
+
Vx
2 Vx
At node 1,
V1 V1 V2
2
8 j6
200 45 (29 j3) V1 (4 j3) V2
(1)
At node 2,
V1 V2
V
V2
2Vx 2
,
8 j6
- j 4 j5 j2
(104 j3) V1 (12 j41) V2
12 j41
V1
V
104 j3 2
(2)
4
-j
j5
-j2
445
(12 j41)
V (4 j3) V2
104 j3 2
200 45 (14.2189.17) V2
20045
V2
14.2189.17
Substituting (2) into (1),
200 45 (29 j3)
- j2
- j2
- 6 j8
V2
V2
V2
4 j5 j2
4 j3
25
10 233.13
200 45
Vo
25
14.2189.17
Vo
Vo 5.63189 V
where Vx V1
+
Vo
Chapter 10, Solution 19.
We have a supernode as shown in the circuit below.
j2
V1
2
Notice that
V2
+
4
-j4
Vo
V3
0.2 V o
Vo V1 .
At the supernode,
V3 V2 V2 V1 V1 V3
4
- j4 2
j2
0 (2 j2) V1 (1 j) V2 (-1 j2) V3
At node 3,
V1 V3 V3 V2
0.2V1
j2
4
(0.8 j2) V1 V2 (-1 j2) V3 0
(2)
Subtracting (2) from (1),
0 1.2V1 j V2
(3)
But at the supernode,
V1 12 0 V2
or
V2 V1 12
(4)
Substituting (4) into (3),
0 1.2V1 j (V1 12)
j12
V1
Vo
1.2 j
Vo
1290
1.56239.81
Vo 7.68250.19 V
(1)
Chapter 10, Solution 20.
The circuit is converted to its frequency-domain equivalent circuit as shown below.
R
V m 0
Let
Z jL ||
1
jC
+
+
L
C
jL
1
jC
If
and
Vo A , then
A
Vo
1
jC
jL
1 2 LC
jL
jL
1 2 LC
Vm
V
jL
R (1 2 LC) jL m
R
1 2 LC
L Vm
L
90 tan -1
R (1 2 LC)
R 2 (1 2 LC) 2 2 L2
Z
Vo
V
RZ m
Vo
jL
L Vm
R 2 (1 2 LC) 2 2 L2
90 tan -1
L
R (1 2 LC)
Chapter 10, Solution 21.
(a)
Vo
Vi
1
jC
R jL
At 0 ,
1
jC
As ,
At
(b)
Vo
Vi
1
LC
jL
At 0 ,
As ,
At
1
LC
,
1
1 2 LC jRC
Vo 1
1
Vi 1
Vo
0
Vi
Vo
Vi
,
R jL
1
jC
1
jRC
1
-j L
R C
LC
2 LC
1 2 LC jRC
Vo
0
Vi
Vo 1
1
Vi 1
Vo
Vi
1
jRC
1
LC
j L
R C
Chapter 10, Solution 22.
Consider the circuit in the frequency domain as shown below.
R1
R2
Vs
1
jC
+
jL
Let
Z (R 2 jL) ||
+
Vo
1
jC
1
(R jL)
R 2 jL
jC 2
Z
1
1 jR 2 2 LC
R 2 jL
jC
R 2 jL
Vo
1 2 LC jR 2 C
Z
R 2 jL
Vs Z R 1
R1
1 2 LC jR 2 C
Vo
R 2 jL
2
Vs R 1 R 2 LCR 1 j (L R 1 R 2 C)
Chapter 10, Solution 23.
V Vs
V
jCV 0
1
R
jL
j C
V
jRCV
2LC 1
jRCV Vs
1 2LC jRC jRC j3RLC2
V Vs
2
1 LC
V
(1 2 LC )V s
1 2 LC jRC ( 2 2 LC )
Chapter 10, Solution 24.
Design a problem to help other students to better understand mesh analysis.
Although there are many ways to work this problem, this is an example based on the
same kind of problem asked in the third edition.
Problem
Use mesh analysis to find V o in the circuit in Prob. 10.2.
Solution
Consider the circuit as shown below.
2
+
o
40 V
I1
+
_
j4
–j5
Vo
I2
–
For mesh 1,
4 (2 j 5) I1 j 5 I1
For mesh 2,
(1)
0 j 5I 1 ( j 4 j 5) I 2
I1
1
I2
5
(2)
Substituting (2) into (1),
1
4 (2 j 5) I 2 j 5I 2
5
I2
1
0.1 j
V o = j4I 2 = j4/(0.1+j) = j4/(1.00499 84.29°) = 3.98 5.71° V
Chapter 10, Solution 25.
2
10 cos(2t )
100
6 sin(2t )
6 - 90 -j6
2H
jL j4
1
1
0.25 F
- j2
jC j (2)(1 4)
The circuit is shown below.
4
j4
Io
100 V
+
-j2
I1
For loop 1,
- 10 (4 j2) I 1 j2 I 2 0
5 (2 j) I 1 j I 2
(1)
For loop 2,
j2 I 1 ( j4 j2) I 2 (- j6) 0
I1 I 2 3
I2
+
6-90 V
(2)
In matrix form (1) and (2) become
2 j j I 1 5
1 1 I 3
2
2 (1 j) ,
I o I1 I 2
Therefore,
1 5 j3 ,
2 1 j3
4
1 2
1 j 1.414245
2 (1 j )
i o ( t ) 1.4142cos(2t + 45) A
Chapter 10, Solution 26.
0.4 H
1 F
j L j10 x 0.4 j400
3
1
j C
1
j1000
6
j10 x10
3
The circuit becomes that shown below.
2 k
–j1000
Io
100o
+
_
+
_
I1
j400
–j20
I2
For loop 1,
10 (12000 j 400) I1 j 400 I 2 0
1 (200 j 40) I1 j 40 I 2 (1)
For loop 2,
j 20 ( j 400 j1000) I 2 j 400 I1 0
12 40 I1 60 I 2
(2)
In matrix form, (1) and (2) become
1 200 j 40 j 40 I1
12 40
60 I 2
Solving this leads to
I 1 =0.0025-j0.0075, I 2 = -0.035+j0.005
I o = I 1 – I 2 = 0.0375 – j0.0125 = 39.5 –18.43° mA
i o (t) = 39.5cos(103t–18.43°) mA
Chapter 10, Solution 27.
For mesh 1,
For mesh 2,
- 40 30 ( j10 j20) I 1 j20 I 2 0
4 30 - j I 1 j2 I 2
(1)
50 0 (40 j20) I 2 j20 I 1 0
5 - j2 I 1 (4 j2) I 2
(2)
From (1) and (2),
430 - j
j2 I 1
5 - j2 - (4 j2) I
2
-2 4 j 4.472116.56
1 -(4 30)(4 j2) j10 21.01211.8
2 - j5 8120 4.44 154.27
I1
I2
1
4.69895.24 A
2
992.837.71 mA
Chapter 10, Solution 28.
1
1
j0.25
jC j1x 4
The frequency-domain version of the circuit is shown below, where
1H
V1 100 o ,
jL j4,
1F
V2 20 30 o .
1
j4
j4
1
-j0.25
+
+
V1
I1
1
I2
V2
-
-
V1 100 o ,
V2 20 30 o
Applying mesh analysis,
10 (2 j3.75)I1 (1 j0.25)I 2
(1)
20 30 o (1 j0.25)I1 (2 j3.75)I 2
(2)
From (1) and (2), we obtain
10
2 j3.75 1 j0.25 I1
17.32 j10 1 j0.25 2 j3.75 I 2
Solving this leads to
I1 2.741 41.07 o ,
I 2 4.11492 o
Hence,
i 1 (t) = 2.741cos(4t–41.07˚)A, i 2 (t) = 4.114cos(4t+92˚)A.
Chapter 10, Solution 29.
Using Fig. 10.77, design a problem to help other students better understand mesh analysis.
Although there are many ways to work this problem, this is an example based on the same kind
of problem asked in the third edition.
Problem
By using mesh analysis, find I 1 and I 2 in the circuit depicted in Fig. 10.77.
Figure 10.77
Solution
For mesh 1,
(5 j5) I 1 (2 j) I 2 30 20 0
30 20 (5 j5) I 1 (2 j) I 2
(1)
For mesh 2,
(5 j3 j6) I 2 (2 j) I 1 0
0 - (2 j) I 1 (5 j3) I 2
(2)
From (1) and (2),
3020 5 j5 - (2 j) I 1
0 - (2 j) 5 - j3 I
2
37 j6 37.489.21
1 (30 20)(5.831 - 30.96) 175 - 10.96
2 (30 20)(2.356 26.56) 67.0846.56
I1
I2
1
4.67–20.17 A
2
1.7937.35 A
Chapter 10, Solution 30.
300m H
200m H
400m H
50 F
j L j100 x 300 x10
3
j L j100 x 400 x10
3
j L j100 x 200 x10
3
j20
j40
1
j200
6
j100 x 50 x10
The circuit becomes that shown below.
