Chapter 1, Solution 1
(a) q = 6.482x1017 x [-1.602x10-19 C] = -0.10384 C
(b) q = 1. 24x1018 x [-1.602x10-19 C] = -0.19865 C
(c) q = 2.46x1019 x [-1.602x10-19 C] = -3.941 C
(d) q = 1.628x1020 x [-1.602x10-19 C] = -26.08 C
Chapter 1, Solution 2
(a)
(b)
(c)
(d)
(e)
i = dq/dt = 3 mA
i = dq/dt = (16t + 4) A
i = dq/dt = (-3e-t + 10e-2t) nA
i=dq/dt = 1200S cos 120S t pA
i =dq/dt = e 4t (80 cos 50 t 1000 sin 50 t ) P A
Chapter 1, Solution 3
(a) q(t)
(b) q(t)
(c) q(t)
(d)
q(t)
³ i(t)dt q(0) (3t 1) C
³ (2t s) dt q(v) (t 5t) mC
³ 20 cos 10t S / 6 q(0) (2sin(10t S / 6) 1)P C
2
10e -30t
( 30 sin 40t - 40 cos t)
900 1600
e - 30t (0.16cos40 t 0.12 sin 40t) C
-30t
³ 10e sin 40t q(0)
Chapter 1, Solution 4
q
³ idt ³ 5sin 6 ʌ t dt
5
1 cos 0.06S
6S
10
5
cos 6ʌ t
6S
0
4.698 mC
Chapter 1, Solution 5
q
³ idt ³ e
-2t
1
- e -2t
2
dt mC
1
(1 e 4 ) mC
2
490 µC
Chapter 1, Solution 6
(a) At t = 1ms, i
dq
dt
80
2
(b) At t = 6ms, i
dq
dt
0 mA
(c) At t = 10ms, i
dq
dt
80
4
Chapter 1, Solution 7
i
dq
dt
ª25A,
«- 25A,
«
«¬ 25A,
0t2
2t6
6t8
which is sketched below:
40 mA
- 20 mA
2
0
Chapter 1, Solution 8
q
10 u 1
10 u 1 15 µC
2
³ idt
Chapter 1, Solution 9
(a) q
(b)
q
(c) q
1
³ idt ³ 10 dt
10 C
0
5 u1·
§
10 u 1 ¨10 ¸ 5 u1
0
2 ¹
©
15 10 25 22.5 C
3
³ idt
5
³ idt
0
10 10 10
30 C
Chapter 1, Solution 10
q
ixt
8 x10 3 x15 x10 6
120 P C
Chapter 1, Solution 11
q = it = 85 x10-3 x 12 x 60 x 60 = 3,672 C
E = pt = ivt = qv = 3672 x1.2 = 4406.4 J
Chapter 1, Solution 12
For 0 < t < 6s, assuming q(0) = 0,
t
q (t )
³
0
t
idt q (0 )
³ 3tdt 0
0
At t=6, q(6) = 1.5(6)2 = 54
For 6 < t < 10s,
1.5t 2
t
³
q (t )
t
idt q (6 )
6
³ 18dt 54
18 t 54
6
At t=10, q(10) = 180 –– 54 = 126
For 10<t<15s,
t
t
10
10
³ idt q(10) ³ (12)dt 126
q (t )
12t 246
At t=15, q(15) = -12x15 + 246 = 66
For 15<t<20s,
t
³ 0dt q(15)
q (t )
66
15
Thus,
q (t )
­
1.5t 2 C, 0 < t < 6s
°
° 18 t 54 C, 6 < t < 10s
®
°12t 246 C, 10 < t < 15s
°
66 C, 15 < t < 20s
¯
The plot of the charge is shown below.
140
120
100
q(t)
80
60
40
20
0
0
5
10
t
15
20
Chapter 1, Solution 13
w
2
2
0
0
³ vidt ³ 1200 cos
2
4 t dt
2
1200 ³ ( 2 cos 8t - 1)dt (since, cos 2 x
2 cos 2x - 1)
0
2
§2
·
1200¨ sin 8t t ¸
©8
¹0
- 2.486 kJ
§1
·
1200¨ sin 16 2 ¸
©4
¹
Chapter 1, Solution 14
(a)
q
1
³ idt ³ 10 1 - e
0
10 1 2e
(b)
-0.5t
-0.5
2
10 t 2e -0.5t
dt
1
0
2.131 C
p(t) = v(t)i(t)
p(1) = 5cos2 ˜ 10(1- e-0.5) = (-2.081)(3.935)
= -8.188 W
Chapter 1, Solution 15
(a)
q
³ idt ³
2
0
-2t
3e dt
1 .5 e 1
-2
(b)
v
p
(c) w
3 2t
e
2
2
0
1.297 C
5di
6e 2t ( 5) 30e -2t
dt
vi 90 e 4 t W
³ pdt
3
-90³ e -4t dt
0
90 -4t
e
4
3
22.5 J
0
Chapter 1, Solution 16
i(t)
w
ª
0t2
« 25t mA
«100 - 25t mA 2 t 4 ,
«¬
1
v(t)
2
0 t 1
ª 10t V
« 10 V
1 t 3
«
«¬40 - 10t V 3 t 4
3
4
2
3
³ v(t)i(t)dt ³ 10 (25t)dt ³ 10( 25t)dt ³ 10(100 25t)dt ³ ( 40 10t)(100 - 25t)mJ
0
1
1
3
2
4
§
250 3
250
t2 ·
t 250¨¨ 4 t - ¸¸ ³ 250( 4 t) 2 dt
3
2 1
2 ¹2 3
0
©
4
§
250 250
9
t2 ·
§
·
2
( 3) 250¨12 8 2 ¸ 250¨¨16 t - 4t ¸¸
3
2
2
3 ¹3
©
¹
©
916.7 mJ
Chapter 1, Solution 17
6 p = 0 o -205 + 60 + 45 + 30 + p3 = 0
p3 = 205 –– 135 = 70 W
Thus element 3 receives 70 W.
Chapter 1, Solution 18
p1 = 30(-10) = -300 W
p2 = 10(10) = 100 W
p3 = 20(14) = 280 W
p4 = 8(-4) = -32 W
p5 = 12(-4) = -48 W
Chapter 1, Solution 19
¦p
0

o
4 I s 2 x6 13 x 2 5 x10
0

o
Is
3A
Chapter 1, Solution 20
Since 6 p = 0
-30u6 + 6u12 + 3V0 + 28 + 28u2 - 3u10 = 0
72 + 84 + 3V0 = 210 or 3V0 = 54
V0 = 18 V
Chapter 1, Solution 21
i
'q
't
§ photon · 1 § electron ·
4 u 1011 ¨
¸¸ ˜ 1. 6 u 1019 ( C / electron)
¸ ˜ ¨¨
© sec ¹ 8 © photon ¹
4
u 1011 u 1. 6 u 10 19 C/s 0.8 u 10 -8 C/s 8 nA
8
Chapter 1, Solution 22
It should be noted that these are only typical answers.
(a)
(b)
(c)
(d)
(e)
(f)
(g)
(h)
Light bulb
Radio set
TV set
Refrigerator
PC
PC printer
Microwave oven
Blender
60 W, 100 W
4W
110 W
700 W
120 W
18 W
1000 W
350 W
Chapter 1, Solution 23
(a) i
(b) w
p
v
pt
1500
120
12.5 W
1. 5 u 103 u 45 u 60 ˜ J
(c) Cost = 1.125 u 10 = 11.25 cents
1.5 u
45
kWh
60
1.125 kWh
Chapter 1, Solution 24
p = vi = 110 x 8 = 880 W
Chapter 1, Solution 25
4
Cost 1.2 kW u hr u 30 u 9 cents/kWh 21.6 cents
6
Chapter 1, Solution 26
0. 8A ˜ h
80 mA
10h
(b) p = vi = 6 u 0.08 = 0.48 W
(c) w = pt = 0.48 u 10 Wh = 0.0048 kWh
(a) i
Chapter 1, Solution 27
(a) Let T 4h 4 u 36005
q
( b) W
T
³ idt
³ 3dt
3T 3 u 4 u 3600 43.2 kC
0
T
T
0
0
³ pdt ³ vidt ³
0 . 5t ·
§
( 3) ¨10 ¸dt
3600 ¹
©
4u3600
§
0. 25t 2 ·
¸
3¨¨10t 3600 ¸¹ 0
©
475.2 kJ
( c)
W
Cost
3>40 u 3600 0. 25 u 16 u 3600@
475.2 kWs, (J Ws)
475.2
kWh u 9 cent 1.188 cents
3600
Chapter 1, Solution 28
(a) i
P
V
30
120
0.25 A
pt 30 u 365 u 24 Wh 262.8 kWh
$0.12 u 262.8 $31.54
( b) W
Cost
Chapter 1, Solution 29
(20 40 15 45)
§ 30 ·
hr 1.8 kW¨ ¸hr
60
© 60 ¹
2.4 0.9 3.3 kWh
Cost 12 cents u 3.3 39.6 cents
w
pt
1. 2kW
Chapter 1, Solution 30
Energy = (52.75 –– 5.23)/0.11 = 432 kWh
Chapter 1, Solution 31
Total energy consumed = 365(4 +8) W
Cost = $0.12 x 365 x 12 = $526.60
Chapter 1, Solution 32
(20 40 15 45)
§ 30 ·
hr 1.8 kW¨ ¸hr
60
© 60 ¹
2.4 0.9 3.3 kWh
Cost 12 cents u 3.3 39.6 cents
w
pt
1. 2kW
Chapter 1, Solution 33
i
dq
oq
dt
³ idt
2000 u 3 u 10 3
6C
Chapter 1, Solution 34
(b) Energy =
¦ pt
= 200 x 6 + 800 x 2 + 200 x 10 + 1200 x 4 + 200 x 2
= 10,000 kWh
(c) Average power = 10,000/24 = 416.67 W
Chapter 1, Solution 35
( a) W
³ p( t ) dt
400 u 6 1000 u 2 200 u 12 u 1200 u 2 400 u 2
7200 2800
( b)
10.4 kW
24 h
10.4 kWh
433.3 W/h
Chapter 1, Solution 36
(a)
i
( b) t
160A ˜ h
4A
40
160Ah 160, 000h
0.001A 24h / day
6,667 days
Chapter 1, Solution 37
q 5 u 10 20 1. 602 u 10 19 80. 1 C
W qv 80. 1 u 12 901.2 J
Chapter 1, Solution 38
P = 10 hp = 7460 W
W = pt = 7460 u 30 u 60 J = 13.43 u 106 J
Chapter 1, Solution 39
p
vi o i
p
v
2 u 10 3
120
16.667 A
Chapter 2, Solution 1
v = iR
i = v/R = (16/5) mA = 3.2 mA
Chapter 2, Solution 2
p = v2/R o
R = v2/p = 14400/60 = 240 ohms
Chapter 2, Solution 3
R = v/i = 120/(2.5x10-3) = 48k ohms
Chapter 2, Solution 4
(a)
(b)
i = 3/100 = 30 mA
i = 3/150 = 20 mA
Chapter 2, Solution 5
n = 9; l = 7; b = n + l –– 1 = 15
Chapter 2, Solution 6
n = 12;
l = 8;
b = n + l ––1 = 19
Chapter 2, Solution 7
7 elements or 7 branches and 4 nodes, as indicated.
30 V
1
20 :
2
3
++++ -
2A
30 :
60 :
4
40 :
10 :
Chapter 2, Solution 8
12 A
a
i1
b
8A
i3
i2
12 A
c
At node a,
At node c,
At node d,
9A d
8 = 12 + i1
9 = 8 + i2
9 = 12 + i3
i1 = - 4A
i2 = 1A
i3 = -3A
Chapter 2, Solution 9
Applying KCL,
i1 + 1 = 10 + 2
1 + i2 = 2 + 3
i2 = i3 + 3
i1 = 11A
i2 = 4A
i3 = 1A
Chapter 2, Solution 10
2
4A
1
-2A
i2
i1
3
3A
At node 1,
At node 3,
4 + 3 = i1
3 + i2 = -2
i1 = 7A
i2 = -5A
Chapter 2, Solution 11
Applying KVL to each loop gives
-8 + v1 + 12 = 0
-12 - v2 + 6 = 0
10 - 6 - v3 = 0
-v4 + 8 - 10 = 0
v1 = 4v
v2 = -6v
v3 = 4v
v4 = -2v
Chapter 2, Solution 12
+ 15v -
loop 2
–– 25v +
+
20v
-
+ 10v +
v1
-
loop 1
For loop 1,
For loop 2,
For loop 3,
+ v2 -
loop 3
-20 -25 +10 + v1 = 0
-10 +15 -v2 = 0
-v1 +v2 +v3 = 0
+
v3
-
v1 = 35v
v2 = 5v
v3 = 30v
Chapter 2, Solution 13
2A
1
I2
7A
2
3
I4
4
4A
I1
3A
I3
At node 2,
3 7 I2
0

o
I2
10 A
At node 1,
I1 I 2

o
2
2 I2
I1
12 A
At node 4,
2
I4 4

o
24
I4
2 A
At node 3,
7 I4
I3

o
I3
72
5A
10 A,
I3
5 A,
I4
Hence,
I1
12 A,
I2
2 A
Chapter 2, Solution 14
+
3V
-
+
I3
4V
+
V1
V3 -
- V4
For mesh 1,
V4 2 5
0

o
V4
7V
For mesh 2,
4 V3 V4
0

o
V3
4 7
11V
0

o
V1
V3 3
8V
0

o
V2
V1 2
For mesh 3,
3 V1 V3
For mesh 4,
V1 V2 2
6V
Thus,
V1
8V ,
V2
6V ,
V3
11V ,
I4
2V -
+
I2
+
+
V4
7V
+
V2
+
I1
5V
-
Chapter 2, Solution 15
+
+
+
12V
-
1
- 8V +
v2
-
v1
-
3
+
2
v3
10V
+
-
For loop 1,
8 12 v2

o
0
v2
4V
For loop 2,
v3 8 10
0

o
v3
18V
0

o
v1
6V
For loop 3,
v1 12 v3
Thus,
v1
6V ,
v2
4V ,
v3
18V
Chapter 2, Solution 16
+ v1 -
6V
loop 1
+
-
+-
12V
10V
+-
+
v1
-
loop 2
+ v2 -
Applying KVL around loop 1,
––6 + v1 + v1 –– 10 –– 12 = 0
v1 = 14V
Applying KVL around loop 2,
12 + 10 –– v2 = 0
v2 = 22V
Chapter 2, Solution 17
+ v1 -
24V
+
loop 1
-
+
v3
-
-
v2
+
loop 2
-+
12V
It is evident that v3 = 10V
Applying KVL to loop 2,
v2 + v3 + 12 = 0
v2 = -22V
Applying KVL to loop 1,
-24 + v1 - v2 = 0
v1 = 2V
Thus,
v1 = 2V, v2 = -22V, v3 = 10V
Chapter 2, Solution 18
Applying KVL,
-30 -10 +8 + I(3+5) = 0
8I = 32
I = 4A
-Vab + 5I + 8 = 0
Vab = 28V
+
-
10V
Chapter 2, Solution 19
Applying KVL around the loop, we obtain
-12 + 10 - (-8) + 3i = 0
i = -2A
Power dissipated by the resistor:
p 3: = i2R = 4(3) = 12W
Power supplied by the sources:
p12V = 12 (- -2) = 24W
p10V = 10 (-2) = -20W
p8V = (- -2) = -16W
Chapter 2, Solution 20
Applying KVL around the loop,
-36 + 4i0 + 5i0 = 0
i0 = 4A
Chapter 2, Solution 21
Apply KVL to obtain
10 :
-45 + 10i - 3V0 + 5i = 0
+ v0 -
But v0 = 10i,
-45 + 15i - 30i = 0
P3 = i2R = 9 x 5 = 45W
i = -3A
45V
+
+
-
5:
3v0
Chapter 2, Solution 22
4:
+ v0 6:
10A
2v0
At the node, KCL requires that
v0
10 2 v 0 = 0
4
v0 = ––4.444V
The current through the controlled source is
i = 2V0 = -8.888A
and the voltage across it is
v = (6 + 4) i0 = 10
v0
4
11.111
Hence,
p2 vi = (-8.888)(-11.111) = 98.75 W
Chapter 2, Solution 23
8//12 = 4.8, 3//6 = 2, (4 + 2)//(1.2 + 4.8) = 6//6 = 3
The circuit is reduced to that shown below.
ix
1:
+
6A
2:
vx
3:
Applying current division,
ix
2
(6 A)
2 1 3
2 A,
vx
1i x
2V
The current through the 1.2- : resistor is 0.5ix = 1A. The voltage across the 12- :
resistor is 1 x 4.8 = 4.8 V. Hence the power is
p
v2
R
4.8 2
12
1.92W
Chapter 2, Solution 24
(a)
(b)
I0 =
Vs
R1 R2
V0
D I0 R3 R4 = V0
Vs
DR3 R4
R1 R2 R3 R4
R1 R 2
˜
R3 R4
R3 R4
If R1 = R2 = R3 = R4 = R,
V0
VS
D R
˜
2R 2
D
4
10
Chapter 2, Solution 25
V0 = 5 x 10-3 x 10 x 103 = 50V
Using current division,
I20
DV0
5
(0.01x50)
5 20
0.1 A
V20 = 20 x 0.1 kV = 2 kV
p20 = I20 V20 = 0.2 kW
D = 40
Chapter 2, Solution 26
V0 = 5 x 10-3 x 10 x 103 = 50V
Using current division,
I20
5
(0.01x50)
5 20
0.1 A
V20 = 20 x 0.1 kV = 2 kV
p20 = I20 V20 = 0.2 kW
Chapter 2, Solution 27
Using current division,
i1 =
4
(20)
46
8A
i2 =
6
(20)
46
12 A
Chapter 2, Solution 28
We first combine the two resistors in parallel
15 10
6:
We now apply voltage division,
v1 =
14
(40)
14 6
v2 = v3 =
Hence,
20 V
6
(40)
14 6
12 V
v1 = 28 V, v2 = 12 V, vs = 12 V
Chapter 2, Solution 29
The series combination of 6 : and 3 : resistors is shorted. Hence
i2 = 0 = v2
v1 = 12, i1 =
12
4
3A
Hence v1 = 12 V, i1 = 3 A, i2 = 0 = v2
Chapter 2, Solution 30
8:
i1
i
9A
By current division, i
i1
96
3A, v
+
v
-
6:
4i1
12
(9)
6 12
4:
6A
4 x 3 = 12 V
p6 = 12R = 36 x 6 = 216 W
Chapter 2, Solution 31
The 5 : resistor is in series with the combination of 10 (4 6)
Hence by the voltage division principle,
v
5
(20V)
55
10 V
by ohm's law,
i
v
46
10
4 6
1A
pp = i2R = (1)2(4) = 4 W
5: .
Chapter 2, Solution 32
We first combine resistors in parallel.
20 30
20 x30
50
12 :
10 40
10x 40
50
8:
Using current division principle,
8
i1 i 2
(20) 8A, i 3 i 4
8 12
i1
20
(8)
50
3.2 A
i2
30
(8)
50
4.8 A
i3
10
(12)
50
2.4A
i4
40
(12)
50
9.6 A
12
(20) 12A
20
Chapter 2, Solution 33
Combining the conductance leads to the equivalent circuit below
i
+
v
-
9A
1S
i
4S
4S
6x3
25 and 25 + 25 = 4 S
9
Using current division,
6 S 3S
i
1
1
1
2
(9)
6 A, v = 3(1) = 3 V
9A
+
v
-
1S
2S
Chapter 2, Solution 34
By parallel and series combinations, the circuit is reduced to the one below:
Thus i1 =
28
86
8:
i1
10 x15
6:
10 ( 2 13 )
25
15 x15
15 (4 6)
6:
25
12 (6 6) 6:
28V
+
v1
-
+
-
6:
2 A and v1 = 6i1 = 12 V
We now work backward to get i2 and v2.
i1 = 2A
8:
6:
1A
1A
28V
+
12V
-
+
-
8:
i1 = 2A
6:
+
6V
-
12 :
6:
4:
1A
0.6A
1A
28V
Thus, v2 =
+
12V
-
+
-
13
(3 ˜ 6)
15
v2
13
3 ˜ 12, i2 =
12 :
+
6V
-
+
15 :
3.6V
0.24
p2 = i2R = (0.24)2 (2) = 0.1152 W
i1 = 2 A, i2 = 0.24 A, v1 = 12 V, v2 = 3.12 V, p2 = 0.1152 W
Chapter 2, Solution 35
i
70 :
50V
+
-
a
+
V1
i1 -
30 :
I0
+
20 :
i2
b
V0 5 :
-
-
6:
Combining the versions in parallel,
70 30
i=
70x30
100
50
21 4
21: ,
20 15
20x 5
25
4:
2A
vi = 21i = 42 V, v0 = 4i = 8 V
v
v
i1 = 1 0.6 A, i2 = 2 0.4 A
70
20
At node a, KCL must be satisfied
i1 = i2 + I0
0.6 = 0.4 + I0
I0 = 0.2 A
Hence v0 = 8 V and I0 = 0.2A
Chapter 2, Solution 36
The 8-: resistor is shorted. No current flows through the 1-: resistor. Hence v0
is the voltage across the 6: resistor.
I0 =
4
2 3 16
v0 = I0 3 6
4
4
2I 0
1A
2V
Chapter 2, Solution 37
Let I = current through the 16: resistor. If 4 V is the voltage drop across the 6 R
combination, then 20 - 4 = 16 V in the voltage drop across the 16: resistor.
16
Hence, I =
1 A.
16
20
6R
R = 12 :
4= 6R
But I =
1
6R
16 6 R
Chapter 2, Solution 38
Let I0 = current through the 6: resistor. Since 6: and 3: resistors are in parallel.
6I0 = 2 x 3
R0 = 1 A
The total current through the 4: resistor = 1 + 2 = 3 A.
Hence
vS = (2 + 4 + 2 3 ) (3 A) = 24 V
I=
vS
10
2.4 A
Chapter 2, Solution 39
(a)
Req = R 0
0
(c)
R R
R
2 2
Req = (R R ) (R R ) 2R 2R
(d)
Req = 3R (R R R )
(b)
Req = R R R R
3
3Rx R
2
=
3
3R R
2
(e)
Req = R 2R 3R
= 3R
2
R
3
R
1
3R (R R )
2
R
§ R ˜ 2R ·
¨
¸
© 3R ¹
2
3Rx R
6
3
R
2
11
3R R
3
3R
Chapter 2, Solution 40
Req = 3 4 (2 6 3)
I=
3 2 5:
10
Re q
10
5
2A
Chapter 2, Solution 41
Let R0 = combination of three 12: resistors in parallel
1
Ro
1
1
1
12 12 12
R eq
30 60 (10 R 0 R )
50
30 Ro = 4
60(14 R )
74 R
30 60 (14 R )
74 + R = 42 + 3R
or R = 16 :
Chapter 2, Solution 42
5 (8 12)
5x 20
25
(a)
Rab = 5 (8 20 30)
(b)
Rab = 2 4 (5 3) 8 5 10 (6 4)
4:
24 455
Chapter 2, Solution 43
(a)
(b)
Rab = 5 20 10 40
60 20 30
5x 20 400
25
50
1
1 ·
§ 1
¸
¨ © 60 20 30 ¹
Rab = 80 (10 10)
1
80 20
100
48
60
6
16 :
10:
12 :
2 2 2.5
6.5 :
Chapter 2, Solution 44
(a) Convert T to Y and obtain
20 x 20 20 x10 10 x 20
10
800
R2
40 : R3
20
R1
800
10
80 :
The circuit becomes that shown below.
R1
a
R3
R2
5:
b
R1//0 = 0,
Rab
R3//5 = 40//5 = 4.444 :
R2 / /(0 4.444)
40 / /4.444
4:
(b) 30//(20+50) = 30//70 = 21 :
Convert the T to Y and obtain
20 x10 10 x 40 40 x 20 1400
35:
40
40
1400
1400
140 :
R2
70 : , R3
20
10
The circuit is reduced to
that shown below.
15:
R1
R1
11 :
R2
R3
21 :
30 :
21 :
Combining the resistors in parallel
R1//15 =35//15=10.5, 30//R2=30//70 = 21
leads to the circuit below.
10.5 :
11 :
140 :
21 :
21 :
21 :
Coverting the T to Y leads to the circuit below.
10.5 :
11 :
R4
R5
R6
R4
21x140 140 x 21 21x 21
21
R5
6321
140
6321
21
301:
21 :
R6
45.15
10.5//301 = 10.15, 301//21 = 19.63
R5//(10.15 +19.63) = 45.15//29.78 = 17.94
Rab
11 17 .94
28.94:
Chapter 2, Solution 45
(a) 10//40 = 8, 20//30 = 12, 8//12 = 4.8
Rab
5 50 4.8
59.8 :
(b) 12 and 60 ohm resistors are in parallel. Hence, 12//60 = 10 ohm. This 10 ohm
and 20 ohm are in series to give 30 ohm. This is in parallel with 30 ohm to give
30//30 = 15 ohm. And 25//(15+10) = 12.5. Thus
Rab
5 12.8 15
32.5:
Chapter 2, Solution 46
(a)
30x 70
60 20
40 100
80
Rab = 30 70 40 60 20
76 :
21 40 15
(b)
The 10-:, 50-:, 70-:, and 80-: resistors are shorted.
20 30
20x30
50
12:
40 60
40x 60
100
24
Rab = 8 + 12 + 24 + 6 + 0 + 4 = 54 :
Chapter 2, Solution 47
5 20
63
5x 20
25
6x3
9
4:
2:
10 :
8:
a
b
4:
Rab = 10 + 4 + 2 + 8 = 24 :
2:
Chapter 2, Solution 48
(a)
Ra =
(b)
Ra
Rb
R 1R 2 R 2 R 3 R 3 R 1
R3
Ra = Rb = Rc = 30 :
100 100 100
10
30
30x 20 30x50 20x 50 3100
103.3:
30
30
3100
3100
155:, R c
62:
20
50
Ra = 103.3 :, Rb = 155 :, Rc = 62 :
Chapter 2, Solution 49
(a)
R1 =
(b)
R1
R2
R3
RaRc
12 12
Ra Rb Rc
36
R1 = R2 = R3 = 4 :
4:
60x30
18:
60 30 10
60 x10
6:
100
30x10
3:
100
R1 = 18:, R2 = 6:, R3 = 3:
Chapter 2, Solution 50
Using R ' = 3RY = 3R, we obtain the equivalent circuit shown below:
30mA
3R
3R
3R
R
R
30mA
3R
3R/2
3RxR 3
R
4R
4
3R (3RxR ) /(4R ) 3 /(4R )
3R R
3
3Rx R
3 ·
3
§3
2
3R ¨ R R ¸ 3R R
3
4 ¹
2
©4
3R R R
2
800 x 10-3 = (30 x 10-3)2 R
P = I2 R
R = 889 :
Chapter 2, Solution 51
30 30 15: and 30 20
(a)
30 x 20 /(50) 12:
Rab = 15 (12 12) 15x 24 /(39) 9.31 :
a
a
30 :
30 :
30 :
30 :
20 :
12 :
15 :
12 :
20 :
b
b
Converting the T-subnetwork into its equivalent ' network gives
(b)
Ra'b' = 10x20 + 20x5 + 5x10/(5) = 350/(5) = 70 :
Rb'c' = 350/(10) = 35:, Ra'c' = 350/(20) = 17.5 :
Also
30 70
30 x 70 /(100)
21: and 35/(15) = 35x15/(50) = 10.5
Rab = 25 + 17.5 (21 10.5)
25 17.5 31.5
Rab = 36.25 :
30 :
30 :
a
25 :
10 :
5:
b
20 :
a
15 :
25 :
a’’
17.5 :
b
70 :
b’’
35 :
c’’
15 :
c’’
Chapter 2, Solution 52
(a) We first convert from T to ' .
100 :
a
100 :
a
100 :
100 :
100 :
100 :
b
100 :
200 :
100 :
100 :
200 :
b
R1
100x 200 200x 200 200 x100
100
100 :
100 :
100 :
100 :
100 :
100 :
80000
100
R3
R2
800:
R2 = R3 = 80000/(200) = 400
100x 400
But
100 400
80:
500
We connect the ' to Y.
100 :
a
b
100 :
a
100 :
100 :
80 :
100 :
100 :
80 :
800 :
80 x800
80 80 800
80x80 20
:
Rb =
960
3
Ra = Rc =
64,000
960
b
100 :
100 :
Rb
100 :
100 :
Rc
400
:
3
We convert T to ' .
a
500/3 :
100 :
320/3 :
b
100 :
500/3 :
500/3 :
a
R2’’
R1’’
R3’’
b
Ra
500/3 :
R1
320
320
100 x
3
3
320
3
100 x100 100 x
R 1'
R '2
R 13
94,000 /(3)
100
940 /(30) 500 /(3)
94,000 /(3)
320 /(3)
293.75:
313.33
940 /(3) x500 /(3)
1440 /(3)
108.796
293.75x 217.6
511.36
Rab = 293.75 (2 x108.796)
125 :
Converting the Ts to ' s, we have the equivalent circuit below.
(b)
100 :
100 :
a
100 :
300 :
300 :
100 :
300 :
100 :
100 :
100 :
100 :
a
100 :
300 :
300 :
b
300 :
100 :
100 :
100 :
100 :
100 :
b
300 100
300 x100 /(400)
Rab = 100 + 100 300 100
75, 300 (75 75)
200 100 x 300 /(400)
Rab = 2.75 :
100 :
300 :
100 :
300 :
300 :
100 :
300 x150 /(450) 100
100 :
Chapter 2, Solution 53
(a)
Converting one ' to T yields the equivalent circuit below:
30 :
a’’
4:
20 :
a
60 :
c’’
5:
b’’
b
20 :
80 :
40 x10
10 x50
4:, R b 'n
5:, R c 'n
40 10 50
100
Rab = 20 + 80 + 20 + (30 4) (60 5) 120 34 65
40x50
100
Ra'n =
20:
Rab = 142.32 :
(a) We combine the resistor in series and in parallel.
30 (30 30)
30x 60
90
20:
We convert the balanced ' s to Ts as shown below:
a
30 :
30 :
a
10 :
30 :
30 :
20 :
10 :
30 :
b
30 :
10 :
10 :
10 :
b
Rab = 10 + (10 10) (10 20 10) 10
20 20 40
Rab = 33.33 :
Chapter 2, Solution 54
(a) Rab
50 100 / /(150 100 150 )
(b) Rab
60 100 / /(150 100 150 )
10 :
50 100 / /400
130 :
60 100 / /400
140 :
20 :
Chapter 2, Solution 55
We convert the T to ' .
I0
a
24 V
+
-
I0
20 :
40 :
60 :
10 :
20 :
50 :
a
140 :
+
35 :
-
70 :
70 :
b
b
Req
R 1 R 2 R 2 R 3 R 3 R 1 20 x 40 40 x10 10 x 20
R3
40
Rac = 1400/(10) = 140:, Rbc = 1400/(40) = 35:
70 70 35 and 140 160 140x60/(200) = 42
Rab =
Req = 35 (35 42)
60 :
24 V
Req
1400
40
35:
24.0625:
I0 = 24/(Rab) = 0.9774A
Chapter 2, Solution 56
We need to find Req and apply voltage division. We first tranform the Y network to ' .
30 :
+
100 V
-
16 :
15 :
35 :
12 :
30 :
16 :
10 :
20 :
Req
15x10 10x12 12x15 450
37.5:
12
12
Rac = 450/(10) = 45:, Rbc = 450/(15) = 30:
Rab =
Combining the resistors in parallel,
+
100 V
35 :
Req
a
37.5 :
30 :
45 :
c
b
20 :
30||20 = (600/50) = 12 :,
37.5||30 = (37.5x30/67.5) = 16.667 :
35||45 = (35x45/80) = 19.688 :
Req = 19.688||(12 + 16.667) = 11.672:
By voltage division,
v =
11.672
100 = 42.18 V
11.672 16
Chapter 2, Solution 57
4: a
2:
27 :
1:
18 :
36 :
b
d
10 :
c
e
7:
14 :
28 :
f
6x12 12x8 8x 6 216
18 :
12
12
Rac = 216/(8) = 27:, Rbc = 36 :
4x 2 2x8 8x 4 56
Rde =
7:
8
8
Ref = 56/(4) = 14:, Rdf = 56/(2) = 28 :
Rab =
Combining resistors in parallel,
10 28
27 3
280
7.368:, 36 7
38
27 x 3
2.7:
30
36x 7
43
5.868:
4:
4:
18 :
5.868 :
7.568 :
R an
R bn
R cn
R eq
1.829 :
2.7 :
3.977 :
0.5964 :
14 :
7.568 :
14 :
18x 2.7
18x 2.7
1.829 :
18 2.7 5.867 26.567
18x5.868
3.977 :
26.567
5.868x 2.7
0.5904 :
26.567
4 1.829 (3.977 7.368) (0.5964 14)
12.21 :
5.829 11.346 14.5964
i = 20/(Req) = 1.64 A
Chapter 2, Solution 58
The resistor of the bulb is 120/(0.75) = 160:
40 :
2.25 A
+ 90 V - 0.75 A
VS
+
-
160 :
1.5 A
+
80 :
120
Once the 160: and 80: resistors are in parallel, they have the same voltage 120V.
Hence the current through the 40: resistor is
40(0.75 + 1.5) = 2.25 x 40 = 90
Thus
vs = 90 + 120 = 210 V
Chapter 2, Solution 59
Total power p = 30 + 40 + 50 + 120 W = vi
or i = p/(v) = 120/(100) = 1.2 A
Chapter 2, Solution 60
p = iv
i = p/(v)
i30W = 30/(100) = 0.3 A
i40W = 40/(100) = 0.4 A
i50W = 50/(100) = 0.5 A
Chapter 2, Solution 61
There are three possibilities
(a)
Use R1 and R2:
R = R 1 R 2 80 90
42.35:
2
p=i R
i = 1.2A + 5% = 1.2 r 0.06 = 1.26, 1.14A
p = 67.23W or 55.04W, cost = $1.50
(b)
Use R1 and R3:
R = R 1 R 3 80 100
44.44 :
2
p = I R = 70.52W or 57.76W, cost = $1.35
(c)
Use R2 and R3:
R = R 2 R 3 90 100
47.37:
2
p = I R = 75.2W or 61.56W, cost = $1.65
Note that cases (b) and (c) give p that exceed 70W that can be supplied.
Hence case (a) is the right choice, i.e.
R1 and R2
Chapter 2, Solution 62
pA = 110x8 = 880 W,
pB = 110x2 = 220 W
Energy cost = $0.06 x 360 x10 x (880 + 220)/1000 = $237.60
Chapter 2, Solution 63
Use eq. (2.61),
Im
2 x10 3 x100
Rn =
0.04:
Rm
I Im
5 2 x10 3
In = I - Im = 4.998 A
p = I 2n R (4.998) 2 (0.04) 0.9992 # 1 W
Chapter 2, Solution 64
When Rx = 0, i x
R=
10A
110
11 :
10
When Rx is maximum, ix = 1A
R Rx
i.e., Rx = 110 - R = 99 :
Thus, R = 11 :,
Rx = 99 :
110
1
Chapter 2, Solution 65
Rn
Vfs
Rm
I fs
50
1 k:
10mA
4 k:
Chapter 2, Solution 66
20 k:/V = sensitivity =
1
I fs
1
k: / V 50 PA
20
V
The intended resistance Rm = fs 10(20k: / V) 200k:
I fs
Vfs
50 V
(a)
Rn
Rm
200 k: 800 k:
i fs
50 PA
i.e., Ifs =
(b)
p = I fs2 R n
(50 PA) 2 (800 k:)
2 mW
110 :
Chapter 2, Solution 67
(a)
By current division,
i0 = 5/(5 + 5) (2 mA) = 1 mA
V0 = (4 k:) i0 = 4 x 103 x 10-3 = 4 V
(b)
4k 6k
i '0
v '0
2.4k:. By current division,
5
(2mA) 1.19 mA
1 2.4 5
(2.4 k:)(1.19 mA) 2.857 V
(c)
% error =
(d)
4k 30 k:
v 0 v '0
1.143
x 100% =
x100 28.57%
v0
4
3.6 k:. By current division,
5
(2mA) 1.042mA
1 3.6 5
v '0 (3.6 k:)(1.042 mA) 3.75V
i '0
% error =
v v '0
x100%
v0
0.25x100
4
6.25%
Chapter 2, Solution 68
24 60:
(a)
40
(b)
4
0.1 A
16 24
4
i'
0.09756 A
16 1 24
0.1 0.09756
% error =
x100%
0.1
i=
(c)
2.44%
Chapter 2, Solution 69
With the voltmeter in place,
R2 Rm
V0
VS
R1 R S R 2 R m
where Rm = 100 k: without the voltmeter,
R2
VS
V0
R1 R 2 R S
100
k:
101
(a)
When R2 = 1 k:, R m R 2
(b)
100
V0 = 101 (40) 1.278 V (with)
100
101 30
1
V0 =
(40) 1.29 V (without)
1 30
1000
When R2 = 10 k:, R 2 R m
9.091k:
110
9.091
V0 =
(40) 9.30 V (with)
9.091 30
10
V0 =
(40) 10 V (without)
10 30
When R2 = 100 k:, R 2 R m 50k:
(c)
50
(40) 25 V (with)
50 30
100
V0 =
(40) 30.77 V (without)
100 30
V0
Chapter 2, Solution 70
(a) Using voltage division,
12
(25) 15V
12 8
10
vb
(25) 10V
10 15
va vb 15 10 5V
va
vab
(b)
+
25 V
-
15k :
8k :
a
b
10k :
12k :
o
va
0,
vb
10V ,
vab
va vb
0 10
10V
Chapter 2, Solution 71
R1
iL
Vs +
−
RL
Given that vs = 30 V, R1 = 20 :, IL = 1 A, find RL.
vs
i L ( R1 R L )

o
RL
vs
R1
iL
30
20
1
10:
Chapter 2, Solution 72
The system can be modeled as shown.
12A
+
9V
-
R
R
R
xxx
The n parallel resistors R give a combined resistance of R/n. Thus,
9
12 x
R
n

o
n
12 xR
9
12 x15
9
20
Chapter 2, Solution 73
By the current division principle, the current through the ammeter will be
one-half its previous value when
R = 20 + Rx
65 = 20 + Rx
Rx = 45 :
Chapter 2, Solution 74
With the switch in high position,
6 = (0.01 + R3 + 0.02) x 5
R3 = 1.17 :
At the medium position,
6 = (0.01 + R2 + R3 + 0.02) x 3
R2 + R3 = 1.97
or R2 = 1.97 - 1.17 = 0.8 :
At the low position,
6 = (0.01 + R1 + R2 + R3 + 0.02) x 1
R1 = 5.97 - 1.97 = 4 :
R1 + R2 + R3 = 5.97
Chapter 2, Solution 75
100 :
R
VS
M
12 :
+
-
(a) When Rx = 0, then
t
Im = Ifs =
R Rm
R2 =
E2
2
Rm=
100 19.9k:
I fs
0.1x10 3
I fs
0.05mA
2
E
2
(R R m )
20k:
Rx =
Im
0.05x10 3
(b) For half-scale deflection, Im =
Im =
E
R Rm Rx
20 k:
Chapter 2, Solution 76
For series connection, R = 2 x 0.4: = 0.8:
V 2 (120) 2
p
18 k: (low)
R
0.8
For parallel connection, R = 1/2 x 0.4: = 0.2:
V 2 (120) 2
p=
72 kW (high)
R
0.2
Chapter 2, Solution 77
(a)
5 : = 10 10
20 20 20 20
i.e., four 20 : resistors in parallel.
(b)
311.8 = 300 + 10 + 1.8 = 300 + 20 20 1.8
i.e., one 300: resistor in series with 1.8: resistor and
a parallel combination of two 20: resistors.
(c)
40k: = 12k: + 28k: = 24 24k 56k 50k
i.e., Two 24k: resistors in parallel connected in series with two
50k: resistors in parallel.
(d)
42.32k: = 42l + 320
= 24k + 28k = 320
= 24k = 56k 56k 300 20
i.e., A series combination of 20: resistor, 300: resistor, 24k:
resistor and a parallel combination of two 56k: resistors.
Chapter 2, Solution 78
The equivalent circuit is shown below:
R
VS
+
+
V0
-
V0 =
(1 D)R
VS
R (1 D)R
V0
VS
(1 D)R
(1-D)R
-
(1 D)R 0 VS
Chapter 2, Solution 79
Since p = v2/R, the resistance of the sharpener is
R = v2/(p) = 62/(240 x 10-3) = 150:
I = p/(v) = 240 mW/(6V) = 40 mA
Since R and Rx are in series, I flows through both.
IRx = Vx = 9 - 6 = 3 V
Rx = 3/(I) = 3/(40 mA) = 3000/(40) = 75 :
Chapter 2, Solution 80
The amplifier can be modeled as a voltage source and the loudspeaker as a resistor:
V
+
V
R1
-
+
Case 1
Hence p
V2 p2
,
R p1
R2
-
Case 2
R1
R2
p2
R1
p1
R2
10
(12)
4
30 W
Chapter 2, Solution 81
Let R1 and R2 be in k:.
R eq
R1 R 2 5
V0
VS
5 R2
(1)
(2)
5 R 2 R1
5 R1
From (1) and (2), 0.05
2 = 5 R2
40
From (1), 40 = R1 + 2
5R 2
or R2 = 3.33 k:
5 R2
R1 = 38 k:
Thus R1 = 38 k:, R2 = 3.33 k:
Chapter 2, Solution 82
(a)
10 :
40 :
10 :
80 :
1
2
R12
R12 = 80 + 10 (10 40)
80 50
6
88.33 :
(b)
3
10 :
10 :
20 :
40 :
R13
80 :
1
R13 = 80 + 10 (10 40) 20 100 10 50 108.33 :
4
(c)
20 :
10 :
R14
10 :
40 :
80 :
1
R14 = 80 0 (10 40 10) 20
80 0 20
100 :
Chapter 2, Solution 83
The voltage across the tube is 2 x 60 mV = 0.06 V, which is negligible
compared with 24 V. Ignoring this voltage amp, we can calculate the
current through the devices.
p1
V1
p
I2 = 2
V2
I1 =
45mW
9V
5mA
480mW
24
20mA
60 mA
i2 = 20 mA
iR1
24 V
R1
+
-
i1 = 5 mA
R2
iR2
By applying KCL, we obtain
I R1
60 20
40 mA and I R 2
Hence, I R1 R1 = 24 - 9 = 15 V
IR2 R 2
9V
R2
R1
9V
35mA
40 5
35 mA
15V
40mA
257.14 :
375 :
Chapter 3, Solution 1.
40 :
v1
v2
8:
6A
2:
10 A
At node 1,
6 = v1/(8) + (v1 - v2)/4
48 = 3v1 - 2v2
(1)
40 = v1 - 3v2
(2)
At node 2,
v1 - v2/4 = v2/2 + 10
Solving (1) and (2),
v1 = 9.143V, v2 = -10.286 V
v12
P8: =
8
9.143
8
10.45 W
2
v v2
P4: = 1
4
v12
P2: =
2
2
94.37 W
10.286
2
2
52.9 W
Chapter 3, Solution 2
At node 1,
v v2
v1 v1
6 1
10
5
2
At node 2,
v2
v v2
36 1
4
2
Solving (1) and (2),
v1 = 0 V, v2 = 12 V
60 = - 8v1 + 5v2
36 = - 2v1 + 3v2
(1)
(2)
Chapter 3, Solution 3
Applying KCL to the upper node,
10 =
v0 vo vo
v
2 0
10 20 30
60
i1 =
v0
10
4 A , i2 =
v0
20
v0 = 40 V
2 A, i3 =
v0
30
1.33 A, i4 =
v0
60
i3
i4
67 mA
Chapter 3, Solution 4
2A
v1
i1
4A
5:
i2
10 :
v2
10 :
5:
At node 1,
4 + 2 = v1/(5) + v1/(10)
v1 = 20
At node 2,
5 - 2 = v2/(10) + v2/(5)
v2 = 10
i1 = v1/(5) = 4 A, i2 = v1/(10) = 2 A, i3 = v2/(10) = 1 A, i4 = v2/(5) = 2 A
Chapter 3, Solution 5
Apply KCL to the top node.
30 v 0 20 v 0
2k
6k
v0
4k
v0 = 20 V
5A
Chapter 3, Solution 6
i1 + i2 + i3 = 0
v 2 12 v 0 v 0 10
4
6
2
0
or v0 = 8.727 V
Chapter 3, Solution 7
At node a,
10 Va Va Va Vb
(1)

o 10 6Va 3Vb
30
15
10
At node b,
Va Vb 12 Vb 9 Vb
0

o
24 2Va 7Vb
10
20
5
Solving (1) and (2) leads to
Va = -0.556 V, Vb = -3.444V
(2)
Chapter 3, Solution 8
3:
i1
v1
i3
5:
i2
+
V0
3V
2:
+
––
+ 4V0
––
––
1:
v1 v1 3 v1 4 v 0
0
5
1
5
2
8
v0
v1 so that v1 + 5v1 - 15 + v1 - v1 0
5
5
or v1 = 15x5/(27) = 2.778 V, therefore vo = 2v1/5 = 1.1111 V
i1 + i2 + i3 = 0
But
Chapter 3, Solution 9
3:
i1
v1
+ v0 ––
12V
6:
i3
i2
+
+
v1
––
8:
––
+
––
2v0
At the non-reference node,
v1 v1 2 v 0
8
6
12 v1
3
(1)
But
-12 + v0 + v1 = 0
v0 = 12 - v1
(2)
Substituting (2) into (1),
12 v1
3
v1 3v1 24
8
6
v0 = 3.652 V
Chapter 3, Solution 10
At node 1,
v 2 v1
1
4
v1
8
32 = -v1 + 8v2 - 8v0
1:
4A
v1
8:
i0
2i0
v0
v2
2:
4:
(1)
At node 0,
4
v0
2I 0 and I 0
2
v1
8
16 = 2v0 + v1
(2)
v2 = v1
(3)
At node 2,
2I0 =
v 2 v1 v 2
and I 0
1
4
v1
8
From (1), (2) and (3), v0 = 24 V, but from (2) we get
v
4 o
2 2 24 2 6 = - 4 A
io =
4
2
Chapter 3, Solution 11
4:
i1 v i2
3:
i3
10 V
+
––
5A
6:
Note that i2 = -5A. At the non-reference node
10 v
5
4
i1 =
10 v
4
v
6
v = 18
-2 A, i2 = -5 A
Chapter 3, Solution 12
10 :
v1
20 :
50 :
v2
i3
24 V
+
––
40 :
5A
At node 1,
v1 v 2 v1 0
20
40
24 v 1
10
At node 2, 5 v1 v 2
20
v2
50
96 = 7v1 - 2v2
500 = -5v1 + 7v2
(1)
(2)
Solving (1) and (2) gives,
v1 = 42.87 V, v2 = 102.05 V
v1
v
i1
1.072 A, v2 = 2
40
50
2.041 A
Chapter 3, Solution 13
At node number 2, [(v2 + 2) –– 0]/10 + v2/4 = 3 or v2 = 8 volts
But, I = [(v2 + 2) –– 0]/10 = (8 + 2)/10 = 1 amp and v1 = 8x1 = 8volts
Chapter 3, Solution 14
5A
v0
v1
1:
8:
2:
4:
40 V
20 V
––
+
+
––
At node 1,
v1 v 0
5
2
40 v 0
1
At node 0,
v1 v 0
5
2
v 0 v 0 20
4
8
Solving (1) and (2), v0 = 20 V
v1 + v0 = 70
4v1 - 7v0 = -20
(1)
(2)
Chapter 3, Solution 15
5A
v0
v1
1:
2:
4:
40 V
8:
20 V
+
––
Nodes 1 and 2 form a supernode so that v1 = v2 + 10
At the supernode, 2 + 6v1 + 5v2 = 3 (v3 - v2)
At node 3, 2 + 4 = 3 (v3 - v2)
(1)
2 + 6v1 + 8v2 = 3v3
v3 = v2 + 2
2 + 6v2 + 60 + 8v2 = 3v2 + 6
v1 = v2 + 10 =
v2 =
54
11
i0 = 6vi = 29.45 A
v12
R
2
§ 54 ·
¨ ¸ 6
© 11 ¹
v12 G
144.6 W
2
2
2
P55 = v G
§ 56 ·
¸ 5
¨
© 11 ¹
P35 = v L v 3
2
G
129.6 W
( 2) 2 3
12 W
(2)
(3)
Substituting (1) and (3) into (2),
P65 =
––
+
56
11
Chapter 3, Solution 16
2S
i0
2A
8S
v2
v1
+
1S
v0
4S
v3
13 V
––
+
––
At the supernode,
2 = v1 + 2 (v1 - v3) + 8(v2 –– v3) + 4v2, which leads to 2 = 3v1 + 12v2 - 10v3
(1)
But
v1 = v2 + 2v0 and v0 = v2.
Hence
v1 = 3v2
v3 = 13V
(2)
(3)
Substituting (2) and (3) with (1) gives,
v1 = 18.858 V, v2 = 6.286 V, v3 = 13 V
Chapter 3, Solution 17
i0
4:
2:
10 :
60 V
60 V
8:
+
––
3i0
60 v1 v1 v1 v 2
4
8
2
60 v 2 v1 v 2
At node 2, 3i0 +
10
2
At node 1,
120 = 7v1 - 4v2
(1)
0
60 v1
.
4
But i0 =
Hence
3 60 v1 60 v 2 v1 v 2
4
10
2
0
Solving (1) and (2) gives v1 = 53.08 V. Hence i0 =
1020 = 5v1 - 12v2
60 v1
4
(2)
1.73 A
Chapter 3, Solution 18
–– +
v2
v1
2:
5A
v3
2:
8:
4:
10 V
+
+
v1
v3
––
(a)
At node 2, in Fig. (a), 5 =
At the supernode,
––
(b)
v 2 v1 v 2 v3
2
2
v 2 v1 v 2 v 3
2
2
From Fig. (b), - v1 - 10 + v3 = 0
10 = - v1 + 2v2 - v3
v1 v 3
4
8
v3 = v1 + 10
Solving (1) to (3), we obtain v1 = 10 V, v2 = 20 V = v3
40 = 2v1 + v3
(1)
(2)
(3)
Chapter 3, Solution 19
At node 1,
V1 V3 V1 V2 V1
2
8
4
At node 2,
5
3
V1 V2 V2 V2 V3
8
2
4
At node 3,
12 V3

o

o
V1 V3 V2 V3
8
2
4
From (1) to (3),
3
§ 7 1 4 ·§ V1 ·
¨
¸¨ ¸
¨ 1 7 2 ¸¨V2 ¸
¨4
2 7 ¸¹¨© V3 ¸¹
©
Using MATLAB,
V
A 1 B
ª 10 º
« 4.933 »
«
»
«¬12.267 »¼
16

o

o

o
V1
(1)
V1 7V2 2V3
0
0
§ 16 ·
¨
¸
¨ 0 ¸
¨ 36 ¸
©
¹
7V1 V2 4V3
AV
10 V, V2
36
(2)
4V1 2V2 7V3 (3)
B
4.933 V, V3
12.267 V
Chapter 3, Solution 20
Nodes 1 and 2 form a supernode; so do nodes 1 and 3. Hence
V1 V2 V3
0

o V1 4V2 V3 0
(1)
4
1
4
.
V1
4:
.
V2
1:
2:
V3
4:
Between nodes 1 and 3,
V1 12 V3 0

o V3 V1 12
Similarly, between nodes 1 and 2,
V1 V2 2i
But i V3 / 4 . Combining this with (2) and (3) gives
. V2
6 V1 / 2
(2)
(3)
(4)
Solving (1), (2), and (4) leads to
V1
3V, V2
4.5V, V3
15V
Chapter 3, Solution 21
4 k:
v1
2 k:
v3
3v0
+
3v0
v2
+
v0
3 mA
––
1 k:
+
+
+
v3
v2
––
––
(b)
(a)
Let v3 be the voltage between the 2k: resistor and the voltage-controlled voltage source.
At node 1,
v1 v 2 v1 v 3
3x10 3
12 = 3v1 - v2 - 2v3
(1)
4000
2000
At node 2,
v1 v 2 v1 v 3 v 2
3v1 - 5v2 - 2v3 = 0
(2)
4
2
1
Note that v0 = v2. We now apply KVL in Fig. (b)
- v3 - 3v2 + v2 = 0
From (1) to (3),
v1 = 1 V, v2 = 3 V
v3 = - 2v2
(3)
Chapter 3, Solution 22
At node 1,
12 v 0
2
At node 2, 3 +
v v0
v1
3 1
4
8
v1 v 2
8
24 = 7v1 - v2
(1)
v 2 5v 2
1
But, v1 = 12 - v1
Hence, 24 + v1 - v2 = 8 (v2 + 60 + 5v1) = 4 V
456 = 41v1 - 9v2
(2)
Solving (1) and (2),
v1 = - 10.91 V, v2 = - 100.36 V
Chapter 3, Solution 23
At the supernode, 5 + 2 =
v1 v 2
10 5
70 = v1 + 2v2
(1)
v2 = v1 + 8
(2)
Considering Fig. (b), - v1 - 8 + v2 = 0
Solving (1) and (2),
v1 = 18 V, v2 = 26 V
v1
v2
5A
–– +
2A
10 :
5:
8V
+
+
v1
v2
––
(a)
––
(b)
Chapter 3, Solution 24
6mA
1 k:
2 k:
V1
+
30V
-
3 k:
V2
io
-
4 k:
5 k:
At node 1,
30 V 1
V V V2

o 96 7V1 2V2
6 1 1
1
4
2
At node 2,
(15 V 2) V2 V2 V1
6

o 30 15V1 31V2
3
5
2
Solving (1) and (2) gives V1=16.24. Hence
io = V1/4 = 4.06 mA
Chapter 3, Solution 25
20V
+
i0
(2)
v0
2:
10V
––
1:
(1)
40V
+
––
4:
+
––
2:
Using nodal analysis,
20 v 0 40 v 0 10 v 0
1
2
2
i0 =
20 v 0
= 0A
1
v0 0
4
v0 = 20V
15V
+
Chapter 3, Solution 26
At node 1,
V V3 V1 V2
15 V1

o
45 7V1 4V2 2V3
3 1
20
10
5
At node 2,
V1 V2 4 I o V2 V2 V3
5
5
5
V1 V3
. Hence, (2) becomes
But I o
10
0 7V1 15V2 3V3
At node 3,
V V3 10 V3 V2 V3
3 1
0

o
10 V1 2V2 5V3
10
5
5
Putting (1), (3), and (4) in matrix form produces
§ 7 4 2 ·§ V1 · § 45 ·
¨
¸¨ ¸ ¨
¸

o
AV B
¨ 7 15 3 ¸¨V2 ¸ ¨ 0 ¸
¨1
2
5 ¸¹¨© V3 ¸¹ ¨© 10 ¸¹
©
Using MATLAB leads to
§ 9.835 ·
¨
¸
1
V A B ¨ 4.982 ¸
¨ 1.96 ¸
©
¹
Thus,
V1 9.835 V, V2 4.982 V, V3 1.95 V
Chapter 3, Solution 27
At node 1,
2 = 2v1 + v1 –– v2 + (v1 –– v3)4 + 3i0, i0 = 4v2. Hence,
At node 2,
2 = 7v1 + 11v2 –– 4v3
v1 –– v2 = 4v2 + v2 –– v3
(1)
0 = –– v1 + 6v2 –– v3
At node 3,
2v3 = 4 + v2 –– v3 + 12v2 + 4(v1 –– v3)
(2)
(1)
(2)
(3)
(4)
or
–– 4 = 4v1 + 13v2 –– 7v3
(3)
In matrix form,
ª7 11 4º ª v 1 º
«1 6 1 » « v »
»« 2 »
«
¬«4 13 7¼» ¬« v 3 ¼»
7
'
1 6
4
v1 =
176, ' 1
1
7
13
7
'2
4
11
4
2
110
176
v3 =
2
11
4
0
6
1
4
13
7
7
1 0
1
4 4 7
'1
'
ª 2 º
« 0 »
« »
¬« 4¼»
66,
'3
0.625V, v2 =
'3
'
286
176
11
2
1 6 0
4 13 4
'2
'
66
176
110
286
0.375V
1.625V.
v1 = 625 mV, v2 = 375 mV, v3 = 1.625 V.
Chapter 3, Solution 28
At node c,
Vd Vc Vc Vb Vc

o 0 5Vb 11Vc 2Vd
(1)
10
4
5
At node b,
Va 45 Vb Vc Vb Vb

o
45 Va 4Vb 2Vc
(2)
8
4
8
At node a,
Va 30 Vd Va Va 45 Vb
0

o 30 7Va 2Vb 4Vd (3)
4
16
8
At node d,
Va 30 Vd Vd Vd Vc

o 150 5Va 2Vc 7Vd
(4)
4
20
10
In matrix form, (1) to (4) become
§ 0 5 11 2 ·§ Va · § 0 ·
¨
¸¨ ¸ ¨
¸
¨ 1 4 2 0 ¸¨ Vb ¸ ¨ 45 ¸

o
¨ 7 2 0 4 ¸¨ V ¸ ¨ 30 ¸
c
¨
¸¨ ¸ ¨
¸
¨ 5 0 2 7 ¸¨V ¸ ¨ 150 ¸
©
¹© d ¹ ©
¹
We use MATLAB to invert A and obtain
V
A1B
AV
B
§ 10.14 ·
¨
¸
¨ 7.847 ¸
¨ 1.736 ¸
¨
¸
¨ 29.17 ¸
©
¹
Thus,
Va
10.14 V, Vb
7.847 V, Vc
1.736 V, Vd
29.17 V
Chapter 3, Solution 29
At node 1,
5 V1 V4 2V1 V1 V2 0

o
5 4V1 V2 V4
At node 2,
V1 V2 2V2 4(V2 V3 ) 0

o 0 V1 7V2 4V3
At node 3,
6 4(V2 V3 ) V3 V4

o 6 4V2 5V3 V4
At node 4,
2 V3 V4 V1 V4 3V4

o
2 V1 V3 5V4
In matrix form, (1) to (4) become
§ 4 1 0 1·§ V1 · § 5 ·
¨
¸¨ ¸ ¨ ¸
¨ 1 7 4 0 ¸¨V2 ¸ ¨ 0 ¸
AV B

o
¨ 0 4 5 1¸¨ V ¸ ¨ 6 ¸
3
¨
¸¨ ¸ ¨ ¸
¨ 1 0 1 5 ¸¨V ¸ ¨ 2 ¸
©
¹© 4 ¹ © ¹
Using MATLAB,
V
A 1 B
§ 0.7708 ·
¨
¸
¨ 1.209 ¸
¨ 2.309 ¸
¨
¸
¨ 0.7076 ¸
©
¹
i.e.
V1
0.7708 V, V2
1.209 V, V3
2.309 V, V4
0.7076 V
(1)
(2)
(3)
(4)
Chapter 3, Solution 30
v2
40 :
I0
100 V
+
120 V
20 :
v1
10 :
v0
1
2
4v0
––
–– +
+
––
2I0
80 :
At node 1,
v1 v 2
40
But, vo = 120 + v2
100 v 1 4 v o v 1
10
20
(1)
v2 = vo –– 120. Hence (1) becomes
7v1 –– 9vo = 280
(2)
At node 2,
Io + 2Io =
vo 0
80
§ v 120 v o ·
3¨ 1
¸
40
¹
©
or
6v1 –– 7vo = -720
from (2) and (3),
ª7 9 º ª v 1 º
«6 7» « v »
¬
¼¬ o ¼
'
'1
280 9
720 7
7 9
6 7
8440 ,
vo
80
(3)
ª 280 º
« 720»
¬
¼
49 54
'2
5
7 280
6 720
6720
v1 =
'1
'
8440
5
1688, vo =
'2
'
6720
1344 V
5
Io = -5.6 A
Chapter 3, Solution 31
1:
+ v0 ––
v2
v1
1A
2v0
v3
2:
i0
4:
1:
10 V
4:
At the supernode,
1 + 2v0 =
v1 v 2 v1 v 3
4
1
1
(1)
But vo = v1 –– v3. Hence (1) becomes,
4 = -3v1 + 4v2 +4v3
At node 3,
2vo +
or
v2
4
v2 = v1 + v3
v1 = 4 V, v2 = 4 V, v3 = 0 V.
10 v 3
2
20 = 4v1 + v2 –– 2v3
At the supernode, v2 = v1 + 4io. But io =
Solving (2) to (4) leads to,
v1 v 3 (2)
(3)
v3
. Hence,
4
(4)
+
––
Chapter 3, Solution 32
5 k:
v1
v3
v2
+
10 k:
4 mA
10 V
20 V
–– +
+ ––
+
loop 1
v1
12 V
––
+
loop 2
––
v3
––
(b)
(a)
We have a supernode as shown in figure (a). It is evident that v2 = 12 V, Applying KVL
to loops 1and 2 in figure (b), we obtain,
-v1 –– 10 + 12 = 0 or v1 = 2 and -12 + 20 + v3 = 0 or v3 = -8 V
Thus,
v1 = 2 V, v2 = 12 V, v3 = -8V.
Chapter 3, Solution 33
(a) This is a non-planar circuit because there is no way of redrawing the circuit
with no crossing branches.
(b) This is a planar circuit. It can be redrawn as shown below.
4:
3:
12 V
+
5:
––
1:
2:
Chapter 3, Solution 34
(a)
This is a planar circuit because it can be redrawn as shown below,
7:
2:
1:
3:
6:
10 V
5:
+
––
4:
(b)
This is a non-planar circuit.
Chapter 3, Solution 35
30 V
20 V
+
––
+
––
i1
2 k:
+
i2
v0
––
5 k:
4 k:
Assume that i1 and i2 are in mA. We apply mesh analysis. For mesh 1,
-30 + 20 + 7i1 –– 5i2 = 0 or 7i1 –– 5i2 = 10
(1)
For mesh 2,
-20 + 9i2 –– 5i1 = 0 or -5i1 + 9i2 = 20
Solving (1) and (2), we obtain, i2 = 5.
v0 = 4i2 = 20 volts.
(2)
Chapter 3, Solution 36
10 V
4:
i1
12 V
+ ––
i2
I1
+
I2
6:
––
i3
2:
Applying mesh analysis gives,
12 = 10I1 –– 6I2
-10 = -6I1 + 8I2
ª 6 º
« 5»
¬ ¼
or
'
5 3
3 4
11, '1
I1
'1
'
ª 5 3º ª I 1 º
« 3 4 » «I »
¬
¼¬ 2 ¼
6 3
5 4
9 I
, 2
11
5
6
3 5
9, ' 2
'2
'
7
7
11
i1 = -I1 = -9/11 = -0.8181 A, i2 = I1 –– I2 = 10/11 = 1.4545 A.
vo = 6i2 = 6x1.4545 = 8.727 V.
Chapter 3, Solution 37
3:
3V
+
v0
––
5:
2:
i1
1:
+
––
i2
4v0
+
––
Applying mesh analysis to loops 1 and 2, we get,
6i1 –– 1i2 + 3 = 0 which leads to i2 = 6i1 + 3
(1)
-1i1 + 6i2 –– 3 + 4v0 = 0
(2)
But, v0 = -2i1
(3)
Using (1), (2), and (3) we get i1 = -5/9.
Therefore, we get v0 = -2i1 = -2(-5/9) = 1.111 volts
Chapter 3, Solution 38
3:
6:
+ v0 ––
12 V
+
––
i1
8:
2v0
i2
+
––
We apply mesh analysis.
12 = 3 i1 + 8(i1 –– i2) which leads to 12 = 11 i1 –– 8 i2
(1)
-2 v0 = 6 i2 + 8(i2 –– i1) and v0 = 3 i1 or i1 = 7 i2
(2)
From (1) and (2), i1 = 84/69 and v0 = 3 i1 = 3x89/69
v0 = 3.652 volts
Chapter 3, Solution 39
For mesh 1,
But I x
10 2 I x 10 I 1 6 I 2
I 1 I 2 . Hence,
10 12 I 1 12 I 2 10 I 1 6 I 2

o 5
For mesh 2,
12 8I 2 6 I 1 0

o 6 3I 1 4 I 2
Solving (1) and (2) leads to
I 1 0.8 A, I 2 -0.9A
0
4I1 2I 2
(1)
(2)
Chapter 3, Solution 40
2 k:
30V
+
i2
2 k:
i1
––
6 k:
6 k:
i3
4 k:
4 k:
Assume all currents are in mA and apply mesh analysis for mesh 1.
30 = 12i1 –– 6i2 –– 4i3
15 = 6i1 –– 3i2 –– 2i3
(1)
0 = -3i1 + 7i2 –– i3
(2)
0 = -2i1 –– i2 + 5i3
(3)
for mesh 2,
0 = - 6i1 + 14i2 –– 2i3
for mesh 2,
0 = -4i1 –– 2i2 + 10i3
Solving (1), (2), and (3), we obtain,
io = i1 = 4.286 mA.
Chapter 3, Solution 41
10 :
i1
6V
2:
+ ––
1:
i2
4:
8V
i3
+
––
i
i2
i3
0
5:
For loop 1,
6 = 12i1 –– 2i2
3 = 6i1 –– i2
(1)
For loop 2,
-8 = 7i2 –– 2i1 –– i3
(2)
For loop 3,
-8 + 6 + 6i3 –– i2 = 0
2 = 6i3 –– i2
(3)
We put (1), (2), and (3) in matrix form,
ª 6 1 0º ª i 1 º
« 2 7 1 » «i »
«
»« 2 »
«¬0 1 6»¼ «¬i 3 »¼
6
'
1 0
2 7 1
0
6 3 0
234, ' 2
1 6
At node 0, i + i2 = i3 or i = i3 –– i2 =
2 8 1
240
0 2 6
6
'3
ª 3º
«8 »
« »
«¬2»¼
1 3
2 7 8
0 1 2
'3 '2
'
38
38 240
= 1.188 A
234
Chapter 3, Solution 42
For mesh 1,
12 50 I 1 30 I 2 0

o 12 50 I 1 30 I 2
(1)
For mesh 2,
8 100 I 2 30 I 1 40 I 3 0

o 8 30 I 1 100 I 2 40 I 3
For mesh 3,
(3)
6 50 I 3 40 I 2 0

o 6 40 I 2 50 I 3
Putting eqs. (1) to (3) in matrix form, we get
0 ·§ I 1 ·
§ 50 30
¨
¸¨ ¸
¨ 30 100 40 ¸¨ I 2 ¸
¨ 0
40 50 ¸¹¨© I 3 ¸¹
©
§12 ·
¨ ¸
¨8¸
¨6¸
© ¹

o
AI
(2)
B
Using Matlab,
§ 0.48 ·
¨
¸
¨ 0.40 ¸
¨ 0.44 ¸
©
¹
1
I
A B
i.e. I1 = 0.48 A, I2 = 0.4 A, I3 = 0.44 A
Chapter 3, Solution 43
20 :
a
80 V
+
i1
––
30 :
+
i3
30 :
20 :
80 V
+
i2
––
20 :
30 :
Vab
––
b
For loop 1,
80 = 70i1 –– 20i2 –– 30i3
8 = 7i1 –– 2i2 –– 3i3
(1)
For loop 2,
80 = 70i2 –– 20i1 –– 30i3
8 = -2i1 + 7i2 –– 3i3
(2)
0 = -30i1 –– 30i2 + 90i3
0 = i1 + i2 –– 3i3
(3)
For loop 3,
Solving (1) to (3), we obtain i3 = 16/9
Io = i3 = 16/9 = 1.778 A
Vab = 30i3 = 53.33 V.
Chapter 3, Solution 44
6V
+
2:
i3
4:
i2
1:
6V
5:
i1
3A
i1
i2
Loop 1 and 2 form a supermesh. For the supermesh,
6i1 + 4i2 - 5i3 + 12 = 0
(1)
For loop 3,
-i1 –– 4i2 + 7i3 + 6 = 0
(2)
Also,
i 2 = 3 + i1
(3)
Solving (1) to (3), i1 = -3.067, i3 = -1.3333; io = i1 –– i3 = -1.7333 A
+
––
Chapter 3, Solution 45
4:
30V
+
i3
i4
2:
6:
i1
––
8:
i2
3:
1:
For loop 1,
30 = 5i1 –– 3i2 –– 2i3
(1)
For loop 2,
10i2 - 3i1 –– 6i4 = 0
(2)
For the supermesh,
6i3 + 14i4 –– 2i1 –– 6i2 = 0
(3)
But
i4 –– i3 = 4 which leads to i4 = i3 + 4
(4)
Solving (1) to (4) by elimination gives i = i1 = 8.561 A.
Chapter 3, Solution 46
For loop 1,
12 11i1 8i2 0

o
For loop 2,
8i1 14i2 2vo 0
But vo 3i1 ,
11i1 8i2
(1)
12
8i1 14i2 6i1 0

o i1 7i2
Substituting (2) into (1),
77i2 8i2 12

o i 2 0.1739 A and i1
(2)
7i 2
1.217 A
Chapter 3, Solution 47
First, transform the current sources as shown below.
- 6V +
2:
V1
8:
4:
V2
I3
V3
4:
+
20V
-
I1
8:
2:
I2
+
12V
-
For mesh 1,
20 14 I 1 2 I 2 8I 3 0

o 10 7 I 1 I 2 4 I 3
For mesh 2,
12 14 I 2 2 I 1 4 I 3 0

o 6 I 1 7 I 2 2 I 3
For mesh 3,
6 14 I 3 4 I 2 8I 1 0

o 3 4 I 1 2 I 2 7 I 3
Putting (1) to (3) in matrix form, we obtain
§ 7 1 4 ·§ I 1 · § 10 ·
¨
¸¨ ¸ ¨ ¸

o
AI B
¨ 1 7 2 ¸¨ I 2 ¸ ¨ 6 ¸
¨ 4 2 7 ¸¨ I ¸ ¨ 3 ¸
©
¹© 3 ¹ © ¹
Using MATLAB,
ª 2 º
1
I A B ««0.0333»»
«¬1.8667 »¼
But

o I 1
2.5, I 2
0.0333, I 3
20 V

o V1 20 4 I1 10 V
4
V2 2( I1 I 2 ) 4.933 V
Also,
V3 12
I2

o V3 12 8I 2 12.267V
8
I1
1.8667
(1)
(2)
(3)
Chapter 3, Solution 48
We apply mesh analysis and let the mesh currents be in mA.
3k :
I4
4k :
2k :
Io
1k :
+
12 V
-
I1
5k :
I2
+
8V
-
I3
10k :
6V
+
For mesh 1,
12 8 5I 1 I 2 4 I 4 0

o 4 5I 1 I 2 4 I 4
(1)
For mesh 2,
8 13I 2 I 1 10 I 3 2 I 4 0

o 8 I 1 13I 2 10 I 3 2 I 4 (2)
For mesh 3,
(3)
6 15I 3 10 I 2 5I 4 0

o 6 10 I 2 15I 3 5I 4
For mesh 4,
4 I 1 2 I 2 5I 3 14 I 4 0
(4)
Putting (1) to (4) in matrix form gives
1
4 ·§ I 1 · § 4 ·
0
§ 5
¨
¸¨ ¸ ¨ ¸
¨ 1 13 10 2 ¸¨ I 2 ¸ ¨ 8 ¸
AI B

o
¨ 0 10 15 5 ¸¨ I ¸ ¨ 6 ¸
3
¨
¸¨ ¸ ¨ ¸
¨ 4 2 5 14 ¸¨ I ¸ ¨ 0 ¸
©
¹© 4 ¹ © ¹
Using MATLAB,
§ 7.217 ·
¨
¸
8
.
087
¨
¸
I A 1 B ¨
7.791 ¸
¨
¸
¨ 6 ¸
©
¹
The current through the 10k : resistor is Io= I2 –– I3 = 0.2957 mA
Chapter 3, Solution 49
3:
i3
2:
1:
i1
2:
16 V
i2
+
––
2i0
i1
i2
0
(a)
2:
1:
2:
+
i1
v0
+
or
––
v0
––
i2
16V
+
––
(b)
For the supermesh in figure (a),
3i1 + 2i2 –– 3i3 + 16 = 0
(1)
At node 0,
i2 –– i1 = 2i0 and i0 = -i1 which leads to i2 = -i1
(2)
For loop 3,
-i1 ––2i2 + 6i3 = 0 which leads to 6i3 = -i1
(3)
Solving (1) to (3), i1 = (-32/3)A, i2 = (32/3)A, i3 = (16/9)A
i0 = -i1 = 10.667 A, from fig. (b), v0 = i3-3i1 = (16/9) + 32 = 33.78 V.
Chapter 3, Solution 50
i1
4:
2:
i3
10 :
8:
60 V
+
i2
––
3i0
i3
i2
For loop 1,
16i1 –– 10i2 –– 2i3 = 0 which leads to 8i1 –– 5i2 –– i3 = 0
(1)
For the supermesh, -60 + 10i2 –– 10i1 + 10i3 –– 2i1 = 0
or
-6i1 + 5i2 + 5i3 = 30
(2)
Also, 3i0 = i3 –– i2 and i0 = i1 which leads to 3i1 = i3 –– i2
(3)
Solving (1), (2), and (3), we obtain i1 = 1.731 and i0 = i1 = 1.731 A
Chapter 3, Solution 51
5A
i1
8:
2:
i3
1:
i2
40 V
+
––
4:
+
v0
20V
––
+
For loop 1,
i1 = 5A
(1)
For loop 2,
-40 + 7i2 –– 2i1 –– 4i3 = 0 which leads to 50 = 7i2 –– 4i3
(2)
For loop 3,
-20 + 12i3 –– 4i2 = 0 which leads to 5 = - i2 + 3 i3
(3)
Solving with (2) and (3),
And,
i2 = 10 A, i3 = 5 A
v0 = 4(i2 –– i3) = 4(10 –– 5) = 20 V.
Chapter 3, Solution 52
+
v0 2 :
i2
––
VS
+
––
8:
3A
i2
i1
i3
4:
i3
+
––
2V0
For mesh 1,
2(i1 –– i2) + 4(i1 –– i3) –– 12 = 0 which leads to 3i1 –– i2 –– 2i3 = 6
(1)
For the supermesh, 2(i2 –– i1) + 8i2 + 2v0 + 4(i3 –– i1) = 0
But v0 = 2(i1 –– i2) which leads to -i1 + 3i2 + 2i3 = 0
(2)
For the independent current source, i3 = 3 + i2
(3)
Solving (1), (2), and (3), we obtain,
i1 = 3.5 A, i2 = -0.5 A, i3 = 2.5 A.
Chapter 3, Solution 53
+
v0 2 :
i2
––
VS
+
––
8:
3A
i2
i1
i3
4:
i3
+
––
2V0
For mesh 1,
2(i1 –– i2) + 4(i1 –– i3) –– 12 = 0 which leads to 3i1 –– i2 –– 2i3 = 6
(1)
For the supermesh, 2(i2 –– i1) + 8i2 + 2v0 + 4(i3 –– i1) = 0
But v0 = 2(i1 –– i2) which leads to -i1 + 3i2 + 2i3 = 0
(2)
For the independent current source, i3 = 3 + i2
(3)
Solving (1), (2), and (3), we obtain,
i1 = 3.5 A, i2 = -0.5 A, i3 = 2.5 A.
Chapter 3, Solution 54
Let the mesh currents be in mA. For mesh 1,
12 10 2 I 1 I 2 0

o
2 2I1 I 2
(1)
For mesh 2,
10 3I 2 I 1 I 3 0

o 10 I 1 3I 2 I 3
For mesh 3,
(3)
12 2 I 3 I 2 0

o 12 I 2 2 I 3
Putting (1) to (3) in matrix form leads to
§ 2 1 0 ·§ I 1 ·
¨
¸¨ ¸
¨ 1 3 1¸¨ I 2 ¸
¨ 0 1 2 ¸¨ I ¸
©
¹© 3 ¹
Using MATLAB,
I
1
A B
§2·
¨ ¸
¨10 ¸
¨12 ¸
© ¹
ª 5.25 º
« 8 .5 »
«
»
«¬10.25»¼

o

o I 1
Chapter 3, Solution 55
i1
4A
5.25 mA, I 2
1A
I2
1A
i2
I4
2:
i3
12 :
a
I3
d
I4
4A
8.5 mA, I 3
c
+
6:
I1
B
10 V
I2
b
AI
4:
+ ––
8V
(2)
I3
0
10.25 mA
It is evident that I1 = 4
For mesh 4,
12(I4 –– I1) + 4(I4 –– I3) –– 8 = 0
For the supermesh
At node c,
(1)
(2)
6(I2 –– I1) + 10 + 2I3 + 4(I3 –– I4) = 0
or -3I1 + 3I2 + 3I3 –– 2I4 = -5
(3)
I2 = I3 + 1
(4)
Solving (1), (2), (3), and (4) yields, I1 = 4A, I2 = 3A, I3 = 2A, and I4 = 4A
At node b,
i1 = I2 –– I1 = -1A
At node a,
i2 = 4 –– I4 = 0A
At node 0,
i3 = I4 –– I3 = 2A
Chapter 3, Solution 56
+ v1 ––
2:
2:
12 V
+
––
i2
2:
2:
i1
2:
i3
+
v2
––
For loop 1, 12 = 4i1 –– 2i2 –– 2i3 which leads to 6 = 2i1 –– i2 –– i3
(1)
For loop 2, 0 = 6i2 ––2i1 –– 2 i3 which leads to 0 = -i1 + 3i2 –– i3
(2)
For loop 3, 0 = 6i3 –– 2i1 –– 2i2 which leads to 0 = -i1 –– i2 + 3i3
(3)
In matrix form (1), (2), and (3) become,
ª 2 1 1º ª i1 º
« 1 3 1» «i »
«
»« 2 »
¬« 1 1 3 ¼» ¬«i 3 ¼»
2
1 1
2
' = 1 3 1
1 1 3
2
ª 6º
« 0»
« »
¬«0¼»
6 1
8, '2 = 1 3 1
1 0
24
3
1 6
'3 = 1 3 0 24 , therefore i2 = i3 = 24/8 = 3A,
1 1 0
v1 = 2i2 = 6 volts, v = 2i3 = 6 volts
Chapter 3, Solution 57
Assume R is in kilo-ohms.
V2 4k:x18mA 72V ,
V1 100 V2 100 72 28V
Current through R is
3
3
28
iR
io ,
V1 i R R

o
(18) R
3 R
3 R
This leads to R = 84/26 = 3.23 k :
Chapter 3, Solution 58
30 :
i2
30 :
10 :
i1
10 :
i3
+
––
120 V
30 :
For loop 1, 120 + 40i1 –– 10i2 = 0, which leads to -12 = 4i1 –– i2
(1)
For loop 2, 50i2 –– 10i1 –– 10i3 = 0, which leads to -i1 + 5i2 –– i3 = 0
(2)
For loop 3, -120 –– 10i2 + 40i3 = 0, which leads to 12 = -i2 + 4i3
(3)
Solving (1), (2), and (3), we get, i1 = -3A, i2 = 0, and i3 = 3A
Chapter 3, Solution 59
40 :
I0
–– +
120 V
i2
10 :
20 :
i1
100V +
4v0
––
+
i3
+
––
v0
80 :
––
2I0
i2
For loop 1, -100 + 30i1 –– 20i2 + 4v0 = 0, where v0 = 80i3
or 5 = 1.5i1 –– i2 + 16i3
i3
(1)
For the supermesh, 60i2 –– 20i1 –– 120 + 80i3 –– 4 v0 = 0, where v0 = 80i3
or 6 = -i1 + 3i2 –– 12i3
(2)
Also, 2I0 = i3 –– i2 and I0 = i2, hence, 3i2 = i3
(3)
From (1), (2), and (3),
3
' = 1
0
2
32
3
12
3
1
ª 3 2 32 º
« 1 3 12»
«
»
3
1 ¼»
¬« 0
3
5, '2 = 1
0
10
32
6
12
0
1
ª i1 º
«i »
« 2»
«¬i 3 »¼
ª10º
«6»
« »
«¬ 0 »¼
3
28, '3 = 1
I0 = i2 = '2/' = -28/5 = -5.6 A
v0 = 8i3 = (-84/5)80 = -1344 volts
0
2 10
3
6
3
0
84
Chapter 3, Solution 60
0.5i0
4:
v1
10 V
1:
10 V
8:
v2
2:
+
––
i0
At node 1, (v1/1) + (0.5v1/1) = (10 –– v1)/4, which leads to v1 = 10/7
At node 2, (0.5v1/1) + ((10 –– v2)/8) = v2/2 which leads to v2 = 22/7
P1: = (v1)2/1 = 2.041 watts, P2: = (v2)2/2 = 4.939 watts
P4: = (10 –– v1)2/4 = 18.38 watts, P8: = (10 –– v2)2/8 = 5.88 watts
Chapter 3, Solution 61
v1
is
20 :
v2
10 :
i0
+
v0
––
30 :
––
+ 5v0
At node 1, is = (v1/30) + ((v1 –– v2)/20) which leads to 60is = 5v1 –– 3v2
But v2 = -5v0 and v0 = v1 which leads to v2 = -5v1
Hence, 60is = 5v1 + 15v1 = 20v1 which leads to v1 = 3is, v2 = -15is
i0 = v2/50 = -15is/50 which leads to i0/is = -15/50 = -0.3
40 :
(1)
Chapter 3, Solution 62
4 k:
100V +
8 k:
A
i1
––
B
i2
2 k:
i3
+
––
40 V
We have a supermesh. Let all R be in k:, i in mA, and v in volts.
For the supermesh, -100 +4i1 + 8i2 + 2i3 + 40 = 0 or 30 = 2i1 + 4i2 + i3
(1)
At node A,
i 1 + 4 = i2
(2)
At node B,
i2 = 2i1 + i3
(3)
Solving (1), (2), and (3), we get i1 = 2 mA, i2 = 6 mA, and i3 = 2 mA.
Chapter 3, Solution 63
10 :
A
5:
50 V
+
––
i1
i2
+
––
For the supermesh, -50 + 10i1 + 5i2 + 4ix = 0, but ix = i1. Hence,
50 = 14i1 + 5i2
At node A, i1 + 3 + (vx/4) = i2, but vx = 2(i1 –– i2), hence, i1 + 2 = i2
Solving (1) and (2) gives i1 = 2.105 A and i2 = 4.105 A
vx = 2(i1 –– i2) = -4 volts and ix = i2 –– 2 = 4.105 amp
(1)
(2)
4ix
Chapter 3, Solution 64
i1
50 :
i2 10 :
+
A
i0
i1
10 :
i2
+
––
4i0
i3
40 :
100V +
––
2A
0.2V0
i1
B
i3
20i2 –– 10i1 + 4i0 = 0
For mesh 2,
(1)
But at node A, io = i1 –– i2 so that (1) becomes i1 = (7/12)i2
(2)
For the supermesh, -100 + 50i1 + 10(i1 –– i2) –– 4i0 + 40i3 = 0
or
50 = 28i1 –– 3i2 + 20i3
(3)
At node B,
i3 + 0.2v0 = 2 + i1
(4)
But,
v0 = 10i2 so that (4) becomes i3 = 2 –– (17/12)i2
(5)
Solving (1) to (5), i2 = -0.674,
v0 = 10i2 = -6.74 volts,
i0 = i1 - i2 = -(5/12)i2 = 0.281 amps
Chapter 3, Solution 65
For mesh 1,
For mesh 2,
For mesh 3,
For mesh 4,
For mesh 5,
12 12 I 1 6 I 2 I 4
0 6 I 1 16 I 2 8I 3 I 4 I 5
9 8I 2 15I 3 I 5
6 I1 I 2 5I 4 2 I 5
10 I 2 I 3 2 I 4 8I 5
(1)
(2)
(3)
(4)
(5)
Casting (1) to (5) in matrix form gives
1
0 ·§ I 1 · §12 ·
§ 12 6 0
¸¨ ¸ ¨ ¸
¨
¨ 6 16 8 1 1 ¸¨ I 2 ¸ ¨ 0 ¸
¨ 0 8 15 0 1 ¸¨ I ¸ ¨ 9 ¸