1
j C
j30
20
j40
j20
10
+
o
12090
+
_
I1
j30
–j200
I2
vo
I 3`
-
+
_
800o
For mesh 1,
120 90o (20 j30) I1 j 30 I 2 0
j120 (20 j 30) I1 j 30 I 2
For mesh 2,
j 30 I1 ( j 30 j 40 j 200) I 2 j 200 I 3 0
0 3I1 13I 2 20 I 3
For mesh 3,
80 j200I 2 (10 j180)I 3 0 8 j20I 2 (1 j18)I 3
(3)
We put (1) to (3) in matrix form.
0 I1 j12
2 j3 j3
3 13
20 I 2 0
0
j20 1 j18 I 3 8
This is an excellent candidate for MATLAB.
>> Z=[(2+3i),-3i,0;-3,-13,20;0,20i,(1-18i)]
Z=
2.0000 + 3.0000i
0 - 3.0000i
0
(1)
(2)
-3.0000
0
-13.0000
20.0000
0 +20.0000i 1.0000 -18.0000i
>> V=[12i;0;-8]
V=
0 +12.0000i
0
-8.0000
>> I=inv(Z)*V
I=
2.0557 + 3.5651i
0.4324 + 2.1946i
0.5894 + 1.9612i
V o = –j200(I 2 – I 3 ) = –j200(–0.157+j0.2334) = 46.68 + j31.4 = 56.2633.93˚
v o = 56.26cos(100t + 33.93˚) V.
Chapter 10, Solution 31.
Consider the network shown below.
80
100120 V
+
For loop 1,
For loop 2,
For loop 3,
From (2),
I1
Io
-j40
j60
I2
-j40
20
I3
+
60-30 V
- 100120 (80 j40) I1 j40 I 2 0
10 20 4 (2 j) I 1 j4 I 2
(1)
j40 I 1 ( j60 j80) I 2 j40 I 3 0
0 2 I1 I 2 2 I 3
(2)
60 - 30 (20 j40) I 3 j40 I 2 0
- 6 - 30 j4 I 2 2 (1 j2) I 3
(3)
2 I 3 I 2 2 I1
Substituting this equation into (3),
- 6 - 30 -2 (1 j2) I 1 (1 j2) I 2
(4)
From (1) and (4),
10120 4 (2 j)
j4 I 1
- 6 - 30 - 2 (1 j2) 1 j2 I
2
2
8 j4
- j4
32 j20 37.7432
- 2 j4 1 j2
8 j4 10120
-4.928 j82.11 82.2593.44
- 2 j4 - 6 - 30
Io I2
2
2.17961.44 A
Chapter 10, Solution 32.
Consider the circuit below.
j4
2
4-30 V
For mesh 1,
where
Hence,
Io
+
Vo
I1
+
3 Vo
I2
(2 j4) I 1 2 (4 - 30) 3 Vo 0
Vo 2 (4 - 30 I 1 )
(2 j4) I 1 8 - 30 6 (4 - 30 I 1 ) 0
4 - 30 (1 j) I 1
I 1 2 2 15
or
Io
3 Vo
3
(2)(4 - 30 I 1 )
- j2 - j2
I o j3 (4 - 30 2 2 15)
I o 8.48515 A
Vo
- j2 I o
5.657-75 V
3
-j2
5A
Chapter 10, Solution 33.
Consider the circuit shown below.
I4
2
j
I
-j20 V
For mesh 1,
+
I1
-j2
I2
2I
j20 (2 j2) I 1 j2 I 2 0
(1 j) I 1 j I 2 - j10
(1)
For the supermesh,
( j j2) I 2 j2 I 1 4 I 3 j I 4 0
Also,
For mesh 4,
4
I3
(2)
I 3 I 2 2 I 2 (I 1 I 2 )
I 3 2 I1 I 2
(3)
I4 5
(4)
Substituting (3) and (4) into (2),
(8 j2) I 1 (- 4 j) I 2 j5
(5)
Putting (1) and (5) in matrix form,
1 j
j I 1 - j10
8 j2 4 j I j5
2
-3 j5 ,
I I1 I 2
1 -5 j40 ,
2 -15 j85
1 2 10 j45
- 3 j5
7.90643.49 A
Chapter 10, Solution 34.
The circuit is shown below.
Io
5
I2
8
4090 V
For mesh 1,
+
3A
-j2
10
I1
20
I3
j4
- j40 (18 j2) I 1 (8 j2) I 2 (10 j4) I 3 0
For the supermesh,
(13 j2) I 2 (30 j19) I 3 (18 j2) I 1 0
Also,
I2 I3 3
Adding (1) and (2) and incorporating (3),
- j40 5 (I 3 3) (20 j15) I 3 0
3 j8
I3
1.46538.48
5 j3
I o I 3 1.46538.48 A
j15
(1)
(2)
(3)
4
Chapter 10, Solution 35.
j2
Consider the circuit shown below.
I3
8
1
-j3
10
20 V
+
I1
-j4 A
For the supermesh,
- 20 8 I 1 (11 j8) I 2 (9 j3) I 3 0
Also,
For mesh 3,
I2
-j5
(1)
I 1 I 2 j4
(2)
(13 j) I 3 8 I 1 (1 j3) I 2 0
(3)
Substituting (2) into (1),
(19 j8) I 2 (9 j3) I 3 20 j32
(4)
Substituting (2) into (3),
- (9 j3) I 2 (13 j) I 3 j32
(5)
From (4) and (5),
19 j8 - (9 j3) I 2 20 j32
- (9 j3) 13 j I j32
3
167 j69 ,
I2
2 324 j148
2 324 j148 356.2 - 24.55
167 j69 180.69 - 22.45
I 2 1.971–2.1 A
Chapter 10, Solution 36.
Consider the circuit below.
j4
490 A
-j3
+
2
I1
Vo
2
I2
+
2
I3
20 A
Clearly,
I 1 4 90 j4
and
I 3 -2
For mesh 2,
(4 j3) I 2 2 I 1 2 I 3 12 0
(4 j3) I 2 j8 4 12 0
- 16 j8
-3.52 j0.64
I2
4 j3
Thus,
Vo 2 (I 1 I 2 ) (2)(3.52 j4.64) 7.04 j9.28
Vo 11.64852.82 V
120 V
Chapter 10, Solution 37.
I1
+
120 90 o V
-
Ix
Z=80-j35
Z
I2
Iz
o
120 30 V
+
Iy
Z
I3
For mesh x,
ZI x ZI z j120
(1)
ZI y ZI z 12030 o 103.92 j60
(2)
ZI x ZI y 3ZI z 0
(3)
For mesh y,
For mesh z,
Putting (1) to (3) together leads to the following matrix equation:
0
(80 j35) I x
j120
(80 j35)
0
(80 j35) (80 j35) I y 103.92 j60
(80 j35) (80 j35) (240 j105) I
0
z
Using MATLAB, we obtain
- 0.2641 j2.366
I inv(A) * B - 2.181 - j0.954
- 0.815 j1.1066
I 1 I x 0.2641 j 2.366 2.38 96.37 o A
I 2 I y I x 1.9167 j1.4116 2.38143.63 o A
I 3 I y 2.181 j 0.954 2.38 23.63 o A
AI B
Chapter 10, Solution 38.
Consider the circuit below.
Io
2
I1
20 A
j2
1
I2
-j4
40 A
I3
+
I4
1090 V
1
A
Clearly,
For mesh 2,
I1 2
(1)
(2 j4) I 2 2 I 1 j4 I 4 10 90 0
(2)
Substitute (1) into (2) to get
(1 j2) I 2 j2 I 4 2 j5
For the supermesh,
(1 j2) I 3 j2 I 1 (1 j4) I 4 j4 I 2 0
j4 I 2 (1 j2) I 3 (1 j4) I 4 j4
At node A,
I3 I4 4
Substituting (4) into (3) gives
j2 I 2 (1 j) I 4 2 (1 j3)
From (2) and (5),
1 j2 j2 I 2 2 j5
j2 1 j I 2 j6
4
3 j3 ,
1 9 j11
- 1 - (9 j11) 1
(-10 j)
3 j3
3
I o 3.35174.3 A
Io -I2
(3)
(4)
(5)
Chapter 10, Solution 39.
For mesh 1,
(28 j15)I1 8I 2 j15I 3 1264 o
(1)
8I1 (8 j9)I 2 j16I 3 0
(2)
j15I1 j16I 2 (10 j)I 3 0
(3)
For mesh 2,
For mesh 3,
In matrix form, (1) to (3) can be cast as
8
j15 I1 1264 o
(28 j15)
(8 j9) j16 I 2 0
8
j15
j16 (10 j) I 3 0
or
AI B
Using MATLAB,
I = inv(A)*B
I 1 0.128 j 0.3593 381.4109.6° mA
I 2 0.1946 j 0.2841 344.3124.4° mA
I 3 0.0718 j 0.1265 145.5–60.42° mA
I x I 1 I 2 0.0666 j 0.0752 100.548.5° mA
381.4109.6° mA, 344.3124.4° mA, 145.5–60.42° mA, 100.548.5° mA
Chapter 10, Solution 40.
Let I o = I o1 + I o2 , where I o1 is due to the dc source and I o2 is due to the ac source. For I o1 ,
consider the circuit in Fig. (a).
Clearly,
4
2
I o1
+
8V
(a)
I o1 = 8/2 = 4 A
For I o2 , consider the circuit in Fig. (b).
4
2
I o2
100 V
j4
+
(b)
If we transform the voltage source, we have the circuit in Fig. (c), where 4 || 2 4 3 .
I o2
2.50 A
4
2
j4
(c)
By the current division principle,
43
Io2
(2.50)
4 3 j4
I o 2 0.25 j 0.75 0.79 - 71.56
Thus,
I o 2 0.79 cos(4t 71.56) A
Therefore,
I o = I o1 + I o2 = [4 + 0.79cos(4t–71.56)] A
Chapter 10, Solution 41.
We apply superposition principle. We let
vo = v1 + v2
where v 1 and v 2 are due to the sources 6cos2t and 4sin4t respectively. To find v 1 ,
consider the circuit below.
-j2
+
+
_
60o
2
V1
–
1/ 4F
1
j C
1
j2
j2 x1/ 4
(6) = 3+j3 = 4.243 45°
Thus,
v 1 (t) = 4.243cos(2t+45°) volts.
To get v 2 (t), consider the circuit below,
–j
+
40o
+
_
2
V2
–
1/ 4F
1
j C
1
j1
j4 x1/ 4
or
v 2 (t) = 3.578sin(4t+25.56°) volts.
Hence,
v o = [4.243cos(2t+45˚) + 3.578sin(4t+25.56˚)] volts.
Chapter 10, Solution 42.
Using Fig. 10.87, design a problem to help other students to better understand the
superposition theorem.
Although there are many ways to work this problem, this is an example based on the
same kind of problem asked in the third edition.
Problem
Solve for I o in the circuit of Fig. 10.87.
j10
o
200 V
+
_
50
60
Io
–j40
+
_
3045o V
Figure 10.87 For Prob. 10.42.
Solution
Let I o I1 I 2
where I 1 and I 2 are due to 20<0o and 30<45o sources respectively. To get I 1 , we use the
circuit below.
I1
j10
60
o
200 V
+
_
50
–j40
Let Z 1 = -j40//60 = 18.4615 –j27.6927, Z 2 = j10//50=1.9231 + j9.615
Transforming the voltage source to a current source leads to the circuit below.
I1
Z2
Z1
–j2
Using current division,
Z2
I1
( j 2) 0.6217 j 0.3626
Z1 Z 2
To get I 2 , we use the circuit below.
j10
I2
50
60
–j40
+
_
After transforming the voltage source, we obtain the circuit below.
I2
Z2
Z1
0.545o
Using current division,
Z1
I2
(0.5 45o ) 0.5275 j 0.3077
Z1 Z 2
3045o V
Hence, I o = I 1 + I 2 = 0.0942+j0.0509 = 109 30° mA.
Chapter 10, Solution 43.
Let I x I 1 I 2 , where I 1 is due to the voltage source and
I 2 is due to the current source.
2
5 cos(2t 10)
510
10 cos(2t 60)
10 - 60
4H
1
F
8
jL j8
1
1
-j4
jC j (2)(1 / 8)
For I 1 , consider the circuit in Fig. (a).
-j4
3
I1
j8
+
10-60 V
(a)
I1
10 - 60 10 - 60
3 j8 j4
3 j4
For I 2 , consider the circuit in Fig. (b).
510 A
-j4
3
j8
(b)
I2
- j8
- j40 10
(510)
3 j8 j4
3 j4
I x I1 I 2
1
(10 - 60 j4010)
3 j4
49.51 - 76.04
9.902 - 129.17
Ix
553.13
Therefore,
i x 9.902 cos(2t – 129.17) A
I2
Chapter 10, Solution 44.
Let v x v1 v 2 , where v 1 and v 2
respectively.
For v 1 , 6 , 5 H
are due to the current source and voltage source
jL j30
The frequency-domain circuit is shown below.
20
16
Is
Let Z 16 //(20 j30)
+
V1
-
16(20 j30)
11.8 j3.497 12.3116.5 o
36 j30
V1 I s Z (1210 o )(12.3116.5 o ) 147.726.5 o
For v 2 , 2 , 5 H
j30
jL j10
v1 147.7 cos(6 t 26.5 o ) V
The frequency-domain circuit is shown below.
20
16
j10
+
V2
+
Vs
-
-
Using voltage division,
-
V2
Thus,
16(500 o )
16
21.41 15.52 o
Vs
36 j10
16 20 j10
v 2 21.41sin(2t 15.52 o ) V
v x [147.7cos(6t+26.5°)+21.41sin(2t–15.52°)] V
Chapter 10, Solution 45.
Let i i1 i2 , where i 1 and i 2 are due to 16cos(10t +30o) and 6sin4t sources respectively.
To find i 1 , consider the circuit below.
I1
20
+
_
16 30o V
jX
X
L 10 x 300 x103 3
Type equation here.
.47°
= 0.7913
i 1 (t) = 791.1cos(10t+
.47°) mA.
To find i 2 (t), consider the circuit below,
I2
20
+
_
60o V
jX
X
L 4 x 300 x103 1.2
= 0.2995 176.57° or
i 2 (t) = 299.5sin(4t+176.57°) mA.
Thus,
i(t) = i 1 (t) + i 2 (t) = [791.1cos(10t+21.47°) + 299.5sin(4t+176.57°)] mA.
Chapter 10, Solution 46.
Let v o v1 v 2 v 3 , where v1 , v 2 , and v 3 are respectively due to the 10-V dc source, the ac
current source, and the ac voltage source. For v1 consider the circuit in Fig. (a).
6
2H
+
1/12 F
+
v1
10 V
(a)
The capacitor is open to dc, while the inductor is a short circuit. Hence,
v1 10 V
For v 2 , consider the circuit in Fig. (b).
2
2H
jL j4
1
1
1
F
- j6
jC j (2)(1 / 12)
12
6
-j6
+
40 A
V2
(b)
Applying nodal analysis,
V
V
V 1 j j
4 2 2 2 V2
6 - j6 j4 6 6 4
V2
Hence,
24
21.4526.56
1 j0.5
v 2 21.45 sin( 2 t 26.56) V
For v 3 , consider the circuit in Fig. (c).
3
j4
2H
jL j6
1
1
1
- j4
F
12
jC j (3)(1 / 12)
6
120 V
+
j6
+
-j4
V3
(c)
At the non-reference node,
12 V3 V3 V3
6
- j4 j6
12
V3
10.73 - 26.56
1 j0.5
Hence,
v 3 10.73 cos(3t 26.56) V
Therefore,
v o [10+21.45sin(2t+26.56)+10.73cos(3t–26.56)] V
Chapter 10, Solution 47.
Let i o i1 i 2 i 3 , where i1 , i 2 , and i 3 are respectively due to the 24-V dc source, the
ac voltage source, and the ac current source. For i1 , consider the circuit in Fig. (a).
1
24 V
1/6 F
+
2H
i1
2
4
(a)
Since the capacitor is an open circuit to dc,
24
4A
i1
42
For i 2 , consider the circuit in Fig. (b).
1
2H
jL j2
1
1
- j6
F
6
jC
1
j2
-j6
I2
10-30 V
+
2
I1
I2
4
(b)
For mesh 1,
For mesh 2,
- 10 - 30 (3 j6) I 1 2 I 2 0
10 - 30 3 (1 2 j) I 1 2 I 2
(1)
0 -2 I 1 (6 j2) I 2
I 1 (3 j) I 2
(2)
Substituting (2) into (1)
Hence,
10 - 30 13 j15 I 2
I 2 0.504 19.1
i 2 0.504 sin( t 19.1) A
For i 3 , consider the circuit in Fig. (c).
3
2H
jL j6
1
1
1
F
- j2
jC j (3)(1 / 6)
6
1
-j2
j6
I3
2
20 A
4
(c)
2 || (1 j2)
2 (1 j2)
3 j2
Using current division,
2 (1 j2)
(20)
2 (1 j2)
3 j2
I3
2 (1 j2)
13 j3
4 j6
3 j2
I 3 0.3352 - 76.43
Hence
i 3 0.3352 cos(3t 76.43) A
Therefore,
i o [4 + 0.504 sin(t + 19.1) + 0.3352 cos(3t – 76.43)] A
Chapter 10, Solution 48.
Let i O i O1 i O 2 i O 3 , where i O1 is due to the ac voltage source, i O 2 is due to the dc
voltage source, and i O3 is due to the ac current source. For i O1 , consider the circuit in
Fig. (a).