o
¸¨ 3 ¸ ¨ ¸
¨
5 2 ¸¨ I 4 ¸ ¨ 6 ¸
¨ 1 1 0
¨ 0 1 1 2 8 ¸¨ I ¸ ¨10 ¸
¹© 5 ¹ © ¹
©
Using MATLAB leads to
§ 1.673 ·
¨
¸
¨ 1.824 ¸
I A 1 B ¨ 1.733 ¸
¨
¸
¨ 2.864 ¸
¨ 2.411 ¸
©
¹
Thus,
I 1 1.673 A, I 2 1.824 A, I 3 1.733 A, I 4
AI
B
1.864 A, I 5
2.411 A
Chapter 3, Solution 66
Consider the circuit below.
2 k:
+
20V
-
2 k:
1 k:
I1
1 k:
+
10V
-
I2
1 k:
Io
2 k:
1 k:
I3
12V
+
We use mesh analysis. Let the mesh currents be in mA.
For mesh 1, 20 4 I 1 I 2 I 3
For mesh 2, 10 I 1 4 I 2 I 4
For mesh 3, 12 I 1 4 I 3 I 4
For mesh 4, 12 I 2 I 3 4 I 4
2 k:
I4
(1)
(2)
(3)
(4)
In matrix form, (1) to (4) become
§ 4 1 1 0 ·§ I 1 · § 20 ·
¨
¸¨ ¸ ¨
¸
¨ 1 4 0 1¸¨ I 2 ¸ ¨ 10 ¸
¨1 0
4 1¸¨ I 3 ¸ ¨ 12 ¸
¨
¸¨ ¸ ¨
¸
¨ 0 1 1 4 ¸¨ I ¸ ¨ 12 ¸
©
¹© 4 ¹ ©
¹
Using MATLAB,
I
A 1 B

o
AI
B
§ 5.5 ·
¨
¸
¨ 1.75 ¸
¨ 3.75 ¸
¨
¸
¨ 2.5 ¸
©
¹
Thus,
Io
I 3
3.75 mA
Chapter 3, Solution 67
G11 = (1/1) + (1/4) = 1.25, G22 = (1/1) + (1/2) = 1.5
G12 = -1 = G21, i1 = 6 –– 3 = 3, i2 = 5-6 = -1
Hence, we have,
ª1.25 1º
« 1 1.5»
¼
¬
1
ª v1 º
«v »
¬ 2¼
ª1.25 1º ª v 1 º
« 1 1.5» « v »
¬
¼¬ 2 ¼
ª3º
« 1»
¬ ¼
1 ª1.5 1 º
, where ' = [(1.25)(1.5)-(-1)(-1)] = 0.875
' «¬ 1 1.25»¼
ª1.7143 1.1429º ª 3 º
«1.1429 1.4286» « 1»
¬
¼¬ ¼
ª3(1.7143) 1(1.1429) º
«3(1.1429) 1(1.4286)»
¬
¼
Clearly v1 = 4 volts and v2 = 2 volts
ª 4º
« 2»
¬ ¼
Chapter 3, Solution 68
By inspection, G11 = 1 + 3 + 5 = 8S, G22 = 1 + 2 = 3S, G33 = 2 + 5 = 7S
G12 = -1, G13 = -5, G21 = -1, G23 = -2, G31 = -5, G32 = -2
i1 = 4, i2 = 2, i3 = -1
We can either use matrix inverse as we did in Problem 3.51 or use Cramer’’s Rule.
Let us use Cramer’’s rule for this problem.
First, we develop the matrix relationships.
ª 8 1 5º ª v 1 º
« 1 3 2» « v »
«
»« 2 »
«¬ 5 2 7 »¼ «¬ v 3 »¼
8
'
1
1 5
3
5 2
8
'2
4
2
4
34, ' 1
2
1 5
3
1 2
7
5
1 2 2
5 1 7
ª4º
«2»
« »
«¬ 1»¼
8
109, ' 3
1
2
85
7
4
1 3
2
5 2 1
87
v1 = '1/' = 85/34 = 3.5 volts, v2 = '2/' = 109/34 = 3.206 volts
v3 = '3/' = 87/34 = 2.56 volts
Chapter 3, Solution 69
Assume that all conductances are in mS, all currents are in mA, and all voltages
are in volts.
G11 = (1/2) + (1/4) + (1/1) = 1.75, G22 = (1/4) + (1/4) + (1/2) = 1,
G33 = (1/1) + (1/4) = 1.25, G12 = -1/4 = -0.25, G13 = -1/1 = -1,
G21 = -0.25, G23 = -1/4 = -0.25, G31 = -1, G32 = -0.25
i1 = 20, i2 = 5, and i3 = 10 –– 5 = 5
The node-voltage equations are:
1 ºª v1 º
ª 1.75 0.25
« 0.25
1
0.25» « v 2 »
«
»« »
0.25 1.25 ¼» ¬« v 3 ¼»
¬« 1
ª20º
«5»
« »
¬« 5 ¼»
Chapter 3, Solution 70
G11 = G1 + G2 + G4, G12 = -G2, G13 = 0,
G22 = G2 + G3, G21 = -G2, G23 = -G3,
G33 = G1 + G3 + G5, G31 = 0, G32 = -G3
i1 = -I1, i2 = I2, and i3 = I1
Then, the node-voltage equations are:
ªG 1 G 2 G 4
«
G2
«
0
«¬
G2
G1 G 2
G3
ºª v1 º
»«v »
G3
»« 2 »
G 1 G 3 G 5 »¼ «¬ v 3 »¼
0
ª I 1 º
« I »
« 2 »
«¬ I 1 »¼
Chapter 3, Solution 71
R11 = 4 + 2 = 6, R22 = 2 + 8 + 2 = 12, R33 = 2 + 5 = 7,
R12 = -2, R13 = 0, R21 = -2, R23 = -2, R31 = 0, R32 = -2
v1 = 12, v2 = -8, and v3 = -20
Now we can write the matrix relationships for the mesh-current equations.
ª 6 2 0 º ª i1 º
« 2 12 2» «i »
«
»« 2 »
«¬ 0 2 7 »¼ «¬i 3 »¼
ª 12 º
«8»
«
»
«¬ 20»¼
Now we can solve for i2 using Cramer’’s Rule.
'
6
2
0
2
12
2
0
2
7
452, ' 2
6
12
0
2
8
2
0
20
7
408
i2 = '2/' = -0.9026, p = (i2)2R = 6.52 watts
Chapter 3, Solution 72
R11 = 5 + 2 = 7, R22 = 2 + 4 = 6, R33 = 1 + 4 = 5, R44 = 1 + 4 = 5,
R12 = -2, R13 = 0 = R14, R21 = -2, R23 = -4, R24 = 0, R31 = 0,
R32 = -4, R34 = -1, R41 = 0 = R42, R43 = -1, we note that Rij = Rji for
all i not equal to j.
v1 = 8, v2 = 4, v3 = -10, and v4 = -4
Hence the mesh-current equations are:
0 º ª i1 º
ª 7 2 0
« 2 6 4 0 » «i »
«
»« 2 »
« 0 4 5 1» «i 3 »
«
»« »
0 1 5 ¼ ¬i 4 ¼
¬ 0
ª 8 º
« 4 »
«
»
« 10»
«
»
¬ 4¼
Chapter 3, Solution 73
R11 = 2 + 3 +4 = 9, R22 = 3 + 5 = 8, R33 = 1 + 4 = 5, R44 = 1 + 1 = 2,
R12 = -3, R13 = -4, R14 = 0, R23 = 0, R24 = 0, R34 = -1
v1 = 6, v2 = 4, v3 = 2, and v4 = -3
Hence,
ª 9 3 4 0 º ª i1 º
« 3 8
0
0 »» ««i 2 »»
«
« 4 0
6 1» «i3 »
«
»« »
0 1 2 ¼ ¬i 4 ¼
¬0
ª6º
«4»
« »
«2»
« »
¬ 3¼
Chapter 3, Solution 74
R11 = R1 + R4 + R6, R22 = R2 + R4 + R5, R33 = R6 + R7 + R8,
R44 = R3 + R5 + R8, R12 = -R4, R13 = -R6, R14 = 0, R23 = 0,
R24 = -R5, R34 = -R8, again, we note that Rij = Rji for all i not equal to j.
ª V1 º
« V »
2»
The input voltage vector is = «
« V3 »
«
»
¬ V4 ¼
ªR 1 R 4 R 6
«
R4
«
«
R6
«
0
¬
R4
R6
R2 R4 R5
0
0
R5
R6 R7 R8
R8
º ª i1 º
» «i »
R5
»« 2 »
R8
» «i 3 »
»« »
R 3 R 5 R 8 ¼ ¬i 4 ¼
0
ª V1 º
« V »
2»
«
« V3 »
«
»
¬ V4 ¼
Chapter 3, Solution 75
* Schematics Netlist *
R_R4
R_R2
R_R1
R_R3
R_R5
V_V4
v_V3
v_V2
v_V1
$N_0002 $N_0001 30
$N_0001 $N_0003 10
$N_0005 $N_0004 30
$N_0003 $N_0004 10
$N_0006 $N_0004 30
$N_0003 0 120V
$N_0005 $N_0001 0
0 $N_0006 0
0 $N_0002 0
i3
i1
i2
Clearly, i1 = -3 amps, i2 = 0 amps, and i3 = 3 amps, which agrees with the answers in
Problem 3.44.
Chapter 3, Solution 76
* Schematics Netlist *
I_I2
R_R1
R_R3
R_R2
F_F1
VF_F1
R_R4
R_R6
I_I1
R_R5
0 $N_0001 DC 4A
$N_0002 $N_0001 0.25
$N_0003 $N_0001 1
$N_0002 $N_0003 1
$N_0002 $N_0001 VF_F1 3
$N_0003 $N_0004 0V
0 $N_0002 0.5
0 $N_0001 0.5
0 $N_0002 DC 2A
0 $N_0004 0.25
Clearly, v1 = 625 mVolts, v2 = 375 mVolts, and v3 = 1.625 volts, which agrees with
the solution obtained in Problem 3.27.
Chapter 3, Solution 77
* Schematics Netlist *
R_R2
I_I1
I_I3
R_R3
R_R1
I_I2
0 $N_0001 4
$N_0001 0 DC 3A
$N_0002 $N_0001 DC 6A
0 $N_0002 2
$N_0001 $N_0002 1
0 $N_0002 DC 5A
Clearly, v1 = 4 volts and v2 = 2 volts, which agrees with the answer obtained in Problem
3.51.
Chapter 3, Solution 78
The schematic is shown below. When the circuit is saved and simulated the node
voltages are displaced on the pseudocomponents as shown. Thus,
V1
.
3V, V2
4.5V, V3
15V,
Chapter 3, Solution 79
The schematic is shown below. When the circuit is saved and simulated, we obtain the
node voltages as displaced. Thus,
Va
5.278 V, Vb
10.28 V, Vc
0.6944 V, Vd
Chapter 3, Solution 80
* Schematics Netlist *
H_H1
VH_H1
I_I1
V_V1
R_R4
R_R1
R_R2
R_R5
R_R3
$N_0002 $N_0003 VH_H1 6
0 $N_0001 0V
$N_0004 $N_0005 DC 8A
$N_0002 0 20V
0 $N_0003 4
$N_0005 $N_0003 10
$N_0003 $N_0002 12
0 $N_0004 1
$N_0004 $N_0001 2
26.88 V
Clearly, v1 = 84 volts, v2 = 4 volts, v3 = 20 volts, and v4 = -5.333 volts
Chapter 3, Solution 81
Clearly, v1 = 26.67 volts, v2 = 6.667 volts, v3 = 173.33 volts, and v4 = -46.67 volts
which agrees with the results of Example 3.4.
This is the netlist for this circuit.
* Schematics Netlist *
R_R1
R_R2
R_R3
R_R4
R_R5
I_I1
V_V1
E_E1
0 $N_0001 2
$N_0003 $N_0002 6
0 $N_0002 4
0 $N_0004 1
$N_0001 $N_0004 3
0 $N_0003 DC 10A
$N_0001 $N_0003 20V
$N_0002 $N_0004 $N_0001 $N_0004 3
Chapter 3, Solution 82
2i0
+ v0 ––
3 k:
1
2 k:
2
+
3v0
3
6 k:
4
4A
4 k:
8 k:
0
This network corresponds to the Netlist.
100V +
––
Chapter 3, Solution 83
The circuit is shown below.
20 :
1
70 :
2i02
3
+ v0 ––
20 V
50 :
+
––
2
2 k:
1
30 :
2A
3 k:
3v0
+
3
0
4
6 k:
4A
4 k:
8 k:
100V +
0
When the circuit is saved and simulated, we obtain v2 = -12.5 volts
Chapter 3, Solution 84
From the output loop, v0 = 50i0x20x103 = 106i0
(1)
From the input loop, 3x10-3 + 4000i0 –– v0/100 = 0
(2)
From (1) and (2) we get, i0 = 0.5PA and v0 = 0.5 volt.
Chapter 3, Solution 85
The amplifier acts as a source.
Rs
+
Vs
-
RL
For maximum power transfer,
RL
Rs
9:
––
Chapter 3, Solution 86
Let v1 be the potential across the 2 k-ohm resistor with plus being on top. Then,
[(0.03 –– v1)/1k] + 400i = v1/2k
(1)
Assume that i is in mA. But, i = (0.03 –– v1)/1
(2)
Combining (1) and (2) yields,
v1 = 29.963 mVolts and i = 37.4 nA, therefore,
v0 = -5000x400x37.4x10-9 = -74.8 mvolts
Chapter 3, Solution 87
v1 = 500(vs)/(500 + 2000) = vs/5
v0 = -400(60v1)/(400 + 2000) = -40v1 = -40(vs/5) = -8vs,
Therefore, v0/vs = -8
Chapter 3, Solution 88
Let v1 be the potential at the top end of the 100-ohm resistor.
(vs –– v1)/200 = v1/100 + (v1 –– 10-3v0)/2000
(1)
For the right loop, v0 = -40i0(10,000) = -40(v1 –– 10-3)10,000/2000,
or, v0 = -200v1 + 0.2v0 = -4x10-3v0
Substituting (2) into (1) gives,
(vs + 0.004v1)/2 = -0.004v0 + (-0.004v1 –– 0.001v0)/20
This leads to 0.125v0 = 10vs or (v0/vs) = 10/0.125 = -80
(2)
Chapter 3, Solution 89
vi = VBE + 40k IB
(1)
5 = VCE + 2k IC
(2)
If IC = EIB = 75IB and VCE = 2 volts, then (2) becomes 5 = 2 +2k(75IB)
which leads to IB = 20 PA.
Substituting this into (1) produces, vi = 0.7 + 0.8 = 1.5 volts.
2 k:
IB
40 k:
vi
+
VBE
+
-
––
5v
+
-
Chapter 3, Solution 90
1 k:
100 k:
vs
i1
i2
+
+
VBE
+
-
IB
500 :
IE
VCE
––
––
+
18V
+
-
V0
––
For loop 1, -vs + 10k(IB) + VBE + IE (500) = 0 = -vs + 0.7 + 10,000IB + 500(1 + E)IB
which leads to vs + 0.7 = 10,000IB + 500(151)IB = 85,500IB
But, v0 = 500IE = 500x151IB = 4 which leads to IB = 5.298x10-5
Therefore, vs = 0.7 + 85,500IB = 5.23 volts
Chapter 3, Solution 91
We first determine the Thevenin equivalent for the input circuit.
RTh = 6||2 = 6x2/8 = 1.5 k: and VTh = 2(3)/(2+6) = 0.75 volts
5 k:
IC
1.5 k:
0.75 V
IB
+
VBE
+
-
i1
i2
+
VCE
––
––
9V
+
400 :
+
-
V0
IE
––
For loop 1, -0.75 + 1.5kIB + VBE + 400IE = 0 = -0.75 + 0.7 + 1500IB + 400(1 + E)IB
IB = 0.05/81,900 = 0.61 PA
v0 = 400IE = 400(1 + E)IB = 49 mV
For loop 2, -400IE –– VCE –– 5kIC + 9 = 0, but, IC = EIB and IE = (1 + E)IB
VCE = 9 –– 5kEIB –– 400(1 + E)IB = 9 –– 0.659 = 8.641 volts
Chapter 3, Solution 92
I1
5 k:
10 k:
VC
IB
IC
+
+
VBE
4 k:
IE
VCE
––
––
+
V0
––
12V
+
-
I1 = IB + IC = (1 + E)IB and IE = IB + IC = I1
Applying KVL around the outer loop,
4kIE + VBE + 10kIB + 5kI1 = 12
12 –– 0.7 = 5k(1 + E)IB + 10kIB + 4k(1 + E)IB = 919kIB
IB = 11.3/919k = 12.296 PA
Also, 12 = 5kI1 + VC which leads to VC = 12 –– 5k(101)IB = 5.791 volts
Chapter 3, Solution 93
1:
4:
v1
i1
24V
+
––
3v0
i
2:
2:
+
8:
2:
v2 i3
3v0
i
i2
4:
+
+
+
+
v0
v1
v2
––
––
(a)
––
(b)
From (b), -v1 + 2i –– 3v0 + v2 = 0 which leads to i = (v1 + 3v0 –– v2)/2
At node 1 in (a), ((24 –– v1)/4) = (v1/2) + ((v1 +3v0 –– v2)/2) + ((v1 –– v2)/1), where v0 = v2
or 24 = 9v1 which leads to v1 = 2.667 volts
At node 2, ((v1 –– v2)/1) + ((v1 + 3v0 –– v2)/2) = (v2/8) + v2/4, v0 = v2
v2 = 4v1 = 10.66 volts
Now we can solve for the currents, i1 = v1/2 = 1.333 A, i2 = 1.333 A, and
i3 = 2.6667 A.
Chapter 4, Solution 1.
1:
+
1V
8 (5 3)
io
1
i
2
4: , i
1
10
5:
i
1
1 4
io
8:
3:
1
5
0.1A
Chapter 4, Solution 2.
1
A
2
6 ( 4 2)
3:, i1
i2
1
i1
2
1
, vo
4
2i o
io
0.5V
5:
4:
i1
io
i2
1A
8:
6:
2:
If is = 1PA, then vo = 0.5PV
Chapter 4, Solution 3.
R
3R
io
3R
Vs
3R
+
+
R
vo
1V
+
3R
(a)
(b)
1.5R
(a) We transform the Y sub-circuit to the equivalent ' .
3R 2
4R
R 3R
3
3
3
R, R R
4
4
4
3
R
2
vs
independent of R
2
io = vo/(R)
vo
When vs = 1V, vo = 0.5V, io = 0.5A
(b)
(c)
When vs = 10V, vo = 5V, io = 5A
When vs = 10V and R = 10:,
vo = 5V, io = 10/(10) = 500mA
Chapter 4, Solution 4.
If Io = 1, the voltage across the 6: resistor is 6V so that the current through the 3:
resistor is 2A.
2:
2A
1A
2:
3A
3A
i1
+
3:
6:
Is
4:
2:
4:
v1
(a)
36
2: , vo = 3(4) = 12V, i1
(b)
vo
4
3A.
Hence Is = 3 + 3 = 6A
If
Is = 6A
Is = 9A
Io = 1
Io = 6/(9) = 0.6667A
Is
Chapter 4, Solution 5.
2:
+
Vs
If vo = 1V,
vo
6:
V1
Vs
If vs =
3:
v1
10
3
6:
§1·
¨ ¸ 1 2V
©3¹
10
§2·
2¨ ¸ v1
3
©3¹
vo = 1
Then vs = 15
vo =
3
x15
10
4.5V
Chapter 4, Solution 6
Let RT
k
Vo
Vs
R2 // R3
RT
RT R1
R2 R3
, then Vo
R2 R3
R2 R3
R2 R3
R2 R3
R1
R2 R3
RT
Vs
RT R1
R2 R3
R1 R2 R2 R3 R3 R1
6:
Chapter 4, Solution 7
We find the Thevenin equivalent across the 10-ohm resistor. To find VTh, consider the
circuit below.
3Vx
5:
5:
+
+
15 :
4V
-
VTh
6:
+
Vx
-
From the figure,
15
(4) 3V
15 5
consider the circuit below:
0,
Vx
To find RTh,
VTh
3Vx
5:
5:
V1
V2
+
15 :
4V
-
+
At node 1,
V V V2
4 V1
3V x 1 1
,
5
15
5
At node 2,
Vx
1A
6:
Vx
6 x1 6
-

o
258
3V2 7V1
(1)
V1 V2
0

o V1 V2 95
5
Solving (1) and (2) leads to V2 = 101.75 V
2
VTh
V2
9
RTh
101.75:,
p max
1
4 RTh 4 x101.75
1 3V x (2)
22.11 mW
Chapter 4, Solution 8.
Let i = i1 + i2,
where i1 and iL are due to current and voltage sources respectively.
6:
i2
i1
6:
4: 5A
20V
+
4:
(a)
i1 =
6
(5)
64
(b)
3A, i 2
20
64
2A
Thus i = i1 + i2 = 3 + 2 = 5A
Chapter 4, Solution 9.
Let i x
i x1 i x 2
where i x1 is due to 15V source and i x 2 is due to 4A source,
12 :
i
ix1
15V
+
10 :
(a)
40:
-4A
ix2
12:
10:
(b)
40:
For ix1, consider Fig. (a).
10||40 = 400/50 = 8 ohms, i = 15/(12 + 8) = 0.75
ix1 = [40/(40 + 10)]i = (4/5)0.75 = 0.6
For ix2, consider Fig. (b).
12||40 = 480/52 = 120/13
ix2 = [(120/13)/((120/13) + 10)](-4) = -1.92
ix = 0.6 –– 1.92 = -1.32 A
p = vix = ix2R = (-1.32)210 = 17.43 watts
Chapter 4, Solution 10.
Let vab = vab1 + vab2 where vab1 and vab2 are due to the 4-V and the 2-A sources
respectively.
3vab1
10 :
10 :
+
3vab2
+
+
4V
+
vab1
+
2A
(a)
(b)
For vab1, consider Fig. (a). Applying KVL gives,
- vab1 –– 3 vab1 + 10x0 + 4 = 0, which leads to vab1 = 1 V
For vab2, consider Fig. (b). Applying KVL gives,
vab = 1 + 5 = 6 V
vab2
vab2 –– 3vab2 + 10x2 = 0, which leads to vab2 = 5
Chapter 4, Solution 11.
Let i = i1 + i2, where i1 is due to the 12-V source and i2 is due to the 4-A source.
6:
io
i1
12V
+
2:
3:
(a)
4A
i2
6:
2:
3:
ix2
2:
4A
2:
(b)
For i1, consider Fig. (a).
2||3 = 2x3/5 = 6/5, io = 12/(6 + 6/5) = 10/6
i1 = [3/(2 + 3)]io = (3/5)x(10/6) = 1 A
For i2, consider Fig. (b),
6||3 = 2 ohm, i2 = 4/2 = 2 A
i = 1+2 = 3A
Chapter 4, Solution 12.
Let vo = vo1 + vo2 + vo3, where vo1, vo2, and vo3 are due to the 2-A, 12-V, and 19-V
sources respectively. For vo1, consider the circuit below.
2A
5:
6:
3:
+ vo1 2A
4:
12 :
io 5 :
+ vo1 5:
6||3 = 2 ohms, 4||12 = 3 ohms. Hence,
io = 2/2 = 1, vo1 = 5io = 5 V
For vo2, consider the circuit below.
6:
12V
+
5:
4:
6:
+ vo2 3:
12 :
12V
+
5:
+
+ vo2 v1
3:
3:
3||8 = 24/11, v1 = [(24/11)/(6 + 24/11)]12 = 16/5
vo2 = (5/8)v1 = (5/8)(16/5) = 2 V
For vo3, consider the circuit shown below.
5:
4:
+ vo3 6:
3:
12 :
5:
+
+ vo3 19V
2:
12 :
4:
+
v2
+
19V
7||12 = (84/19) ohms, v2 = [(84/19)/(4 + 84/19)]19 = 9.975
v = (-5/7)v2 = -7.125
vo = 5 + 2 –– 7.125 = -125 mV
Chapter 4, Solution 13
Let
io i1 i2 i3 ,
where i1, i2, and i3 are the contributions to io due to 30-V, 15-V, and 6-mA sources
respectively. For i1, consider the circuit below.
1 k:
2 k:
+
30V
-
3 k:
i1
4 k:
5 k:
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.
i1
30 mA
4 k:
0.7949 k :
Using current division,
0.7949
(30mA)
4.7949
i1
4.973 mA
For i2, consider the circuit below.
1 k:
2 k:
3 k:
i2
-
4 k:
5 k:
15V
+
After successive source transformation and resistance combinations, we obtain the circuit
below:
2.42mA
i2
4 k:
0.7949 k :
Using current division,
i2
0.7949
(2.42mA)
4.7949
0.4012 mA
For i3, consider the circuit below.
6mA
1 k:
2 k:
3 k:
i3
4 k:
5 k:
After successive source transformation and resistance combinations, we obtain the circuit
below:
3.097mA
i3
4 k:
0.7949
(3.097mA)
4.7949
i3
io
i1 i2 i3
0.7949 k :
0.5134 mA
Thus,
4.058 mA
Chapter 4, Solution 14.
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.
6:
4:
2:
+
+
20V
vo1
3:
6||(4 + 2) = 3 ohms, vo1 = (½)20 = 10 V
For vo2, consider the circuit below.
6:
4:
6:
4V
2:
2:
+
+
1A
4:
+
3:
vo2
vo2
3:
3||6 = 2 ohms, vo2 = [2/(4 + 2 + 2)]4 = 1 V
For vo3, consider the circuit below.
6:
2A
4:
2A
2:
3:
+
3:
3:
vo3
vo3 +
6||(4 + 2) = 3, vo3 = (-1)3 = -3
vo = 10 + 1 –– 3 = 8 V
Chapter 4, Solution 15.
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.
io
20V
+
1:
i1
2:
3:
4:
4||(3 + 1) = 2 ohms, Then io = [20/(2 + 2)] = 5 A, i1 = io/2 = 2.5 A
For i3, consider the circuit below.
+
2:
vo’’
1:
4:
i3
+
3:
16V
2||(1 + 3) = 4/3, vo’’ = [(4/3)/((4/3) + 4)](-16) = -4
i3 = vo’’/4 = -1
For i2, consider the circuit below.
2:
1:
2A
(4/3):
i2
4:
3:
2||4 = 4/3, 3 + 4/3 = 13/3
Using the current division principle.
i2 = [1/(1 + 13/2)]2 = 3/8 = 0.375
i = 2.5 + 0.375 - 1 = 1.875 A
p = i2R = (1.875)23 = 10.55 watts
1:
2A
i2
3:
Chapter 4, Solution 16.
Let io = io1 + io2 + io3, where io1, io2, and io3 are due to
the 12-V, 4-A, and 2-A sources. For io1, consider the circuit below.
4:
io1
12V
+
3:
10 :
2:
5:
10||(3 + 2 + 5) = 5 ohms, io1 = 12/(5 + 4) = (12/9) A
4A
For io2, consider the circuit below.
3:
io2
4:
2:
5:
10:
i1
2 + 5 + 4||10 = 7 + 40/14 = 69/7
i1 = [3/(3 + 69/7)]4 = 84/90, io2 =[-10/(4 + 10)]i1 = -6/9
For io3, consider the circuit below.
3:
io3
2:
i2
4:
10 :
5: 2A
3 + 2 + 4||10 = 5 + 20/7 = 55/7
i2 = [5/(5 + 55/7)]2 = 7/9, io3 = [-10/(10 + 4)]i2 = -5/9
io = (12/9) –– (6/9) –– (5/9) = 1/9 = 111.11 mA
Chapter 4, Solution 17.
Let vx = vx1 + vx2 + vx3, where vx1,vx2, and vx3 are due to the 90-V, 6-A, and 40-V
sources. For vx1, consider the circuit below.
30 :
10 :
+
90V
+
vx1
60 :
20 :
30 :
io 10 :
+
vx1
3A
20 :
12 :
20||30 = 12 ohms, 60||30 = 20 ohms
By using current division,
io = [20/(22 + 20)]3 = 60/42, vx1 = 10io = 600/42 = 14.286 V
For vx2, consider the circuit below.
10 : i ’’
o
+
30 :
10 : i ’’
o
vx2 + vx2 60 : 6A
30 :
20 :
6A
20 :
12 :
io’’ = [12/(12 + 30)]6 = 72/42, vx2 = -10io’’ = -17.143 V
For vx3, consider the circuit below.
10 :
+
30 :
60 :
vx3
10 :
10 :
30 :
+
40V
+
vx3
20 :
io””
7.5:
io”” = [12/(12 + 30)]2 = 24/42, vx3 = -10io”” = -5.714
vx = 14.286 –– 17.143 –– 5.714 = -8.571 V
4A
Chapter 4, Solution 18.
Let ix = i1 + i2, where i1 and i2 are due to the 10-V and 2-A sources respectively. To
obtain i1, consider the circuit below.
2:
1:
i1
10V
+
5i1
4:
i1
1:
+
10V
10i1
2:
+
4:
-10 + 10i1 + 7i1 = 0, therefore i1 = (10/17) A
For i2, consider the circuit below.
i2
1:
2: i
o
+
10i2
2A
4:
io
1:
+
2V
10i2
2:
+
4:
-2 + 10i2 + 7io = 0, but i2 + 2 = io. Hence,
-2 + 10i2 +7i2 + 14 = 0, or i2 = (-12/17) A
vx = 1xix = 1(i1 + i2) = (10/17) –– (12/17) = -2/17 = -117.6 mA
Chapter 4, Solution 19.
Let vx = v1 + v2, where v1 and v2 are due to the 4-A and 6-A sources respectively.
v1
ix
ix
v2
+
2:
4A
8:
v1
+
2:
6A 8:
+
+
4ix
4ix
(a)
(b)
v2
To find v1, consider the circuit in Fig. (a).
v1/8 = 4 + (-4ix –– v1)/2
-ix = (-4ix –– v1)/2 and we have -2ix = v1. Thus,
But,
v1/8 = 4 + (2v1 –– v1)/8, which leads to v1 = -32/3
To find v2, consider the circuit shown in Fig. (b).
v2/2 = 6 + (4ix –– v2)/8
But ix = v2/2 and 2ix = v2. Therefore,
v2/2 = 6 + (2v2 –– v2)/8 which leads to v2 = -16
Hence,
vx = ––(32/3) –– 16 = -26.67 V
Chapter 4, Solution 20.
Transform the voltage sources and obtain the circuit in Fig. (a). Combining the 6-ohm
and 3-ohm resistors produces a 2-ohm resistor (6||3 = 2). Combining the 2-A and 4-A
sources gives a 6-A source. This leads to the circuit shown in Fig. (b).
i
i
6:
2A
2:
3: 4A
2:
(a)
From Fig. (b),
2:
6A
(b)
i = 6/2 = 3 A
Chapter 4, Solution 21.
To get io, transform the current sources as shown in Fig. (a).
io
6:
3:
+
12V
+
i
6V 2 A
6:
3:
+
vo 2 A
(a)
(b)
From Fig. (a),
-12 + 9io + 6 = 0, therefore io = 666.7 mA
To get vo, transform the voltage sources as shown in Fig. (b).
i = [6/(3 + 6)](2 + 2) = 8/3
vo = 3i = 8 V
Chapter 4, Solution 22.
We transform the two sources to get the circuit shown in Fig. (a).
5:
+ 10V
5:
4:
10:
2A
(a)
i
1A
10:
4:
10:
2A
(b)
We now transform only the voltage source to obtain the circuit in Fig. (b).
10||10 = 5 ohms, i = [5/(5 + 4)](2 –– 1) = 5/9 = 555.5 mA
Chapter 4, Solution 23
If we transform the voltage source, we obtain the circuit below.
8:
10 :
6:
3:
5A
3A
3//6 = 2-ohm. Convert the current sources to voltages sources as shown below.
10 :
8:
+
+
10V
-
30V
-
Applying KVL to the loop gives
30 10 I (10 8 2) 0

o
p VI
2:
I 2R
8W
I
1A
Chapter 4, Solution 24
Convert the current source to voltage source.
16 :
1:
4:
5:
+
+
48 V
10 :
+
12 V
-
-
Vo
-
Combine the 16-ohm and 4-ohm resistors and convert both voltages sources to current
Sources. We obtain the circuit below.
1:
20 :
2.4A
5:
2.4A
10 :
Combine the resistors and current sources.
20//5 = (20x5)/25 = 4 : , 2.4 + 2.4 = 4.8 A
Convert the current source to voltage source. We obtain the circuit below.
4:
+
19.2V
Using voltage division,
Vo
10
(19.2) 12.8 V
10 4 1
1:
+
Vo
-
10 :
Chapter 4, Solution 25.
Transforming only the current source gives the circuit below.
18 V
9:
+
12V
+
5:
i
4:
vo
+
+
30 V
+
2:
30 V
Applying KVL to the loop gives,
(4 + 9 + 5 + 2)i –– 12 –– 18 –– 30 –– 30 = 0
20i = 90 which leads to i = 4.5
vo = 2i = 9 V
Chapter 4, Solution 26.
Transform the voltage sources to current sources. The result is shown in Fig. (a),
30||60 = 20 ohms,
30||20 = 12 ohms
10 :
+ vx 3A
30:
60:
30:
6A
20:
(a)
20 :
+
10 :
+ vx 60V
i
(b)
12 :
+
96V
2A
Combining the resistors and transforming the current sources to voltage sources, we
obtain the circuit in Fig. (b). Applying KVL to Fig. (b),
42i –– 60 + 96 = 0, which leads to i = -36/42
vx = 10i = -8.571 V
Chapter 4, Solution 27.
Transforming the voltage sources to current sources gives the circuit in Fig. (a).
10||40 = 8 ohms
Transforming the current sources to voltage sources yields the circuit in Fig. (b).
Applying KVL to the loop,
-40 + (8 + 12 + 20)i + 200 = 0 leads to i = -4
vx 12i = -48 V
12 :
+ vx 5A
10:
40:
8A
20:
2A
(a)
8:
+
12 :
+ vx 40V
i
(b)
20 :
+
200V
Chapter 4, Solution 28.
Transforming only the current sources leads to Fig. (a). Continuing with source
transformations finally produces the circuit in Fig. (d).
io
12 V
+
4:
3:
12 V
2:
10 V
5:
+
+
10:
(a)
io
+
4:
12V
10 :
+
10:
22 V
(b)
io
+
4:
12V
io
10:
10:
(c)
2.2A
+
12V
4:
5:
io
(d)
Applying KVL to the loop in fig. (d),
-12 + 9io + 11 = 0, produces io = 1/9 = 111.11 mA
11V
+
Chapter 4, Solution 29.
Transform the dependent voltage source to a current source as shown in
Fig. (a). 2||4 = (4/3) k ohms
4 k:
2 k:
2vo
(4/3) k:
+
1.5vo
3 mA
1 k:
i
3 mA
+
1 k:
+
vo
vo
(a)
(b)
It is clear that i = 3 mA which leads to vo = 1000i = 3 V
If the use of source transformations was not required for this problem, the actual answer
could have been determined by inspection right away since the only current that could
have flowed through the 1 k ohm resistor is 3 mA.
Chapter 4, Solution 30
Transform the dependent current source as shown below.
ix
+
12V
-
24 :
60 :
30 :
10 :
+
-
7ix
Combine the 60-ohm with the 10-ohm and transform the dependent source as shown
below.
24 :
ix
+
12V
-
30 :
70 :
0.1ix
Combining 30-ohm and 70-ohm gives 30//70 = 70x30/100 = 21-ohm. Transform the
dependent current source as shown below.
24 :
ix
21 :
+
12V
-
+
-
2.1ix
Applying KVL to the loop gives
45i x 12 2.1i x
0

o
12
47.1
ix
254.8 mA
Chapter 4, Solution 31.
Transform the dependent source so that we have the circuit in
Fig. (a). 6||8 = (24/7) ohms. Transform the dependent source again to get the circuit in
Fig. (b).
3:
+
12V
+
vx
8:
vx/3
6:
(a)
3:
+
12V
+
vx
(24/7) :
i
(b)
+
(8/7)vx
From Fig. (b),
vx = 3i, or i = vx/3.
Applying KVL,
-12 + (3 + 24/7)i + (24/21)vx = 0
12 = [(21 + 24)/7]vx/3 + (8/7)vx, leads to vx = 84/23 = 3.625 V
Chapter 4, Solution 32.
As shown in Fig. (a), we transform the dependent current source to a voltage source,
15 :
10 :
5ix
+
60V
+
50 :
40 :
(a)
15 :
60V
+
50 :
50 :
0.1ix
(b)
ix
60V
+
15 :
25 :
ix
(c)
2.5ix
In Fig. (b), 50||50 = 25 ohms. Applying KVL in Fig. (c),
-60 + 40ix –– 2.5ix = 0, or ix = 1.6 A
Chapter 4, Solution 33.
(a)
RTh = 10||40 = 400/50 = 8 ohms
VTh = (40/(40 + 10))20 = 16 V
(b)
RTh = 30||60 = 1800/90 = 20 ohms
2 + (30 –– v1)/60 = v1/30, and v1 = VTh
120 + 30 –– v1 = 2v1, or v1 = 50 V
VTh = 50 V
Chapter 4, Solution 34.
To find RTh, consider the circuit in Fig. (a).
10 :
10 :
20 :
RTh
40 :
3A
+
v1
20 :
v2
+
40V
40 :
VTh
(a)
(b)
RTh = 20 + 10||40 = 20 + 400/50 = 28 ohms
To find VTh, consider the circuit in Fig. (b).
At node 1,
At node 2,
(40 –– v1)/10 = 3 + [(v1 –– v2)/20] + v1/40, 40 = 7v1 –– 2v2
3 + (v1- v2)/20 = 0, or v1 = v2 –– 60
Solving (1) and (2),
v1 = 32 V, v2 = 92 V, and VTh = v2 = 92 V
(1)
(2)
Chapter 4, Solution 35.
To find RTh, consider the circuit in Fig. (a).
RTh = Rab = 6||3 + 12||4 = 2 + 3 =5 ohms
To find VTh, consider the circuit shown in Fig. (b).
RTh
a
6:
b
3:
12 :
4:
(a)
2A
6:
+
v1
v2 4 :
+ VTh
+
12V v1
+
3:
12:
v2
+
19V
At node 1,
(b)
2 + (12 –– v1)/6 = v1/3, or v1 = 8
At node 2,
(19 –– v2)/4 = 2 + v2/12, or v2 = 33/4
But,
-v1 + VTh + v2 = 0, or VTh = v1 –– v2 = 8 –– 33/4 = -0.25
a +
vo
b
10 :
RTh = 5 :
+
VTh = (-1/4)V
vo = VTh/2 = -0.25/2 = -125 mV
Chapter 4, Solution 36.
Remove the 30-V voltage source and the 20-ohm resistor.
a
RTh
10:
a
10:
+
+
40:
VTh
40:
50V
b
b
(a)
(b)
From Fig. (a),
RTh = 10||40 = 8 ohms
From Fig. (b),
VTh = (40/(10 + 40))50 = 40V
8:
+
i
a
12 :
40V
+
30V
b
(c)
The equivalent circuit of the original circuit is shown in Fig. (c). Applying KVL,
30 –– 40 + (8 + 12)i = 0, which leads to i = 500mA
Chapter 4, Solution 37
RN is found from the circuit below.
20 :
a
40 :
12 :
b
R N 12 //( 20 40) 10:
IN is found from the circuit below.
2A
20 :
a
40 :
+
120V
-
12 :
IN
b
Applying source transformation to the current source yields the circuit below.
20 :
40 :
+ 80 V -
+
120V
-
Applying KVL to the loop yields
120 80 60 I N 0

o
IN
IN
40 / 60
0.6667 A
Chapter 4, Solution 38
We find Thevenin equivalent at the terminals of the 10-ohm resistor. For RTh, consider
the circuit below.
1:
4:
5:
RTh
16 :
RTh 1 5 //( 4 16) 1 4
For VTh, consider the circuit below.
V1
4:
5:
1:
V2
5:
3A
+
16 :
VTh
+
12 V
At node 1,
V1 V1 V2
3

o
48 5V1 4V2
16
4
At node 2,
V1 V2 12 V2
0

o
48 5V1 9V2
4
5
Solving (1) and (2) leads to
VTh V2 19.2
-
(1)
(2)
Thus, the given circuit can be replaced as shown below.
5:
+
19.2V
-
+
Vo
-
10 :
Using voltage division,
Vo
10
(19.2) 12.8 V
10 5
Chapter 4, Solution 39.
To find RTh, consider the circuit in Fig. (a).
10 :
3vab10 : a
+
+
10 :
+
io
+
1V v1 50V
+ VTh
+
40 :4V
+
8 A 2A
v2
a
+
b
(b)
(a)
(b)
1 –– 3 + 10io = 0, or io = 0.4
RTh = 1/io = 2.5 ohms
To find VTh, consider the circuit shown in Fig. (b).
[(4 –– v)/10] + 2 = 0, or v = 24
But,
vab = VTh
40V
b
-
+
3vab
20
:
v
+
v = VTh + 3vab = 4VTh = 24, which leads to VTh = 6 V
Chapter 4, Solution 40.
To find RTh, consider the circuit in Fig. (a).
10 :
RTh
a
b
40:
(a)
20 :
RTh = 10||40 + 20 = 28 ohms
To get VTh, consider the circuit in Fig. (b). The two loops are independent. From loop 1,
v1 = (40/50)50 = 40 V
For loop 2,
But,
-v2 + 20x8 + 40 = 0, or v2 = 200
VTh + v2 –– v1 = 0,
VTh = v1 = v2 = 40 –– 200 = -160 volts
This results in the following equivalent circuit.
28 :
+
-160V
+
vx
12 :
vx = [12/(12 + 28)](-160) = -48 V
Chapter 4, Solution 41
To find RTh, consider the circuit below
14 :
a
6:
5:
b
RTh 5 //(14 6) 4: R N
Applying source transformation to the 1-A current source, we obtain the circuit below.
6:
14 :
- 14V +
VTh
a
+
6V
5:
3A
b
At node a,
14 6 VTh
6 14
VTh
RTh
IN
Thus,
RTh R N
3
VTh
5
(8) / 4
4:,

o
VTh
8 V
2 A
8V,
VTh
2 A
IN
Chapter 4, Solution 42.
To find RTh, consider the circuit in Fig. (a).
20 :
10 :
30 :
a
20 :
30 : 30 :
b
a
10 :
10:
(a)
10 :
b
10 :
10 :
(b)
20||20 = 10 ohms. Transform the wye sub-network to a delta as shown in Fig. (b).
10||30 = 7.5 ohms. RTh = Rab = 30||(7.5 + 7.5) = 10 ohms.
To find VTh, we transform the 20-V and the 5-V sources. We obtain the circuit shown in
Fig. (c).
a
10 :
+
b
+
10 :
10 :
i1
30V
10 V
10 :
+
10 :
i2
+
50V
(c)
For loop 1,
-30 + 50 + 30i1 –– 10i2 = 0, or -2 = 3i1 –– i2
(1)
For loop 2,
-50 –– 10 + 30i2 –– 10i1 = 0, or 6 = -i1 + 3i2
(2)
Solving (1) and (2),
i1 = 0, i2 = 2 A
Applying KVL to the output loop, -vab –– 10i1 + 30 –– 10i2 = 0, vab = 10 V
VTh = vab = 10 volts
Chapter 4, Solution 43.
To find RTh, consider the circuit in Fig. (a).
RTh
a
10:
b
10:
5:
(a)
10 :
+
a
+
50V va
b
+ VTh
10 :
+
vb
(b)
RTh = 10||10 + 5 = 10 ohms
5:
2A
To find VTh, consider the circuit in Fig. (b).
vb = 2x5 = 10 V, va = 20/2 = 10 V
But,
-va + VTh + vb = 0, or VTh = va –– vb = 0 volts
Chapter 4, Solution 44.
(a)
For RTh, consider the circuit in Fig. (a).
RTh = 1 + 4||(3 + 2 + 5) = 3.857 ohms
For VTh, consider the circuit in Fig. (b). Applying KVL gives,
10 –– 24 + i(3 + 4 + 5 + 2), or i = 1
VTh = 4i = 4 V
3:
3:
1:
a
+
+
2:
2:
i
b
5:
+
VTh
b
10V
5:
(b)
(a)
(b)
a
4:
24V
RTh
4:
1:
For RTh, consider the circuit in Fig. (c).
3:
1:
4:
3:
b
24V
2:
RTh
5:
1:
4:
vo
+
+
2:
5:
2A
c
(c)
b
VTh
c
(d)
RTh = 5||(2 + 3 + 4) = 3.214 ohms
To get VTh, consider the circuit in Fig. (d). At the node, KCL gives,
[(24 –– vo)/9] + 2 = vo/5, or vo = 15
VTh = vo = 15 V
Chapter 4, Solution 45.
For RN, consider the circuit in Fig. (a).
6:
6:
6:
4:
RN
4A
6:
(a)
IN
4:
(b)
RN = (6 + 6)||4 = 3 ohms
For IN, consider the circuit in Fig. (b). The 4-ohm resistor is shorted so that 4-A current
is equally divided between the two 6-ohm resistors. Hence,
IN = 4/2 = 2 A
Chapter 4, Solution 46.
(a)
RN = RTh = 8 ohms. To find IN, consider the circuit in Fig. (a).
10 :
20V
+
60 :
4:
Isc
2A
(a)
(b)
IN = Isc = 20/10 = 2 A
(b)
To get IN, consider the circuit in Fig. (b).
IN = Isc = 2 + 30/60 = 2.5 A
30 :
IN 30V
+
Chapter 4, Solution 47
Since VTh = Vab = Vx, we apply KCL at the node a and obtain
30 VTh VTh
2VTh

o VTh
12
60
To find RTh, consider the circuit below.
12 :
150 / 126 1.19 V
Vx
a
2Vx
60 :
1A
At node a, KCL gives
V V
1 2V x x x

o V x 60 / 126 0.4762
60 12
Vx
VTh
RTh
IN
0.4762:,
1.19 / 0.4762 2.5
RTh
1
Thus,
VTh 1.19V ,
RTh R N 0.4762:,
I N 2 .5 A
Chapter 4, Solution 48.
To get RTh, consider the circuit in Fig. (a).
10Io
10Io
+
2:
+
1A
2A
(a)
From Fig. (a),
VTh
4:
V
4:
+
Io
+
Io
(b)
Io = 1,
2:
6 –– 10 –– V = 0, or V = -4
RN = RTh = V/1 = -4 ohms
To get VTh, consider the circuit in Fig. (b),
Io = 2, VTh = -10Io + 4Io = -12 V
IN = VTh/RTh = 3A
Chapter 4, Solution 49.
RN = RTh = 28 ohms
To find IN, consider the circuit below,
3A
10 :
40V
At the node,
+
vo
20 :
io
Isc = IN
40 :
(40 –– vo)/10 = 3 + (vo/40) + (vo/20), or vo = 40/7
io = vo/20 = 2/7, but IN = Isc = io + 3 = 3.286 A
Chapter 4, Solution 50.
From Fig. (a), RN = 6 + 4 = 10 ohms
6:
6:
Isc = IN
4:
(a)
From Fig. (b),
2A
4:
+
12V (b)
2 + (12 –– v)/6 = v/4, or v = 9.6 V
-IN = (12 –– v)/6 = 0.4, which leads to IN = -0.4 A
Combining the Norton equivalent with the right-hand side of the original circuit produces
the circuit in Fig. (c).
i
10 :
0.4A
5:
4A
(c)
i = [10/(10 + 5)] (4 –– 0.4) = 2.4 A
Chapter 4, Solution 51.
(a)
From the circuit in Fig. (a),
RN = 4||(2 + 6||3) = 4||4 = 2 ohms
RTh
6:
VTh
+
4:
6:
3:
2:
120V
4:
+
3:
(a)
6A
2:
(b)
For IN or VTh, consider the circuit in Fig. (b). After some source transformations, the
circuit becomes that shown in Fig. (c).
+ VTh
2:
40V
+
4:
i
2:
12V
+
(c)
Applying KVL to the circuit in Fig. (c),
-40 + 8i + 12 = 0 which gives i = 7/2
VTh = 4i = 14 therefore IN = VTh/RN = 14/2 = 7 A
(b)
To get RN, consider the circuit in Fig. (d).
RN = 2||(4 + 6||3) = 2||6 = 1.5 ohms
6:
4:
2:
i
+
3:
RN
2:
VTh
(d)
12V
+
(e)
To get IN, the circuit in Fig. (c) applies except that it needs slight modification as in
Fig. (e).
i = 7/2, VTh = 12 + 2i = 19, IN = VTh/RN = 19/1.5 = 12.667 A
Chapter 4, Solution 52.
For RTh, consider the circuit in Fig. (a).
a
3 k:
Io
20Io
2 k:
RTh
b
(a)
3 k:
6V
+
a
+
Io
20Io
2 k:
VTh
b
(b)
For Fig. (a), Io = 0, hence the current source is inactive and
RTh = 2 k ohms
For VTh, consider the circuit in Fig. (b).
Io = 6/3k = 2 mA
VTh = (-20Io)(2k) = -20x2x10-3x2x103 = -80 V
Chapter 4, Solution 53.
To get RTh, consider the circuit in Fig. (a).
0.25vo
0.25vo
2:
2:
a
+
6:
3:
+
vo
2:
1A
1/2
a
1/2
vo
vab
b
b
(a)
(b)
From Fig. (b),
vo = 2x1 = 2V, -vab + 2x(1/2) +vo = 0
vab = 3V
RN = vab/1 = 3 ohms
To get IN, consider the circuit in Fig. (c).
0.25vo
6:
2:
a
+
18V
+
3:
vo
Isc = IN
b
(c)
[(18 –– vo)/6] + 0.25vo = (vo/2) + (vo/3) or vo = 4V
But,
(vo/2) = 0.25vo + IN, which leads to IN = 1 A
+
1A
Chapter 4, Solution 54
To find VTh =Vx, consider the left loop.
3 1000io 2V x 0

o
For the right loop,
V x 50 x 40i o 2000io
Combining (1) and (2),
3 1000io 4000io 3000io
Vx
2000io

o
2
3 1000io 2V x
(1)
(2)

o
VTh
io
1mA
2
To find RTh, insert a 1-V source at terminals a-b and remove the 3-V independent
source, as shown below.
1 k:
ix
.
+
2Vx
-
io
Vx
1,
io
ix
40io 1
ix
RTh
Vx
50
2V x
1000
40io
+
1V
-
50 :
2mA
80mA 1 / 0.060
+
Vx
-
1
A
50
-60mA
16.67:
Chapter 4, Solution 55.
To get RN, apply a 1 mA source at the terminals a and b as shown in Fig. (a).
a
I
vab/1000
8 k:
+
80I
+
50 k:
vab
b
(a)
1mA
We assume all resistances are in k ohms, all currents in mA, and all voltages in volts. At
node a,
(vab/50) + 80I = 1
(1)
Also,
-8I = (vab/1000), or I = -vab/8000
(2)
From (1) and (2),
(vab/50) –– (80vab/8000) = 1, or vab = 100
RN = vab/1 = 100 k ohms
To get IN, consider the circuit in Fig. (b).
8 k:
2V
+
a
I
vab/1000
+
80I
+
50 k:
IN
vab
b
(b)
Since the 50-k ohm resistor is shorted,
IN = -80I, vab = 0
Hence,
8i = 2 which leads to I = (1/4) mA
IN = -20 mA
Chapter 4, Solution 56.
We first need RN and IN.
1:
20V
2:
a
2A
+
1:
4:
4:
RN
b
(a)
2:
IN
(b)
+
16V
To find RN, consider the circuit in Fig. (a).
RN = 1 + 2||4 = (7/3) ohms
To get IN, short-circuit ab and find Isc from the circuit in Fig. (b). The current source can
be transformed to a voltage source as shown in Fig. (c).
vo
20V
+
1:
4:
a
2V
2:
+
+
IN
i
16V
RN
IN
(c)
(d)
3:
b
(20 –– vo)/2 = [(vo + 2)/1] + [(vo + 16)/4], or vo = 16/7
IN = (vo + 2)/1 = 30/7
From the Norton equivalent circuit in Fig. (d),
i = RN/(RN + 3), IN = [(7/3)/((7/3) + 3)](30/7) = 30/16 = 1.875 A
Chapter 4, Solution 57.
To find RTh, remove the 50V source and insert a 1-V source at a –– b, as shown in Fig. (a).
2:
B
a
A
i
+
3:
vx
6:
0.5vx
(a)
10 :
+
1V
b
We apply nodal analysis. At node A,
At node B,
i + 0.5vx = (1/10) + (1 –– vx)/2, or i + vx = 0.6
(1 –– vo)/2 = (vx/3) + (vx/6), and vx = 0.5
(1)
(2)
From (1) and (2),
i = 0.1 and
RTh = 1/i = 10 ohms
To get VTh, consider the circuit in Fig. (b).
3:
2:
v1
v2
a
+
50V
+
+
6:
vx
10 : VTh
0.5vx
b
(b)
At node 1,
(50 –– v1)/3 = (v1/6) + (v1 –– v2)/2, or 100 = 6v1 –– 3v2
(3)
At node 2,
0.5vx + (v1 –– v2)/2 = v2/10, vx = v1, and v1 = 0.6v2
(4)
From (3) and (4),
v2 = VTh = 166.67 V
IN = VTh/RTh = 16.667 A
RN = RTh = 10 ohms
Chapter 4, Solution 58.
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 RN =
infinity. IN can be found by solving for Isc.
ib
VS
R1
+
E ib
vo
R2
Writing the node equation at node vo,
ib + Eib = vo/R2 = (1 + E)ib
Isc
But
ib = (Vs –– vo)/R1
vo = Vs –– ibR1
Vs –– ibR1 = (1 + E)R2ib, or ib = Vs/(R1 + (1 + E)R2)
Isc = IN = -Eib = -EVs/(R1 + (1 + E)R2)
Chapter 4, Solution 59.
RTh = (10 + 20)||(50 + 40) 30||90 = 22.5 ohms
To find VTh, consider the circuit below.
i1
i2
10 :
20 :
+ VTh
8A
50 :
40 :
i1 = i2 = 8/2 = 4, 10i1 + VTh –– 20i2 = 0, or VTh = 20i2 ––10i1 = 10i1 = 10x4
VTh = 40V, and IN = VTh/RTh = 40/22.5 = 1.7778 A
Chapter 4, Solution 60.
The circuit can be reduced by source transformations.
2A
18 V
12 V
+ + 10 V
+ 10 :
5:
2A
10 :
a
b
3A
5:
2A
3A
a
3.333:
b
Norton Equivalent Circuit
10 V
3.333:
a
+ Thevenin Equivalent Circuit
Chapter 4, Solution 61.
To find RTh, consider the circuit in Fig. (a).
Let
R = 2||18 = 1.8 ohms,
RTh = 2R||R = (2/3)R = 1.2 ohms.
To get VTh, we apply mesh analysis to the circuit in Fig. (d).
2:
a
6:
6:
2:
2:
6:
b
(a)
b
2:
a
18 :
1.8 :
2:
18 :
a
2:
18 :
1.8 :
1.8 :
RTh
b
b
(b)
(c)
2:
a
6:
12V
6:
i3
+
12V
+
+
VTh
6:
2:
i1
+
i2
2:
12V
b
(d)
-12 –– 12 + 14i1 –– 6i2 –– 6i3 = 0, and 7 i1 –– 3 i2 –– 3i3 = 12
(1)
12 + 12 + 14 i2 –– 6 i1 –– 6 i3 = 0, and -3 i1 + 7 i2 –– 3 i3 = -12
(2)
14 i3 –– 6 i1 –– 6 i2 = 0, and
(3)
-3 i1 –– 3 i2 + 7 i3 = 0
This leads to the following matrix form for (1), (2) and (3),
ª 7 3 3º ª i1 º
« 3 7 3» «i »
«
»« 2 »
«¬ 3 3 7 »¼ «¬i 3 »¼
ª 12 º
« 12»
«
»
«¬ 0 »¼
3 3
7
'
3 7 3
3 3 7
7
'2
100 ,
3
12
3 12 3
3 0
7
120
i2 = '/'2 = -120/100 = -1.2 A
VTh = 12 + 2i2 = 9.6 V, and IN = VTh/RTh = 8 A
Chapter 4, Solution 62.
Since there are no independent sources, VTh = 0 V
To obtain RTh, consider the circuit below.
0.1io
ix
2
+
vo
1
40 :
10 :
v1
io
2vo
VS
+
20 :
+ At node 2,
At node 1,
ix + 0.1io = (1 –– v1)/10, or 10ix + io = 1 –– v1
(1)
(v1/20) + 0.1io = [(2vo –– v1)/40] + [(1 –– v1)/10]
(2)
But io = (v1/20) and vo = 1 –– v1, then (2) becomes,
1.1v1/20 = [(2 –– 3v1)/40] + [(1 –– v1)/10]
2.2v1 = 2 –– 3v1 + 4 –– 4v1 = 6 –– 7v1
or
v1 = 6/9.2
(3)
From (1) and (3),
10ix + v1/20 = 1 –– v1
10ix = 1 –– v1 –– v1/20 = 1 –– (21/20)v1 = 1 –– (21/20)(6/9.2)
ix = 31.52 mA, RTh = 1/ix = 31.73 ohms.
Chapter 4, Solution 63.
Because there are no independent sources, IN = Isc = 0 A
RN can be found using the circuit below.
10 :
+
vo
20 :
Applying KCL at node 1,
3:
v1
0.5vo
io
+
1V
0.5vo + (1 –– v1)/3 = v1/30, but vo = (20/30)v1
Hence, 0.5(2/3)(30)v1 + 10 –– 10v1 =v1, or v1 = 10 and io = (1 –– v1)/3 = -3
RN = 1/io = -1/3 = -333.3 m ohms
Chapter 4, Solution 64.
With no independent sources, VTh = 0 V. To obtain RTh, consider the circuit
shown below.
4:
1:
vo
io
ix
+
––
2:
+
1V
10ix
ix = [(1 –– vo)/1] + [(10ix –– vo)/4], or 2vo = 1 + 3ix
But ix = vo/2. Hence,
2vo = 1 + 1.5vo, or vo = 2, io = (1 –– vo)/1 = -1
Thus, RTh = 1/io = -1 ohm
(1)
Chapter 4, Solution 65
At the terminals of the unknown resistance, we replace the circuit by its Thevenin
equivalent.
12
RTh 2 4 // 12 2 3 5:,
VTh
(32) 24 V
12 4
Thus, the circuit can be replaced by that shown below.
5:
Io
+
24 V
-
+
Vo
-
Applying KVL to the loop,
24 5I o Vo
0

o
Vo
24 5I o
Chapter 4, Solution 66.
We first find the Thevenin equivalent at terminals a and b. We find RTh using the circuit
in Fig. (a).
2:
10V
+
3:
2:
a
b
+
3:
a
VTh
5:
b
RTh
+
5:
20V
i
30V
(a)
+
(b)
RTh = 2||(3 + 5) = 2||8 = 1.6 ohms
By performing source transformation on the given circuit, we obatin the circuit in (b).
We now use this to find VTh.
10i + 30 + 20 + 10 = 0, or i = -5
VTh + 10 + 2i = 0, or VTh = 2 V
p = VTh2/(4RTh) = (2)2/[4(1.6)] = 625 m watts
Chapter 4, Solution 67.
We need to find the Thevenin equivalent at terminals a and b.
From Fig. (a),
RTh = 4||6 + 8||12 = 2.4 + 4.8 = 7.2 ohms
From Fig. (b),
10i1 –– 30 = 0, or i1 = 3
+
4:
4:
6:
8:
6:
30V
+ RTh
12 :
i1
VTh
+
12 :
+
i2
8:
(a)
(b)
20i2 + 30 = 0, or i2 = 1.5, VTh = 6i1 + 8i2 = 6x3 –– 8x1.5 = 6 V
For maximum power transfer,
p = VTh2/(4RTh) = (6)2/[4(7.2)] = 1.25 watts
Chapter 4, Solution 68.
This is a challenging problem in that the load is already specified. This now becomes a
"minimize losses" style problem. When a load is specified and internal losses can be
adjusted, then the objective becomes, reduce RThev as much as possible, which will result
in maximum power transfer to the load.
+
-
+
-
Removing the 10 ohm resistor and solving for the Thevenin Circuit results in:
RTh = (Rx20/(R+20)) and a Voc = VTh = 12x(20/(R +20)) + (-8)
As R goes to zero, RTh goes to zero and VTh goes to 4 volts, which produces the
maximum power delivered to the 10-ohm resistor.
P = vi = v2/R = 4x4/10 = 1.6 watts
Notice that if R = 20 ohms which gives an RTh = 10 ohms, then VTh becomes -2 volts and
the power delivered to the load becomes 0.1 watts, much less that the 1.6 watts.
It is also interesting to note that the internal losses for the first case are 122/20 = 7.2 watts
and for the second case are = to 12 watts. This is a significant difference.
Chapter 4, Solution 69.
We need the Thevenin equivalent across the resistor R. To find RTh, consider the circuit
below.
22 k: v1
+
10 k:
vo
40 k:
3vo
30 k:
Assume that all resistances are in k ohms and all currents are in mA.
1mA
10||40 = 8, and 8 + 22 = 30
1 + 3vo = (v1/30) + (v1/30) = (v1/15)
15 + 45vo = v1
But vo = (8/30)v1, hence,
15 + 45x(8v1/30) v1, which leads to v1 = 1.3636
RTh = v1/1 = -1.3636 k ohms
To find VTh, consider the circuit below.
10 k: vo 22 k:
v1
+
100V
+
+
40 k:
vo
3vo
30 k:
VTh
(100 –– vo)/10 = (vo/40) + (vo –– v1)/22
(1)
[(vo –– v1)/22] + 3vo = (v1/30)
(2)
Solving (1) and (2),
v1 = VTh = -243.6 volts
p = VTh2/(4RTh) = (243.6)2/[4(-1363.6)] = -10.882 watts
Chapter 4, Solution 70
We find the Thevenin equivalent across the 10-ohm resistor. To find VTh, consider the
circuit below.
3Vx
5:
5:
+
+
15 :
4V
-
VTh
6:
+
Vx
-
From the figure,
15
(4) 3V
15 5
consider the circuit below:
0,
Vx
To find RTh,
VTh
3Vx
5:
5:
V2
V1
+
15 :
4V
-
+
At node 1,
V V V2
4 V1
3V x 1 1
,
5
15
5
Vx
1A
6:
Vx
6 x1 6
-

o
258
3V2 7V1
(1)
At node 2,
V V2
1 3V x 1
0

o V1 V2 95
5
Solving (1) and (2) leads to V2 = 101.75 V
2
VTh
V2
RTh
101.75:,
p max
1
4 RTh
(2)
9
4 x101.75
22.11 mW
Chapter 4, Solution 71.
We need RTh and VTh at terminals a and b. To find RTh, we insert a 1-mA source at the
terminals a and b as shown below.
10 k:
a
+
3 k:
vo
+
1 k:
120vo
40 k:
1mA
b
Assume that all resistances are in k ohms, all currents are in mA, and all voltages are in
volts. At node a,
1 = (va/40) + [(va + 120vo)/10], or 40 = 5va + 480vo
(1)
The loop on the left side has no voltage source. Hence, vo = 0. From (1), va = 8 V.
RTh = va/1 mA = 8 kohms
To get VTh, consider the original circuit. For the left loop,
vo = (1/4)8 = 2 V
For the right loop,
vR = VTh = (40/50)(-120vo) = -192
The resistance at the required resistor is
R = RTh = 8 kohms
p = VTh2/(4RTh) = (-192)2/(4x8x103) = 1.152 watts
Chapter 4, Solution 72.
(a)
RTh and VTh are calculated using the circuits shown in Fig. (a) and (b)
respectively.
From Fig. (a),
RTh = 2 + 4 + 6 = 12 ohms
From Fig. (b),
-VTh + 12 + 8 + 20 = 0, or VTh = 40 V
4:
6:
2:
4:
12V
6:
+
2:
+
RTh
VTh
8V
20V
(a)
(b)
i = VTh/(RTh + R) = 40/(12 + 8) = 2A
(c)
For maximum power transfer,
(d)
p = VTh2/(4RTh) = (40)2/(4x12) = 33.33 watts.
RL = RTh = 12 ohms
Chapter 4, Solution 73
Find the Thevenin’’s equivalent circuit across the terminals of R.
25 :
RTh
20 :
RTh
+ (b)
10 :
10 // 20 25 // 5
325 / 30 10.833:
+
5:
10 :
+
60 V
-
+ VTh -
+
Va
25 :
+
20 :
5:
20
(60) 40,
30
Va VTh Vb 0
Va
2
p max
VTh
4 RTh
Vb
-
5
(60) 10
30

o VTh Va Vb
Vb
30 2
4 x10.833
40 10
30 V
20.77 W
Chapter 4, Solution 74.
When RL is removed and Vs is short-circuited,
RTh = R1||R2 + R3||R4 = [R1 R2/( R1 + R2)] + [R3 R4/( R3 + R4)]
RL = RTh = (R1 R2 R3 + R1 R2 R4 + R1 R3 R4 + R2 R3 R4)/[( R1 + R2)( R3 + R4)]
When RL is removed and we apply the voltage division principle,
Voc = VTh = vR2 –– vR4
= ([R2/(R1 + R2)] –– [R4/(R3 + R4)])Vs = {[(R2R3) –– (R1R4)]/[(R1 + R2)(R3 + R4)]}Vs
pmax = VTh2/(4RTh)
= {[(R2R3) –– (R1R4)]2/[(R1 + R2)(R3 + R4)]2}Vs2[( R1 + R2)( R3 + R4)]/[4(a)]
where a = (R1 R2 R3 + R1 R2 R4 + R1 R3 R4 + R2 R3 R4)
pmax =
[(R2R3) –– (R1R4)]2Vs2/[4(R1 + R2)(R3 + R4) (R1 R2 R3 + R1 R2 R4 + R1 R3 R4 + R2 R3 R4)]
Chapter 4, Solution 75.
We need to first find RTh and VTh.
R
R
R
R
R
R
RTh
+
1V
(a)
vo
+
2V
+
+
3V
VTh
(b)
Consider the circuit in Fig. (a).
(1/RTh) = (1/R) + (1/R) + (1/R) = 3/R
RTh = R/3
From the circuit in Fig. (b),
((1 –– vo)/R) + ((2 –– vo)/R) + ((3 –– vo)/R) = 0
vo = 2 = VTh
For maximum power transfer,
RL = RTh = R/3
Pmax = [(VTh)2/(4RTh)] = 3 mW
RTh = [(VTh)2/(4Pmax)] = 4/(4xPmax) = 1/Pmax = R/3
R = 3/(3x10-3) = 1 k ohms
Chapter 4, Solution 76.
Follow the steps in Example 4.14. The schematic and the output plots are shown below.
From the plot, we obtain,
V = 92 V [i = 0, voltage axis intercept]
R = Slope = (120 –– 92)/1 = 28 ohms
Chapter 4, Solution 77.
(a)
The schematic is shown below. We perform a dc sweep on a current source, I1,
connected between terminals a and b. We label the top and bottom of source I1 as 2 and
1 respectively. We plot V(2) –– V(1) as shown.
VTh = 4 V [zero intercept]
RTh = (7.8 –– 4)/1 = 3.8 ohms
(b)
Everything remains the same as in part (a) except that the current source, I1, is
connected between terminals b and c as shown below. We perform a dc sweep on
I1 and obtain the plot shown below. From the plot, we obtain,
V = 15 V [zero intercept]
R = (18.2 –– 15)/1 = 3.2 ohms
Chapter 4, Solution 78.
The schematic is shown below. We perform a dc sweep on the current source, I1,
connected between terminals a and b. The plot is shown. From the plot we obtain,
VTh = -80 V [zero intercept]
RTh = (1920 –– (-80))/1 = 2 k ohms
Chapter 4, Solution 79.
After drawing and saving the schematic as shown below, we perform a dc sweep on I1
connected across a and b. The plot is shown. From the plot, we get,
V = 167 V [zero intercept]
R = (177 –– 167)/1 = 10 ohms
Chapter 4, Solution 80.
The schematic in shown below. We label nodes a and b as 1 and 2 respectively. We
perform dc sweep on I1. In the Trace/Add menu, type v(1) –– v(2) which will result in the
plot below. From the plot,
VTh = 40 V [zero intercept]
RTh = (40 –– 17.5)/1 = 22.5 ohms [slope]
Chapter 4, Solution 81.
The schematic is shown below. We perform a dc sweep on the current source, I2,
connected between terminals a and b. The plot of the voltage across I2 is shown below.
From the plot,
VTh = 10 V [zero intercept]
RTh = (10 –– 6.4)/1 = 3.4 ohms.
Chapter 4, Solution 82.
VTh = Voc = 12 V, Isc = 20 A
RTh = Voc/Isc = 12/20 = 0.6 ohm.
0.6 :
i
12V
i = 12/2.6 ,
+
2:
p = i2R = (12/2.6)2(2) = 42.6 watts
Chapter 4, Solution 83.
VTh = Voc = 12 V, Isc = IN = 1.5 A
RTh = VTh/IN = 8 ohms, VTh = 12 V, RTh = 8 ohms
Chapter 4, Solution 84
Let the equivalent circuit of the battery terminated by a load be as shown below.
RTh
IL
+
+
VTh
VL
-
RL
-
For open circuit,
R L f,

o VTh Voc
When RL = 4 ohm, VL=10.5,
IL
VL
RL
10.8 / 4
VL
10.8 V
2 .7
But
VTh
VL I L RTh

o
RTh
VTh V L
IL
12 10.8
2 .7
0.4444:
Chapter 4, Solution 85
(a) Consider the equivalent circuit terminated with R as shown below.
RTh
a
+
VTh
-
+
Vab
-
R
b
Vab
R
VTh
R RTh

o
6
10
VTh
10 RTh
or
60 6 RTh 10VTh
where RTh is in k-ohm.
(1)
Similarly,
30

o
VTh
30 RTh
Solving (1) and (2) leads to
360 12 RTh
12
VTh
24 V, RTh
(b) Vab
(2)
30VTh
30k:
20
(24)
20 30
9.6 V
Chapter 4, Solution 86.
We replace the box with the Thevenin equivalent.
RTh
+
VTh
+
i
R
v
VTh = v + iRTh
When i = 1.5, v = 3, which implies that VTh = 3 + 1.5RTh
(1)
When i = 1, v = 8, which implies that VTh = 8 + 1xRTh
(2)
From (1) and (2), RTh = 10 ohms and VTh = 18 V.
(a)
When R = 4, i = VTh/(R + RTh) = 18/(4 + 10) = 1.2857 A
(b)
For maximum power, R = RTH
Pmax = (VTh)2/4RTh = 182/(4x10) = 8.1 watts
Chapter 4, Solution 87.
(a)
im = 9.975 mA
im = 9.876 mA
+
Is
vm
Rs Rm
Is
Rs
Rs Rm
(a)
(b)
From Fig. (a),
vm = Rmim = 9.975 mA x 20 = 0.1995 V
From Fig. (b),
Is = 9.975 mA + (0.1995/Rs)
(1)
vm = Rmim = 20x9.876 = 0.19752 V
Is = 9.876 mA + (0.19752/2k) + (0.19752/Rs)
= 9.975 mA + (0.19752/Rs)
Solving (1) and (2) gives,
Rs = 8 k ohms,
Is = 10 mA
(b)
im’’ = 9.876 mA
Is
Rs
Rs Rm
(b)
8k||4k = 2.667 k ohms
im’’ = [2667/(2667 + 20)](10 mA) = 9.926 mA
Chapter 4, Solution 88
To find RTh, consider the circuit below.
A
RTh
30k :
RTh
5k :
B
20k :
10k :
30 10 20 // 5 44k:
(2)
To find VTh , consider the circuit below.
5k :
A
B
io
30k :
+
20k :
4mA
60 V
-
10k :
VA
30 x 4 120,
VB
20
(60)
25
48,
VTh
V A VB
72 V
Chapter 4, Solution 89
It is easy to solve this problem using Pspice.
(a) The schematic is shown below. We insert IPROBE to measure the desired ammeter
reading. We insert a very small resistance in series IPROBE to avoid problem. After the
circuit is saved and simulated, the current is displaced on IPROBE as 99.99PA .
(b) By interchanging the ammeter and the 12-V voltage source, the schematic is shown
below. We obtain exactly the same result as in part (a).
Chapter 4, Solution 90.
Rx = (R3/R1)R2 = (4/2)R2 = 42.6, R2 = 21.3
which is (21.3ohms/100ohms)% = 21.3%
Chapter 4, Solution 91.
Rx = (R3/R1)R2
(a)
Since 0 < R2 < 50 ohms, to make 0 < Rx < 10 ohms requires that when R2
= 50 ohms, Rx = 10 ohms.
10 = (R3/R1)50 or R3 = R1/5
so we select R1 = 100 ohms and R3 = 20 ohms
(b)
For 0 < Rx < 100 ohms
100 = (R3/R1)50, or R3 = 2R1
So we can select R1 = 100 ohms and R3 = 200 ohms
Chapter 4, Solution 92.
For a balanced bridge, vab = 0. We can use mesh analysis to find vab. Consider the
circuit in Fig. (a), where i1 and i2 are assumed to be in mA.
2 k:
3 k:
220V
+
a
i1
+
6 k:
i2
b
vab
5 k:
10 k:
0
(a)
220 = 2i1 + 8(i1 –– i2) or 220 = 10i1 –– 8i2 (1)
From (1) and (2),
0 = 24i2 –– 8i1 or i2 = (1/3)i1
(2)
i1 = 30 mA and i2 = 10 mA
Applying KVL to loop 0ab0 gives
5(i2 –– i1) + vab + 10i2 = 0 V
Since vab = 0, the bridge is balanced.
When the 10 k ohm resistor is replaced by the 18 k ohm resistor, the gridge becomes
unbalanced. (1) remains the same but (2) becomes
Solving (1) and (3),
0 = 32i2 –– 8i1, or i2 = (1/4)i1
(3)
i1 = 27.5 mA, i2 = 6.875 mA
vab = 5(i1 –– i2) –– 18i2 = -20.625 V
VTh = vab = -20.625 V
To obtain RTh, we convert the delta connection in Fig. (b) to a wye connection shown in
Fig. (c).
2 k:
3 k:
R2
6 k:
a
RTh
R1
b
5 k:
6 k:
a RTh
R3
18 k:
(b)
18 k:
(c)
R1 = 3x5/(2 + 3 + 5) = 1.5 k ohms, R2 = 2x3/10 = 600 ohms,
R3 = 2x5/10 = 1 k ohm.
RTh = R1 + (R2 + 6)||(R3 + 18) = 1.5 + 6.6||9 = 6.398 k ohms
RL = RTh = 6.398 k ohms
Pmax = (VTh)2/(4RTh) = (20.625)2/(4x6.398) = 16.622 mWatts
Chapter 4, Solution 93.
ix
VS
+
Rs
Ro
ix
+
ERoix
-Vs + (Rs + Ro)ix + ERoix = 0
ix = Vs/(Rs + (1 + E)Ro)
b
Chapter 4, Solution 94.
(a)
Vo/Vg = Rp/(Rg + Rs + Rp)
(1)
Req = Rp||(Rg + Rs) = Rg
Rg = Rp(Rg + Rs)/(Rp + Rg + Rs)
RgRp + Rg2 + RgRs = RpRg + RpRs
RpRs = Rg(Rg + Rs)
From (1),
(2)
Rp/D = Rg + Rs + Rp
Rg + Rs = Rp((1/D) –– 1) = Rp(1 - D)/D
(1a)
Combining (2) and (1a) gives,
Rs = [(1 - D)/D]Req
= (1 –– 0.125)(100)/0.125 = 700 ohms
From (3) and (1a),
Rp(1 - D)/D = Rg + [(1 - D)/D]Rg = Rg/D
Rp = Rg/(1 - D) = 100/(1 –– 0.125) = 114.29 ohms
(b)
RTh
I
VTh
+
RL
VTh = Vs = 0.125Vg = 1.5 V
RTh = Rg = 100 ohms
I = VTh/(RTh + RL) = 1.5/150 = 10 mA
(3)
Chapter 4, Solution 95.
Let 1/sensitivity = 1/(20 k ohms/volt) = 50 PA
For the 0 –– 10 V scale,
Rm = Vfs/Ifs = 10/50 PA = 200 k ohms
For the 0 –– 50 V scale,
Rm = 50(20 k ohms/V) = 1 M ohm
RTh
VTh
+
Rm
VTh = I(RTh + Rm)
(a)
A 4V reading corresponds to
I = (4/10)Ifs = 0.4x50 PA = 20 PA
VTh = 20 PA RTh + 20 PA 250 k ohms
= 4 + 20 PA RTh
(b)
(1)
A 5V reading corresponds to
I = (5/50)Ifs = 0.1 x 50 PA = 5 PA
VTh = 5 PA x RTh + 5 PA x 1 M ohm
From (1) and (2)
VTh = 5 + 5 PA RTh
(2)
0 = -1 + 15 PA RTh which leads to RTh = 66.67 k ohms
From (1),
VTh = 4 + 20x10-6x(1/(15x10-6)) = 5.333 V
Chapter 4, Solution 96.
(a)
The resistance network can be redrawn as shown in Fig. (a),
10 :
8:
10 :
RTh
9V
+
i1
40:
+
i2
60 :
8:
R
VTh
10 :
+
VTh
+
Vo
(a)
(b)
RTh = 10 + 10 + 60||(8 + 8 + 10||40) = 20 + 60||24 = 37.14 ohms
Using mesh analysis,
-9 + 50i1 - 40i2 = 0
116i2 –– 40i1 = 0 or i1 = 2.9i2
From (1) and (2),
(1)
(2)
i2 = 9/105
VTh = 60i2 = 5.143 V
From Fig. (b),
Vo = [R/(R + RTh)]VTh = 1.8
R/(R + 37.14) = 1.8/5.143 which leads to R = 20 ohms
(b)
R = RTh = 37.14 ohms
Imax = VTh/(2RTh) = 5.143/(2x37.14) = 69.23 mA
Chapter 4, Solution 97.
4 k:
12V
+
B
+
4 k:
VTh
E
R
RTh = R1||R2 = 6||4 = 2.4 k ohms
VTh = [R2/(R1 + R2)]vs = [4/(6 + 4)](12) = 4.8 V
Chapter 4, Solution 98.
The 20-ohm, 60-ohm, and 14-ohm resistors form a delta connection which needs to be
connected to the wye connection as shown in Fig. (b),
20 :
30 :
30 :
R2
R1
14 :
a
60 :
a
b
RTh
b
R3
(a)
(b)
R1 = 20x60/(20 + 60 + 14) = 1200/94 = 12.97 ohms
R2 = 20x14/94 = 2.98 ohms
R3 = 60x14/94 = 8.94 ohms
RTh = R3 + R1||(R2 + 30) = 8.94 + 12.77||32.98 = 18.15 ohms
To find VTh, consider the circuit in Fig. (c).
IT
30 :
20 :
I1
14 :
b
a
60 :
+
IT
16 V
+ (c)
VTh
RTh
IT = 16/(30 + 15.74) = 350 mA
I1 = [20/(20 + 60 + 14)]IT = 94.5 mA
VTh = 14I1 + 30IT = 11.824 V
I40 = VTh/(RTh + 40) = 11.824/(18.15 + 40) = 203.3 mA
P40 = I402R = 1.654 watts
Chapter 5, Solution 1.
(a)
(b)
(c)
Rin = 1.5 M:
Rout = 60 :
A = 8x104
Therefore AdB = 20 log 8x104 = 98.0 dB
Chapter 5, Solution 2.
v0 = Avd = A(v2 - v1)
= 105 (20-10) x 10-6 = 0.1V
Chapter 5, Solution 3.
v0 = Avd = A(v2 - v1)
= 2 x 105 (30 + 20) x 10-6 = 10V
Chapter 5, Solution 4.
v0 = Avd = A(v2 - v1)
v
4
v2 - v1 = 0
A 2 x10 5
20PV
If v1 and v2 are in mV, then
v2 - v1 = -20 mV = 0.02
1 - v1 = -0.02
v1 = 1.02 mV
Chapter 5, Solution 5.
I
R0
Rin
vd
+
vi
+
-
+
-
Avd
+
v0
-
-vi + Avd + (Ri - R0) I = 0
But
(1)
vd = RiI,
-vi + (Ri + R0 + RiA) I = 0
vd =
vi R i
R 0 (1 A)R i
(2)
-Avd - R0I + v0 = 0
v0 = Avd + R0I = (R0 + RiA)I =
v0
vi
#
R 0 RiA
R 0 (1 A)R i
10 9
˜ 10 4
1 10 5
(R 0 R i A) v i
R 0 (1 A)R i
100 10 4 x10 5
˜ 10 4
5
100 (1 10 )
100,000
100,001
0.9999990
Chapter 5, Solution 6.
vi
+ -
R0
I
vd
+
Rin
+
-
Avd
+
vo
-
(R0 + Ri)R + vi + Avd = 0
But
vd = RiI,
vi + (R0 + Ri + RiA)I = 0
I=
vi
R 0 (1 A)R i
(1)
-Avd - R0I + vo = 0
vo = Avd + R0I = (R0 + RiA)I
Substituting for I in (1),
§ R 0 R iA ·
¸¸ vi
v0 = ¨¨
R
(
1
A
)
R
i ¹
© 0
6
50 2 x10 x 2 x10 5 ˜ 10 3
= 50 1 2x10 5 x 2 x10 6
#
200,000 x 2 x10 6
mV
200,001x 2 x10 6
v0 = -0.999995 mV
Chapter 5, Solution 7.
100 k:
10 k:
VS
+
-
Rout = 100 :
1
2
+
Vd
-
Rin
+
-
AVd
+
Vout
-
At node 1,
(VS –– V1)/10 k = [V1/100 k] + [(V1 –– V0)/100 k]
10 VS –– 10 V1 = V1 + V1 –– V0
which leads to V1 = (10VS + V0)/12
At node 2,
(V1 –– V0)/100 k = (V0 –– AVd)/100
But Vd = V1 and A = 100,000,
V1 –– V0 = 1000 (V0 –– 100,000V1)
0= 1001V0 –– 100,000,001[(10VS + V0)/12]
0 = -83,333,334.17 VS - 8,332,333.42 V0
which gives us (V0/ VS) = -10 (for all practical purposes)
If VS = 1 mV, then V0 = -10 mV
Since V0 = A Vd = 100,000 Vd, then Vd = (V0/105) V = -100 nV
Chapter 5, Solution 8.
(a)
If va and vb are the voltages at the inverting and noninverting terminals of the op
amp.
va = v b = 0
1mA =
0 v0
2k
v0 = -2V
(b)
10 k:
2V
+
ia
va
2V
+
vb
1V
+
+
vo
-
-
(a)
2 k:
+
va
-
10 k:
+-
+
ia
(b)
vo
-
Since va = vb = 1V and ia = 0, no current flows through the 10 k: resistor. From Fig. (b),
-va + 2 + v0 = 0
va = va - 2 = 1 - 2 = -1V
Chapter 5, Solution 9.
(a)
Let va and vb be respectively the voltages at the inverting and noninverting
terminals of the op amp
va = vb = 4V
At the inverting terminal,
1mA =
4 v0
2k
v0 = 2V
(b)
1V
+-
+
+
vb
vo
-
-
Since va = vb = 3V,
-vb + 1 + vo = 0
vo = vb - 1 = 2V
Chapter 5, Solution 10.
Since no current enters the op amp, the voltage at the input of the op amp is vs.
Hence
§ 10 ·
vs = v o ¨
¸
© 10 10 ¹
vo
2
vo
=2
vs
Chapter 5, Solution 11.
8 k:
2 k:
3V
vb =
5 k:
+
a
io
+
b
+
10 k:
4 k:
vo
10
(3)
10 5
2V
At node a,
3 va
2
va vo
8
12 = 5va –– vo
But va = vb = 2V,
vo = -2V
12 = 10 –– vo
––io =
va vo 0 vo
8
4
22 2
8
4
1mA
i o = -1mA
Chapter 5, Solution 12.
4 k:
1 k:
1.2V
+
a
b
+
4 k:
2 k:
+
vo
4
vo
42
At node b,
vb =
At node a,
1 .2 v a
1
2
vo
3
va vo
2
, but va = vb = v o
4
3
4.8 - 4 x
2
vo
3
2
vo vo
3
va = vb =
2
vo
3
9.6
7
is =
1 .2 v a
1
2
vo
3
vo =
3x 4.8
7
2.0570V
1 .2
7
§ 1.2 ·
p = vsis = 1.2 ¨
¸
© 7 ¹
-205.7 mW
Chapter 5, Solution 13.
10 k:
a
b
1V
+
io
100 k: i2
i1
4 k:
90 k:
By voltage division,
90
va =
(1) 0.9V
100
vo
50
vb =
vo
3
150
v0
But va = vb
0 .9
3
v
v
io = i1 + i2 = o o
10k 150k
+
50 k:
vo = 2.7V
0.27mA + 0.018mA = 288 PA
+
vo
Chapter 5, Solution 14.
Transform the current source as shown below. At node 1,
10 v1
5
v1 v 2 v1 v o
20
10
10 k:
5 k:
10 k:
20 k:
v1
10V
vo
v2
+
+
+
vo
But v2 = 0. Hence 40 - 4v1 = v1 + 2v1 - 2vo
At node 2,
v1 v 2
20
v2 vo
,
10
v2
From (1) and (2), 40 = -14vo - 2vo
40 = 7v1 - 2vo
0 or v1 = -2vo
(1)
(2)
vo = -2.5V
Chapter 5, Solution 15
(a) Let v1 be the voltage at the node where the three resistors meet. Applying
KCL at this node gives
§ 1
v1 v1 vo
1 · vo
¸¸ (1)
is
v1 ¨¨
R2
R3
R
R
R3
3 ¹
© 2
At the inverting terminal,
0 v1
(2)
is

o v1 i s R1
R1
Combining (1) and (2) leads to
§
§
v
vo
RR ·
R
R ·

o
¨¨ R1 R3 1 3 ¸¸
i s ¨¨1 1 1 ¸¸ o
R3
is
R2 ¹
©
© R2 R3 ¹
(b) For this case,
vo
20 x 40 ·
§
¨ 20 40 ¸ k:
is
25 ¹
©
- 92 k:
Chapter 5, Solution 16
10k :
5k :
ix
va
vb
iy
+
vo
2k :
+
0.5V
-
8k :
Let currents be in mA and resistances be in k : . At node a,
0 .5 v a v a v o

o 1 3v a vo
5
10
(1)
But
8
10
vo

o vo
va
(2)
82
8
Substituting (2) into (1) gives
10
8
1 3v a v a

o v a
8
14
Thus,
0 .5 v a
ix
1 / 70 mA 14.28 PA
5
v o vb v o v a
10
0 .6 8
iy
0 .6 ( v o v a ) 0 .6 ( v a v a )
x mA 85.71 PA
2
10
8
4 14
va
vb
Chapter 5, Solution 17.
(a)
(b)
(c)
G=
vo
vi
vo
vi
vo
vi
R2
R1
80
= -16
5
2000
5
12
5
-400
-2.4
Chapter 5, Solution 18.
Converting the voltage source to current source and back to a voltage source, we have the
circuit shown below:
20
k:
3
10 20
1 M:
(20/3) k: 50 k:
+
+
2vi/3
+
vo
vo
1000 2v i
˜
20 3
50 3
vo
v1
200
17
-11.764
Chapter 5, Solution 19.
We convert the current source and back to a voltage source.
24
(4/3) k:
(2/3)V
+
4 k:
0V
4
3
10 k:
+
vo
5 k:
vo
io
10k § 2 ·
¨ ¸ -1.25V
§ 4· ©3¹
¨ 4x ¸k
© 3¹
vo vo 0
-0.375mA
5k
10k
Chapter 5, Solution 20.
8 k:
4 k:
9V
a
+
4 k:
vs
2 k:
b
+
+
+
vo
At node a,
9 va
4
va vo va vb
8
4
18 = 5va –– vo - 2vb
(1)
At node b,
va vb
4
vb vo
2
va = 3vb - 2vo
But vb = vs = 0; (2) becomes va = ––2vo and (1) becomes
-18 = -10vo –– vo
vo = -18/(11) = -1.6364V
(2)
Chapter 5, Solution 21.
Eqs. (1) and (2) remain the same. When vb = vs = 3V, eq. (2) becomes
va = 3 x 3 - 2v0 = 9 - 2vo
Substituting this into (1), 18 = 5 (9-2vo) –– vo –– 6 leads to
vo = 21/(11) = 1.909V
Chapter 5, Solution 22.
Av = -Rf/Ri = -15.
If Ri = 10k:, then Rf = 150 k:.
Chapter 5, Solution 23
At the inverting terminal, v=0 so that KCL gives
vs 0
R1
0 0 vo
R2
Rf
o
vo
vs
Rf
R1
Chapter 5, Solution 24
v1
Rf
R2
R1
- vs +
+
+
R3
R4
vo
-
v2
We notice that v1 = v2. Applying KCL at node 1 gives
v1 (v1 v s ) v1 vo
R1
R2
Rf
0

o
§ 1
·
¨ 1 1 ¸v1 v s
¨R R
R f ¸¹
R2
2
© 1
vo
Rf
(1)
Applying KCL at node 2 gives
v1 v1 v s
0

o v1
R3
R4
Substituting (2) into (1) yields
vo
R3
vs
R3 R4
ª§ R
R
R ·§ R3
R f «¨ 3 3 4 ¸¨¨
«¬¨© R1 R f R2 ¸¹© R3 R4
(2)
· 1 º
¸¸ » v s
¹ R2 »¼
i.e.
k
ª§ R
R
R ·§ R3
R f «¨ 3 3 4 ¸¨¨
«¬¨© R1 R f R2 ¸¹© R3 R4
· 1 º
¸¸ »
¹ R2 »¼
Chapter 5, Solution 25.
vo = 2 V
+ +
+
va
vo
-va + 3 + vo = 0 which leads to va = vo + 3 = 5 V.
Chapter 5, Solution 26
+
vb
+
0.4V
-
+
2k :
8k :
vo
-
vb
0.4
8
vo
8 2
0.8vo

o
Hence,
io
vo
5k
0 .5
5k
0.1 mA
vo
0.4 / 0.8
0.5 V
io
5k :
Chapter 5, Solution 27.
(a)
Let va be the voltage at the noninverting terminal.
va = 2/(8+2) vi = 0.2vi
§ 1000 ·
v 0 ¨1 ¸ v a 10.2v i
20 ¹
©
G = v0/(vi) = 10.2
(b)
vi = v0/(G) = 15/(10.2) cos 120St = 1.471 cos 120St V
Chapter 5, Solution 28.
+
+
At node 1,
0 v1
10k
v1 v o
50k
But v1 = 0.4V,
-5v1 = v1 –– vo, leads to
vo = 6v1 = 2.4V
Alternatively, viewed as a noninverting amplifier,
vo = (1 + (50/10)) (0.4V) = 2.4V
io = vo/(20k) = 2.4/(20k) = 120 PA
Chapter 5, Solution 29
R1
va
vb
+
vi
-
va
R2
vi ,
R1 R2
But v a
Or
vo
vi
R2
vb

o
vb
+
R2
R1
R1
vo
R1 R2
R2
R1
Chapter 5, Solution 30.
The output of the voltage becomes
vo = vi = 12
30 20 12k:
By voltage division,
vx
12
(1.2)
12 60
0.2V
ix
p
v 2x
R
0.04
20k
2PW
vx
20k
0 .2
20k
vo
-
R1
vo
R1 R2
R2
vi
R1 R2
+
10PA
Chapter 5, Solution 31.
After converting the current source to a voltage source, the circuit is as shown below:
12 k:
3 k:
1
6 k: v
o
v1
12 V
+
2
+
vo
6 k:
At node 1,
12 v1
3
v1 v o v1 v o
6
12
v1 v o
6
vo 0
6
48 = 7v1 - 3vo
(1)
At node 2,
ix
v1 = 2vo
(2)
From (1) and (2),
vo
ix
48
11
vo
6k
0.7272mA
Chapter 5, Solution 32.
Let vx = the voltage at the output of the op amp. The given circuit is a non-inverting
amplifier.
§ 50 ·
¨1 ¸ (4 mV) = 24 mV
© 10 ¹
60 30 20k:
vx
By voltage division,
vo =
20
vo
20 20
vo
2
ix =
vx
20 20 k
24mV
40k
p=
144x10 6
60x10 3
v o2
R
12mV
600nA
204nW
Chapter 5, Solution 33.
After transforming the current source, the current is as shown below:
1 k:
4 k:
vi
va
+
4V
+
2 k:
vo
3 k:
This is a noninverting amplifier.
vo
§ 1·
¨1 ¸ v i
© 2¹
3
vi
2
Since the current entering the op amp is 0, the source resistor has a OV potential drop.
Hence vi = 4V.
vo
3
(4)
2
6V
Power dissipated by the 3k: resistor is
v o2
R
ix
36
3k
12mW
va vo
R
46
1k
-2mA
Chapter 5, Solution 34
v1 vin v1 vin
R1
R2
0
(1)
but
va
R3
vo
R3 R 4
(2)
Combining (1) and (2),
v1 va R1
R
v 2 1 va
R2
R2
§
R ·
v a ¨¨1 1 ¸¸
© R2 ¹
v1 R1
v2
R2
R ·
R 3v o §
¨¨1 1 ¸¸
R3 R 4 © R 2 ¹
v1 vo
vO =
0
R1
v2
R2
·
R3 R 4 §
R
¨¨ v1 1 v 2 ¸¸
R2 ¹
§
R ·
R 3 ¨¨1 1 ¸¸ ©
© R2 ¹
R3 R 4
( v1R 2 v 2 )
R 3 ( R1 R 2 )
Chapter 5, Solution 35.
Av
vo
vi
1
Rf
Ri
10
If Ri = 10k:, Rf = 90k:
Rf = 9Ri
Chapter 5, Solution 36
VTh
But
VTh
Vab
vs
Vab
R1
Vab . Thus,
R1 R2
R
R1 R2
v s (1 2 )v s
R1
R1
To get RTh, apply a current source Io at terminals a-b as shown below.
v1
+
-
v2
a
+
R2
R1
vo
io
b
Since the noninverting terminal is connected to ground, v1 = v2 =0, i.e. no current passes
through R1 and consequently R2 . Thus, vo=0 and
vo
RTh
0
io
Chapter 5, Solution 37.
vo
º
ªR
R
R
« f v1 f v 2 f v 3 »
R3 ¼
R2
¬ R1
30
30
ª 30
º
« (1) (2) (3)»
20
30
¬ 10
¼
vo = -3V
Chapter 5, Solution 38.
vo
º
ªR
R
R
R
« f v1 f v 2 f v 3 f v 4 »
R4 ¼
R3
R2
¬ R1
50
50
50
ª 50
º
« (10) (20) (50) (100)»
20
10
50
¬ 25
¼
= -120mV
Chapter 5, Solution 39
This is a summing amplifier.
Rf
Rf ·
§ Rf
50
50
§ 50
·
¨¨
v1 v2 v3 ¸¸ ¨ (2) v 2 (1) ¸
R2
R3 ¹
20
50
© 10
¹
© R1
Thus,
vo 16.5 9 2.5v 2

o v 2 3 V
vo
9 2.5v 2
Chapter 5, Solution 40
R1
va
R2
+
v1
-
+
R3
+
v2
-
vb
-
+
Rf
+
v3
-
R
vo
-
Applying KCL at node a,
v1 v a v 2 v a v3 v a
R1
R2
R3
0

o
v1 v 2 v3
R1 R2 R3
va (
1
1
1
) (1)
R1 R2 R3
But
va
R
vo
R Rf
vb
(2)
Substituting (2) into (1)gives
v1 v 2 v3
R1 R2 R3
Rvo
1
1
1
( )
R R f R1 R2 R3
or
vo
R Rf
R
(
v1 v 2 v3
1
1
1
) /( )
R1 R2 R3 R1 R2 R3
Chapter 5, Solution 41.
Rf/Ri = 1/(4)
Ri = 4Rf = 40k:
The averaging amplifier is as shown below:
v1
v2
v3
v4
Chapter 5, Solution 42
Rf
1
R1
3
10 k:
R1 = 40 k:
10 k:
R2 = 40 k:
R3 = 40 k:
R4 = 40 k:
+
vo
Chapter 5, Solution 43.
In order for
vo
·
§ Rf
R
R
R
¨¨
v1 f v 2 f v 3 f v 4 ¸¸
R2
R3
R4 ¹
© R1
vo
to become
1
v1 v 2 v 3 v 4
4
Ri
1
Rf
4
4
Rf
Ri
12
4
3k:
Chapter 5, Solution 44.
R4
R3
v1
v2
a
R1
+
b
R2
At node b,
v b v1 v b v 2
R1
R2
At node a,
0 va
R3
vb
0
va vo
R4
va
But va = vb. We set (1) and (2) equal.
vo
1 R4 / R3
R 2 v1 R 1 v1
R1 R 2
or
vo =
R3 R4
R 2 v1 R 1 v1
R 3 R1 R 2
v1 v 2
R1 R 2
1
1
R1 R 2
vo
1 R4 / R3
vo
(1)
(2)
Chapter 5, Solution 45.
This can be achieved as follows:
vo
R
ª R
º
v2 »
«
v1 R/2 ¼
¬R / 3
º
ªR
R
« f v1 f v 2 »
R2 ¼
¬ R1
i.e. Rf = R, R1 = R/3, and R2 = R/2
Thus we need an inverter to invert v1, and a summer, as shown below (R<100k:).
R
v1
R
R
+
R/3
-v1
v2
R/2
+
vo
Chapter 5, Solution 46.
v1 1
Rf
R
R
1
( v 2 ) v 3
v1 x ( v 2 ) f v 3
3 3
2
R1
R2
R3
i.e. R3 = 2Rf, R1 = R2 = 3Rf. To get -v2, we need an inverter with Rf = Ri. If Rf = 10k:,
a solution is given below.
vo
10 k:
v2
v1
10 k:
+
30 k:
30 k:
10 k:
-v2
v3
20 k:
+
vo
Chapter 5, Solution 47.
If a is the inverting terminal at the op amp and b is the noninverting terminal,
then,
vb
3
(8)
3 1
6V, v a
vb
which leads to vo = ––2 V and io =
6V and at node a,
10 v a
2
va vo
4
v o (v a v o )
= ––0.4 –– 2 mA = ––2.4 mA
5k
4k
Chapter 5, Solution 48.
Since the op amp draws no current from the bridge, the bridge may be treated separately
v1
as follows:
i1
+ i2
v2
For loop 1, (10 + 30) i1 = 5
i1 = 5/(40) = 0.125PA
For loop 2, (40 + 60) i2 = -5
i2 = -0.05PA
But, 10i + v1 - 5 = 0
60i + v2 + 5 = 0
v1 = 5 - 10i = 3.75mV
v2 = -5 - 60i = -2mV
As a difference amplifier,
R2
vo
v 2 v1
R1
= 23mV
80
>3.75 (2)@mV
20
Chapter 5, Solution 49.
R1 = R3 = 10k:, R2/(R1) = 2
i.e.
R2 = 2R1 = 20k: = R4
Verify:
R 2 1 R1 / R 2
R
v 2 2 v1
R1 1 R 3 / R 4
R1
vo
2
(1 0.5)
v 2 2 v1
1 0.5
2 v 2 v1
Thus, R1 = R3 = 10k:, R2 = R4 = 20k:
Chapter 5, Solution 50.
(a)
We use a difference amplifier, as shown below:
v1
R1
R2
+
v2
vo
(b)
R1
R2
R2
v 2 v1 2 v 2 v1 , i.e. R2/R1 = 2
R1
If R1 = 10 k: then R2 = 20k:
We may apply the idea in Prob. 5.35.
v0
vo
2 v1 2 v 2
R
ª R
º
v2 »
«
v1 R/2 ¼
¬R / 2
ªR
º
R
« f v1 f v 2 »
R2 ¼
¬ R1
i.e. Rf = R, R1 = R/2 = R2
We need an inverter to invert v1 and a summer, as shown below. We may let R = 10k:.
R
v1
R
R
+
R/2
-v1
v2
R/2
+
vo
Chapter 5, Solution 51.
We achieve this by cascading an inverting amplifier and two-input inverting summer as
shown below:
R
R
R
R
v2
va
+
R
vo
v1
+
Verify:
But
vo = -va - v1
va = -v2. Hence
vo = v2 - v1.
Chapter 5, Solution 52
A summing amplifier shown below will achieve the objective. An inverter is inserted to
invert v2. Let R = 10 k : .
R/2
R
v1
R/5
v3
R
+
v4
vo
R
R
v2
-
R/4
+
Chapter 5, Solution 53.
(a)
v1
R2
R1
va
vb
v2
R1
+
vo
R2
At node a,
At node b,
vb
R2
v2
R1 R 2
But va = vb. Setting (1) and (2) equal gives
R 2 v1 R 1 v o
R2
v2
R1 R 2
R1 R 2
(1)
(2)
v 2 v1
vo
vi
R1
vo
R2
vi
R2
R1
(b)
v1
vi
+ v2
R1/2 v
A
R1/2
R2
va
Rg
R1/2
R1/2
vb
vB
+
R2
+
vo
At node A,
v1 v A v B v A
R1 / 2
Rg
or
v1 v A At node B,
v2 vB
R1 / 2
or
v2 vB vA va
R1 / 2
R1
vB vA
2R g
vA va
(1)
vB vA vB vb
R1 / 2
Rg
R1
(v B v A )
2R g
vB vb
(2)
Subtracting (1) from (2),
v 2 v1 v B v A 2R 1
vB vA
2R g
Since, va = vb,
v 2 v1
2
§
·
¨1 R 1 ¸ v B v A
¨ 2R ¸
g ¹
©
vi
2
vB vA vb va
vi
˜
2
vB vA
or
1
R
1 1
2R g
(3)
But for the difference amplifier,
R2
vB vA
R1 / 2
R1
vo
vB vA
2R 2
vo
or
Equating (3) and (4),
(c)
R1
vo
2R 2
vi
˜
2
vo
vi
1
R
1 1
2R g
R2
˜
R1
v1 v a
R1
At node a,
v1 v a
v2 vb
At node b,
(4)
1
R
1 1
2R g
va vA
R2 /2
2R 1
2R 1
vA
va R2
R2
2R 1
2R 1
vB
vb R2
R2
(1)
(2)
Since va = vb, we subtract (1) from (2),
v 2 v1
or
vB vA
2R 1
(v B v A )
R2
R2
vi
2R 1
vi
2
(3)
At node A,
va vA vB vA
R2 /2
Rg
va vA vA vo
R/2
R2
vB vA
2R g
vA vo
(4)
At node B,
vb vB vB vA
R/2
Rg
vB 0
R/2
R2
vB vA
2R g
vb vB vB
(5)
Subtracting (5) from (4),
v B v A R2
v B v A
Rg
§
R
2 v B v A ¨1 2
¨ 2R
g
©
Combining (3) and (6),
R 2 §¨
R ·
vi 1 2 ¸
¨ 2R ¸
R1
g ¹
©
vo
vi
R2
R1
§
¨1 R 2
¨ 2R
g
©
v A v B v o
·
¸
¸
¹
v o
(6)
v o
·
¸
¸
¹
Chapter 5, Solution 54.
(a)
(b)
But
A0 = A1A2A3 = (-30)(-12.5)(0.8) = 300
A = A1A2A3A4 = A0A4 = 300A4
20Log10 A
60dB
Log10 A
3
A = 103 = 1000
A4 = A/(300) = 3.333
Chapter 5, Solution 55.
Let A1 = k, A2 = k, and A3 = k/(4)
A = A1A2A3 = k3/(4)
20Log10 A 42
Thus
Log10 A 2.1
A = 102 ˜1 = 125.89
k3 = 4A = 503.57
k = 3 503.57 7.956
A1 = A2 = 7.956, A3 = 1.989
Chapter 5, Solution 56.
There is a cascading system of two inverting amplifiers.
vo
io
(a)
(b)
12 § 12 ·
¨
¸v s
4 © 6 ¹
vs
3v s mA
2k
6v s
When vs = 12V, io = 36mA
When vs = 10 cos 377t V, io = 30 cos 377t mA
Chapter 5, Solution 57
The first stage is a difference amplifier. Since R1/R2 = R3/R4,
v oc
R2
( v 2 v1 )
R1
100
(1 4) 10 mA
50
The second stage is a non-inverter.
vo
R· c
§
¨1 ¸ v o
© 40 ¹
R·
§
¨1 ¸10 mA
© 40 ¹
40 mV(given)
Which leads to,
R = 120 kȍ
Chapter 5, Solution 58.
By voltage division, the input to the voltage follower is:
v1
3
(0.6)
3 1
vo
10
10
v1 v1
2
5
io
0 vo
4k
Thus
0.45V
7 v1
0.7875mA
3.15
Chapter 5, Solution 59.
Let a be the node between the two op amps.
va = vo
The first stage is a summer
10
10
vs vo
5
20
1.5vs = -2vs
va
or
vo
vs
2
1 .5
vo
-1.333
Chapter 5, Solution 60.
Transform the current source as shown below:
4 k:
10 k:
5 k:
+
5is
v1
+
io
3 k:
+
3:
2 k:
Assume all currents are in mA. The first stage is a summer
v1
By voltage division,
10
10
5i s v o
5
4
10i s 2.5v o
(1)
v1
3
vo
33
1
vo
2
(2)
Alternatively, we notice that the second stage is a non-inverter.
vo
§ 1 ·
¸ v1
¨
©3 3¹
2v1
From (1) and (2),
10i s 2.5v o
0.5v o
vo
2i o
10i s
3
3vo = 10is
io
is
5
3
1.667
Chapter 5, Solution 61.
Let v01 be the voltage at the left end of R5. The first stage is an inverter, while the
second stage is a summer.
v 01
v0
v1 =
R2
v1
R1
R
R
4 v 01 4 v 2
R5
R3
R 2R 4
R
v1 4 v 2
R 1R 5
R3
Chapter 5, Solution 62.
Let v1 = output of the first op amp
v2 = output of the second op amp
The first stage is a summer
v1
R2
R
v i –– 2 v o
R1
Rf
The second stage is a follower. By voltage division
(1)
vo
R4
v1
R3 R4
v2
v1
R3 R4
vo
R4
(2)
From (1) and (2),
§
¨¨1 ©
§
¨¨1 ©
vo
vi
·
¸¸ v o
¹
R3 R2
R4 Rf
R3
R4
R2
˜
R1
R
R2
vi 2 vo
Rf
R1
·
¸¸ v o
¹
R2
vi
R1
1
R3 R2
R4 R4
R 2R 4
R1 R 2 R 3 R 4
1
Chapter 5, Solution 63.
The two op amps are summer. Let v1 be the output of the first op amp. For the first
stage,
v1
R2
R
vi 2 vo
R1
R3
(1)
For the second stage,
vo
R4
R
v1 4 v i
R5
Ro
(2)
Combining (1) and (2),
§ R2 ·
R §R ·
R
¸¸ v i 4 ¨¨ 2 ¸¸ v o 4 v i
¨¨
R5 © R3 ¹
R6
© R1 ¹
§ R R · §R R
R ·
v o ¨¨1 2 4 ¸¸ ¨¨ 2 4 4 ¸¸ v i
© R 3 R 5 ¹ © R 1R 5 R 6 ¹
vo
R4
R5
vo
vi
R 2R 4 R 4
R 1R 3 R 6
R R
1 2 4
R 3R 5
Chapter 5, Solution 64
G4
G
G1
+
G3
1
G
+
0V
vs
2
0V +
v
G2
+
vo
-
-
At node 1, v1=0 so that KCL gives
G1v s G4 vo
Gv
(1)
At node 2,
G2 v s G3 v o Gv
From (1) and (2),
G1v s G4 v o G2 v s G3 vo
or
vo G1 G2
v s G3 G 4
(2)
o

(G1 G2 )v s
(G3 G4 )vo
Chapter 5, Solution 65
The output of the first op amp (to the left) is 6 mV. The second op amp is an
inverter so that its output is
30
(6mV) -18 mV
10
The third op amp is a noninverter so that
vo ' vo '
40
vo
40 8

o
vo
48
v o ' 21.6 mV
40
Chapter 5, Solution 66.
vo
100 § 40 ·
100
110
(6) (2)
¨ ¸(4) 25
20 © 20 ¹
10
24 40 20 -4V
Chapter 5, Solution 67.
80 § 80 ·
80
¨ ¸(0.5) (0.2)
40 © 20 ¹
20
3.2 0.8 2.4V
vo = Chapter 5, Solution 68.
If Rq = f, the first stage is an inverter.
Va
15
(10)
5
30mV
when Va is the output of the first op amp.
The second stage is a noninverting amplifier.
vo
§ 6·
¨1 ¸ v a
© 2¹
(1 3)(30) -120mV
Chapter 5, Solution 69.
In this case, the first stage is a summer
va
15
15
(10) v o
5
10
30 1.5v o
For the second stage,
vo
7v o
§ 6·
¨1 ¸ v a
© 2¹
120
4 30 1.5v o
4v a
vo
120
7
-17.143mV
Chapter 5, Solution 70.
The output of amplifier A is
vA
30
30
(10) (2)
10
10
9
The output of amplifier B is
vB
20
20
(3) (4)
10
10
14
40 k:
vA
vB
20 k:
a
60 k:
+
b
vo
10 k:
vb
60
(14)
60 10
At node a,
vA va
20
2V
va vo
40
But va = vb = -2V, 2(-9+2) = -2-vo
Therefore,
vo = 12V
Chapter 5, Solution 71
20k :
5k :
100k :
40k :
+
+
2V
-
v2
10k :
80k :
+
20k :
+
vo
-
10k :
+
v1
+
3V
-
v1
vo
+
-
30k :
v3
50k :
20
50
(2) 8, v3 (1 )v1
5
30
100 ·
§ 100
¨
v2 v3 ¸ (20 10) 10 V
80 ¹
© 40
3,
v2
8
Chapter 5, Solution 72.
Since no current flows into the input terminals of ideal op amp, there is no voltage
drop across the 20 k: resistor. As a voltage summer, the output of the first op
amp is
v01 = 0.4
The second stage is an inverter
150
v 01
100
2.5(0.4)
v2
-1V
Chapter 5, Solution 73.
The first stage is an inverter. The output is
50
v 01 (1.8) 9V
10
The second stage is
v 2 v 01 -9V
Chapter 5, Solution 74.
Let v1 = output of the first op amp
v2 = input of the second op amp.
The two sub-circuits are inverting amplifiers
v1
v2
io
100
(0.6) 6V
10
32
(0.4) 8V
1.6
v1 v 2
68
20k
20k
100 PA
Chapter 5, Solution 75.
The schematic is shown below. Pseudo-components VIEWPOINT and IPROBE are
involved as shown to measure vo and i respectively. Once the circuit is saved, we click
Analysis | Simulate. The values of v and i are displayed on the pseudo-components as:
i = 200 PA
(vo/vs) = -4/2 = -2
The results are slightly different than those obtained in Example 5.11.
Chapter 5, Solution 76.
The schematic is shown below. IPROBE is inserted to measure io. Upon simulation, the
value of io is displayed on IPROBE as
io = -374.78 PA
Chapter 5, Solution 77.
The schematic is shown below. IPROBE is inserted to measure io. Upon simulation, the
value of io is displayed on IPROBE as
io = -374.78 PA
Chapter 5, Solution 78.
The circuit is constructed as shown below. We insert a VIEWPOINT to display vo.
Upon simulating the circuit, we obtain,
vo = 667.75 mV
Chapter 5, Solution 79.
The schematic is shown below. A pseudo-component VIEWPOINT is inserted to display
vo. After saving and simulating the circuit, we obtain,
vo = -14.61 V
Chapter 5, Solution 80.
The schematic is shown below. VIEWPOINT is inserted to display vo. After simulation,
we obtain,
vo = 12 V
Chapter 5, Solution 81.
The schematic is shown below. We insert one VIEWPOINT and one IPROBE to
measure vo and io respectively. Upon saving and simulating the circuit, we obtain,
vo = 343.37 mV
io = 24.51 PA
Chapter 5, Solution 82.
The maximum voltage level corresponds to
11111 = 25 –– 1 = 31
Hence, each bit is worth
(7.75/31) = 250 mV
Chapter 5, Solution 83.
The result depends on your design. Hence, let RG = 10 k ohms, R1 = 10 k ohms, R2 =
20 k ohms, R3 = 40 k ohms, R4 = 80 k ohms, R5 = 160 k ohms, R6 = 320 k ohms,
then,
-vo = (Rf/R1)v1 + --------- + (Rf/R6)v6
= v1 + 0.5v2 + 0.25v3 + 0.125v4 + 0.0625v5 + 0.03125v6
(a)
|vo| = 1.1875 = 1 + 0.125 + 0.0625 = 1 + (1/8) + (1/16) which implies,
[v1 v2 v3 v4 v5 v6] = [100110]
(b)
|vo| = 0 + (1/2) + (1/4) + 0 + (1/16) + (1/32) = (27/32) = 843.75 mV
(c)
This corresponds to [1 1 1 1 1 1].
|vo| = 1 + (1/2) + (1/4) + (1/8) + (1/16) + (1/32) = 63/32 = 1.96875 V
Chapter 5, Solution 84.
For (a), the process of the proof is time consuming and the results are only approximate,
but close enough for the applications where this device is used.
(a)
The easiest way to solve this problem is to use superposition and to solve
for each term letting all of the corresponding voltages be equal to zero.
Also, starting with each current contribution (ik) equal to one amp and
working backwards is easiest.
2R
v1
+
R
R
R
2R
ik
v2
+
2R
v3
+
2R
v4
+
R
For the first case, let v2 = v3 = v4 = 0, and i1 = 1A.
Therefore,
v1 = 2R volts or i1 = v1/(2R).
Second case, let v1 = v3 = v4 = 0, and i2 = 1A.
Therefore,
v2 = 85R/21 volts or i2 = 21v2/(85R). Clearly this is not
(1/4th), so where is the difference? (21/85) = 0.247 which is a really
good approximation for 0.25. Since this is a practical electronic circuit,
the result is good enough for all practical purposes.
Now for the third case, let v1 = v2 = v4 = 0, and i3 = 1A.
Therefore,
v3 = 8.5R volts or i3 = v3/(8.5R). Clearly this is not
(1/8th), so where is the difference? (1/8.5) = 0.11765 which is a really
good approximation for 0.125. Since this is a practical electronic circuit,
the result is good enough for all practical purposes.
Finally, for the fourth case, let v1 = v2 = v4 = 0, and i3 = 1A.
Therefore,
v4 = 16.25R volts or i4 = v4/(16.25R). Clearly this is not
th
(1/16 ), so where is the difference? (1/16.25) = 0.06154 which is a
really good approximation for 0.0625. Since this is a practical electronic
circuit, the result is good enough for all practical purposes.
Please note that a goal of a lot of electronic design is to come up with
practical circuits that are economical to design and build yet give the
desired results.
(b)
If Rf = 12 k ohms and R = 10 k ohms,
-vo = (12/20)[v1 + (v2/2) + (v3/4) + (v4/8)]
= 0.6[v1 + 0.5v2 + 0.25v3 + 0.125v4]
For
[v1 v2 v3 v4] = [1 0 11],
|vo| = 0.6[1 + 0.25 + 0.125] = 825 mV
For
[v1 v2 v3 v4] = [0 1 0 1],
|vo| = 0.6[0.5 + 0.125] = 375 mV
Chapter 5, Solution 85.
Av = 1 + (2R/Rg) = 1 + 20,000/100 = 201
Chapter 5, Solution 86.
vo = A(v2 –– v1) = 200(v2 –– v1)
(a)
(b)
vo = 200(0.386 –– 0.402) = -3.2 V
vo = 200(1.011 –– 1.002) = 1.8 V
Chapter 5, Solution 87.
The output, va, of the first op amp is,
Also,
va = (1 + (R2/R1))v1
(1)
vo = (-R4/R3)va + (1 + (R4/R3))v2
(2)
Substituting (1) into (2),
vo = (-R4/R3) (1 + (R2/R1))v1 + (1 + (R4/R3))v2
Or,
If
vo = (1 + (R4/R3))v2 –– (R4/R3 + (R2R4/R1R3))v1
R4 = R1 and R3 = R2, then,
vo = (1 + (R4/R3))(v2 –– v1)
which is a subtractor with a gain of (1 + (R4/R3)).
Chapter 5, Solution 88.
We need to find VTh at terminals a –– b, from this,
vo = (R2/R1)(1 + 2(R3/R4))VTh = (500/25)(1 + 2(10/2))VTh
= 220VTh
Now we use Fig. (b) to find VTh in terms of vi.
a
a
30 k:
20 k:
20 k:
vi
vi
30 k:
+
40 k:
80 k:
40 k:
b
80 k:
b
(a)
(b)
va = (3/5)vi, vb = (2/3)vi
VTh = vb –– va (1/15)vi
(vo/vi) = Av = -220/15 = -14.667
Chapter 5, Solution 89.
If we use an inverter, R = 2 k ohms,
(vo/vi) = -R2/R1 = -6
R = 6R = 12 k ohms
Hence the op amp circuit is as shown below.
12 k:
2 k:
vi
+
+
+
vo
Chapter 5, Solution 90.
Transforming the current source to a voltage source produces the circuit below,
At node b,
vb = (2/(2 + 4))vo = vo/3
20 k:
5 k: a
5is
+
b
+
4 k:
io
2 k:
At node a,
But va = vb = vo/3.
+
vo
(5is –– va)/5 = (va –– vo)/20
20is –– (4/3)vo = (1/3)vo –– vo, or is = vo/30
io = [(2/(2 + 4))/2]vo = vo/6
io/is = (vo/6)/(vo/30) = 5
Chapter 5, Solution 91.
+
vo
R2
R1
is
i2
i1
But
io
io = i1 + i2
(1)
i1 = is
(2)
R1 and R2 have the same voltage, vo, across them.
R1i1 = R2i2, which leads to i2 = (R1/R2)i1
(3)
Substituting (2) and (3) into (1) gives,
io = is(1 + R1/R2)
io/is = 1 + (R1/R2) = 1 + 8/1 = 9
Chapter 5, Solution 92
The top op amp circuit is a non-inverter, while the lower one is an inverter. The
output at the top op amp is
v1 = (1 + 60/30)vi = 3vi
while the output of the lower op amp is
v2 = -(50/20)vi = -2.5vi
Hence,
vo = v1 –– v2 = 3vi + 2.5vi = 5.5vi
vo/vi = 5.5
Chapter 5, Solution 93.
R3
R1 v
a
vb
+
io
+
R4
vi
+
R2
vL
iL
RL
+
vo
At node a,
(vi –– va)/R1 = (va –– vo)/R3
vi –– va = (R1/R2)(va –– vo)
vi + (R1/R3)vo = (1 + R1/R3)va
(1)
But va = vb = vL. Hence, (1) becomes
vi = (1 + R1/R3)vL –– (R1/R3)vo
(2)
io = vo/(R4 + R2||RL), iL = (RL/(R2 + RL))io = (R2/(R2 + RL))(vo/( R4 + R2||RL))
Or,
vo = iL[(R2 + RL)( R4 + R2||RL)/R2
(3)
But,
vL = iLRL
(4)
Substituting (3) and (4) into (2),
vi = (1 + R1/R3) iLRL –– R1[(R2 + RL)/(R2R3)]( R4 + R2||RL)iL
= [((R3 + R1)/R3)RL –– R1((R2 + RL)/(R2R3)(R4 + (R2RL/(R2 + RL))]iL
= (1/A)iL
Thus,
A =
1
§
§ R RL
R ·
¨¨ 1 1 ¸¸ R L R 1 ¨¨ 2
R3 ¹
©
© R 2R 3
·§
R 2RL
¸¸¨¨ R 4 R2 RL
¹©
·
¸¸
¹
Chapter 6, Solution 1.
i
C
dv
dt
5 2e 3t 6 e 3 t
10(1 - 3t)e-3t A
p = vi = 10(1-3t)e-3t ˜ 2t e-3t = 20t(1 - 3t)e-6t W
Chapter 6, Solution 2.
1 2
Cv1
2
1
w2 = Cv12
2
w1
'w
1
(40)(120) 2
2
1
(40)(80) 2
2
w1 w 2
20 120 2 80 2
160 kW
Chapter 6, Solution 3.
i=C
dv
dt
40x10 3
280 160
5
480 mA
Chapter 6, Solution 4.
v
1 t
idt v(0)
C ³o
1
6 sin 4 tdt 1
2³
= 1 - 0.75 cos 4t
Chapter 6, Solution 5.
1 t
idt v(0)
C ³o
For 0 < t < 1, i = 4t,
t
1
v
4t dt + 0 = 100t2 kV
6 ³o
20x10
v(1) = 100 kV
v=
For 1 < t < 2, i = 8 - 4t,
t
1
v
(8 4t )dt v(1)
20x10 6 ³1
= 100 (4t - t2 - 3) + 100 kV
ª100t 2 kV,
0 t 1
v (t) = «
2
«¬100(4t t 2)kV, 1 t 2
Thus
Chapter 6, Solution 6.
dv
30x10 6 x slope of the waveform.
dt
For example, for 0 < t < 2,
i
C
10
2x10 3
10
dv
30x10 6 x
150mA
i= C
dt
2x10 3
Thus the current i is sketched below.
dv
dt
i(t) (mA)
150
4
8
2
6
t (msec)
10
-150
Chapter 6, Solution 7.
v
1
idt v( t o )
C³
=
1
50x10 3
t
³ 4tx10
o
2t 2
10 0.04k2 + 10 V
50
3
dt 10
12
Chapter 6, Solution 8.
(a) i
i ( 0)
C
2
dv
dt
100 ACe 100t 600 BCe 600t
100 AC 600 BC
v (0 ) v (0 )

o 50
Solving (2) and (3) leads to
A=61, B=-11
(b) Energy
1 2
Cv (0)
2

o
(1)
A 6B
5
A B
1
x 4 x10 3 x 2500
2
(2)
(3)
5J
(c ) From (1),
i
100 x61x 4 x10 3 e 100t 600 x11x 4 x10 3 e 600t
Chapter 6, Solution 9.
v(t) =
1 t
6 1 e t dt 0 12 t e t V
³
o
12
v(2) = 12(2 + e-2) = 25.62 V
p = iv = 12 (t + e-t) 6 (1-e-t) = 72(t-e-2t)
p(2) = 72(2-e-4) = 142.68 W
Chapter 6, Solution 10
dv
dt
2 x10 3
dv
dt
i
C
v
­ 16t , 0 t 1Ps
°
® 16, 1 t 3 Ps
°64 - 16t, 3 t 4 Ps
¯
24.4e 100t 26.4e 600t A
dv
dt
­ 16 x10 6 , 0 t 1Ps
°
® 0, 1 t 3 Ps
°- 16x10 6 , 3 t 4 Ps
¯
i (t )
­ 32 kA, 0 t 1Ps
°
® 0, 1 t 3 Ps
°- 32 kA, 3 t 4Ps
¯
Chapter 6, Solution 11.
1 t
idt v(0)
C ³o
For 0 < t < 1,
v=
t
1
40 x10 3 dt 10t kV
6 ³o
4x10
v(1) = 10 kV
v
For 1 < t < 2,
v
1 t
vdt v(1) 10kV
C ³1
For 2 < t < 3,
1
4x10 6
= -10t + 30kV
v
t
³ (40x10
2
3
)dt v(2)
Thus
0 t 1
ª10 t ˜ kV,
«
v(t) = «10kV,
1 t 2
«¬ 10 t 30kV, 2 t 3
Chapter 6, Solution 12.
i
dv
3x10 3 x 60(4S)( sin 4S t)
dt
= - 0.7e S sin 4St A
C
P = vi = 60(-0.72)S cos 4S t sin 4S t = -21.6S sin 8S t W
W=
³
t
o
1
pdt
³ 8 21.6S sin 8S t dt
o
21.6S
cos 8S
=
8S
1/ 8
o
= -5.4J
Chapter 6, Solution 13.
Under dc conditions, the circuit becomes that shown below:
i1
10 :
50 :
i2
20 :
+
+
v1
30 :
v2
60V
+
i2 = 0, i1 = 60/(30+10+20) = 1A
v1 = 30i2 = 30V, v2 = 60-20i1 = 40V
Thus, v1 = 30V, v2 = 40V
Chapter 6, Solution 14.
(a) Ceq = 4C = 120 mF
1
C eq
(b)
4
C
4
30
Ceq = 7.5 mF
Chapter 6, Solution 15.
In parallel, as in Fig. (a),
v1 = v2 = 100
+
+
100V
+
v1
C1
+
v2
C2
100V
+
v1 C1
v2
(a)
+
(b)
C2
1 2 1
Cv
x 20x10 6 x100 2
2
2
1
w30 = x30x10 6 x100 2 0.15J
2
w20 =
(b)
0.1J
When they are connected in series as in Fig. (b):
C2
V
C1 C 2
v1
w20 =
30
x100
50
1
x30x10 6 x 60 2
2
w30 =
1
x30x10 6 x 40 2
2
60, v2 = 40
36 mJ
24 mJ
Chapter 6, Solution 16
C eq
14 Cx80
C 80
30

o
C
20 PF
Chapter 6, Solution 17.
(a)
(b)
(c)
4F in series with 12F = 4 x 12/(16) = 3F
3F in parallel with 6F and 3F = 3+6+3 = 12F
4F in series with 12F = 3F
i.e. Ceq = 3F
Ceq = 5 + [6 || (4 + 2)] = 5 + (6 || 6) = 5 + 3 = 8F
3F in series with 6F = (3 x 6)/9 = 6F
1
1 1 1
1
C eq 2 6 3
Ceq = 1F
Chapter 6, Solution 18.
For the capacitors in parallel
C1eq = 15 + 5 + 40 = 60 PF
Hence
1
C eq
1
1
1
20 30 60
1
10
Ceq = 10 PF
Chapter 6, Solution 19.
We combine 10-, 20-, and 30- P F capacitors in parallel to get 60 P F. The 60 - P F
capacitor in series with another 60- P F capacitor gives 30 P F.
30 + 50 = 80 P F, 80 + 40 = 120 P F
The circuit is reduced to that shown below.
12
120
12
80
120- P F capacitor in series with 80 P F gives (80x120)/200 = 48
48 + 12 = 60
60- P F capacitor in series with 12 P F gives (60x12)/72 = 10 P F
Chapter 6, Solution 20.
3 in series with 6 = 6x3/(9) = 2
2 in parallel with 2 = 4
4 in series with 4 = (4x4)/8 = 2
The circuit is reduced to that shown below:
20
1
6
8
2
6 in parallel with 2 = 8
8 in series with 8 = 4
4 in parallel with 1 = 5
5 in series with 20 = (5x20)/25 = 4
Thus Ceq = 4 mF
Chapter 6, Solution 21.
4PF in series with 12PF = (4x12)/16 = 3PF
3PF in parallel with 3PF = 6PF
6PF in series with 6PF = 3PF
3PF in parallel with 2PF = 5PF
5PF in series with 5PF = 2.5PF
Hence Ceq = 2.5PF
Chapter 6, Solution 22.
Combining the capacitors in parallel, we obtain the equivalent circuit shown below:
a
b
40 PF
60 PF
30 PF
20 PF
Combining the capacitors in series gives C1eq , where
1
C1eq
1
1
1
60 20 30
1
10
Thus
Ceq = 10 + 40 = 50 PF
C1eq = 10PF
Chapter 6, Solution 23.
(a)
(b)
3PF is in series with 6PF
v4PF = 1/2 x 120 = 60V
v2PF = 60V
3
v6PF =
(60) 20V
63
v3PF = 60 - 20 = 40V
3x6/(9) = 2PF
Hence w = 1/2 Cv2
w4PF = 1/2 x 4 x 10-6 x 3600 = 7.2mJ
w2PF = 1/2 x 2 x 10-6 x 3600 = 3.6mJ
w6PF = 1/2 x 6 x 10-6 x 400 = 1.2mJ
w3PF = 1/2 x 3 x 10-6 x 1600 = 2.4mJ
Chapter 6, Solution 24.
20PF is series with 80PF = 20x80/(100) = 16PF
14PF is parallel with 16PF = 30PF
(a)
v30PF = 90V
v60PF = 30V
v14PF = 60V
80
v20PF =
x 60 48V
20 80
v80PF = 60 - 48 = 12V
(b)
1 2
Cv
2
w30PF = 1/2 x 30 x 10-6 x 8100 = 121.5mJ
w60PF = 1/2 x 60 x 10-6 x 900 = 27mJ
w14PF = 1/2 x 14 x 10-6 x 3600 = 25.2mJ
w20PF = 1/2 x 20 x 10-6 x (48)2 = 23.04mJ
w80PF = 1/2 x 80 x 10-6 x 144 = 5.76mJ
Since w =
Chapter 6, Solution 25.
(a) For the capacitors in series,
Q1 = Q2
C1v1 = C2v2
v1
v2
C2
C1
vs = v1 + v2 =
C2
v2 v2
C1
C1 C 2
v2
C1
v2
C2
vs
C1 C 2
Similarly, v1
(b) For capacitors in parallel
Q2
C2
C
Qs = Q1 + Q2 = 1 Q 2 Q 2
C2
v1 = v2 =
Q1
C1
C1 C 2
Q2
C2
or
Q2 =
Q1
i=
C2
C1 C 2
C1
Qs
C1 C 2
dQ
dt
i1
C1
is ,
C1 C 2
i2
C2
is
C1 C 2
Chapter 6, Solution 26.
(a)
Ceq = C1 + C2 + C3 = 35PF
(b)
Q1 = C1v = 5 x 150PC = 0.75mC
Q2 = C2v = 10 x 150PC = 1.5mC
Q3 = C3v = 20 x 150 = 3mC
(c)
w=
1
C eq v 2
2
1
x35x150 2 PJ = 393.8mJ
2
C1
vs
C1 C 2
Chapter 6, Solution 27.
1
C eq
(a)
Ceq =
(b)
(c)
1
1
1
C1 C 2 C 3
20
PF
7
1 1
1
5 10 20
7
20
2.857PF
Since the capacitors are in series,
20
Q1 = Q2 = Q3 = Q = Ceqv =
x 200PV 0.5714mV
7
1
1 20
w = C eq v 2
x x 200 2 PJ 57.143mJ
2
2 7
Chapter 6, Solution 28.
We may treat this like a resistive circuit and apply delta-wye transformation, except that
R is replaced by 1/C.
Cb
50 PF
Cc
20 PF
Ca
§ 1 ·§ 1 · § 1 ·§ 1 · § 1 ·§ 1 ·
¨ ¸¨ ¸ ¨ ¸¨ ¸ ¨ ¸¨ ¸
1
© 10 ¹© 40 ¹ © 10 ¹© 30 ¹ © 30 ¹© 40 ¹
1
Ca
30
3
1
1
2
=
40 10 40 10
Ca = 5PF
1
1
1
400 300 1200
1
10
Cb = 15PF
1
C6
2
30
1
1
1
1
400 300 1200
1
Cc
40
Cc = 3.75PF
4
15
Cb in parallel with 50PF = 50 + 15 = 65PF
Cc in series with 20PF = 23.75PF
65x 23.75
65PF in series with 23.75PF =
17.39PF
88.75
17.39PF in parallel with Ca = 17.39 + 5 = 22.39PF
Hence Ceq = 22.39PF
Chapter 6, Solution 29.
(a)
C in series with C = C/(2)
C/2 in parallel with C = 3C/2
3C
in series with C =
2
3
3C
2
C
5
2
Cx
3C
5
C
C
in parallel with C = C + 3
5
5
(b)
2C
Ceq
2C
1
C eq
1
1
2C 2C
Ceq = C
1
C
1.6 C
Chapter 6, Solution 30.
1 t
idt i(0)
C ³o
For 0 < t < 1, i = 60t mA,
10 3 t
vo
60tdt 0 10 t 2 kV
3x10 6 ³o
vo(1) = 10kV
vo =
For 1< t < 2, i = 120 - 60t mA,
10 3 t
vo =
(120 60t )dt v o (1)
3x10t 6 ³1
= [40t –– 10t2 ] 1 10kV
= 40t –– 10t2 - 20
ª10t 2 kV,
0 t 1
v o (t) «
2
«¬40t 10 t 20kV, 1 t 2
Chapter 6, Solution 31.
i s (t)
ª20 tmA,
«20mA,
«
«¬ 50 10 t ,
0 t 1
1 t 3
3 t 5
Ceq = 4 + 6 = 10PF
1 t
v
idt v(0)
C eq ³o
For 0 < t < 1,
v
10 3
10x10 6
For 1 < t < 3,
10 3
v
10
2 t 1kV
For 3 < t < 5,
10 3
v
10
t
t
³ 20t dt + 0 = t
o
³ 20dt v(1)
1
t
2
2( t 1) 1kV
³ 10(t 5)dt v(3)
3
kV
t 2 5 3t 5kV
ª t 2 kV,
0 t 1
«
1 t 3
«2t 1kV,
2
« t 5t 11kV, 3 t 5
¬
v( t )
dv
dv
6x10 6
dt
dt
0 t 1
ª12 tmA,
«12mA,
1 t 3
«
«¬12 30mA, 3 t 5
i1
i1
t 2 5t 11kV
C1
dv
dv
4x10 6
dt
dt
0 t 1
ª8tmA,
«8mA,
1 t 3
«
«¬8t 20mA, 3 t 5
C2
Chapter 6, Solution 32.
(a)
Ceq = (12x60)/72 = 10 P F
t
v1
10 3
30e 2t dt v1 (0)
6 ³
12 x10 0
v2
10 3
30e 2t dt v 2 (0)
6 ³
60 x10 0
t
(b)
t
1250e 2t
250e 2t
0
t
0
1250e 2t 1300
50
20
250e 2t 230
At t=0.5s,
v1
1250e 1 1300
w12 PF
w20 PF
w40 PF
840.15,
1
x12 x10 6 x(840.15) 2
2
1
x 20 x10 6 x(138.03) 2
2
1
x 40 x10 6 x(138.03) 2
2
v2
250e 1 230
4.235 J
0.1905 J
0.381 J
138.03
Chapter 6, Solution 33
Because this is a totally capacitive circuit, we can combine all the capacitors using
the property that capacitors in parallel can be combined by just adding their
values and we combine capacitors in series by adding their reciprocals.
3F + 2F = 5F
1/5 +1/5 = 2/5 or 2.5F
The voltage will divide equally across the two 5F capacitors. Therefore, we get:
VTh = 7.5 V, CTh = 2.5 F
Chapter 6, Solution 34.
i = 6e-t/2
di
§1·
v L
10 x10 3 (6)¨ ¸e t / 2
dt
©2¹
-t/2
= -30e mV
v(3) = -300e-3/2 mV = -0.9487 mV
p = vi = -180e-t mW
p(3) = -180e-3 mW = -0.8 mW
Chapter 6, Solution 35.
v
L
di
dt
L
V
'i / 't
60 x10 3
0.6 /(2)
200 mH
Chapter 6, Solution 36.
v
di 1
x10 3 (12)(2)( sin 2 t )V
dt 4
= - 6 sin 2t mV
L
p = vi = -72 sin 2t cos 2t mW
But 2 sin A cos A = sin 2A
p = -36 sin 4t mW
Chapter 6, Solution 37.
v
di
dt
L
12 x10 3 x 4(100) cos100t
= 4.8 cos 100t V
p = vi = 4.8 x 4 sin 100t cos 100t = 9.6 sin 200t
w=
t
11 / 200
o
o
³ pdt ³
9.6 sin 200 t
9.6
/ 200
cos 200t 11
J
o
200
48(cos S 1)mJ 96 mJ
Chapter 6, Solution 38.
v
L
di
dt
40x10 3 e 2 t 2te 2 t dt
= 40(1 2t )e 2 t mV, t ! 0
Chapter 6, Solution 39
v
i
di

o i
dt
L
1
200x10
1 t
³ idt i(0)
L 0
t
(3t 2
3 ³ 0
5( t 3 t 2 4t )
t
0
2t 4)dt 1
1
i(t) = 5t3 + 5t2 + 20t + 1 A
Chapter 6, Solution 40
di
dt
20 x10 3
di
dt
v
L
i
­ 10t , 0 t 1 ms
°
® 20 - 10t, 1 t 3 ms
°- 40 10t, 3 t 4 ms
¯
di
dt
v
­ 10 x10 3 ,
°
3
®- 10x10 ,
°10x10 3 ,
¯
­ 200 V,
°
®- 200 V,
° 200 V,
¯
0 t 1 ms
1 t 3 ms
3 t 4 ms
0 t 1 ms
1 t 3 ms
3 t 4 ms
which is sketched below.
v(t) V
200
0
-200
1
2
3
4
t(ms)
Chapter 6, Solution 41.
i
1 t
vdt i(0)
L ³0
§1· t
2 t
¨ ¸ ³o 20 1 2 dt 0.3
©2¹
§ 1
·
= 10¨ t e 2t ¸ to 0. 3 10t 5e 2t 4. 7 A
© 2
¹
At t = ls, i = 10 - 4.7 + 5e-2 = 5.977 A
w
1 2
L i = 35.72J
2
Chapter 6, Solution 42.
i
1 t
vdt i(0)
L ³o
For 0 < t < 1, i
1 t
v( t )dt 1
5 ³o
10 t
dt 1 2t 1 A
5 ³0
For 1 < t < 2, i = 0 + i(1) = 1A
1
10dt i(2)
5³
= 2t - 3 A
For 2 < t < 3, i =
2t 2t 1
For 3 < t < 4, i = 0 + i(3) = 3 A
1 t
10dt i(4)
5 ³4
= 2t - 5 A
For 4 < t < 5, i =
Thus, i (t )
ª 2t 1A,
«
«1A,
« 2t 3 A,
«
«3 A,
«¬ 2t 5,
0 t 1
1 t 2
2t3
3t 4
4t 5
2 t 4t 3
Chapter 6, Solution 43.
2
1
1
Li( t ) Li (f)
2
2
t
w = L ³ idt
f
1
x80 x10 3 x 60x10 3 0
2
= 144 PJ
Chapter 6, Solution 44.
1 t
vdt i t o
L ³t o
i
1 t
(4 10 cos 2t )dt 1
5 ³o
= 0.8t + sin 2t -1
Chapter 6, Solution 45.
i(t) =
1 t
v( t ) i(0)
L ³o
For 0 < t < 1, v = 5t
i
1
10x10 3
t
³ 5t dt + 0
o
= 0.25t2 kA
For 1 < t < 2, v = -10 + 5t
i
1
10x10 3
t
³ (10 5t )dt i(1)
1
t
³ (0.5t 1)dt 0.25kA
1
= 1 - t + 0.25t2 kA
i( t )
ª0.25t 2 kA,
0 t 1
«
2
«¬1 t 0.25t kA, 1 t 2
Chapter 6, Solution 46.
Under dc conditions, the circuit is as shown below:
2:
iL
+
vC
4:
3A
By current division,
iL
4
(3)
42
wL
1 2
L iL
2
1§1· 2
¨ ¸(2)
2©2¹
wc
1
C v c2
2
1
(2)( v)
2
2A, vc = 0V
1J
0J
Chapter 6, Solution 47.
Under dc conditions, the circuit is equivalent to that shown below:
R
+
2:
5A
iL
2
(5)
R2
10
, vc
R2
Ri L
10R
R2
vC
iL
1 2
Cv c
2
1 2
wL
Li1
2
If wc = wL,
wc
100R 2
(R 2) 2
100
2x10 3 x
(R 2) 2
80x10 6 x
100R 2
80x10 x
(Rx 2) 2
6
2x10 3 x100
(R 2) 2
80 x 10-3R2 = 2
R = 5:
Chapter 6, Solution 48.
Under dc conditions, the circuit is as shown below:
4:
iL1
+
+
30V
+
vC1
i L1
i L2
v C1
6i L1
vC 2
0V
30
46
18V
3A
iL2
vC2
6:
Chapter 6, Solution 49.
(a)
L eq
5 6 1 4 4
5 6 3 7H
(b)
L eq
12 1 6 6
12 4
(c)
L eq
4 236
44
3H
2H
Chapter 6, Solution 50.
L eq
10 5 4 12 3 6
= 10 + 5||(3 + 2) = 10 + 2.5 = 12.5 mH
Chapter 6, Solution 51.
1
L
1
1
1
60 20 30
L eq
10 25 10
1
10
L = 10 mH
10x35
45
= 7.778 mH
Chapter 6, Solution 52.
3//2//6 = 1H, 4//12 = 3H
After the parallel combinations, the circuit becomes that shown below.
3H
a
1H
1H
Lab = (3+1)//1 = (4x1)/5 = 0.8 H
b
Chapter 6, Solution 53.
L eq
6 10 8 >5 (8 12) 6 (8 4)@
16 8 (4 4) 16 4
Leq = 20 mH
Chapter 6, Solution 54.
L eq
4 (9 3) 10 0 6 12
4 12 (0 4)
43
Leq = 7H
Chapter 6, Solution 55.
(a) L//L = 0.5L, L + L = 2L
Leq
L 2 L // 0.5L
L
2 Lx0.5 L
2 L 0.5L
1.4 L
(b) L//L = 0.5L, L//L + L//L = L
Leq = L//L = 0.5L
Chapter 6, Solution 56.
1 L
3 3
L
Hence the given circuit is equivalent to that shown below:
LLL
L
L/3
L/3
L
L eq
2 ·
§
L ¨L L¸
3 ¹
©
5
Lx L
3
5
L L
3
5
L
8
Chapter 6, Solution 57.
Let v
L eq
di
dt
(1)
di
v2
dt
i2 = i –– i1
i = i1 + i2
di
di
v2
v 2 3 1 or 1
dt
dt
3
and
di
di
v2 2 5 2 0
dt
dt
di
di
v2 2 5 2
dt
dt
Incorporating (3) and (4) into (5),
di
v
di
di
di
v2 2 5 5 1 7 5 2
dt
dt
dt
dt
3
di
§ 5·
v 2 ¨1 ¸ 7
dt
© 3¹
35 di
v2
8 dt
v
v1 v 2
4
Substituting this into (2) gives
v
4
di 35 di
dt 8 dt
67 di
8 dt
Comparing this with (1),
L eq
67
8
8.375H
(2)
(3)
(4)
(5)
Chapter 6, Solution 58.
v
L
di
dt
3
di
dt
3 x slope of i(t).
Thus v is sketched below:
v(t) (V)
6
t (s)
1
2
3
4
5
6
7
-6
Chapter 6, Solution 59.
(a) v s
di
dt
v1
v1
(b)
L1 L 2
di
dt
vs
L1 L 2
di
di
L1 , v 2 L 2
dt
dt
L1
L2
vs , vL
vs
L1 L 2
L1 L 2
vi
v2
L1
di1
dt
i s i1 i 2
di s di1 di 2
dt
dt
dt
i1
1
vdt
L1 ³
L2
di 2
dt
L L2
v
v
v 1
L1 L 2
L1 L 2
L1 L 2 di s
L2
1
dt
is
³
L1 L1 L 2 dt
L1 L 2
1
L2
1
vdt
L2 ³
i2
L1 L 2 di s
dt
L
dt
1
2
L1
is
L1 L 2
³L
Chapter 6, Solution 60
Leq
di
dt
15
8
3 // 5
15 d
4e 2t
8 dt
vo
Leq
io
I
vo (t )dt io (0)
L ³0
15e 2t
t
t
2
1
(15)e 2t
5 ³0
2 1.5e 2t
t
0.5 1.5e 2t A
0
Chapter 6, Solution 61.
(a)
is = i1 + i2
i s (0) i1 (0) i 2 (0)
6 4 i 2 ( 0)
i2(0) = 2mA
(b) Using current division:
20
2.4e-2t mA
i1
i s 0.4 6e 2 t
30 20
i 2 i s i1 3.6e-2t mA
30 x 20
12mH
(c) 30 20
50
di
d
v1 L
10x10 3
6e 2 t x10 3 -120e-2t PV
dt
dt
di
d
v2 L
12x10 3
6e 2 t x10 3 -144e-2t PV
dt
dt
(d)
w 10 mH
0.8e 4 t
1
x30x10 3 36e 4 t x10 6
2
t
1
2
PJ
= 24.36nJ
1
w 30 mH
x30 x10 3 5.76e 4 t x10 6 t
2
= 11.693nJ
1
w 20 mH
x 20x10 3 12.96e 4 t x10 6
2
= 17.54 nJ
1/ 2
t 1/ 2
Chapter 6, Solution 62.
(a)
v
Leq
Leq
25 20 // 60
di
dt

o
i
25 20 x60
80
40 mH
1
v(t )dt i (0)
Leq ³
Using current division,
60
3
1
i1
i
i, i 2
i
80
4
4
3
i1 (0)
i (0)

o 0.75i (0)
4
0.01
t
10 3
12e 3t dt i (0)
40 x10 3 ³0

o
1
(0.1e 3t 0.08667) A - 25e -3t 21.67 mA
4
i2 (0) 25 21.67 3.33 mA
i2
3
(0.1e 3t 0.08667) A
4
- 25e -3t 21.67 mA
(b) i1
i2
- 75e -3t 65 mA
Chapter 6, Solution 63.
We apply superposition principle and let
vo
v1 v 2
where v1 and v2 are due to i1 and i2 respectively.
di1
dt
v1
L
v2
di
L 2
dt
2
di1
dt
di
2 2
dt
­ 2,
®
¯ 2,
­ 4,
°
® 0,
° 4,
¯
0t 3
3t 6
0t 2
2t 4
4t6
i (0)
0.01333
0.1(e 3t 1) i (0)
v1
v2
2
4
0
3
6
t
0
-2
2
4
6
-4
Adding v1 and v2 gives vo, which is shown below.
vo(t) V
6
2
0
2 3
4
6
t (s)
-2
-6
Chapter 6, Solution 64.
(a) When the switch is in position A,
i=-6 =i(0)
When the switch is in position B,
i (f) 12 / 4 3,
W L / R 1/ 8
i (t )
i (f) [i (0) i (f)]e t / L
3 9e 8t A
(b) -12 + 4i(0) + v=0, i.e. v=12 –– 4i(0) = 36 V
(c) At steady state, the inductor becomes a short circuit so that
v= 0 V
t
Chapter 6, Solution 65.
1
1
L1i12
x5x (4) 2 40 W
2
2
1
w 20
(20)(2) 2 40 W
2
(b) w = w5 + w20 = 80 W
dv
(c) i1 L1
5 200 50e 200 t x10 3
dt
= -50e-200tA
(a)
w5
i2
L2
i2
L2
dv
20(200) 50e 200 t x10 3
dt
= -200e-200tA
dv
dt
20(200) 50e 200 t x10 3
= -200e-200t A
(d)
i = i1 + i2 = -250e-200t A
Chapter 6, Solution 66.
20 16 60 40
L eq
v
i
i
L
36 24
60mH
di
dt
1 t
vdt i(0)
L ³o
t
1
12 sin 4t dt + 0 mA
3 ³o
60x10
50 cos 4t ot 50(1 - cos 4t) mA
60 40
24mH
d
di
24x10 3 (50)(1 cos 4t )mV
dt
dt
= 4.8 sin 4t mV
v
L
Chapter 6, Solution 67.
1
vi dt, RC = 50 x 103 x 0.04 x 10-6 = 2 x 10-3
RC ³
10 3
vo
10 sin 50t dt
2 ³
vo = 100 cos 50t mV
vo
Chapter 6, Solution 68.
1
vi dt + v(0), RC = 50 x 103 x 100 x 10-6 = 5
³
RC
1 t
vo = ³ 10dt 0 2t
5 o
The op amp will saturate at vo = r 12
vo
-12 = -2t
t = 6s
Chapter 6, Solution 69.
RC = 4 x 106 x 1 x 10-6 = 4
vo
1
v i dt
RC ³
1
v i dt
4³
For 0 < t < 1, vi = 20, v o
For 1 < t < 2, vi = 10, v o
-5t mV
1 t
10dt v(1)
4 ³1
= -2.5t - 2.5mV
For 2 < t < 4, vi = - 20, v o
1 t
20dt v(2)
4 ³2
= 5t - 17.5 mV
For 4 < t < 5m, vi = -10, v o
For 5 < t < 6, vi = 20, v o
1 t
20dt
4 ³o
1 t
10dt v(4)
4 ³4
= 2.5t - 7.5 mV
1 t
20dt v(5)
4 ³5
= - 5t + 30 mV
2.5( t 1) 5
5( t 2) 7.5
2.5( t 4) 2.5
5( t 5) 5
Thus vo(t) is as shown below:
5
25
0
1
2
3
4
5
6
7
5
Chapter 6, Solution 70.
One possibility is as follows:
1
50
RC
1
Let R = 100 k:, C
50 x100 x10 3
0.2PF
Chapter 6, Solution 71.
By combining a summer with an integrator, we have the circuit below:
+
1
1
1
v1dt v 2 dt v 2 dt
³
³
R 1C
R 2C
R 2C ³
For the given problem, C = 2PF,
vo
R1C = 1
R2C = 1/(4)
R3C = 1/(10)
R1 = 1/(C) = 1006/(2) = 500 k:
R2 = 1/(4C) = 500k:/(4) = 125 k:
R3 = 1/(10C) = 50 k:
Chapter 6, Solution 72.
The output of the first op amp is
v1
1
1
v i dt = 3
³
RC
10x10 x 2 x10 6
t
³ idt
o
100 t
2
= - 50t
vo
1
1
v i dt = 3
³
RC
20x10 x 0.5x10 6
t
³ (50t )dt
o
= 2500t2
At t = 1.5ms,
v o 2500(1.5) 2 x10 6
5.625 mV
Chapter 6, Solution 73.
Consider the op amp as shown below:
Let va = vb = v
At node a,
0v
R
v vo
R
2v - vo = 0
(1)
R
R
a
R
v
+
R
v
vo
b
vi
At node b,
vi v
R
+
v vo
dv
C
R
dt
+
C
2v v o RC
vi
dv
dt
(2)
Combining (1) and (2),
v i v o v o RC dv o
2 dt
or
2
v i dt
RC ³
vo
showing that the circuit is a noninverting integrator.
Chapter 6, Solution 74.
RC = 0.01 x 20 x 10-3 sec
dv i
dt
vo
RC
vo
ª 2V,
« 2V,
«
«¬ 2V,
0.2
dv
m sec
dt
0 t 1
1 t 3
3 t 4
Thus vo(t) is as sketched below:
vo(t) (V)
2
t (ms)
1
-2
2
3
Chapter 6, Solution 75.
v0
RC
dv i
, RC
dt
vo
2.5
d
(12t )
dt
250 x10 3 x10x10 6
2.5
-30 mV
Chapter 6, Solution 76.
vo
vo
dv i
dt
dv
0.5 i
dt
RC
, RC = 50 x 103 x 10 x 10-6 = 0.5
ª 10, 0 t 5
«5,
5t 5
¬
The input is sketched in Fig. (a), while the output is sketched in Fig. (b).
vo(t) (V)
vi(t) (V)
5
5
t (ms)
0
5
10
t (ms)
15
0
5
10
(a)
-10
(b)
Chapter 6, Solution 77.
i = iR + i C
vi 0
R
0 v0
d
0 vo
C
dt
RF
R F C 10 6 x10 6
1
15
dv ·
§
¨ v o o ¸
dt ¹
©
Hence v i
Thus vi is obtained from vo as shown below:
––dvo(t)/dt
–– vo(t) (V)
4
4
t (ms)
t (ms)
0
1
2
3
0
4
1
2
3
4
-4
-4
vi(t) (V)
8
t (ms)
-4
1
2
3
4
-8
Chapter 6, Solution 78.
d 2 vo
dt
10 sin 2 t 2dv o
vo
dt
Thus, by combining integrators with a summer, we obtain the appropriate analog
computer as shown below:
2vo
t=0
+
C
C
R
R
+
d2vo/dt
2
R
R
+
-dvo/dt
+
vo
R
d2vo/dt
2
R
R/2
+
dvo/dt
R
R
+
sin2t
R/10
+
-sin2t
Chapter 6, Solution 79.
We can write the equation as
dy
f (t ) 4 y (t )
dt
which is implemented by the circuit below.
1V
t=0
C
R
R
R
R/4
dy/dt
+
+
-y
R
f(t)
R
+
dy/dt
Chapter 6, Solution 80.
From the given circuit,
d 2 vo
dt 2
f (t) 1000k:
1000k: dv o
vo 5000k:
200k: dt
or
d 2 vo
dv
5 o 2v o
2
dt
dt
f (t)
Chapter 6, Solution 81
We can write the equation as
d 2v
dt 2
5v 2 f (t )
which is implemented by the circuit below.
C
C
R
R
2
2
d v/dt
+
R
R/5
-
-dv/dt
+
v
+
R/2
f(t)
d2v/dt2
Chapter 6, Solution 82
The circuit consists of a summer, an inverter, and an integrator. Such circuit is shown
below.
10R
R
R
R
+
+
vo
R
C=1/(2R)
R
+
+
vs
-
Chapter 6, Solution 83.
Since two 10PF capacitors in series gives 5PF, rated at 600V, it requires 8 groups in
parallel with each group consisting of two capacitors in series, as shown below:
+
600
Answer: 8 groups in parallel with each group made up of 2 capacitors in series.
Chapter 6, Solution 84.
'I
'q
't
'I x 't = 'q
'q = 0.6 x 4 x 10-6
= 2.4PC
C
2.4 x106
150nF
(36 20)
'q
'v
Chapter 6, Solution 85.
It is evident that differentiating i will give a waveform similar to v. Hence,
di
v L
dt
But,
i
ª4 t ,0 t 1
«8 4 t ,1 t 2
¬
v
L
v
ª5mV,0 t 1
« 5mV,1 t 2
¬
di
dt
Thus, 4L = 5 x 10-3
ª4L,0 t 1
« 4L,1 t 2
¬
L = 1.25 mH in a 1.25 mH inductor
Chapter 6, Solution 86.
(a) For the series-connected capacitor
Cs =
1
1 1
1
.... C C
C
C
8
For the parallel-connected strings,
C eq
10C s
10C s
8
10 x
1000
PF 1250PF
3
(b)
vT = 8 x 100V = 800V
w
1
C eq v T2
2
= 400J
1
1250 x10 6 (800) 2
2
Chapter 7, Solution 1.
Applying KVL to Fig. 7.1.
1 t
³ i dt Ri 0
C -f
Taking the derivative of each term,
i
di
R
0
C
dt
di
dt
or
i
RC
Integrating,
§ i( t ) ·
¸
ln¨
© I0 ¹
-t
RC
i( t )
I 0 e - t RC
v( t )
Ri( t )
RI 0 e - t RC
v(t )
V0e- t RC
or
Chapter 7, Solution 2.
W R th C
where R th is the Thevenin equivalent at the capacitor terminals.
R th
W
120 || 80 12
60 u 0.5 u 10 -3
60 :
30 ms
Chapter 7, Solution 3.
5k:,
(a) RTh
10 // 10
(b) RTh
20 //(5 25) 8
W
5 x10 3 x 2 x10 6
RTh C
20:,
W
RTh C
Chapter 7, Solution 4.
W
R eq C eq
where C eq
C1C 2
,
C1 C 2
W
R eq
R 1R 2
R1 R 2
R 1 R 2 C1C 2
( R 1 R 2 )(C 1 C 2 )
20 x0.3
10 ms
6s
Chapter 7, Solution 5.
v( t ) v(4) e -(t -4) W
where v(4) 24 ,
W RC (20)(0.1)
-(t - 4) 2
v( t ) 24 e
v(10) 24 e -6 2 1.195 V
2
Chapter 7, Solution 6.
vo
v ( 0)
2
(24)
10 2
v( t )
voe t / W , W
v( t )
4e 12.5t V
RC
4V
2
25
40 x106 x 2 x103
Chapter 7, Solution 7.
v( t ) v(0) e - t W ,
W R th C
where R th is the Thevenin resistance across the capacitor. To determine R th , we insert a
1-V voltage source in place of the capacitor as shown below.
8:
i2
i
i1
+
0.5 V
10 :
+
v=1
i1
1
10
i
i1 i 2
R th
W
1
i
0.1 ,
0.1 i2
1
16
13
80
80
13
80
8
u 0.1
R th C
13
13
-13t 8
v( t ) 20 e
V
1 0.5
8
1
16
+
1V
Chapter 7, Solution 8.
(a)
W
RC
-i
C
dv
dt
- 0.2 e -4t
R
1
4
1
4C
C (10)(-4) e-4t

o C
50 :
1
0.25 s
4
1
1
w C (0)
CV02
(5 u 10 -3 )(100) 250 mJ
2
2
1 1
1
wR
u CV02
CV02 1 e -2t 0 W
2 2
2
1
o e -8t 0
0.5 1 e -8t 0 
2
8t 0
or
e
2
1
t0
ln (2) 86.6 ms
8
W
(b)
(c)
(d)
RC
Chapter 7, Solution 9.
v( t )
v(0) e- t W ,
R eq
2 8 || 8 6 || 3
W
5 mF
R eq C
v( t )
(0.25)(8)
20 e -t 2 V
W
242
2
R eq C
8:
Chapter 7, Solution 10.
io
15 :
i
10 :
iT
+
10 mF
4:
v
(10)(3)
2A
15
i.e. if i(0) 3 A , then i o (0) 2 A
i T (0) i(0) i o (0) 5 A
v(0) 10i(0) 4i T (0) 30 20 50 V
across the capacitor terminals.
15 i o
10 i 
o i o
R th
4 10 || 15
W
R th C
v( t )
iC
iC
4 6 10 :
(10)(10 u 10 -3 )
-t W
0.1
-10t
v(0) e
50 e
dv
C
(10 u 10 -3 )(-500 e -10t )
dt
- 5 e -10t A
By applying the current division principle,
15
i( t )
( - i ) -0.6 i C 3 e -10t A
10 15 C
Chapter 7, Solution 11.
Applying KCL to the RL circuit,
1
v
v dt 0
³
L
R
Differentiating both sides,
v 1 dv
0 
o
L R dt
v A e -Rt L
dv R
v
dt L
0
If the initial current is I 0 , then
v(0) I 0 R A
v
i
i
i
i( t )
I 0 R e -t W ,
W
L
R
1 t
³ v(t ) dt
L -f
- W I 0 R -t W t
e -f
L
- I 0 R e -t W
I 0 e -t W
Chapter 7, Solution 12.
When t < 0, the switch is closed and the inductor acts like a short circuit to dc. The 4 :
resistor is short-circuited so that the resulting circuit is as shown in Fig. (a).
3:
12 V
i(0-)
+
4:
(a)
2H
(b)
12
4A
3
Since the current through an inductor cannot change abruptly,
i(0) i(0 ) i(0 ) 4 A
i (0 )
When t > 0, the voltage source is cut off and we have the RL circuit in Fig. (b).
L 2
W
0.5
R 4
Hence,
i( t ) i(0) e - t W 4 e -2t A
Chapter 7, Solution 13.
L
R th
where R th is the Thevenin resistance at the terminals of the inductor.
W
R th
W
70 || 30 80 || 20
2 u 10
37
-3
21 16
37 :
81.08 Ps
Chapter 7, Solution 14
Converting the wye-subnetwork to delta gives
16 :
R2
80mH
R1
R3
30 :
R1
10 x 20 20 x50 50 x10
20
1700 / 20
85:,
R2
1700
50
34: ,
R3
1700
10
30//170 = (30x170)/200 = 25.5 : , 34//16=(34x16)/50 =10.88 :
RTh
85 //( 25.5 10.88)
85 x36.38
121.38
25.476:,
W
L
RTh
80 x10 3
25.476
3.14 ms
170:
Chapter 7, Solution 15
(a) RTh
12 10 // 40
20:,
W
(b) RTh
40 // 160 8
40:,
W
L
RTh
L
RTh
5 / 20
0.25s
(20 x10 3 ) / 40
0.5 ms
Chapter 7, Solution 16.
W
L eq
R eq
(a)
L eq
W
(b)
L and R eq
R2 R 1R 3
R1 R 3
R 2 (R 1 R 3 ) R 1 R 3
R1 R 3
L( R 1 R 3 )
R 2 (R 1 R 3 ) R 1 R 3
where L eq
W
L1 L 2
R 1R 2
and R eq R 3 L1 L 2
R1 R 2
L1L 2 (R 1 R 2 )
(L 1 L 2 ) ( R 3 ( R 1 R 2 ) R 1 R 2 )
Chapter 7, Solution 17.
i( t )
i(0) e - t W ,
i( t )
2 e -16t
di
dt
vo (t )
3i L
vo (t )
- 2 e -16t V
W
L
R eq
14
4
1
16
6 e-16t (1 4)(-16) 2 e-16t
R 3 (R 1 R 2 ) R 1 R 2
R1 R 2
Chapter 7, Solution 18.
If v( t )
0 , the circuit can be redrawn as shown below.
+
0.4 H
Req
vo(t)
i(t)
R eq
2 || 3
6
,
5
W
i(0) e - t W e -3t
di - 2
v o ( t ) -L
(-3) e -3t
dt
5
L
R
2 5
u
5 6
1
3
i( t )
1.2 e -3t V
Chapter 7, Solution 19.
i
1V
+
10 :
i1
i1
i2
i/2
i2
40 :
To find R th we replace the inductor by a 1-V voltage source as shown above.
10 i1 1 40 i 2 0
i i2 i 2
and
i i1
But
i1 2 i 2 2 i
i.e.
1
10 i 1 20 i 0 
o i
30
R th
W
i( t )
1
i
L
R th
30 :
6
30
2 e -5t A
0.2 s
Chapter 7, Solution 20.
(a).
1