2000
50 cos(2000t )
500
40 mH
jL j (2000)(40 10 -3 ) j80
1
1
- j25
jC j (2000)(20 10 -6 )
20 F
I
500 V
-j25
I O1
80
+
j80
(a)
100
60
80 || (60 100) 160 3
50
30
I
160 3 j80 j25 32 j33
- 80 I
-1
10180
I
80 160 3
4645.9
0.217 134.1
i O1 0.217 cos(2000 t 134.1) A
Using current division,
I O1
I O1
Hence,
For i O 2 , consider the circuit in Fig. (b).
i O2
80
100
60
+
(b)
24 V
i O2
24
0 .1 A
80 60 100
For i O3 , consider the circuit in Fig. (c).
4000
2 cos(4000t )
2 0
40 mH
20 F
jL j (4000)(40 10 -3 ) j160
1
1
- j12.5
jC j (4000)(20 10 -6 )
-j12.5
I2
80
j160
I O3
I3
20 A
60
I1
(c)
`
For mesh 1,
I1 2
(1)
For mesh 2,
(80 j160 j12.5) I 2 j160 I 1 80 I 3 0
Simplifying and substituting (1) into this equation yields
(8 j14.75) I 2 8 I 3 j32
(2)
For mesh 3,
240 I 3 60 I 1 80 I 2 0
Simplifying and substituting (1) into this equation yields
I 2 3 I 3 1.5
(3)
100
Substituting (3) into (2) yields
(16 j44.25) I 3 12 j54.125
12 j54.125
1.17827.38
I3
16 j44.25
Hence,
Therefore,
I O 3 - I 3 -1.17827.38
i O 3 -1.1782 sin( 4000t 7.38) A
i O {0.1 + 0.217 cos(2000t + 134.1) – 1.1782 sin(4000t + 7.38)} A
Chapter 10, Solution 49.
8 sin( 200t 30)
830, 200
5 mH
1 mF
jL j (200)(5 10 -3 ) j
1
1
- j5
jC j (200)(1 10 -3 )
After transforming the current source, the circuit becomes that shown in the figure below.
5
4030 V
I
3
I
j
+
-j5
40 30
40 30
4.47256.56
5 3 j j5
8 j4
i [4.472sin(200t+56.56)] A
Chapter 10, Solution 50.
Using Fig. 10.95, design a problem to help other students to better understand source
transformation.
Although there are many ways to work this problem, this is an example based on the same kind
of problem asked in the third edition.
Problem
Use source transformation to find v o in the circuit in Fig. 10.95.
Figure 10.95
Solution
5 cos(10 5 t )
50, 10 5
0.4 mH
0.2 F
jL j (10 5 )(0.4 10 -3 ) j40
1
1
- j50
5
jC j (10 )(0.2 10 -6 )
After transforming the voltage source, we get the circuit in Fig. (a).
j40
20
0.250
-j50
(a)
Let
and
Z 20 || - j 50
- j100
2 j5
Vs (0.250) Z
- j25
2 j5
80
+
Vo
With these, the current source is transformed to obtain the circuit in Fig.(b).
j40
Z
Vs
80
+
+
(b)
By voltage division,
- j25
80
80
Vs
- j100
2 j5
Z 80 j40
80 j40
2 j5
8 (- j25)
Vo
3.615 - 40.6
36 j42
v o 3.615 cos(105 t – 40.6) V
Vo
Therefore,
Vo
Chapter 10, Solution 51.
There are many ways to create this problem, here is one possible solution. Let V 1 =
40 30° V, X L = 10 Ω, X C = 20 Ω, R 1 = R 2 = 80 Ω, and V 2 = 50 V.
If we let the voltage across the capacitor be equal to V x , then
I o = [V x /(–j20)] + [(V x –50)/80] = (0.0125+j0.05)V x – 0.625 = (0.051539 75.96°)V x –
0.625.
The following circuit is obtained by transforming the voltage sources.
Vx
4-60 V
j10
-j20
40
V x = (4-60+1.25)/(–j0.1+j0.05+0.025) = (2–j3.4641+1.25)/(0.025–j0.05)
= (3.25–j3.4641)/( 0.025–j0.05) = (4.75 –46.826°)/(0.055902 –63.435°)
= 84.97 16.609° V.
Therefore,
1.250 A
I o = (0.051539 75.96°)(84.97 16.609°) – 0.625 = 4.3793 92.569° – 0.625
= –0.196291+j4.3749 – 0.625 = –0.821291+j4.3749 = 4.451 100.63° A.
Chapter 10, Solution 52.
We transform the voltage source to a current source.
600
6 j12
Is
2 j4
The new circuit is shown in Fig. (a).
-j2
Ix
2
I s = 6 – j12
4
6
j4
590 A
-j3
(a)
Let
6 (2 j4)
2.4 j1.8
8 j4
Vs I s Z s (6 j12)(2.4 j1.8) 36 j18 18 (2 j)
Z s 6 || (2 j4)
With these, we transform the current source on the left hand side of the circuit to a voltage
source. We obtain the circuit in Fig. (b).
Zs
-j2
Ix
Vs
4
+
-j3
(b)
Let
Z o Z s j2 2.4 j0.2 0.2 (12 j)
Vs
18 (2 j)
Io
15.517 j6.207
Z o 0.2 (12 j)
j5 A
With these, we transform the voltage source in Fig. (b) to a current source. We obtain the circuit
in Fig. (c).
Ix
Io
4
Zo
-j3
j5 A
(c)
Using current division,
Zo
2.4 j0.2
Ix
(I o j5)
(15.517 j1.207)
Z o 4 j3
6.4 j3.2
I x 5 j1.5625 5.23817.35 A
Chapter 10, Solution 53.
We transform the voltage source to a current source to obtain the circuit in Fig. (a).
-j3
4
50 A
Let
j4
j2
+
2
Vo
-j2
(a)
Z s 4 || j2
j8
0.8 j1.6
4 j2
Vs (50) Z s (5)(0.8 j1.6) 4 j8
With these, the current source is transformed so that the circuit becomes that shown in
Fig. (b).
Zs
Vs
-j3
j4
2
+
-j2
+
Vo
(b)
Let
Z x Z s j3 0.8 j1.4
V
4 j8
Ix s
3.0769 j4.6154
Z s 0.8 j1.4
With these, we transform the voltage source in Fig. (b) to obtain the circuit in Fig. (c).
j4
Ix
Zx
2
(c)
Let
Z y 2 || Z x
1.6 j2.8
0.8571 j0.5714
2.8 j1.4
-j2
+
Vo
Vy I x Z y ( 3.0769 j4.6154) (0.8571 j0.5714) j5.7143
With these, we transform the current source to obtain the circuit in Fig. (d).
j4
Zy
Vy
-j2
+
+
Vo
(d)
Using current division,
Vo
- j2 ( j5.7143)
- j2
Vy
(3.529 – j5.883) V
Z y j4 j2
0.8571 j0.5714 j4 j2
Chapter 10, Solution 54.
50 x( j 30)
13.24 j 22.059
50 j 30
We convert the current source to voltage source and obtain the circuit below.
50 //( j 30)
40
+
V s =115.91 –j31.06V
13.24 – j22.059
j20
+
-
I
134.95-j74.912 V
V
-
+
-
Applying KVL gives
-115.91 + j31.058 + (53.24-j2.059)I -134.95 + j74.912 = 0
or I
250.86 j105.97
4.7817 j1.8055
53.24 j 2.059
But Vs (40 j20)I V 0
V Vs (40 j20)I
V 115.91 j 31.05 (40 j 20)( 4.7817 j1.8055) 124.06 154 o V
which agrees with the result in Prob. 10.7.
Chapter 10, Solution 55.
(a)
To find Z th , consider the circuit in Fig. (a).
j20
10
-j10
Z th
(a)
Z N Z th 10 j20 || (- j10) 10
( j20)(- j10)
j20 j10
10 j20 22.36-63.43
To find Vth , consider the circuit in Fig. (b).
j20
5030 V
+
10
+
-j10
V th
(b)
Vth
IN
(b)
- j10
(50 30) -5030 V
j20 j10
Vth
- 50 30
2.236273.4 A
Z th 22.36 - 63.43
To find Z th , consider the circuit in Fig. (c).
-j5
8
j10
(c)
Z th
Z N Z th j10 || (8 j5)
( j10)(8 j5)
1026
j10 8 j5
To obtain Vth , consider the circuit in Fig. (d).
-j5
Io
40 A
8
j10
(d)
By current division,
32
8
Io
(40)
8 j5
8 j10 j5
Vth j10 I o
IN
j320
33.9258 V
8 j5
Vth 33.92 58
3.39232 A
10 26
Z th
+
V th
Chapter 10, Solution 56.