o R 50L
50
di
-v L
dt
-50t
- 150 e
L(30)(-50) e -50t 
o L
W
R
(b).
(c).
(d).
i.e.
L
R
0.1 H
5:
50L
L
1
20 ms
R 50
1 2
1
w
L i ( 0)
(0.1)(30) 2 45 J
2
2
Let p be the fraction
1
1
L I0 ˜ p
L I 1 e -2t 0 W
2
2 0
p 1 e -(2)(10) 50 1 e -0.4 0.3296
p 33%
W
Chapter 7, Solution 21.
The circuit can be replaced by its Thevenin equivalent shown below.
Rth
+
Vth
Vth
80
(60)
80 40
R th
I
2H
40 V
80
R
3
Vth
40
R th 80 3 R
40 || 80 R
i ( 0)
i(f)
2
1 § 40 ·
¸
w
(2)¨
2 © R 80 3 ¹
40
40
o R
1 
R 80 3
3
R 13.33 :
1 2
LI
2
1
Chapter 7, Solution 22.
i( t )
i(0) e - t W ,
W
L
R eq
R eq
5 || 20 1 5 : ,
i( t )
10 e -2.5t A
W
2
5
Using current division, the current through the 20 ohm resistor is
5
-i
io
(-i)
-2 e -2.5t
5 20
5
v( t )
20 i o
- 40 e -2.5t V
Chapter 7, Solution 23.
Since the 2 : resistor, 1/3 H inductor, and the (3+1) : resistor are in parallel,
they always have the same voltage.
2
2
1.5 
o i(0) -1.5
2 3 1
The Thevenin resistance R th at the inductor’’s terminals is
13 1
L
4
R th 2 || (3 1)
,
W
3
R th 4 3 4
-i
i( t )
vL
vo
vx
i(0) e - t W -1.5 e -4t , t ! 0
di
vo L
-1.5(-4)(1/3) e -4t
dt
2 e -4t V , t ! 0
1
v
3 1 L
0.5 e -4t V , t ! 0
Chapter 7, Solution 24.
(a) v( t )
- 5 u(t)
(b) i( t )
-10 > u ( t ) u ( t 3)@ 10> u ( t 3) u ( t 5)@
= - 10 u(t ) 20 u(t 3) 10 u(t 5)
( t 1) > u ( t 1) u ( t 2)@ > u ( t 2) u ( t 3)@
(4 t ) > u ( t 3) u ( t 4)@
= ( t 1) u ( t 1) ( t 2) u ( t 2) ( t 3) u ( t 3) ( t 4) u ( t 4)
= r(t 1) r(t 2) r(t 3) r(t 4)
(c) x ( t )
(d) y( t )
2 u (-t ) 5 > u ( t ) u ( t 1)@
= 2 u(-t ) 5 u(t ) 5 u(t 1)
Chapter 7, Solution 25.
v(t) = [u(t) + r(t –– 1) –– r(t –– 2) –– 2u(t –– 2)] V
Chapter 7, Solution 26.
(a)
v1 ( t )
v1 ( t )
u ( t 1) u ( t ) > u ( t 1) u ( t )@
u(t 1) 2 u(t ) u(t 1)
(b)
v 2 (t)
v 2 (t)
v 2 (t)
( 4 t ) > u ( t 2) u ( t 4) @
-( t 4) u ( t 2) ( t 4) u ( t 4)
2 u(t 2) r(t 2) r(t 4)
(c)
v 3 (t)
v 3 (t)
2 > u(t 2) u(t 4)@ 4 > u(t 4) u(t 6)@
2 u(t 2) 2 u(t 4) 4 u(t 6)
(d)
v 4 (t)
v 4 (t)
v 4 (t)
-t > u ( t 1) u ( t 2)@ -t u(t 1) t u ( t 2)
(-t 1 1) u ( t 1) ( t 2 2) u ( t 2)
- r(t 1) u(t 1) r(t 2) 2 u(t 2)
Chapter 7, Solution 27.
v(t) is sketched below.
v(t)
2
1
0
-1
1
2
3
4
t
Chapter 7, Solution 28.
i(t) is sketched below.
i(t)
1
0
1
3
2
4
t
-1
Chapter 7, Solution 29
x(t)
(a)
3.679
0
(b)
1
t
y(t)
27.18
0
t
(c)
z (t )
cos 4tG (t 1)
cos 4G (t 1)
0.6536G (t 1) , which is sketched below.
z(t)
0
1
t
-0.653 G (t )
Chapter 7, Solution 30.
(a)
³
4 t 2 G( t 1) dt
(b)
³
cos(2St ) G( t 0.5) dt
10
0
f
-f
4t 2
t 1
4
cos(2St )
t 0.5
cos S
-1
Chapter 7, Solution 31.
(a)
(b)
e
³ > e G(t 2)@ dt e
³ > 5 G(t ) e G(t ) cos 2St G(t )@ dt
f
- 4t 2
- 4t 2
-16
t 2
-f
f
-t
-f
112 u 10 -9
5 e - t cos(2St )
Chapter 7, Solution 32.
t
(a)
1
4
(b)
³ 1dO
5
³ (t 6)
1
2
O
t
t 1
1
1
³ r (t 1)dt
0
(c )
t
³ u (O )dO
1
4
0
1
³ 0dt ³ (t 1)dt
G (t 2)dt
(t 6) 2
t2
t 14 4.5
2
t 2
16
t 0
5 11 7
Chapter 7, Solution 33.
i( t )
1 t
³ v(t ) dt i(0)
L 0
i( t )
10 -3
10 u 10 -3
i( t )
2 u( t 2 ) A
³ 20 G(t 2) dt 0
t
0
Chapter 7, Solution 34.
d
>u ( t 1) u ( t 1)@ G( t 1)u ( t 1) dt
u ( t 1)G( t 1) G( t 1) x 1 0 x G( t 1)
(a)
d
>r ( t 6) u ( t 2)@ u ( t 6)u ( t 2) dt
r ( t 6)G( t 2) u ( t 6) x 1 0 x G( t 2)
(b)
G( t 1)
u ( t 6)
d
>sin 4t u (t 3)@ 4 cos 4t u ( t 3) sin 4tG( t 3)
dt
4 cos 4t u ( t 3) sin 4x3G( t 3)
(c)
4 cos 4t u ( t 3) 0.5366G( t 3)
Chapter 7, Solution 35.
(a)
v( t )
v( t )
A e -5t 3 , v(0)
- 2 e -5t 3 V
A
-2
(b)
v( t )
v( t )
A e 2t 3 , v(0)
5 e 2t 3 V
A
5
Chapter 7, Solution 36.
(a)
(b)
v( t ) A B e-t , t ! 0
A 1,
v(0) 0 1 B
v( t )
1 e -t V , t ! 0
or
B -1
v( t ) A B e t 2 , t ! 0
A -3 ,
v(0) -6 -3 B
v( t ) - 3 1 e t 2 V , t ! 0
or
B
-3
Chapter 7, Solution 37.
Let v = vh + vp, vp =10.
x
1
vh 4 v

o
0
h
Ae t / 4
vh
v 10 Ae 0.25t
v(0) 2 10 A
v 10 8e 0.25t
(a) W

o
A
8
4s
(b) v(f) 10 V
(c ) v 10 8e 0.25t
Chapter 7, Solution 38
Let i = ip +ih
x
i h 3ih
Let i p
ku (t ),
x
ip
0,
0

o
3ku (t )
2u (t )
ih
Ae 3t u (t )

o
k
2
3
2
u (t )
3
ip
2
( Ae 3t )u (t )
3
i
If i(0) =0, then A + 2/3 = 0, i.e. A=-2/3. Thus
i
2
(1 e 3t )u (t )
3
Chapter 7, Solution 39.
(a)
Before t = 0,
v( t )
1
(20)
4 1
4V
After t = 0,
v( t ) v(f) > v(0) v(f)@ e - t W
W RC (4)(2) 8 , v(0) 4 ,
v( t ) 20 (8 20) e - t 8
v( t ) 20 12 e -t 8 V
v(f)
20
Before t = 0, v v1 v 2 , where v1 is due to the 12-V source and v 2 is
due to the 2-A source.
v1 12 V
To get v 2 , transform the current source as shown in Fig. (a).
v 2 -8 V
Thus,
v 12 8 4 V
(b)
After t = 0, the circuit becomes that shown in Fig. (b).
2F
+
v2
2F
4:
+
8V
12 V
+
3:
3:
(a)
(b)
v( t )
v(f)
v( t )
v( t )
v(f) > v(0) v(f)@ e - t W
12 ,
v(0) 4 ,
W
-t 6
12 (4 12) e
12 8 e -t 6 V
RC
(2)(3)
6
Chapter 7, Solution 40.
(a)
(b)
Before t = 0, v
12 V .
After t = 0, v( t ) v(f) > v(0) v(f)@ e - t W
v(f) 4 ,
v(0) 12 ,
W RC
-t 6
v( t ) 4 (12 4) e
v( t ) 4 8 e - t 6 V
(2)(3)
12 V .
Before t = 0, v
After t = 0, v( t ) v(f) > v(0) v(f)@ e - t W
After transforming the current source, the circuit is shown below.
t=0
2:
4:
+
12 V
v(f) 12 ,
v(0) 12 ,
v 12 V
Chapter 7, Solution 41.
v(0)
R eq C
v( t )
0,
30
(12) 10
16
v(f)
(6 || 30)(1)
(6)(30)
36
5
v(f) > v(0) v(f)@ e - t W
v( t ) 10 (0 10) e - t 5
v( t )
6
10 ( 1 e -0.2t ) V
W
RC
5F
(2)(5) 10
Chapter 7, Solution 42.
(a)
v o (f) > v o (0) v o (f)@ e - t W
4
v o (f)
(12) 8
v o (0) 0 ,
42
4
W R eq C eq , R eq 2 || 4
3
4
W
(3) 4
3
v o ( t ) 8 8 e -t 4
v o (t)
v o (t)
(b)
8 ( 1 e -0.25t ) V
For this case, v o (f)
v o (0) e -t W
4
v o (0)
(12)
42
v o ( t ) 8 e -t 12 V
0 so that
v o (t)
W
8,
RC
(4)(3) 12
Chapter 7, Solution 43.
Before t = 0, the circuit has reached steady state so that the capacitor acts like an open
circuit. The circuit is equivalent to that shown in Fig. (a) after transforming the voltage
source.
vo
vo
,
0.5i 2 i
40
80
vo
1 vo
320
2

o v o
64
Hence,
2 80
40
5
vo
i
0.8 A
80
After t = 0, the circuit is as shown in Fig. (b).
v C (t)
v C (0) e - t W ,
W
R th C
To find R th , we replace the capacitor with a 1-V voltage source as shown in Fig. (c).
0.5i
vC
i
1V
+
0.5i
(c)
80 :
vC
80
1
,
80
1 80
R th
i o 0.5
v C (0) 64 V
i
v C (t)
0.5 i
io
0.5 i
160 : ,
W
0.5
80
R th C
480
64 e - t 480
dv C
§ 1 ·
¸ 64 e - t 480
-i C -C
-3 ¨
©
dt
480 ¹
i( t ) 0.8 e -t 480 A
Chapter 7, Solution 44.
R eq
v( t )
6 || 3
2 :,
W
RC
v(f) > v(0) v(f)@ e - t W
4
Using voltage division,
3
3
v(0)
(30) 10 V ,
v(f)
(12)
3 6
3 6
Thus,
v( t ) 4 (10 4) e - t 4 4 6 e - t 4
§ - 1·
dv
i( t ) C
(2)(6) ¨ ¸ e - t 4 - 3 e -0.25t A
©4¹
dt
4V
Chapter 7, Solution 45.
For t < 0, v s
5 u(t)
0 
o v(0)
For t > 0, v s
5,
v(f)
4
(5)
4 12
R eq
7 4 || 12 10 ,
v( t )
v(f) > v(0) v(f)@ e - t W
v( t )
1.25 ( 1 e -t 5 ) V
i( t )
W
dv
dt
§ 1 ·§ - 5 ·§ - 1 · - t 5
¨ ¸¨ ¸¨ ¸ e
© 2 ¹© 4 ¹© 5 ¹
i( t )
0.125 e -t 5 A
C
0
5
4
R eq C
(10)(1 2)
5
Chapter 7, Solution 46.
W
RTh C
v(t )
(2 6) x0.25
2 s,
v(0)
v(f) [v(0) v(f)]e t / W
v (f )
0,
6i s
30(1 e t / 2 ) V
Chapter 7, Solution 47.
For t < 0, u ( t )
For 0 < t < 1,
v(0)
v( t )
v( t )
For t > 1,
0,
u ( t 1)
0,
v(0)
0
W RC (2 8)(0.1) 1
0,
v(f) (8)(3) 24
v(f) > v(0) v(f)@ e - t W
24 1 e - t
v(1)
24 1 e -1
15.17
- 6 v(f) - 24 0 
o v(f)
v( t ) 30 (15.17 30) e -(t-1)
v( t ) 30 14.83 e -(t-1)
30
Thus,
v( t )
­ 24 1 e - t V ,
0t1
®
-(t -1)
V,
t !1
¯ 30 14.83 e
Chapter 7, Solution 48.
For t < 0,
For t > 0,
R th
u (-t) 1 ,
u (-t) 0 ,
20 10 30 ,
v(0) 10 V
v(f) 0
W R th C
v( t )
v( t )
v(f) > v(0) v(f)@ e - t W
10 e -t 3 V
i( t )
C
i( t )
§ - 1·
dv
(0.1) ¨ ¸10 e - t 3
©3¹
dt
- 1 -t 3
e A
3
(30)(0.1)
3
6 x5
30
Chapter 7, Solution 49.
For 0 < t < 1, v(0) 0 ,
R eq 4 6 10 ,
v(f) (2)(4) 8
W R eq C (10)(0.5)
v(f) > v(0) v(f)@ e - t W
8 1 e -t 5 V
v( t )
v( t )
For t > 1,
v(1) 8 1 e -0.2 1.45 ,
v( t ) v(f) > v(1) v(f)@ e -( t 1) W
v( t ) 1.45 e -( t 1) 5 V
v(f)
5
0
Thus,
­ 8 1 e -t 5 V , 0 t 1
®
- ( t 1 ) 5
V,
t !1
¯ 1.45 e
v( t )
Chapter 7, Solution 50.
For the capacitor voltage,
v( t ) v(f) > v(0) v(f)@ e- t W
v(0) 0
For t < 0, we transform the current source to a voltage source as shown in Fig. (a).
1 k:
1 k:
+
30 V
+
v
(a)
2
(30) 15 V
2 11
R th (1 1) || 2 1 k:
1
1
W R th C 10 3 u u 10 -3
4
4
-4t
v( t ) 15 1 e , t ! 0
v(f)
2 k:
We now obtain i x from v(t). Consider Fig. (b).
iT 1 k:
v
ix
1 k:
30 mA
1/4 mF
(b)
30 mA i T
v
dv
C
R3
dt
ix
But
iT
1
u 10 -3 (-15)(-4) e -4t A
4
i T (t)
7.5 1 e -4t mA i T (t)
7.5 1 e -4t mA
i x (t)
30 7.5 7.5 e -4t mA
i x ( t ) 7.5 3 e -4t mA , t ! 0
Thus,
Chapter 7, Solution 51.
Consider the circuit below.
t=0
R
+
VS
+
i
L
v
After the switch is closed, applying KVL gives
di
VS Ri L
dt
§
VS ·
di
¸
or
L
-R ¨ i ©
dt
R¹
di
-R
dt
i VS R
L
Integrating both sides,
2 k:
or
§
V · i(t ) - R
ln ¨ i S ¸ I 0
t
©
R¹
L
§ i VS R · - t
¸
ln ¨
© I0 VS R ¹ W
i VS R
e- t W
I0 VS R
i( t )
VS §
VS · -t W
¸e
¨ I0 R ©
R¹
which is the same as Eq. (7.60).
Chapter 7, Solution 52.
20
2 A,
i(f) 2 A
10
i( t ) i(f) > i(0) i(f)@ e- t W
i(0)
i( t )
2A
Chapter 7, Solution 53.
(a)
(b)
25
5A
3 2
After t = 0,
i( t ) i(0) e - t W
L 4
W
2,
i(0)
R 2
i( t ) 5 e - t 2 A
Before t = 0,
i
5
Before t = 0, the inductor acts as a short circuit so that the 2 : and 4 :
resistors are short-circuited.
i( t ) 6 A
After t = 0, we have an RL circuit.
i( t )
i(0) e - t W ,
i( t )
6 e -2t 3 A
W
L
R
3
2
Chapter 7, Solution 54.
(a)
Before t = 0, i is obtained by current division or
4
i( t )
(2) 1 A
44
After t = 0,
i( t ) i(f) > i(0) i(f)@ e- t W
L
W
,
R eq 4 4 || 12 7 :
R eq
W
3.5
7
i(0) 1 ,
1
2
i(f)
4 || 12
(2)
4 4 || 12
3
(2)
43
6
7
6 §
6·
¨ 1 ¸ e -2 t
7 ©
7¹
1
i( t )
6 e - 2t A
7
10
2A
Before t = 0, i( t )
23
After t = 0,
R eq 3 6 || 2 4.5
i( t )
(b)
L
2
4
R eq 4.5 9
i(0) 2
To find i(f) , consider the circuit below, at t = when the inductor
becomes a short circuit,
v
W
i
10 V
2:
+
24 V
+
2H
6:
10 v 24 v v

o v
2
6
3
v
i(f)
3A
3
i( t ) 3 (2 3) e -9 t 4
i( t ) 3 e - 9 t 4 A
3:
9
Chapter 7, Solution 55.
For t < 0, consider the circuit shown in Fig. (a).
0.5 H
io
3:
24 V
io
+
+
0.5 H
i
+
4io
8:
2:
v
(a)
3i o 24 4i o
v( t )
4i o
0 
o i o
96 V
20 V
+
v
+
(b)
24
v
i
2
For t > 0, consider the circuit in Fig. (b).
i( t ) i(f) > i(0) i(f)@ e- t W
20
i(f)
i(0) 48 ,
2A
8 2
L
0.5
R th 2 8 10 : , W
R th 10
48 A
1
20
i( t ) 2 (48 2) e -20t 2 46 e -20t
v( t ) 2 i( t ) 4 92 e -20t V
Chapter 7, Solution 56.
R eq
6 20 || 5 10 : ,
W
L
R
0.05
i( t ) i(f) > i(0) i(f)@ e- t W
i(0) is found by applying nodal analysis to the following circuit.
2:
5:
i
vx
12 :
2A
6:
+
20 :
0.5 H
+
20 V
v
20 v x v x v x v x
5
12 20 6
vx
i ( 0)
2A
6
2
Since 20 || 5
i(f)
i( t )
v( t )
v( t )

o v x
12
4,
4
(4) 1.6
46
1.6 (2 1.6) e- t 0.05 1.6 0.4 e-20t
di 1
L
(0.4) (-20) e -20t
dt 2
- 4 e -20t V
Chapter 7, Solution 57.
At t 0 , the circuit has reached steady state so that the inductors act like short
circuits.
6:
30 V
+
30
30
6 5 || 20 10
i 1 ( 0 ) 2 .4 A ,
i
i
3,
i 2 ( 0)
i1
i2
5:
20 :
20
(3)
25
0 .6 A
i1
2.4 ,
i2
0 .6
For t > 0, the switch is closed so that the energies in L1 and L 2 flow through the
closed switch and become dissipated in the 5 : and 20 : resistors.
L1 2.5 1
i1 ( t ) i1 (0) e - t W1 ,
W1
R1
5
2
i1 ( t )
2.4 e -2t A
i2 (t)
i 2 (0) e - t W 2 ,
i2 (t)
0.6 e -5t A
L2
R2
W2
4
20
1
5
Chapter 7, Solution 58.
For t < 0,
v o (t)
For t > 0,
i(0) 10 ,
R th
0
1 3 4 : ,
20
5
1 3
L 14 1
R th
4 16
i(f)
W
i( t ) i(f) > i(0) i(f)@ e- t W
i( t ) 5 1 e-16t A
di
1
15 1 e -16t (-16)(5) e-16t
dt
4
-16t
15 5 e V
vo (t ) 3i L
vo (t )
Chapter 7, Solution 59.
Let I be the current through the inductor.
i(0) 0
For t < 0,
vs 0 ,
For t > 0,
R eq
4 6 || 3 6 ,
2
(3) 1
24
i( t ) i(f) > i(0) i(f)@ e- t W
i( t ) 1 e-4t
i(f)
vo (t )
vo (t )
di
(1.5)(-4)(-e- 4t )
dt
6 e -4t V
L
W
L
R eq
1 .5
6
0.25
Chapter 7, Solution 60.
Let I be the inductor current.
For t < 0,
u(t) 0 
o i(0)
For t > 0,
R eq
5 || 20
4 :,
0
W
L
R eq
8
4
2
i(f) 4
i( t ) i(f) > i(0) i(f)@ e- t W
i( t ) 4 1 e - t 2
§ - 1·
di
(8)(-4)¨ ¸ e - t 2
©2¹
dt
16 e -0.5t V
v( t )
L
v( t )
Chapter 7, Solution 61.
The current source is transformed as shown below.
4:
20u(-t) + 40u(t)
+
0.5 H
i( t )
i( t )
12 1
,
i(0) 5 ,
4
8
i(f) > i(0) i(f)@ e - t W
10 5 e -8t A
v( t )
L
W
v( t )
L
R
di
dt
20 e -8t
§1·
¨ ¸(-5)(-8) e -8t
©2¹
V
Chapter 7, Solution 62.
L
2
1
R eq 3 || 6
For 0 < t < 1, u ( t 1)
W
0 so that
i(f) 10
i(f)
i(0)
0,
i( t )
1
1 e -t
6
1
6
1
1 e -1 0.1054
6
1 1 1
i(f)
3 6 2
i( t ) 0.5 (0.1054 0.5) e-(t -1)
i( t ) 0.5 0.3946 e-(t -1)
i(1)
For t > 1,
Thus,
­°
1
1 e -t A
0t1
®
6
°̄ 0.5 0.3946 e -(t -1) A
t!1
i( t )
Chapter 7, Solution 63.
For t < 0,
u (- t ) 1 ,
i(0)
10
5
For t > 0,
u (-t) 0 ,
i(f)
0
4 :,
W
L
R th
R th
5 || 20
i( t )
i( t )
i(f) > i(0) i(f)@ e - t W
2 e -8t A
v( t )
L
v( t )
2
0.5
4
1
8
di § 1 ·
¨ ¸(-8)(2) e-8t
dt © 2 ¹
- 8 e -8t V
Chapter 7, Solution 64.
Let i be the inductor current.
For t < 0, the inductor acts like a short circuit and the 3 : resistor is shortcircuited so that the equivalent circuit is shown in Fig. (a).
6:
10 :
+
3:
(a)
6:
i
10 :
+
io
v
i
3:
2:
(b)
i
10
6
i(0)
For t > 0,
1.667 A
2 3 || 6
R th
4 :,
W
L
R th
4
4
1
To find i(f) , consider the circuit in Fig. (b).
10 v v v
10

o v
6
3 2
6
v 5
i i(f)
2 6
i( t ) i(f) > i(0) i(f)@ e - t W
5 §10 5 · - t 5
i( t )
1 e -t A
¨ ¸e
6 © 6 6¹
6
v o is the voltage across the 4 H inductor and the 2 : resistor
v o (t)
v o (t)
§5·
di 10 10 - t
e (4)¨ ¸(-1) e - t
©6¹
dt 6 6
1.667 1 e -t V
10 10 - t
e
6 6
2i L
Chapter 7, Solution 65.
Since v s 10 > u ( t ) u ( t 1)@ , this is the same as saying that a 10 V source is
turned on at t = 0 and a -10 V source is turned on later at t = 1. This is shown in
the figure below.
vs
10
1
t
-10
For 0 < t < 1, i(0)
R th
5 || 20
0,
4,
i(f)
W
10
5
L
R th
i( t ) i(f) > i(0) i(f)@ e- t W
2
2
4
1
2
i( t ) 2 1 e -2t A
i(1) 2 1 e-2 1.729
For t > 1,
i(f)
0
since vs
0
i( t ) i(1) e- ( t 1) W
i( t ) 1.729 e-2( t 1) A
Thus,
i( t )
­ 2 1 e - 2t A 0 t 1
®
t!1
¯ 1.729 e - 2( t 1) A
Chapter 7, Solution 66.
Following Practice Problem 7.14,
v( t ) VT e - t W
VT
v ( 0)
v( t ) -4 e -50t
v o ( t ) - v( t )
i o (t)
W
-4 ,
RfC
(10 u 103 )(2 u 10- 6 )
1
50
4 e -50t , t ! 0
v o (t)
Ro
4
e -50t
10 u 10 3
0.4 e -50t mA , t ! 0
Chapter 7, Solution 67.
The op amp is a voltage follower so that v o
v as shown below.
R
R
+
vo
v1
+
R
vo
C
vo
At node 1,
v o v1
R
v1 0 v1 v o
R
R

o v1
At the noninverting terminal,
dv
v v1
C o o
0
dt
R
dv
2
RC o v o v1 v o v o
3
dt
dv o
v
o
dt
3RC
v o ( t ) VT e - t 3RC
VT
v o (t)
vo (0)
5V,
W
3RC
2
v
3 o
1
vo
3
(3)(10 u 103 )(1 u 10- 6 )
3
100
5 e -100t 3 V
Chapter 7, Solution 68.
This is a very interesting problem and has both an important ideal solution as well as an
important practical solution. Let us look at the ideal solution first. Just before the switch
closes, the value of the voltage across the capacitor is zero which means that the voltage
at both terminals input of the op amp are each zero. As soon as the switch closes, the
output tries to go to a voltage such that the input to the op amp both go to 4 volts. The
ideal op amp puts out whatever current is necessary to reach this condition. An infinite
(impulse) current is necessary if the voltage across the capacitor is to go to 8 volts in zero
time (8 volts across the capacitor will result in 4 volts appearing at the negative terminal
of the op amp). So vo will be equal to 8 volts for all t > 0.
What happens in a real circuit? Essentially, the output of the amplifier portion of the op
amp goes to whatever its maximum value can be. Then this maximum voltage appears
across the output resistance of the op amp and the capacitor that is in series with it. This
results in an exponential rise in the capacitor voltage to the steady-state value of 8 volts.
vC(t) = Vop amp max(1 –– e-t/(RoutC)) volts, for all values of vC less than 8 V,
= 8 V when t is large enough so that the 8 V is reached.
Chapter 7, Solution 69.
Let v x be the capacitor voltage.
v x ( 0) 0
For t < 0,
For t > 0, the 20 k: and 100 k: resistors are in series since no current enters the
op amp terminals. As t o f , the capacitor acts like an open circuit so that
20 100
48
v x (f)
(4)
20 100 10
13
R th 20 100 120 k: ,
W R th C (120 u 103 )(25 u 10-3 ) 3000
v x (t)
v x (t)
vo (t )
v x (f) > v x (0) v x (f)@ e- t W
48
1 e - t 3000
13
100
v (t)
120 x
40
1 e -t 3000 V
13
Chapter 7, Solution 70.
Let v = capacitor voltage.
For t < 0, the switch is open and v(0)
0.
For t > 0, the switch is closed and the circuit becomes as shown below.
1
+
2
vS
+
+
vo
v
C
R
v1 v 2 v s
0 vs
dv
C
R
dt
where v v s v o 
o v o
From (1),
dv
dt
v
vs
0
RC
-1
³ v dt v(0)
RC s
Since v is constant,
(1)
(2)
vs v
- t vs
RC
(3)
RC
v
From (3),
vo
vo
(20 u 10 3 )(5 u 10 -6 ) 0.1
- 20 t
mV -200 t mV
0.1
v s v 20 200 t
20 ( 1 10t ) mV
Chapter 7, Solution 71.
Let v = voltage across the capacitor.
Let v o = voltage across the 8 k: resistor.
For t < 2, v
0 so that v(2)
0.
For t > 2, we have the circuit shown below.
10 k:
10 k:
20 k:
+
+
4V
+
100 mF
v
+
io
8 k:
vo
Since no current enters the op amp, the input circuit forms an RC circuit.
W RC (10 u 10 3 )(100 u 10 -3 ) 1000
v( t ) v(f) > v(2) v(f)@ e -( t 2 ) W
v( t ) 4 1 e -( t 2 ) 1000
As an inverter,
- 10k
v 2 e -( t 2 ) 1000 1
vo
20k
vo
io
0.25 e -( t 2 ) 1000 1 A
8
Chapter 7, Solution 72.
The op amp acts as an emitter follower so that the Thevenin equivalent circuit is
shown below.
C
+
3u(t)
Hence,
v
io
+
R
v( t ) v(f) > v(0) v(f)@ e - t W
v(0) -2 V , v(f) 3 V , W RC
v( t ) 3 (-2 - 3) e -10t 3 5 e -10t
io
io
(10 u 10 3 )(10 u 10 -6 )
0.1
dv
(10 u 10 -6 )(-5)(-10) e -10t
dt
0.5 e -10t mA , t ! 0
C
Chapter 7, Solution 73.
Consider the circuit below.
Rf
v1
R1
v2
+
v1
+
C
v
v3
+
+
vo
At node 2,
v1 v 2
R1
At node 3,
C
dv
dt
(1)
C
But v 3
v3 vo
Rf
dv
dt
0 and v
v1 v
R1
(2)
v2 v3
v 2 . Hence, (1) becomes
dv
dt
dv
v1 v R 1C
dt
v1
dv
v
or
dt R 1C R 1C
which is similar to Eq. (7.42). Hence,
­
vT
t0
v( t ) ®
-t W
t!0
¯ v1 v T v1 e
where v T
C
v(0) 1 and v1
W
4
R 1C (10 u 10 )(20 u 10 -6 )
­ 1
t0
v( t ) ®
-5t
t!0
¯4 3 e
3
0.2
From (2),
vo
dv
(20 u 10 3 )(20 u 10 -6 )(15 e -5t )
dt
-6 e -5t , t ! 0
vo
- 6 e -5t u(t ) V
vo
-R f C
Chapter 7, Solution 74.
Let v = capacitor voltage.
Rf
v1
R1
v2
+
v1
+
C
v
v3
+
+
vo
For t < 0,
For t > 0,
is
v( t )
v(0) 0
i s 10 PA . Consider the circuit below.
C
dv v
dt R
v(f) > v(0) v(f)@ e - t W
It is evident from the circuit that
W RC (2 u 10 6 )(50 u 10 3 )
(1)
(2)
0.1
Rf
C
is
+
R
is
+
vo
At steady state, the capacitor acts like an open circuit so that i s passes through R.
Hence,
v(f) i s R (10 u 10 6 )(50 u 10 3 ) 0.5 V
Then,
v( t )
But
is
0.5 1 e -10t V
0 vo
Rf

o v o
(3)
-i s R f
Combining (1), (3), and (4), we obtain
- Rf
dv
vo
v RfC
R
dt
-1
dv
vo
v (10 u 10 3 )(2 u 10 -6 )
5
dt
-10t
-2
v o -0.1 0.1e (2 u 10 )(0.5) - 10 e -10t
vo
0.2 e -10t 0.1
vo
0.1 2 e -10t 1 V
(4)
Chapter 7, Solution 75.
Let v1 = voltage at the noninverting terminal.
Let v 2 = voltage at the inverting terminal.
For t > 0,
v1 v 2 v s 4
0 vs
i o , R 1 20 k:
R1
vo -ioR
v
dv
C ,
R2
dt
Also, i o
i.e.
- vs
R1
R2
(1)
10 k: , C
dv
v
C
dt
R2
(2)
This is a step response.
v( t ) v(f) > v(0) v(f)@ e - t W ,
where W
2 PF
R 2C
(10 u 10 3 )(2 u 10 -6 )
v(0) 1
1
50
At steady state, the capacitor acts like an open circuit so that i o passes through
R 2 . Hence, as t o f
- vs
v(f)
io
R2
R1
- R2
- 10
(4) -2
v(f)
vs
i.e.
20
R1
v( t )
v( t )
-2 (1 2) e -50t
-2 3 e -50t
vs vo
But
v
or
vo
vs v
vo
6 3 e -50 t V
io
or
io
4 2 3 e -50 t
- vs
-4
R 1 20k
dv
v
C
dt
R2
-0.2 mA
- 0.2 mA
Chapter 7, Solution 76.
The schematic is shown below. For the pulse, we use IPWL and enter the corresponding
values as attributes as shown. By selecting Analysis/Setup/Transient, we let Print Step =
25 ms and Final Step = 2 s since the width of the input pulse is 1 s. After saving and
simulating the circuit, we select Trace/Add and display ––V(C1:2). The plot of V(t) is
shown below.
Chapter 7, Solution 77.
The schematic is shown below. We click Marker and insert Mark Voltage Differential at
the terminals of the capacitor to display V after simulation. The plot of V is shown
below. Note from the plot that V(0) = 12 V and V(f) = -24 V which are correct.
Chapter 7, Solution 78.
(a)
When the switch is in position (a), the schematic is shown below. We insert
IPROBE to display i. After simulation, we obtain,
i(0) = 7.714 A
from the display of IPROBE.
(b)
When the switch is in position (b), the schematic is as shown below. For inductor
I1, we let IC = 7.714. By clicking Analysis/Setup/Transient, we let Print Step = 25 ms
and Final Step = 2 s. After Simulation, we click Trace/Add in the probe menu and
display I(L1) as shown below. Note that i(f) = 12A, which is correct.
Chapter 7, Solution 79.
When the switch is in position 1, io(0) = 12/3 = 4A. When the switch is in position 2,
RTh
4
80 / 3
i o (f ) 0.5 A,
RTh (3 5) // 4 8 / 3, W
53
L
io (t )
io (f) [io (0) io (f)]e t / W
0.5 4.5e 3t / 80 A
Chapter 7, Solution 80.
(a) When the switch is in position A, the 5-ohm and 6-ohm resistors are shortcircuited so that
i1 (0)
i 2 ( 0) v o ( 0)
0
but the current through the 4-H inductor is iL(0) =30/10 = 3A.
(b) When the switch is in position B,
RTh
i L (t )
(c) i1 (f)
30
10 5
3 // 6
W
RTh
L
2/4
i L (f) [i L (0) i L (f)]e t / W
i 2 (f )
2 A,
vo (t )
2 :,
L
di L
dt
3
i L (f )
9

o
0 .5
0 3e t / 0.5
3e 2t A
0A
v o (f )
0V
Chapter 7, Solution 81.
The schematic is shown below. We use VPWL for the pulse and specify the attributes as
shown. In the Analysis/Setup/Transient menu, we select Print Step = 25 ms and final
Step = 3 S. By inserting a current marker at one termial of LI, we automatically obtain
the plot of i after simulation as shown below.
Chapter 7, Solution 82.
W
RC 
o R
W
C
3 u 10 -3
100 u 10 -6
30 :
Chapter 7, Solution 83.
v(f) 120,
v(t )
v(0)
0,
W
34 x10 6 x15 x10 6
RC
v(f) [v(0) v(f)]e t / W
85.6 120(1 e t / 510 )

o
Solving for t gives
t 510 ln 3.488
510s
637.16 s
speed = 4000m/637.16s = 6.278m/s
Chapter 7, Solution 84.
Let Io be the final value of the current. Then
I o (1 e t / W ),
i (t )
0.6 I o
I o (1 e 50t )
W
R/L

o
0.16 / 8 1 / 50
t
1
1
ln
50 0.4
18.33 ms.
Chapter 7, Solution 85.
(a)
W
(4 u 106 )(6 u 10-6 )
RC
24 s
Since v( t ) v(f) > v(0) v(f)@ e - t W
v( t 1 ) v(f) > v(0) v(f)@ e - t1 W
v( t 2 ) v(f) > v(0) v(f)@ e- t 2 W
Dividing (1) by (2),
v( t1 ) v(f)
v( t 2 ) v(f)
t0
§ v( t ) v(f) ·
¸
W ln ¨ 1
© v( t 2 ) v(f) ¹
§ 75 120 ·
¸ 24 ln (2) 16.63 s
24 ln ¨
© 30 120 ¹
Since t 0 t , the light flashes repeatedly every
W RC 24 s
t0
(b)
t 2 t1
e( t 2 t1 ) W
(1)
(2)
Chapter 7, Solution 86.
v( t )
v(f) > v(0) v(f)@ e- t W
v(f) 12 ,
v(0) 0
-t W
v( t ) 12 1 e
v( t 0 ) 8 12 1 e- t 0 W
8
o e- t 0
1 e- t 0 W 
12
t 0 W ln (3)
1
3
W
For R 100 k: ,
W RC (100 u 103 )(2 u 10-6 )
t 0 0.2 ln (3) 0.2197 s
For R 1 M: ,
W RC (1 u 106 )(2 u 10-6 )
t 0 2 ln (3) 2.197 s
0.2 s
2s
Thus,
0.2197 s t 0 2.197 s
Chapter 7, Solution 87.
Let i be the inductor current.
For t < 0,
i (0 )
120
100
1.2 A
For t > 0, we have an RL circuit
L
50
W
i(f)
0.1 ,
R 100 400
i( t ) i(f) > i(0) i(f)@ e - t W
i( t ) 1.2 e -10t
At t = 100 ms = 0.1 s,
i(0.1) 1.2 e -1
0
0.441 A
which is the same as the current through the resistor.
Chapter 7, Solution 88.
(a)
W RC (300 u 10 3 )(200 u 10 -12 ) 60 Ps
As a differentiator,
T ! 10 W 600 Ps 0.6 ms
Tmin 0.6 ms
i.e.
(b)
W RC 60 Ps
As an integrator,
T 0.1W 6 Ps
Tmax 6 Ps
i.e.
Chapter 7, Solution 89.
Since W 0.1 T 1 Ps
L
1 Ps
R
L R u 10 -6
(200 u 10 3 )(1 u 10 -6 )
L 200 mH
Chapter 7, Solution 90.
We determine the Thevenin equivalent circuit for the capacitor C s .
Rs
v th
v,
R th R s || R p
Rs Rp i
Rth
Vth
+
Cs
The Thevenin equivalent is an RC circuit. Since
Rs
1
1
v th
o
vi 
10
10 R s R p
Rs
Also,
1
R
9 p
6
9
2
M:
3
W
where R th
Cs
15 Ps
6 (2 3)
R p || R s
62 3
R th C s
W
R th
15 u 10 -6
0.6 u 10 6
0.6 M:
25 pF
Chapter 7, Solution 91.
12
240 mA ,
i(f)
50
i( t ) i(f) > i(0) i(f)@ e - t W
i( t ) 240 e - t W
L 2
W
R R
i( t 0 ) 10 240 e - t 0 W
i o (0)
e t0
W
24 
o t 0
0
W ln (24)
t0
5
1.573
ln (24) ln (24)
2
R
1.271 :
1.573
W
2
R
Chapter 7, Solution 92.
­ 10
° 10 -3 0 t t R
4 u 10 -9 ˜ ® 2 u- 10
i
°
tR t tD
¯ 5 u 10 -6
­ 20 PA
0 t 2 ms
i( t ) ®
¯- 8 mA 2 ms t 2 ms 5 Ps
which is sketched below.
dv
C
dt
i(t)
5 Ps
20 PA
t
2 ms
-8 mA
(not to scale)
Chapter 8, Solution 1.
(a)
At t = 0-, the circuit has reached steady state so that the equivalent circuit is
shown in Figure (a).
6:
VS
+
6:
6:
+
+
vL
10 H
(a)
v
10 PF
(b)
i(0-) = 12/6 = 2A, v(0-) = 12V
At t = 0+, i(0+) = i(0-) = 2A, v(0+) = v(0-) = 12V
(b)
For t > 0, we have the equivalent circuit shown in Figure (b).
vL = Ldi/dt or di/dt = vL/L
Applying KVL at t = 0+, we obtain,
vL(0+) –– v(0+) + 10i(0+) = 0
vL(0+) –– 12 + 20 = 0, or vL(0+) = -8
Hence,
di(0+)/dt = -8/2 = -4 A/s
Similarly,
iC = Cdv/dt, or dv/dt = iC/C
iC(0+) = -i(0+) = -2
dv(0+)/dt = -2/0.4 = -5 V/s
(c)
As t approaches infinity, the circuit reaches steady state.
i(f) = 0 A, v(f) = 0 V
Chapter 8, Solution 2.
(a)
At t = 0-, the equivalent circuit is shown in Figure (a).
25 k:
20 k:
iR
+
80V
iL
+
60 k: v
(a)
25 k:
20 k:
iL
iR
80V
+
(b)
60||20 = 15 kohms, iR(0-) = 80/(25 + 15) = 2mA.
By the current division principle,
iL(0-) = 60(2mA)/(60 + 20) = 1.5 mA
At t = 0+,
vC(0-) = 0
vC(0+) = vC(0-) = 0
iL(0+) = iL(0-) = 1.5 mA
80 = iR(0+)(25 + 20) + vC(0-)
iR(0+) = 80/45k = 1.778 mA
But,
iR = i C + i L
1.778 = iC(0+) + 1.5 or iC(0+) = 0.278 mA
(b)
vL(0+) = vC(0+) = 0
But,
vL = LdiL/dt and diL(0+)/dt = vL(0+)/L = 0
diL(0+)/dt = 0
Again, 80 = 45iR + vC
0
But,
= 45diR/dt + dvC/dt
dvC(0+)/dt = iC(0+)/C = 0.278 mohms/1 PF = 278 V/s
Hence,
diR(0+)/dt = (-1/45)dvC(0+)/dt = -278/45
diR(0+)/dt = -6.1778 A/s
Also, iR = iC + iL
diR(0+)/dt = diC(0+)/dt + diL(0+)/dt
-6.1788 = diC(0+)/dt + 0, or diC(0+)/dt = -6.1788 A/s
(c)
As t approaches infinity, we have the equivalent circuit in Figure
(b).
iR(f) = iL(f) = 80/45k = 1.778 mA
iC(f) = Cdv(f)/dt = 0.
Chapter 8, Solution 3.
At t = 0-, u(t) = 0. Consider the circuit shown in Figure (a). iL(0-) = 0, and vR(0-) =
0. But, -vR(0-) + vC(0-) + 10 = 0, or vC(0-) = -10V.
(a)
At t = 0+, since the inductor current and capacitor voltage cannot change abruptly,
the inductor current must still be equal to 0A, the capacitor has a voltage equal to
––10V. Since it is in series with the +10V source, together they represent a direct
short at t = 0+. This means that the entire 2A from the current source flows
through the capacitor and not the resistor. Therefore, vR(0+) = 0 V.
(b)
At t = 0+, vL(0+) = 0, therefore LdiL(0+)/dt = vL(0+) = 0, thus, diL/dt = 0A/s,
iC(0+) = 2 A, this means that dvC(0+)/dt = 2/C = 8 V/s. Now for the value of
dvR(0+)/dt. Since vR = vC + 10, then dvR(0+)/dt = dvC(0+)/dt + 0 = 8 V/s.
40 :
40 :
+
vC
+
vR
+
+
10 :
vR
+
10V
2A
iL
vC
10 :
+
(a)
10V
(b)
(c)
As t approaches infinity, we end up with the equivalent circuit shown in
Figure (b).
iL(f) = 10(2)/(40 + 10) = 400 mA
vC(f) = 2[10||40] ––10 = 16 –– 10 = 6V
vR(f) = 2[10||40] = 16 V
Chapter 8, Solution 4.
(a)
At t = 0-, u(-t) = 1 and u(t) = 0 so that the equivalent circuit is shown in
Figure (a).
i(0-) = 40/(3 + 5) = 5A, and v(0-) = 5i(0-) = 25V.
i(0+) = i(0-) = 5A
Hence,
v(0+) = v(0-) = 25V
3:
i
40V
+
+
v
5:
(a)
3:
0.25 H
+ vL iC
i
+
40V
iR
0.1F
4A
5:
(b)
(b)
iC = Cdv/dt or dv(0+)/dt = iC(0+)/C
For t = 0+, 4u(t) = 4 and 4u(-t) = 0. The equivalent circuit is shown in Figure (b).
Since i and v cannot change abruptly,
iR = v/5 = 25/5 = 5A, i(0+) + 4 =iC(0+) + iR(0+)
5 + 4 = iC(0+) + 5 which leads to iC(0+) = 4
dv(0+)/dt = 4/0.1 = 40 V/s
Chapter 8, Solution 5.
(a)
For t < 0, 4u(t) = 0 so that the circuit is not active (all initial conditions = 0).
iL(0-) = 0 and vC(0-) = 0.
For t = 0+, 4u(t) = 4. Consider the circuit below.
iL
A
i
4A
+
4 : vC
1H
iC +
0.25F
vL +
6:
v
Since the 4-ohm resistor is in parallel with the capacitor,
i(0+) = vC(0+)/4 = 0/4 = 0 A
Also, since the 6-ohm resistor is in series with the inductor,
v(0+) = 6iL(0+) = 0V.
(b)
di(0+)/dt = d(vR(0+)/R)/dt = (1/R)dvR(0+)/dt = (1/R)dvC(0+)/dt
= (1/4)4/0.25 A/s = 4 A/s
v = 6iL or dv/dt = 6diL/dt and dv(0+)/dt = 6diL(0+)/dt = 6vL(0+)/L = 0
Therefore dv(0+)/dt = 0 V/s
(c)
As t approaches infinity, the circuit is in steady-state.
i(f) = 6(4)/10 = 2.4 A
v(f) = 6(4 –– 2.4) = 9.6 V
Chapter 8, Solution 6.
(a)
Let i = the inductor current. For t < 0, u(t) = 0 so that
i(0) = 0 and v(0) = 0.
For t > 0, u(t) = 1. Since, v(0+) = v(0-) = 0, and i(0+) = i(0-) = 0.
vR(0+) = Ri(0+) = 0 V
Also, since v(0+) = vR(0+) + vL(0+) = 0 = 0 + vL(0+) or vL(0+) = 0 V.
(1)
(b)
Since i(0+) = 0,
iC(0+) = VS/RS
But,
iC = Cdv/dt which leads to dv(0+)/dt = VS/(CRS)
(2)
From (1),
dv(0+)/dt = dvR(0+)/dt + dvL(0+)/dt
vR = iR or dvR/dt = Rdi/dt
(3)
(4)
But,
vL = Ldi/dt, vL(0+) = 0 = Ldi(0+)/dt and di(0+)/dt = 0
From (4) and (5),
dvR(0+)/dt = 0 V/s
From (2) and (3),
dvL(0+)/dt = dv(0+)/dt = Vs/(CRs)
(5)
(c)
As t approaches infinity, the capacitor acts like an open circuit, while the inductor
acts like a short circuit.
vR(f) = [R/(R + Rs)]Vs
vL(f) = 0 V
Chapter 8, Solution 7.
s2 + 4s + 4 = 0, thus s1,2 =
4 r 4 2 4x 4
= -2, repeated roots.
2
v(t) = [(A + Bt)e-2t], v(0) = 1 = A
dv/dt = [Be-2t] + [-2(A + Bt)e-2t]
dv(0)/dt = -1 = B –– 2A = B –– 2 or B = 1.
Therefore, v(t) = [(1 + t)e-2t] V
Chapter 8, Solution 8.
s2 + 6s + 9 = 0, thus s1,2 =
6 r 6 2 36
= -3, repeated roots.
2
i(t) = [(A + Bt)e-3t], i(0) = 0 = A
di/dt = [Be-3t] + [-3(Bt)e-3t]
di(0)/dt = 4 = B.
Therefore, i(t) = [4te-3t] A
Chapter 8, Solution 9.
s2 + 10s + 25 = 0, thus s1,2 =
10 r 10 10
= -5, repeated roots.
2
i(t) = [(A + Bt)e-5t], i(0) = 10 = A
di/dt = [Be-5t] + [-5(A + Bt)e-5t]
di(0)/dt = 0 = B –– 5A = B –– 50 or B = 50.
Therefore, i(t) = [(10 + 50t)e-5t] A
Chapter 8, Solution 10.
s2 + 5s + 4 = 0, thus s1,2 =
5 r 25 16
= -4, -1.
2
v(t) = (Ae-4t + Be-t), v(0) = 0 = A + B, or B = -A
dv/dt = (-4Ae-4t - Be-t)
dv(0)/dt = 10 = –– 4A –– B = ––3A or A = ––10/3 and B = 10/3.
Therefore, v(t) = (––(10/3)e-4t + (10/3)e-t) V
Chapter 8, Solution 11.
s2 + 2s + 1 = 0, thus s1,2 =
2r 44
= -1, repeated roots.
2
v(t) = [(A + Bt)e-t], v(0) = 10 = A
dv/dt = [Be-t] + [-(A + Bt)e-t]
dv(0)/dt = 0 = B –– A = B –– 10 or B = 10.
Therefore, v(t) = [(10 + 10t)e-t] V
Chapter 8, Solution 12.
(a)
Overdamped when C > 4L/(R2) = 4x0.6/400 = 6x10-3, or C > 6 mF
(b)
Critically damped when C = 6 mF
(c)
Underdamped when C < 6mF
Chapter 8, Solution 13.
Let R||60 = Ro. For a series RLC circuit,
Zo =
1
LC
=
1
0.01x 4
= 5
For critical damping, Zo = D = Ro/(2L) = 5
or Ro = 10L = 40 = 60R/(60 + R)
which leads to R = 120 ohms
Chapter 8, Solution 14.
This is a series, source-free circuit. 60||30 = 20 ohms
D = R/(2L) = 20/(2x2) = 5 and Zo =
1
LC
=
1
0.04
= 5
Zo = D leads to critical damping
i(t) = [(A + Bt)e-5t], i(0) = 2 = A
v = Ldi/dt = 2{[Be-5t] + [-5(A + Bt)e-5t]}
v(0) = 6 = 2B –– 10A = 2B –– 20 or B = 13.
Therefore, i(t) = [(2 + 13t)e-5t] A
Chapter 8, Solution 15.
This is a series, source-free circuit. 60||30 = 20 ohms
D = R/(2L) = 20/(2x2) = 5 and Zo =
1
LC
Zo = D leads to critical damping
=
1
0.04
i(t) = [(A + Bt)e-5t], i(0) = 2 = A
v = Ldi/dt = 2{[Be-5t] + [-5(A + Bt)e-5t]}
v(0) = 6 = 2B –– 10A = 2B –– 20 or B = 13.
Therefore, i(t) = [(2 + 13t)e-5t] A
= 5
Chapter 8, Solution 16.
At t = 0, i(0) = 0, vC(0) = 40x30/50 = 24V
For t > 0, we have a source-free RLC circuit.
D = R/(2L) = (40 + 60)/5 = 20 and Zo =
1
LC
=
1
3
10 x 2.5
Zo = D leads to critical damping
i(t) = [(A + Bt)e-20t], i(0) = 0 = A
di/dt = {[Be-20t] + [-20(Bt)e-20t]},
but di(0)/dt = -(1/L)[Ri(0) + vC(0)] = -(1/2.5)[0 + 24]
B = -9.6 or i(t) = [-9.6te-20t] A
Hence,
Chapter 8, Solution 17.
i(0)
I0
di(0)
dt
0, v(0)
V0
4 x15
60
1
(RI0 V0 ) 4(0 60)
L
1
1
10
Zo
LC
1 1
4 25
R
10
D
20, which is ! Zo .
2L 2 1
4
s
D r D 2 Zo2
20 r 300
240
20 r 10 3
A1e 2.68t A 2e 37.32 t
di(0)
i(0) 0 A1 A 2 ,
2.68A1 37.32A 2
dt
This leads to A1 6.928 A 2
2.68, 37.32
i( t )
i( t )
6.928 e 37.32 t e 268t
Since, v( t )
1 t
³ i( t )dt 60, we get
C 0
v(t) = (60 + 64.53e-2.68t –– 4.6412e-37.32t) V
240
= 20
Chapter 8, Solution 18.
When the switch is off, we have a source-free parallel RLC circuit.
1
Zo
1
0.25 x1
LC
D Zo

o
D
2,
1
2 RC
0.5
underdamped case Z d
Zo2 D 2
4 0.25
1.936
Io(0) = i(0) = initial inductor current = 20/5 = 4A
Vo(0) = v(0) = initial capacitor voltage = 0 V
v(t ) e Dt ( A1 cos Z d t A2 sin Z d t )
v(0) =0 = A1
dv
dt
e 0.5Dt (0.5)( A1 cos1.936t A2 sin 1.936t ) e 0.5Dt (1.936 A1 sin 1.936t 1.936 A2 cos1.936t )
dv(0)
dt
Thus,
v(t )
e 0.5Dt ( A1 cos1.936t A2 sin 1.936t )
(Vo RI o )
RC
( 0 4)
1
4
0.5 A1 1.936 A2

o
2.066
A2
2.066e 0.5t sin 1.936t
Chapter 8, Solution 19.
For t < 0, the equivalent circuit is shown in Figure (a).
10 :
i
+
120V
+
i
+
v
L
(a)
i(0) = 120/10 = 12, v(0) = 0
v
C
(b)
For t > 0, we have a series RLC circuit as shown in Figure (b) with R = 0 = D.
1
Zo =
1
=
LC
4
= 0.5 = Zd
i(t) = [Acos0.5t + Bsin0.5t], i(0) = 12 = A
v = -Ldi/dt, and -v/L = di/dt = 0.5[-12sin0.5t + Bcos0.5t],
which leads to -v(0)/L = 0 = B
Hence,
i(t) = 12cos0.5t A and v = 0.5
However, v = -Ldi/dt = -4(0.5)[-12sin0.5t] = 24sin0.5t V
Chapter 8, Solution 20.
For t < 0, the equivalent circuit is as shown below.
2:
i
12
+ vC
+
v(0) = -12V and i(0) = 12/2 = 6A
For t > 0, we have a series RLC circuit.
D = R/(2L) = 2/(2x0.5) = 2
Zo = 1/ LC
1 / 0 .5 x 1 4
2 2
Since D is less than Zo, we have an under-damped response.
Zd
Zo2 D 2
84
2
i(t) = (Acos2t + Bsin2t)e-2t
i(0) = 6 = A
di/dt = -2(6cos2t + Bsin2t)e-2t + (-2x6sin2t + 2Bcos2t)e-Dt
di(0)/dt = -12 + 2B = -(1/L)[Ri(0) + vC(0)] = -2[12 –– 12] = 0
Thus, B = 6 and i(t) = (6cos2t + 6sin2t)e-2t A
Chapter 8, Solution 21.
By combining some resistors, the circuit is equivalent to that shown below.
60||(15 + 25) = 24 ohms.
12 :
24V
6:
t=0
i
3H
+
24 :
+
(1/27)F
v
At t = 0-,
i(0) = 0, v(0) = 24x24/36 = 16V
For t > 0, we have a series RLC circuit.
R = 30 ohms, L = 3 H, C = (1/27) F
D = R/(2L) = 30/6 = 5
Zo
1 / LC
1 / 3x1 / 27 = 3, clearly D > Zo (overdamped response)
s1,2 = D r D 2 Zo2
5 r 5 2 3 2 = -9, -1
v(t) = [Ae-t + Be-9t], v(0) = 16 = A + B
(1)
i = Cdv/dt = C[-Ae-t - 9Be-9t]
i(0) = 0 = C[-A –– 9B] or A = -9B
From (1) and (2),
B = -2 and A = 18.
Hence,
v(t) = (18e-t –– 2e-9t) V
(2)
Chapter 8, Solution 22.
D = 20 = 1/(2RC) or RC = 1/40
Zd
50
(1)
Zo2 D 2 which leads to 2500 + 400 = Zo2 = 1/(LC)
Thus, LC 1/2900
(2)
In a parallel circuit, vC = vL = vR
But,
iC = CdvC/dt or iC/C = dvC/dt
= -80e-20tcos50t –– 200e-20tsin50t + 200e-20tsin50t –– 500e-20tcos50t
= -580e-20tcos50t
iC(0)/C = -580 which leads to C = -6.5x10-3/(-580) = 11.21 PF
R = 1/(40C) = 106/(2900x11.21) = 2.23 kohms
L = 1/(2900x11.21) = 30.76 H
Chapter 8, Solution 23.
Let Co = C + 0.01. For a parallel RLC circuit,
D = 1/(2RCo), Zo = 1/ LC o
D = 1 = 1/(2RCo), we then have Co = 1/(2R) = 1/20 = 50 mF
Zo = 1/ 0.5x 0.5 = 6.32 > D (underdamped)
Co = C + 10 mF = 50 mF or 40 mF
Chapter 8, Solution 24.
For t < 0, u(-t) 1, namely, the switch is on.
v(0) = 0, i(0) = 25/5 = 5A
For t > 0, the voltage source is off and we have a source-free parallel RLC circuit.
D = 1/(2RC) = 1/(2x5x10-3) = 100
Zo = 1/ LC
1 / 0.1x10 3 = 100
Zo = D (critically damped)
v(t) = [(A1 + A2t)e-100t]
v(0) = 0 = A1
dv(0)/dt = -[v(0) + Ri(0)]/(RC) = -[0 + 5x5]/(5x10-3) = -5000
But,
dv/dt = [(A2 + (-100)A2t)e-100t]
Therefore, dv(0)/dt = -5000 = A2 –– 0
v(t) = -5000te-100t V
Chapter 8, Solution 25.
In the circuit in Fig. 8.76, calculate io(t) and vo(t) for t>0.
1H
2:
30V
io(t)
+
t=0, note this is a
+
make before break
switch so the
inductor current is
not interrupted.
Figure 8.78
8:
For Problem 8.25.
For t > 0, we have a source-free parallel RLC circuit.
D = 1/(2RC) = ¼
1 / 1x 1 4
2
Since D is less than Zo, we have an under-damped response.
Zd
Zo2 D 2
4 (1 / 16)
vo(t)
At t = 0-, vo(0) = (8/(2 + 8)(30) = 24
Zo = 1/ LC
(1/4)F
1.9843
vo(t) = (A1cosZdt + A2sinZdt)e-Dt
vo(0) = 24 = A1 and io(t) = C(dvo/dt) = 0 when t = 0.
dvo/dt = -D(A1cosZdt + A2sinZdt)e-Dt + (-ZdA1sinZdt + ZdA2cosZdt)e-Dt
at t = 0, we get dvo(0)/dt = 0 = -DA1 + ZdA2
Thus, A2 = (D/Zd)A1 = (1/4)(24)/1.9843 = 3.024
vo(t) = (24cosZdt + 3.024sinZdt)e-t/4 volts
Chapter 8, Solution 26.
s2 + 2s + 5 = 0, which leads to s1,2 =
2 r 4 20
= -1rj4
2
i(t) = Is + [(A1cos4t + A2sin4t)e-t], Is = 10/5 = 2
i(0) = 2 = = 2 + A1, or A1 = 0
di/dt = [(A2cos4t)e-t] + [(-A2sin4t)e-t] = 4 = 4A2, or A2 = 1
i(t) = 2 + sin4te-t A
Chapter 8, Solution 27.
s2 + 4s + 8 = 0 leads to s =
4 r 16 32
2
2 r j2
v(t) = Vs + (A1cos2t + A2sin2t)e-2t
8Vs = 24 means that Vs = 3
v(0) = 0 = 3 + A1 leads to A1 = -3
dv/dt = -2(A1cos2t + A2sin2t)e-2t + (-2A1sin2t + 2A2cos2t)e-2t
0 = dv(0)/dt = -2A1 +2A2 or A2 = A1 = -3
v(t) = [3 –– 3(cos2t + sin2t)e-2t] volts
Chapter 8, Solution 28.
The characteristic equation is s2 + 6s + 8 with roots
6 r 36 32
s1, 2
4,2
2
Hence,
i (t )
I s Ae 2t Be 4t
8I s
12

o
i (0)
0

o
Is
1.5
0 1.5 A B
(1)
di
2 Ae 2t 4 Be 4t
dt
di(0)
2 2 A 4 B

o 0 1 A 2 B
dt
Solving (1) and (2) leads to A=-2 and B=0.5.
(2)
i (t ) 1.5 2e 2t 0.5e 4t A
Chapter 8, Solution 29.
(a)
s2 + 4 = 0 which leads to s1,2 = rj2 (an undamped circuit)
v(t) = Vs + Acos2t + Bsin2t
4Vs = 12 or Vs = 3
v(0) = 0 = 3 + A or A = -3
dv/dt = -2Asin2t + 2Bcos2t
dv(0)/dt = 2 = 2B or B = 1, therefore v(t) = (3 –– 3cos2t + sin2t) V
(b)
s2 + 5s + 4 = 0 which leads to s1,2 = -1, -4
i(t) = (Is + Ae-t + Be-4t)
4Is = 8 or Is = 2
i(0) = -1 = 2 + A + B, or A + B = -3
(1)
di/dt = -Ae-t - 4Be-4t
di(0)/dt = 0 = -A –– 4B, or B = -A/4
From (1) and (2) we get A = -4 and B = 1
i(t) = (2 –– 4e-t + e-4t) A
s2 + 2s + 1 = 0, s1,2 = -1, -1
(c)
v(t) = [Vs + (A + Bt)e-t], Vs = 3.
v(0) = 5 = 3 + A or A = 2
dv/dt = [-(A + Bt)e-t] + [Be-t]
dv(0)/dt = -A + B = 1 or B = 2 + 1 = 3
v(t) = [3 + (2 + 3t)e-t] V
Chapter 8, Solution 30.
s1
500
s1 s 2
D D 2 Z o ,
2
1300
2D
800
s2

o
D
650
D D 2 Z o
2
R
2L
Hence,
L
s1 s 2
300
2 D 2 Zo
C
R
2D
2
200
2 x650

o
1
(632.45) 2 L
153.8 mH
Zo
623.45
16.25PF
1
LC
(2)
Chapter 8, Solution 31.
For t = 0-, we have the equivalent circuit in Figure (a). For t = 0+, the equivalent
circuit is shown in Figure (b). By KVL,
v(0+) = v(0-) = 40, i(0+) = i(0-) = 1
By KCL, 2 = i(0+) + i1 = 1 + i1 which leads to i1 = 1. By KVL, -vL + 40i1 + v(0+)
= 0 which leads to vL(0+) = 40x1 + 40 = 80
vL(0+) = 80 V,
40 :
i
vC(0+) = 40 V
10 :
i1 40 :
+
+
+
v
50V
+
v
vL
10 :
0.5H
(a)
50V
(b)
Chapter 8, Solution 32.
For t = 0-, the equivalent circuit is shown below.
2A
i
+
v
6:
i(0-) = 0, v(0-) = -2x6 = -12V
For t > 0, we have a series RLC circuit with a step input.
D = R/(2L) = 6/2 = 3, Zo = 1/ LC
s = 3 r 9 25
1 / 0.04
3 r j4
Thus, v(t) = Vf + [(Acos4t + Bsin4t)e-3t]
+
where Vf = final capacitor voltage = 50 V
v(t) = 50 + [(Acos4t + Bsin4t)e-3t]
v(0) = -12 = 50 + A which gives A = -62
i(0) = 0 = Cdv(0)/dt
dv/dt = [-3(Acos4t + Bsin4t)e-3t] + [4(-Asin4t + Bcos4t)e-3t]
0 = dv(0)/dt = -3A + 4B or B = (3/4)A = -46.5
v(t) = {50 + [(-62cos4t –– 46.5sin4t)e-3t]} V
Chapter 8, Solution 33.
We may transform the current sources to voltage sources. For t = 0-, the equivalent
circuit is shown in Figure (a).
10 :
i
i
+
30V
1H
+
v
+
5:
v
10 :
30V
4F
+
(a)
(b)
i(0) = 30/15 = 2 A, v(0) = 5x30/15 = 10 V
For t > 0, we have a series RLC circuit.
D = R/(2L) = 5/2 = 2.5
Zo
1 / LC
1 / 4 = 0.25, clearly D > Zo (overdamped response)
s1,2 = D r D 2 Z 2o
2.5 r 6.25 0.25 = -4.95, -0.05
v(t) = Vs + [A1e-4.95t + A2e-0.05t], v = 20.
v(0) = 10 = 20 + A1 + A2
(1)
i(0) = Cdv(0)/dt or dv(0)/dt = 2/4 = 1/2
Hence,
½ = -4.95A1 –– 0.05A2
From (1) and (2),
A1 = 0, A2 = -10.
(2)
v(t) = {20 –– 10e-0.05t} V
Chapter 8, Solution 34.
Before t = 0, the capacitor acts like an open circuit while the inductor behaves like a short
circuit.
i(0) = 0, v(0) = 20 V
For t > 0, the LC circuit is disconnected from the voltage source as shown below.
Vx
+
i
(1/16)F
(¼) H
This is a lossless, source-free, series RLC circuit.
D = R/(2L) = 0, Zo = 1/ LC = 1/
1
1
= 8, s = rj8
16 4
Since D is less than Zo, we have an underdamped response. Therefore,
i(t) = A1cos8t + A2sin8t where i(0) = 0 = A1
di(0)/dt = (1/L)vL(0) = -(1/L)v(0) = -4x20 = -80
However, di/dt = 8A2cos8t, thus, di(0)/dt = -80 = 8A2 which leads to A2 = -10
Now we have
i(t) = -10sin8t A
Chapter 8, Solution 35.
At t = 0-, iL(0) = 0, v(0) = vC(0) = 8 V
For t > 0, we have a series RLC circuit with a step input.
D = R/(2L) = 2/2 = 1, Zo = 1/ LC = 1/ 1 / 5 =
s1,2 = D r D 2 Z 2o
5
1 r j2
v(t) = Vs + [(Acos2t + Bsin2t)e-t], Vs = 12.
v(0) = 8 = 12 + A or A = -4, i(0) = Cdv(0)/dt = 0.
But dv/dt = [-(Acos2t + Bsin2t)e-t] + [2(-Asin2t + Bcos2t)e-t]
0
= dv(0)/dt = -A + 2B or 2B = A = -4 and B = -2
v(t) = {12 –– (4cos2t + 2sin2t)e-t V.
Chapter 8, Solution 36.
For t = 0-, 3u(t) = 0. Thus, i(0) = 0, and v(0) = 20 V.
For t > 0, we have the series RLC circuit shown below.
10 :
i
10 :
5H
+
15V
+
2:
20 V
0.2 F
+ v
D = R/(2L) = (2 + 5 + 1)/(2x5) = 0.8
Zo = 1/ LC = 1/ 5x 0.2 = 1
s1,2 = D r D 2 Z2o
0.8 r j0.6
v(t) = Vs + [(Acos0.6t + Bsin0.6t)e-0.8t]
Vs = 15 + 20 = 35V and v(0) = 20 = 35 + A or A = -15
i(0) = Cdv(0)/dt = 0
But dv/dt = [-0.8(Acos0.6t + Bsin0.6t)e-0.8t] + [0.6(-Asin0.6t + Bcos0.6t)e-0.8t]
0
= dv(0)/dt = -0.8A + 0.6B which leads to B = 0.8x(-15)/0.6 = -20
v(t) = {35 –– [(15cos0.6t + 20sin0.6t)e-0.8t]} V
i = Cdv/dt = 0.2{[0.8(15cos0.6t + 20sin0.6t)e-0.8t] + [0.6(15sin0.6t –– 20cos0.6t)e-0.8t]}
i(t) = [(5sin0.6t)e-0.8t] A
Chapter 8, Solution 37.
For t = 0-, the equivalent circuit is shown below.
+
i2
6:
6:
6:
v(0)
30V
+
i1
10V
+
18i2 –– 6i1 = 0 or i1 = 3i2
(1)
-30 + 6(i1 –– i2) + 10 = 0 or i1 –– i2 = 10/3
(2)
From (1) and (2).
i1 = 5, i2 = 5/3
i(0) = i1 = 5A
-10 –– 6i2 + v(0) = 0
v(0) = 10 + 6x5/3 = 20
For t > 0, we have a series RLC circuit.
R = 6||12 = 4
Zo = 1/ LC = 1/ (1 / 2)(1 / 8) = 4
D = R/(2L) = (4)/(2x(1/2)) = 4
D = Zo, therefore the circuit is critically damped
v(t) = Vs +[(A + Bt)e-4t], and Vs = 10
v(0) = 20 = 10 + A, or A = 10
i = Cdv/dt = -4C[(A + Bt)e-4t] + C[(B)e-4t]
i(0) = 5 = C(-4A + B) which leads to 40 = -40 + B or B = 80
i(t) = [-(1/2)(10 + 80t)e-4t] + [(10)e-4t]
i(t) = [(5 –– 40t)e-4t] A
Chapter 8, Solution 38.
At t = 0-, the equivalent circuit is as shown.
2A
+
i
10 :
v
i1
5:
10 :
i(0) = 2A, i1(0) = 10(2)/(10 + 15) = 0.8 A
v(0) = 5i1(0) = 4V
For t > 0, we have a source-free series RLC circuit.
R = 5||(10 + 10) = 4 ohms
Zo = 1/ LC = 1/ (1 / 3)(3 / 4) = 2
D = R/(2L) = (4)/(2x(3/4)) = 8/3
s1,2 = D r D 2 Z 2o
-4.431, -0.903
i(t) = [Ae-4.431t + Be-0.903t]
i(0) = A + B = 2
(1)
di(0)/dt = (1/L)[-Ri(0) + v(0)] = (4/3)(-4x2 + 4) = -16/3 = -5.333
Hence, -5.333 = -4.431A –– 0.903B
(2)
From (1) and (2), A = 1 and B = 1.
i(t) = [e-4.431t + e-0.903t] A
Chapter 8, Solution 39.
For t = 0-, the equivalent circuit is shown in Figure (a). Where 60u(-t) = 60 and
30u(t) = 0.
30 :
60V
+
+ v 20 :
(a)
30 :
0.5F
0.25H
20 : 30V
(b)
v(0) = (20/50)(60) = 24 and i(0) = 0
+
For t > 0, the circuit is shown in Figure (b).
R = 20||30 = 12 ohms
Zo = 1/ LC = 1/ (1 / 2)(1 / 4) =
8
D = R/(2L) = (12)/(0.5) = 24
Since D > Zo, we have an overdamped response.
s1,2 = D r D 2 Z 2o
-47.833, -0.167
v(t) = Vs + [Ae-47.833t + Be-0.167t], Vs = 30
Thus,
v(0) = 24 = 30 + A + B or -6 = A + B
(1)
i(0) = Cdv(0)/dt = 0
But,
dv(0)/dt = -47.833A –– 0.167B = 0
B = -286.43A
From (1) and (2),
(2)
A = 0.021 and B = -6.021
v(t) = 30 + [0.021e-47.833t –– 6.021e-0.167t] V
Chapter 8, Solution 40.
At t = 0-, vC(0) = 0 and iL(0) = i(0) = (6/(6 + 2))4 = 3A
For t > 0, we have a series RLC circuit with a step input as shown below.
i
0.02 F
2H
+
6:
v
14 :
24V
12V
+ Zo = 1/ LC = 1/ 2 x 0.02 = 5
D = R/(2L) = (6 + 14)/(2x2) = 5
+
Since D = Zo, we have a critically damped response.
v(t) = Vs + [(A + Bt)e-5t], Vs = 24 –– 12 = 12V
v(0) = 0 = 12 + A or A = -12
i = Cdv/dt = C{[Be-5t] + [-5(A + Bt)e-5t]}
i(0) = 3 = C[-5A + B] = 0.02[60 + B] or B = 90
Thus, i(t) = 0.02{[90e-5t] + [-5(-12 + 90t)e-5t]}
i(t) = {(3 –– 9t)e-5t} A
Chapter 8, Solution 41.
At t = 0-, the switch is open. i(0) = 0, and
v(0) = 5x100/(20 + 5 + 5) = 50/3
For t > 0, we have a series RLC circuit shown in Figure (a). After source
transformation, it becomes that shown in Figure (b).
10 H
4:
5A
20 :
5:
1H
i
+
10 PF
20V
+
0.04F
v
(a)
(b)
Zo = 1/ LC = 1/ 1x1 / 25 = 5
D = R/(2L) = (4)/(2x1) = 2
s1,2 = D r D 2 Z 2o
Thus,
-2 r j4.583
v(t) = Vs + [(AcosZdt + BsinZdt)e-2t],
where Zd = 4.583 and Vs = 20
v(0) = 50/3 = 20 + A or A = -10/3
i(t) = Cdv/dt = C(-2) [(AcosZdt + BsinZdt)e-2t] + CZd[(-AsinZdt + BcosZdt)e-2t]
i(0) = 0 = -2A + ZdB
B = 2A/Zd = -20/(3x4.583) = -1.455
i(t) = C{[(0cosZdt + (-2B - ZdA)sinZdt)]e-2t}
= (1/25){[(2.91 + 15.2767) sinZdt)]e-2t}
i(t) = {0.7275sin(4.583t)e-2t} A
Chapter 8, Solution 42.
For t = 0-, we have the equivalent circuit as shown in Figure (a).
i(0) = i(0) = 0, and v(0) = 4 –– 12 = -8V
4V
+
1:
5:
6:
12V
+ i
1H
+
v(0)
+
+
12V
v
0.04F
(a)
(b)
For t > 0, the circuit becomes that shown in Figure (b) after source transformation.
Zo = 1/ LC = 1/ 1x1 / 25 = 5
D = R/(2L) = (6)/(2) = 3
s1,2 = D r D 2 Z 2o
Thus,
-3 r j4
v(t) = Vs + [(Acos4t + Bsin4t)e-3t], Vs = -12
v(0) = -8 = -12 + A or A = 4
i = Cdv/dt, or i/C = dv/dt = [-3(Acos4t + Bsin4t)e-3t] + [4(-Asin4t + Bcos4t)e-3t]
i(0) = -3A + 4B or B = 3
v(t) = {-12 + [(4cos4t + 3sin4t)e-3t]} A
Chapter 8, Solution 43.
For t>0, we have a source-free series RLC circuit.
D
R
2L

o
Zd
Zo2 D 2
Zo
1
R
2 x8 x0.5 8:
Zo

o
30

o
LC
2DL
C
1
900 64
Z oL
1
836 x0.5
1
1
2
836
2.392 mF
Chapter 8, Solution 44.
D
R
2L
Zo ! D
1000
2 x1
500,

o
Zo
LC
100 x10
10 4
9
underdamped.
Chapter 8, Solution 45.
Zo = 1/ LC = 1/ 1x 0.5 =
2
D = R/(2L) = (1)/(2x2x0.5) = 0.5
Since D < Zo, we have an underdamped response.
s1,2 = D r D 2 Z 2o
Thus,
-0.5 r j1.323
i(t) = Is + [(Acos1.323t + Bsin1.323t)e-0.5t], Is = 4
i(0) = 1 = 4 + A or A = -3
v = vC = vL = Ldi(0)/dt = 0
di/dt = [1.323(-Asin1.323t + Bcos1.323t)e-0.5t] + [-0.5(Acos1.323t + Bsin1.323t)e-0.5t]
di(0)/dt = 0 = 1.323B –– 0.5A or B = 0.5(-3)/1.323 = -1.134
Thus,
i(t) = {4 –– [(3cos1.323t + 1.134sin1.323t)e-0.5t]} A
Chapter 8, Solution 46.
For t = 0-, u(t) = 0, so that v(0) = 0 and i(0) = 0.
For t > 0, we have a parallel RLC circuit with a step input, as shown below.
+
i
8mH
5PF
v
2 k:
6mA
D = 1/(2RC) = (1)/(2x2x103 x5x10-6) = 50
Zo = 1/ LC = 1/ 8x10 3 x 5x10 6 = 5,000
Since D < Zo, we have an underdamped response.
s1,2 = D r D 2 Zo2 # -50 r j5,000
Thus,
i(t) = Is + [(Acos5,000t + Bsin5,000t)e-50t], Is = 6mA
i(0) = 0 = 6 + A or A = -6mA
v(0) = 0 = Ldi(0)/dt
di/dt = [5,000(-Asin5,000t + Bcos5,000t)e-50t] + [-50(Acos5,000t + Bsin5,000t)e-50t]
di(0)/dt = 0 = 5,000B –– 50A or B = 0.01(-6) = -0.06mA
Thus,
i(t) = {6 –– [(6cos5,000t + 0.06sin5,000t)e-50t]} mA
Chapter 8, Solution 47.
At t = 0-, we obtain,
iL(0) = 3x5/(10 + 5) = 1A
and vo(0) = 0.
For t > 0, the 20-ohm resistor is short-circuited and we have a parallel RLC circuit
with a step input.
D = 1/(2RC) = (1)/(2x5x0.01) = 10
Zo = 1/ LC = 1/ 1x 0.01 = 10
Since D = Zo, we have a critically damped response.
s1,2 = -10
i(t) = Is + [(A + Bt)e-10t], Is = 3
Thus,
i(0) = 1 = 3 + A or A = -2
vo = Ldi/dt = [Be-10t] + [-10(A + Bt)e-10t]
vo(0) = 0 = B –– 10A or B = -20
Thus, vo(t) = (200te-10t) V
Chapter 8, Solution 48.
For t = 0-, we obtain i(0) = -6/(1 + 2) = -2 and v(0) = 2x1 = 2.
For t > 0, the voltage is short-circuited and we have a source-free parallel RLC
circuit.
D = 1/(2RC) = (1)/(2x1x0.25) = 2
Zo = 1/ LC = 1/ 1x 0.25 = 2
Since D = Zo, we have a critically damped response.
s1,2 = -2
Thus,
i(t) = [(A + Bt)e-2t], i(0) = -2 = A
v = Ldi/dt = [Be-2t] + [-2(-2 + Bt)e-2t]
vo(0) = 2 = B + 4 or B = -2
Thus, i(t) = [(-2 - 2t)e-2t] A
and v(t) = [(2 + 4t)e-2t] V
Chapter 8, Solution 49.
For t = 0-, i(0) = 3 + 12/4 = 6 and v(0) = 0.
For t > 0, we have a parallel RLC circuit with a step input.
D = 1/(2RC) = (1)/(2x5x0.05) = 2
Zo = 1/ LC = 1/ 5x 0.05 = 2
Since D = Zo, we have a critically damped response.
s1,2 = -2
i(t) = Is + [(A + Bt)e-2t], Is = 3
Thus,
i(0) = 6 = 3 + A or A = 3
v = Ldi/dt or v/L = di/dt = [Be-2t] + [-2(A + Bt)e-2t]
v(0)/L = 0 = di(0)/dt = B –– 2x3 or B = 6
Thus, i(t) = {3 + [(3 + 6t)e-2t]} A
Chapter 8, Solution 50.
For t = 0-, 4u(t) = 0, v(0) = 0, and i(0) = 30/10 = 3A.
For t > 0, we have a parallel RLC circuit.
i
+
3A
10 :
10 mF
6A
40 :
v
Is = 3 + 6 = 9A and R = 10||40 = 8 ohms
D = 1/(2RC) = (1)/(2x8x0.01) = 25/4 = 6.25
Zo = 1/ LC = 1/ 4x 0.01 = 5
Since D > Zo, we have a overdamped response.
s1,2 = D r D 2 Z o2
-10, -2.5
10 H
i(t) = Is + [Ae-10t] + [Be-2.5t], Is = 9
Thus,
i(0) = 3 = 9 + A + B or A + B = -6
di/dt = [-10Ae-10t] + [-2.5Be-2.5t],
v(0) = 0 = Ldi(0)/dt or di(0)/dt = 0 = -10A –– 2.5B or B = -4A
Thus, A = 2 and B = -8
Clearly,
i(t) = { 9 + [2e-10t] + [-8e-2.5t]} A
Chapter 8, Solution 51.
Let i = inductor current and v = capacitor voltage.
At t = 0, v(0) = 0 and i(0) = io.
For t > 0, we have a parallel, source-free LC circuit (R = f).
D = 1/(2RC) = 0 and Zo = 1/ LC which leads to s1,2 = r jZo
v = AcosZot + BsinZot, v(0) = 0 A
iC = Cdv/dt = -i
dv/dt = ZoBsinZot = -i/C
dv(0)/dt = ZoB = -io/C therefore B = io/(ZoC)
v(t) = -(io/(ZoC))sinZot V where Zo =
LC
Chapter 8, Solution 52.
D
Zd
300
1
2 RC
Zo2 D 2
(1)
400