(a)
To find Z th , consider the circuit in Fig. (a).
j4
6
-j2
Z th
(a)
Z N Z th 6 j4 || (- j2) 6
( j4)(- j2)
6 j4
j4 j2
= 7.211-33.69
By placing short circuit at terminals a-b, we obtain,
I N 20 A
Vth Z th I th (7.211 - 33.69) (20) 14.422-33.69 V
(b)
To find Z th , consider the circuit in Fig. (b).
j10
30
60
-j5
(b)
30 || 60 20
(- j5)(20 j10)
20 j5
= 5.423-77.47
Z N Z th - j5 || (20 j10)
Z th
To find Vth and I N , we transform the voltage source and combine the 30
and 60 resistors. The result is shown in Fig. (c).
j10
445 A
20
a
-j5
(c)
IN
2
20
(445) (2 j)(445)
5
20 j10
= 3.57818.43 A
Vth Z th I N (5.423 - 77.47) (3.57818.43)
= 19.4-59 V
IN
b
Chapter 10, Solution 57.
Using Fig. 10.100, design a problem to help other students to better understand Thevenin and
Norton equivalent circuits.
Although there are many ways to work this problem, this is an example based on the same kind
of problem asked in the third edition.
Problem
Find the Thevenin and Norton equivalent circuits for the circuit shown in
Fig. 10.100.
Figure 10.100
Solution
To find Z th , consider the circuit in Fig. (a).
5
-j10
2
j20
Z th
(a)
( j20)(5 j10)
5 j10
18 j12 21.633-33.7
Z N Z th 2 j20 || (5 j10) 2
To find Vth , consider the circuit in Fig. (b).
5
60120 V
+
-j10
2
+
j20
V th
(b)
Vth
j20
j4
(60 120)
(60120)
5 j10 j20
1 j2
= 107.3146.56 V
IN
Vth 107.3146.56
4.961-179.7 A
Z th 21.633 - 33.7
Chapter 10, Solution 58.
Consider the circuit in Fig. (a) to find Z eq .
8
-j6
j10
Z eq
(a)
Z eq j10 || (8 j 6)
( j10)(8 j 6)
5 (2 j )
8 j4
= 11.1826.56
Consider the circuit in Fig. (b) to find V Thev .
Io
8
545 A
j10
-j6
(b)
Io
4 j3
8 j6
(545)
(545)
4 j2
8 j6 j10
VThev j10 I o
( j10)(4 j 3)(545)
(2)(2 j )
= 55.971.56 V
+
V Thev
Chapter 10, Solution 59.
Calculate the output impedance of the circuit shown in Fig. 10.102.
–j2 Ω
+
0.2V o
j40 Ω
Vo
10
Figure 10.102
For Prob. 10.59.
Solution
Since there are no independent sources, we need to inject a current, best value is
to make it 1 amp, into the terminals on the right and then to determine the voltage
at the terminals.
–j2 Ω
V1
+
0.2V o
j40 Ω
Vo
10
a
1A
b
Clearly V o = –(–j2) = j2 and V 1 = (0.2V o + 1)j40 = (1+j0.4)j40 = –16+j40 V.
Next, V ab = 10 – j2 – 16 + j40 = –6+j38 = 38.47 98.97° V or
Z eq = (–6+j38) Ω.
Chapter 10, Solution 60.
(a)
To find Z eq , consider the circuit in Fig. (a).
10
-j4
a
j5
4
Z eq
b
(a)
Z eq 4 || (- j 4 10 || j 5) 4 || (- j 4 2 j 4)
Z eq 4 || 2
= 1.333
To find VThev , consider the circuit in Fig. (b).
10
200 V
+
-j4
V1
V2
j5
40 A
4
+
V Thev
(b)
At node 1,
20 V1 V1 V1 V2
10
j5
- j4
(1 j0.5) V1 j2.5 V2 20
(1)
V1 V2 V2
- j4
4
V1 (1 j) V2 j16
(2)
At node 2,
4
Substituting (2) into (1) leads to
28 j16 (1.5 j3) V2
28 j16
8 j5.333
V2
1.5 j3
VThev V2 9.61533.69 V
Therefore,
(b)
To find Z eq , consider the circuit in Fig. (c).
Z eq
10
c
-j4
j5
d
4
(c)
j10
Z eq - j 4 || (4 10 || j 5) - j 4 || 4
2 j
Z eq - j 4 || (6 j 4)
- j4
(6 j 4) (2.667 – j4)
6
To find VThev ,we will make use of the result in part (a).
V2 8 j5.333 (8 3 ) (3 j2)
V1 (1 j) V2 j16 j16 (8 3) (5 j)
VThev V1 V2 16 3 j8 9.61456.31 V
Chapter 10, Solution 61.
Find the Thevenin equivalent at terminals a-b of the circuit in Fig. 10.104.
V 1 and V 2
4
a
Ix
+
-j3
o
20 A
1.5I x
V oc
I sc
b
Figure 10.104
For Prob. 10.61.
Solution
Step 1.
First we solve for the open circuit voltage using the above circuit and
writing two node equations. Then we solve for the short circuit current which
only need one node equation. For being able to solve for V oc , we need to solve
these three equations,
–2 + [(V 1 –0)/(–j3)] + [(V 1 –V oc )/4] = 0 and
[(V oc –V 1 )/4] – 1.5I x = 0 where I x = [(V 1 –0)/(–j3)].
To solve for I sc , all we need to do is to solve these three equations,
–2 + [(V 2 –0)/(–j3)] + [(V 2 –0)/4] = 0, I sc = [V 2 /4] + 1.5I x , and
I x = [V 2 /–j3].
Finally, V Thev = V oc and Z eq = V oc /I sc .
Step 2.
Now all we need to do is to solve for the unknowns. For V oc ,
I x = j0.33333V 1 and (0.25+(1.5)(j0.33333))V 1 = 0.25V oc or
(0.25+j0.5)V 1 = (0.55902
.4 °)V 1 = 0.25V oc or
V 1 = (0.44721 –
.4 °)V oc which leads to,
(0.25+j0.33333)V 1 – 0.25V oc = 2
.
= (0.41666
= (0.186335 –
°)(0.44721 –
.4 °)V oc – 0.25V oc
. °)V oc – 0.25V oc = (0.183333–0.25–j0.033333)V oc
= (–0.066667–j0.033333)V oc = (0.074536 –
.4
°)V oc = 2 or
V oc = V Thev = 26.83 153.44° V = (–24+j12) V.
Now for I sc ,
I sc = [V 2 /4] + 1.5I x = (0.25+(1.5)(j0.33333))V 2 = (0.25+j0.5)V 2 .
[(V 2 –0)/(–j3)] + [(V 2 –0)/4] = 2 = (0.25+j0.3333)V 2
= (0.41667 53.13°)V 2 = 2 or V 2 = 4.8 –53.13°
I sc = (0.25+j0.5)V 2 = (0.55901 63.435°)(4.8 –53.13°)
= 2.6832
.
°A
Finally,
Z eq = V oc /I sc = 26.833 153.435°/2.6832
= 10
4 .
.
°
° Ω or = (–8+j6) Ω.
Chapter 10, Solution 62.
First, we transform the circuit to the frequency domain.
12 cos( t )
120, 1
2H
1
F
4
1
F
8
jL j2
1
- j4
jC
1
- j8
jC
To find Z eq , consider the circuit in Fig. (a).
3 Io
Io
4
Vx
j2
1
Ix
2
-j4
-j8
+
1V
(a)
At node 1,
Thus,
At node 2,
Vx Vx
1 Vx
3Io
,
j2
4 - j4
where I o
Vx 2 Vx 1 Vx
- j4
4
j2
Vx 0.4 j0.8
I x 3Io
1 1 Vx
- j8
j2
I x (0.75 j0.5) Vx j
I x -0.1 j0.425
Z eq
3
8
1
-0.5246 j 2.229 2.29 - 103.24
Ix
- Vx
4
To find VThev , consider the circuit in Fig. (b).
3 Io
Io
4
j2
V1
1
120 V
+
V2
2
-j4
-j8
+
V Thev
(b)
At node 1,
At node 2,
12 V1
V
V V2
12 V1
3Io 1 1
, where I o
4
4
- j4
j2
24 (2 j) V1 j2 V2
(1)
V1 V2
V2
3Io
j2
- j8
72 (6 j4) V1 j3 V2
(2)
From (1) and (2),
24 2 j - j2 V1
72 6 j4 - j3 V
2
-5 j6 ,
Vth V2
Thus,
Therefore,
2
3.073 - 219.8
2 - j24
2
(2)(3.073 - 219.8)
Vth
2 Z th
1.4754 j2.229
6.146 - 219.8
2.3 - 163.3
Vo
2.673 - 56.5
Vo
v o 2.3cos(t–163.3) V
Chapter 10, Solution 63.
Transform the circuit to the frequency domain.
4 cos(200t 30)
430, 200
10 H
5 F
jL j (200)(10) j2 k
1
1
- j k
jC j (200)(5 10 -6 )
Z N is found using the circuit in Fig. (a).
j2 k
-j k
2 k
ZN
(a)
Z N - j 2 || j2 - j 1 j 1 k
We find I N using the circuit in Fig. (b).