o
Zo
400 2 300 2
264.575
From (2),
C
1
(264.575) 2 x50 x10 3
285.71PF
From (1),
R
1
2DC
1
(3500)
2 x300
5.833:
1
LC
(2)
Chapter 8, Solution 53.
C1
+
+
vS
v1
R2
i1
+
R1
C2
i2
vo
i2 = C2dvo/dt
(1)
i1 = C1dv1/dt
(2)
0 = R2i2 + R1(i2 –– i1) +vo
(3)
Substituting (1) and (2) into (3) we get,
0 = R2C2dvo/dt + R1(C2dvo/dt –– C1dv1/dt)
(4)
Applying KVL to the outer loop produces,
vs = v1 + i2R2 + vo = v1 + R2C2dvo/dt + vo, which leads to
v1 = vs –– vo –– R2C2dvo/dt
(5)
Substituting (5) into (4) leads to,
0 = R1C2dvo/dt + R1C2dvo/dt –– R1C1(dvs/dt –– dvo/dt –– R2C2d2vo/dt2)
Hence, (R1C1R2C2)(d2vo/dt2) + (R1C1 + R2C2 +R1C2)(dvo/dt) = R1C1(dvs/dt)
Chapter 8, Solution 54.
Let i be the inductor current.
v
dv
0.5
4
dt
di
v 2i dt
Substituting (1) into (2) gives
i
(1)
(2)
v dv 1 dv 1 d 2 v
2 dt 4 dt 2 dt 2
v
s 2 2.5s 3
v

o
0
d 2v
dv
2.5 3v
2
dt
dt

o
s
0
1.25 r j1.199
Ae 1.25t cos1.199t Be 1.25t sin 1.199t
v(0) = 2=A. Let w=1.199
dv
dt
1.25( Ae 1.25t cos wt Be 1.25t sin wt ) w( Ae 1.25t sin wt Be 1.25t cos wt )
dv(0)
dt
0
v
1.25 A Bw

o
B
1.25 X 2
1.199
2.085
2e 1.25t cos1.199t 2.085e 1.25t sin 1.199t V
Chapter 8, Solution 55.
At the top node, writing a KCL equation produces,
i/4 +i = C1dv/dt, C1 = 0.1
5i/4 = C1dv/dt = 0.1dv/dt
i = 0.08dv/dt
But,
(1)
v = (2i (1 / C 2 ) ³ idt ) , C2 = 0.5
or
-dv/dt = 2di/dt + 2i
(2)
Substituting (1) into (2) gives,
-dv/dt = 0.16d2v/dt2 + 0.16dv/dt
0.16d2v/dt2 + 0.16dv/dt + dv/dt = 0, or d2v/dt2 + 7.25dv/dt = 0
Which leads to s2 + 7.25s = 0 = s(s + 7.25) or s1,2 = 0, -7.25
v(t) = A + Be-7.25t
(3)
v(0) = 4 = A + B
(4)
From (1),
i(0) = 2 = 0.08dv(0+)/dt or dv(0+)/dt = 25
But,
dv/dt = -7.25Be-7.25t, which leads to,
dv(0)/dt = -7.25B = 25 or B = -3.448 and A = 4 –– B = 4 + 3.448 = 7.448
Thus, v(t) = {7.45 –– 3.45e-7.25t} V
Chapter 8, Solution 56.
For t < 0, i(0) = 0 and v(0) = 0.
For t > 0, the circuit is as shown below.
4:
i
6:
i
0.04F
+
20
io
0.25H
Applying KVL to the larger loop,
-20 +6io +0.25dio/dt + 25 ³ (i o i)dt = 0
Taking the derivative,
6dio/dt + 0.25d2io/dt2 + 25(io + i) = 0
For the smaller loop,
4 + 25 ³ (i i o )dt = 0
Taking the derivative,
25(i + io) = 0 or i = -io
From (1) and (2)
6dio/dt + 0.25d2io/dt2 = 0
This leads to, 0.25s2 + 6s = 0 or s1,2 = 0, -24
(1)
(2)
io(t) = (A + Be-24t) and io(0) = 0 = A + B or B = -A
As t approaches infinity, io(f) = 20/10 = 2 = A, therefore B = -2
Thus, io(t) = (2 - 2e-24t) = -i(t) or i(t) = (-2 + 2e-24t) A
Chapter 8, Solution 57.
(a)
Let v = capacitor voltage and i = inductor current. At t = 0-, the switch is
closed and the circuit has reached steady-state.
v(0-) = 16V and i(0-) = 16/8 = 2A
At t = 0+, the switch is open but, v(0+) = 16 and i(0+) = 2.
We now have a source-free RLC circuit.
R 8 + 12 = 20 ohms, L = 1H, C = 4mF.
D = R/(2L) = (20)/(2x1) = 10
Zo = 1/ LC = 1/ 1x (1 / 36) = 6
Since D > Zo, we have a overdamped response.
s1,2 = D r D 2 Zo2
-18, -2
Thus, the characteristic equation is (s + 2)(s + 18) = 0 or s2 + 20s +36 = 0.
(b)
i(t) = [Ae-2t + Be-18t] and i(0) = 2 = A + B
To get di(0)/dt, consider the circuit below at t = 0+.
i
12 :
+
+
(1/36)F
v
8:
vL
1H
-v(0) + 20i(0) + vL(0) = 0, which leads to,
(1)
-16 + 20x2 + vL(0) = 0 or vL(0) = -24
Ldi(0)/dt = vL(0) which gives di(0)/dt = vL(0)/L = -24/1 = -24 A/s
Hence -24 = -2A –– 18B or 12 = A + 9B
From (1) and (2),
(2)
B = 1.25 and A = 0.75
i(t) = [0.75e-2t + 1.25e-18t] = -ix(t) or ix(t) = [-0.75e-2t - 1.25e-18t] A
v(t) = 8i(t) = [6e-2t + 10e-18t] A
Chapter 8, Solution 58.
(a) Let i =inductor current, v = capacitor voltage i(0) =0, v(0) = 4
dv(0)
dt
[v(0) Ri(0)]
RC
(4 0)
0.5
8 V/s
(b) For t t 0 , the circuit is a source-free RLC parallel circuit.
D
1
2 RC
Zd
1
1,
2 x0.5 x1
Z 2o D 2
1
Zo
LC
1
0.25 x1
2
4 1 1.732
Thus,
v(t ) e t ( A1 cos1.732t A2 sin 1.732t )
v(0) = 4 = A1
dv
dt
e t A1 cos1.732t 1.732e t A1 sin 1.732t e t A2 sin 1.732t 1.732e t A2 cos1.732t
dv(0)
dt
v(t )
8
A1 1.732 A2
e t (4 cos 1.732t 2.309 sin 1.732t ) V

o
A2
2.309
Chapter 8, Solution 59.
Let i = inductor current and v = capacitor voltage
v(0) = 0, i(0) = 40/(4+16) = 2A
For t>0, the circuit becomes a source-free series RLC with
R
16
1
1
2, Z o
2,

o D Z o 2
2L 2 x4
4 x1 / 16
LC
i (t ) Ae 2t Bte 2t
i(0) = 2 = A
di
2 Ae 2t Be 2t 2 Bte 2t
dt
1
1
di (0)
2 A B [ Ri(0) v(0)]

o 2 A B (32 0),
dt
L
4
D
i (t )
2e 2t 4te 2t
t
v
v
B
1
idt v(0)
C ³0
t
t
32 ³ e 2t dt 64 ³ te 2t dt
0
16e 2t
0
t
0
64 2t
e (2t 1)
4
t
0
32te 2t V
Chapter 8, Solution 60.
At t = 0-, 4u(t) = 0 so that i1(0) = 0 = i2(0)
(1)
Applying nodal analysis,
4 = 0.5di1/dt + i1 + i2
Also,
i2 = [1di1/dt –– 1di2/dt]/3 or 3i2 = di1/dt –– di2/dt
Taking the derivative of (2), 0 = d2i1/dt2 + 2di1/dt + 2di2/dt
From (2) and (3),
(2)
(3)
(4)
di2/dt = di1/dt –– 3i2 = di1/dt –– 3(4 –– i1 –– 0.5di1/dt)
= di1/dt –– 12 + 3i1 + 1.5di1/dt
Substituting this into (4),
d2i1/dt2 + 7di1/dt + 6i1 = 24 which gives s2 + 7s + 6 = 0 = (s + 1)(s + 6)
4
Thus, i1(t) = Is + [Ae-t + Be-6t], 6Is = 24 or Is = 4
i1(t) = 4 + [Ae-t + Be-6t] and i1(0) = 4 + [A + B]
(5)
i2 = 4 –– i1 –– 0.5di1/dt = i1(t) = 4 + -4 - [Ae-t + Be-6t] –– [-Ae-t - 6Be-6t]
= [-0.5Ae-t + 2Be-6t] and i2(0) = 0 = -0.5A + 2B
From (5) and (6),
(6)
A = -3.2 and B = -0.8
i1(t) = {4 + [-3.2e-t –– 0.8e-6t]} A
i2(t) = [1.6e-t –– 1.6e-6t] A
Chapter 8, Solution 61.
For t > 0, we obtain the natural response by considering the circuit below.
1H
a
iL
+
4:
vC
0.25F
6:
At node a,
vC/4 + 0.25dvC/dt + iL = 0
(1)
But,
vC = 1diL/dt + 6iL
(2)
Combining (1) and (2),
(1/4)diL/dt + (6/4)iL + 0.25d2iL/dt2 + (6/4)diL/dt + iL = 0
d2iL/dt2 + 7diL/dt + 10iL = 0
s2 + 7s + 10 = 0 = (s + 2)(s + 5) or s1,2 = -2, -5
Thus, iL(t) = iL(f) + [Ae-2t + Be-5t],
where iL(f) represents the final inductor current = 4(4)/(4 + 6) = 1.6
iL(t) = 1.6 + [Ae-2t + Be-5t] and iL(0) = 1.6 + [A+B] or -1.6 = A+B
diL/dt = [-2Ae-2t - 5Be-5t]
(3)
and diL(0)/dt = 0 = -2A –– 5B or A = -2.5B
(4)
From (3) and (4), A = -8/3 and B = 16/15
iL(t) = 1.6 + [-(8/3)e-2t + (16/15)e-5t]
v(t) = 6iL(t) = {9.6 + [-16e-2t + 6.4e-5t]} V
vC = 1diL/dt + 6iL = [ (16/3)e-2t - (16/3)e-5t] + {9.6 + [-16e-2t + 6.4e-5t]}
vC = {9.6 + [-(32/3)e-2t + 1.0667e-5t]}
i(t) = vC/4 = {2.4 + [-2.667e-2t + 0.2667e-5t]} A
Chapter 8, Solution 62.
This is a parallel RLC circuit as evident when the voltage source is turned off.
D = 1/(2RC) = (1)/(2x3x(1/18)) = 3
Zo = 1/ LC = 1/ 2x1 / 18 = 3
Since D = Zo, we have a critically damped response.
s1,2 = -3
Let v(t) = capacitor voltage
Thus, v(t) = Vs + [(A + Bt)e-3t] where Vs = 0
But -10 + vR + v = 0 or vR = 10 –– v
Therefore vR = 10 –– [(A + Bt)e-3t] where A and B are determined from initial
conditions.
Chapter 8, Solution 63.
R
v1
vs
+
R
vo
v2
C
C
At node 1,
v s v1
R
C
dv1
dt
(1)
At node 2,
dv
v2 vo
C o
dt
R
As a voltage follower, v1 v 2
dv
v v o RC o
dt
and (1) becomes
dv
v s v RC
dt
Substituting (3) into (4) gives
vs
vo RC
(2)
v . Hence (2) becomes
(3)
(4)
dvo
dv
d 2 vo
RC o R 2 C 2
dt
dt
dt 2
or
R 2C 2
d 2 vo
dv
2 RC o vo
2
dt
dt
vs
Chapter 8, Solution 64.
C2
R2
R1
vs
1
v1
C1
2
+
vo
At node 1,
(vs –– v1)/R1 = C1 d(v1 –– 0)/dt or vs = v1 + R1C1dv1/dt
At node 2,
C1dv1/dt = (0 –– vo)/R2 + C2d(0 –– vo)/dt
or
From (1) and (2),
or
(1)
––R2C1dv1/dt = vo + C2dvo/dt
(2)
(vs –– v1)/R1 = C1 dv1/dt = -(1/R2)(vo + C2dvo/dt)
v1 = vs + (R1/R2)(vo + C2dvo/dt)
(3)
Substituting (3) into (1) produces,
vs = vs + (R1/R2)(vo + C2dvo/dt) + R1C1d{vs + (R1/R2)(vo + C2dvo/dt)}/dt
= vs + (R1/R2)(vo)+ (R1C2/R2) dvo/dt) + R1C1dvs/dt + (R1R1C1/R2)dvo/dt
+ (R12 C1C2/R2)[d2vo/dt2]
Simplifying we get,
d2vo/dt2 + [(1/ R1C1) + (1/ C2)]dvo/dt + [1/(R1C1C2)](vo) = - [R2/(R1C2)]dvs/dt
Chapter 8, Solution 65.
At the input of the first op amp,
(vo –– 0)/R = Cd(v1 –– 0)
(1)
At the input of the second op amp,
(-v1 –– 0)/R = Cdv2/dt
(2)
Let us now examine our constraints. Since the input terminals are essentially at ground,
then we have the following,
vo = -v2 or v2 = -vo
(3)
Combining (1), (2), and (3), eliminating v1 and v2 we get,
d 2 vo § 1 ·
¨
¸v o
dt 2 © R 2 C 2 ¹
d 2vo
100 v o
dt 2
0
Which leads to s2 –– 100 = 0
Clearly this produces roots of ––10 and +10.
And, we obtain,
vo(t) = (Ae+10t + Be-10t)V
At t = 0, vo(0+) = –– v2(0+) = 0 = A + B, thus B = ––A
This leads to vo(t) = (Ae+10t –– Ae-10t)V. Now we can use v1(0+) = 2V.
From (2), v1 = ––RCdv2/dt = 0.1dvo/dt = 0.1(10Ae+10t + 10Ae-10t)
v1(0+) = 2 = 0.1(20A) = 2A or A = 1
Thus, vo(t) = (e+10t –– e-10t)V
It should be noted that this circuit is unstable (clearly one of the poles lies in the righthalf-plane).
Chapter 8, Solution 66.
C2
vS
R2
R1
2
+
––
1
vo
R4
C1
Note that the voltage across C1 is
R3
v2 = [R3/(R3 + R4)]vo
This is the only difference between this problem and Example 8.11, i.e. v = kv, where
k = [R3/(R3 + R4)].
At node 1,
(vs –– v1)/R1 = C2[d(v1 –– vo)/dt] + (v1 –– v2)/R2
vs/R1 = (v1/R1) + C2[d(v1)/dt] –– C2[d(vo)/dt] + (v1 –– kvo)/R2
(1)
At node 2,
(v1 –– kvo)/R2 = C1[d(kvo)/dt]
or
v1 = kvo + kR2C1[d(vo)/dt]
(2)
Substituting (2) into (1),
vs/R1 = (kvo/R1) + (kR2C1/R1)[d(vo)/dt] + kC2[d(vo)/dt] + kR2C1C2[d2(vo)/dt2] –– (kvo/R2)
+ kC1[d(vo)/dt] –– (kvo/R2) + C2[d(vo)/dt]
We now rearrange the terms.
[d2(vo)/dt2] + [(1/C2R1) + (1/ R2C2) + (1/R2C1) –– (1/ kR2C1)][d(vo)/dt] + [vo/(R1R2C1C2)]
= vs/(kR1R2C1C2)
If R1 = R2 10 kohms, C1 = C2 = 100 PF, R3 = 20 kohms, and R4 = 60 kohms,
k = [R3/(R3 + R4)] = 1/3
R1R2C1C2 = 104 x104 x10-4 x10-4 = 1
(1/C2R1) + (1/ R2C2) + (1/R2C1) –– (1/ kR2C1) = 1 + 1 + 1 –– 3 = 3 –– 3 = 0
Hence,
[d2(vo)/dt2] + vo = 3vs = 6, t > 0, and s2 + 1 = 0, or s1,2 = rj
vo(t) = Vs + [Acost + B sint], Vs = 6
vo(0) = 0 = 6 + A or A = ––6
dvo/dt = ––Asint + Bcost, but dvo(0)/dt = 0 = B
Hence,
vo(t) = 6(1 –– cost)u(t) volts.
Chapter 8, Solution 67.
At node 1,
v in v1
R1
At node 2,
C2
C1
d ( v 1 0)
dt
d ( v1 v o )
d ( v 1 0)
C2
dt
dt
0 vo
dv1
, or
R2
dt
(1)
vo
C2R 2
(2)
From (1) and (2),
v in v1
v1
v in v
dv
R 1C1 dv o
R 1 C1 o R 1 o
R2
dt
C 2 R 2 dt
v
dv
R 1C1 dv o
R 1 C1 o R 1 o
R2
dt
C 2 R 2 dt
(3)
C1
R2
R1
C2
1
vin
v1
2
0V
+
vo
From (2) and (3),
vo
C2R 2
dv1
dt
d 2 v o R 1 dv o
dv in R 1C1 dv o
R 1 C1
dt
C 2 R 2 dt
R 2 dt
dt 2
d 2 vo
vo
1 § 1
1 · dv o
¸¸
¨¨
2
R 2 © C1 C 2 ¹ dt
C1 C 2 R 2 R 1
dt
1 dv in
R 1C1 dt
But C1C2R1R2 = 10-4 x10-4 x104 x104 = 1
1
R2
§ 1
1 ·
¸¸
¨¨
© C1 C 2 ¹
2
R 2 C1
d 2 vo
dv
2 o vo
2
dt
dt
2
10 x10 4
4
dv in
dt
2
Which leads to s2 + 2s + 1 = 0 or (s + 1)2 = 0 and s = ––1, ––1
Therefore,
vo(t) = [(A + Bt)e-t] + Vf
As t approaches infinity, the capacitor acts like an open circuit so that
Vf = vo(f) = 0
vin = 10u(t) mV and the fact that the initial voltages across each capacitor is 0
means that vo(0) = 0 which leads to A = 0.
vo(t) = [Bte-t]
dv o
= [(B –– Bt)e-t]
dt
dv o (0 )
v (0 )
o
dt
C2R 2
From (2),
(4)
0
From (1) at t = 0+,
1 0
R1
C 1
dv o (0)
dv o (0 )
which leads to
dt
dt
Substituting this into (4) gives B = ––1
Thus,
v(t) = ––te-tu(t) V
1
C1 R 1
1
Chapter 8, Solution 68.
The schematic is as shown below. The unit step is modeled by VPWL as shown. We
insert a voltage marker to display V after simulation. We set Print Step = 25 ms and
final step = 6s in the transient box. The output plot is shown below.
Chapter 8, Solution 69.
The schematic is shown below. The initial values are set as attributes of L1 and C1. We
set Print Step to 25 ms and the Final Time to 20s in the transient box. A current marker
is inserted at the terminal of L1 to automatically display i(t) after simulation. The result
is shown below.
Chapter 8, Solution 70.
The schematic is shown below.
After the circuit is saved and simulated, we obtain the capacitor voltage v(t) as shown
below.
Chapter 8, Solution 71.
The schematic is shown below. We use VPWL and IPWL to model the 39 u(t) V and 13
u(t) A respectively. We set Print Step to 25 ms and Final Step to 4s in the Transient
box. A voltage marker is inserted at the terminal of R2 to automatically produce the plot
of v(t) after simulation. The result is shown below.
Chapter 8, Solution 72.
When the switch is in position 1, we obtain IC=10 for the capacitor and IC=0 for the
inductor. When the switch is in position 2, the schematic of the circuit is shown below.
When the circuit is simulated, we obtain i(t) as shown below.
Chapter 8, Solution 73.
(a)
For t < 0, we have the schematic below. When this is saved and simulated, we
obtain the initial inductor current and capacitor voltage as
iL(0) = 3 A and vc(0) = 24 V.
(b)
For t > 0, we have the schematic shown below. To display i(t) and v(t), we
insert current and voltage markers as shown. The initial inductor current and capacitor
voltage are also incorporated. In the Transient box, we set Print Step = 25 ms and the
Final Time to 4s. After simulation, we automatically have io(t) and vo(t) displayed as
shown below.
Chapter 8, Solution 74.
10 :
5:
+
20 V
-
2F
4H
Hence the dual circuit is shown below.
2H
20A
0.1 :
4F
0.2 :
Chapter 8, Solution 75.
The dual circuit is connected as shown in Figure (a). It is redrawn in Figure (b).
0.1 :
12V
+
10 :
12A
24A
0.5 F
24V
0.25 :
4:
10 H
10 H
10 PF
(a)
0.1 :
2F
0.5 H
24A
12A
0.25 :
(b)
+
Chapter 8, Solution 76.
The dual is obtained from the original circuit as shown in Figure (a). It is redrawn in
Figure (b).
0.1 :
0.05 :
1/3 :
10 :
20 :
60 A
30 :
120 A
+ –– +
60 V
2V
120 V
+ 4H
1F
1H
2A
4F
(a)
0.05 :
60 A
120 A
1H
0.1 :
1/30 :
1/4 F 2V
(b)
+
Chapter 8, Solution 77.
The dual is constructed in Figure (a) and redrawn in Figure (b).
–– +
5A
5V
2:
1/3 :
1/2 :
1F
1:
1/4 H
1H
3:
12V
1:
1/4 F
+
12 A
(a)
1:
2:
1/4 F
1/3 :
12 A
1H
5V
+
(b)
Chapter 8, Solution 78.
The voltage across the igniter is vR = vC since the circuit is a parallel RLC type.
vC(0) = 12, and iL(0) = 0.
D = 1/(2RC) = 1/(2x3x1/30) = 5
Zo
1 / 60 x10 3 x1 / 30 = 22.36
1 / LC
D < Zo produces an underdamped response.
s1, 2
D r D 2 Z o2 = ––5 r j21.794
vC(t) = e-5t(Acos21.794t + Bsin21.794t)
(1)
vC(0) = 12 = A
dvC/dt = ––5[(Acos21.794t + Bsin21.794t)e-5t]
+ 21.794[(––Asin21.794t + Bcos21.794t)e-5t]
(2)
dvC(0)/dt = ––5A + 21.794B
But,
dvC(0)/dt = ––[vC(0) + RiL(0)]/(RC) = ––(12 + 0)/(1/10) = ––120
Hence,
––120 = ––5A + 21.794B, leads to B (5x12 –– 120)/21.794 = ––2.753
At the peak value, dvC(to)/dt = 0, i.e.,
0
= A + Btan21.794to + (A21.794/5)tan21.794to –– 21.794B/5
(B + A21.794/5)tan21.794to = (21.794B/5) –– A
tan21.794to = [(21.794B/5) –– A]/(B + A21.794/5) = ––24/49.55 = ––0.484
Therefore,
21.7945to = |––0.451|
to = |––0.451|/21.794 = 20.68 ms
Chapter 8, Solution 79.
For critical damping of a parallel RLC circuit,
D
Zo
C
L
4R 2
1
2 RC

o
1
LC
Hence,
0.25
4 x144
434 PF
Chapter 8, Solution 80.
t1 = 1/|s1| = 0.1x10-3 leads to s1 = ––1000/0.1 = ––10,000
t2 = 1/|s2| = 0.5x10-3 leads to s1 = ––2,000
s1
D D 2 Zo2
s2
D D 2 Z o2
s1 + s2 = ––2D = ––12,000, therefore D = 6,000 = R/(2L)
L = R/12,000 = 60,000/12,000 = 5H
s2
D D 2 Zo2 = ––2,000
D D 2 Zo2 = 2,000
6,000 D 2 Zo2 = 2,000
D 2 Zo2 = 4,000
D2 –– Zo2 = 16x106
Zo2 = D2 –– 16x106 = 36x106 –– 16x106
Zo = 103 20
1 / LC
C = 1/(20x106x5) = 10 nF
Chapter 8, Solution 81.
t = 1/D = 0.25 leads to D = 4
D 1/(2RC) or,
But,
C = 1/(2DR) = 1/(2x4x200) = 625 PF
Zd
Zo2
Zd2 D 2
Zo2 D 2
(2S4x10 3 ) 2 16 # (2S4 x10 3 0 2 = 1/(LC)
This results in L = 1/(64S2x106x625x10-6) = 2.533 PH
Chapter 8, Solution 82.
For t = 0-, v(0) = 0.
For t > 0, the circuit is as shown below.
R1
a
+
+
C1
vo
R2
C2
v
At node a,
(vo –– v/R1 = (v/R2) + C2dv/dt
vo = v(1 + R1/R2) + R1C2 dv/dt
60 = (1 + 5/2.5) + (5x106 x5x10-6)dv/dt
60 = 3v + 25dv/dt
v(t) = Vs + [Ae-3t/25]
where
3Vs = 60 yields Vs = 20
v(0) = 0 = 20 + A or A = ––20
v(t) = 20(1 –– e-3t/25)V
Chapter 8, Solution 83.
i = iD + Cdv/dt
(1)
––vs + iR + Ldi/dt + v = 0
(2)
Substituting (1) into (2),
vs = RiD + RCdv/dt + Ldi/dt + LCd2v/dt2 + v = 0
LCd2v/dt2 + RCdv/dt + RiD + Ldi/dt = vs
d2v/dt2 + (R/L)dv/dt + (R/LC)iD + (1/C)di/dt = vs/LC
Chapter 9, Solution 1.
Z = 103 rad/s
(a)
angular frequency
(b)
frequency
f =
Z
= 159.2 Hz
2S
(c)
period
T =
1
= 6.283 ms
f
(d)
Since sin(A) = cos(A –– 90q),
vs = 12 sin(103t + 24q) = 12 cos(103t + 24q –– 90q)
vs in cosine form is
vs = 12 cos(103t –– 66q) V
(e)
vs(2.5 ms) = 12 sin((10 3 )(2.5 u 10 -3 ) 24q)
= 12 sin(2.5 + 24q) = 12 sin(143.24q + 24q)
= 2.65 V
Chapter 9, Solution 2.
(a)
amplitude = 8 A
(b)
Z = 500S = 1570.8 rad/s
(c)
f =
(d)
Is = 8‘-25q A
Is(2 ms) = 8 cos((500S )(2 u 10 -3 ) 25q)
= 8 cos(S 25q) = 8 cos(155q)
= -7.25 A
Z
= 250 Hz
2S
Chapter 9, Solution 3.
(a)
4 sin(Zt –– 30q) = 4 cos(Zt –– 30q –– 90q) = 4 cos(Zt –– 120q)
(b)
-2 sin(6t) = 2 cos(6t + 90q)
(c)
-10 sin(Zt + 20q) = 10 cos(Zt + 20q + 90q) = 10 cos(Zt + 110q)
Chapter 9, Solution 4.
(a)
v = 8 cos(7t + 15q) = 8 sin(7t + 15q + 90q) = 8 sin(7t + 105q)
(b)
i = -10 sin(3t –– 85q) = 10 cos(3t –– 85q + 90q) = 10 cos(3t + 5q)
Chapter 9, Solution 5.
v1 = 20 sin(Zt + 60q) = 20 cos(Zt + 60q 90q) = 20 cos(Zt 30q)
v2 = 60 cos(Zt 10q)
This indicates that the phase angle between the two signals is 20q and that v1 lags
v2.
Chapter 9, Solution 6.
(a)
v(t) = 10 cos(4t –– 60q)
i(t) = 4 sin(4t + 50q) = 4 cos(4t + 50q –– 90q) = 4 cos(4t –– 40q)
Thus, i(t) leads v(t) by 20q.
(b)
v1(t) = 4 cos(377t + 10q)
v2(t) = -20 cos(377t) = 20 cos(377t + 180q)
Thus, v2(t) leads v1(t) by 170q.
(c)
x(t) = 13 cos(2t) + 5 sin(2t) = 13 cos(2t) + 5 cos(2t –– 90q)
X = 13‘0q + 5‘-90q = 13 –– j5 = 13.928‘-21.04q
x(t) = 13.928 cos(2t –– 21.04q)
y(t) = 15 cos(2t –– 11.8q)
phase difference = -11.8q + 21.04q = 9.24q
Thus, y(t) leads x(t) by 9.24q.
Chapter 9, Solution 7.
If f(I) = cosI + j sinI,
df
dI
-sinI j cos I
df
f
j dI
j (cos I j sin I)
j f (I )
Integrating both sides
ln f = jI + ln A
f = AejI = cosI + j sinI
f(0) = A = 1
i.e. f(I) = ejI = cosI + j sinI
Chapter 9, Solution 8.
(a)
(b)
(c)
15‘45q
15‘45q
+ j2 =
+ j2
5‘ - 53.13q
3 j4
= 3‘98.13q + j2
= -0.4245 + j2.97 + j2
= -0.4243 + j4.97
(2 + j)(3 –– j4) = 6 –– j8 + j3 + 4 = 10 –– j5 = 11.18‘-26.57q
8‘ - 20q
(-5 j12)(10)
8‘ - 20q
10
+
+
=
11.18‘ - 26.57q
25 144
(2 j)(3 - j4) - 5 j12
= 0.7156‘6.57q 0.2958
j0.71
= 0.7109 + j0.08188 0.2958 j0.71
= 0.4151 j0.6281
10 + (8‘50q)(13‘-68.38q) = 10+104‘-17.38q
= 109.25 –– j31.07
Chapter 9, Solution 9.
(3 j4)(5 j8)
3 j4
= 2+
25 64
5 j8
15 j24 j20 32
= 2+
89
= 1.809 + j0.4944
(a)
2+
(b)
4‘-10q +
1 j2
2.236 ‘ - 63.43q
= 4‘-10q +
3‘6q
3‘6q
= 4‘-10q + 0.7453‘-69.43q
= 3.939 –– j0.6946 + 0.2619 –– j0.6978
= 4.201 –– j1.392
(c)
8‘10q 6 ‘ - 20q 7.879 j1.3892 5.638 j2.052
=
9‘80q 4‘50q
1.5628 j8.863 2.571 j3.064
13.533‘ - 2.81q
13.517 j0.6629
=
=
5.886‘99.86q
1.0083 j5.799
= 2.299‘-102.67q
= -0.5043 –– j2.243
Chapter 9, Solution 10.
(a) z1 6 j8, z 2 8.66 j 5, and z 3
z1 z 2 z 3 10.66 j19.93
(b)
z1 z 2
z3
4 j 6.9282
9.999 j 7.499
Chapter 9, Solution 11.
(a)
(b)
(c)
z 1 z 2 = (-3 + j4)(12 + j5)
= -36 –– j15 + j48 –– 20
= -56 + j33
z1
- 3 j4 (-3 j4)(12 j5)
=
= -0.3314 + j0.1953
=
z2
144 25
12 j5
z 1 z 2 = (-3 + j4) + (12 + j5) = 9 + j9
z 1 z 2 = (-3 + j4) –– (12 + j5) = -15 –– j
z1 z 2
9 (1 j)
- 9 (1 j)(15 - j)
- 9 (16 j14)
=
=
=
z1 z 2
15 2 12
226
- (15 j)
= -0.6372 –– j0.5575
Chapter 9, Solution 12.
(a)
z 1 z 2 = (-3 + j4)(12 + j5)
= -36 –– j15 + j48 –– 20
= -56 + j33
z1
(-3 j4)(12 j5)
- 3 j4
=
=
= -0.3314 + j0.1953
z2
144 25
12 j5
(b)
z 1 z 2 = (-3 + j4) + (12 + j5) = 9 + j9
z 1 z 2 = (-3 + j4) –– (12 + j5) = -15 –– j
z1 z 2
9 (1 j)
- 9 (1 j)(15 - j)
- 9 (16 j14)
=
=
=
2
2
z1 z 2
15 1
226
- (15 j)
= -0.6372 –– j0.5575
(c)
Chapter 9, Solution 13.
(a) (0.4324 j 0.4054) (0.8425 j 0.2534)
(b)
50‘ 30 o
24‘150 o
1.2749 j 0.1520
2.0833
(c) (2+j3)(8-j5) ––(-4) = 35 +j14
Chapter 9, Solution 14.
(a)
3 j14
15 j11
(b)
(62.116 j 231.82 138.56 j80)(60 j80)
(67 j84)(16.96 j10.5983)
(c) 2 j 4
2
0.5751 j 0.5116
(260 j120)
256.4 j 200.89
24186 6944.9
246.06 j 2134.7
1.922 j11.55
Chapter 9, Solution 15.
(a)
(b)
(c)
10 j6 2 j3
= -10 –– j6 + j10 –– 6 + 10 –– j15
-5
-1 j
= -6 –– j11
20‘ 30q - 4‘ - 10q
= 60‘15q + 64‘-10q
16‘0q
3‘45q
= 57.96 + j15.529 + 63.03 –– j11.114
= 120.99 –– j4.415
1 j j 0
j
1 j
1
j 1 j
1 j j
j
1
= 1 1 0 1 0 j2 (1 j) j2 (1 j)
0
j
= 1 1 (1 j 1 j)
= 1 –– 2 = -1
Chapter 9, Solution 16.
(a)
-10 cos(4t + 75q) = 10 cos(4t + 75q 180q)
= 10 cos(4t 105q)
The phasor form is 10‘-105q
(b)
5 sin(20t –– 10q) = 5 cos(20t –– 10q –– 90q)
= 5 cos(20t –– 100q)
The phasor form is 5‘-100q
(c)
4 cos(2t) + 3 sin(2t) = 4 cos(2t) + 3 cos(2t –– 90q)
The phasor form is 4‘0q + 3‘-90q = 4 –– j3 = 5‘-36.87q
Chapter 9, Solution 17.
(a)
Let A = 8‘-30q + 6‘0q
= 12.928 –– j4
= 13.533‘-17.19q
a(t) = 13.533 cos(5t + 342.81q)
(b)
We know that -sinD = cos(D + 90q).
Let B = 20‘45q + 30‘(20q + 90q)
= 14.142 + j14.142 –– 10.261 + j28.19
= 3.881 + j42.33
= 42.51‘84.76q
b(t) = 42.51 cos(120St + 84.76q)
(c)
Let C = 4‘-90q + 3‘(-10q –– 90q)
= -j4 –– 0.5209 –– j2.954
= 6.974‘265.72q
c(t) = 6.974 cos(8t + 265.72q)
Chapter 9, Solution 18.
(a)
v1 ( t ) = 60 cos(t + 15q)
(b)
V2 = 6 + j8 = 10‘53.13q
v 2 ( t ) = 10 cos(40t + 53.13q)
(c)
i1 ( t ) = 2.8 cos(377t –– S/3)
(d)
I 2 = -0.5 –– j1.2 = 1.3‘247.4q
i 2 ( t ) = 1.3 cos(103t + 247.4q)
Chapter 9, Solution 19.
(a)
3‘10q 5‘-30q = 2.954 + j0.5209 –– 4.33 + j2.5
= -1.376 + j3.021
= 3.32‘114.49q
Therefore,
3 cos(20t + 10q) –– 5 cos(20t –– 30q) = 3.32 cos(20t +
114.49q)
(b)
4‘-90q + 3‘-45q = -j40 + 21.21 –– j21.21
= 21.21 –– j61.21
= 64.78‘-70.89q
Therefore,
40 sin(50t) + 30 cos(50t –– 45q) = 64.78 cos(50t –– 70.89q)
(c)
Using sinD = cos(D 90q),
20‘-90q + 10‘60q 5‘-110q = -j20 + 5 + j8.66 + 1.7101 + j4.699
= 6.7101 –– j6.641
= 9.44‘-44.7q
Therefore,
20 sin(400t) + 10 cos(400t + 60q) –– 5 sin(400t –– 20q)
= 9.44 cos(400t –– 44.7q)
Chapter 9, Solution 20.
(a) V
4‘ 60 o 90 o 5‘40 o
3.464 j 2 3.83 j 3.2139
8.966‘ 4.399 o
Hence,
8.966 cos(377t 4.399 o )
v
(b) I
10‘0 o jZ 8‘20 o 90 o ,
Z
5 , i.e. I
10 40‘20 o
49.51‘16.04 o
49.51 cos(5t 16.04 o )
i
Chapter 9, Solution 21.
(a) F
5‘15 o 4‘ 30 o 90 o
f (t )
(b) G
8‘ 90 o 4‘50 o
i.e. H
8.324 cos(30t 34.86 o )
2.571 j 4.9358
g (t )
(c) H
6.8296 j 4.758 8.3236‘34.86 o
5.565‘ 62.49 o
5.565 cos(t 62.49 o )
1
10‘0 o 5‘ 90 o , Z
jZ
0.25‘ 90 o 0.125‘ 180 o
40
j 0.25 0.125
0.2795‘ 116.6 o
0.2795 cos(40t 116.6 o )
h(t )
Chapter 9, Solution 22.
t
dv
Let f(t) = 10v(t ) 4 2 ³ v(t )dt
dt
f
2V
F 10V jZ 4V , Z 5, V
jZ
F
10V j 20V j 0.4V
20‘ 30 o
(10 j19.6)(17.32 j10)
f (t )
440.1‘ 92.97 o
440.1 cos(5t 92.97 o )
Chapter 9, Solution 23.
(a)
v(t) = 40 cos(Zt –– 60q)
(b)
V = -30‘10q + 50‘60q
= -4.54 + j38.09
= 38.36‘96.8q
v(t) = 38.36 cos(Zt + 96.8q)
(c)
I = j6‘-10q = 6‘(90q 10q) = 6‘80q
i(t) = 6 cos(Zt + 80q)
(d)
2
+ 10‘-45q = -j2 + 7.071 –– j7.071
j
= 11.5‘-52.06q
i(t) = 11.5 cos(Zt –– 52.06q)
I =
Chapter 9, Solution 24.
(a)
V
10‘0q, Z 1
jZ
V (1 j) 10
10
V
5 j5 7.071‘45q
1 j
Therefore,
v(t) = 7.071 cos(t + 45q)
V
(b)
4V
20‘(10q 90q), Z
jZ
§
4·
V ¨ j4 5 ¸ 20 ‘ - 80q
j4 ¹
©
20‘ - 80q
3.43‘ - 110.96q
V
5 j3
Therefore,
v(t) = 3.43 cos(4t –– 110.96q)
jZV 5V 4
Chapter 9, Solution 25.
(a)
2jZI 3I 4‘ - 45q, Z 2
I (3 j4) 4‘ - 45q
4‘ - 45q 4‘ - 45q
0.8‘ - 98.13q
I
3 j4
5‘53.13q
Therefore,
i(t) = 0.8 cos(2t –– 98.13q)
(b)
I
jZI 6I 5‘22q, Z 5
jZ
(- j2 j5 6) I 5‘22q
5‘22q
5‘22q
I
0.745‘ - 4.56q
6 j3 6.708‘26.56q
Therefore,
i(t) = 0.745 cos(5t –– 4.56q)
10
Chapter 9, Solution 26.
I
1‘0q, Z 2
jZ
§
1·
I ¨ j2 2 ¸ 1
j2 ¹
©
1
0.4‘ - 36.87q
I
2 j1.5
Therefore,
i(t) = 0.4 cos(2t –– 36.87q)
jZI 2I Chapter 9, Solution 27.
V
110‘ - 10q, Z 377
jZ
§
j100 ·
¸ 110‘ - 10q
V ¨ j377 50 ©
377 ¹
V (380.6‘82.45q) 110‘ - 10q
V 0.289 ‘ - 92.45q
jZV 50V 100
Therefore, v(t) = 0.289 cos(377t –– 92.45q).
Chapter 9, Solution 28.
v s (t)
R
i( t )
110 cos(377 t )
8
13.75 cos(377t) A.
Chapter 9, Solution 29.
Z
V
Therefore
IZ
1
jZC
1
j (10 )(2 u 10 -6 )
6
(4‘25q)(0.5‘ - 90q)
- j 0.5
2 ‘ - 65q
v(t) = 2 sin(106t –– 65q) V.
Chapter 9, Solution 30.
Z jZL j (500)(4 u 10 -3 ) j2
V 60 ‘ - 65q
I
30‘ - 155q
Z
2‘90q
Therefore,
i(t) = 30 cos(500t –– 155q) A.
Chapter 9, Solution 31.
Thus,
i(t) = 10 sin(Zt + 30q) = 10 cos(Zt + 30q 90q) = 10 cos(Zt 60q)
I = 10‘-60q
v(t) = -65 cos(Zt + 120q) = 65 cos(Zt + 120q 180q) = 65 cos(Zt 60q)
Thus,
V = 65‘-60q
Z
V
I
65‘ - 60q
10‘ - 60q
6.5 :
Since V and I are in phase, the element is a resistor with R = 6.5 :.
Chapter 9, Solution 32.
V = 180‘10q,
Z
V
I
I = 12‘-30q,
Z = 2
180‘10q
15‘40q 11.49 j 9.642 :
12‘ - 30q
One element is a resistor with R = 11.49 :.
The other element is an inductor with ZL = 9.642 or
L = 4.821 H.
Chapter 9, Solution 33.
110
v 2R v 2L
vL
110 2 v 2R
vL
110 2 85 2
69.82 V
Chapter 9, Solution 34.
vo
Z
1
ZC
0 if ZL

o Z
1
(5 u 10 3 )(2 u 10 3 )
1
LC
100 rad/s
Chapter 9, Solution 35.
Vs 5‘0q
jZL j (2)(1) j2
1
1
- j2
jZC j (2)(0.25)
j2
j2
Vs
5‘0q (1‘90q)(5‘0q)
2 j2 j2
2
Thus, v o ( t ) 5 cos(2t + 90q) = -5 sin(2t) V
Vo
5‘90q
Chapter 9, Solution 36.
Let Z be the input impedance at the source.
10 PF
jZL

o
100 mH
j 200 x100 x10 3
1
jZ C

o
1
j10 x10 6 x 200
j 20
j 500
1000//-j500 = 200 ––j400
1000//(j20 + 200 ––j400) = 242.62 ––j239.84
Z
2242.62 j 239.84
2255‘ 6.104 o
I
60‘ 10 o
2255‘ 6.104 o
i
266.1 cos(200t 3.896 o )
26.61‘ 3.896 o mA
Chapter 9, Solution 37.
Let Z1
-j,
Then,
Therefore,
jZL
j (5)(1)
1
jZC
1
j (5)(0.2)
j5
-j
(2)( j5)
2 j5
j10
2 j5
Z2
2 || j5
Ix
Z2
I ,
Z1 Z 2 s
where I s
Ix
j10
2 j5
(2)
j10
- j
2 j5
j20
5 j8
i x (t)
2.12 sin(5t + 32q) A
2‘0q
2.12 ‘32q
Chapter 9, Solution 38.
1
F 
o
6
(a)
1
jZC
1
j (3)(1 / 6)
- j2
- j2
(10 ‘45q) 4.472‘ - 18.43q
4 j2
Hence, i(t) = 4.472 cos(3t –– 18.43q) A
I
V 4I (4)(4.472‘ - 18.43q) 17.89‘ - 18.43q
Hence, v(t) = 17.89 cos(3t –– 18.43q) V
1
F 
o
12
(b)
3H 
o
1
jZC
jZL
1
j (4)(1 / 12)
j (4)(3)
- j3
j12
V 50‘0q
10‘36.87q
Z 4 j3
Hence, i(t) = 10 cos(4t + 36.87q) A
I
j12
(50‘0q) 41.6 ‘33.69q
8 j12
Hence, v(t) = 41.6 cos(4t + 33.69q) V
V
Chapter 9, Solution 39.
Z
8 j5 || (- j10)
I
V
Z
40 ‘0q
8 j10
I1
- j10
I
j5 j10
I2
j5
I
- j5
Therefore,
-I
8
( j5)(- j10)
j5 j10
20
6.403‘51.34q
2I
8 j10
3.124‘ - 51.34q
6.248‘ - 51.34q
3.124‘128.66q
i1 ( t )
6.248 cos(120St –– 51.34q) A
i 2 (t)
3.124 cos(120St + 128.66q) A
Chapter 9, Solution 40.
(a)
For Z 1 ,
1H 
o
jZL
j (1)(1)
j
1
1
0.05 F 
o
- j20
jZC j (1)(0.05)
- j40
Z j 2 || (- j20) j 1.98 j0.802
2 j20
V
4 ‘0q
4‘0q
1.872 ‘ - 22.05q
Z 1.98 j0.802 2.136‘22.05q
Hence, i o ( t ) 1.872 cos(t –– 22.05q) A
Io
(b)
For Z 5 ,
1H 
o
jZL
0.05 F 
o
Z
j (5)(1)
j5
1
jZC
1
- j4
j (5)(0.05)
- j4
j5 1.6 j4.2
1 j2
j5 2 || (- j4)
4‘0q
4‘0q
V
0.89‘ - 69.14q
Z 1.6 j4 4.494‘69.14q
Hence, i o ( t ) 0.89 cos(5t –– 69.14q) A
Io
(c)
For Z 10 ,
1H 
o jZL
0.05 F 
o
Z
j (10)(1)
j10
1
jZC
j10 2 || (- j2)
1
- j2
j (10)(0.05)
- j4
1 j9
j10 2 j2
V 4‘0q
4 ‘0q
0.4417 ‘ - 83.66q
Z 1 j9 9.055‘83.66q
Hence, i o ( t ) 0.4417 cos(10t –– 83.66q) A
Io
Chapter 9, Solution 41.
Z 1,
1H 
o
1F 
o
jZL
j (1)(1)
j
1
jZC
1
j (1)(1)
-j
Z 1 (1 j) || (- j) 1 Thus,
I
Vs
Z
V
(- j)(1 j) I
10
,
2 j
Ic
- j1
1
2 j
(1 j) I
(1 j)(10)
2 j
(1 j) I
6.325‘ - 18.43q
v(t) = 6.325 cos(t –– 18.43q) V
Chapter 9, Solution 42.
Z
200
50 PF 
o
1
jZC
0.1 H 
o
jZL
50 || -j100
Vo
1
j (200)(50 u 10 -6 )
j (200)(0.1)
(50)(-j100)
50 j100
- j100
1 - j2
j20
(60‘0q)
j20 30 40 j20
Thus, v o ( t )
17.14 sin(200t + 90q) V
or
v o (t)
17.14 cos(200t) V
- j100
j20
40 j20
j20
(60‘0q) 17.14 ‘90q
70
Chapter 9, Solution 43.
Z 2
1H 
o jZL
j (2)(1)
j2
1
jZC
1
j (2)(1)
- j0.5
1F 
o
j2 j0.5
I
j2 j0.5 1
Io
j1.5
4‘0q 3.328‘33.69q
1 j1.5
3.328 cos(2t + 33.69q) A
Thus, i o ( t )
Chapter 9, Solution 44.
Z 200
10 mH 
o jZL
5 mF 
o
1
jZC
Y
1 1
1
4 j2 3 j
Z
1
Y
I
6‘0q
5 Z
1
0.55 j0.4
j (200)(10 u 10 -3 )
1
j (200)(5 u 10 -3 )
0.25 j0.5 3 j
10
j2
-j
0.55 j0.4
1.1892 j0.865
6‘0q
6.1892 j0.865
0.96 ‘ - 7.956q
Thus, i(t) = 0.96 cos(200t –– 7.956q) A
Chapter 9, Solution 45.
We obtain I o by applying the principle of current division twice.
I
I2
I2
Io
-j2 :
Z2
Z1
(a)
(b)
Z1
- j2 ,
I2
Z1
I
Z1 Z 2
Io
- j2
I
2 - j2 2
Z2
j4 (-j2) || 2
- j2
(5‘0q)
- j2 1 j3
§ - j ·§ - j10 ·
¨
¸¨
¸
©1 - j ¹© 1 j ¹
- 10
11
j4 - j4
2 - j2
- j10
1 j
-5 A
Chapter 9, Solution 46.
is
Let
5 cos(10 t 40q) 
o I s
5‘40q
0.1 F 
o
1
jZC
1
j (10)(0.1)
-j
0.2 H 
o
jZL
j (10)(0.2)
j2
j8
4 j2
0.8 j1.6 ,
Z1
4 || j2
Io
Z1
I
Z1 Z 2 s
Io
(1.789‘63.43q)(5‘40q)
3.847 ‘8.97q
Thus, i o ( t )
2:
Z2
0.8 j1.6
(5‘40q)
3.8 j0.6
2.325‘94.46q
2.325 cos(10t + 94.46q) A
3 j
1 j3
Chapter 9, Solution 47.
First, we convert the circuit into the frequency domain.
Ix
j4
+
5‘0
Ix
2:
5
j10(20 j4)
2
j10 20 j4
-j10
5
2 4.588 j8.626
20 :
5
10.854‘ 52.63q
is(t) = 0.4607cos(2000t +52.63) A
Chapter 9, Solution 48.
Converting the circuit to the frequency domain, we get:
10 :
V1 30 :
Ix
20‘-40
+
We can solve this using nodal analysis.
j20
-j20
0.4607‘52.63q
V1 20‘ 40q V1 0
V 0
0
1
10
j20
30 j20
V1(0.1 j0.05 0.02307 j0.01538) 2‘ 40q
V1
Ix
ix
2‘40q
15.643‘ 24.29q
0.12307 j0.03462
15.643‘ 24.29q
0.4338‘9.4q
30 j20
0.4338 sin(100 t 9.4q) A
Chapter 9, Solution 49.
2 j2 || (1 j)
ZT
2
( j2)(1 j)
1 j
I
Ix
4
1:
j2 :
Ix
I
j2
j2 1 1 j
I
j2 x
Vs
v s (t)
I
j
1 j
j4
-j :
j2
I,
1 j
where I x
0.5‘0q
1
2
1 j
1 j
(4)
1 j 1.414‘ - 45q
j4
j
1.414 sin(200t –– 45q) V
I ZT
Chapter 9, Solution 50.
Since Ȧ = 100, the inductor = j100x0.1 = j10 ȍ and the capacitor = 1/(j100x10-3)
= -j10ȍ.
j10
5‘40
Ix
+
-j10
20 :
vx
Using the current dividing rule:
Vx
j10
5‘40q
j10 20 j10
20I x 50‘ 50q
vx
50 cos(100t 50q) V
Ix
j2.5‘40q
2.5‘ 50q
Chapter 9, Solution 51.
0.1 F 
o
1
jZC
1
j (2)(0.1)
- j5
0.5 H 
o
jZL
j (2)(0.5)
j
The current I through the 2-: resistor is
Is
1
I
Is
,
1 j5 j 2
3 j4
I s (10)(3 j4) 50‘ - 53.13q
Therefore,
i s (t)
where I 10 ‘0q
50 cos(2t –– 53.13q) A
Chapter 9, Solution 52.
5 || j5
Z1
j25
5 j5
10 ,
j5
1 j
Z2
2.5 j2.5
- j5 2.5 j2.5
2.5 j2.5
I2
IS
Z1
Z2
I2
Z1
I
Z1 Z 2 s
Vo
I 2 (2.5 j2.5)
§ 4
8‘30q ¨
©5 10
I
12.5 j2.5 s
·
¸ I (2.5)(1 j)
j¹ s
(8‘30q)(5 j)
10 (1 j)
Is
4
I
5 j s
10 (1 j)
I
5 j s
2.884‘-26.31q A
Chapter 9, Solution 53.
Convert the delta to wye subnetwork as shown below.
Z1
Io
Z2
2:
Z3
+
10 :
8:
60‘ 30 V
o
-
Z
Z1
j 2 x4
10 j 2
0.1532 j 0.7692,
Z3
12
10 j 2
1.1538 j 0.2308
( Z 3 8) //( Z 2 10)
Z
Io
j6 x4
10 j 2
0.4615 j 2.3077,
(9.1538 j 0.2308) //(9.5385 j 2.3077)
2 Z 1 4.726 j 0.6062
60‘ 30 o
Z
Z2
6.878 j 0.163
60‘ 30 o
6.88‘ 1.3575
o
8.721‘ 28.64 o A
4.726 j 0.6062
Chapter 9, Solution 54.
Since the left portion of the circuit is twice as large as the right portion, the
equivalent circuit is shown below.
+ Vs
+
2Z
+
V1
V2
Vs
I o (1 j) 2 (1 j)
2V1 4 (1 j)
V1 V2 6 (1 j)
Vs
8.485‘-45q V
Z
V1
V2
Chapter 9, Solution 55.
12 :
-j20 V
I
Z
I1
+
I2
+
-j4 :
Vo
I1
Vo
j8
I2
I 1 (Z j8)
- j4
I
I1 I 2
4
j8
-j0.5
(-j0.5)(Z j8)
- j4
-j0.5 Z
j
8
Z
j
8
Z
j0.5
8
- j20 12 I I 1 (Z j8)
§Z j· - j
- j20 12 ¨ ¸ (Z j8)
© 8 2¹ 2
j8 :
- 4 - j26
Z
§3
1·
Z¨ j ¸
©2
2¹
- 4 - j26
3
1
j
2
2
26.31‘261.25q
16.64‘279.68q
1.5811‘ - 18.43q
Z = 2.798 –– j16.403 :
Chapter 9, Solution 56.
3H

o
jZ L
j 30
3F

o
1
jZ C
j / 30
1
jZ C

o
1.5F
j
15
j
j 30 15
j 30 x
j 30 //( j / 15)
Z
j / 15
j
//(2 j 0.06681)
30
j 0.06681
j 0.033(2 j 0.06681)
j 0.033 2 j 0.06681
Chapter 9, Solution 57.

o
2H

o
1F
jZL
j2
1
jZ C
j
Z 1 j2 //( 2 j) 1 Y
1
Z
j2(2 j)
j2 2 j
0.3171 j0.1463 S
2.6 j1.2
6 j 333 m:
Chapter 9, Solution 58.
(a)
10 mF 
o
1
jZC
1
j (50)(10 u 10 -3 )
- j2
10 mH 
o
jZL
j (50)(10 u 10 -3 )
j0.5
Z in
Z in
Z in
Z in
(b)
j0.5 1 || (1 j2)
1 j2
j0.5 2 j2
j0.5 0.25 (3 j)
0.75 + j0.25 :
0.4 H 
o
jZL
j (50)(0.4)
0.2 H 
o
jZL
1
jZC
j (50)(0.2) j10
1
- j20
j (50)(1 u 10 -3 )
1 mF 
o
j20
For the parallel elements,
1
1
1
1
Z p 20 j10 - j20
Zp
10 j10
Then,
Z in
10 + j20 + Z p = 20 + j30 :
Chapter 9, Solution 59.
Z eq
6 (1 j2) || (2 j4)
Z eq
6
Z eq
6 2.308 j1.5385
Z eq
8.308 –– j1.5385 :
(1 j2)(2 j4)
(1 j2) (2 j4)
Chapter 9, Solution 60.
Z
(25 j15) (20 j 50) //(30 j10)
25 j15 26.097 j 5.122
51.1 j 9.878:
Chapter 9, Solution 61.
All of the impedances are in parallel.
1
Z eq
1
1
1
1
1 j 1 j2 j5 1 j3
1
Z eq
(0.5 j0.5) (0.2 j0.4) (- j0.2) (0.1 j0.3)
Z eq
1
0.8 j0.4
0.8 j0.4
1 + j0.5 :
Chapter 9, Solution 62.
2 mH 
o
jZL j (10 u 10 3 )(2 u 10 -3 ) j20
1
1
1 PF 
o
- j100
3
jZC j (10 u 10 )(1 u 10 -6 )
50 :
+
+
1‘0q A
V
j20 :
Vin
+
-j100 :
V
(1‘0q)(50)
50
Vin
Vin
(1‘0q)(50 j20 j100) (2)(50)
50 j80 100 150 j80
Z in
Vin
1‘0q
150 –– j80 :
2V
Chapter 9, Solution 63.
First, replace the wye composed of the 20-ohm, 10-ohm, and j15-ohm impedances with
the corresponding delta.
200 j150 j300
20 j45
10
200 j450
30 j13.333, z3
j15
z1
z2
8:
200 j450
20
––j12 :
10 j22.5
––j16 :
z2
10 :
z1
ZT
z3
10 :
––j16 :
Now all we need to do is to combine impedances.
z 2 (10 j16)
(30 j13.333)(10 j16)
40 j29.33
z3 (10 j16)
21.70 j3.821
ZT
8.721 j8.938
8 j12 z1 (8.721 j8.938 21.7 j3.821)
Chapter 9, Solution 64.
ZT
I
j10(6 j8)
19 j5:
6 j2
30‘90q
0.3866 j1.4767 1.527‘104.7q A
ZT
4
34.69 j6.93:
Chapter 9, Solution 65.
ZT
2 (4 j6) || (3 j4)
ZT
2
ZT
6.83 + j1.094 : = 6.917‘9.1q :
I
V
ZT
(4 j6)(3 j4)
7 j2
120 ‘10q
6.917 ‘9.1q
17.35‘0.9q A
Chapter 9, Solution 66.
(20 j5)(40 j10)
60 j5
ZT
(20 j5) || (40 j10)
ZT
14.069 –– j1.172 : = 14.118‘-4.76q
I
V
ZT
60‘90q
14.118‘ - 4.76q
170
(12 j)
145
4.25‘94.76q
I
I1
I2
20 :
j10 :
+
I1
40 j10
I
60 j5
I2
20 j5
I
60 j5
Vab
8 j2
I
12 j
4 j
I
12 j
-20 I 1 j10 I 2
Vab
Vab
- (160 j40)
10 j40
I
I
12 j
12 j
Vab
- 150
I
12 j
Vab
(12.457 ‘175.24q)(4.25‘97.76q)
Vab
52.94‘273q V
(-12 j)(150)
I
145
Chapter 9, Solution 67.
(a)
20 mH 
o
jZL j (10 3 )(20 u 10 -3 ) j20
1
1
12.5 PF 
o
- j80
3
jZC j (10 )(12.5 u 10 -6 )
Z in
Z in
Z in
Yin
(b)
60 j20 || (60 j80)
( j20)(60 j80)
60 60 j60
63.33 j23.33 67.494 ‘20.22q
1
Z in
0.0148‘-20.22q S
10 mH 
o
20 PF 
o
30 || 60
Z in
Z in
Z in
Yin
jZL
1
jZC
j (10 3 )(10 u 10 -3 )
1
j (10 3 )(20 u 10 -6 )
j10
- j50
20
- j50 20 || (40 j10)
(20)(40 j10)
- j50 60 j10
13.5 j48.92 50.75‘ - 74.56q
1
Z in
0.0197‘74.56q S = 5.24 + j18.99 mS
Chapter 9, Solution 68.
Yeq
1
1
1
5 j2 3 j - j4
Yeq
(0.1724 j0.069) (0.3 j0.1) ( j0.25)
Yeq
0.4724 + j0.219 S
Chapter 9, Solution 69.
1
Yo
1
1
4 - j2
Yo
4
1 j2
1
(1 j2)
4
(4)(1 j2)
5
0.8 j1.6
Yo j 0.8 j0.6
1
Yo c
1
1 1
1
1 - j3 0.8 j0.6
Yo c
1.8 j0.933
Yo c
0.4932‘ - 27.41q
Yo c j5
(1) ( j0.333) (0.8 j0.6)
2.028‘27.41q
0.4378 j0.2271
0.4378 j4.773
1
Yeq
1
1
2 0.4378 j4.773
1
Yeq
0.5191 j0.2078
Yeq
0.5191 j0.2078
0.3126
0.5 0.4378 j4.773
22.97
1.661 + j0.6647 S
Chapter 9, Solution 70.
Make a delta-to-wye transformation as shown in the figure below.
a
Zan
Zbn
Zeq
n
Zcn
c
b
8:
2:
-j5 :
(10)(15 j10)
15 j5
Z an
(- j10)(10 j15)
5 j10 10 j15
Z bn
(5)(10 j15)
15 j5
Z cn
(5)(- j10)
15 j5
Z eq
Z an (Z bn 2) || (Z cn 8 j5)
Z eq
7 j9 (6.5 j3.5) || (7 j8)
Z eq
7 j9 Z eq
7 j9 5.511 j0.2
Z eq
12.51 j9.2
4.5 j3.5
-1 j3
(6.5 j3.5)(7 j8)
13.5 j4.5
15.53‘-36.33q :
7 j9
Chapter 9, Solution 71.
We apply a wye-to-delta transformation.
j4 :
Zab
b
a
Zac
Zbc
Zeq
-j2 :
1:
c
Z ab
2 j2 j4
j2
Z ac
2 j2
2
1 j
Z bc
2 j2
-j
-2 j2
j4 || Z ab
1 || Z ac
1 || (1 j)
1 j
( j4)(1 j)
1 j3
j4 || (1 j)
j4 || Z ab 1 || Z ac
1
Z eq
2 j2
j2
(1)(1 j)
2 j
1.6 j0.8
0.6 j0.2
2.2 j0.6
1
1
1
- j2 - 2 j2 2.2 j0.6
j0.5 0.25 j0.25 0.4231 j0.1154
0.173 j0.3654
Z eq
0.4043‘64.66q
2.473‘-64.66q : = 1.058 –– j2.235 :
Chapter 9, Solution 72.
Transform the delta connections to wye connections as shown below.
a
j2 :
j2 :
-j18 :
-j9 :
j2 :
R1
R2
R3
b
- j9 || - j18
R1
- j6 ,
(20)(20)
20 20 10
8 :,
R2
Z ab
j2 ( j2 8) || (j2 j6 4) 4
Z ab
4 j2 (8 j2) || (4 j4)
Z ab
4 j2 Z ab
4 j2 3.567 j1.4054
Z ab
7.567 + j0.5946 :
(8 j2)(4 j4)
12 - j2
(20)(10)
50
4:,
R3
(20)(10)
50
4:
Chapter 9, Solution 73.
Transform the delta connection to a wye connection as in Fig. (a) and then
transform the wye connection to a delta connection as in Fig. (b).
a
j2 :
j2 :
-j18 :
-j9 :
j2 :
R1
R2
R3
b
Z1
Z2
Z3
( j8)(- j6)
48
- j4.8
j8 j8 j6 j10
Z1 -j4.8
( j8)( j8) - 64
j6.4
j10
j10
(2 Z1 )(4 Z 2 ) (4 Z 2 )(Z 3 ) (2 Z1 )(Z 3 )
(2 j4.8)(4 j4.8) (4 j4.8)( j6.4) (2 j4.8)( j6.4)
Za
Zb
Zc
46.4 j9.6
1.5 j7.25
j6.4
46.4 j9.6
3.574 j6.688
4 j4.8
46.4 j9.6
1.727 j8.945
2 j4.8
j6 || Z b
- j4 || Z a
(6‘90q)(7.583‘61.88q)
07407 j3.3716
3.574 j12.688
(-j4)(1.5 j7.25)
0.186 j2.602
1.5 j11.25
46.4 j9.6
j12 || Z c
(12‘90q)(9.11‘79.07q)
1.727 j20.945
0.5634 j5.1693
Z eq
( j6 || Z b ) || (- j4 || Z a j12 || Z c )
Z eq
(0.7407 j3.3716) || (0.7494 j2.5673)
Z eq
1.508‘75.42q : = 0.3796 + j1.46 :
Chapter 9, Solution 74.
One such RL circuit is shown below.
20 :
V
20 :
+
+
j20 :
Vi = 1‘0q
j20 :
Vo
Z
We now want to show that this circuit will produce a 90q phase shift.
Z
j20 || (20 j20)
V
Z
V
Z 20 i
Vo
j20
V
20 j20
( j20)(20 j20)
20 j40
4 j12
(1‘0q)
24 j12
§ j ·§ 1
¨
¸¨ (1 ©1 j ¹© 3
·
j) ¸
¹
- 20 j20
1 j2
4 (1 j3)
1 j3
6 j3
1
(1 j)
3
j
3
0.3333‘90q
This shows that the output leads the input by 90q.
Chapter 9, Solution 75.
Since cos(Zt ) sin(Zt 90q) , we need a phase shift circuit that will cause the
output to lead the input by 90q. This is achieved by the RL circuit shown
below, as explained in the previous problem.
10 :
10 :
+
+
j10 :
Vi
j10 :
Vo
This can also be obtained by an RC circuit.
Chapter 9, Solution 76.
1
ZC
Let Z = R –– jX, where X
|Z|
R 2 X2

o
C
1
2SfC
| Z |2 R 2
X
1
2SfX
1162
1
2Sx 60x95.394
662
95.394
27.81PF
Chapter 9, Solution 77.
(a)
- jX c
V
R jX c i
1
where X c
ZC
Vo
Vo
Vi
- j3.979
5 - j3.979
Vo
Vi
3.979
Vo
Vi
25 15.83
1
(2S)(2 u 10 6 )(20 u 10 -9 )
3.979
5 3.979
2
2
3.979
‘(-90q tan -1 (3.979 5))
‘(-90q 38.51q)
0.6227 ‘ - 51.49q
Therefore, the phase shift is 51.49q lagging
(b)
T
-45q
45q
-90q tan -1 (X c R )
tan -1 (X c R ) 
o R
Z
2Sf
f
1
2SRC
Xc
1
ZC
1
RC
1
(2S)(5)(20 u 10 -9 )
1.5915 MHz
Chapter 9, Solution 78.
8+j6
R
Z
-jX
Z
R //>8 j (6 X )@
R[8 j (6 X )]
R 8 j (6 X )
5
i.e 8R + j6R –– jXR = 5R + 40 + j30 ––j5X
Equating real and imaginary parts:
8R = 5R + 40 which leads to R=13.33ȍ
6R-XR =30-5 which leads to X=4.125ȍ.
Chapter 9, Solution 79.
(a)
Consider the circuit as shown.
20 :
V2
40 :
V1
30 :
+
Vi
+
j10 :
j30 :
j60 :
Vo
Z2
Z1
Z1
j30 || (30 j60)
Z2
j10 || (40 Z1 )
Let Vi
V2
V2
V1
V1
Vo
Vo
Vo
( j30)(30 j60)
3 j21
30 j90
( j10)(43 j21)
1.535 j8.896
43 j31
1‘0q .
Z2
(9.028‘80.21q)(1‘0q)
Vi
21.535 j8.896
Z 2 20
0.3875‘57.77q
Z1
3 j21
V2
V
Z1 40
43 j21 2
0.1718‘113.61q
(21.213‘81.87q)(0.3875‘57.77q)
47.85‘26.03q
j60
j2
2
V1
V1
(2 j)V1
30 j60
1 j2
5
(0.8944‘26.56q)(0.1718‘113.6q)
0.1536‘140.2q
Therefore, the phase shift is 140.2q
(b)
The phase shift is leading.
(c)
If Vi 120 V , then
Vo (120)(0.1536‘140.2q) 18.43‘140.2q V
and the magnitude is 18.43 V.
Chapter 9, Solution 80.
200 mH 
o
Vo
(a)
jZL
j75.4
V
R 50 j75.4 i
j (2S)(60)(200 u 10 -3 )
j75.4 :
j75.4
(120‘0q)
R 50 j75.4
100 : ,
j75.4
(120 ‘0q)
150 j75.4
53.89‘63.31q V
When R
Vo
Vo
9.028‘80.21q
(75.4‘90q)(120‘0q)
167.88‘26.69q
(b)
Vo
Vo
(c)
0 :,
j75.4
(120‘0q)
50 j75.4
100‘33.55q V
When R
(75.4‘90q)(120 ‘0q)
90.47 ‘56.45q
To produce a phase shift of 45q, the phase of Vo = 90q + 0q D = 45q.
Hence, D = phase of (R + 50 + j75.4) = 45q.
For D to be 45q,
R + 50 = 75.4
Therefore,
R = 25.4 :
Chapter 9, Solution 81.
Z2
R2 Let
Z1
Zx
Z3
Z
Z1 2
Rx 1
jZC x
Rx
R3
R
R1 2
1
Cx
§R3 · § 1 ·
¨ ¸¨ ¸ 
o C x
© R1 ¹ © C2 ¹
R1 ,
1
,
jZC 2
Z3
R1
C
R2 s
1200
(600)
400
1.8 k:
R1
C
R3 2
§ 400 ·
¨
¸(0.3 u 10 -6 )
© 1200 ¹
§ 100 ·
¨
¸(40 u 10 -6 )
© 2000 ¹
2 PF
Chapter 9, Solution 83.
Lx
R2
L
R1 s
Rx R3 §
1 ·
¨R 2 ¸
R1 ©
jZC 2 ¹
Chapter 9, Solution 82.
Cx
R 3 , and Z x
§ 500 ·
¨
¸(250 u 10 -3 )
©1200 ¹
104.17 mH
0.1 PF
1
.
jZC x
Chapter 9, Solution 84.
Let
Z1
Z1
Since Z x
1
,
jZC s
R1
jZC s
1
R1 jZC s
Z2
R 1 ||
R2,
Z3
R 3 , and Z x
R x jZL x .
R1
jZR 1C s 1
Z3
Z ,
Z1 2
R x jZL x
R 2R 3
jZR 1C s 1
R1
R 2R 3
(1 jZR 1C s )
R1
Equating the real and imaginary components,
R 2R 3
Rx
R1
ZL x
R 2R 3
(ZR 1C s ) implies that
R1
L x R 2 R 3Cs
Given that R 1
Rx
Lx
40 k: , R 2
1 .6 k : , R 3
4 k: , and C s
0.45 PF
R 2 R 3 (1.6)(4)
k: 0.16 k: 160 :
R1
40
R 2 R 3 C s (1.6)(4)(0.45) 2.88 H
Chapter 9, Solution 85.
Let
Z2
Z1
R1 ,
Z4
R4
jZR 4 C 4 1
Since Z 4
Z3
Z
Z1 2
1
,
jZC 2
- jR 4
ZR 4 C 4 j
R2 
o Z1 Z 4
Z2Z3 ,
Z3
R 3 , and Z 4
R 4 ||
1
.
jZC 4
§
j ·
¸
R 3 ¨R 2 ZC 2 ¹
©
jR 3
- jR 4 R 1 (ZR 4 C 4 j)
R 3R 2 2
2 2
Z R 4C4 1
ZC 2
- jR 4 R 1
ZR 4 C 4 j
Equating the real and imaginary components,
R 1R 4
R 2R 3
2
Z R 24 C 24 1
(1)
2
R3
ZR 1 R 4 C 4
2
2 2
Z R 4 C 4 1 ZC 2
(2)
Dividing (1) by (2),
1
ZR 4 C 4
1
R 2C2R 4C4
Z2
Z
f
ZR 2 C 2
2Sf
1
R 2C2 R 4C4
1
2S R 2 R 4 C 2 C 4
Chapter 9, Solution 86.
Y
1
1
1
240 j95 - j84
Y
4.1667 u 10 -3 j0.01053 j0.0119
Z
1
Y
1000
4.1667 j1.37
Z = 228‘-18.2q :
1000
4.3861‘18.2q
Chapter 9, Solution 87.
1
jZC
50 -j
(2S)(2 u 10 3 )(2 u 10 -6 )
Z1
50 Z1
50 j39.79
Z2
80 jZL
Z2
80 j125.66
Z3
100
1
Z
1
1
1
Z1 Z 2 Z 3
1
Z
1
1
1
100 50 j39.79 80 j125.66
1
Z
10 -3 (10 12.24 j9.745 3.605 j5.663)
80 j (2S)(2 u 10 3 )(10 u 10 -3 )
(25.85 j4.082) u 10 -3
26.17 u 10 -3 ‘8.97q
Z = 38.21‘-8.97q :
Chapter 9, Solution 88.
(a)
(b)
Z - j20 j30 120 j20
Z = 120 –– j10 :
1
1
would cause the capacitive
ZC 2Sf C
impedance to double, while ZL 2Sf L would cause the inductive
impedance to halve. Thus,
Z - j40 j15 120 j40
Z = 120 –– j65 :
If the frequency were halved,
Chapter 9, Solution 89.
Z in
§
1 ·
¸
jZL || ¨ R jZC ¹
©
Z in
§
1 ·
¸
jZL ¨ R jZC ¹
©
1
R jZL jZC
Z in
L
jZL R
C
§
1 ·
¸
R j¨ZL ©
ZC ¹
·§
§
§L
1 ··
¸¸
¨ jZL R ¸¨ R j¨ZL ¹©
©
©C
ZC ¹¹
§
1 ·
¸
R ¨ZL ©
ZC ¹
2
2
To have a resistive impedance, Im(Z in )
§ L ·§
1 ·
¸
ZL R 2 ¨ ¸¨ZL © C ¹©
ZC ¹
ZR 2 C ZL Z2 R 2 C 2
L
0 . Hence,
0
1
ZC
Z2 LC 1
Z2 R 2 C 2 1
Z2 C
(1)
Ignoring the +1 in the numerator in (1),
L
R 2C
(200) 2 (50 u 10 -9 )
2 mH
Chapter 9, Solution 90.
Let
Vs
145‘0q ,
I
Vs
80 R jX
X
jZL
145‘0q
80 R jX
j (2S)(60) L
j377 L
V1
80 I
(80)(145)
80 R jX
50
(80)(145)
80 R jX
Vo
(R jX) I
110
(1)
(R jX)(145‘0q)
80 R jX
(R jX)(145)
80 R jX
(2)
From (1) and (2),
50
80
110
R jX
R jX
R 2 X2
From (1),
§11 ·
(80) ¨ ¸
©5¹
30976
(3)
(80)(145)
50
80 R jX
6400 160R R 2 X 2
160R R 2 X 2
232
53824
47424
(4)
Subtracting (3) from (4),
160R 16448 
o R 102.8 :
From (3),
X 2 30976 10568 20408
X 142.86
377 L 
o L
0.3789 H
Chapter 9, Solution 91.
Z in
1
R || jZL
jZC
Z in
-j
jZLR
ZC R jZL
- j Z 2 L2 R jZLR 2
ZC
R 2 Z 2 L2
To have a resistive impedance, Im(Z in )
Hence,
-1
ZLR 2
0
2
ZC R Z2 L2
ZLR 2
R 2 Z2 L2
1
ZC
C
where Z
0.
R 2 Z2 L2
Z2 LR 2
2S f
2S u 10 7
C
9 u 10 4 (4S 2 u 1014 )(400 u 10 12 )
(4S 2 u 1014 )(20 u 10 6 )(9 u 10 4 )
C
9 16S 2
nF
72S 2
C = 235 pF
Chapter 9, Solution 92.
(a) Z o
Z
Y
(b) J
ZY
100‘75 o
450‘48 o x10 6
471.4‘13.5 o :
100‘75 o x 450‘48 o x10 6
0.2121‘61.5 o
Chapter 9, Solution 93.
Z
Z
Z
Zs 2 ZA ZL
(1 0.8 23.2) j(0.5 0.6 18.9)
25 j20
115‘0q
32.02 ‘38.66q
IL
VS
Z
IL
3.592‘-38.66q A
Chapter 10, Solution 1.
Z 1
10 cos( t 45q) 
o 10‘ - 45q
5 sin( t 30q) 
o 5‘ - 60q
1H 
o
1F 
o
jZL
1
jZC
j
-j
The circuit becomes as shown below.
3:
10‘-45q V
Vo
+
j:
2 Io
+
5‘-60q V
Applying nodal analysis,
(10‘ - 45q) Vo (5‘ - 60q) Vo Vo
3
j
-j
j10‘ - 45q 15‘ - 60q j Vo
Vo 10 ‘ - 45q 15‘ - 150q 15.73‘247.9q
Therefore,
v o ( t ) 15.73 cos(t + 247.9q) V
Chapter 10, Solution 2.
Z 10
4 cos(10t S 4) 
o 4‘ - 45q
20 sin(10 t S 3) 
o 20 ‘ - 150q
1H 
o
jZL
0.02 F 
o
j10
1
jZC
1
j 0.2
- j5
The circuit becomes that shown below.
10 :
Vo
Io
20‘-150q V
+
j10 :
4‘-45q A
-j5 :
Applying nodal analysis,
Vo Vo
(20‘ - 150q) Vo
4‘ - 45q
10
j10 - j5
20 ‘ - 150q 4‘ - 45q 0.1(1 j) Vo
Vo
j10
Io
Therefore,
2 ‘ - 150q 4 ‘ - 45q
2.816 ‘150.98q
j (1 j)
i o ( t ) 2.816 cos(10t + 150.98q) A
Chapter 10, Solution 3.
Z 4
2 cos(4t ) 
o 2‘0q
16 sin(4 t ) 
o 16‘ - 90q -j16
2H 
o
jZL j8
1
1
1 12 F 
o
jZC j (4)(1 12)
- j3
The circuit is shown below.
4:
-j16 V
+
-j3 :
Vo
1:
j8 :
2‘0q A
6:
Applying nodal analysis,
- j16 Vo
2
4 j3
Vo
Vo
1 6 j8
§
- j16
1
1 ·
¸V
2 ¨1 4 j3
© 4 j3 6 j8 ¹ o
Vo
Therefore,
3.92 j2.56
1.22 j0.04
4.682‘ - 33.15q
1.2207 ‘1.88q
3.835‘ - 35.02q
3.835 cos(4t –– 35.02q) V
v o (t)
Chapter 10, Solution 4.
16 sin(4 t 10q) 
o 16‘ - 10q, Z
1H 
o
jZL
0.25 F 
o
Ix
16‘-10q V
4
j4
1
jZC
1
j (4)(1 4)
j4 :
V1
-j
-j :
+
+
0.5 Ix
1:
Vo
(16‘ - 10q) V1 1
Ix
j4
2
V1
1 j
But
Ix
So,
(16‘ - 10q) V1
j4
3 ((16‘ - 10q) V1 )
j8
V1
1 j
V1
48‘ - 10q
- 1 j4
Using voltage division,
1
48‘ - 10q
Vo
V1
1 j
(1 - j)(-1 j4)
Therefore,
v o (t)
8.232‘ - 69.04q
8.232 sin(4t –– 69.04q) V
Chapter 10, Solution 5.
Let the voltage across the capacitor and the inductor be Vx and we get:
Vx 0.5I x 10‘30q Vx Vx
0
4
j2 j3
(3 j6 j4)Vx 1.5I x
30‘30q but I x
Vx
j2
j0.5Vx
Combining these equations we get:
(3 j2 j0.75)Vx
Ix
j0.5
30‘30q
3 j1.25
30‘30q or Vx
30‘30q
3 j1.25
4.615‘97.38q A
Chapter 10, Solution 6.
Let Vo be the voltage across the current source. Using nodal analysis we get:
Vo 4Vx
Vo
3
20
20 j10
0 where Vx
20
Vo
20 j10
Combining these we get:
Vo
4Vo
Vo
3
20 20 j10
20 j10
Vo
60 j30
or Vx
2 j0.5
0 o (1 j0.5 3)Vo
20(3)
2 j0.5
60 j30
29.11‘––166 V.
Chapter 10, Solution 7.
At the main node,
120‘ 15 o V
40 j20
6‘30 o V
V
j30 50
115.91 j31.058
5.196 j3
40 j20

o
§
1
j
1·
¸¸
V¨¨
© 40 j20 30 50 ¹
3.1885 j4.7805
0.04 j0.0233
V
124.08‘ 154 o V
Chapter 10, Solution 8.
Z
200,

o
100mH
50PF

o
jZ L
1
jZC
j200x 0.1
j20
1
j100
j200x 50x10 6
The frequency-domain version of the circuit is shown below.
0.1 Vo
40 :
V1
6‘15
o
20 :
+
Vo
-
Io
V2
-j100 :
j20 :
At node 1,
V1
V1
V V2
1
20 j100
40
5.7955 j1.5529 (0.025 j 0.01)V1 0.025V2
6‘15 o 0.1V1
or
(1)
At node 2,
V1 V2
40
From (1) and (2),
0.1V1 V2
j20
o
0
3V1 (1 j2)V2
§ (5.7955 j1.5529) ·
¸¸
¨¨
0
¹
©
ª(0.025 j0.01) 0.025º§ V1 ·
¨ ¸
«
3
(1 j2) »¼¨© V2 ¸¹
¬
or
(2)
AV
B
Using MATLAB,
leads to V1
V = inv(A)*B
70.63 j127.23,
V2
Io
V1 V2
40
110.3 j161.09
7.276‘ 82.17 o
Thus,
i o (t)
7.276 cos(200 t 82.17 o ) A
Chapter 10, Solution 9.
10 cos(10 3 t ) 
o 10 ‘0q, Z 10 3
10 mH 
o
jZL
j10
50 PF 
o
1
jZC
1
j (10 )(50 u 10 -6 )
3
- j20
Consider the circuit shown below.
20 :
V1
-j20 :
V2
j10 :
Io
10‘0q V
+
20 :
+
4 Io
30 :
Vo
At node 1,
10 V1
20
10
V1 V1 V2
20
- j20
(2 j) V1 jV2
(1)
At node 2,
V1 V2
- j20
(4)
(-4 j) V1
V1
V1
V2
, where I o
20 30 j10
V1
has been substituted.
20
(0.6 j0.8) V2
0.6 j0.8
V2
-4 j
(2)
Substituting (2) into (1)
(2 j)(0.6 j0.8)
10
V2 jV2
-4 j
or
V2
170
0.6 j26.2
Vo
30
V
30 j10 2
Therefore,
v o (t)
3
170
˜
3 j 0.6 j26.2
6.154 ‘70.26q
6.154 cos(103 t + 70.26q) V
Chapter 10, Solution 10.
50 mH
2PF