-j k
430 A
j2 k
2 k
(b)
j2 || 2 1 j
By the current division principle,
1 j
(4 30) 5.657 75
IN
1 j j
Therefore,
i N (t) = 5.657 cos(200t + 75) A
Z N 1 k
IN
Chapter 10, Solution 64.
Z N is obtained from the circuit in Fig. (a).
60
40
ZN
-j30
j80
(a)
Z N (60 40) || ( j80 j30) 100 || j50
Z N 20 j40 44.7263.43
(100)( j50)
100 j50
To find I N , consider the circuit in Fig. (b).
60
I1
40
I2
-j30
IN
360 A
Is
j80
(b)
I s 360
For mesh 1,
100 I 1 60 I s 0
I 1 1.860
For mesh 2,
( j80 j30) I 2 j80 I s 0
I 2 4.860
I N = I 2 – I 1 = 360 A
Chapter 10, Solution 65.
Using Fig. 10.108, design a problem to help other students to better understand Norton’s
theorem.
Although there are many ways to work this problem, this is an example based on the same kind
of problem asked in the third edition.
Problem
Compute i o in Fig. 10.108 using Norton's theorem.
Figure 10.108
Solution
5 cos(2 t )
50, 2
4H
1
F
4
1
F
2
jL j (2)(4) j8
1
1
- j2
jC j (2)(1 / 4)
1
1
-j
jC j (2)(1 / 2)
To find Z N , consider the circuit in Fig. (a).
2
-j2
ZN
(a)
-j
Z N - j || (2 j2)
- j (2 j2) 1
(2 j10)
2 j3
13
To find I N , consider the circuit in Fig. (b).
2
50 V
+
-j2
IN
50
j5
-j
-j
IN
(b)
The Norton equivalent of the circuit is shown in Fig. (c).
Io
ZN
IN
j8
(c)
Using current division,
ZN
(1 13)(2 j10)( j5) 50 j10
IN
Io
(1 13)(2 j10) j8 2 j94
Z N j8
I o 0.1176 j0.5294 0542 - 77.47
Therefore, i o 542 cos(2t – 77.47) mA
Chapter 10, Solution 66.
10
0.5 H
jL j (10)(0.5) j5
1
1
- j10
10 mF
jC j (10)(10 10 -3 )
To find Z th , consider the circuit in Fig. (a).
-j10
10
Vx
+
j5
Vo
2 Vo
1A
(a)
1 2 Vo
Vx
Vx
,
j5 10 j10
19 Vx
V
- 10 j10
1
x
Vx
10 j10
j5
21 j2
Z N Z th
where Vo
10Vx
10 j10
Vx
14.142 135
670129.56 m
1
21.0955.44
To find Vth and I N , consider the circuit in Fig. (b).
120 V
-j10
-j2 A
10
+
+
Vo
j5
I
(b)
where
(10 j10 j5) I (10)(- j2) j5 (2 Vo ) 12 0
Vo (10)(- j2 I )
+
2 Vo
V th
Thus,
(10 j105) I -188 j20
188 j20
I
- 10 j105
Vth j5 (I 2 Vo ) j5 (19I j40) j95 I 200
j 95 (188 j 20)
(95 90)(189.066.07)
Vth
200
200
- 10 j105
105.4895.44
170.28 179.37 200 170.27 j1.8723 200 29.73 j1.8723
Vth 29.79–3.6 V
IN
Vth 29.79 3.6
44.46–133.16 A
Z th 0.67129.56
Chapter 10, Solution 67.
Z N Z Th 10 //(13 j 5) 12 //( 8 j 6)
10(13 j 5) 12(8 j 6)
11.243 j1.079
23 j 5
20 j 6
Va
Vb
10
(6045 o ) 13.78 j21.44,
23 j5
(8 j6)
(6045 o ) 12.069 j26.08
20 j6
VTh Va Vb 1.711 j 4.64 4.945 69.76 o V,
IN
VTh 4.945 69.76
437.8 75.24 o mA
11.295 5.48
Z Th
Chapter 10, Solution 68.
1H
jL j10x1 j10
1
1
1
j2
F
1
20
j C
j10 x
20
We obtain V Th using the circuit below.
Io
4
a
+
+
6<0o
-
V o /3
+
-
-j2 j10
b
j10( j2)
j2.5
j10 j2
Vo 4I o x ( j2.5) j10I o
1
6 4I o Vo 0
3
j10 //( j2)
(1)
(2)
Combining (1) and (2) gives
Io
6
,
4 j10 / 3
Vo
4I o
VTh Vo j10I o
j60
11.52 50.19 o
4 j10 / 3
v Th 11.52 sin(10t 50.19 o )
To find R Th, we insert a 1-A source at terminals a-b, as shown below.
Io
4
a
+
+
-
V o /3
-j2 j10
Vo
4I o
-
1
4I o Vo 0
3
V
Io o
12
1 4I o
Vo Vo
j2 j10
Combining the two equations leads to
Vo
1
1.2293 j1.4766
0.333 j0.4
V
Z Th o 1.2293 1.477
1
1<0o
Chapter 10, Solution 69.
This is an inverting op amp so that
Vo - Z f
-R
–jRC
1 jC
Vs
Zi
When Vs Vm and 1 RC ,
1
RC Vm - j Vm Vm - 90
Vo - j
RC
Therefore,
v o ( t ) Vm sin(t 90) - V m cos(t)
Chapter 10, Solution 70.
Using Fig. 10.113, design a problem to help other students to better understand op amps in AC
circuits.
Although there are many ways to work this problem, this is an example based on the same kind
of problem asked in the third edition.
Problem
The circuit in Fig. 10.113 is an integrator with a feedback resistor. Calculate v o (t) if v s =
2 cos 4 104t V.
Figure 10.113
Solution
This may also be regarded as an inverting amplifier.
2 cos(4 10 4 t )
2 0, 4 10 4
1
1
10 nF
- j2.5 k
4
jC j (4 10 )(10 10 -9 )
Vo - Z f
Vs
Zi
where Z i 50 k and Z f 100k || (- j2.5k )
Thus,
If Vs 2 0 ,
Vo - (- j2)
Vs 40 j
- j100
k .
40 j
Vo
Therefore,
j4
490
0.191.43
40 j 40.01 - 1.43
v o ( t ) 100 cos(4x104 t + 91.43) mV
Chapter 10, Solution 71.
8 cos(2t 30 o )
830 o
1
1
0. 5F
j1M
jC j2x 0.5x10 6
At the inverting terminal,
Vo 830 o Vo 830 o 830 o
j1000k
10k
2k
Vo (1 j100) 830 800 60 4000 60
Vo
6.928 j4 2400 j4157 4800 59.9
4829.53o
1 j100
100 89.43
v o (t) = 48cos(2t+29.53o) V
Chapter 10, Solution 72.
4 cos(10 4 t )
4 0, 10 4
1
1
1 nF
- j100 k
4
jC j (10 )(10 -9 )
Consider the circuit as shown below.
50 k
40 V
+
-j100 k
At the noninverting node,
4 Vo
Vo
50
- j100
Io
Therefore,
Vo
Vo
+
Vo
Io
100 k
4
1 j0.5
Vo
4
mA 35.78 - 26.56 A
100k (100)(1 j0.5)
i o ( t ) 35.78cos(104t–26.56) A
Chapter 10, Solution 73.
As a voltage follower, V2 Vo
C1 10 nF
1
1
-j20 k
3
jC1 j (5 10 )(10 10 -9 )
1
1
C 2 20 nF
-j10 k
3
jC 2 j (5 10 )(20 10 -9 )
Consider the circuit in the frequency domain as shown below.
-j20 k
I s 10 k
20 k
V1
VS
+
-j10 k
Z in
At node 1,
Vs V1 V1 Vo V1 Vo
10
- j20
20
2 Vs (3 j)V1 (1 j)Vo
(1)
At node 2,
V1 Vo Vo 0
20
- j10
V1 (1 j2)Vo
(2)
Substituting (2) into (1) gives
2 Vs j6Vo
or
1
Vo -j Vs
3
2
1
V1 (1 j2)Vo j Vs
3
3
V2
+
Io
Vo
Vs V1 (1 3)(1 j)
Vs
10k
10k
1 j
30k
Is
Is
Vs
Z in
Vs 30k
15 (1 j) k
Is 1 j
Z in 21.21–45 k
Chapter 10, Solution 74.
Zi R1
Zf R 2
1
,
jC1
1
jC 2
R2
1
C 1 j R 2 C 2
V
jC 2
- Zf
1
Av o
1
Vs
Zi
C 2 1 jR 1C1
R1
jC1
Av –
At 0 ,
C1
C2
As ,
Av –
At
1
,
R 1 C1
C 1 j R 2 C 2 R 1C1
A v – 1
1 j
C2
1
,
R 2C2
C
1 j
A v – 1
C 2 1 j R 1C1 R 2 C 2
At
R2
R1
Chapter 10, Solution 75.