o

o
jZ L
1
jZC
j2000x50 x10 3
1
j2000 x 2x10 6
j100,
Z
j250
Consider the frequency-domain equivalent circuit below.
V1
36<0o
2k :
-j250
j100
V2
0.1V1
4k :
2000
At node 1,
36
V1
V
V V2
1 1
2000 j100 j250

o
(0.0005 j0.006)V1 j0.004V2
36
(1)
At node 2,
V1 V2
j250
0.1V1 V2
4000
o
(0.1 j0.004)V1 (0.00025 j0.004)V2 (2)
0
Solving (1) and (2) gives
Vo
V2
535.6 j893.5 8951.1‘93.43o
vo (t) = 8.951 sin(2000t +93.43o) kV
Chapter 10, Solution 11.
cos(2t ) 
o 1‘0q, Z
2
8 sin( 2t 30q) 
o 8‘ - 60q
1H 
o
jZL
j2
12F 
o
1
jZC
1
j (2)(1 2)
-j
2H 
o
jZL
j4
14F 
o
1
jZC
1
j (2)(1 4)
- j2
Consider the circuit below.
2 Io
2 Io
2 Io
2 -j :
2
2 Io
2
I
-j :
2
2 Io
At node 1,
(8‘ - 60q) V1
2
8‘ - 60q
V1 V1 V2
-j
j2
(1 j) V1 j V2
(1)
At node 2,
1
V1 V2 (8‘ - 60q) V2
j2
j4 j2
0
V2 4 ‘ - 60q j 0.5 V1
Substituting (2) into (1),
1 8‘ - 60q 4 ‘30q (1 j1.5) V1
Therefore,
V1
1 8‘ - 60q 4‘30q
1 j1.5
Io
V1
-j
i o (t)
1 8‘ - 60q 4 ‘30q
1.5 j
(2)
5.024‘ - 46.55q
5.024 cos(2t –– 46.55q)
Chapter 10, Solution 12.
20 sin(1000t ) 
o 20 ‘0q, Z 1000
10 mH 
o
jZL
j10
50 PF 
o
1
jZC
1
j (10 )(50 u 10 -6 )
3
- j20
The frequency-domain equivalent circuit is shown below.
2 Io
V1
10 :
V2
Io
20‘0q A
20 :
-j20 :
j10 :
At node 1,
V1 V1 V2
,
20
10
20
2Io Io
V2
j10
20
2V2 V1 V1 V2
j10 20
10
400
3V1 (2 j4) V2
where
(1)
At node 2,
2V2 V1 V2
j10
10
V2
V
2
- j20 j10
j2 V1 (-3 j2) V2
V1 (1 j1.5) V2
or
Substituting (2) into (1),
400 (3 j4.5) V2 (2 j4) V2
Therefore,
V2
400
1 j0.5
Io
V2
j10
40
j (1 j0.5)
(2)
(1 j0.5) V2
35.74 ‘ - 116.6q
35.74 sin(1000t –– 116.6q) A
i o (t)
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 :
40‘30º V
+
18 :
j6 :
+
Vx
3:
50‘0º V
+
Vx 40‘30q Vx Vx 50
j2
3
18 j6
0
which leads to Vx = 29.36‘62.88 A.
Chapter 10, Solution 14.
At node 1,
0 V1 0 V1 V2 V1
- j2
10
j4
- (1 j2.5) V1 j2.5 V2
20‘30q
173.2 j100
(1)
At node 2,
V2 V2 V2 V1
j2 - j5
j4
- j5.5 V2 j2.5 V1
20‘30q
173.2 j100
(2)
Equations (1) and (2) can be cast into matrix form as
ª1 j2.5 j2.5 ºª V1 º ª - 200 ‘30qº
« j2.5
- j5.5»¼«¬ V2 »¼ «¬ 200 ‘30q »¼
¬
'
'1
'2
1 j2.5 j2.5
j2.5
- j5.5
20 j5.5
- 200 ‘30q j2.5
200 ‘30q - j5.5
1 j2.5 - 200‘30q
j2.5
200‘30q
V1
'1
'
28.93‘135.38q
V2
'2
'
49.18‘124.08q
20.74‘ - 15.38q
j3 (200‘30q)
600‘120q
(200 ‘30q)(1 j5) 1020‘108.7q
Chapter 10, Solution 15.
We apply nodal analysis to the circuit shown below.
5A
2:
V2
I
+
-j20 V
j:
V1
-j2 :
2I
4:
At node 1,
- j20 V1
2
- 5 j10
5
V1 V1 V2
- j2
j
(0.5 j0.5) V1 j V2
At node 2,
5 2I V1 V2
j
where I
V1
- j2
V2
V2
,
4
5
V1
0.25 j
(2)
Substituting (2) into (1),
- 5 j10 (1 j) V1
j5
0.25 j
-10 j20 ( 2 ‘ - 45q) V1
V1
0.5 (1 j) V1
j40
1 j4
-10 j20 15.81‘313.5q
160 j40
17 17
(1)
I
V1
- j2
I
7.906‘43.49q A
(0.5‘90q)(15.81‘313.5q)
Chapter 10, Solution 16.
At node 1,
V1 V1 V2 V1 V2
20
10
- j5
j40 (3 j4) V1 (2 j4) V2
j2
At node 2,
V1 V2 V1 V2
V2
1 j
10
- j5
j10
10 (1 j) - (1 j2) V1 (1 j) V2
Thus,
ª j40
«10 (1 ¬
º ª 3 j4 - 2 (1 j2) ºª V1 º
j) »¼ «¬ - (1 j2)
1 j »¼«¬ V2 »¼
'
3 j4 - 2 (1 j2)
- (1 j2)
1 j
'1
j40
- 2 (1 j2)
10 (1 j)
1 j
'2
V1
V2
3 j4
j40
- (1 j2) 10 (1 j)
'1
'
'2
'
5 j 5.099 ‘ - 11.31q
60 j100 116.62 ‘120.96q
-90 j110 142.13‘129.29q
22.87‘132.27q V
27.87‘140.6q V
Chapter 10, Solution 17.
Consider the circuit below.
j4 :
2:
Io
+
100‘20q V
1:
V1
V2
3:
-j2 :
At node 1,
100‘20q V1
j4
100 ‘20q
V1 V1 V2
3
2
V1
(3 j10) j2 V2
3
(1)
At node 2,
100‘20q V2 V1 V2
1
2
V2
- j2
100 ‘20q -0.5 V1 (1.5 j0.5) V2
From (1) and (2),
ª100‘20qº ª - 0.5
0.5 (3 j) ºª V1 º
«100‘20q» «1 j10 3
- j2 »¼«¬ V2 »¼
¬
¼ ¬
'
'1
'2
- 0.5
1.5 j0.5
1 j10 3
- j2
100‘20q 1.5 j0.5
100‘20q
- j2
- 0.5
100‘20q
1 j10 3 100‘20q
0.1667 j4.5
-55.45 j286.2
-26.95 j364.5
(2)
V1
'1
'
64.74 ‘ - 13.08q
V2
'2
'
81.17 ‘ - 6.35q
Io
V1 V2
2
Io
9.25‘-162.12q
'1 ' 2
2'
- 28.5 j78.31
0.3333 j 9
Chapter 10, Solution 18.
Consider the circuit shown below.
8:
V1
j6 : V
2
4:
j5 :
+
4‘45q A
2:
+
2 Vx
Vx
-j :
-j2 :
At node 1,
4‘45q
200 ‘45q
V1 V1 V2
2
8 j6
(29 j3) V1 (4 j3) V2
(1)
At node 2,
V1 V2
2Vx
8 j6
(104 j3) V1
V2
V2
,
- j 4 j5 j2
where Vx
V1
(12 j41) V2
12 j41
V
104 j3 2
Substituting (2) into (1),
V1
200‘45q (29 j3)
Vo
(2)
(12 j41)
V (4 j3) V2
104 j3 2
200 ‘45q
(14.21‘89.17q) V2
V2
200‘45q
14.21‘89.17q
Vo
- j2
V
4 j5 j2 2
Vo
10‘233.13q
200‘45q
˜
25
14.21‘89.17q
Vo
5.63‘189q V
- 6 j8
V2
25
- j2
V
4 j3 2
Chapter 10, Solution 19.
We have a supernode as shown in the circuit below.
j2 :
V1
V2
4:
V3
+
2:
-j4 :
Vo
0.2 Vo
Vo V1 .
Notice that
At the supernode,
V3 V2 V2 V1 V1 V3
4
- j4 2
j2
0
(2 j2) V1 (1 j) V2 (-1 j2) V3
(1)
At node 3,
0.2V1 V1 V3
j2
V3 V2
4
(0.8 j2) V1 V2 (-1 j2) V3
Subtracting (2) from (1),
0
(2)
0 1.2V1 j V2
But at the supernode,
V1 12 ‘0q V2
V2 V1 12
or
Substituting (4) into (3),
0 1.2V1 j (V1 12)
V1
j12
1.2 j
Vo
12‘90q
1.562‘39.81q
Vo
7.682‘50.19q V
(3)
(4)
Vo
Chapter 10, Solution 20.
The circuit is converted to its frequency-domain equivalent circuit as shown below.
R
+
Vm‘0q
+
jZL
Vo
1
jZC
Let
Z
Vo
Vo
jZL ||
1
jZC
Z
V
RZ m
L
C
1
jZL jZC
jZL
1 Z2 LC
jZL
1 Z2 LC
Vm
jZL
R
1 Z2 LC
jZL
V
R (1 Z2 LC) jZL m
·
§
ZL
¸
¨90q tan -1
‘
2
R (1 Z LC) ¹
R 2 (1 Z2 LC) 2 Z2 L2 ©
ZL Vm
If
Vo
A
and
I
A‘I , then
ZL Vm
R 2 (1 Z 2 LC) 2 Z 2 L2
90q tan -1
ZL
R (1 Z 2 LC)
Chapter 10, Solution 21.
(a)
Vo
Vi
At Z
1
jZC
R jZL 1
jZC
Vo
Vi
Vo
Vi
0,
As Z o f ,
At Z
(b)
Vo
Vi
At Z
1
LC
Vo
Vi
,
jZL
R jZL 1
LC
1
jZC
,
1
1
1
0
1
jRC ˜
1
-j L
R C
LC
Z2 LC
1 Z2 LC jZRC
Vo
Vi
Vo
Vi
0,
As Z o f ,
At Z
1
1 Z LC jZRC
2
Vo
Vi
0
1
1
1
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
jZC
+
jZL
Let
(R 2 jZL) ||
Z
1
(R jZL)
jZC 2
1
R 2 jZL jZC
Vo
Vs
Vo
1
jZC
Z
Vo
Vs
+
R 2 jZL
1 jZR 2 Z2 LC
R 2 jZL
1 Z2 LC jZR 2 C
Z
R 2 jZL
Z R1
R1 1 Z2 LC jZR 2 C
R 2 jZL
2
R 1 R 2 Z LCR 1 jZ (L R 1 R 2 C)
Chapter 10, Solution 23.
V Vs
V
jZCV
1
R
jZL jZ C
V
jZRCV
Z2LC 1
jZRCV
0
Vs
§ 1 Z2LC jZRC jZRC jZ3RLC2 ·
¨
¸V
2
¨
¸
Z
1
LC
©
¹
Vs
V
(1 Z2 LC)Vs
1 Z2LC jZRC(2 Z2LC)
Chapter 10, Solution 24.
For mesh 1,
Vs
§ 1
1 ·
1
¨
¸ I1 I
jZC 2 2
© jZC1 jZC 2 ¹
(1)
For mesh 2,
§
1 ·
1
¸I
I 1 ¨ R jZL jZC 2
jZC 2 ¹ 2
©
Putting (1) and (2) into matrix form,
1
1
ª 1
º
«
» ª I1 º
ªVs º
jZC1 jZC 2
jZC 2
«
»
«0»
1
1 » «¬I 2 »¼
¬ ¼ «
R jZL «¬
jZ C 2
jZC 2 »¼
(2)
0
'
'1
I1
I2
§ 1
1 ·§
1 ·
1
¸ 2
¨
¸¨ R jZL jZC 2 ¹ Z C1C 2
© jZC1 jZC 2 ¹©
§
1 ·
¸
Vs ¨ R jZL jZC 2 ¹
©
'1
'
'2
'
and
§
1 ·
¸
Vs ¨ R jZL jZC 2 ¹
©
§ 1
1 ·
1
1 ·§
¨
¸¨ R jZL ¸ 2
jZC 2 ¹ Z C1 C 2
© jZC 1 jZC 2 ¹©
Vs
jZC 2
§ 1
1 ·§
1 ·
1
¸ 2
¨
¸¨ R jZL jZC 2 ¹ Z C1 C 2
© jZC 1 jZC 2 ¹©
Chapter 10, Solution 25.
Z 2
10 cos(2t ) 
o 10‘0q
'2
Vs
jZC 2
6 sin(2t ) 
o 6 ‘ - 90q
2H 
o
jZL
0.25 F 
o
-j6
j4
1
jZC
1
j (2)(1 4)
- j2
The circuit is shown below.
4:
j4 :
Io
10‘0q V
+
I1
+
I2
-j2 :
6‘-90q V
For loop 1,
- 10 (4 j2) I 1 j2 I 2
5 (2 j) I 1 j I 2
0
(1)
For loop 2,
j2 I 1 ( j4 j2) I 2 (- j6)
I1 I 2 3
In matrix form (1) and (2) become
ª 2 j j º ª I 1 º ª 5º
« 1 1 » « I » « 3»
¼¬ 2 ¼ ¬ ¼
¬
Therefore,
'
2 (1 j) ,
Io
I1 I 2
i o (t)
0
'1
(2)
5 j3 ,
'1 ' 2
4
2 (1 j)
'
1.414 cos(2t + 45q) A
'2
1 j3
1 j 1.414 ‘45q
Chapter 10, Solution 26.
We apply mesh analysis to the circuit shown below.
For mesh 1,
- 10 40 I 1 20 I 2 0
1 4 I1 2 I 2
For the supermesh,
(20 j20) I 2 20 I 1 (30 j10) I 3
- 2 I 1 (2 j2) I 2 (3 j) I 3
0
(1)
0
(2)
At node A,
I1 I 2
(3)
I2 I3 4Io
Substituting (3) into (4)
I 2 I 3 4 I1 4 I 2
I 3 5 I 2 4 I1
Substituting (5) into (2) gives
0 -(14 j4) I 1 (17 j3) I 2
From (1) and (6),
ª1 º ª
4
- 2 ºª I 1 º
« 0» « - (14 j4) 17 j3»« I »
¬ ¼ ¬
¼¬ 2 ¼
(4)
Io
At node B,
'
'1
(6)
40 j4
1
-2
17 j3 ,
0 17 j3
I3
5 I 2 4 I1
Vo
30 I 3
Therefore,
(5)
5 ' 2 4 '1
'
15 (1 j4)
10 j
v o (t)
4
'2
1
- (14 j4) 0
14 j4
2 j8
40 j4
6.154‘70.25q
6.154 cos(103 t + 70.25q) V
Chapter 10, Solution 27.
For mesh 1,
- 40 ‘30q ( j10 j20) I 1 j20 I 2
4 ‘30q - j I 1 j2 I 2
0
(1)
For mesh 2,
50 ‘0q (40 j20) I 2 j20 I 1 0
5 - j2 I 1 (4 j2) I 2
From (1) and (2),
ª 4‘30qº ª - j
j2 ºª I 1 º
« 5 » « - j2 - (4 j2) »« I »
¬
¼ ¬
¼¬ 2 ¼
'
-2 4 j
4.472‘116.56q
(2)
'1
-(4 ‘30q)(4 j2) j10
'2
- j5 8‘120q
I1
'1
'
4.698‘95.24q A
I2
'2
'
0.9928‘37.71q A
21.01‘211.8q
4.44 ‘154.27q
Chapter 10, Solution 28.
1H

o
jZL
j4,

o
1F
1
jZC
1
j1x 4
j0.25
The frequency-domain version of the circuit is shown below, where
V1
10‘0 o ,
V2
1
20‘ 30 o .
j4
j4
1
-j0.25
+
+
V1
-
I1
V1
10‘0 o ,
V2
1
I2
V2
-
20‘ 30 o
Applying mesh analysis,
10
(2 j3.75)I1 (1 j0.25)I 2
20‘ 30 o
From (1) and (2), we obtain
(1 j0.025)I1 (2 j3.75)I 2
(1)
(2)
10
·
§
¸¸
¨¨
© 17.32 j10 ¹
§ 2 j3.75 1 j0.25 ·§ I1 ·
¸¸¨¨ ¸¸
¨¨
© 1 j0.25 2 j3.75 ¹© I 2 ¹
Solving this leads to
I1
1.3602 j0.9769 1.6747‘ 35.69 o ,
I2
4.1438 j2.111 4.6505‘153o
Hence,
i1
1.675 cos(4t 35.69 o ) A,
i2
4.651cos(46 153o ) A
Chapter 10, Solution 29.
For mesh 1,
(5 j5) I 1 (2 j) I 2 30 ‘20q
30 ‘20q (5 j5) I 1 (2 j) I 2
0
(1)
For mesh 2,
(5 j3 j6) I 2 (2 j) I 1 0
0 - (2 j) I 1 (5 j3) I 2
From (1) and (2),
ª30‘20qº ª 5 j5 - (2 j) ºª I 1 º
« 0 » « - (2 j) 5 - j3 »« I »
¬
¼ ¬
¼¬ 2 ¼
' 37 j6 37.48‘9.21q
' 1 (30 ‘20q)(5.831‘ - 30.96q) 175‘ - 10.96q
' 2 (30 ‘20q)(2.356 ‘26.56q) 67.08‘46.56q
I1
'1
'
4.67‘-20.17q A
I2
'2
'
1.79‘37.35q A
(2)
Chapter 10, Solution 30.
Consider the circuit shown below.
I2
j4 :
10‘0q V
+
1:
2:
Io
I1
I3
3:
-j2 :
For mesh 1,
100 ‘20q
(3 j4) I 1 j4 I 2 3 I 3
(1)
For mesh 2,
- j4 I 1 (3 j4) I 2 j2 I 3
(2)
0 -3 I 1 2 I 2 (5 j2) I 3
Put (1), (2), and (3) into matrix form.
ª3 j4 - j4
- 3 ºª I 1 º
«
»« »
j4
3
j4
j2
«
»« I 2 »
«¬ - 3
-2
5 - j2 »¼«¬ I 3 »¼
(3)
0
For mesh 3,
'
3 j4 - j4
-3
- j4 3 j4 - j2
-3
-2
5 - j2
ª100‘20qº
«
»
0
«
»
«¬
»¼
0
106 j30
'2
3 j4 100‘20q - 3
- j4
0
- j2
-3
0
5 - j2
(100‘20q)(8 j26)
'3
3 j4 - j4 100‘20q
- j4 3 j4
0
-3
-2
0
(100‘20q)(9 j20)
'3 '2
'
Io
I3 I2
Io
5.521‘-76.34q A
(100‘20q)(1 j6)
106 j30
Chapter 10, Solution 31.
Consider the network shown below.
80 :
100‘120q V
+
I1
Io
j60 :
I2
-j40 :
20 :
I3
-j40 :
+
60‘-30q V
For loop 1,
- 100 ‘20q (80 j40) I 1 j40 I 2
10 ‘20q 4 (2 j) I 1 j4 I 2
0
(1)
For loop 2,
j40 I 1 ( j60 j80) I 2 j40 I 3
0 2 I1 I 2 2 I 3
0
(2)
For loop 3,
60 ‘ - 30q (20 j40) I 3 j40 I 2
- 6 ‘ - 30q
0
j4 I 2 2 (1 j2) I 3
(3)
From (2),
2I3
I 2 2 I1
Substituting this equation into (3),
- 6 ‘ - 30q -2 (1 j2) I 1 (1 j2) I 2
(4)
From (1) and (4),
ª 10‘120q º ª 4 (2 j)
j4 ºª I 1 º
« - 6‘ - 30q» « - 2 (1 j2) 1 j2»« I »
¼¬ 2 ¼
¼ ¬
¬
'
'2
Io
8 j4
- j4
- 2 j4 1 j2
32 j20
8 j4 10‘120q
- 2 j4 - 6‘ - 30q
I2
'2
'
37.74‘32q
-4.928 j82.11 82.25‘93.44q
2.179‘61.44q A
Chapter 10, Solution 32.
Consider the circuit below.
j4 :
Io
+
2:
4‘-30q V
Vo
I1
+
3 Vo
For mesh 1,
where
(2 j4) I 1 2 (4‘ - 30q) 3 Vo
Vo 2 (4‘ - 30q I 1 )
0
Hence,
(2 j4) I 1 8‘ - 30q 6 (4 ‘ - 30q I 1 )
4 ‘ - 30q (1 j) I 1
I1
or
2 2 ‘15q
Io
3 Vo
- j2
Io
j3 (4 ‘ - 30q 2 2 ‘15q)
Io
8.485‘15q A
Vo
- j2 I o
3
3
(2)(4‘ - 30q I 1 )
- j2
5.657‘-75q V
Chapter 10, Solution 33.
Consider the circuit shown below.
0
I2
-j2 :
5A
I4
2:
j:
I
-j20 V
+
I1
-j2 :
I2
j20 (2 j2) I 1 j2 I 2
(1 j) I 1 j I 2 - j10
0
2I
I3
4:
For mesh 1,
(1)
For the supermesh,
( j j2) I 2 j2 I 1 4 I 3 j I 4
0
(2)
Also,
I 3 I 2 2 I 2 (I 1 I 2 )
I 3 2 I1 I 2
(3)
For mesh 4,
I4
(4)
5
Substituting (3) and (4) into (2),
(8 j2) I 1 (- 4 j) I 2
(5)
j5
Putting (1) and (5) in matrix form,
ª 1 j
j ºª I 1 º ª - j10 º
«8 j2 4 j»« I » « j5 »
¼
¬
¼¬ 2 ¼ ¬
'
-3 j5 ,
I
I1 I 2
'1
'1 ' 2
'
-5 j40 ,
10 j45
- 3 j5
'2
-15 j85
7.906‘43.49q A
Chapter 10, Solution 34.
The circuit is shown below.
Io
I2
5:
3A
20 :
8:
40‘90q V
+
-j2 :
I3
10 :
j15 :
I1
j4 :
For mesh 1,
- j40 (18 j2) I 1 (8 j2) I 2 (10 j4) I 3
For the supermesh,
(13 j2) I 2 (30 j19) I 3 (18 j2) I 1
0
0
(1)
(2)
Also,
I2
I3 3
Adding (1) and (2) and incorporating (3),
- j40 5 (I 3 3) (20 j15) I 3
3 j8
I3
1.465‘38.48q
5 j3
I o I 3 1.465‘38.48q A
(3)
0
Chapter 10, Solution 35.
Consider the circuit shown below.
4:
j2 :
I3
8:
1:
-j3 :
10 :
+
20 V
I1
-j4 A
I2
-j5 :
For the supermesh,
- 20 8 I 1 (11 j8) I 2 (9 j3) I 3
0
(1)
Also,
I1
I 2 j4
(2)
For mesh 3,
(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),
ª 19 j8 - (9 j3) ºª I 2 º ª 20 j32 º
« - (9 j3) 13 j »« I » « j32 »
¼
¼¬ 3 ¼ ¬
¬
'
I2
I2
167 j69 ,
' 2 324 j148
'
167 j69
1.971‘-2.1q A
'2
324 j148
356.2‘ - 24.55q
180.69‘ - 22.45q
Chapter 10, Solution 36.
Consider the circuit below.
j4 :
-j3 :
+
I1
4‘90q A
2:
I2
Vo
+
2:
2:
I3
2‘0q A
Clearly,
I1
4 ‘90q
and
j4
I3
-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)
Vo
11.648‘52.82q V
7.04 j9.28
Chapter 10, Solution 37.
I1
+
120‘ 90 o V
-
Ix
Z
Z=80-j35 :
I2
Iy
Iz
12‘0q V
120‘ 30 o V
+
Z
I3
For mesh x,
ZI x ZI z
j120
ZI y ZI z
120‘30 o
(1)
For mesh y,
103.92 j60
(2)
For mesh z,
ZI x ZI y 3ZI z
(3)
0
Putting (1) to (3) together leads to the following matrix equation:
0
(80 j35) ·§ I x ·
§ (80 j35)
¨
¸¨ ¸
0
(80 j35) (80 j35) ¸¨ I y ¸
¨
¨ (80 j35) (80 j35) (240 j105) ¸¨ I ¸
©
¹© z ¹
j120
§
·
¨
¸
¨ 103.92 j60 ¸
¨
¸
0
©
¹

o
Using MATLAB, we obtain
I
inv(A) * B
§ - 1.9165 j1.4115 ·
¨
¸
¨ - 2.1806 - j0.954 ¸
¨ - 1.3657 j0.1525 ¸
©
¹
1.9165 j1.4115
I1
Ix
I2
Iy Ix
I3
I y
2.3802‘143.6 o A
0.2641 j2.3655
2.1806 j0.954
Chapter 10, Solution 38.
Consider the circuit below.
2.3802‘ 96.37 o A
2.3802‘23.63o A
AI
B
Io
I1
2‘0q A
I2
2:
j2 :
1:
+
-j4 :
4‘0q A
I3
I4
10‘90q V
1:
A
Clearly,
I1
(1)
2
For mesh 2,
(2 j4) I 2 2 I 1 j4 I 4 10 ‘90q 0
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
(2)
(3)
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),
ª1 j2 j2 ºª I 2 º ª 2 j5º
« j2 1 j»« I » « 2 j6»
¬
¼¬ 4 ¼ ¬
¼
'
3 j3 ,
Io
-I2
Io
'1
- ' 1 - (9 j11)
'
3 j3
3.35‘174.3q A
(4)
(5)
9 j11
1
(-10 j)
3
Chapter 10, Solution 39.
For mesh 1,
(28 j15)I1 8I 2 j15I 3
12‘64 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
j15 ·§ I1 ·
8
§ (28 j15)
¸¨ ¸
¨
(8 j9) j16 ¸¨ I 2 ¸
¨ 8
¨
j15
j16 (10 j) ¸¹¨© I 3 ¸¹
©
§12‘64 o ·
¸
¨
¨ 0 ¸
¸
¨
¨ 0 ¸
¹
©
Using MATLAB,
I = inv(A)*B
0.3814‘109.6 o A
I1
0.128 j0.3593
I2
0.1946 j0.2841 0.3443‘124.4 o A
I3
0.0718 j0.1265
0.1455‘ 60.42 o A
or
AI
B
Ix
I1 I 2
0.0666 j0.0752
0.1005‘48.5 o A
Chapter 10, Solution 40.
Let i O i O1 i O 2 , where i O1 is due to the dc source and i O 2 is due to the ac source. For
i O1 , consider the circuit in Fig. (a).
4:
2:
iO1
+
8V
(a)
Clearly,
i O1
82
4A
For i O 2 , consider the circuit in Fig. (b).
4:
2:
IO2
10‘0q V
+
j4 :
(b)
If we transform the voltage source, we have the circuit in Fig. (c), where 4 || 2
IO2
2.5‘0q A
4:
(c)
By the current division principle,
43
I O2
(2.5‘0q)
4 3 j4
I O 2 0.25 j0.75 0.79‘ - 71.56q
2:
j4 :
4 3 :.
Thus,
Therefore,
0.79 cos(4t 71.56q) A
i O2
iO
i O1 i O 2
4 + 0.79 cos(4t –– 71.56q) A
Chapter 10, Solution 41.
Let vx = v1 + v2.
For v1 we let the DC source equal zero.
5:
+
+
––
20‘0
1:
––j
V1
V1 20 V1 V1
5
j 1
0 which simplifies to (1j 5 5 j)V1
100 j
V1 = 2.56‘––39.8 or v1 = 2.56sin(500t –– 39.8) V
Setting the AC signal to zero produces:
5:
1:
+
V2
+
––
6V
The 1-ohm resistor in series with the 5-ohm resistor creating a simple voltage divider
yielding:
v2 = (5/6)6 = 5 V.
vx = {2.56sin(500t –– 39.8) + 5} V.
Chapter 10, Solution 42.
Let ix = i1 + i2, where i1 and i2 which are generated by is and vs respectively. For i1 we let
is = 6sin2t A becomes Is = 6‘0, where Ȧ =2.
2 j4
1 j2
6 12
3 j2 2 j4
5 j2
i1= 4.983sin(2t –– 41.63) A
I1
––j4
3.724 j3.31 4.983‘ 41.63q
2:
i1
3:
is
j2
For i2, we transform vs = 12cos(4t –– 30) into the frequency domain and get
Vs = 12‘––30.
Thus, I 2
12‘ 30q
2 j2 3 j4
5.385‘8.2q or i2 = 5.385cos(4t + 8.2) A
––j2
2:
i2
3:
Vs
+
j4
ix = [5.385cos(4t + 8.2) + 4.983sin(2t –– 41.63)] A.
Chapter 10, Solution 43.
Let i O i O1 i O 2 , where i O1 is due to the dc source and i O 2 is due to the ac source. For
i O1 , consider the circuit in Fig. (a).
4:
2:
iO1
+
8V
(a)
Clearly,
i O1
82
4A
For i O 2 , consider the circuit in Fig. (b).
4:
2:
IO2
10‘0q V
+
j4 :
(b)
If we transform the voltage source, we have the circuit in Fig. (c), where 4 || 2
IO2
2.5‘0q A
4:
2:
(c)
By the current division principle,
43
I O2
(2.5‘0q)
4 3 j4
I O 2 0.25 j0.75 0.79‘ - 71.56q
Thus,
i O 2 0.79 cos(4t 71.56q) A
Therefore,
i O i O1 i O 2 4 + 0.79 cos (89)(4t –– 71.56q) A
j4 :
4 3 :.
Chapter 10, Solution 44.
Let v x v1 v 2 , where v1 and v2 are due to the current source and voltage source
respectively.
For v1 , Z 6 , 5 H

o
jZ L
j30
The frequency-domain circuit is shown below.
20 :
+
V1
-
16 :
Is
Let Z 16 //( 20 j30)
V1
Is Z
16(20 j30)
36 j30
11.8 j3.497 12.31‘16.5 o
(12‘10 o )(12.31‘16.5 o ) 147.7‘26.5 o
For v2 , Z 2 , 5 H

o
j30
jZL

o
v1
147.7 cos(6 t 26.5 o ) V
j10
The frequency-domain circuit is shown below.
20 :
16 :
-
j10
+
V2
-
+
Vs
-
Using voltage division,
V2
16
Vs
16 20 j10
16(50‘0 o )
36 j10
21.41‘ 15.52 o

o
v2
21.41sin(2t 15.52 o ) V
Thus,
vx
147.7 cos(6 t 26.5 o ) 21.41sin( 2 t 15.52 o ) V
Chapter 10, Solution 45.
Let I o I 1 I 2 , where I 1 is due to the voltage source and I 2 is due to the current
source. For I 1 , consider the circuit in Fig. (a).
10 :
IT
I1
20‘-150q V
+
j10 :
-j5 :
(a)
j10 || - j5 - j10
20 ‘ - 150q
IT
10 j10
Using current division,
- j5
I1
I
j10 j5 T
2‘ - 150q
1 j
- j5 2 ‘ - 150q
˜
j5
1 j
- (1 j) ‘ - 150q
For I 2 , consider the circuit in Fig. (b).
I2
10 :
j10 :
(b)
10 || - j5
Using current division,
- j10
2 j
-j5 :
4‘-45q A
I2
- j10 (2 j)
(4‘ - 45q)
- j10 (2 j) j10
Io
I1 I 2
Io
-2.462 j1.366
Therefore,
-2 (1 j) ‘ - 45q
- 2 ‘ - 105q 2 2 ‘0q
2.816‘150.98q
2.816 cos(10t + 150.98q) A
io
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).
Z 2
2H 
o jZL j4
1
1
1
F 
o
12
jZC j (2)(1 / 12)
- j6
+
6:
-j6 :
4‘0q A
V2
(b)
Applying nodal analysis,
V2 V2 V2
4
6 - j6 j4
V2
24
1 j0.5
§1 j j ·
¨ ¸ V2
©6 6 4¹
21.45‘26.56q
j4 :
Hence,
v2
21.45 sin( 2 t 26.56q) V
For v 3 , consider the circuit in Fig. (c).
Z 3
2H 
o jZL j6
1
1
1
F 
o
12
jZC j (3)(1 / 12)
- j4
6:
12‘0q V
j6 :
+
+
-j4 :
V3
(c)
At the non-reference node,
12 V3 V3 V3
6
- j4 j6
12
V3
10.73‘ - 26.56q
1 j0.5
Hence,
v 3 10.73 cos(3t 26.56q) V
Therefore,
vo
10 + 21.45 sin(2t + 26.56q) + 10.73 cos(3t –– 26.56q) 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
+
2:
(a)
Since the capacitor is an open circuit to dc,
2H
i1
4:
i1
24
42
4A
For i 2 , consider the circuit in Fig. (b).
Z 1
2H 
o jZL j2
1
1
F 
o
- j6
6
jZC
1:
j2 :
-j6 :
I2
10‘-30q V
+
I1
2:
I2
4:
(b)
For mesh 1,
- 10 ‘ - 30q (3 j6) I 1 2 I 2
10 ‘ - 30q 3 (1 2 j) I 1 2 I 2
(1)
0
For mesh 2,
0 -2 I 1 (6 j2) I 2
I 1 (3 j) I 2
(2)
Substituting (2) into (1)
10 ‘ - 30q 13 j15 I 2
I 2 0.504 ‘19.1q
Hence,
i 2 0.504 sin( t 19.1q) A
For i 3 , consider the circuit in Fig. (c).
Z 3
2H 
o jZL j6
1
1
1
F 
o
jZC j (3)(1 / 6)
6
1:
- j2
j6 :
-j2 :
I3
2:
(c)
2‘0q A
4:
2 || (1 j2)
2 (1 j2)
3 j2
Using current division,
2 (1 j2)
˜ (2‘0q)
2 (1 j2)
3 j2
I3
2 (1 j2)
13 j3
4 j6 3 j2
I 3 0.3352 ‘ - 76.43q
Hence
i 3 0.3352 cos(3t 76.43q) A
Therefore,
io
4 + 0.504 sin(t + 19.1q) + 0.3352 cos(3t –– 76.43q) 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).
Z 2000
50 cos(2000t ) 
o 50‘0q
o
40 mH 
jZL
1
jZC
o
20 PF 
I
50‘0q V
j (2000)(40 u 10 -3 )
1
j (2000)(20 u 10 -6 )
-j25 :
80 :
(a)
80 || (60 100) 160 3
50
30
I
160 3 j80 j25 32 j33
Using current division,
- j25
IO1
+
j80 :
j80
60 :
100 :
I O1
I O1
Hence,
- 80 I
-1
10‘180q
I
80 160 3
46‘45.9q
0.217 ‘134.1q
i O1 0.217 cos(2000 t 134.1q) A
For i O 2 , consider the circuit in Fig. (b).
iO2
80 :
100 :
60 :
+
24 V
(b)
24
80 60 100
i O2
0.1 A
For i O3 , consider the circuit in Fig. (c).
Z 4000
2 cos(4000t ) 
o 2‘0q
o
40 mH 
20 PF 
o
jZL
1
jZC
j (4000)(40 u 10 -3 )
1
j (4000)(20 u 10 -6 )
j160
- j12.5
-j12.5 :
I2
IO3
80 :
j160 :
I3
2‘0q A
I1
100 :
60 :
(c)
For mesh 1,
I1
For mesh 2,
2
(1)
(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
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
Substituting (3) into (2) yields
(16 j44.25) I 3 12 j54.125
12 j54.125
1.1782‘7.38q
I3
16 j44.25
(3)
I O3
- I 3 -1.1782‘7.38q
i O 3 -1.1782 sin( 4000t 7.38q) A
iO
0.1 + 0.217 cos(2000t + 134.1q) –– 1.1782 sin(4000t + 7.38q) A
Hence,
Therefore,
(2)
Chapter 10, Solution 49.
8 sin( 200t 30q) 
o 8‘30q, Z
5 mH 
o
1 mF 
o
jZL
1
jZC
200
j (200)(5 u 10 ) j
1
- j5
j (200)(1 u 10 -3 )
-3
After transforming the current source, the circuit becomes that shown in the figure below.
5:
40‘30q V
I
i
3:
I
+
40 ‘30q
40 ‘30q
4.472‘56.56q
5 3 j j5
8 j4
4.472 sin(200t + 56.56q) A
j:
-j5 :
Chapter 10, Solution 50.
50 cos(10 5 t ) 
o 50 ‘0q, Z 10 5
0.4 mH 
o
0.2 PF 
o
jZL
1
jZC
j (10 5 )(0.4 u 10 -3 )
1
5
j (10 )(0.2 u 10 -6 )
j40
- j50
After transforming the voltage source, we get the circuit in Fig. (a).
j40 :
+
20 :
2.5‘0q A
-j50 :
80 :
Vo
(a)
Let
Z
and
20 || - j 50
Vs
- j100
2 j5
(2.5‘0q) Z
- j250
2 j5
With these, the current source is transformed to obtain the circuit in Fig.(b).
j40 :
Z
Vs
+
+
80 :
Vo
(b)
By voltage division,
Vo
Vo
Therefore,
80
V
Z 80 j40 s
80
- j100
80 j40
2 j5
8 (- j250)
36.15‘ - 40.6q
36 j42
v o 36.15 cos(105 t –– 40.6q) V
˜
- j250
2 j5
Chapter 10, Solution 51.
The original circuit with mesh currents and a node voltage labeled is shown below.
Io
j10 :
4‘-60q V
-j20 :
40 :
1.25‘0q A
The following circuit is obtained by transforming the voltage sources.
Io
j10 :
4‘-60q V
-j20 :
40 :
Use nodal analysis to find Vx .
§ 1
1
1·
4 ‘ - 60q 1.25‘0q ¨
¸ Vx
© j10 - j20 40 ¹
3.25 j3.464 (0.025 j0.05) Vx
3.25 j3.464
Vx
81.42 j24.29 84.97 ‘16.61q
0.025 j0.05
Thus, from the original circuit,
40 ‘30q Vx (34.64 j20) (81.42 j24.29)
I1
j10
j10
- 46.78 j4.29
I1
-0.429 j4.678 4.698‘95.24q A
j10
I2
I2
Vx 50 ‘0q 31.42 j24.29
40
40
0.7855 j0.6072 0.9928‘37.7q
0.9928‘37.7q A
Chapter 10, Solution 52.
We transform the voltage source to a current source.
60‘0q
6 j12
Is
2 j4
1.25‘0q A
The new circuit is shown in Fig. (a).
-j2 :
Ix
2:
Is = 6 –– j12 A
4:
6:
j4 :
5‘90q A
-j3 :
(a)
Let
6 (2 j4)
2.4 j1.8
8 j4
(6 j12)(2.4 j1.8) 36 j18 18 (2 j)
Zs
6 || (2 j4)
Vs
Is Zs
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:
+
j5 A
-j3 :
(b)
Let
Zo
Io
Z s j2 2.4 j0.2 0.2 (12 j)
Vs
18 (2 j)
15.517 j6.207
Z o 0.2 (12 j)
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 :
(c)
j5 A
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 5.238‘17.35q A
Chapter 10, Solution 53.
We transform the voltage source to a current source to obtain the circuit in Fig. (a).
-j3 :
j4 :
+
4:
5‘0q A
j2 :
2:
Vo
-j2 :
(a)
Let
j8
0.8 j1.6
4 j2
(5‘0q) Z s (5)(0.8 j1.6)
Zs
4 || j2
Vs
4 j8
With these, the current source is transformed so that the circuit becomes that shown in
Fig. (b).
-j3 :
Zs
Vs
j4 :
+
+
2:
-j2 :
Vo
(b)
Let
Z s j3 0.8 j1.4
Vs
4 j8
3.0769 j4.6154
Z s 0.8 j1.4
Zx
Ix
With these, we transform the voltage source in Fig. (b) to obtain the circuit in Fig. (c).
j4 :
+
Ix
Zx
2:
-j2 :
Vo
(c)
Let
Zy
2 || Z x
Vy
Ix Zy
1.6 j2.8
0.8571 j0.5714
2.8 j1.4
(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
Vy
Z y j4 j2
- j2 ( j5.7143)
0.8571 j0.5714 j4 j2
(3.529 –– j5.883) V
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 :
+
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
53.24 j 2.059
4.7817 j1.8055
But V (40 j20)I V

o
0
V
Vs (40 j20)I
V 115.91 j31.05 (40 j20)(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 :
Zth
-j10 :
(a)
ZN
Z th
( j20)(- j10)
j20 j10
22.36‘-63.43q :
10 j20 || (- j10) 10 10 j20
To find Vth , consider the circuit in Fig. (b).
j20 :
10 :
+
50‘30q V
+
-j10 :
Vth
(b)
Vth
IN
- j10
(50‘30q)
j20 j10
Vth
Z th
- 50 ‘30q
22.36 ‘ - 63.43q
-50‘30q V
2.236‘273.4q A
To find Z th , consider the circuit in Fig. (c).
(b)
-j5 :
8:
Zth
j10 :
(c)
ZN
j10 || (8 j5)
Z th
( j10)(8 j5)
j10 8 j5
10‘26q :
To obtain Vth , consider the circuit in Fig. (d).
-j5 :
Io
4‘0q A
8:
j10 :
+
Vth
(d)
By current division,
8
Io
(4‘0q)
8 j10 j5
Vth
j10 I o
IN
Vth
Z th
32
8 j5
j320
33.92‘58q V
8 j5
33.92 ‘58q
10 ‘26q
3.392‘32q A
Chapter 10, Solution 56.
To find Z th , consider the circuit in Fig. (a).
(a)
j4 :
6:
Zth
-j2 :
(a)
ZN
Z th
( j4)(- j2)
j4 j2
= 7.211‘-33.69q :
6 j4 || (- j2)
6
6 j4
By placing short circuit at terminals a-b, we obtain,
I N 2‘0q A
Vth
Z th I th
(7.211‘ - 33.69q) (2‘0q)
14.422‘-33.69q V
To find Z th , consider the circuit in Fig. (b).
(b)
j10 :
30 :
60 :
-j5 :
(b)
30 || 60
ZN
Z th
20
(- j5)(20 j10)
20 j5
= 5.423‘-77.47q :
- j5 || (20 j10)
Zth
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 :
4‘45q A
20 :
a
IN
-j5 :
b
(c)
IN
Vth
20
(4‘45q)
20 j10
= 3.578‘18.43q A
2
(2 j)(4‘45q)
5
Z th I N (5.423‘ - 77.47q) (3.578‘18.43q)
= 19.4‘-59q V
Chapter 10, Solution 57.
To find Z th , consider the circuit in Fig. (a).
5:
-j10 :
2:
Zth
j20 :
(a)
ZN
Z th
( j20)(5 j10)
5 j10
21.633‘-33.7q :
2 j20 || (5 j10)
18 j12
2
To find Vth , consider the circuit in Fig. (b).
5:
-j10 :
2:
+
60‘120q V
+
j20 :
Vth
(b)
Vth
IN
j20
(60 ‘120q)
5 j10 j20
= 107.3‘146.56q V
Vth
Z th
107.3‘146.56q
21.633‘ - 33.7q
j4
(60‘120q)
1 j2
4.961‘-179.7q A
Chapter 10, Solution 58.
Consider the circuit in Fig. (a) to find Z th .
8:
Zth
j10 :
-j6 :
(a)
Z th
( j10)(8 j6)
8 j4
= 11.18‘26.56q :
j10 || (8 j6)
5 (2 j)
Consider the circuit in Fig. (b) to find Vth .
Io
+
8:
5‘45q A
j10 :
-j6 :
(b)
Io
Vth
8 j6
(5‘45q)
8 j6 j10
j10 I o
4 j3
(5‘45q)
4 j2
( j10)(4 j3)(5‘45q)
(2)(2 j)
55.9‘71.56q V
Vth
Chapter 10, Solution 59.
The frequency-domain equivalent circuit is shown in Fig. (a). Our goal is to find Vth and
Z th across the terminals of the capacitor as shown in Figs. (b) and (c).
3:
10‘-45q V
j:
j:
a
+
+
3:
+
-j :
Vo
Zth
5‘-60q A
b
(b)
(a)
3:
j:
Zth
a
+
10‘-45q V
+
Vth
+
5‘-60q A
Vth
+
+
Vo
(d)
(c)
From Fig. (b),
Z th
3 || j
j3
3 j
3
(1 j3)
10
From Fig.(c),
10‘ - 45q Vth 5‘ - 60q Vth
3
j
10 ‘ - 45q 15‘30q
Vth
1 j3
From Fig. (d),
0
-j :
b
-j
V
10‘ - 45q 15‘30q
Z th j th
15.73‘247.9q V
Vo
Vo
Therefore,
vo
15.73 cos(t + 247.9q) V
Chapter 10, Solution 60.
(a)
To find Z th , consider the circuit in Fig. (a).
10 :
-j4 :
a
j5 :
Zth
4:
b
(a)
Z th
4 || (- j4 10 || j5)
Z th
4 || 2
4 || (- j4 2 j4)
1.333 :
To find Vth , consider the circuit in Fig. (b).
10 :
V1
-j4 :
V2
+
20‘0q V
+
j5 :
4‘0q A
4:
Vth
(b)
At node 1,
20 V1 V1 V1 V2
10
j5
- j4
(1 j0.5) V1 j2.5 V2 20
(1)
At node 2,
V1 V2 V2
- j4
4
V1 (1 j) V2 j16
(2)
4
Substituting (2) into (1) leads to
28 j16 (1.5 j3) V2
28 j16
8 j5.333
V2
1.5 j3
Therefore,
Vth
(b)
V2
9.615‘33.69q V
To find Z th , consider the circuit in Fig. (c).
Zth
c
d
10 :
-j4 :
j5 :
4:
(c)
Z th
- j4 || (4 10 || j5)
Z th
- j4 || (6 j4)
§
j10 ·
¸
- j4 || ¨ 4 2 j¹
©
- j4
(6 j4)
6
2.667 –– j4 :
To find Vth ,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)
Vth
V1 V2
16 3 j8
9.614‘56.31q V
Chapter 10, Solution 61.
First, we need to find Vth and Z th across the 1 : resistor.
4:
-j3 :
j8 :
6:
Zth
(a)
From Fig. (a),
Z th
(4 j3) || (6 j8)
Z th
4.472‘-10.3q :
4:
+
-j16 V
(4 j3)(6 j8)
10 j5
-j3 :
j8 :
+
2A
Vth
(b)
From Fig. (b),
- j16 Vth
Vth
2
4 j3
6 j8
3.92 j2.56
Vth
20.93‘ - 43.45q
0.22 j0.4
Vo
Vo
Therefore,
Vth
20.93‘ - 43.45q
1 Z th
5.46 ‘ - 8.43q
3.835‘ - 35.02q
vo
3.835 cos(4t –– 35.02q) V
4.4 j0.8
6:
Chapter 10, Solution 62.
First, we transform the circuit to the frequency domain.
12 cos( t ) 
o 12‘0q, Z 1
2H 
o
1
F 
o
4
1
F 
o
8
jZL
1
jZC
1
jZC
j2
- j4
- j8
To find Z th , consider the circuit in Fig. (a).
3 Io
Io
4:
Vx
j2 :
1
Ix
2
-j4 :
-j8 :
+
1V
(a)
At node 1,
Vx Vx
3Io
4 - j4
Thus,
1 Vx
,
j2
where I o
Vx 2 Vx 1 Vx
- j4
4
j2
Vx 0.4 j0.8
At node 2,
I x 3Io
1 1 Vx
- j8
j2
Ix
(0.75 j0.5) Vx j
Ix
-0.1 j0.425
Z th
1
Ix
3
8
-0.5246 j2.229
2.29‘ - 103.24q :
- Vx
4
To find Vth , consider the circuit in Fig. (b).
3 Io
4:
Io
j2 :
V1
V2
1
12‘0q V
+
2
-j4 :
-j8 :
+
Vth
(b)
At node 1,
12 V1
V
V V2
,
3Io 1 1
4
- j4
j2
24 (2 j) V1 j2 V2
where I o
(1)
At node 2,
V1 V2
V2
3Io
j2
- j8
72 (6 j4) V1 j3 V2
(2)
From (1) and (2),
ª 24º ª 2 j - j2º ª V1 º
« 72 » « 6 j4 - j3» « V »
¼¬ 2¼
¬ ¼ ¬
'
-5 j6 ,
Vth
Thus,
Vo
Vo
Therefore,
V2
'2
'2
'
3.073‘ - 219.8q
2
(2)(3.073‘ - 219.8q)
Vth
2 Z th
1.4754 j2.229
6.146‘ - 219.8q
2.3‘ - 163.3q
2.673‘ - 56.5q
vo
2.3 cos(t –– 163.3q) V
- j24
12 V1
4
Chapter 10, Solution 63.
Transform the circuit to the frequency domain.
4 cos(200t 30q) 
o 4‘30q, Z
10 H 
o
5 PF 
o
jZL
1
jZC
200
j (200)(10) j2 k:
1
- j k:
j (200)(5 u 10 -6 )
Z N is found using the circuit in Fig. (a).
-j k:
j2 k:
ZN
2 k:
(a)
ZN
- j 2 || j2
- j 1 j 1 k:
We find I N using the circuit in Fig. (b).
-j k:
4‘30q A
j2 k:
(b)
j2 || 2 1 j
By the current division principle,
1 j
IN
(4 ‘30q)
1 j j
5.657 ‘75q
Therefore,
iN
ZN
5.657 cos(200t + 75q) A
1 k:
2 k:
IN
Chapter 10, Solution 64.
Z N is obtained from the circuit in Fig. (a).
60 :
40 :
ZN
-j30 :
j80 :
(a)
ZN
(60 40) || ( j80 j30) 100 || j50
ZN
20 j40
(100)( j50)
100 j50
44.72‘63.43q :
To find I N , consider the circuit in Fig. (b).
60 :
j80 :
(b)
Is
3‘60q
For mesh 1,
100 I 1 60 I s 0
I 1 1.8‘60q
For mesh 2,
( j80 j30) I 2 j80 I s
I 2 4.8‘60q
IN
I1 I 2
3‘60q A
40 :
I2
-j30 :
IN
Is
3‘60q A
I1
0
Chapter 10, Solution 65.
5 cos(2 t ) 
o 5‘0q, Z
4H 
o
1
F 
o
4
1
F 
o
2
jZL
1
jZC
1
jZC
2
j (2)(4) j8
1
- j2
j (2)(1 / 4)
1
-j
j (2)(1 / 2)
To find Z N , consider the circuit in Fig. (a).
2:
ZN
-j2 :
-j :
(a)
ZN
- j || (2 j2)
- j (2 j2)
2 j3
1
(2 j10)
13
To find I N , consider the circuit in Fig. (b).
5‘0q V
2:
+ -j2 :
-j :
IN
(b)
IN
5‘0q
-j
j5
The Norton equivalent of the circuit is shown in Fig. (c).
Io
IN
ZN
(c)
j8 :
Using current division,
ZN
(1 13)(2 j10)( j5) 50 j10
Io
IN
Z N j8
(1 13)(2 j10) j8 2 j94
I o 0.1176 j0.5294 0542‘ - 77.47q
0.542 cos(2t –– 77.47q) A
Therefore, i o
Chapter 10, Solution 66.
Z 10
0.5 H 
o
jZL j (10)(0.5) j5
1
1
10 mF 
o
jZC j (10)(10 u 10 -3 )
-j10 :
- j10
Vx
+
10 :
j5 :
Vo
2 Vo
1A
(a)
To find Z th , consider the circuit in Fig. (a).
Vx
Vx
,
j5 10 j10
Vx
19 Vx
1

o Vx
10 j10
j5
1 2 Vo
ZN
Z th
Vx
1
where Vo
10Vx
10 j10
- 10 j10
21 j2
14.142 ‘135q
21.095‘5.44q
0.67‘129.56q :
To find Vth and I N , consider the circuit in Fig. (b).
12‘0q V
-j10 :
+
+
+
-j2 A
10 :
Vo
j5 :
I
2 Vo
Vth
(b)
(10 j10 j5) I (10)(- j2) j5 (2 Vo ) 12
Vo (10)(- j2 I )
where
Thus,
0
(10 j105) I -188 j20
188 j20
I
- 10 j105
Vth
Vth
Vth
IN
j5 (I 2 Vo ) j5 (21I j40) j105 I 200
j105 (188 j20)
200 -11.802 j2.076
- 10 j105
11.97‘170q V
Vth
Z th
11.97 ‘170q
0.67 ‘129.56q
17.86‘40.44q A
Chapter 10, Solution 67.
10(13 j5) 12(8 j6)
11.243 j1.079:
23 j5
20 j6
10
(8 j6)
Va
(60‘45 o ) 13.78 j21.44,
Vb
(60‘45 o ) 25.93 j454.37:
23 j5
20 j6
VTh
VTh Va Vb 433.1‘ 1.599 o V,
IN
38.34‘ 97.09 o A
Z Th
ZN
Z Th
10 //(13 j5) 12 //(8 j6)
Chapter 10, Solution 68.
1H

o
jZL j10x1 j10
1
1
1
F

o
j2
1
20
jZ C
j10 x
20
We obtain VTh using the circuit below.
4:
Io
a
+
+
6<0o
-
+
-
Vo/3
-j2 j10
Vo
4Io
j10( j2)
j2.5
j10 j2
Vo 4I o x ( j2.5) j10I o
1
6 4I o Vo 0
3
b
j10 //( j2)
(1)
(2)
Combining (1) and (2) gives
Io
6
,
4 j10 / 3
VTh
v Th
j10I o
Vo
j60
4 j10 / 3
11.52‘ 50.19 o
11.52 sin(10 t 50.19 o )
To find RTh, we insert a 1-A source at terminals a-b, as shown below.
4:
Io
a
+
Vo/3
1
4I o Vo
3
1 4I o
0
o
Vo Vo
j2 j10
+
-
Io
-j2 j10
4Io
V
o
12
Vo
-
1<0o
Combining the two equations leads to
1
0.333 j0.4
Vo
1.2293 j1.4766
Z Th
Vo
1
1.2293 1.477:
Chapter 10, Solution 69.
This is an inverting op amp so that
Vo - Z f
-R
-jZRC
Vs
Zi
1 jZC
Vm and Z 1 RC ,
1
Vo - j ˜
˜ RC ˜ Vm
RC
When Vs
- j Vm
Vm ‘ - 90q
Therefore,
Vm sin(Zt 90q)
v o (t)
- Vm cos(Zt)
Chapter 10, Solution 70.
This may also be regarded as an inverting amplifier.
2 cos(4 u 10 4 t ) 
o 2 ‘0q, Z 4 u 10 4
1
1
10 nF 
o
- j2.5 k:
4
jZC j (4 u 10 )(10 u 10 -9 )
Vo
Vs
- Zf
Zi
where Z i
50 k: and Z f
Vo
Vs
Thus,
If Vs
100k || (- j2.5k )
- j100
k: .
40 j
- j2
40 j
2 ‘0q ,
Vo
Therefore,
v o (t)
- j4
40 j
4 ‘ - 90q
40.01‘ - 1.43q
0.1‘ - 88.57q
0.1 cos(4x104 t –– 88.57q) V
Chapter 10, Solution 71.
8 cos(2t 30 o )
o 8‘30 o
1
1
0. 5PF

o
jZC j2x 0.5x10 6
At the inverting terminal,
Vo 8‘30 o Vo 8‘30 o
j1k
10k
Vo
8‘30 o
2k
j1k:

o
Vo (0.1 j)
8‘30(0.6 j)
(6.9282 j4)(0.6 j)
9.283‘4.747 o
0.1 j
vo(t) = 9.283cos(2t + 4.75o) V
Chapter 10, Solution 72.
4 cos(10 4 t ) 
o 4 ‘0q, Z 10 4
1
1
1 nF 
o
- j100 k:
4
jZC j (10 )(10 -9 )
Consider the circuit as shown below.
50 k:
4‘0q V
+
Therefore,
i o (t)
+
-j100 k:
At the noninverting node,
4 Vo
Vo
50
- j100
Io
Vo
Vo
100k

o Vo
4
mA
(100)(1 j0.5)
Vo
Io
100 k:
4
1 j0.5
35.78‘ - 26.56q PA
35.78 cos(104 t –– 26.56q) PA
Chapter 10, Solution 73.
As a voltage follower, V2
C1
Vo
1
jZC1
1
20 nF 
o
jZC 2
10 nF 
o
C2
1
-j20 k:
j (5 u 10 )(10 u 10 -9 )
1
-j10 k:
3
j (5 u 10 )(20 u 10 -9 )
3
Consider the circuit in the frequency domain as shown below.
-j20 k:
Is 10 k:
20 k:
V1
VS
+
V2
+
Io
Vo
-j10 k:
Zin
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
Substituting (2) into (1) gives
2 Vs
j6Vo
or
V1
(1 j2)Vo
Is
Vs V1
10k
(2)
Vo
1
-j Vs
3
§2
1·
¨ j ¸ Vs
©3
3¹
(1 3)(1 j)
Vs
10k
1 j
30k
Is
Vs
Vs 30k
15 (1 j) k
Is 1 j
21.21‘45q k:
Z in
Z in
Chapter 10, Solution 74.
R1 Zi
Vo
Vs
Av
1
,
jZC1
- Zf
Zi
Zf
1
jZC 2
1
R1 jZC1
R2 At Z 0 ,
1
jZC 2
§ C 1 · § 1 jZR 2 C 2 ·
¸
¨ ¸¨
© C 2 ¹ © 1 jZR 1C 1 ¹
C1
C2
Av
As Z o f ,
R2 Av
R2
R1
At Z
1
,
R 1 C1
Av
§ C1 · § 1 j R 2 C 2 R 1C1 ·
¨ ¸¨
¸
1 j
¹
© C2 ¹ ©
At Z
1
,
R 2C2
Av
§ C1 · §
·
1 j
¨ ¸¨
¸
© C 2 ¹ © 1 j R 1C1 R 2 C 2 ¹
Chapter 10, Solution 75.
Z
C1
2 u 10 3
C2
o
1 nF 
1
jZC1
1
j (2 u 10 )(1 u 10 -9 )
3
-j500 k:
Consider the circuit shown below.
100 k:
-j500 k:
-j500 k:
V2
V1
VS
+
20 k:
+
100 k:
+
Vo
20 k:
At node 1,
Vs V1 Vo V1 V1 V2
- j500
100
- j500
Vs (2 j5) V1 j5 Vo V2
(1)
V1 V2 V2
- j500
100
V1 (1 j5) V2
(2)
At node 2,
But
V2
Vo
2
R3
V
R3 R4 o
From (2) and (3),
1
V1
˜ (1 j5) Vo
2
Substituting (3) and (4) into (1),
1
1
Vs
˜ (2 j5)(1 j5) Vo j5 Vo Vo
2
2
1
Vs
˜ (26 j25) Vo
2
Vo
Vs
2
26 j25
Chapter 10, Solution 76.
0.0554‘43.88q
(3)
(4)
Let the voltage between the -jk : capacitor and the 10k : resistor be V1.
2‘30 o V1
j4k
2‘30 o
Also,
V1 Vo
10k
V1 Vo V1 Vo
10k
20k

o
(1)
(1 j0.6)V1 j0.6Vo
Vo
j2k
o
V1
(1 j5)Vo
(2)
Solving (2) into (1) yields
Vo
0.047 j0.3088
0.3123‘ 81.34 o V
Chapter 10, Solution 77.
Consider the circuit below.
R3
2
R1
1
+
VS
At node 1,
V1
V1
C2
+
C1
At node 2,
V1
+
Vo
Vs V1
jZC V1
R1
Vs (1 jZR 1C1 ) V1
0 V1
R3
R2
V1 Vo
jZC 2 (V1 Vo )
R2
§R3
·
jZC 2 R 3 ¸
(Vo V1 ) ¨
©R2
¹
(1)
§
·
1
¨1 ¸ V1
© (R 3 R 2 ) jZC 2 R 3 ¹
From (1) and (2),
§
·
Vs
R2
¨1 ¸
Vo
1 jZR 1C1 © R 3 jZC 2 R 2 R 3 ¹
(2)
Vo
Vo
Vs
R 2 R 3 jZC 2 R 2 R 3
(1 jZR 1C 1 ) ( R 3 jZC 2 R 2 R 3 )
Chapter 10, Solution 78.
2 sin(400t ) 
o 2‘0q, Z 400
1
1
0.5 PF 
o
- j5 k:
jZC j (400)(0.5 u 10 -6 )
1
1
0.25 PF 
o
- j10 k:
jZC j (400)(0.25 u 10 -6 )
Consider the circuit as shown below.
20 k:
10 k: V
1
2‘0q V
+
-j5 k:
V2
+
Vo
40 k:
-j10 k:
10 k:
20 k:
At node 1,
2 V1
V1
V V2 V1 Vo
1
10
- j10
- j5
20
4 (3 j6) V1 j4 V2 Vo
(1)
V1 V2 V2
j5
10
V1 (1 j0.5) V2
(2)
At node 2,
But
V2
20
V
20 40 o
1
V
3 o
(3)
From (2) and (3),
1
V1
˜ (1 j0.5) Vo
3
Substituting (3) and (4) into (1) gives
4
(4)
§
1·
¨1 j ¸ Vo
©
6¹
1
4
(3 j6) ˜ ˜ (1 j0.5) Vo j Vo Vo
3
3
Vo
24
6 j
3.945‘9.46q
Therefore,
3.945 sin(400t + 9.46q) V
v o (t)
Chapter 10, Solution 79.
5 cos(1000t ) 
o 5‘0q, Z 1000
0.1 PF 
o
1
jZC
1
j (1000)(0.1 u 10 -6 )
- j10 k:
0.2 PF 
o
1
jZC
1
j (1000)(0.2 u 10 -6 )
- j5 k:
Consider the circuit shown below.
20 k:
-j10 k:
10 k:
Vs = 5‘0q V
+
+
40 k:
V1
Since each stage is an inverter, we apply Vo
-j5 k:
+
- Zf
V to each stage.
Zi i
+
Vo
Vo
- 40
V
- j15 1
(1)
and
V1
- 20 || (- j10)
Vs
10
(2)
From (1) and (2),
§ - j8 ·§ - (20)(-j10) ·
¸ 5‘0q
¸¨
Vo ¨
© 10 ¹© 20 j10 ¹
Vo
Therefore,
16 (2 j)
35.78‘26.56q
35.78 cos(1000t + 26.56q) V
v o (t)
Chapter 10, Solution 80.
4 cos(1000t 60q) 
o 4‘ - 60q, Z 1000
0.1 PF 
o
1
jZC
1
j (1000)(0.1 u 10 -6 )
- j10 k:
0.2 PF 
o
1
jZC
1
j (1000)(0.2 u 10 -6 )
- j5 k:
The two stages are inverters so that
§ 20
20 ·§ - j5 ·
¸
˜ (4‘ - 60q) Vo ¨
V ¸¨
50 o ¹© 10 ¹
© - j10
-j
-j 2
˜ ( j2) ˜ (4‘ - 60q) ˜ Vo
2
2 5
(1 j 5) Vo
Vo
Therefore,
4‘ - 60q
4‘ - 60q
1 j 5
v o (t)
3.922 ‘ - 71.31q
3.922 cos(1000t –– 71.31q) V
Chapter 10, Solution 81.
The schematic is shown below. The pseudocomponent IPRINT is inserted to print the
value of Io in the output. We click Analysis/Setup/AC Sweep and set Total Pts. = 1,
Start Freq = 0.1592, and End Freq = 0.1592. Since we assume that w = 1. The output
file includes:
FREQ
1.592 E-01
IM(V_PRINT1)
1.465 E+00
IP(V_PRINT1)
7.959 E+01
Io = 1.465‘79.59o A
Thus,
Chapter 10, Solution 82.
The schematic is shown below. We insert PRINT to print Vo 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
VM($N_0001)
7.684 E+00
VP($N_0001)
5.019 E+01
which means that
Vo = 7.684‘50.19o V
Chapter 10, Solution 83.
The schematic is shown below. The frequency is f
Z / 2S
1000
2S
159.15
When the circuit is saved and simulated, we obtain from the output file
FREQ
1.592E+02
VM(1)
6.611E+00
Thus,
VP(1)
-1.592E+02
vo = 6.611cos(1000t –– 159.2o) V
Chapter 10, Solution 84.
The schematic is shown below. We set PRINT to print Vo 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
Vo = 1.664‘-146.4o V
Namely,
Chapter 10, Solution 85.
The schematic is shown below. We let Z
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
Vo
2.228‘ 167.5 V
-1.646
Chapter 10, Solution 86.
We insert three pseudocomponent PRINTs at nodes 1, 2, and 3 to print V1, V2, and V3,
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
VP($N_0002)
3.000
E+01
VP($N_0003)
E+01
-8.483
FREQ
VM($N_0001)
1.592 E-01
1.082 E+02
VP($N_0001)
1.254
E+02
Therefore,
V1 = 60‘30o V V2 = 236.7‘-84.83o V V3 = 108.2‘125.4o V
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
VP($N_0004)
1.696
E+02
VP($N_0001)
E+02
-1.386
FREQ
VM($N_0003)
1.592 E-01
2.270 E+00
VP($N_0003)
-1.524
E+02
Therefore,
V1 = 15.91‘169.6o V V2 = 5.172‘-138.6o V V3 = 2.27‘-152.4o V
Chapter 10, Solution 88.
The schematic is shown below. We insert IPRINT and PRINT to print Io and Vo in the
output file. Since w = 4, f = w/2S = 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
Vo = 34.96‘12.6o V, Io = 0.8912‘-88.7o A
Therefore,
vo = 34.96 cos(4t + 12.6o)V,
io = 0.8912cos(4t - 88.7o )A
Chapter 10, Solution 89.
Consider the circuit below.
R1
Vin
R2
2
1
R3
Vin
C
4
R4
3
+
Iin
+
+
Vin
At node 1,
0 Vin
R1
Vin V2
R2
- Vin V2
R2
V
R 1 in
V2 Vin
R3
Vin V4
1 jZC
At node 3,
(1)
- Vin V4
Vin V2
jZCR 3
(2)
From (1) and (2),
- R2
V
jZCR 3 R 1 in
- Vin V4
Thus,
I in
Vin V4
R4
Z in
Vin
I in
where
R2
V
jZCR 3 R 1 R 4 in
jZCR 1R 3 R 4
R2
L eq
jZL eq
R 1R 3 R 4C
R2
Chapter 10, Solution 90.
Let
Z4
R ||
1
jZC
R
1 jZRC
Z3
R
1
jZC
1 jZRC
jZC
Consider the circuit shown below.
Z3
Vi
+
R1
+
Z4
Vo
R2
Vo
Z4
R2
Vi V
Z3 Z 4
R1 R 2 i
Vo
Vi
R
R2
1 jZC
R
1 jZRC R 1 R 2
1 jZC
jZC
jZRC
R2
2
jZRC (1 jZRC)
R1 R 2
Vo
Vi
R2
jZRC
2 2
1 Z R C j3ZRC R 1 R 2
2
For Vo and Vi to be in phase,
1 Z2 R 2 C 2
Z
1
RC
or
Vo
must be purely real. This happens when
Vi
0
2Sf
f
1
2SRC
At this frequency,
Av
Vo
Vi
R2
1
3 R1 R 2
Chapter 10, Solution 91.
(a)
Let
V2 voltage at the noninverting terminal of the op amp
output voltage of the op amp
Vo
Zp
10 k:
Zs
R jZL As in Section 10.9,
Ro
1
jZC
Zp
Ro
V2
Vo
Zs Zp
V2
Vo
ZCR o
ZC (R R o ) j (Z2 LC 1)
R R o jZL j
ZC
For this to be purely real,
1
Zo2 LC 1 0 
o Zo
fo
fo
(b)
LC
1
1
2S LC
2S (0.4 u 10 -3 )(2 u 10 -9 )
180 kHz
At oscillation,
Zo CR o
V2
Vo Zo C (R R o )
Ro
R Ro
This must be compensated for by
Vo
80
1
5
Av
V2
20
Ro
R Ro
1
5

o R
4R o
40 k:
Chapter 10, Solution 92.
Let
V2 voltage at the noninverting terminal of the op amp
output voltage of the op amp
Vo
Zs
Ro
Zp
jZL ||
1
|| R
jZC
As in Section 10.9,
V2
Vo
Zp
Zs Zp
1
1
1
jZC R
jZL
ZRL
ZL jR (Z2 LC 1)
ZRL
Ro ZL jR (Z2 LC 1)
ZRL
ZL jR (Z2 LC 1)
V2
Vo
ZRL
ZRL ZR o L jR o R (Z2 LC 1)
For this to be purely real,
Zo2 LC 1 
o f o
(a)
At Z Zo ,
V2
Vo
1
2S LC
Zo RL
Zo RL Zo R o L
R
R Ro
This must be compensated for by
Vo
Rf
1000k
1
Av
1
V2
Ro
100k
11
Hence,
R
R Ro
(b)
fo
fo
1

o R o
11
10R
100 k:
1
2S (10 u 10 -6 )(2 u 10 -9 )
1.125 MHz
Chapter 10, Solution 93.
As shown below, the impedance of the feedback is
jZL
1
jZC2
ZT
1
jZC1
1
jZC1
§
1 ·
¸
|| ¨ jZL jZC 2 ¹
©
ZT
-j §
-j ·
¨ jZL ¸
ZC1 ©
ZC 2 ¹
-j
-j
jZL ZC1
ZC 2
ZT
1
ZLC 2
Z
j (C1 C 2 Z2 LC1C 2 )
In order for Z T to be real, the imaginary term must be zero; i.e.
C1 C 2 Zo2 LC1C 2 0
C1 C 2
1
Zo2
LC1C 2
LC T
1
fo
2S LC T
Chapter 10, Solution 94.
If we select C1
C2
C1 C 2
C1 C 2
CT
Since f o
1
2S LC T
L
Xc
20 nF
C1
2
10 nF
,
1
(2Sf ) 2 C T
1
ZC 2
1
(4S )(2500 u 10 6 )(10 u 10 -9 )
2
1
(2S)(50 u 10 3 )(20 u 10 -9 )
We may select R i 20 k: and R f t R i , say R f
Thus,
C1 C 2 20 nF,
L 10.13 mH
159 :
20 k: .
Rf
Ri
Chapter 10, Solution 95.
First, we find the feedback impedance.
C
ZT
L2
L1
10.13 mH
20 k:
ZT
§
1 ·
¸
jZL1 || ¨ jZL 2 jZC ¹
©
ZT
§
j ·
¸
jZL1 ¨ jZL 2 ©
ZC ¹
j
jZL1 jZL 2 ZC
Z2 L1C (1 ZL 2 )
j (Z2 C (L1 L 2 ) 1)
In order for Z T to be real, the imaginary term must be zero; i.e.
Zo2 C (L1 L 2 ) 1 0
Zo
fo
2Sf o
1
C ( L1 L 2 )
1
2S C (L 1 L 2 )
Chapter 10, Solution 96.
(a)
Consider the feedback portion of the circuit, as shown below.
jZL
Vo
V2
+
V1
R
R
jZL
V
R jZL 1

o V1
Applying KCL at node 1,
Vo V1 V1
V1
jZL
R R jZL
Vo V1
·
§1
1
¸
jZL V1 ¨ © R R jZL ¹
V2
jZL
R jZL
V2
jZL
(1)
Vo
§ j2ZRL Z2 L2 ·
¸
V1 ¨1 R (R jZL) ¹
©
(2)
From (1) and (2),
§ R jZL ·§ j2ZRL Z2 L2 ·
¸V
¸¨1 Vo ¨
R (R jZL) ¹ 2
© jZL ¹©
(b)
Vo
V2
R 2 jZRL j2ZRL Z2 L2
jZRL
V2
Vo
1
R Z2 L2
3
jZRL
V2
Vo
1
3 j (ZL R R ZL )
2
V2
must be real,
Vo
Since the ratio
Zo L
R
R
Zo L
Zo L
(c)
0
R2
Zo L
Zo
2Sf o
fo
R
2S L
R
L
When Z Zo
V2 1
Vo 3
This must be compensated for by A v
R2
Av 1
3
R1
R 2 2 R1
3 . But
Chapter 11, Solution 1.
v( t ) 160 cos(50t )
i( t ) -20 sin(50t 30q)
i( t ) 20 cos(50t 60q)
2 cos(50t 30q 180q 90q)
p( t ) v( t ) i( t ) (160)(20) cos(50t ) cos(50t 60q)
p( t ) 1600 > cos(100 t 60q) cos(60q) @ W
p( t ) 800 1600 cos(100t 60q) W
P
1
V I cos(T v Ti )
2 m m
P
800 W
1
(160)(20) cos(60q)
2
Chapter 11, Solution 2.
First, transform the circuit to the frequency domain.
30 cos(500t ) 
o 30 ‘0q ,
Z 500
0.3 H 
o
o
20PF 
jZL
1
jZC
I
j150
-j
(500)(20)(10 -6 )
I2
- j100
-j100 :
I1
30‘0q V
I1
i1 ( t )
I2
+
30‘0q
j150
j150 :
0.2‘ 90q
0.2 cos(500 t 90q)
30‘0q
200 j100
0.3
2 j
200 :
- j0.2
0.2 sin(500 t )
0.1342‘26.56q
0.12 j0.06
i 2 (t)
0.1342 cos(500 t 25.56q)
I I 1 I 2 0.12 j0.14 0.1844 ‘ - 49.4q
i( t ) 0.1844 cos(500t 35q)
For the voltage source,
p( t ) v( t ) i( t ) [ 30 cos(500t ) ] u [ 0.1844 cos(500t 35q) ]
At t
2 s,
p
p
p
5.532 cos(1000) cos(1000 35q)
(5.532)(0.5624)(0.935)
2.91 W
For the inductor,
p( t )
At t
2 s,
p
p
p
v( t ) i( t ) [ 30 cos(500t ) ] u [ 0.2 sin(500t ) ]
6 cos(1000) sin(1000)
(6)(0.5624)(0.8269)
2.79 W
For the capacitor,
Vc I 2 (- j100) 13.42‘ - 63.44q
p( t ) v( t ) i( t ) [13.42 cos(500 63.44q) ] u [ 0.1342 cos(500t 25.56q)
At t
2 s,
p 18 cos(1000 63.44q) cos(1000 26.56q)
p (18)(0.991)(0.1329)
p 2.37 W
For the resistor,
VR 200 I 2 26.84 ‘25.56q
p( t ) v( t ) i( t ) [ 26.84 cos(500t 26.56q) ] u [ 0.1342 cos(500t 26.56q) ]
At t
2 s,
p
p
p
3.602 cos 2 (1000 25.56q)
(3.602)(0.1329 2
0.0636 W
Chapter 11, Solution 3.
Z
10 cos(2t 30q) 
o 10‘30q ,
1H 
o
jZL
j2
1
jZC
0.25 F 
o
-j2
4:
I
2
I1
2:
I2
+
10‘30q V
j2 || (2 j2)
I
j2 :
( j2)(2 j2)
2
10 ‘30q
4 2 j2
I1
j2
I
2
I2
2 j2
I
2
2 j2
1.581‘11.565q
j I 1.581‘101.565q
2.236 ‘56.565q
For the source,
S
V I*
1
(10‘30q)(1.581‘ - 11.565q)
2
S 7.905‘18.43q 7.5 j2.5
The average power supplied by the source = 7.5 W
For the 4-: resistor, the average power absorbed is
1 2
1
P
I R
(1.581) 2 (4) 5 W
2
2
For the inductor,
1
1
2
S
I2 ZL
(2.236) 2 ( j2) j5
2
2
The average power absorbed by the inductor = 0 W
-j2 :
For the 2-: resistor, the average power absorbed is
1
1
2
P
I1 R
(1.581) 2 (2) 2.5 W
2
2
For the capacitor,
1
2
I1 Z c
2
S
1
(1.581) 2 (- j2)
2
- j2.5
The average power absorbed by the capacitor = 0 W
Chapter 11, Solution 4.
20 :
50 V
+
10 :
I1
-j10 :
I2
j5 :
For mesh 1,
50 (20 j10) I 1 j10 I 2
5 (2 j) I 1 j I 2
(1)
For mesh 2,
0
0
(10 j5 j10) I 2 j10 I 1
(2 j) I 2 j2 I 1
(2)
In matrix form,
ª5º ª 2 j
j ºª I 1 º
« 0» « j2 2 j»« I »
¼¬ 2 ¼
¬ ¼ ¬
'
5 j4 ,
I1
'1
'
5 (2 j)
5 j4
1.746‘12.1q
I2
'2
'
- j10
5 - j4
1.562 ‘128.66q
S
1
V I 1*
2
'1
For the source,
43.65‘ - 12.1q
5 (2 j) ,
'2
-j10
The average power supplied 43.65 cos(12.1q)
42.68 W
For the 20-: resistor,
1
2
P
I 1 R 30.48 W
2
For the inductor and capacitor,
P 0W
For the 10-: resistor,
1
2
P
I2 R
2
12.2 W
Chapter 11, Solution 5.
Converting the circuit into the frequency domain, we get:
1:
8‘––40
I1:
P1:
2:
+
––j2
j6
8‘ 40q
j6(2 j2)
1
j6 2 j2
1.6828‘ 25.38q
1.6828 2
1 1.4159 W
2
P3H = P0.25F = 0
I 2:
P2:
j6
1.6828‘ 25.38q
j6 2 j2
2.258 2
2
2
5.097 W
2.258
Chapter 11, Solution 6.
20 :
50 V
+
10 :
I1
I2
-j10 :
j5 :
For mesh 1,
(4 j2) I 1 j2 (4 ‘60q) 4 Vo
Vo 2 (4 ‘60q I 2 )
0
(1)
(2)
For mesh 2,
(2 j) I 2 2 (4‘60q) 4Vo 0
Substituting (2) into (3),
(2 j) I 2 8‘60q 8 (4 ‘60q I 2 )
I2
(3)
0
40‘60q
10 j
Hence,
§
40‘60q · - j8‘60q
¸
2 ¨ 4 ‘60q 10 j ¹
10 j
©
Substituting this into (1),
§14 j ·
j32 ‘60q
¸
(4 j2) I 1 j8‘60q ( j8‘60q) ¨
10 j
©10 j ¹
Vo
I1
(4‘60q)(1 j14)
21 j8
P4
1
2
I1 R
2
2.498‘125.06q
1
(2.498) 2 (4)
2
12.48 W
Chapter 11, Solution 7.
20 :
50 V
+
I1
10 :
-j10 :
I2
j5 :
Applying KVL to the left-hand side of the circuit,
8‘20q 4 I o 0.1Vo
Applying KCL to the right side of the circuit,
V
V1
8Io 1 0
j5 10 j5
10
V
10 j5 1
But,
Vo
Hence,
8Io 
o V1
Vo
10 j5
Vo j50
10
(1)
10 j5
Vo
10
0
I o j0.025 Vo
Substituting (2) into (1),
8‘20q 0.1 Vo (1 j)
Vo
80‘20q
1 j
I1
Vo
10
P
1
2
I1 R
2
10
2
(2)
‘ - 25q
§ 1 ·§100 ·
¨ ¸¨
¸(10)
© 2 ¹© 2 ¹
250 W
Chapter 11, Solution 8.
We apply nodal analysis to the following circuit.
V1 Io -j20 :
V2
I2
j10 :
6‘0q A
0.5 Io
40 :
At node 1,
6
At node 2,
V1 V1 V2
V1
j10
- j20
j120 V2
(1)
V2
40
0 .5 I o I o
V1 V2
- j20
But,
Io
Hence,
1.5 (V1 V2 )
- j20
3V1
V2
40
(3 j) V2
(2)
Substituting (1) into (2),
j360 3V2 3V2 j V2
V2
j360
6 j
I2
V2
40
P
0
360
(-1 j6)
37
9
(-1 j6)
37
1
2
I2 R
2
2
1§ 9 ·
¨
¸ (40)
2 © 37 ¹
Chapter 11, Solution 9.
Vo
§ 6·
¨1 ¸ Vs
© 2¹
P10
Vo2
R
(4)(2)
64
mW
10
8 V rms
6.4 mW
The current through the 2 -k: resistor is
Vs
1 mA
2k
P2
I2R
2 mW
P6
I2R
6 mW
Similarly,
43.78 W
Chapter 11, Solution 10.
No current flows through each of the resistors. Hence, for each resistor,
P 0 W.
Chapter 11, Solution 11.
Z 377 ,
R 10 4 ,
C 200 u 10 -9
ZRC (377)(10 4 )(200 u 10 -9 ) 0.754
tan -1 (ZRC)
37.02q
10k
Z ab
‘ - 37.02q
1 (0.754) 2
i( t ) 2 sin(377 t 22q)
I 2 ‘ - 68q
6.375‘ - 37.02q k:
2 cos(377 t 68q) mA
2
S
I
2
rms
Z ab
§ 2 u 10 -3 ·
¸ (6.375‘ - 37.02q) u 10 3
¨
©
2 ¹
S 12.751‘ - 37.02q mVA
P
S cos(37.02)
10.181 mW
Chapter 11, Solution 12.
(a)
We find Z Th using the circuit in Fig. (a).
Zth
8:
-j2 :
Z Th
8 || -j2
ZL
Z *Th
(a)
(8)(-j2) 8
(1 j4)
8 j2 17
0.471 j1.882 :
0.471 j1.882
We find VTh using the circuit in Fig. (b).
Io
+
8:
-j2 :
Vth
4‘0q A
(b)
Io
VTh
- j2
(4‘0q)
8 j2
8 Io
- j64
8 j2
2
Pmax
(b)
VTh
8RL
2
§ 64 ·
¸
¨
© 68 ¹
(8)(0.471)
15.99 W
We obtain Z Th from the circuit in Fig. (c).
5:
-j3 :
j2 :
4:
Zth
(c)
Z Th
ZL
j2 5 || (4 j3)
Z *Th
j2 2.5 j1.167 :
(5)(4 j3)
9 j3
2.5 j1.167
Chapter 11, Solution 13.
(a)
We find Z Th at the load terminals using the circuit in Fig. (a).
j100 :
80 :
Zth
-j40 :
(a)
(b)
Z Th
-j40 || (80 j100)
ZL
Z *Th
(-j40)(80 j100)
80 j60
51.2 j1.6
51.2 j1.6 :
We find VTh at the load terminals using Fig. (b).
j100 :
Io
+
3‘20q A
80 :
-j40 :
Vth
(b)
Io
VTh
80
(3‘20q)
80 j100 j40
- j40 I o
(8)(3‘20q)
8 j6
(- j40)(24‘20q)
8 j6
2
Pmax
VTh
8RL
2
§ 40
·
¨ ˜ 24¸
© 10
¹
(8)(51.2)
22.5 W
From Fig.(d), we obtain VTh using the voltage division principle.
5:
-j3 :
j2 :
10‘30q V
+
4:
+
Vth
(d)
§ 4 j3 ·
§ 4 j3 ·§ 10
·
¸(10‘30q) ¨
¨
¸¨ ‘30q¸
¹
© 9 j3 ¹
© 3 j ¹© 3
VTh
Pmax
VTh
§ 5 10 ·
¨
˜ ¸
© 10 3 ¹
(8)(2.5)
2
8RL
2
1.389 W
Chapter 11, Solution 14.
I
j24 :
––j10 :
16 :
VTh
10 :
40‘90º A
j8 :
Z Th
Z
j10 Z Th
(10 j24)(16 j8)
10 j24 16 j8
8.245 j2.3:
+
ZTh
_
j10 8.245 j7.7
8.245 j2.3:
10
j40(16 j8)
10 j24 16 j8
173.55‘65.66q 71.53 j158.12 V
VTh
I(16 j8)
2
VTh
Pmax
2
2
8.245
(2x8.245) 2
I 2rms 8.245
456.6 W
Chapter 11, Solution 15.
To find Z Th , insert a 1-A current source at the load terminals as shown in Fig. (a).
1:
-j :
1
2
+
2 Vo
j:
Vo
1A
(a)
At node 1,
Vo Vo
1
j
At node 2,
1 2 Vo
V2 Vo
-j
V2 Vo
-j