2 10 3
C1 C 2 1 nF
1
1
-j500 k
3
jC1 j (2 10 )(1 10 -9 )
Consider the circuit shown below.
100 k
Let V s = 10V.
-j500 k
-j500 k
V2
+
V1
VS
100 k
+
40 k
Vb
20 k
+
Vo
At node 1,
[(V 1 –10)/(–j500k)] + [(V 1 –V o )/105] + [(V 1 –V 2 )/(–j500k)] = 0
or (1+j0.4)V 1 – j0.2V 2 – V o = j2
(1)
[(V 2 –V 1 )/(–j500k)] + [(V 2 –0)/100k] + 0 = 0 or
–j0.2V 1 + (1+j0.2)V 2 = 0 or V 1 = [–(1+j0.2)/(–j0.2)]V 2
= (1–j5)V 2
(2)
At node 2,
At node b,
Vb =
R3
V
Vo o V 2
3
R3 R4
From (2) and (3),
V 1 = (0.3333–j1.6667)V o
Substituting (3) and (4) into (1),
(1+j0.4)(0.3333–j1.6667)V o – j0.06667V o – V o = j2
(1+j0.4)(0.3333–j1.6667) = (1.07721.8˚)(1.6997–78.69˚)
= 1.8306–56.89˚ = 1–j1.5334
(3)
(4)
(1–1+j(–1.5334–0.06667))V o = (–j1.6001)V o = 1.6001–90˚
Therefore,
V o = 290˚/(1.6001–90˚) = 1.2499180˚
Since V s = 10,
V o /V s = 0.12499180˚.
Chapter 10, Solution 76.
Let the voltage between the -jk capacitor and the 10k resistor be V 1.
230 o V1 V1 Vo V1 Vo
j4k
10k
20k
230 o (1 j0.6)V1 j0.6Vo
= 1.7321+j1
Also,
V1 Vo
V
o
j2k
10k
V1 (1 j5)Vo
(1)
(2)
Solving (2) into (1) yields
230 (1 j0.6)(1 j5)Vo j0.6Vo (1 3 j0.6 j5 j6)Vo
= (4+j5)V o
230
Vo
0.3124 21.34 o V
6.40351.34
= 312.4–21.34˚ mV
I o = (V 1 –V o )/20k = V o /(–j4k) = (0.3124/4k)(–21.43+90)˚
= 78.168.57˚ µA
We can easily check this answer using MATLAM. Using equations (1) and (2) we can
identify the following matrix equations:
YV = I where
>> Y=[1-0.6i,0.6i;1,-1-0.5i]
Y=
1.0000 - 0.6000i
0 + 0.6000i
1.0000
-1.0000 - 5.0000i
>> I=[1.7321+1i;0]
I=
1.7321 + 1.0000i
0
>> V=inv(Y)*I
V=
0.8593 + 1.3410i
0.2909 - 0.1137i = V o = 312.3–21.35˚ mV. The answer checks.
Chapter 10, Solution 77.
Consider the circuit below.
R3
2
R1
1
+
VS
At node 1,
At node 2,
V1
+
C2
R2
V1
C1
Vs V1
jC V1
R1
Vs (1 jR 1C1 ) V1
+
Vo
(1)
0 V1 V1 Vo
jC 2 (V1 Vo )
R3
R2
R3
jC 2 R 3
V1 (Vo V1 )
R2
1
V1
Vo 1
(R 3 R 2 ) jC 2 R 3
Vs
R2
1
1 jR 1C1 R 3 jC 2 R 2 R 3
From (1) and (2),
Vo
Vo
R 2 R 3 jC 2 R 2 R 3
Vs (1 jR 1C 1 ) ( R 3 jC 2 R 2 R 3 )
(2)
Chapter 10, Solution 78.
2 sin(400t )
20, 400
1
1
0.5 F
- j5 k
jC j (400)(0.5 10 -6 )
1
1
- j10 k
0.25 F
jC j (400)(0.25 10 -6 )
Consider the circuit as shown below.
20 k
10 k V
1
20 V
+
-j5 k
V2
+
Vo
40 k
-j10 k
10 k
20 k
At node 1,
At node 2,
But
V
V V2 V1 Vo
2 V1
1 1
10
- j10
- j5
20
4 (3 j6) V1 j4 V2 Vo
(1)
V1 V2 V2
j5
10
V1 (1 j0.5) V2
(2)
V2
(3)
20
1
Vo Vo
20 40
3
From (2) and (3),
1
V1 (1 j0.5) Vo
3
Substituting (3) and (4) into (1) gives
1
4
1
4 (3 j6) (1 j0.5) Vo j Vo Vo 1 j Vo
3
3
6
24
3.945 9.46
Vo
6 j
Therefore,
v o ( t ) 3.945sin(400t–9.46) V
(4)
Chapter 10, Solution 79.
0.5 cos(1000 t )
0.50, 1000
1
1
- j10 k
0.1 F
jC j (1000)(0.1 10 -6 )
1
1
0.2 F
- j5 k
jC j (1000)(0.2 10 -6 )
Consider the circuit shown below.
20 k
-j10 k
10 k
V s = 0.50
+
+
40 k
-j5 k
Since each stage is an inverter, we apply Vo
Vo
and
V1
- Zf
V to each stage.
Zi i
- 40
V1
- j5
(1)
- 20 || (- j10)
Vs
10
(2)
From (1) and (2),
- j8 - ( 20)(-j10)
Vo
0.50
10 20 j10
Vo 1.6 ( 2 j) 35.7826.56
Therefore,
+
V1
v o ( t ) 3.578cos(1000t + 26.56) V
+
Vo
Chapter 10, Solution 80.
4 cos(1000t 60)
4 - 60, 1000
1
1
0.1 F
- j10 k
jC j (1000)(0.1 10 -6 )
1
1
- j5 k
0.2 F
jC j (1000)(0.2 10 -6 )
The two stages are inverters so that
20
20 - j5
Vo
V
(4 - 60)
50 o 10
- j10
-j 2
-j
( j2) (4 - 60) Vo
2 5
2
(1 j 5) Vo 4 - 60
4 - 60
3.922 - 71.31
Vo
1 j 5
Therefore,
v o ( t ) 3.922 cos(1000t – 71.31) V
1
Chapter 10, Solution 81.
We need to get the capacitance and inductance corresponding to –j2 and j4 .
1
1
j2
C
0.5 F
X c 1x 2
X
j4
L L 4H
The schematic is shown below.
When the circuit is simulated, we obtain the following from the output file.
FREQ
VM(5)
VP(5)
1.592E-01 1.127E+01 -1.281E+02
From this, we obtain
V o = 11.27128.1o V.
Chapter 10, Solution 82.
The schematic is shown below. We insert PRINT to print V o in the output file. For AC
Sweep, we set Total Pts = 1, Start Freq = 0.1592, and End Freq = 0.1592. After
simulation, we print out the output file which includes:
FREQ
1.592 E-01
which means that
VM($N_0001)
7.684 E+00
V o = 7.68450.19o V
VP($N_0001)
5.019 E+01
Chapter 10, Solution 83.
The schematic is shown below. The frequency is f / 2
1000
159.15
2
When the circuit is saved and simulated, we obtain from the output file
FREQ
1.592E+02
VM(1)
6.611E+00
VP(1)
-1.592E+02
Thus,
v o = 6.611cos(1000t – 159.2o) V
Chapter 10, Solution 84.
The schematic is shown below. We set PRINT to print V o in the output file. In AC
Sweep box, we set Total Pts = 1, Start Freq = 0.1592, and End Freq = 0.1592. After
simulation, we obtain the output file which includes:
FREQ
VM($N_0003)
1.592 E-01
1.664 E+00
VP($N_0003)
E+02
Namely,
V o = 1.664–146.4o V
–1.646
Chapter 10, Solution 85.
Using Fig. 10.127, design a problem to help other students to better understand performing AC
analysis with PSpice.
Although there are many ways to solve this problem, this is an example based on the same kind
of problem asked in the third edition.
Problem
Use PSpice to find V o in the circuit of Fig. 10.127. Let R 1 = 2 Ω, R 2 = 1 Ω, R 3 = 1 Ω, R 4 = 2 Ω,
I s = 20˚ A, X L = 1 Ω, and X C = 1 Ω.
Solution
The schematic is shown below. We let 1 rad/s so that L=1H and C=1F.
When the circuit is saved and simulated, we obtain from the output file
FREQ
1.591E-01
VM(1)
2.228E+00
VP(1)
-1.675E+02
From this, we conclude that
V o = 2.228–167.5° V.
Chapter 10, Solution 86.
The schematic is shown below. We insert three pseudocomponent PRINTs at nodes 1, 2,
and 3 to print V 1 , V 2 , and V 3 , into the output file. Assume that w = 1, we set Total Pts =
1, Start Freq = 0.1592, and End Freq = 0.1592. After saving and simulating the circuit,
we obtain the output file which includes:
FREQ
VM($N_0002)
1.592 E-01
6.000 E+01
FREQ
VM($N_0003)
1.592 E-01
2.367 E+02
FREQ
VM($N_0001)
1.592 E-01
1.082 E+02
VP($N_0002)
3.000
E+01
VP($N_0003)
-8.483
E+01
VP($N_0001)
E+02
Therefore,
V 1 = 6030o V V 2 = 236.7-84.83o V V 3 = 108.2125.4o V
1.254
Chapter 10, Solution 87.