o Vo

o 1
Substituting (1) into (2),
1 j V2 (2 j)( j) V2
V2
1
1 j
VTh
V2
1
ZL
Z *Th
1 j
2
(1 j) V2
0.5 j0.5
0.5 j0.5 :
j V2
j V2 (2 j) Vo
(1)
(2)
We now obtain VTh from Fig. (b).
1:
12‘0q V
-j :
+
+
+
j:
Vo
2 Vo
Vth
(b)
12 Vo
1
- 12
1 j
Vo
j
2 Vo Vo
Vo (- j u 2 Vo ) VTh
VTh
0
(12)(1 j2)
1 j
-(1 j2)Vo
2
VTh
Pmax
2
8RL
§12 5 ·
¸¸
¨¨
© 2 ¹
(8)(0.5)
90 W
Chapter 11, Solution 16.
Z
4,
1H

o
jZL
j 4,

o
1 / 20F
1
jZC
1
j 4 x1 / 20
j5
We find the Thevenin equivalent at the terminals of ZL. To find VTh, we use the circuit
shown below.
0.5Vo
2:
4:
V1
V2
+
+
10<0o
-
+
Vo
-
-j5
j4
VTh
-
At node 1,
V1
V V2
10 V1
0.25V1 1
2
j5
4
At node 2,
V1 V2
0.25V1
4
V2
j4

o

o
0
5 V1 (1 j 0.2) 0.25V2
0.5V1 V2 (0.25 j 0.25)
Solving (1) and (2) leads to
VTh V2 6.1947 j 7.0796 9.4072‘48.81o
Chapter 11, Solution 17.
We find R Th at terminals a-b following Fig. (a).
-j10 :
30 :
a
b
40 :
j20 :
(a)
(30)( j20) (40)(-j10)
30 j20
40 j10
Z Th
30 || j20 40 || (- j10)
Z Th
Z Th
9.23 j13.85 2.353 j9.41
11.583 j4.44 :
ZL
Z *Th
11.583 j4.44 :
We obtain VTh from Fig. (b).
I1
I2
-j10 :
30 :
j5 A
+ VTh 40 :
j20 :
(b)
(1)
(2)
Using current division,
30 j20
I1
( j5)
70 j10
I2
-1.1 j2.3
40 j10
( j5) 1.1 j2.7
70 j10
30 I 2 j10 I 1
VTh
VTh
Pmax
10 j70
2
8RL
5000
(8)(11.583)
53.96 W
Chapter 11, Solution 18.
We find Z Th at terminals a-b as shown in the figure below.
40 :
40 :
a
j20 :
Zth
b
Z Th
j20 40 || 40 80 || (-j10)
Z Th
21.23 j10.154
ZL
-j10 :
80 :
Z *Th
j20 20 21.23 j10.15 :
Chapter 11, Solution 19.
At the load terminals,
Z Th
- j2 6 || (3 j)
Z Th
2.049 j1.561
RL
Z Th
2.576 :
-j2 (6)(3 j)
9 j
(80)(-j10)
80 j10
To get VTh , let Z
6 || (3 j)
2.049 j0.439 .
By transforming the current sources, we obtain
VTh (4 ‘0q) Z 8.196 j1.756
VTh
Pmax
2
8RL
70.258
20.608
3.409 W
Chapter 11, Solution 20.
Combine j20 : and -j10 : to get
j20 || -j10 -j20
To find Z Th , insert a 1-A current source at the terminals of R L , as shown in Fig. (a).
40 :
Io
V1
4 Io
V2
+ -j20 :
-j10 :
1A
(a)
At the supernode,
1
Also,
V1
V
V
1 2
40 - j20 - j10
40
(1 j2) V1 j4 V2
V1
V2 4 I o ,
1.1 V1
V2

o V1
Substituting (2) into (1),
40
§V ·
(1 j2) ¨ 2 ¸ j4 V2
© 1 .1 ¹
(1)
where I o
V2
1 .1
- V1
40
(2)
V2
44
1 j6.4
Z Th
V2
1
RL
Z Th
1.05 j6.71 :
6.792 :
To find VTh , consider the circuit in Fig. (b).
Io
40 :
V1
4 Io
V2
+ +
+
120‘0q V
-j20 :
-j10 :
Vth
(b)
At the supernode,
120 V1
40
120
Also,
V1
V
2
- j20 - j10
(1 j2) V1 j4 V2
V1
V2 4 I o ,
V1
V2 12
1 .1
Substituting (4) into (3),
109.09 j21.82
VTh
Pmax
V2
where I o
8RL
2
120 V1
40
(4)
(0.9091 j5.818) V2
109.09 j21.82
0.9091 j5.818
VTh
(3)
(18.893) 2
(8)(6.792)
18.893‘ - 92.43q
6.569 W
Chapter 11, Solution 21.
We find Z Th at terminals a-b, as shown in the figure below.
100 :
-j10 :
a
40 :
Zth
50 :
j30 :
b
Z Th
50 || [ - j10 100 || (40 j30) ]
(100)(40 j30)
140 j30
where 100 || (40 j30)
Z Th
50 || (31.707 j4.634)
Z Th
19.5 j1.73
RL
Z Th
31.707 j14.634
(50)(31.707 j4.634)
81.707 j4.634
19.58 :
Chapter 11, Solution 22.
i (t )
I
2
rms
4 sin t ,
1
0t S
S
16 sin
S³
2
tdt
0
I rms
8
2.828 A
16 § t sin 2t ·
¨ ¸
S ©2
4 ¹
S
0
16 S
( 0)
S 2
8
Chapter 11, Solution 23.
­15, 0 t 2
®
¯ 5, 2 t 6
v( t )
> ³ 15
2
Vrms
1
6
Vrms
9.574 V
2
2
0
dt ³2 5 2 dt
6
@
550
6
Chapter 11, Solution 24.
T
2,
­ 5, 0 t 1
®
¯- 5, 1 t 2
v( t )
>³ 5
2
Vrms
1
2
Vrms
5V
1
0
2
dt ³1 (-5) 2 dt
2
@
25
[1 1]
2
25
Chapter 11, Solution 25.
>
1 T 2
1 1
2
3
f
(
t
)
dt
(4) 2 dt ³ 1 0dt ³2 4 2 dt
³
³
0
0
T
3
1
32
[16 0 16]
3
3
2
f rms
f rms
32
3
3.266
@
Chapter 11, Solution 26.
T
4,
­5 0 t2
®
¯10 2 t 4
v( t )
>³ 5
2
Vrms
1
4
Vrms
7.906 V
2
2
0
dt ³2 (10) 2 dt
4
@
1
[50 200 ]
4
Chapter 11, Solution 27.
5,
T
i( t )
2
rms
1 5 2
³ t dt
5 0
I rms
2.887 A
I
t, 0 t 5
1 t3
˜
5 3
5
0
125
15
Chapter 11, Solution 28.
2
Vrms
1
5
> ³ (4t )
2
0
V
1 16 t 3
˜
5 3
Vrms
2.92 V
2
rms
P
2
Vrms
R
2
2
0
8.533
2
dt ³2 0 2 dt
5
16
(8)
15
@
8.533
4.267 W
8.333
62.5
Chapter 11, Solution 29.
20 ,
T
­ 20 2t 5 t 15
®
¯- 40 2t 15 t 25
i( t )
>³
@
2
I eff
1
20
2
I eff
25
1 ª 15
2
(
100
20
t
t
)
dt
( t 2 40 t 400) dt º
³
³
»¼
«
5
15
5¬
I
15
5
(20 2 t ) 2 dt ³15 (-40 2t) 2 dt
25
1ª§
t 3 · 15
2
¨
« 100 t 10 t ¸ 5
5¬ ©
3¹
2
eff
· 25 º
§ t3
¨ 20 t 2 400 t ¸ 15
»
¹ ¼
©3
2
I eff
1
[83.33 83.33 ] 33.332
5
I eff
5.773 A
2
I eff
R
P
400 W
Chapter 11, Solution 30.
v( t )
­t 0t2
®
¯- 1 2 t 4
>³ t
2
Vrms
1
4
Vrms
1.08 V
2
2
0
dt ³2 (-1) 2 dt
4
@
º
1 ª8
2 » 1.1667
«
¼
4¬3
Chapter 11, Solution 31.
2
V
2
rms
Vrms
1
v(t )dt
2 ³0
2.944 V
1
2
º
1ª
2
2
« ³ (2t ) dt ³ (4) dt »
2 ¬0
1
¼
1 ª4
º
16»
«
2 ¬3
¼
8.6667
Chapter 11, Solution 32.
I 2rms
I
2
rms
I rms
2
1ª 1
(10t 2 ) 2 dt ³ 0 dt º
³
»¼
1
2 «¬ 0
50 ³0 t dt
1
4
t5
50 ˜
5
1
0
10
3.162 A
Chapter 11, Solution 33.
i( t )
­ 10
0 t 1
°
® 20 10t 1 t 2
° 0
2t3
¯
I 2rms
1
3
> ³ 10
1
dt ³1 (20 10t ) 2 dt 0
2
2
0
3 I 2rms
100 100³1 (4 4t t 2 ) dt
I rms
133.33
3
2
@
100 (100)(1 3) 133.33
6.667 A
Chapter 11, Solution 34.
>
1 T 2
1 2
3
f ( t )dt
(3t ) 2 dt ³ 2 6 2 dt
³
³
0
0
T
3
2
º
ª
1 « 9t 3
36» 20
»
3« 3
0
»¼
«¬
2
f rms
f rms
20
4.472
@
Chapter 11, Solution 35.
> ³ 10
2
Vrms
1
6
2
Vrms
1
[100 400 1800 400 100 ]
6
Vrms
21.6 V
1
0
2
dt ³1 20 2 dt ³2 30 2 dt ³4 20 2 dt ³5 10 2 dt
6
5
4
2
@
466.67
Chapter 11, Solution 36.
(a) Irms = 10 A
2
(b) V
2
§ 3 ·
4 ¨
¸
© 2¹
36
64 2
2
rms
(c)
I rms
(d)
Vrms
25 16
2
2

o
Vrms
16 9
2
4.528 V (checked)
9.055 A
4.528 V
Chapter 11, Solution 37.
i
i1 i2 i3
I rms
8 4 sin(t 10 o ) 6 cos(2t 30 o )
I 21rms I 2 2 rms I 2 3rms
64 16 36
2
2
90
9.487 A
Chapter 11, Solution 38.
0.5 H 
o
Z
R jX L
jZL
j (2S )(50)(0.5)
30 j157.08
j157.08
S
Apparent power
V
2
(210) 2
30 j157.08
Z*
(210) 2
160
S
275.6 VA
§
§157.08 ··
¸¸
cos ¨ tan -1 ¨
© 36 ¹¹
©
pf
cos T
pf
0.1876 (lagging)
cos(79.19q)
Chapter 11, Solution 39.
ZT
j4 || (12 j8)
( j4)(12 j8)
12 j4
ZT
0.4 (3 j11)
4.56 ‘74.74q
pf
cos(74.74q)
0.2631
Chapter 11, Solution 40.
At node 1,
120‘30 o V1
20
V1 V1 V2
j30
50

o
103.92 j60
V1 (1.4 j0.6667) 0.4V2
(1)
At node 2,
V1 V2
50
V2
V
2
10 j 40

o
0
V1 (6 j1.25)V2
Solving (1) and (2) leads to
V1 = 45.045 + j66.935, V2 = 9.423 + j9.193
(2)
(a) Pj 30 :
(b) I
0
P j 40 :
1 | V2 | 2
2 R
P10 :
V 2 rms
R
P50 :
1 | V1 V2 | 2
R
2
P20 :
1 | 120‘30 o V1 | 2
R
2
120‘30 o V1
20
S
1
Vs I x
2
173.3 / 20
4603.1 / 100
8.665 W
46.03 W
3514 / 40
2.944 j 0.3467,
142.5 j106.3,
87.86 W
Vs
120‘30 o
103.92 j 60
S | S | 177.8 VA
(c ) pf = 142.5/177.8 = 0.8015 (leading).
Chapter 11, Solution 41.
(a)
- j2 || ( j5 j2)
ZT
pf
(b)
4 j6
-j2 || -j3
0.5547 (leading)
( j2)(4 j)
4 j3
Z 1 || (0.64 j1.52 j)
pf
-j6
7.211‘ - 56.31q
cos(-56.31q)
j2 || (4 j)
(-j2)(-j3)
j
cos(21.5q)
0.64 j1.52
0.64 j0.44
1.64 j0.44
0.9304 (lagging)
0.4793‘21.5q
Chapter 11, Solution 42.
cos T 
o T 30.683q
pf
0.86
Q
S sin T 
o S
S
V I*

o I *
Q
sin T
9.798 kVA
9.798 u 10 3 ‘30.683q
220
S
V
2 u 44.536
Peak current
5
sin(30.683q)
44.536 ‘30.683q
62.98 A
Apparent power = S = 9.798 kVA
Chapter 11, Solution 43.
(a) Vrms
(b) P
V 21rms V 2 2 rms V 2 3rms
V 2 rms
R
30 / 10
25 9 1
2 2
30
5.477 V
3W
Chapter 11, Solution 44.
0.65
pf
S
cosT
S (cosT j sin T )

o
T
49.46 o
50(0.65 j 0.7599)
32.5 j 38 kVA
Thus,
Average power = 32.5 kW, Reactive power = 38 kVAR
Chapter 11, Solution 45.
(a) V 2 rms
20 2 I rms
(b) P Vrms I rms
60 2
2
2200
12 
o
0.5 2
2
Vrms
1.125
46.9 V
1.061A
49.74 W
Chapter 11, Solution 46.
(a)
S
S
V I * (220‘30q)(0.5‘ - 60q) 110‘ - 30q
95.26 j55 VA
Apparent power = 110 VA
Real power = 95.26 W
Reactive power = 55 VAR
pf is leading because current leads voltage
(b)
S
S
V I * (250‘ - 10q)(6.2 ‘25q) 1550‘15q
1497.2 j401.2 VA
Apparent power =1550 VA
Real power = 1497.2 W
Reactive power = 401.2 VAR
pf is lagging because current lags voltage
(c)
S
S
V I * (120‘0q)(2.4‘15q)
278.2 j74.54 VA
288‘15q
Apparent power = 288 VA
Real power = 278.2 W
Reactive power = 74.54 VAR
pf is lagging because current lags voltage
(d)
S
S
V I * (160 ‘45q)(8.5‘ - 180q) 1360‘ - 135q
- 961.7 j961.7 VA
Apparent power = 1360 VA
Real power = - 961.7 W
Reactive power = - 961.7 VAR
pf is leading because current leads voltage
Chapter 11, Solution 47.
(a)
112 ‘10q ,
I 4‘ - 50q
1
S
V I * 224‘60q 112 j194 VA
2
Average power = 112 W
V
Reactive power =194 VAR
(b)
V
S
160 ‘0q ,
I 25‘45q
1
V I * 200‘ - 45q 141.42 j141.42 VA
2
Average power = 141.42 W
Reactive power = - 141.42 VAR
(c)
S
2
V
Z*
(80) 2
128‘30q
50‘ - 30q
90.51 j64 VA
Average power = 90.51 W
Reactive power = 64 VAR
(d)
S
2
I Z
(100)(100‘45q)
Average power = 7.071 kW
Reactive power = 7.071 kVAR
7.071 j7.071 kVA
Chapter 11, Solution 48.
(a)
S
P jQ
(b)
pf
cos T
Q
S sin T 
o S
P
S cos T
S
4129 j2000 VA
Q
S sin T 
o sin T
T
48.59 ,
P
S cos T
S
396.9 j450 VA
(c)
(d)
S
P
V
2
Z
269 j150 VA
0.9 
o T
25.84q
Q
sin T
2000
sin(25.84q)
4129.48
pf
(600)(0.6614)
(220) 2
40
Q 450
S 600
0.6614
1210
S cos T 
o cos T
Q
S sin T
S
1000 j681.2 VA
P
S
1000
1210
681.25
Chapter 11, Solution 49.
S
S
4
sin(cos -1 (0.86)) kVA
0.86
4 j2.373 kVA
4 j
0.75
396.86
T 34.26q
(a)
4588.31
0.8264
(b)
pf
P
S
1.6
0.8
2
cos T 
o sin T
0.6
S 1.6 j2 sin T 1.6 j1.2 kVA
(c)
S Vrms I *rms (208‘20q)(6.5‘50q) VA
S 1.352 ‘70q 0.4624 j1.2705 kVA
2
(d)
V
(120) 2
14400
S
*
Z
40 j60 72.11‘ - 56.31q
S 199.7 ‘56.31q 110.77 j166.16 VA
Chapter 11, Solution 50.
(a)
P jQ 1000 j
S
1000
sin(cos -1 (0.8))
0.8
S 1000 j750
But,
S
*
Vrms
Z
(b)
2
Z*
2
S
(220) 2
1000 j750
Z
30.98 j23.23 :
S
I rms Z
Z
30.98 j23.23
2
1500 j2000
(12) 2
S
Z
(c)
Vrms
I rms
*
2
Vrms
2
S
Z 1.6 ‘60q
V
2
2S
10.42 j13.89 :
(120) 2
(2)(4500 ‘60q)
0.8 j1.386 :
1.6 ‘ - 60q
Chapter 11, Solution 51.
(a)
(b)
ZT
2 (10 j5) || (8 j6)
ZT
2
ZT
8.152 j0.768
pf
cos(5.382q)
S
(10 j5)(8 j6)
18 j
1
V I*
2
V
2
110 j20
18 j
8.188‘5.382q
0.9956 (lagging)
2
(16) 2
(2)(8.188‘ - 5.382q)
2 Z*
S 15.63‘5.382q
P
S cos T
15.56 W
(c)
Q
S sin T
1.466 VAR
(d)
S
S
(e)
S 15.63‘5.382q
15.63 VA
15.56 j1.466 VA
Chapter 11, Solution 52.
SB
2000
0.6 2000 j1500
0 .8
3000 x 0.4 j3000 x 0.9165 1200 j2749
SC
1000 j500
SA
2000 j
S S A S B SC
(a)
pf
(b)
S
4200 j749
4200
4200 2 749 2
Vrms I rms 
o I rms
Irms = 35.55‘––55.11 A.
0.9845 leading.
4200 j749
120‘45q
35.55‘ 55.11q
Chapter 11, Solution 53.
S = SA + SB + SC = 4000(0.8––j0.6) + 2400(0.6+j0.8) + 1000 + j500
= 5640 + j20 = 5640‘0.2
I rms
(a)
2 x 66.46‘29.88q
I
(b)
SB
S SC
A
Vrms
Vrms
S
Vrms
5640‘0.2q
120‘30q
2
66.46‘ 29.8q
93.97‘29.8q A
pf = cos(0.2) § 1.0 lagging.
Chapter 11, Solution 54.
(a)
P jQ 1000 j
S
1000
sin(cos -1 (0.8))
0.8
S 1000 j750
But,
S
*
Vrms
Z
(b)
2
Z*
2
S
(220) 2
1000 j750
Z
30.98 j23.23 :
S
I rms Z
Z
30.98 j23.23
2
1500 j2000
(12) 2
S
Z
(c)
Vrms
I rms
*
2
Vrms
2
S
Z 1.6 ‘60q
V
2
2S
10.42 j13.89 :
(120) 2
(2)(4500 ‘60q)
0.8 j1.386 :
1.6 ‘ - 60q
Chapter 11, Solution 55.
We apply mesh analysis to the following circuit.
-j20 :
j10 :
I3
+
40‘0q V rms
I1
20 :
I2
+
40 (20 j20) I1 20 I 2
2 (1 j) I1 I 2
For mesh 2, - j50 (20 j10) I 2 20 I1
- j5 -2 I1 (2 j) I 2
Putting (1) and (2) in matrix form,
ª 2 º ª1 j - 1 ºª I1 º
« - j5» « - 2 2 j»« I »
¬ ¼ ¬
¼¬ 2 ¼
50‘90q V rms
For mesh 1,
'
1 j ,
I1
'1
'
4 j3
1 j
I2
'2
'
- 1 j5
1 j
(2)
'1
I 3 I1 I 2
For the 40-V source,
S
(1)
4 j3 ,
1
(7 j)
2
2 j3 3.605‘ - 56.31q
(3.5 j0.5) (2 j3) 1.5 j3.5
§1
·
-(40) ¨ ˜ (7 j) ¸
©2
¹
-V I 1*
I1
2
Zc
- 140 j20 VA
- j250 VA
For the resistor,
S
I3
2
R
290 VA
For the inductor,
S
I2
2
For the j50-V source,
S V I *2
ZL
-1 j5
3.535‘8.13q
For the capacitor,
S
'2
j130 VA
( j50)(2 j3)
- 150 j100 VA
3.808‘66.8q
Chapter 11, Solution 56.
- j2 || 6
(6)(- j2)
6 j2
3 j4 (-j2) || 6
0.6 j1.8
3.6 j2.2
The circuit is reduced to that shown below.
Io
+
2‘30q A
3.6 + j2.2 :
5:
Vo
Io
3.6 j2.2
(2‘30q)
8.6 j2.2
Vo
5Io
S
1
V I*
2 o s
S
4.75‘17.08q
0.95‘47.08q
4.75‘47.08q
1
˜ (4.75‘47.08q)(2‘ - 30q)
2
4.543 j1.396 VA
Chapter 11, Solution 57.
Consider the circuit as shown below.
4:
24‘0q V
+
Vo
-j1 :
V1
2:
+
1:
j2 :
V2
At node o,
24 Vo
4
Vo Vo V1
1
-j
2 Vo
(5 j4) Vo j4 V1
24
At node 1,
Vo V1
2 Vo
-j
(1)
V1
j2
(2 j4) Vo
V1
(2)
Substituting (2) into (1),
24 (5 j4 j8 16) Vo
- 24
,
11 j4
Vo
V1
(-24)(2 - j4)
11 j4
The voltage across the dependent source is
V2 V1 (2)(2 Vo ) V1 4 Vo
- 24
˜ (2 j4 4)
11 j4
V2
S
1
V2 I *
2
S
(-24)(6 j4) - 24
˜
11 j4
11 - j4
S
25.23 j16.82 VA
(-24)(6 j4)
11 j4
1
V2 (2 Vo* )
2
§ 576 ·
¨
¸(6 j4)
© 137 ¹
Chapter 11, Solution 58.
Ix -j3 k:
8 mA
4 k:
From the left portion of the circuit,
0.2
Io
0.4 mA
500
20 I o
8 mA
j1 k:
10 k:
From the right portion of the circuit,
4
Ix
(8 mA)
4 10 j j3
2
16
mA
7 j
(16 u 10 -3 ) 2
˜ (10 u 10 3 )
50
S
Ix
S
51.2 mVA
R
Chapter 11, Solution 59.
Consider the circuit below.
Ix -j3 k:
8 mA
4
j1 k:
4 k:
240 Vo
50
Vo
Vo
- j20 40 j30
88
(0.36 j0.38) Vo
Vo
88
0.36 j0.38
I1
Vo
- j20
I2
Vo
40 j30
10 k:
168.13‘ - 46.55q
8.41‘43.45q
3.363‘ - 83.42q
Reactive power in the inductor is
1
1
2
S
I2 ZL
˜ (3.363) 2 ( j30)
2
2
Reactive power in the capacitor is
1
1
2
S
I1 Z c
˜ (8.41) 2 (- j20)
2
2
j169.65 VAR
- j707.3 VAR
Chapter 11, Solution 60.
S1
20 j
20
sin(cos -1 (0.8))
0.8
S2
16 j
16
sin(cos -1 (0.9)) 16 j7.749
0.9
S S1 S 2
But
S
36 j22.749
Vo I *
Vo
S
6
pf
cos(32.29q)
20 j15
42.585‘32.29q
6 Vo
7.098 ‘ 32.29q
0.8454 (lagging)
Chapter 11, Solution 61.
Consider the network shown below.
I2
Io
S2
I1
+
Vo
So
S1
S3
S2
1.2 j0.8 kVA
S3
4 j
Let
S4
S2 S3
But
S4
1
V I*
2 o 2
4
sin(cos -1 (0.9))
0.9
4 j1.937 kVA
5.2 j1.137 kVA
I
*
2
(2)(5.2 j1.137) u 10 3
100 ‘90q
2S4
Vo
22.74 j104
I2
22.74 j104
Similarly,
S1
2 j
But
S1
1
Vo I 1*
2
I 1*
2 S1
Vo
I1
-40 j40
Io
I1 I 2
So
1
Vo I *o
2
So
1
˜ (100‘90q)(145‘ - 96.83q) VA
2
So
7.2 j0.862 kVA
2
sin(cos -1 (0.707))
0.707
(4 j4) u 10 3
j100
2 j2 kVA
-40 j40
-17.26 j144 145‘96.83q
Chapter 11, Solution 62.
Consider the circuit below
0.2 + j0.04 :
I
I2
0.3 + j0.15 :
I1
Vs
+
+
+
V1
V2
But
But
15
sin(cos -1 (0.8)) 15 j11.25
0.8
S2
15 j
S2
V2 I *2
I *2
S2
V2
I2
V1
V1
V1
0.125 j0.09375
V2 I 2 (0.3 j0.15)
120 (0.125 j0.09375)(0.3 j0.15)
120.02 j0.0469
S1
10 j
S1
V1 I 1*
I 1*
S1
V1
15 j11.25
120
10
sin(cos -1 (0.9)) 10 j4.843
0.9
11.111‘25.84q
120.02 ‘0.02q
I 1 0.093‘ - 25.82q 0.0837 j0.0405
I I 1 I 2 0.2087 j0.053
Vs V1 I (0.2 j0.04)
Vs
Vs
Vs
(120.02 j0.0469) (0.2087 j0.053)(0.2 j0.04)
120.06 j0.0658
120.06‘0.03q V
Chapter 11, Solution 63.
Let
S
S1 S 2 S 3 .
S1
12 j
12
sin(cos -1 (0.866)) 12 j6.929
0.866
S2
16 j
16
sin(cos -1 (0.85)) 16 j9.916
0.85
(20)(0.6)
j20 15 j20
sin(cos -1 (0.6)
S3
S
1
V I *o
2
43 j22.987
44 j22.98
110
I *o
2S
V
Io
0.4513‘ - 27.58q A
Chapter 11, Solution 64.
I2
I1
8:
+
Is
120‘0º V
j12
Is + I2 = I1 or Is = I1 –– I2
I1
But,
S
120
8 j12
4.615 j6.923
VI 2 
o I 2
or I 2
S
V
2500 j400
120
20.83 j3.333
20.83 j3.333
Is = I1 –– I2 = ––16.22 –– j10.256 = 19.19‘––147.69 A.
Chapter 11, Solution 65.
C 1 nF 
o
1
jZC
At the noninverting terminal,
4‘0q Vo
Vo
100
- j100
4
Vo
2
v o (t)
-j
10 u 10 -9
-j100 k:
4

o Vo
4
1 j
‘ - 45q
4
2
P
2
Vrms
R
P
80 PW
cos(10 4 t 45q)
2
§ 4 1 · § 1 ·
¨
¸ ¨
¸W
˜
© 2 2 ¹ © 50 u 10 3 ¹
Chapter 11, Solution 66.
As an inverter,
- Zf
Vo
V
Zi s
Io
Vo
mA
6 j2
- (2 j4)
˜ (4 ‘45q)
4 j3
- (2 j4)(4‘45q)
mA
(6 - j2)(4 j3)
The power absorbed by the 6-k: resistor is
2
2
1
1 §¨ 20 u 4 ·¸
-6
3
P
I R
˜¨
¸ u 10 u 6 u 10
2 o
2 © 40 u 5 ¹
P
0.96 mW
Chapter 11, Solution 67.
Z
2,
3H
j 50
10 j 5
10 //( j 5)
jZ L

o
j 6,
0.1F

o
1
jZ C
1
j 2 x0.1
j5
2 j4
The frequency-domain version of the circuit is shown below.
Z2=2-j4 :
Z1 =8+j6 :
I1
+
+
+
0.6‘20 o V
Io
Z3
Vo
-
(a) I 1
S
0.6‘20 o 0
8 j6
1
Vs I *1
2
(b) Vo
0.5638 j 0.2052
8 j6
0.06‘ 16.87 o
(0.3‘20 o )(0.06‘ 16.87 o ) 14.4 j10.8 mVA 18‘36.86 o mVA
Z2
Vs ,
Z1
P
Io
1
| I o |2 R
2
Vo
Z3
( 2 j 4)
(0.6‘20 o )
12(8 j 6)
0.5(0.0224) 2 (12)
2.904 mW
0.0224‘99.7 o
12:
Chapter 11, Solution 68.
Let
where
Hence,
S
SR SL Sc
SR
PR jQ R
1 2
I R j0
2 o
SL
PL jQ L
1
0 j I o2 ZL
2
Sc
Pc jQ c
1
1
0 j I o2 ˜
ZC
2
S
§
1 2ª
1 ·º
¸
I o « R j¨ZL ©
2 ¬
ZC ¹»¼
Chapter 11, Solution 69.
(a)
Given that Z 10 j12
12
10
tan T
pf
(b)
S
cos T
V
2
2 Z*

o T 50.19q
0.6402
(120) 2
(2)(10 j12)
295.12 j354.09
The average power absorbed
(c)
P
Re(S)
295.1 W
For unity power factor, T1 0q , which implies that the reactive power due
to the capacitor is Q c 354.09
But
C
Qc
2 Qc
ZV2
V2
2 Xc
1
ZC V 2
2
(2)(354.09)
(2S )(60)(120) 2
130.4 PF
Chapter 11, Solution 70.
pf cos T 0.8 
o
sin T
Q S sin T (880)(0.6) 528
0.6
If the power factor is to be unity, the reactive power due to the capacitor is
Q c Q 528 VAR
But
C
Q
2
Vrms
Xc
1
ZC V 2
2
(2)(528)
(2S)(50)(220) 2

o C
2 Qc
ZV2
69.45 PF
Chapter 11, Solution 71.
P1
Q1
S1
106.065 j106.065,
S
S1 S 2
Qc
150 x0.7071 106.065,
S2
Q2
P(tan T 1 tan T 2 ) 172.735(tan 17.98 o 0)
Qc
ZV 2 rms
56.058
2Sx60 x120 2
pf
56.058
10.33 PF
T1
T2
cos -1 (0.76) 40.54q
cos -1 (0.9) 25.84q
Qc
Qc
Qc
P (tan T1 tan T 2 )
(40)[ tan(40.54q) tan(25.84q) ] kVAR
14.84 kVAR
C
Qc
2
Z Vrms
0 .8 S
cos17.98 o
Chapter 11, Solution 72.
(a)
P2
0 .8
66.67 j 50
172.735 j 56.06 181.6‘17.98 o ,
C
Q2
,
0 .6
S2
50,
14840
(2S )(60)(120) 2
2.734 mF
0.9512
50
0 .6
66.67
(b)
T1
Qc
C
40.54q ,
T2
0q
(40)[ tan(40.54q) 0 ] kVAR
Qc
34210
2
Z Vrms (2S)(60)(120) 2
34.21 kVAR
6.3 mF
Chapter 11, Solution 73.
(a)
(b)
(c)
S 10 j15 j22 10 j7 kVA
S
S
10 2 7 2
S
V I*

o I *
I
41.667 j29.167
T1
Qc
Qc
C
(d)
§7·
tan -1 ¨ ¸
©10 ¹
12.21 kVA
S
V
10,000 j7,000
240
50.86‘ - 35q A
35q ,
T2
cos -1 (0.96) 16.26q
P1 [ tan T1 tan T 2 ] 10 [ tan(35q) - tan(16.26q) ]
4.083 kVAR
Qc
2
Z Vrms
4083
(2S )(60)(240) 2
S2
P2 jQ 2 ,
Q2
Q1 Q c
S2
10 j2.917 kVA
S2
But
P2
7 4.083
P1
188.03 PF
10 kW
2.917 kVAR
V I *2
10,000 j2917
240
I *2
S2
V
I2
41.667 j12.154
43.4‘ - 16.26q A
Chapter 11, Solution 74.
(a)
T1
cos -1 (0.8)
S1
P1
cos T1
36.87q
24
0 .8
30 kVA
Q1 S1 sin T1 (30)(0.6) 18 kVAR
S1 24 j18 kVA
(b)
T2
cos -1 (0.95) 18.19q
S2
P2
cos T 2
Q2
S2
S 2 sin T 2 13.144 kVAR
40 j13.144 kVA
40
0.95
S
S1 S 2
T
§ 31.144 ·
¸
tan -1 ¨
© 64 ¹
42.105 kVA
64 j31.144 kVA
25.95q
pf
cos T
T2
25.95q ,
Qc
P [ tan T 2 tan T1 ]
C
Qc
2
Z Vrms
0.8992
T1
0q
64 [ tan(25.95q) 0 ] 31.144 kVAR
31,144
(2S )(60)(120) 2
5.74 mF
Chapter 11, Solution 75.
(a)
(b)
(c)
S1
2
V
Z
*
1
(240) 2
80 j50
S2
(240) 2
120 j70
S3
(240) 2
60
5760
8 j5
5760
12 j7
517.75 j323.59 VA
358.13 j208.91 VA
960 VA
S
S1 S 2 S 3
T
§ 114.68 ·
¸
tan -1 ¨
©1835.88 ¹
1835.88 j114.68 VA
3.574q
pf
cos T
Qc
Qc
P [ tan T 2 tan T1 ] 18.35.88[ tan(3.574q) 0 ]
114.68 VAR
C
Qc
2
Z Vrms
0.998
114.68
(2S )(50)(240) 2
6.336 PF
Chapter 11, Solution 76.
The wattmeter reads the real power supplied by the current source. Consider the
circuit below.
4:
12‘0q V
-j3 :
+
3‘30q Vo
j2 :
12 Vo
4 j3
Vo Vo
j2
8
8:
3‘30q A
Vo
S
36.14 j23.52
2.28 j3.04
1
V I*
2 o o
0.7547 j11.322 11.347 ‘86.19q
1
˜ (11.347 ‘86.19q)(3‘ - 30q)
2
S 17.021‘56.19q
P
Re(S)
9.471 W
Chapter 11, Solution 77.
The wattmeter measures the power absorbed by the parallel combination of 0.1 F
and 150 :.
120 cos(2t ) 
o 120‘0q ,
Z 2
4H 
o
jZL j8
1
0.1 F 
o
-j5
jZC
Consider the following circuit.
6:
120‘0q V
Z 15 || (-j5)
j8 :
I
+
Z
(15)(-j5)
15 j5
I
120
(6 j8) (1.5 j4.5)
S
1
V I*
2
1 2
I Z
2
14.5‘ - 25.02q
1
˜ (14.5) 2 (1.5 j4.5)
2
S 157.69 j473.06 VA
The wattmeter reads
P Re(S) 157.69 W
1.5 j4.5
Chapter 11, Solution 78.
The wattmeter reads the power absorbed by the element to its right side.
Z 4
2 cos(4t ) 
o 2‘0q ,
1H 
o
jZL
1
o
F 
12
j4
1
jZC
-j3
Consider the following circuit.
10 :
20‘0q V
I
+
Z
5 j4 4 || - j3 5 j4 Z
6.44 j2.08
I
20
16.44 j2.08
S
1 2
I Z
2
P
Re(S)
Z
(4)(- j3)
4 j3
1.207 ‘ - 7.21q
1
˜ (1.207) 2 (6.44 j2.08)
2
4.691 W
Chapter 11, Solution 79.
The wattmeter reads the power supplied by the source and partly absorbed by the 40- :
resistor.
Z 100,
j100x10x10 3

o
10 mH
j,
500PF

o
1
jZ C
1
j100x500 x10 6
The frequency-domain circuit is shown below.
20
40
I
Io
j
V1
V2
+1
2 Io
o
10<0
-j20
At node 1,
10 V1
V V2 V1 V2
2I o 1
40
j
20
10 (7 j40)V1 (6 j40)V2
3(V1 V2 ) V1 V2
20
j
o
(1)
At node 2,
V1 V 2 V1 V 2
j
20
V2
j 20

o
0
(20 j )V1 (19 j )V 2
Solving (1) and (2) yields V1 = 1.5568 ––j4.1405
I
10 V1
40
0.8443 j 0.4141,
S
1 x
VI
2
P = Re(S) = 4.222 W.
Chapter 11, Solution 80.
(a)
I
V
Z
110
6.4
17.19 A
4.2216 j 2.0703
(2)
j20
(b)
V2
Z
S
cos T
(110) 2
6 .4
1890.625
0.825 
o T 34.41q
pf
S cos T 1559.76 # 1.6 kW
P
Chapter 11, Solution 81.
4017 3246
kWh consumed
771 kWh
The electricity bill is calculated as follows :
(a)
Fixed charge = $12
(b)
First 100 kWh at $0.16 per kWh = $16
(c)
Next 200 kWh at $0.10 per kWh = $20
(d)
The remaining energy (771 –– 300) = 471 kWh
at $0.06 per kWh = $28.26.
Adding (a) to (d) gives $76.26
Chapter 11, Solution 82.
(a)
P1
5,000,
P2
30,000 x0.82
S S1 S 2
Q1
0
24,600,
Q2
(P1 P2 ) j(Q1 Q 2 )
30,000 sin(cos 1 0.82) 17,171
29,600 j17,171
S | S | 34.22 kVA
(b)
Q = 17.171 kVAR
(c )
pf
P
S
Qc
P(tan T1 tan T 2 )
29,600
34,220
0.865
>
29,600 tan(cos 1 0.865) tan(cos 1 0.9)
(d)
C
Qc
ZV 2 rms
2833
2Sx60 x 240 2
130.46P F
@
2833 VAR
Chapter 11, Solution 83.
1
VI
2
(a) S
1
(210‘60 o )(8‘ 25 o )
2
S cosT
P
840 cos 35 o
840‘35 o
688.1 W
(b) S = 840 VA
(c) Q
S sin T
(d) pf
P/S
840 sin 35 o
cos 35 o
481.8 VAR
0.8191 (lagging)
Chapter 11, Solution 84.
(a)
Maximum demand charge 2,400 u 30 $72,000
Energy cost $0.04 u 1,200 u 10 3 $48,000
Total charge = $120,000
(b)
To obtain $120,000 from 1,200 MWh will require a flat rate of
$120,000
per kWh $0.10 per kWh
1,200 u 10 3
Chapter 11, Solution 85.

o
j 2Sx60 x15 x10 3 j 5.655
(a) 15 mH
We apply mesh analysis as shown below.
I1
+
Ix
120<0o V
-
10 :
In
30 :
Iz
120<0o V
10 :
+
Iy
-
j5.655 :
I2
For mesh x,
(1)
120 = 10 Ix - 10 Iz
For mesh y,
(2)
120 = (10+j5.655) Iy - (10+j5.655) Iz
For mesh z,
(3)
0 = -10 Ix ––(10+j5.655) Iy + (50+j5.655) Iz
Solving (1) to (3) gives
Ix =20, Iy =17.09-j5.142, Iz =8
Thus,
I1 =Ix =20 A
I2 =-Iy =-17.09+j5.142 = 17.85‘163.26 o A
In =Iy - Ix =-2.091 ––j5.142 = 5.907‘ 119.5 o A
(b)
1
(120) I x x
2
S1
S1 S 2
S
60 x 20 1200,
S2
1
(120) I x y
2
1025.5 j 308.5
2225.5 j 308.5 VA
(c ) pf = P/S = 2225.5/2246.8 = 0.9905
Chapter 11, Solution 86.
For maximum power transfer
Z L Z *Th 
o Z i Z Th Z *L
Z L R jZL 75 j (2S)(4.12 u 10 6 )(4 u 10 -6 )
Z L 75 j103.55 :
75 j103.55 :
Zi
Chapter 11, Solution 87.
Z
R r jX
VR
IR

o R
2
Z
R 2 X2
X 2.5377 k:
T
pf
§X·
tan -1 ¨ ¸
©R ¹
cos T
VR
I

o X 2
§ 2.5377 ·
¸
tan -1 ¨
© 1.6 ¹
0.5333
80
50 u 10 -3
Z
2
1 .6 k :
R2
57.77q
(3) 2 (1.6) 2
Chapter 11, Solution 88.
(a)
(b)
S
(110)(2 ‘55q)
P
S cos T
S
S
220‘55q
220 cos(55q)
126.2 W
220 VA
Chapter 11, Solution 89.
(a)
Apparent power
12 kVA
S
P S cos T (12)(0.78) 9.36 kW
Q S sin T 12 sin(cos -1 (0.78)) 7.51 kVAR
S
(b)
P jQ
9.36 j7.51 kVA
2
S
V
Z*
Z
34.398 j27.6 :

o Z
V
S
*
2
(210) 2
(9.36 j7.51) u 10 3
Chapter 11, Solution 90
Original load :
P1
2000 kW ,
S1
P1
cos T1
Q1
S1 sin T1
Additional load :
P2
cos T1
0.85 
o T1
31.79q
0.8 
o T 2
36.87q
2352.94 kVA
300 kW ,
S2
P2
cos T 2
Q2
S 2 sin T 2
1239.5 kVAR
cos T 2
375 kVA
225 kVAR
Total load :
S S1 S 2 (P1 P2 ) j (Q1 Q 2 )
P 2000 300 2300 kW
Q 1239.5 225 1464.5 kVAR
P jQ
The minimum operating pf for a 2300 kW load and not exceeding the kVA rating of the
generator is
P
2300
cos T
0.9775
S1 2352.94
T 12.177q
or
The maximum load kVAR for this condition is
Q m S1 sin T 2352.94 sin(12.177q)
Q m 496.313 kVAR
The capacitor must supply the difference between the total load kVAR ( i.e. Q ) and the
permissible generator kVAR ( i.e. Q m ). Thus,
Q c Q Q m 968.2 kVAR
Chapter 11, Solution 91
P
S cos T
pf
cos T
P
S
Q
S sin T
220(15) sin(35.09q) 1897.3
2700
(220)(15)
0.8182
When the power is raised to unity pf, T1
C
Qc
2
Z Vrms
1897.3
(2S )(60)(220) 2
104 PF
0q and Q c
Q 1897.3
Chapter 11, Solution 92
(a)
Apparent power drawn by the motor is
P
60
Sm
80 kVA
cos T 0.75
Qm
S2 P 2
(80) 2 (60) 2
Total real power
P Pm Pc PL
60 0 20
Total reactive power
Q Qm Qc QL
Total apparent power
S
P2 Q2
(b)
pf
(c)
I
P
S
S
V
80
86.51
86510
550
80 kW
52.915 20 0
86.51 kVA
0.9248
157.3 A
Chapter 11, Solution 93
(a)
P1
(5)(0.7457)
S1
P1
pf
3.7285
0.8
3.7285 kW
4.661 kVA
Q1 S1 sin(cos -1 (0.8)) 2.796 kVAR
S 1 3.7285 j2.796 kVA
P2 1.2 kW ,
S 2 1.2 j0 kVA
P3
S3
Q2
(10)(120) 1.2 kW ,
1.2 j0 kVA
52.915 kVAR
0 VAR
Q3
0 VAR
32.91 kVAR
cos T 4
0.6 
o sin T 4
Q4
1.6 kVAR ,
S4
Q4
sin T 4
P4
S4
S 4 cos T 4 (2)(0.6) 1.2 kW
1.2 j1.6 kVA
S
S
2 kVA
S1 S 2 S 3 S 4
7.3285 j1.196 kVA
Total real power = 7.3285 kW
Total reactive power = 1.196 kVAR
(b)
T
pf
§ 1.196 ·
¸
tan -1 ¨
© 7.3285 ¹
cos T
9.27q
0.987
Chapter 11, Solution 94
cos T1 0.7 
o T1 45.57q
S1 1 MVA 1000 kVA
P1 S1 cos T1 700 kW
Q1 S1 sin T1 714.14 kVAR
For improved pf,
cos T 2 0.95 
o T 2
P2 P1 700 kW
S2
P2
cos T 2
Q2
S 2 sin T 2
700
0.95
18.19q
736.84 kVA
230.08 kVAR
0.8
P1 = P2 = 700 kW
T1
T2
Q2
S2
S1
Qc
Q1
(a)
Reactive power across the capacitor
Q c Q1 Q 2 714.14 230.08
484.06 kVAR
Cost of installing capacitors $30 u 484.06
(b)
$14,521.80
Substation capacity released S1 S 2
1000 736.84 263.16 kVA
Saving in cost of substation and distribution facilities
$120 u 263.16 $31,579.20
(c)
Yes, because (a) is greater than (b). Additional system capacity obtained
by using capacitors costs only 46% as much as new substation and
distribution facilities.
Chapter 11, Solution 95
(a)
Zs
ZL
Source impedance
Load impedance
R s jXc
R L jX 2
For maximum load transfer
Z L Z *s 
o R s
Xc
or
XL
Z

o
1
LC
1
ZC
2S f
R L , Xc
ZL
XL
f
1
2S LC
2S (80 u 10 -3 )(40 u 10 -9 )
Vs2
4RL
P
(b)
1
(4.6) 2
(4)(10)
529 mW
2.814 kHz
(since Vs is in rms)
Chapter 11, Solution 96
ZTh
+
VTh
(a)
VTh
Z Th
146 V, 300 Hz
40 j8 :
ZL
Z *Th
VTh
P
(b)
ZL
40 j8 :
2
(146) 2
(8)(40)
8 R Th
66.61 W
Chapter 11, Solution 97
ZT
I
P
(2)(0.1 j) (100 j20) 100.2 j22 :
Vs
ZT
240
100.2 j22
2
I RL
100 I
2
(100)(240) 2
(100.2) 2 (22) 2
547.3 W
Chapter 12, Solution 1.
(a)
If Vab
400 , then
Van
(b)
400
3
231‘ - 30q V
‘ - 30q
Vbn
231‘ - 150q V
Vcn
231‘ - 270q V
For the acb sequence,
Vab Van Vbn
Vab
Vp ‘0q Vp ‘120q
§ 1
3·
Vp ¨¨1 j ¸¸
2 ¹
© 2
Vp 3‘ - 30q
i.e. in the acb sequence, Vab lags Van by 30q.
Hence, if Vab
Van
400 , then
400
3
‘30q
231‘30q V
Vbn
231‘150q V
Vcn
231‘ - 90q V
Chapter 12, Solution 2.
Since phase c lags phase a by 120q, this is an acb sequence.
Vbn
160‘(30q 120q)
160‘150q V
Chapter 12, Solution 3.
Since Vbn leads Vcn by 120q, this is an abc sequence.
Van
208‘(130q 120q)
208‘ 250q V
Chapter 12, Solution 4.
Vbc
Vca ‘120q
208‘140q V
Vab
Vbc ‘120q
208‘260q V
Van
Vbn
208‘260q
Vab
3 ‘30q
3 ‘30q
120‘230q V
120‘110q V
Van ‘ - 120q
Chapter 12, Solution 5.
This is an abc phase sequence.
Vab
or
Van 3 ‘30q
Van
420‘0q
Vab
3 ‘30q
Vbn
Van ‘ - 120q
Vcn
Van ‘120q
3 ‘30q
242.5‘ - 30q V
242.5‘ - 150q V
242.5‘90q V
Chapter 12, Solution 6.
ZY
10 j5 11.18‘26.56q
The line currents are
Van
Ia
ZY
220 ‘0q
11.18‘26.56q
19.68‘ - 26.56q A
Ib
I a ‘ - 120q
19.68‘ - 146.56q A
Ic
I a ‘120q 19.68‘93.44q A
The line voltages are
Vab 200 3 ‘30q
Vbc
381‘ - 90q V
Vca
381‘ - 210q V
The load voltages are
VAN I a Z Y
Van
381‘30q V
220‘0q V
VBN
Vbn
220‘ - 120q V
VCN
Vcn
220‘120q V
Chapter 12, Solution 7.
This is a balanced Y-Y system.
440‘0q V
+
ZY = 6 j8 :
Using the per-phase circuit shown above,
440‘0q
Ia
44‘53.13q A
6 j8
I b I a ‘ - 120q 44‘ - 66.87q A
Ic
I a ‘120q
44‘173.13q A
Chapter 12, Solution 8.
VL
I an
IL
220 V ,
ZY
Vp
VL
ZY
3 ZY
6.918 A
16 j9 :
220
3 (16 j9)
6.918‘ - 29.36q
Chapter 12, Solution 9.
120 ‘0q
20 j15
Ia
Van
ZL ZY
Ib
I a ‘ - 120q
Ic
I a ‘120q
4.8‘ - 36.87q A
4.8‘ - 156.87q A
4.8‘83.13q A
As a balanced system, I n
0A
Chapter 12, Solution 10.
Since the neutral line is present, we can solve this problem on a per-phase basis.
For phase a,
Ia
Van
ZA 2
220 ‘0q
27 j20
Ib
Vbn
ZB 2
220 ‘ - 120q
10 ‘ - 120q
22
Ic
Vcn
ZC 2
220 ‘120q
16.92 ‘97.38q
12 j5
6.55‘36.53q
For phase b,
For phase c,
The current in the neutral line is
I n -(I a I b I c )
or
- In Ia Ib Ic
- In
In
(5.263 j3.9) (-5 j8.66) (-2.173 j16.78)
1.91 j12.02
12.17 ‘ - 81q A
Chapter 12, Solution 11.
Vbc
Van
3 ‘ - 90q
Van
127 ‘100q V
VAB
VBC ‘120q
VAC
VBC ‘ - 120q
If I bB
220‘ - 110q V
I cC
17.32‘90q ,
I AC
-I CA
17.32‘210q
3 ‘ - 30q
3 ‘ - 30q
I BC Z
30 ‘ - 60q
30‘180q
I BC
Z
220‘130q V
30 ‘60q , then
I aA
I AB
3 ‘ - 90q
3 ‘ - 90q
30‘180q ,
I aA
220‘10q
VBC
I CA
17.32 ‘ - 30q
17.32‘150q A
VBC
VBC
I BC
220 ‘0q
17.32 ‘90q
12.7 ‘ - 80q :
Chapter 12, Solution 12.
Convert the delta-load to a wye-load and apply per-phase analysis.
Ia
110‘0q V
ZY
Z'
3
+
20 ‘45q :
ZY
Ib
110‘0q
5.5‘ - 45q A
20‘45q
I a ‘ - 120q 5.5‘ - 165q A
Ic
I a ‘120q
Ia
5.5‘75q A
Chapter 12, Solution 13.
First we calculate the wye equivalent of the balanced load.
ZY = (1/3)Zǻ = 6+j5
Now we only need to calculate the line currents using the wye-wye circuits.
Ia
Ib
Ic
110
6.471‘ 61.93q A
2 j10 6 j5
110‘ 120q
6.471‘178.07q A
8 j15
110‘120q
6.471‘58.07q A
8 j15
Chapter 12, Solution 14.
We apply mesh analysis.
1 j 2:
A
a
+
100‘0 o V
-
ZL
ZL
I3
n
100‘120 o V
+
c
100‘120 o V
+
b
I2
B
I1
-
C
ZL
12 j12:
1 j 2:
1 j 2:
For mesh 1,
100 100‘120 o I 1 (14 j16) (1 j 2) I 2 (12 j12) I 3
or
0
(14 j16) I 1 (1 j 2) I 2 (12 j12) I 3 100 50 j86.6 150 j86.6 (1)
For mesh 2,
100‘120 o 100‘ 120 o I 1 (1 j 2) (12 j12) I 3 (14 j16) I 2 0
or
(1 j 2) I 1 (14 j16) I 2 (12 j12) I 3 50 j86.6 50 j86.6 j173.2 (2)
For mesh 3,
(12 j12) I 1 (12 j12) I 2 (36 j 36) I 3 0
(3)
Solving (1) to (3) gives
I1
3.161 j19.3,
I2
10.098 j16.749,
19.58‘ 99.3 A
o
I aA
I1
I bB
I 2 I1
I cC
I 2
7.392‘159.8 o A
19.56‘58.91o A
I3
4.4197 j12.016
Chapter 12, Solution 15.
Convert the delta load, Z ' , to its equivalent wye load.
Z Ye
Z'
3
8 j10
(12 j5)(8 j10)
20 j5
Zp
Z Y || Z Ye
Zp
7.812 j2.047
ZT
Zp ZL
ZT
8.874 ‘ - 6.78q
8.076‘ - 14.68q
8.812 j1.047
We now use the per-phase equivalent circuit.
Vp
Ia
,
where Vp
Zp ZL
Ia
IL
210
3 (8.874 ‘ - 6.78q)
Ia
13.66 ‘6.78q
13.66 A
Chapter 12, Solution 16.
(a)
I CA
- I AC
10‘(-30q 180q) 10‘150q
This implies that
I AB 10 ‘30q
I BC 10‘ - 90q
(b)
Ia
I AB 3 ‘ - 30q
Ib
17.32‘ - 120q A
Ic
17.32‘120q A
Z'
VAB
I AB
110 ‘0q
10 ‘30q
17.32‘0q A
11‘ - 30q :
210
3
Chapter 12, Solution 17.
Convert the '-connected load to a Y-connected load and use per-phase analysis.
ZL
Van
ZY
Ia
But
I AB
Z'
3
Ia
+
ZY
3 j4
Van
ZY ZL
120 ‘0q
(3 j4) (1 j0.5)
19.931‘ - 48.37q
I AB 3 ‘ - 30q
Ia
19.931‘ - 48.37q
3 ‘ - 30q
11.51‘ - 18.37q A
I BC
11.51‘ - 138.4q A
I CA
11.51‘101.6q A
VAB
VAB
I AB Z ' (11.51‘ - 18.37q)(15‘53.13q)
172.6‘34.76q V
VBC
172.6‘ - 85.24q V
VCA
172.6‘154.8q V
Chapter 12, Solution 18.
VAB
Z'
Van 3 ‘30q
(440 ‘60q)( 3 ‘30q)
12 j9 15‘36.87q
762.1‘90q
762.1‘90q
15‘36.87q
I AB
VAB
Z'
I BC
I AB ‘ - 120q
I CA
I AB ‘120q
50.81‘53.13q A
50.81‘ - 66.87q A
50.81‘173.13q A
Chapter 12, Solution 19.
Z'
30 j10
31.62 ‘18.43q
The phase currents are
Vab
173‘0q
I AB
5.47 ‘ - 18.43q A
Z ' 31.62 ‘18.43q
I BC I AB ‘ - 120q 5.47 ‘ - 138.43q A
I CA
I AB ‘120q
The line currents are
I a I AB I CA
5.47 ‘101.57q A
I AB 3 ‘ - 30q
Ia
5.47 3 ‘ - 48.43q
Ib
I a ‘ - 120q
Ic
I a ‘120q
9.474‘ - 48.43q A
9.474‘ - 168.43q A
9.474‘71.57q A
Chapter 12, Solution 20.
Z'
12 j9 15‘36.87q
The phase currents are
I BC
210‘0q
14‘ - 36.87q A
15‘36.87q
I AB ‘ - 120q 14‘ - 156.87q A
I CA
I AB ‘120q 14‘83.13q A
I AB
The line currents are
I a I AB 3 ‘ - 30q
Ib
I a ‘ - 120q
Ic
I a ‘120q
24.25‘ - 66.87q A
24.25‘ - 186.87q A
24.25‘53.13q A
Chapter 12, Solution 21.
(a)
I AC
I bB
(b)
230‘120q
10 j8
230‘120q
12.806‘38.66q
17.96‘ 98.66q A(rms)
230‘ 120 230‘0q
10 j8
10 j8
17.96‘ 158.66q 17.96‘ 38.66q
16.729 j6.536 14.024 j11.220 30.75 j4.684
31.10‘171.34q A
I BC I BA
I BC I AB
Chapter 12, Solution 22.
Convert the '-connected source to a Y-connected source.
Vp
208
Van
‘ - 30q
‘ - 30q 120 ‘ - 30q
3
3
Convert the '-connected load to a Y-connected load.
Z
(4 j6)(4 j5)
Z Z Y || ' (4 j6) || (4 j5)
3
8 j
Z
5.723 j0.2153
ZL
Van
Ia
+
Z
120 ‘30q
7.723 j0.2153
Ia
Van
ZL Z
Ib
I a ‘ - 120q
Ic
I a ‘120q 15.53‘91.6q A
15.53‘ - 28.4q A
15.53‘ - 148.4q A
Chapter 12, Solution 23.
(a) I AB
Ia
V AB
Z'
208
25‘60 o
I AB 3‘ 30
o
208 3‘ 30 o
25‘60 o
14.411‘ 90 o
I L | I a | 14.41 A
(b) P
P1 P2
3VL I L cosT
§ 208 3 ·
¸ cos 60 o
3 (208)¨¨
¸
© 25 ¹
2.596 kW
Chapter 12, Solution 24.
Convert both the source and the load to their wye equivalents.
Z'
20 ‘30q 17.32 j10
ZY
3
Van
Vab
3
‘ - 30q
240.2‘0q
We now use per-phase analysis.
1+j:
Van
Ia
+
20‘30q :
Ia
Van
(1 j) (17.32 j10)
Ib
I a ‘ - 120q
Ic
I a ‘120q 11.24‘89q A
240.2
21.37 ‘31q
11.24‘ - 151q A
11.24‘ - 31q A
But
I AB
Ia
I AB 3 ‘ - 30q
11.24 ‘ - 31q
3 ‘ - 30q
I BC
I AB ‘ - 120q
I CA
I AB ‘120q
6.489‘ - 1q A
6.489‘ - 121q A
6.489‘119q A
Chapter 12, Solution 25.
Convert the delta-connected source to an equivalent wye-connected source and
consider the single-phase equivalent.
Ia
where
ZY
Ia
440 ‘(10q 30q)
3 ZY
3 j2 10 j8 13 j6 14.32 ‘ - 24q.78q
440 ‘ - 20q
3 (14.32 ‘ - 24.78q)
17.74‘4.78q A
17.74‘ - 115.22q A
Ib
I a ‘ - 120q
Ic
I a ‘120q 17.74‘124.78q A
Chapter 12, Solution 26.
Transform the source to its wye equivalent.
Vp
Van
‘ - 30q 72.17 ‘ - 30q
3
Now, use the per-phase equivalent circuit.
Van
,
Z 24 j15
I aA
Z
28.3‘ - 32q
I aA
72.17 ‘ - 30q
28.3‘ - 32q
I bB
I aA ‘ - 120q
I cC
I aA ‘120q
2.55‘ 2q A
2.55‘ - 118q A
2.55‘122q A
Chapter 12, Solution 27.
Ia
Vab ‘ - 30q
220‘ - 10q
3 ZY
3 (20 j15)
Ia
5.081‘ - 46.87q A
Ib
I a ‘ - 120q
Ic
I a ‘120q
5.081‘ - 166.87q A
5.081‘73.13q A
Chapter 12, Solution 28.
Let
Vab
Ia
IL
400‘0q
Van ‘ - 30q
400‘ - 30q
3 ZY
3 (30 ‘ - 60q)
Ia
7.7 ‘30q
7.7 A
VAN
Ia ZY
Vp
VAN
Van
3
‘ - 30q
230.9 V
230.94‘ - 30q
Chapter 12, Solution 29.
P
3Vp I p cos T ,
VL
Vp
3
,
IL
Ip
3 VL I L cos T
P
P
IL
5000
3 VL cos T
ZY
cos T
240 3 (0.6)
Vp
VL
Ip
3 IL
240
3 (20.05)
20.05
Ip
6.911
0.6 
o T 53.13q
ZY
6.911‘ - 53.13q (leading)
ZY
4.15 j5.53 :
S
P
pf
5000
0.6
Q
S sin T
S
5000 j6667 VA
8333
6667
Chapter 12, Solution 30.
Since this a balanced system, we can replace it by a per-phase equivalent, as
shown below.
+
Vp
-
ZL
S
3S p
3V 2 p
,
Z*p
S
V 2L
Z*p
(208) 2
30‘ 45 o
P
S cosT
VL
Vp
3
1.4421‘45 o kVA
1.02 kW
Chapter 12, Solution 31.
(a)
cosT
Pp
6,000,
Qp
S P sin T
PP
cos T
Sp
0.8,
6 / 0 .8
7.5 kVA
4.5 kVAR
S 3S p 3(6 j 4.5) 18 j13.5 kVA
For delta-connected load, Vp = VL= 240 (rms). But
S
3V 2 p
Z*p
Z*p
3V 2 p
S

o
IL

o
3VL I L cosT
3(240) 2
,
(18 j13.5) x10 3
6000
(b)
Pp
(c )
We find C to bring the power factor to unity
Qc
Qp
4.5 kVA

o
C
Chapter 12, Solution 32.
S
S
IL
3 VL I L ‘T
S
3 VL I L
5000
3 (440)
50 u 10 3
65.61 A
3 x 240 x0.8
Qc
ZV 2 rms
ZP
6.144 j 4.608:
18.04 A
4500
2Sx60 x 240 2
207.2 PF
For a Y-connected load,
Ip
Vp
Z
Z
65.61 ,
IL
Vp
254.03
65.61
Ip
Z ‘T ,
VL
440
3
3
3.872
T
Z
(3.872)(cos T j sin T)
Z
(3.872)(0.6 j0.8)
Z
2.323 j3.098 :
cos -1 (0.6)
Chapter 12, Solution 33.
S
S
3 VL I L ‘T
S
3 VL I L
For a Y-connected load,
VL
IL Ip ,
S
IL
VL
3 Vp
3 Vp I p
Ip
S
3 Vp
3 Vp
254.03
4800
(3)(208)
3 u 208
7.69 A
360.3 V
53.13q
Chapter 12, Solution 34.
Vp
Ia
VL
220
3
3
Vp
200
6.73 A
IL
Ip
S
3 VL I L ‘T
S
6.73‘58q
3 (10 j16)
ZY
3 u 220 u 6.73‘ - 58q
1359 j2174.8 VA
Chapter 12, Solution 35.
(a) This is a balanced three-phase system and we can use per phase equivalent
circuit. The delta-connected load is converted to its wye-connected equivalent
1
Z'
3
Z '' y
(60 j 30) / 3
20 j10
IL
+
Z’’y
230 V
-
(b) S
Zy
Z ' y // Z '' y
IL
230
13.5 j 5.5
Vs I * L
(40 j10) //( 20 j10) 13.5 j 5.5
14.61 j 5.953 A
3.361 j1.368 kVA
(c ) pf = P/S = 0.9261
Z’’’’y
Chapter 12, Solution 36.
S = 1 [0.75 + sin(cos-10.75) ] =0.75 + 0.6614 MVA
(a)
(b) S

o
3V p I * p
I*p
S
3V p
PL | I p | 2 Rl
(c) Vs
VL I p (4 j )
(0.75 j 0.6614) x10 6
3x 4200
(79.36) 2 (4)
4.4381 j 0.21 kV 4.443‘ - 2.709 o kV
Chapter 12, Solution 37.
S
P
pf
S
S‘T
But
IL
S
12
0.6
20
20‘T 12 j16 kVA
3 VL I L ‘T
S
20 u 10 3
3 u 208
3 Ip
2
55.51 A
Zp
For a Y-connected load, I L
Zp
Zp
25.19 kW
S
3 IL
2
Ip .
(12 j16) u 10 3
(3)(55.51) 2
1.298 j1.731 :
59.52 j 52.49
Chapter 12, Solution 38.
As a balanced three-phase system, we can use the per-phase equivalent shown
below.
Ia
110‘0q
(1 j2) (9 j12)
Sp
1
I
2 a
2
110‘0q
10 j14
1
(110) 2
˜
˜ (9 j12)
2 (10 2 14 2 )
ZY
The complex power is
3 (110) 2
S 3S p
˜
˜ (9 j12)
2 296
551.86 j735.81 VA
S
Chapter 12, Solution 39.
Consider the system shown below.
5:
a
100‘120q
+
+
+
c
100‘-120q
100‘0q
A
5:
b
8:
B
I2
4:
-j6 :
I1
j3 :
I3
C
10 :
5:
For mesh 1,
100
(18 j6) I 1 5 I 2 (8 j6) I 3
(1)
For mesh 2,
100 ‘ - 120q 20 I 2 5 I 1 10 I 3
20‘ - 120q - I 1 4 I 2 2 I 3
(2)
For mesh 3,
0
- (8 j6) I 1 10 I 2 (22 j3) I 3
(3)
To eliminate I 2 , start by multiplying (1) by 2,
200 (36 j12) I 1 10 I 2 (16 j12) I 3
(4)
Subtracting (3) from (4),
200 (44 j18) I 1 (38 j15) I 3
(5)
Multiplying (2) by 5 4 ,
25‘ - 120q
(6)
-1.25 I 1 5 I 2 2.5 I 3
Adding (1) and (6),
87.5 j21.65
(16.75 j6) I 1 (10.5 j6) I 3
(7)
In matrix form, (5) and (7) become
º ª 44 j18 - 38 j15 ºª I 1 º
ª
200
«87.5 j12.65» «16.75 j6 - 10.5 j6 »« I »
¼¬ 3 ¼
¼ ¬
¬
'
192.5 j26.25 ,
'1
900.25 j935.2 ,
'3
110.3 j1327.6
I1
'1
'
1298.1‘ - 46.09q
194.28‘ - 7.76q
6.682 ‘ - 38.33q 5.242 j4.144
I3
'3
'
1332.2‘ - 85.25q
194.28‘ - 7.76q
6.857‘ - 77.49q 1.485 j6.694
We obtain I 2 from (6),
I2
1
1
5‘ - 120q I 1 I 3
4
2
I2
I2
(-2.5 j4.33) (1.3104 j1.0359) (0.7425 j3.347)
-0.4471 j8.713
The average power absorbed by the 8-: resistor is
P1
2
I 1 I 3 (8)
2
3.756 j2.551 (8) 164.89 W
The average power absorbed by the 4-: resistor is
P2
2
I 3 (4)
(6.8571) 2 (4) 188.1 W
The average power absorbed by the 10-: resistor is
2
2
P3 I 2 I 3 (10) - 1.9321 j2.019 (10)
78.12 W
Thus, the total real power absorbed by the load is
P P1 P2 P3 431.1 W
Chapter 12, Solution 40.
Transform the delta-connected load to its wye equivalent.
Z'
ZY
7 j8
3
Using the per-phase equivalent circuit above,
100 ‘0q
Ia
8.567 ‘ - 46.75q
(1 j0.5) (7 j8)
For a wye-connected load,
Ip Ia
Ia
2
S
3 Ip
P
Re(S)
8.567
Zp
(3)(8.567) 2 (7 j8)
(3)(8.567) 2 (7)
Chapter 12, Solution 41.
S
But
IL
5 kW
0.8
P
pf
S
S
3 VL
6.25 kVA
3 VL I L
6.25 u 10 3
3 u 400
9.021 A
1.541 kW
Chapter 12, Solution 42.
The load determines the power factor.
40
tan T
1.333 
o T 53.13q
30
cos T
pf
S
§ 7.2 ·
7.2 j¨ ¸(0.8)
© 0.6 ¹
But
Ip
0.6 (leading)
S
3 Ip
S
3Zp
2
Ip
8.944 A
IL
Ip
VL
2
7.2 j9.6 kVA
Zp
(7.2 j9.6) u 10 3
(3)(30 j40)
80
8.944 A
S
12 u 10 3
3 IL
3 (8.944)
774.6 V
Chapter 12, Solution 43.
2
Zp ,
I L for Y-connected loads
S
3 Ip
S
(3)(13.66) 2 (7.812 j2.047)
S
4.373 j1.145 kVA
Ip
Chapter 12, Solution 44.
For a '-connected load,
Vp VL ,
IL
S
3 Ip
3 VL I L
IL
S
(12 2 5 2 ) u 10 3
3 VL
3 (240)
31.273
At the source,
VL' VL I L Z L
VL' 240‘0q (31.273)(1 j3)
VL' 271.273 j93.819
VL'
287.04 V
Also, at the source,
S'
3VL' I *L
3 (271.273 j93.819)(31.273)
S'
T
§ 93.819 ·
¸ 19.078
tan -1 ¨
© 271.273 ¹
pf
cos T
0.9451
Chapter 12, Solution 45.
S
IL
IL
3 VL I L ‘T
S ‘-T
3 VL
,
(635.6) ‘ - T
3 u 440
S
P
pf
450 u 10 3
0.708
834 ‘ - 45q A
At the source,
VL 440 ‘0q I L (0.5 j2)
635.6 kVA
VL
VL
VL
VL
440 (834 ‘ - 45q)(2.062 ‘76q)
440 1719.7 ‘31q
1914.1 j885.7
2.109‘24.83q V
Chapter 12, Solution 46.
For the wye-connected load,
IL Ip ,
VL
3 Vp
S
S
3 Vp I *p
VL
2
Z*
3 Vp
2
3 VL
Z*
S
3V I
S
(3)(110) 2
100
2
121 W
For the delta-connected load,
Vp VL ,
IL
3 Ip ,
*
p p
3
Z*
(110) 2
100
3 Vp
Vp Z
Ip
2
Z*
3 VL
Ip
Vp Z
2
Z*
363 W
This shows that the delta-connected load will deliver three times more average
Z'
.
power than the wye-connected load. This is also evident from Z Y
3
Chapter 12, Solution 47.
pf
S1
0.8 (lagging) 
o T cos -1 (0.8)
250 ‘36.87q 200 j150 kVA
pf
S2
0.95 (leading) 
o T cos -1 (0.95)
300 ‘ - 18.19q 285 j93.65 kVA
36.87q
-18.19q
pf
S3
1.0 
o T
450 kVA
ST
S1 S 2 S 3
ST
cos -1 (1)
0q
935 j56.35
936.7 ‘3.45q kVA
3 VL I L
936.7 u 10 3
IL
39.19 A rms
3 (13.8 u 10 3 )
cos T
pf
cos(3.45q)
0.9982 (lagging)
Chapter 12, Solution 48.
(a) We first convert the delta load to its equivalent wye load, as shown below.
A
A
18-j12 :
ZA
40+j15 :
ZB
ZC
C
B
60 :
ZA
(40 j15)(18 j12)
118 j 3
ZB
60(40 j15).
118 j 3
20.52 j 7.105
ZC
60(18 j12)
118 j 3
8.992 j 6.3303
7.577 j1.923
The system becomes that shown below.
C
B
a
2+j3
A
+
240<0o
240<120
+
c
o
-
-
ZA
I1
240<-120
+
b
I2
ZB
o
ZC
2+j3
B
C
2+j3
We apply KVL to the loops. For mesh 1,
240 240‘ 120 o I 1 (2Z l Z A Z B ) I 2 ( Z B Z l )
or
0
(32.097 j11.13) I 1 (22.52 j10.105) I 2 360 j 207.85
For mesh 2,
240‘120 o 240‘ 120 o I 1 ( Z B Z l ) I 2 (2Z l Z B Z C )
or
(1)
0
(22.52 j10.105) I 1 (33.51 j 6.775) I 2 j 415.69
Solving (1) and (2) gives
I 1 23.75 j 5.328,
I 2 15.165 j11.89
(b)
24.34‘ 12.64 o A,
I bB
I 2 I1
10.81‘ 142.6 o A
I aA
I1
I cC
I 2
Sa
(240‘0 o )(24.34‘12.64 o )
Sb
(240‘ 120 o )(10.81‘142.6 o )
2594.4‘22.6 o
Sb
(240‘120 o )(19.27‘ 141.9 o )
4624.8‘ 21.9 o
S
Sa Sb Sc
19.27‘141.9 o A
5841.6‘12.64 o
12.386 j 0.55 kVA 12.4‘2.54 o kVA
(2)
Chapter 12, Solution 49.
(a) For the delta-connected load, Z p
3V 2 p
Z*p
S
3 x 220 2
(20 j10)
5808 j 2904
(b) For the wye-connected load, Z p
3V 2 p
Z*p
S
20 j10:,
3 x 220 2
3(20 j10)
Vp
VL
220 (rms) ,
6.943‘26.56 o kVA
20 j10:,
Vp
VL / 3 ,
2.164‘26.56 o kVA
Chapter 12, Solution 50.
S S1 S 2
Hence,
S2
S S1
But S 2
Z*p
8(0.6 j 0.8)
S1
3 kVA
1.8 j 6.4 kVA
3V 2 p
,
Z*p
V *L
S2
4.8 j 6.4 kVA,
VL
Vp
3
240 2
(1.8 j 6.4) x10 3