The schematic is shown below. We insert three PRINTs at nodes 1, 2, and 3. We set
Total Pts = 1, Start Freq = 0.1592, End Freq = 0.1592 in the AC Sweep box. After
simulation, the output file includes:
FREQ
VM($N_0004)
1.592 E-01
1.591 E+01
FREQ
VM($N_0001)
1.592 E-01
5.172 E+00
FREQ
VM($N_0003)
1.592 E-01
2.270 E+00
VP($N_0004)
1.696
E+02
VP($N_0001)
-1.386
E+02
VP($N_0003)
E+02
Therefore,
V 1 = 15.91169.6o V V 2 = 5.172-138.6o V V 3 = 2.27-152.4o V
-1.524
Chapter 10, Solution 88.
The schematic is shown below. We insert IPRINT and PRINT to print I o and V o in the
output file. Since w = 4, f = w/2 = 0.6366, we set Total Pts = 1, Start Freq = 0.6366,
and End Freq = 0.6366 in the AC Sweep box. After simulation, the output file includes:
FREQ
VM($N_0002)
6.366 E-01
3.496 E+01
1.261
FREQ
IM(V_PRINT2)
IP
6.366 E-01
8.912 E-01
VP($N_0002)
E+01
(V_PRINT2)
-8.870 E+01
Therefore,
V o = 34.9612.6o V,
v o = 34.96 cos(4t + 12.6o)V,
I o = 0.8912-88.7o A
i o = 0.8912cos(4t – 88.7o )A
Chapter 10, Solution 89.
Consider the circuit below.
R1
R2
V in
2
R3
1
V in
0 Vin Vin V2
R1
R2
R
- Vin V2 2 Vin
R1
V2 Vin Vin V4
R3
1 jC
Vin V2
- Vin V4
jCR 3
(2)
From (1) and (2),
- Vin V4
Thus,
- R2
V
jCR 3 R 1 in
I in
Vin V4
R2
V
R4
jCR 3 R 1 R 4 in
Z in
Vin jCR 1R 3 R 4
jL eq
R2
I in
where
R4
L eq
I in
+
(1)
At node 3,
4
3
+
At node 1,
C
R 1R 3 R 4C
R2
+
V in
Chapter 10, Solution 90.
Let
Z 4 R ||
1
R
jC 1 jRC
1
1 jRC
Z3 R
jC
jC
Consider the circuit shown below.
Z3
Vi
+
+
Z4
Vo
R1
Vo
R2
R2
Z4
Vi
V
R1 R 2 i
Z3 Z 4
R
Vo
R2
1 jC
R
1 jRC R 1 R 2
Vi
1 jC
jC
jRC
R2
2
jRC (1 jRC)
R1 R 2
Vo
R2
jRC
2
2 2
Vi 1 R C j3RC R 1 R 2
For Vo and Vi to be in phase,
1 2 R 2 C 2 0
1
2f
RC
or
f
Vo
must be purely real. This happens when
Vi
1
2RC
At this frequency,
Vo 1
R2
Av
Vi 3 R 1 R 2
Chapter 10, Solution 91.
(a)
Let
V2 voltage at the noninverting terminal of the op amp
Vo output voltage of the op amp
Z p 10 k R o
Z s R jL
1
jC
As in Section 10.9,
Zp
V2
Vo Z s Z p
Ro
R R o jL
CR o
V2
Vo C (R R o ) j ( 2 LC 1)
j
C
For this to be purely real,
o2 LC 1 0
o
fo
1
2 LC
1
LC
1
2 (0.4 10 -3 )(2 10 -9 )
f o 180 kHz
(b)
At oscillation,
o CR o
Ro
V2
Vo o C (R R o ) R R o
This must be compensated for by
Vo
80
Av
1
5
V2
20
Ro
1
R Ro 5
R 4R o 40 k
Chapter 10, Solution 92.
Let
V2 voltage at the noninverting terminal of the op amp
Vo output voltage of the op amp
Zs R o
RL
1
1
Z p jL ||
|| R
1
1
L jR ( 2 LC 1)
jC
jC
R
jL
RL
V2
L jR (2 LC 1)
RL
Vo Z s Z p
Ro
L jR (2 LC 1)
V2
RL
Vo RL R o L jR o R (2 LC 1)
As in Section 10.9,
Zp
For this to be purely real,
o2 LC 1
f o
(a)
1
2 LC
At o ,
o RL
V2
R
Vo o RL o R o L R R o
This must be compensated for by
Vo
Rf
1000k
Av
1
1
11
Ro
V2
100k
Hence,
(b)
fo
R
1
R o 10R 100 k
R R o 11
1
2 (10 10 -6 )(2 10 -9 )
f o 1.125 MHz
Chapter 10, Solution 93.
As shown below, the impedance of the feedback is
jL
1
jC2
ZT
1
jC1
ZT
1
1
|| jL
jC1
jC 2
-j
-j
1
jL
LC 2
C1
C 2
ZT
-j
-j
j (C1 C 2 2 LC1C 2 )
jL
C1
C 2
In order for Z T to be real, the imaginary term must be zero; i.e.
C1 C 2 o2 LC1 C 2 0
C C2
1
o2 1
LC1C 2
LC T
1
fo
2 LC T
Chapter 10, Solution 94.
If we select C1 C 2 20 nF
C1 C 2
C1
CT
10 nF
C1 C 2
2
Since f o
1
2 LC T
L
,
1
1
10.13 mH
2
2
(2f ) C T (4 )(2500 10 6 )(10 10 -9 )
Xc
1
1
159
C 2 ( 2 )(50 10 3 )(20 10 -9 )
We may select R i 20 k and R f R i , say R f 20 k .
Thus,
C1 C 2 20 nF,
L 10.13 mH
R f R i 20 k
Chapter 10, Solution 95.
First, we find the feedback impedance.
C
ZT
L2
L1
1
Z T jL1 || jL 2
jC
j
jL1 jL 2
2 L1C (1 L 2 )
C
ZT
j
j (2 C (L1 L 2 ) 1)
jL1 jL 2
C
In order for Z T to be real, the imaginary term must be zero; i.e.
o2 C ( L 1 L 2 ) 1 0
1
o 2 f o
C (L1 L 2 )
1
fo
2 C ( L1 L2 )
Chapter 10, Solution 96.
(a)
Consider the feedback portion of the circuit, as shown below.
jL
Vo
V2
+
V1
R
V2
R
jL
V
R jL 1
jL
V1
R jL
V2
jL
Applying KCL at node 1,
Vo V1 V1
V1
jL
R R jL
1
1
Vo V1 jL V1
R R jL
j2RL 2 L2
Vo V1 1
R (R jL)
(2)
From (1) and (2),
R jL j2RL 2 L2
V
1
Vo
R (R jL) 2
jL
Vo R 2 jRL j2RL 2 L2
V2
jRL
V2
Vo
1
R 2 L2
3
jRL
2
V2
1
Vo 3 j (L R R L )
(1)
(b)
V2
must be real,
Vo
Since the ratio
o L
R
0
R
o L
o L
R2
o L
o 2 f o
(c)
R
L
When o
V2 1
Vo 3
fo
R
2 L
This must be compensated for by A v 3 . But
Av 1
R2
3
R1
R 2 2 R1
Related papers
Plants used in medicinal practices by the tribal people of Morang District in Koshi Province of Nepal
Pleione, 2023
CALL FOR PAPERS
"YOUNG LINGUISTS' SPECIAL ISSUE-20", JANUARY 2022
ENDANGERED LANGUAGES: TURKIC LANGUAGES, 2022
Standardized Sperm Cryofreezing Technique for Caspian Sea Sturgeons (Huso huso, Acipenser persicus, Acipenser stellatus, Acipenser nudiventris)
Genetics of Aquatic Organisms (Genaqua), 2024
İptal Davasına Konu İmar Planının Dava Devam Ederken Değiştirilmesinin İptal Davasına Etkisi
İstanbul Hukuk Mecmuası, 2023
Clinical and histologic effects from CO2 laser treatment of keloids
Lasers in Medical Science, 2012
Prostate-derived Ets factor, an oncogenic driver in breast cancer
Tumour biology : the journal of the International Society for Oncodevelopmental Biology and Medicine, 2017
La rata en el laberinto o la historia como observatorio jurídico
Anuario De Historia Del Derecho Espanol, 1992
Transient global amnesia after epiduroscopy in a failed back surgery syndrome patient: Case report
Health, 2013
Morning commute with competing modes and distributed demand: User equilibrium, system optimum, and pricing
Transportation Research Part B-methodological, 2012