o

o
S2
Zp
.V 2 L
Z*p
2.346 j8.34:
Chapter 12, Solution 51.
Apply mesh analysis to the circuit as shown below.
Za
+
150‘120q
150‘0q
i1
Zb
+
n
+
150‘-120q
i2
Zc
For mesh 1,
- 150 (Z a Z b ) I 1 Z b I 2 0
150 (18 j) I 1 (12 j9) I 2
(1)
For mesh 2,
- 150 ‘ - 120q (Z b Z c ) I 2 Z b I 1 0
150 ‘ - 120q (27 j9) I 2 (12 j9) I 1
From (1) and (2),
º ª 18 j - 12 j9ºª I 1 º
ª
150
«150‘ - 120q» « - 12 j9 27 j9 »« I »
¼¬ 2 ¼
¼ ¬
¬
'
414 j27 ,
I1
'1
'
5209.5‘43.47q
12.56 ‘47.2q
414.88‘ - 3.73q
I2
'2
'
1211.1‘ - 61.39q
414.88‘ - 3.73q
Ia
I1
Ib
I 2 I1
Ib
5642.3‘235.44q
414.88‘ - 3.73q
Ic
-I2
'1
3780.9 j3583.8 ,
(2)
'2
579.9 j1063.2
2.919 ‘ - 57.66q
12.56‘47.2q A
' 2 '1
'
- 3201 j4647
'
13.6‘239.17q A
2.919‘122.34q A
Chapter 12, Solution 52.
Since the neutral line is present, we can solve this problem on a per-phase basis.
Van 120 ‘120q
6 ‘60q
Ia
20 ‘60q
Z AN
Ib
Vbn
Z BN
120 ‘0q
30 ‘0q
4 ‘0q
Ic
120 ‘ - 120q
40 ‘30q
Vcn
Z CN
3‘ - 150q
Thus,
- In
- In
- In
- In
In
Ia Ib Ic
6 ‘60q 4 ‘0q 3‘ - 150q
(3 j5.196) (4) (-2.598 j1.5)
4.405 j3.696 5.75‘40q
5.75‘ 220q A
Chapter 12, Solution 53.
Vp
250
3
Since we have the neutral line, we can use per-phase equivalent circuit for each
phase.
250‘0q
1
Ia
˜
3.608‘ - 60q A
40‘60q
3
Ib
Ic
- In
- In
In
250‘ - 120q
3
250‘120q
3
˜
˜
1
60‘ - 45q
1
20‘0q
2.406‘ - 75q A
7.217 ‘120q A
Ia Ib Ic
(1.804 j3.125) (0.6227 j2.324) (-3.609 j6.25)
1.1823 j0.801 1.428 ‘ - 34.12q A
Chapter 12, Solution 54.
Consider the circuit shown below.
Ia
a
Vp‘0q
+
IAB
Vp‘120q
+
A
50 :
Vp‘-120q
j50 :
+
c
b
B
C
-j50 :
VAB
Vab
100 u 3 ‘30q
I AB
VAB
Z AB
100 3 ‘30q
50
I BC
VBC
Z BC
100 3 ‘ - 90q
50‘ - 90q
3.464‘0q A
I CA
VCA
Z CA
100 3 ‘150q
50‘90q
3.464‘60q A
3.464‘ 30q A
Chapter 12, Solution 55.
Consider the circuit shown below.
Ia
a
220‘0q
220‘120q
+
+
A
60 + j80
I1
c
220‘-120q
+
b
100 –– j120
Ib
30 + j40
B
C
I2
Ic
For mesh 1,
220 ‘ - 120q 220 ‘0q (160 j40) I 1 (100 j120) I 2
11 11‘ - 120q (8 j2) I 1 (5 j6) I 2
For mesh 2,
220 ‘120q 220 ‘ - 120q (130 j80) I 2 (100 j120) I 1
11‘ - 120q 11‘120q - (5 j6) I 1 (6.5 j4) I 2
0
(1)
0
From (1) and (2),
ª16.5 j9.526º ª 8 j2 - 5 j6 ºª I 1 º
« - j19.053 » « - 5 j6 6.5 - j4 »« I »
¬
¼ ¬
¼¬ 2 ¼
'
55 j15 ,
I1
'1
'
I2
'2
'
Ia
I1
'1
31.04 j99.35 ,
'2
101.55 j203.8
104.08‘ - 72.65q
1.8257 ‘ - 87.91q
57.01‘15.26q
227.7 ‘ - 63.51q
57.01‘15.26q
1.8257 ‘ - 87.91q
3.994 ‘ - 78.77q
(2)
Ib
I 2 I1
Ic
-I2
SA
Ia
SB
Ib
SC
Ic
' 2 '1
'
2.211‘ - 71.23q
3.994‘101.23q
2
2
2
Z AN
(1.8257) 2 (60 j80) 199.99 j266.7
Z BN
(2.211) 2 (100 j120)
Z CN
(3.994) 2 (30 j40)
SA SB SC
S
70.51 j104.45
55 j15
488.9 j586.6
478.6 j638.1
1167.5 j318.2 VA
Chapter 12, Solution 56.
(a)
Consider the circuit below.
a
A
440‘0q + 440‘120q
+
j10 :
I1
b
B
+
440‘-120q
I2
I3
-j5 :
20 :
C
c
For mesh 1,
440‘ - 120q 440‘0q j10 (I 1 I 3 )
I1 I 3
(440)(1.5 j0.866)
j10
76.21‘ - 60q
For mesh 2,
440‘120q 440‘ - 120q 20 (I 2 I 3 )
I3 I2
(440)( j1.732)
20
0
j38.1
For mesh 3,
j10 (I 3 I 1 ) 20 (I 3 I 2 ) j5 I 3
(1)
0
(2)
0
Substituting (1) and (2) into the equation for mesh 3 gives,
(440)(-1.5 j0.866)
I3
152.42‘60q
j5
(b)
From (1),
I1
I 3 76.21‘ - 60q 114.315 j66 132‘30q
From (2),
I2
I 3 j38.1 76.21 j93.9 120.93‘50.94q
132‘30q A
Ia
I1
Ib
I 2 I1
Ic
-I2
-38.105 j27.9
120.9‘230.9q A
2
S AB
I 1 I 3 ( j10)
S BC
I 2 I 3 (20)
S CA
I 3 (-j5)
S
47.23‘143.8q A
2
2
j58.08 kVA
29.04 kVA
(152.42) 2 (-j5)
S AB S BC S CA
-j116.16 kVA
29.04 j58.08 kVA
Real power absorbed = 29.04 kW
(c)
Total complex supplied by the source is
S 29.04 j58.08 kVA
(3)
Chapter 12, Solution 57.
We apply mesh analysis to the circuit shown below.
Ia
+
Va
-
80 j 50:
I1
-
20 j 30:
-
Vc
+
Vb
+
I2
Ib
Ic
(100 j80) I 1 (20 j 30) I 2 Va Vb 165 j 95.263
(20 j 30) I 1 (80 j10) I 2 Vb Vc j190.53
Solving (1) and (2) gives I 1 1.8616 j 0.6084,
I2
Ia
I1
1.9585‘ 18.1o A,
Ic
I 2
1.947‘117.8 o A
60 j 40:
Ib
I 2 I1
(1)
(2)
0.9088 j1.722 .
0.528 j1.1136 1.4656‘ 130.55 o A
Chapter 12, Solution 58.
The schematic is shown below. IPRINT is inserted in the neutral line to measure the
current through the line. In the AC Sweep box, we select Total Ptss = 1, Start Freq. =
0.1592, and End Freq. = 0.1592. After simulation, the output file includes
FREQ
IM(V_PRINT4)
IP(V_PRINT4)
1.592 E––01
1.078 E+01
––8.997 E+01
i.e.
In = 10.78‘––89.97q A
Chapter 12, Solution 59.
The schematic is shown below. In the AC Sweep box, we set Total Pts = 1, Start Freq
= 60, and End Freq = 60. After simulation, we obtain an output file which includes
i.e.
FREQ
VM(1)
VP(1)
6.000 E+01
2.206 E+02
––3.456 E+01
FREQ
VM(2)
VP(2)
6.000 E+01
2.141 E+02
––8.149 E+01
FREQ
VM(3)
VP(3)
6.000 E+01
4.991 E+01
––5.059 E+01
VAN = 220.6‘––34.56q, VBN = 214.1‘––81.49q, VCN = 49.91‘––50.59q V
Chapter 12, Solution 60.
The schematic is shown below. IPRINT is inserted to give Io. We select Total Pts = 1,
Start Freq = 0.1592, and End Freq = 0.1592 in the AC Sweep box. Upon simulation,
the output file includes
from which,
FREQ
IM(V_PRINT4)
IP(V_PRINT4)
1.592 E––01
1.421 E+00
––1.355 E+02
Io = 1.421‘––135.5q A
Chapter 12, Solution 61.
The schematic is shown below. Pseudocomponents IPRINT and PRINT are inserted to
measure IaA and VBN. In the AC Sweep box, we set Total Pts = 1, Start Freq = 0.1592,
and End Freq = 0.1592. Once the circuit is simulated, we get an output file which
includes
FREQ
VM(2)
VP(2)
1.592 E––01
2.308 E+02
––1.334 E+02
FREQ
IM(V_PRINT2)
IP(V_PRINT2)
1.592 E––01
1.115 E+01
3.699 E+01
from which
IaA = 11.15‘37q A, VBN = 230.8‘––133.4q V
Chapter 12, Solution 62.
Because of the delta-connected source involved, we follow Example 12.12. In the AC
Sweep box, we type Total Pts = 1, Start Freq = 60, and End Freq = 60. After
simulation, the output file includes
From which
FREQ
IM(V_PRINT2)
IP(V_PRINT2)
6.000 E+01
5.960 E+00
––9.141 E+01
FREQ
IM(V_PRINT1)
IP(V_PRINT1)
6.000 E+01
7.333 E+07
1.200 E+02
Iab = 7.333x107‘120q A, IbB = 5.96‘––91.41q A
Chapter 12, Solution 63.
Let Z
1 so that L X/Z
20 H, and C
1
ZX
0.0333 F
The schematic is shown below..
.
When the file is saved and run, we obtain an output file which includes the following:
FREQ
1.592E-01
FREQ
1.592E-01
IM(V_PRINT1)IP(V_PRINT1)
1.867E+01
1.589E+02
IM(V_PRINT2)IP(V_PRINT2)
1.238E+01
1.441E+02
From the output file, the required currents are:
I aA
18.67‘158.9 o A, I AC
12.38‘144.1o A
Chapter 12, Solution 64.
We follow Example 12.12. In the AC Sweep box we type Total Pts = 1, Start Freq =
0.1592, and End Freq = 0.1592. After simulation the output file includes
FREQ
IM(V_PRINT1)
IP(V_PRINT1)
1.592 E––01
4.710 E+00
7.138 E+01
FREQ
IM(V_PRINT2)
IP(V_PRINT2)
1.592 E––01
6.781 E+07
––1.426 E+02
FREQ
IM(V_PRINT3)
IP(V_PRINT3)
1.592 E––01
3.898 E+00
––5.076 E+00
FREQ
IM(V_PRINT4)
IP(V_PRINT4)
1.592 E––01
3.547 E+00
6.157 E+01
FREQ
IM(V_PRINT5)
IP(V_PRINT5)
1.592 E––01
1.357 E+00
9.781 E+01
FREQ
IM(V_PRINT6)
IP(V_PRINT6)
1.592 E––01
3.831 E+00
––1.649 E+02
from this we obtain
IaA = 4.71‘71.38q A, IbB = 6.781‘––142.6q A, IcC = 3.898‘––5.08q A
IAB = 3.547‘61.57q A, IAC = 1.357‘97.81q A, IBC = 3.831‘––164.9q A
Chapter 12, Solution 65.
Due to the delta-connected source, we follow Example 12.12. We type Total Pts = 1,
Start Freq = 0.1592, and End Freq = 0.1592. The schematic is shown below. After it
is saved and simulated, we obtain an output file which includes
Thus,
FREQ
IM(V_PRINT1)
IP(V_PRINT1)
1.592 E––01
6.581 E+00
9.866 E+01
FREQ
IM(V_PRINT2)
IP(V_PRINT2)
1.592 E––01
1.140 E+01
––1.113 E+02
FREQ
IM(V_PRINT3)
IP(V_PRINT3)
1.592 E––01
6.581 E+00
3.866 E+01
IaA = 6.581‘98.66q A, IbB = 11.4‘––111.3 A, IcC = 6.581‘38.66q A
Chapter 12, Solution 66.
VL
208
3
3
120 V
(a)
Vp
(b)
Because the load is unbalanced, we have an unbalanced three-phase
system. Assuming an abc sequence,
I1
120 ‘0q
48
I2
120‘ - 120q
40
I3
120‘120q
60
- IN
2.5‘0q A
I1 I 2 I 3
3‘ - 120q A
2‘120q A
§
§
3·
3·
2.5 (3) ¨¨ - 0.5 j ¸¸ (2) ¨¨ - 0.5 j ¸¸
2 ¹
2 ¹
©
©
IN
3
2
j0.866
2.5 A ,
I2
j
0.866‘90q A
Hence,
I1
(c)
(d)
P1
I12 R 1
2
2
3 A,
(2.5) 2 (48)
2
I3
2 A,
300 W
P2
I R2
(3) (40)
360 W
P3
I 32 R 3
(2) 2 (60)
240 W
PT
P1 P2 P3
900 W
Chapter 12, Solution 67.
(a)
The power to the motor is
PT S cos T (260)(0.85)
221 kW
The motor power per phase is
1
P
73.67 kW
Pp
3 T
Hence, the wattmeter readings are as follows:
Wa 73.67 24 97.67 kW
(b)
Wb
73.67 15
Wc
73.67 9
88.67 kW
83.67 kW
The motor load is balanced so that I N
For the lighting loads,
Ia
24,000
120
200 A
Ib
15,000
120
125 A
Ic
9,000
120
If we let
75 A
0.
IN
0.866 A
I a ‘0q 200‘0q A
125‘ - 120q A
75‘120q A
Ia
Ib
Ic
Then,
- IN
Ia Ib Ic
- IN
§
§
3·
3·
200 (125)¨¨ - 0.5 j ¸¸ (75)¨¨ - 0.5 j ¸¸
2 ¹
2 ¹
©
©
- IN
100 86.602 A
IN
132.3 A
Chapter 12, Solution 68.
(a)
S
(b)
P
pf
3 VL I L
3 (330)(8.4)
S cos T 
o pf
4500
4801.24
cos T
4801 VA
P
S
0.9372
(c)
For a wye-connected load,
I p I L 8.4 A
(d)
Vp
VL
330
3
3
190.53 V
Chapter 12, Solution 69.
S1
1.2(0.8 j 0.6)
0.96 j 0.72 MVA,
S
S1 S 2 S 3
Qc
P(tan old tan new )
C
S2
3.26 j 0.603 MVA,
1
x0.1379 x10 6
3
2Sx60 x6.6 2 x10 6
2(0.75 j 0.661) 1.5 1.323 MVA,
pf
P
S
3.26
3.3153
3.26[tan(cos 1 0.9833) tan(cos 1 0.99)
28 mF
S3
0.9833
0.1379 MVA
0.8 MVA
Chapter 12, Solution 70.
PT
P1 P2
QT
P2 P1
tan T
1200 400
800
-400 1200
QT
PT
- 1600
800
-1600
-2 
o T
pf
cos T
Zp
VL
IL
Zp
40 ‘ - 63.43q :
-63.43q
0.4472 (leading)
240
6
40
Chapter 12, Solution 71.
(a)
If Vab
208‘0q , Vbc
208‘0q
20
208‘ - 120q , Vca
I AB
Vab
Z Ab
I BC
Vbc
Z BC
10 2 ‘ - 45q
I CA
Vca
Z CA
208‘120q
16 ‘97.38q
13‘22.62q
208‘120q ,
10.4 ‘0q
208‘ - 120q
14.708‘ - 75q
I aA
I aA
I aA
I AB I CA 10.4‘0q 16‘97.38q
10.4 2.055 j15.867
20.171‘ - 51.87q
I cC
I cC
I CA I BC 16‘97.83q 14.708‘ - 75q
30.64 ‘101.03q
P1
Vab I aA cos(T Vab TIaA )
P1
P2
But
(b)
(208)(20.171) cos(0q 51.87q)
2590 W
Vcb I cC cos(T Vcb T IcC )
Vcb
208‘60q
-Vbc
P2
(208)(30.64) cos(60q 101.03q)
PT
P1 P2
QT
ST
3 (P2 P1 ) 3840.25 VAR
PT jQ T 7398.17 j3840.25 VA
ST
ST
4808 W
7398.17 W
8335 VA
Chapter 12, Solution 72.
From Problem 12.11,
VAB 220 ‘130q V
P1
I aA
and
(220)(30) cos(130q 180q)
30‘180q A
4242 W
VCB -VBC 220‘190q
I cC 30‘ - 60q
P2
(220)(30) cos(190q 60q)
- 2257 W
Chapter 12, Solution 73.
Consider the circuit as shown below.
I1
Ia
240‘-60q V
+
Z
Z
240‘-120q V
Z
+
I2
Ib
Ic
Z 10 j30
31.62‘71.57q
Ia
240‘ - 60q
31.62‘71.57q
7.59‘ - 131.57q
Ib
240 ‘ - 120q
31.62‘71.57q
7.59‘ - 191.57q
I c Z 240‘ - 60q 240 ‘ - 120q
Ic
- 240
31.62‘71.57q
I1
I2
I a I c 13.146‘ - 101.57q
I b I c 13.146‘138.43q
P1
Re > V1 I 1*
P2
Re > V2 I *2
@
0
7.59‘108.43q
Re > (240‘ - 60q)(13.146 ‘101.57q) @
@
Re > (240 ‘ - 120q)(13.146‘ - 138.43q) @
Chapter 12, Solution 74.
Consider the circuit shown below.
Z = 60 j30 :
208‘0q V
208‘-60q V
+
I1
+
I2
Z
Z
For mesh 1,
208
2 Z I1 Z I 2
For mesh 2,
- 208‘ - 60q
2360 W
- Z I1 2 Z I 2
- 632.8 W
In matrix form,
ª
º ª 2 Z - Z ºª I 1 º
208
« - 208‘ - 60q» « - Z 2 Z »« I »
¬
¼ ¬
¼¬ 2 ¼
'1
(208)(1.5 j0.866) Z ,
'
3Z 2 ,
I1
'1
'
(208)(1.5 j0.866)
(3)(60 j30)
I2
'2
'
(208)( j1.732)
(3)(60 j30)
P1
Re > V1 I 1*
P2
Re > V2 (- I 2 ) *
@
'2
1.789‘56.56q
1.79‘116.56q
Re > (208)(1.789‘ - 56.56q) @
@
(208)( j1.732) Z
208.98 W
Re > (208‘ - 60q))(1.79‘63.44q) @
371.65 W
Chapter 12, Solution 75.
(a)
I
V
R
12
600
20 mA
(b)
I
V
R
120
600
200 mA
Chapter 12, Solution 76.
If both appliances have the same power rating, P,
P
I
Vs
P
For the 120-V appliance,
I1
.
120
P
For the 240-V appliance,
I2
.
240
2
Power loss = I R
Since
­ P2 R
° 120 2
® 2
°P R
¯ 240 2
for the 120-V appliance
for the 240-V appliance
1
1
, the losses in the 120-V appliance are higher.
2 !
120
240 2
Chapter 12, Solution 77.
Pg
But
Pg
PT Pload Pline ,
PT
pf
0.85
3600 cos T 3600 u pf
3060 2500 (3)(80)
3060
320 W
Chapter 12, Solution 78.
51
60
0.85 
o T1
Q1
S1 sin T1
(60)(0.5268)
P2
P1
cos T1
cos T 2
31.79q
31.61 kVAR
51 kW
0.95 
o T 2
18.19q
S2
P2
cos T 2
Q2
S 2 sin T 2
16.759 kVAR
Qc
Q1 Q 2
3.61 16.759 14.851 kVAR
53.68 kVA
For each load,
Q c1
C
Qc
3
Q c1
ZV2
4.95 kVAR
4950
(2S )(60)(440) 2
67.82 PF
Chapter 12, Solution 79.
Consider the per-phase equivalent circuit below.
2:
Ia
a
A
+
Van
ZY = 12 + j5 :
n
N
255‘0q
14 j5
Ia
Van
ZY 2
Ib
I a ‘ - 120q
Ic
I a ‘120q 17.15‘100.35q A
17.15‘ - 19.65q A
Thus,
17.15‘ - 139.65q A
(17.15‘ - 19.65q)(13‘22.62q)
VAN
Ia ZY
VBN
VAN ‘ - 120q
VCN
VAN ‘120q
223‘2.97q V
Thus,
223‘ - 117.63q V
223‘122.97q V
Chapter 12, Solution 80.
S
S
But
S
S1 S 2 S 3 6[0.83 j sin(cos 1 0.83)] S 2 8(0.7071 j 0.7071)
10.6368 j 2.31 S 2 kVA
3VL I L ‘T
3 (208)(84.6)(0.8 j 0.6) VA 24.383 j18.287 kVA
From (1) and (2),
S 2 13.746 j 20.6 24.76‘56.28 kVA
Thus, the unknown load is 24.76 kVA at 0.5551 pf lagging.
(1)
(2)
Chapter 12, Solution 81.
pf
S1
0.8 (leading) 
o T1
150 ‘ - 36.87q kVA
pf
S2
1.0 
o T 2
100 ‘0q kVA
pf
0.6 (lagging) 
o T3
S3
200‘53.13q kVA
S4
80 j95 kVA
S
S
-36.87q
0q
S1 S 2 S 3 S 4
420 j165 451.2‘21.45q kVA
S
3 VL I L
SL
17.67 j44.18 kVA
At the source,
S T S S L 437.7 j209.2
S T 485.1‘25.55q kVA
VT
53.13q
ST
3 IL
485.1 u 10 3
3 u 542.7
516 V
Chapter 12, Solution 82.
S1
400(0.8 j 0.6)
320 j 240 kVA,
For the delta-connected load, V L
S2
(2400) 2
3x
10 j8
S2
3
V 2p
Z*p
Vp
1053.7 j842.93 kVA
S S 1 S 2 1.3737 j1.0829 MVA
Let I = I1 + I2 be the total line current. For I1,
S1
3V p I *1 ,
I *1
S1
VL
Vp
3
(320 j 240) x10 3
3VL
I1
76.98 j 57.735
2400
3‘ 30 o
10 j8
273.1 j 289.76
3 (2400)
,
For I2, convert the load to wye.
I2
I
Vs
I p 3‘ 30 o
I1 I 2
350 j 347.5
VL Vline
2400 I (3 j 6)
5.185 j1.405 kV

o
| Vs | 5.372 kV
Chapter 12, Solution 83.
S1
120 x746 x0.95(0.707 j 0.707)
S
S1 S 2
But | S |
60.135 j 60.135 kVA,
S2
140.135 j 60.135 kVA
3VL I L

o
IL
|S|
3VL
152.49 x10 3
3 x 480
183.42 A
80 kVA
Chapter 12, Solution 84.
We first find the magnitude of the various currents.
For the motor,
IL
S
4000
3 VL
440 3
For the capacitor,
Q c 1800
IC
VL
440
For the lighting,
440
Vp
3
I Li
PLi
Vp
5.248 A
4.091 A
254 V
800
254
3.15 A
Consider the figure below.
Ia
a
IC
+
Vab
b
I1
-jXC
Ib
I2
Ic
I3
c
ILi
R
In
n
If Van
Vp ‘0q ,
Vcn
Vab
3 Vp ‘30q
Vp ‘120q
IC
Vab
-j X C
I1
Vab
Z'
where T
4.091‘120q
4.091‘(T 30q)
cos -1 (0.72)
I1
5.249 ‘73.95q
I2
5.249 ‘ - 46.05q
I3
5.249‘193.95q
I Li
Vcn
R
43.95q
3.15‘120q
Thus,
Ia
Ia
I 1 I C 5.249‘73.95q 4.091‘120q
8.608‘93.96q A
Ib
Ib
I 2 I C 5.249‘ - 46.05q 4.091‘120q
9.271‘ - 52.16q A
Ic
Ic
I 3 I Li 5.249‘193.95q 3.15‘120q
6.827 ‘167.6q A
In
- I Li
3.15‘ - 60q A
Chapter 12, Solution 85.
Let
ZY
Vp
P
R
VL
240
3
3
Vp I p
27
2
138.56 V
9 kW
Vp2
R
R
Vp2
(138.56) 2
9000
P
Thus,
2.133 :
2.133 :
ZY
Chapter 12, Solution 86.
Consider the circuit shown below.
1:
a
A
+
120‘0q V rms
I1
24 –– j2 :
1:
n
N
I2
+
120‘0q V rms
15 + j4 :
1:
b
B
For the two meshes,
120 (26 j2) I 1 I 2
120 (17 j4) I 2 I 1
(1)
(2)
In matrix form,
ª120º ª 26 j2
- 1 ºª I 1 º
«120» « - 1
17 j4»¼«¬ I 2 »¼
¬ ¼ ¬
'
449 j70 ,
I1
'1
'
120 u 18.44 ‘12.53q
454.42 ‘8.86q
I2
'2
'
120 u 27.07 ‘ - 4.24q
454.42 ‘8.86q
'1
(120)(18 j4) ,
'2
(120)(27 j2)
4.87 ‘3.67q
7.15‘ - 13.1q
4.87 ‘3.67q A
I aA
I1
I bB
-I2
I nN
I 2 I1
I nN
(120)(9 j6)
449 j70
7.15‘166.9q A
' 2 '1
'
2.856‘ - 42.55q A
Chapter 12, Solution 87.
L 50 mH 
o jZL
Consider the circuit below.
j (2S)(60)(5010 -3 )
j18.85
1:
115 V
+
I1
20 :
2:
15 + j18.85 :
115 V
+
I2
30 :
1:
Applying KVl to the three meshes, we obtain
23 I 1 2 I 2 20 I 3 115
- 2 I 1 33 I 2 30 I 3 115
- 20 I 1 30 I 2 (65 j18.85) I 3
In matrix form,
- 20
ª 23 - 2
º ª I1 º
« - 2 33
» «I »
- 30
«
»« 2»
«¬- 20 - 30 65 j18.85»¼ «¬I 3 »¼
' 12,775 j14,232 ,
' 2 (115)(1825 j471.3) ,
(1)
(2)
(3)
0
ª115º
«115»
« »
«¬ 0 »¼
'1
'3
(115)(1975 j659.8)
(115)(1450)
I1
'1
'
115 u 2082‘18.47q
19214‘48.09q
I2
'2
'
115 u 1884.9 ‘14.48q
11.33‘ - 33.61q
19124 ‘48.09q
In
I 2 I1
S1
V1 I *1
(115)(12.52‘ 29.62q)
S2
V2 I *2
(115)(1.33‘33.61q)
' 2 '1
'
12.52‘ - 29.62q
(115)(-150 j188.5)
12,775 j14,231.75
1.448‘ - 176.6q A
1252 j711.6 VA
1085 j721.2 VA
Chapter 13, Solution 1.
For coil 1, L1 –– M12 + M13 = 6 –– 4 + 2 = 4
For coil 2, L2 –– M21 –– M23 = 8 –– 4 –– 5 = –– 1
For coil 3, L3 + M31 –– M32 = 10 + 2 –– 5 = 7
LT = 4 –– 1 + 7 = 10H
or
LT = L1 + L2 + L3 –– 2M12 –– 2M23 + 2M12
LT = 6 + 8 + 10 = 10H
Chapter 13, Solution 2.
L = L1 + L2 + L3 + 2M12 –– 2M23 2M31
= 10 + 12 +8 + 2x6 –– 2x6 ––2x4
= 22H
Chapter 13, Solution 3.
L1 + L2 + 2M = 250 mH
(1)
L1 + L2 –– 2M = 150 mH
(2)
Adding (1) and (2),
2L1 + 2L2 = 400 mH
But,
L1 = 3L2,, or 8L2 + 400,
and L2 = 50 mH
L1 = 3L2 = 150 mH
From (2),
150 + 50 –– 2M = 150 leads to M = 25 mH
k = M/ L1 L 2
2.5 / 50x150 = 0.2887
Chapter 13, Solution 4.
(a)
For the series connection shown in Figure (a), the current I enters each coil from
its dotted terminal. Therefore, the mutually induced voltages have the same sign as the
self-induced voltages. Thus,
Leq = L1 + L2 + 2M
Is
L1
I1
Vs
L2
I2
+
––
L1
L2
Leq
(a)
(b)
(b)
For the parallel coil, consider Figure (b).
I s = I 1 + I2
and
Zeq = Vs/Is
Applying KVL to each branch gives,
Vs = jZL1I1 + jZMI2
(1)
Vs = jZMI1 + jZ L2I2
(2)
ª Vs º
«V »
¬ s¼
or
ª jZL1
« jZM
¬
jZM º ª I1 º
jZL 2 »¼ «¬I 2 »¼
' = ––Z2L1L2 + Z2M2, '1 = jZVs(L2 –– M), '2 = jZVs(L1 –– M)
I1 = '1/', and I2 = '2/'
Is = I1 + I2 = ('1 + '2)/' = jZ(L1 + L2 –– 2M)Vs/( ––Z2(L1L2 –– M))
Zeq = Vs/Is = jZ(L1L2 –– M)/[jZ(L1 + L2 –– 2M)] = jZLeq
i.e.,
Leq = (L1L2 –– M)/(L1 + L2 –– 2M)
Chapter 13, Solution 5.
(a) If the coils are connected in series,
L
L1 L 2 2M
25 60 2(0.5) 25x 60
123.7 mH
(b) If they are connected in parallel,
L
L1 L 2 M 2
L1 L 2 2M
25x 60 19.36 2
mH
25 60 2x19.36
24.31 mH
Chapter 13, Solution 6.
V1 = (R1 + jZL1)I1 –– jZMI2
V2 = ––jZMI1 + (R2 + jZL2)I2
Chapter 13, Solution 7.
Applying KVL to the loop,
20‘30q = I(––j6 + j8 + j12 + 10 –– j4x2) = I(10 + j6)
where I is the loop current.
I = 20‘30q/(10 + j6)
Vo = I(j12 + 10 –– j4) = I(10 + j8)
= 20‘30q(10 + j8)/(10 + j6) = 22‘37.66q V
Chapter 13, Solution 8.
Consider the current as shown below.
j2
1:
10
+
––
I1
4:
j6
+
j4
I2
-j3
Vo
––
For mesh 1,
10 = (1 + j6)I1 + j2I2
For mesh 2,
(1)
0 = (4 + j4 –– j3)I2 + j2I1
0 = j2I1 +(4 + j)I2
(2)
In matrix form,
ª10º
«0»
¬ ¼
j2
ª1 j6
« j2
4
¬
º ª I1 º
j»¼ «¬ I 2 »¼
' = 2 + j25, and '2 = ––j20
I2 = '2/' = ––j20/(2 + j25)
Vo = ––j3I2 = ––60/(2 + j25) = 2.392‘94.57q
Chapter 13, Solution 9.
Consider the circuit below.
2:
8‘30o
+
––
I1
2:
j4
j4
I2
-j1
-j2V
+
––
For loop 1,
8‘30q = (2 + j4)I1 –– jI2
For loop 2,
((j4 + 2 –– j)I2 –– jI1 + (––j2) = 0
or
Substituting (2) into (1),
(1)
I1 = (3 –– j2)i2 –– 2
8‘30q + (2 + j4)2 = (14 + j7)I2
I2 = (10.928 + j12)/(14 + j7) = 1.037‘21.12q
Vx = 2I2 = 2.074‘21.12q
(2)
Chapter 13, Solution 10.
Consider the circuit below.
jZM
jZL
jZL
Io
I1
Iin‘0o
I2
1/jZC
M k L1 L 2
L2 = L, I1 = Iin‘0q, I2 = Io
Io(jZL + R + 1/(jZC)) –– jZLIin –– (1/(jZC))Iin = 0
Io = j Iin(ZL –– 1/(ZC)) /(R + jZL + 1/(jZC))
Chapter 13, Solution 11.
Consider the circuit below.
V2
R2
+ ––
I3
jZL1
R1
V1
+
––
I1
1/jZC
jZM
jZL2
I2
For mesh 1, V1 = I1(R1 + 1/(jZC)) –– I2(1/jZC)) ––R1I3
For mesh 2,
0 = ––I1(1/(jZC)) + (jZL1 + jZL2 + (1/(jZC)) –– j2ZM)I2 –– jZL1I3 + jZMI3
For mesh 3,
or
––V2 = ––R1I1 –– jZ(L1 –– M)I2 + (R1 + R2 + jZL1)I3
V2 = R1I1 + jZ(L1 –– M)I2 –– (R1 + R2 + jZL1)I3
Chapter 13, Solution 12.
Let Z
j4
1.
j2
+
j6
1V
-
x
j8
j10
I1
I2
x
Applying KVL to the loops,
1 j8 I 1 j 4 I 2
(1)
0 j 4 I 1 j18 I 2
(2)
Solving (1) and (2) gives I1 = -j0.1406. Thus
Z
1
I1
jLeq

o
Leq
1
jI 1
7.111 H
We can also use the equivalent T-section for the transform to find the equivalent
inductance.
Chapter 13, Solution 13.
We replace the coupled inductance with an equivalent T-section and use series and
parallel combinations to calculate Z. Assuming that Z 1,
La L1 M 18 10 8,
Lb L2 M
The equivalent circuit is shown below:
20 10 10,
Lc
M
10
12 :
j8 :
j10 :
2:
j10 :
-j6 :
Z
j4 :
Z=12 +j8 + j14//(2 + j4) = 13.195 + j11.244ȍ
Chapter 13, Solution 14.
To obtain VTh, convert the current source to a voltage source as shown below.
j2
5:
j6 :
j8 :
-j3 :
2:
a
j10 V
+
––
+
VTh
I
––
b
Note that the two coils are connected series aiding.
ZL = ZL1 + ZL2 –– 2ZM
jZL = j6 + j8 –– j4 = j10
Thus,
––j10 + (5 + j10 –– j3 + 2)I + 8 = 0
I = (–– 8 + j10)/ (7 + j7)
But,
––j10 + (5 + j6)I –– j2I + VTh = 0
8V
+
––
VTh = j10 –– (5 + j4)I = j10 –– (5 + j4)(––8 + j10)/(7 + j7)
VTh = 5.349‘34.11q
To obtain ZTh, we set all the sources to zero and insert a 1-A current source at the terminals
a––b as shown below.
j2
5:
j6 :
I1
a
j8 :
-j3 :
+
Vo
1A
2:
I2
––
b
Clearly, we now have only a super mesh to analyze.
(5 + j6)I1 –– j2I2 + (2 + j8 –– j3)I2 –– j2I1 = 0
(5 + j4)I1 + (2 + j3)I2 = 0
(1)
But,
I2 –– I1 = 1 or I2 = I1 –– 1
(2)
Substituting (2) into (1),
(5 + j4)I1 +(2 + j3)(1 + I1) = 0
I1 = ––(2 + j3)/(7 + j7)
Now,
((5 + j6)I1 –– j2I1 + Vo = 0
Vo = ––(5 + j4)I1 = (5 + j4)(2 + j3)/(7 + j7) = (––2 + j23)/(7 + j7) = 2.332‘50q
ZTh = Vo/1 = 2.332‘50q ohms
Chapter 13, Solution 15.
To obtain IN, short-circuit a––b as shown in Figure (a).
20 :
j20 :
20 :
a
j5
+
––
I1
j20 :
j5
j10 :
j10 :
60‘30o
1
+
––
IN
I1
I2
I2
b
(a)
For mesh 1,
a
b
(b)
60‘30q = (20 + j10)I1 + j5I2 –– j10I2
or
For mesh 2,
12‘30q = (4 + j2)I1 –– jI2
(1)
0 = (j20 + j10)I2 –– j5I1 –– j10I1
or
I1 = 2I2
(2)
12‘30q = (8 + j3)I2
Substituting (2) into (1),
IN = I2 = 12‘30q/(8 + j3) = 1.404‘9.44q A
To find ZN, we set all the sources to zero and insert a 1-volt voltage source at terminals a––
b as shown in Figure (b).
For mesh 1,
1 = I1(j10 + j20 –– j5x2) + j5I2
1 = j20I1 + j5I2
For mesh 2,
(3)
0 = (20 + j10)I2 + j5I1 –– j10I1 = (4 + j2)I2 –– jI1
or
Substituting (4) into (3),
I2 = jI1/(4 + j2)
1 = j20I1 + j(j5)I1/(4 + j2) = (––1 + j20.5)I1
I1 = 1/(––1 + j20.5)
ZN = 1/I1 = (––1 + j20.5) ohms
(4)
Chapter 13, Solution 16.
To find IN, we short-circuit a-b.
8:
j:
-j2 :
a
x x
j4 :
+
80‘0 V
o
j6 :
I2
IN
I1
b
80 (8 j 2 j 4) I 1 jI 2 0

o
j 6 I 2 jI 1 0

o
I1 6I 2
(8 j 2) I 1 jI 2
80
(1)
(2)
Solving (1) and (2) leads to
80
IN I2
1.584 j 0.362 1.6246‘ 12.91o A
48 j11
To find ZN, insert a 1-A current source at terminals a-b. Transforming the current source
to voltage source gives the circuit below.
8:
j:
-j2 :
2:
a
x x
j4 :
+
j6 :
I1
2V
I2
b
0
(8 j 2) I 1 jI 2

o
I1
jI 2
8 j2
(3)
2 (2 j 6) I 2 jI 1 0
(4)
Solving (3) and (4) leads to I2 = -0.1055 +j0.2975, Vab=-j6I2 = 1.7853 +0.6332
ZN
Vab
1
1.894‘19.53o :
Chapter 13, Solution 17.
Z = -j6 // Zo
where
144
j30 j2 j5 4
Zo
j20 Z
j6 xZ o
j6 Z o
0.5213 j15.7
0.1989 j9.7:
Chapter 13, Solution 18.
Let Z
1.
L1
5, L2
20, M
k L1 L2
0.5 x10
5
We replace the transformer by its equivalent T-section.
La
L1 ( M )
5 5 10,
Lb
L1 M
20 5
25,
Lc
M
5
We find ZTh using the circuit below.
-j4
j10
j25
j2
-j5
ZTh
4+j6
Z Th
j 27 (4 j ) //( j 6)
j 27 j 6(4 j )
4 j7
We find VTh by looking at the circuit below.
2.215 j 29.12:
-j4
j10
j25
j2
+
-j5
+
VTh
o
120<0
4+j6
-
-
VTh
4 j
(120)
4 j j6
61.37‘ 46.22 o V
Chapter 13, Solution 19.
Let Z
1.
Lb
La
L1 ( M )
L2 M
30 25
40 25
55 H,
65 H
LC
Thus, the T-section is as shown below.
j65 :
j55 :
-j25 :
M
25
Chapter 13, Solution 20.
Transform the current source to a voltage source as shown below.
k=0.5
4:
j10
8:
j10
I3
+
––
j12
-j5
I1
I2
20‘0o
+
––
k = M/ L1 L 2 or M = k L1 L 2
ZM = k ZL1ZL 2 = 0.5(10) = 5
For mesh 1,
j12 = (4 + j10 –– j5)I1 + j5I2 + j5I2 = (4 + j5)I1 + j10I2
For mesh 2,
0 = 20 + (8 + j10 –– j5)I2 + j5I1 + j5I1
––20 = +j10I1 + (8 + j5)I2
From (1) and (2),
(1)
ª j12º
« 20 »
¬ ¼
ª4 j5 j10 º ª I1 º
« j10 8 j5» «I »
¬
¼¬ 2 ¼
' = 107 + j60, '1 = ––60 ––j296, '2 = 40 –– j100
I1 = '1/' = 2.462‘72.18q A
I2 = '2/' = 0.878‘––97.48q A
I3 = I1 –– I2 = 3.329‘74.89q A
i1 = 2.462 cos(1000t + 72.18q) A
i2 = 0.878 cos(1000t –– 97.48q) A
(2)
At t = 2 ms, 1000t = 2 rad = 114.6q
i1 = 0.9736cos(114.6q + 143.09q) = ––2.445
i2 = 2.53cos(114.6q + 153.61q) = ––0.8391
The total energy stored in the coupled coils is
w = 0.5L1i12 + 0.5L2i22 –– Mi1i2
Since ZL1 = 10 and Z = 1000, L1 = L2 = 10 mH, M = 0.5L1 = 5mH
w = 0.5(10)(––2.445)2 + 0.5(10)(––0.8391)2 –– 5(––2.445)(––0.8391)
w = 43.67 mJ
Chapter 13, Solution 21.
For mesh 1, 36‘30q = (7 + j6)I1 –– (2 + j)I2
For mesh 2,
(1)
0 = (6 + j3 –– j4)I2 –– 2I1 jI1 = ––(2 + j)I1 + (6 –– j)I2
Placing (1) and (2) into matrix form,
ª36‘30qº
« 0 »
¬
¼
(2)
ª 7 j6 2 jº ª I1 º
« 2 j 6 j » «I »
¬
¼¬ 2 ¼
' = 48 + j35 = 59.41‘36.1q, '1 = (6 –– j)36‘30q = 219‘20.54q
'2 = (2 + j)36‘30q = 80.5‘56.56q, I1 = '1/' = 3.69‘––15.56q, I2 = '2/' = 1.355‘20.46q
Power absorbed fy the 4-ohm resistor, = 0.5(I2)24 = 2(1.355)2 = 3.672 watts
Chapter 13, Solution 22.
With more complex mutually coupled circuits, it may be easier to show the effects of the
coupling as sources in terms of currents that enter or leave the dot side of the coil. Figure
13.85 then becomes,
-j50
Io
I3
j20Ic
+ j40
j10Ib
j60
+ Ia
j30Ic
+
+
Ix
j30Ib
+
j20Ia
50‘0q V
+
j80
I1
I2
100 :
Ib
+
j10Ia
Note the following,
Ia = I1 –– I3
Ib = I2 –– I1
Ic = I3 –– I2
and
Io = I3
Now all we need to do is to write the mesh equations and to solve for Io.
Loop # 1,
-50 + j20(I3 –– I2) j 40(I1 –– I3) + j10(I2 –– I1) –– j30(I3 –– I2) + j80(I1 –– I2) –– j10(I1 –– I3) = 0
j100I1 –– j60I2 –– j40I3 = 50
Multiplying everything by (1/j10) yields 10I1 –– 6I2 –– 4I3 = - j5
(1)
Loop # 2,
j10(I1 –– I3) + j80(I2––I1) + j30(I3––I2) –– j30(I2 –– I1) + j60(I2 –– I3) –– j20(I1 –– I3) + 100I2 = 0
-j60I1 + (100 + j80)I2 –– j20I3 = 0
(2)
Loop # 3,
-j50I3 +j20(I1 ––I3) +j60(I3 ––I2) +j30(I2 ––I1) ––j10(I2 ––I1) +j40(I3 ––I1) ––j20(I3 ––I2) = 0
-j40I1 –– j20I2 + j10I3 = 0
Multiplying by (1/j10) yields,
-4I1 –– 2I2 + I3 = 0
(3)
Multiplying (2) by (1/j20) yields -3I1 + (4 –– j5)I2 –– I3 = 0
Multiplying (3) by (1/4) yields
(4)
-I1 –– 0.5I2 –– 0.25I3 = 0
(5)
Multiplying (4) by (-1/3) yields I1 –– ((4/3) –– j(5/3))I2 + (1/3)I3 = -j0.5 (7)
Multiplying [(6)+(5)] by 12 yields
(-22 + j20)I2 + 7I3 = 0
(8)
Multiplying [(5)+(7)] by 20 yields
-22I2 –– 3I3 = -j10
(9)
(8) leads to I2 = -7I3/(-22 + j20) = 0.2355‘42.3o = (0.17418+j0.15849)I3
(9) leads to I3 = (j10 –– 22I2)/3, substituting (1) into this equation produces,
I3 = j3.333 + (-1.2273 –– j1.1623)I3
I3 = Io = 1.3040‘63o amp.
or
Chapter 13, Solution 23.
Z = 10
0.5 H converts to jZL1 = j5 ohms
1 H converts to jZL2 = j10 ohms
0.2 H converts to jZM = j2 ohms
25 mF converts to 1/(jZC) = 1/(10x25x10-3) = ––j4 ohms
The frequency-domain equivalent circuit is shown below.
j2
j5
12‘0q
+
I1
j10
––j4
I2
5:
(10)
For mesh 1,
12 = (j5 –– j4)I1 + j2I2 –– (––j4)I2
––j2 = I1 + 6I2
For mesh 2,
0 = (5 + j10)I2 + j2I1 ––(––j4)I1
0 = (5 + j10)I2 + j6I1
From (1),
(1)
(2)
I1 = ––j12 –– 6I2
Substituting this into (2) produces,
I2 = 72/(––5 + j26) = 2.7194‘––100.89q
I1 = ––j12 –– 6 I2 = ––j12 –– 163.17‘––100.89 = 5.068‘52.54q
Hence,
i1 = 5.068cos(10t + 52.54q) A, i2 = 2.719cos(10t –– 100.89q) A.
10t = 10x15x10-3 0.15 rad = 8.59q
At t = 15 ms,
i1 = 5.068cos(61.13q) = 2.446
i2 = 2.719cos(––92.3q) = ––0.1089
w = 0.5(5)(2.446)2 + 0.5(1)(––0.1089)2 –– (0.2)(2.446)(––0.1089) = 15.02 J
Chapter 13, Solution 24.
(a)
k = M/ L1 L 2 = 1/ 4 x 2 = 0.3535
(b)
Z = 4
1/4 F leads to 1/(jZC) = ––j/(4x0.25) = ––j
1||(––j) = ––j/(1 –– j) = 0.5(1 –– j)
1 H produces jZM = j4
4 H produces j16
2 H becomes j8
j4
2:
j8
12‘0q
+
I1
I2
0.5(1––j)
j16
12 = (2 + j16)I1 + j4I2
or
6 = (1 + j8)I1 + j2I2
0 = (j8 + 0.5 –– j0.5)I2 + j4I1 or I1 = (0.5 + j7.5)I2/(––j4)
(1)
(2)
Substituting (2) into (1),
24 = (––11.5 –– j51.5)I2 or I2 = ––24/(11.5 + j51.5) = ––0.455‘––77.41q
Vo = I2(0.5)(1 –– j) = 0.3217‘57.59q
vo = 321.7cos(4t + 57.6q) mV
(c)
From (2),
At t = 2s,
I1 = (0.5 + j7.5)I2/(––j4) = 0.855‘––81.21q
i1 = 0.885cos(4t –– 81.21q) A, i2 = ––0.455cos(4t –– 77.41q) A
4t = 8 rad = 98.37q
i1 = 0.885cos(98.37q –– 81.21q) = 0.8169
i2 = ––0.455cos(98.37q –– 77.41q) = ––0.4249
w = 0.5L1i12 + 0.5L2i22 + Mi1i2
= 0.5(4)(0.8169)2 + 0.5(2)(––.4249)2 + (1)(0.1869)(––0.4249) = 1.168 J
Chapter 13, Solution 25.
m = k L1 L 2 = 0.5 H
We transform the circuit to frequency domain as shown below.
12sin2t converts to 12‘0q, Z = 2
0.5 F converts to 1/(jZC) = ––j
2 H becomes jZL = j4
j1
Io 4 :
1:
a
2:
––j1
12‘0q
+
j2
j2
j4
10 :
b
Applying the concept of reflected impedance,
Zab = (2 –– j)||(1 + j2 + (1)2/(j2 + 3 + j4))
= (2 –– j)||(1 + j2 + (3/45) –– j6/45)
= (2 –– j)||(1 + j2 + (3/45) –– j6/45)
= (2 –– j)||(1.0667 + j1.8667)
=(2 –– j)(1.0667 + j1.8667)/(3.0667 + j0.8667) = 1.5085‘17.9q ohms
Io = 12‘0q/(Zab + 4) = 12/(5.4355 + j0.4636) = 2.2‘––4.88q
io = 2.2sin(2t –– 4.88q) A
Chapter 13, Solution 26.
M = k L1L 2
ZM = k ZL1ZL 2 = 0.6 20x 40 = 17
The frequency-domain equivalent circuit is shown below.
j17
50 :
200‘60q
––j30
I1
+
Io
j20
j40
I2
For mesh 1,
200‘60q = (50 –– j30 + j20)I1 + j17I2 = (50 –– j10)I1 + j17I2
10 :
(1)
For mesh 2,
0 = (10 + j40)I2 + j17I1
In matrix form,
ª200‘60qº
«
»
0
¬
¼
(2)
j17 º ª I1 º
ª50 j10
« j17
10 j40»¼ «¬I 2 »¼
¬
' = 900 + j100, '1 = 2000‘60q(1 + j4) = 8246.2‘136q, '2 = 3400‘––30q
I2 = '2/' = 3.755‘––36.34q
Io = I2 = 3.755‘––36.34q A
Switching the dot on the winding on the right only reverses the direction of Io.
This can be seen by looking at the resulting value of '2 which now becomes
3400‘150q. Thus,
Io = 3.755‘143.66q A
Chapter 13, Solution 27.
Zin = ––j4 + j5 + 9/(12 + j6) = 0.6 + j.07 = 0.922‘49.4q
I1 = 12‘0q/0.922‘49.4q = 13‘––49.4q A
Chapter 13, Solution 28.
We find ZTh by replacing the 20-ohm load with a unit source as shown below.
j10 :
8:
-jX
x x
j12 :
j15 :
I2
+
1V
-
I1
(8 jX j12) I 1 j10 I 2
For mesh 1,
0
For mesh 2,
1 j15I 2 j10 I 1
0
(1)

o
I1
1.5I 2 0.1 j
(2)
Substituting (2) into (1) leads to
1.2 j 0.8 0.1X
I2
12 j8 j1.5 X
Z Th
| Z Th | 20
12 j8 j1.5 X
1.2 j 0.8 0.1X
1
I2
12 2 (8 1.5 X ) 2
(1.2 0.1X ) 0.8
2
2

o
Solving the quadratic equation yields X = 6.425
0 1.75 X 2 72 X 624
Chapter 13, Solution 29.
30 mH becomes jZL = j30x10-3x103 = j30
50 mH becomes j50
Let X = ZM
Using the concept of reflected impedance,
Zin = 10 + j30 + X2/(20 + j50)
I1 = V/Zin = 165/(10 + j30 + X2/(20 + j50))
p = 0.5|I1|2(10) = 320 leads to |I1|2 = 64 or |I1| = 8
8 = |165(20 + j50)/(X2 + (10 + j30)(20 + j50))|
= |165(20 + j50)/(X2 –– 1300 + j1100)|
64 = 27225(400 + 2500)/((X2 –– 1300)2 + 1,210,000)
or
(X2 –– 1300)2 + 1,210,000 = 1,233,633
X = 33.86 or 38.13
If X = 38.127 = ZM
M = 38.127 mH
k = M/ L1 L 2 = 38.127/ 30x50 = 0.984
j38.127
10 :
165‘0q
+
I1
j30
j50
I2
20 :
165 = (10 + j30)I1 –– j38.127I2
(1)
0 = (20 + j50)I2 –– j38.127I1
(2)
ª165º
« 0 »
¬ ¼
In matrix form,
ª 10 j30 j38.127º ª I1 º
« j38.127 20 j50 » « I »
¬
¼¬ 2 ¼
' = 154 + j1100 = 1110.73‘82.03q, '1 = 888.5‘68.2q, '2 = j6291
I1 = '1/' = 8‘––13.81q, I2 = '2/' = 5.664‘7.97q
i1 = 8cos(1000t –– 13.83q), i2 = 5.664cos(1000t + 7.97q)
At t = 1.5 ms, 1000t = 1.5 rad = 85.94q
i1 = 8cos(85.94q –– 13.83q) = 2.457
i2 = 5.664cos(85.94q + 7.97q) = ––0.3862
w = 0.5L1i12 + 0.5L2i22 + Mi1i2
= 0.5(30)(2.547)2 + 0.5(50)(––0.3862)2 –– 38.127(2.547)(––0.3862)
= 130.51 mJ
Chapter 13, Solution 30.
(a)
Zin = j40 + 25 + j30 + (10)2/(8 + j20 –– j6)
= 25 + j70 + 100/(8 + j14) = (28.08 + j64.62) ohms
(b)
jZLa = j30 –– j10 = j20, jZLb = j20 –– j10 = j10, jZLc = j10
Thus the Thevenin Equivalent of the linear transformer is shown below.
j40
25 :
j20
j10
j10
8:
––j6
Zin
Zin = j40 + 25 + j20 + j10||(8 + j4) = 25 + j60 + j10(8 + j4)/(8 + j14)
= (28.08 + j64.62) ohms
Chapter 13, Solution 31.
(a)
La = L1 –– M = 10 H
Lb = L2 –– M = 15 H
Lc = M = 5 H
(b)
L1L2 –– M2 = 300 –– 25 = 275
LA = (L1L2 –– M2)/(L1 –– M) = 275/15 = 18.33 H
LB = (L1L2 –– M2)/(L1 –– M) = 275/10 = 27.5 H
LC = (L1L2 –– M2)/M = 275/5 = 55 H
Chapter 13, Solution 32.
We first find Zin for the second stage using the concept of reflected impedance.
Lb
LB
R
Zin’’
Zin’’ = jZLb + Z2Mb2/(R + jZLb) = (jZLbR - Z2Lb2 + Z2Mb2)/(R + jZLb)
(1)
For the first stage, we have the circuit below.
La
LA
Zin’’
Zin
Zin = jZLa + Z2Ma2/(jZLa + Zin)
= (––Z2La2 + Z2Ma2 + jZLaZin)/( jZLa + Zin)
(2)
Substituting (1) into (2) gives,
( jZL b R Z 2 L2b Z 2 M 2b )
Z L Z M jZL a
R jZ L b
=
2 2
jZL b R Z L b Z 2 M 2b
jZL a R jZ L b
2
=
2
a
2
2
a
––RZ2La2 + Z2Ma2R –– jZ3LbLa + jZ3LbMa2 + jZLa(jZLbR –– Z2Lb2 + Z2Mb2)
jZRLa ––Z2LaLb + jZLbR –– Z2La2 + Z2Mb2
Z2R(La2 + LaLb –– Ma2) + jZ3(La2Lb + LaLb2 –– LaMb2 –– LbMa2)
Zin =
Z2(LaLb +Lb2 –– Mb2) –– jZR(La +Lb)
Chapter 13, Solution 33.
Zin = 10 + j12 + (15)2/(20 + j 40 –– j5) = 10 + j12 + 225/(20 + j35)
= 10 + j12 + 225(20 –– j35)/(400 + 1225)
= (12.769 + j7.154) ohms
Chapter 13, Solution 34.
Insert a 1-V voltage source at the input as shown below.
j6 :
1:
x
+
j12 :
o
1<0 V
-
8:
x
j10 :
I1
j4 :
I2
-j2 :
For loop 1,
1 (1 j10) I 1 j 4 I 2
(1)
For loop 2,
0
(8 j 4 j10 j 2) I 2 j 2 I 1 j 6 I 1

o
0
jI 1 (2 j 3) I 2
(2)
Solving (1) and (2) leads to I1=0.019 ––j0.1068
Z
1
I1
1.6154 j 9.077
9.219‘79.91o :
Alternatively, an easier way to obtain Z is to replace the transformer with its equivalent
T circuit and use series/parallel impedance combinations. This leads to exactly the same
result.
Chapter 13, Solution 35.
For mesh 1,
16
(10 j 4) I 1 j 2 I 2
(1)
For mesh 2,
0
j 2 I 1 (30 j 26) I 2 j12 I 3
(2)
For mesh 3,
0
j12 I 2 (5 j11) I 3
(3)
We may use MATLAB to solve (1) to (3) and obtain
I 1 1.3736 j 0.5385 1.4754‘ 21.41o A
I 2 0.0547 j 0.0549 0.0775‘ 134.85 o A
I 3 0.0268 j 0.0721 0.077‘ 110.41o A
Chapter 13, Solution 36.
Following the two rules in section 13.5, we obtain the following:
(a)
V2/V1 = ––n,
I2/I1 = ––1/n
(b)
V2/V1 = ––n,
I2/I1 = ––1/n
(c)
V2/V1 = n,
I2/I1 = 1/n
(d)
V2/V1 = n,
I2/I1 = ––1/n
(n = V2/V1)
Chapter 13, Solution 37.
V2
V1
(a) n
(b) S1
2400
480
I 1V1
(c ) I 2
S2
50,000
2400
5
I 2V2
50,000

o
I1
50,000
480
104.17 A
20.83 A
Chapter 13, Solution 38.
Zin = Zp + ZL/n2, n = v2/v1 = 230/2300 = 0.1
v2 = 230 V, s2 = v2I2*
I2* = s2/v2 = 17.391‘––53.13q or I2 = 17.391‘53.13q A
ZL = v2/I2 = 230‘0q/17.391‘53.13q = 13.235‘––53.13q
Zin = 2‘10q + 1323.5‘––53.13q
= 1.97 + j0.3473 + 794.1 –– j1058.8
Zin = 1.324‘––53.05q kohms
Chapter 13, Solution 39.
Referred to the high-voltage side,
ZL = (1200/240)2(0.8‘10q) = 20‘10q
Zin = 60‘––30q + 20‘10q = 76.4122‘––20.31q
I1 = 1200/Zin = 1200/76.4122‘––20.31q = 15.7‘20.31q A
Since S = I1v1 = I2v2, I2 = I1v1/v2
= (1200/240)( 15.7‘20.31q) = 78.5‘20.31q A
Chapter 13, Solution 40.
n
N2
N1
500
2000
P
V2
R
60 2
12
1
,
4
n
V2
V1
o
V2
nV1
1
(240)
4
60 V
300 W
Chapter 13, Solution 41.
We reflect the 2-ohm resistor to the primary side.
Zin = 10 + 2/n2,
n = ––1/3
Since both I1 and I2 enter the dotted terminals,
Zin = 10 + 18 = 28 ohms
I1 = 14‘0q/28 = 0.5 A and I2 = I1/n = 0.5/(––1/3) = ––1.5 A
Chapter 13, Solution 42.
10 :
+ x
V1
+
120<0o V
-
I1
-j50 :
1:4
-
x +
V2
+
20 :
-
I2
Applying mesh analysis,
120 10I1 V1
0
(20 j50)I 2 V2
Vo
(1)
(2)
At the terminals of the transformer,
V2
V1
n
I2
I1
1
n
o
4
1
4
V2

o
I1
Substituting (3) and (4) into (1) gives 120
20I 2
(3)
4I 2
(4)
40I 2 0.25V2
(5)
2.4756 j0.6877
Solving (2) and (5) yields I 2
Vo
4V1
51.39‘15.52 o V
Chapter 13, Solution 43.
Transform the two current sources to voltage sources, as shown below.
10 :
+
20 V
+
––
Using mesh analysis,
I1
12 :
1:4
v1
+
v2
I2
12V
+
––
––20 + 10I1 + v1 = 0
20 = v1 + 10I1
12 + 12I2 –– v2 = 0 or 12 = v2 –– 12I2
At the transformer terminal, v2 = nv1 = 4v1
I1 = nI2 = 4I2
(1)
(2)
(3)
(4)
Substituting (3) and (4) into (1) and (2), we get,
Solving (5) and (6) gives
20 = v1 + 40I2
(5)
12 = 4v1 –– 12I2
(6)
v1 = 4.186 V and v2 = 4v = 16.744 V
Chapter 13, Solution 44.
We can apply the superposition theorem. Let i1 = i1’’ + i1”” and i2 = i2’’ + i2””
where the single prime is due to the DC source and the double prime is due to the
AC source. Since we are looking for the steady-state values of i1 and i2,
i1’’ = i2’’ = 0.
For the AC source, consider the circuit below.
R
1:n
+
i1””
+
v1
v2
v2/v1 = ––n,
+
––
i2””
Vn‘0q
I2””/I1”” = ––1/n
But v2 = vm, v1 = ––vm/n or I1”” = vm/(Rn)
I2”” = ––I1””/n = ––vm/(Rn2)
i1(t) = (vm/Rn)cosZt A, and i2(t) = (––vm/(n2R))cosZt A
Hence,
Chapter 13, Solution 45.
48 :
4‘––90
ZL
8
j
ZC
+
8 j4 , n = 1/3
Z
Z
I
ZL
9 Z L 72 j36
n2
4‘ 90q
4‘ 90q
48 72 j36 125.28‘ 16.7q
0.03193‘ 73.3q
We now have some choices, we can go ahead and calculate the current in the second loop
and calculate the power delivered to the 8-ohm resistor directly or we can merely say that
the power delivered to the equivalent resistor in the primary side must be the same as the
power delivered to the 8-ohm resistor. Therefore,
P8:
I2
72
2
0.5098x10 3 72 36.71 mW
The student is encouraged to calculate the current in the secondary and calculate the
power delivered to the 8-ohm resistor to verify that the above is correct.
Chapter 13, Solution 46.
(a)
Reflecting the secondary circuit to the primary, we have the circuit shown below.
Zin
16‘60q
+
I1
+
10‘30q/(––n) = ––5‘30q
Zin = 10 + j16 + (1/4)(12 –– j8) = 13 + j14
––16‘60q + ZinI1 –– 5‘30q = 0 or I1 = (16‘60q + 5‘30q)/(13 + j14)
Hence,
(b)
I1 = 1.072‘5.88q A, and I2 = ––0.5I1 = 0.536‘185.88q A
Switching a dot will not effect Zin but will effect I1 and I2.
I1 = (16‘60q –– 5‘30q)/(13 + j14) = 0.625 ‘25 A
and I2 = 0.5I1 = 0.3125‘25q A
Chapter 13, Solution 47.
0.02 F becomes 1/(jZC) = 1/(j5x0.02) = ––j10
We apply mesh analysis to the circuit shown below.
––j10
I3
10 :
3:1
+
10‘0q
+
––
I1
v1
+
v2
+
I2
vo
2:
For mesh 1,
10 = 10I1 –– 10I3 + v1
(1)
For mesh 2,
v2 = 2I2 = vo
(2)
0 = (10 –– j10)I3 –– 10I1 + v2 –– v1
(3)
v2 = nv1 = v1/3
(4)
I1 = nI2 = I2/3
(5)
From (2) and (4),
v1 = 6I2
(6)
Substituting this into (1),
10 = 10I1 –– 10I3
(7)
For mesh 3,
At the terminals,
Substituting (4) and (6) into (3) yields
From (5), (7), and (8)
0 = ––10I1 –– 4I2 + 10(1 –– j)I3
0 º ª I1 º
0.333
ª 1
« 10
6
10 »» ««I 2 »»
«
«¬ 10
10 j10»¼ «¬ I 3 »¼
4
(8)
ª0º
«10»
« »
«¬ 0 »¼
I2 =
'2
'
100 j100
= 1.482‘32.9q
20 j93.33
vo = 2I2 = 2.963‘32.9q V
(a)
Switching the dot on the secondary side effects only equations (4) and (5).
From (2) and (9),
v2 = ––v1/3
(9)
I1 = ––I2/3
(10)
v1 = ––6I2
Substituting this into (1),
10 = 10I1 –– 10I3 –– 6I2 = (23 –– j5)I1
(11)
Substituting (9) and (10) into (3),
From (10) to (12), we get
0 = ––10I1 + 4I2 + 10(1 –– j)I3
0.333
0 º ª I1 º
ª 1
« 10
6
10 »» ««I 2 »»
«
«¬ 10
4
10 j10»¼ «¬ I 3 »¼
I2 =
'2
'
(12)
ª0º
«10»
« »
«¬ 0 »¼
100 j100
= 1.482‘––147.1q
20 j93.33
vo = 2I2 = 2.963‘––147.1q V
Chapter 13, Solution 48.
We apply mesh analysis.
8:
10 :
2:1
+
+
x
+
o
100‘0 V
-
I1
V1
x
V2
-
Ix
j6 :
I2
-j4 :
100
0
(8 j 4) I 1 j 4 I 2 V1
(1)
(10 j 2) I 2 j 4 I 1 V 2
(2)
But
V2
V1
n
I2
I1
1
2
1
n

o
2

o
(3)
2V2
V1
I1
0.5 I 2
(4)
Substituting (3) and (4) into (1) and (2), we obtain
100 (4 j 2) I 2 2V2
0 (10 j 4) I 2 V2
(1)a
(2)a
Solving (1)a and (2)a leads to I2 = -3.5503 +j1.4793
Ix
I1 I 2
0.5 I 2
1.923‘157.4 o A
Chapter 13, Solution 49.
Z
2,
1
F
20

o
1
jZ C
j10
-j10
Ix
2:
I1
1:3
I2
1
+
12<0o V
-
2
+ x
V1
-
+
V2
x
-
6:
At node 1,
12 V1 V1 V2
I1

o 12 2 I 1 V1 (1 j 0.2) j 0.2V2
2
j10
At node 2,
V V2 V2
I2 1

o 0 6 I 2 j 0.6V1 (1 j 0.6)V2
j10
6
At the terminals of the transformer, V2
3V1 ,
I2
(1)
(2)
1
I1
3
Substituting these in (1) and (2),
12
6 I 2 V1 (1 j 0.8),
0
6 I 2 V1 (3 j 2.4)
Adding these gives V1=1.829 ––j1.463 and
Ix
V1 V2
j10
ix
4V1
j10
0.937‘51.34 o
0.937 cos(2t 51.34 o ) A
Chapter 13, Solution 50.
The value of Zin is not effected by the location of the dots since n2 is involved.
Zin’’ = (6 –– j10)/(n’’)2, n’’ = 1/4
Zin’’ = 16(6 –– j10) = 96 –– j160
Zin = 8 + j12 + (Zin’’ + 24)/n2, n = 5
Zin = 8 + j12 + (120 –– j160)/25 = 8 + j12 + 4.8 –– j6.4
Zin = (12.8 + j5.6) ohms
Chapter 13, Solution 51.
Let Z3 = 36 +j18, where Z3 is reflected to the middle circuit.
ZR’’ = ZL/n2 = (12 + j2)/4 = 3 + j0.5
Zin = 5 –– j2 + ZR’’ = (8 –– j1.5) ohms
I1 = 24‘0q/ZTh = 24‘0q/(8 –– j1.5) = 24‘0q/8.14‘––10.62q = 8.95‘10.62q A
Chapter 13, Solution 52.
For maximum power transfer,
40 = ZL/n2 = 10/n2 or n2 = 10/40 which yields n = 1/2 = 0.5
I = 120/(40 + 40) = 3/2
p = I2R = (9/4)x40 = 90 watts.
Chapter 13, Solution 53.
(a)
The Thevenin equivalent to the left of the transformer is shown below.
8:
20 V
+
The reflected load impedance is ZL’’ = ZL/n2 = 200/n2.
8 = 200/n2 produces n = 5.
For maximum power transfer,
(b)
If n = 10, ZL’’ = 200/10 = 2 and I = 20/(8 + 2) = 2
p = I2ZL’’ = (2)2(2) = 8 watts.
Chapter 13, Solution 54.
(a)
ZTh
VS
+
I1
ZL/n2
For maximum power transfer,
ZTh = ZL/n2, or n2 = ZL/ZTh = 8/128
n = 0.25
(b)
I1 = VTh/(ZTh + ZL/n2) = 10/(128 + 128) = 39.06 mA
(c)
v2 = I2ZL = 156.24x8 mV = 1.25 V
But
v2 = nv1 therefore v1 = v2/n = 4(1.25) = 5 V
Chapter 13, Solution 55.
We reflect Zs to the primary side.
ZR = (500 –– j200)/n2 = 5 –– j2, Zin = Zp + ZR = 3 + j4 + 5 –– j2 = 8 + j2
I1 = 120‘0q/(8 + j2) = 14.552‘––14.04q
Zp
Vp
+
1:n
I1
I2
Zs
Since both currents enter the dotted terminals as shown above,
I2 = ––(1/n)I1 = ––1.4552‘––14.04q = 1.4552‘166q
S2 = |I2|2Zs = (1.4552)(500 –– j200)
P2 = Re(S2) = (1.4552)2(500) = 1054 watts
Chapter 13, Solution 56.
We apply mesh analysis to the circuit as shown below.
2:
1:2
+
+
v1
46V
+
I1
v2
I2
10 :
5:
For mesh 1,
46 = 7I1 –– 5I2 + v1
(1)
For mesh 2,
v2 = 15I2 –– 5I1
(2)
At the terminals of the transformer,
v2 = nv1 = 2v1
(3)
I1 = nI2 = 2I2
(4)
Substituting (3) and (4) into (1) and (2),
Combining (5) and (6),
46 = 9I2 + v1
(5)
v1 = 2.5I2
(6)
46 = 11.5I2 or I2 = 4
P10 = 0.5I22(10) = 80 watts.
Chapter 13, Solution 57.
(a)
ZL = j3||(12 –– j6) = j3(12 –– j6)/(12 –– j3) = (12 + j54)/17
Reflecting this to the primary side gives
Zin = 2 + ZL/n2 = 2 + (3 + j13.5)/17 = 2.3168‘20.04q
I1 = vs/Zin = 60‘90q/2.3168‘20.04q = 25.9‘69.96q A(rms)
I2 = I1/n = 12.95‘69.96q A(rms)
(b)
60‘90q = 2I1 + v1 or v1 = j60 ––2I1 = j60 –– 51.8‘69.96q
v1 = 21.06‘147.44q V(rms)
v2 = nv1 = 42.12‘147.44q V(rms)
vo = v2 = 42.12‘147.44q V(rms)
(c)
S = vsI1* = (60‘90q)(25.9‘––69.96q) = 1554‘20.04q VA
Chapter 13, Solution 58.
Consider the circuit below.
20 :
I3
20 :
1:5
+
80‘0q
+
––
I1
+
v1
v2
+
I2
vo
100 :
For mesh1,
80 = 20I1 –– 20I3 + v1
(1)
For mesh 2,
v2 = 100I2
(2)
For mesh 3,
0 = 40I3 –– 20I1 which leads to I1 = 2I3
(3)
At the transformer terminals, v2 = ––nv1 = ––5v1
(4)
I1 = ––nI2 = ––5I2
From (2) and (4),
(5)
––5v1 = 100I2 or v1 = ––20I2
(6)
Substituting (3), (5), and (6) into (1),
4
= I1 –– I2 –– I 3 = I1 –– (I1/(––5)) –– I1/2 = (7/10)I1
I1 = 40/7, I2 = ––8/7, I3 = 20/7
p20(the one between 1 and 3) = 0.5(20)(I1 –– I3)2 = 10(20/7)2 = 81.63 watts
p20(at the top of the circuit) = 0.5(20)I32 = 81.63 watts
p100 = 0.5(100)I22 = 65.31 watts
Chapter 13, Solution 59.
We apply nodal analysis to the circuit below.
2:
I3
8:
v1 I1 2 : 1
+
20‘0q
+
––
v1
I2
+
v2
v2
4:
20 = 8I1 + V1
(1)
V1 = 2I3 + V2
(2)
V2 = 4I2
(3)
At the transformer terminals,
v2 = 0.5v1
(4)
I1 = 0.5I2
(5)
Solving (1) to (5) gives I1 = 0.833 A, I2 = 1.667 A, I3 = 3.333 A
V1 = 13.33 V, V2 = 6.667 V.
P8: = 0.5(8)|(20 –– V1)/8|2 = 2.778 W
P2: = 0.5(2)I32 = 11.11 W, P4: = 0.5V22/4 = 5.556 W
Chapter 13, Solution 60.
(a)
Transferring the 40-ohm load to the middle circuit,
ZL’’ = 40/(n’’)2 = 10 ohms where n’’ = 2
10||(5 + 10) = 6 ohms
We transfer this to the primary side.
Zin = 4 + 6/n2 = 4 + 96 = 100 ohms, where n = 0.25
I1 = 120/100 = 1.2 A and I2 = I1/n = 4.8 A
4:
1:4
I1
+
120‘0q
+
––
v1
I2
5:
I2’’
+
v2
10 :
10 :
Using current division, I2’’ = (10/25)I2 = 1.92 and I3 = I2’’/n’’ = 0.96 A
(b)
p = 0.5(I3)2(40) = 18.432 watts
Chapter 13, Solution 61.
We reflect the 160-ohm load to the middle circuit.
ZR = ZL/n2 = 160/(4/3)2 = 90 ohms, where n = 4/3
2:
1:5
I1
+
24‘0q
+
––
v1
Io
14 :
Io’’
+
vo
60 :
90 :
14 + 60||90 = 14 + 36 = 50 ohms
We reflect this to the primary side.
ZR’’ = ZL’’/(n’’)2 = 50/52 = 2 ohms when n’’ = 5
I1 = 24/(2 + 2) = 6A
24 = 2I1 + v1 or v1 = 24 –– 2I1 = 12 V
vo = ––nv1 = ––60 V, Io = ––I1 /n1 = ––6/5 = ––1.2
Io‘‘ = [60/(60 + 90)]Io = ––0.48A
I2 = ––Io’’/n = 0.48/(4/3) = 0.36 A
Chapter 13, Solution 62.
(a)
Reflect the load to the middle circuit.
ZL’’ = 8 –– j20 + (18 + j45)/32 = 10 –– j15
We now reflect this to the primary circuit so that
Zin = 6 + j4 + (10 –– j15)/n2 = 7.6 + j1.6 = 7.767‘11.89q, where n =
5/2 = 2.5
I1 = 40/Zin = 40/7.767‘11.89q = 5.15‘––11.89q
S = 0.5vsI1* = (20‘0q)(5.15‘11.89q) = 103‘11.89q VA
(b)
I2 = ––I1/n,
n = 2.5
I3 = ––I2/n’’,
n = 3
I3 = I1/(nn’’) = 5.15‘––11.89q/(2.5x3) = 0.6867‘––11.89q
p = 0.5|I2|2(18) = 9(0.6867)2 = 4.244 watts
Chapter 13, Solution 63.
Reflecting the (9 + j18)-ohm load to the middle circuit gives,
Zin’’ = 7 –– j6 + (9 + j18)/(n’’)2 = 7 –– j6 + 1 + j12 = 8 + j4 when n’’ = 3
Reflecting this to the primary side,
Zin = 1 + Zin’’/n2 = 1 + 2 –– j = 3 –– j, where n = 2
I1 = 12‘0q/(3 –– j) = 12/3.162‘––18.43q = 3.795‘18.43A
I2 = I1/n = 1.8975‘18.43q A
I3 = ––I2/n2 = 632.5‘161.57q mA
Chapter 13, Solution 64.
We find ZTh at the terminals of Z by considering the circuit below.
10 :
1:n
+x
V1
x +
V2
––
+
––
I1
I2
20 :
1V
––
For mesh 1,
30 I 1 20 I 2 V1
0
(1)
For mesh 2,
20 I 1 20 I 2 V2
1
(2)
At the terminals,
V2
nV1 ,
I2
I1
n
Substituting these in (1) and (2) leads to
(20 30n) I 2 V1
20(1 n) I 2 nV1
0,
1
Solving these gives
I2
1
30n 40n 20

o
2
1
I2
Z Th
30n 2 40n 20
7.5
Solving the quadratic equation yields n=0.5 or 0.8333
Chapter 13, Solution 65.
40 :
10 :
I1
-
I2
+x
+
200 V
(rms)
50 :
I2
x
1
+
1:2
+
V1
-
V2
-
V3
-
1:3
I3
x
2
+
V4
-
20 :
x
At node 1,
200 V1
10
V1 V4
I1
40

o
200 1.25V1 0.25V4 10 I 1
(1)
At node 2,
V1 V4
40
V4
I3
20

o
3V4 40 I 3
V1
(2)
At the terminals of the first transformer,
V2
V1
2
I2
I1
1 / 2

o
2V1
V2

o
(3)
2 I 2
I1
(4)
For the middle loop,
V2 50 I 2 V3

o
0
V3
V2 50 I 2
(5)
At the terminals of the second transformer,
V4
V3
3
I3
I2
1 / 3

o
V4

o
(6)
3V3
3 I 3
I2
(7)
We have seven equations and seven unknowns. Combining (1) and (2) leads to
200
3.5V4 10 I 1 50 I 3
But from (4) and (7), I 1
200
2 I 2
2(3I 3 )
6 I 3 . Hence
3.5V4 110 I 3
(8)
From (5), (6), (3), and (7),
V4
3(V2 50 I 2 )
3V2 150 I 2
6V1 450 I 3
Substituting for V1 in (2) gives
V4
6(3V4 40 I 3 ) 450 I 3

o
I3
19
V4
210
(9)
Substituting (9) into (8) yields

o
200 13.452V4
V4
14.87
V 24
20
P
11.05 W
Chapter 13, Solution 66.
v1 = 420 V
(1)
v2 = 120I2
(2)
v1/v2 = 1/4 or v2 = 4v1
(3)
I1/I2 = 4 or I1 = 4 I2
(4)
Combining (2) and (4),
v2 = 120[(1/4)I1] = 30 I1
4v1 = 30I1
4(420) = 1680 = 30I1 or I1 = 56 A
Chapter 13, Solution 67.
(a)
V1
V2
N1 N 2
N2
(b)
S2
I 2V2
(c )
S2
S1
1
0 .4
5,000
I 1V1

o

o
5,000
V2
I2

o
0.4V1
5000
160
I2
0.4 x 400 160 V
31.25 A
5000
400
12.5 A
Chapter 13, Solution 68.
This is a step-up transformer.
I2
+
N1
2 –– j6
I1
10 + j40
+
20‘30q
v2
N2
v1
+
For the primary circuit,
20‘30q = (2 –– j6)I1 + v1
(1)
For the secondary circuit,
v2 = (10 + j40)I2
(2)
At the autotransformer terminals,
v1/v2 = N1/(N1 + N2) = 200/280 = 5/7,
Also,
thus v2 = 7v1/5
(3)
I1/I2 = 7/5 or I2 = 5I1/7
(4)
Substituting (3) and (4) into (2),
v1 = (10 + j40)25I1/49
Substituting that into (1) gives
20‘30q = (7.102 + j14.408)I1
I1 = 20‘30q/16.063‘63.76q = 1.245‘––33.76q A
I2 = 5I1/7 = 0.8893‘––33.76q A
Io = I1 –– I2 = [(5/7) –– 1]I1 = ––2I1/7 = 0.3557‘146.2q A
p = |I2|2R = (0.8893)2(10) = 7.51 watts
Chapter 13, Solution 69.
We can find the Thevenin equivalent.
I2
+
N2
I1
120‘0q
v2
j125 :
75 :
+
VTh
N1
v1
+
+
I1 = I2 = 0
As a step up transformer,
v1/v2 = N1/(N1 + N2) = 600/800 = 3/4
v2 = 4v1/3 = 4(120)/3 = 160‘0q rms = VTh.
To find ZTh, connect a 1-V source at the secondary terminals. We now have a
step-down transformer.
+
v1
j125 :
75 :
I2
I1
1‘0q V
+
v2
+
v1 = 1V, v2 =I2(75 + j125)
But
v1/v2 = (N1 + N2)/N1 = 800/200 which leads to v1 = 4v2 = 1
and v2 = 0.25
I1/I2 = 200/800 = 1/4 which leads to I2 = 4I1
Hence
0.25 = 4I1(75 + j125) or I1 = 1/[16(75 + j125)
ZTh = 1/I1 = 16(75 + j125)
Therefore, ZL = ZTh* = (1.2 –– j2) k:
Since VTh is rms, p = (|VTh|/2)2/RL = (80)2/1200 = 5.333 watts
Chapter 13, Solution 70.
This is a step-down transformer.
30 + j12
I1
+
I2
v1
120‘0q
+
+
v2
20 –– j40
I1/I2 = N2/(N1 + N2) = 200/1200 = 1/6, or I1 = I2/6
(1)
v1/v2 = (N2 + N2)/N2 = 6, or v1 = 6v2
(2)
For the primary loop,
120 = (30 + j12)I1 + v1
(3)
For the secondary loop,
v2 = (20 –– j40)I2
(4)
Substituting (1) and (2) into (3),
120 = (30 + j12)( I2/6) + 6v2
and substituting (4) into this yields
120 = (49 –– j38)I2 or I2 = 1.935‘37.79q
p = |I2|2(20) = 74.9 watts.
Chapter 13, Solution 71.
Zin = V1/I1
But
V1I1 = V2I2, or V2 = I2ZL and I1/I2 = N2/(N1 + N2)
V1 = V2I2/I1 = ZL(I2/I1)I2 = ZL(I2/I1)2I1
Hence
V1/I1 = ZL[(N1 + N2)/N2] 2
Zin = [1 + (N1/N2)] 2ZL
Chapter 13, Solution 72.
(a)
Consider just one phase at a time.
1:n
a
A
B
b
C
c
n = VL/ 3VLp
(b)
7200 /(12470 3 ) = 1/3
The load carried by each transformer is 60/3 = 20 MVA.
Hence
ILp = 20 MVA/12.47 k = 1604 A
ILs = 20 MVA/7.2 k = 2778 A
(c)
The current in incoming line a, b, c is
3I Lp
3x1603.85 = 2778 A
Current in each outgoing line A, B, C is
2778/(n 3 ) = 4812 A
20MVA
Load
Chapter 13, Solution 73.
(a)
This is a three-phase '-Y transformer.
(b)
VLs = nvLp/ 3 = 450/(3 3 ) = 86.6 V, where n = 1/3
As a Y-Y system, we can use per phase equivalent circuit.
Ia = Van/ZY = 86.6‘0q/(8 –– j6) = 8.66‘36.87q
Ic = Ia‘120q = 8.66‘156.87q A
ILp = n 3 ILs
I1 = (1/3) 3 (8.66‘36.87q) = 5‘36.87q
I2 = I1‘––120q = 5‘––83.13q A
(c)
p = 3|Ia|2(8) = 3(8.66)2(8) = 1.8 kw.
Chapter 13, Solution 74.
(a)
This is a '-' connection.
(b)
The easy way is to consider just one phase.
1:n = 4:1 or n = 1/4
n = V2/V1 which leads to V2 = nV1 = 0.25(2400) = 600
i.e. VLp = 2400 V and VLs = 600 V
S = p/cosT = 120/0.8 kVA = 150 kVA
pL = p/3 = 120/3 = 40 kw
4:1
IL
VLp
Ipp
ILs
Ips
VLs
But
pLs = VpsIps
For the '-load,
IL =
Hence,
Ips = 40,000/600 = 66.67 A
ILs =
(c)
3 Ips =
3 Ip and VL = Vp
3 x66.67 = 115.48 A
Similarly, for the primary side
ppp = VppIpp = pps or Ipp = 40,000/2400 = 16.667 A
and
(d)
ILp =
3 Ip = 28.87 A
Since S = 150 kVA therefore Sp = S/3 = 50 kVA
Chapter 13, Solution 75.
(a)
n = VLs/( 3 VLp) 4500/(900 3 ) = 2.887
(b)
S =
3 VLsILs or ILs = 120,000/(900 3 ) = 76.98 A
ILs = ILp/(n 3 ) = 76.98/(2.887 3 ) = 15.395 A
Chapter 13, Solution 76.
(a)
At the load,
VL = 240 V = VAB
VAN = VL/ 3 = 138.56 V
Since S =
3 VLIL then IL = 60,000/(240 3 ) = 144.34 A
1:n
0.05 :
2640V
j0.1 :
A
240V
j0.1 :
0.05 :
B
0.05 :
(b)
j0.1 :
C
Balanced
Load
60kVA
0.85pf
leading
Let VAN = |VAN|‘0q = 138.56‘0q
cosT = pf = 0.85 or T = 31.79q
IAA’’ = IL‘T = 144.34‘31.79q
VA’’N’’ = ZIAA’’ + VAN
= 138.56‘0q + (0.05 + j0.1)(144.34‘31.79q)
= 138.03‘6.69q
VLs = VA’’N’’
(c)
3 = 137.8
3 = 238.7 V
For Y-' connections,
n =
3 VLs/Vps =
3 x238.7/2640 = 0.1569
fLp = nILs/ 3 = 0.1569x144.34/ 3 = 13.05 A
Chapter 13, Solution 77.
(a)
This is a single phase transformer.
V1 = 13.2 kV, V2 = 120 V
n = V2/V1 = 120/13,200 = 1/110, therefore n = 110
(b)
P = VI or I = P/V = 100/120 = 0.8333 A
I1 = nI2 = 0.8333/110 = 7.576 mA
Chapter 13, Solution 78.
The schematic is shown below.
k = M / L1 L 2 = 1 / 6 x 3 = 0.2357
In the AC Sweep box, set Total Pts = 1, Start Freq = 0.1592 and End Freq = 0.1592.
After simulation, the output file includes
From this,
FREQ
IM(V_PRINT1)
IP(V_PRINT1)
1.592 E––01
4.253 E+00
––8.526 E+00
FREQ
IM(V_PRINT2)
IP(V_PRINT2)
1.592 E––01
1.564 E+00
2.749 E+01
I1 = 4.253‘––8.53q A, I2 = 1.564‘27.49q A
The power absorbed by the 4-ohm resistor = 0.5|I|2R = 0.5(1.564)2x4
= 4.892 watts
Chapter 13, Solution 79.
The schematic is shown below.
k1 = 15 / 5000 = 0.2121, k2 = 10 / 8000 = 0.1118
In the AC Sweep box, we type Total Pts = 1, Start Freq = 0.1592, and End Freq =
0.1592. After the circuit is saved and simulated, the output includes
FREQ
IM(V_PRINT1)
IP(V_PRINT1)
1.592 E––01
4.068 E––01
––7.786 E+01
FREQ
IM(V_PRINT2)
IP(V_PRINT2)
1.592 E––01
1.306 E+00
––6.801 E+01
FREQ
IM(V_PRINT3)
IP(V_PRINT3)
1.592 E––01
1.336 E+00
––5.492 E+01
Thus, I1 = 1.306‘––68.01q A, I2 = 406.8‘––77.86q mA, I3 = 1.336‘––54.92q A
Chapter 13, Solution 80.
The schematic is shown below.
k1 = 10 / 40 x80 = 0.1768, k2 = 20 / 40 x 60 = 0.482
k3 = 30 / 80x 60 = 0.433
In the AC Sweep box, we set Total Pts = 1, Start Freq = 0.1592, and End Freq =
0.1592. After the simulation, we obtain the output file which includes
i.e.
FREQ
IM(V_PRINT1)
IP(V_PRINT1)
1.592 E––01
1.304 E+00
6.292 E+01
Io = 1.304‘62.92q A
Chapter 13, Solution 81.
The schematic is shown below.
k1 = 2 / 4x8 = 0.3535, k2 = 1 / 2 x8 = 0.25
In the AC Sweep box, we let Total Pts = 1, Start Freq = 100, and End Freq = 100.
After simulation, the output file includes
FREQ
1.000 E+02
IM(V_PRINT1)
1.0448 E––01
IP(V_PRINT1)
1.396 E+01
FREQ
1.000 E+02
IM(V_PRINT2)
2.954 E––02
IP(V_PRINT2)
––1.438 E+02
FREQ
1.000 E+02
IM(V_PRINT3)
2.088 E––01
IP(V_PRINT3)
2.440 E+01
i.e.
I1 = 104.5‘13.96q mA, I2 = 29.54‘––143.8q mA,
I3 = 208.8‘24.4q mA.
Chapter 13, Solution 82.
The schematic is shown below. In the AC Sweep box, we type Total Pts = 1, Start Freq
= 0.1592, and End Freq = 0.1592. After simulation, we obtain the output file which
includes
FREQ
1.592 E––01
IM(V_PRINT1)
1.955 E+01
IP(V_PRINT1)
8.332 E+01
FREQ
1.592 E––01
IM(V_PRINT2)
6.847 E+01
IP(V_PRINT2)
4.640 E+01
FREQ
1.592 E––01
IM(V_PRINT3)
4.434 E––01
IP(V_PRINT3)
––9.260 E+01
i.e.
V1 = 19.55‘83.32q V, V2 = 68.47‘46.4q V,
Io = 443.4‘––92.6q mA.
Chapter 13, Solution 83.
The schematic is shown below. In the AC Sweep box, we set Total Pts = 1, Start Freq
= 0.1592, and End Freq = 0.1592. After simulation, the output file includes
FREQ
1.592 E––01
IM(V_PRINT1)
1.080 E+00
IP(V_PRINT1)
3.391 E+01
FREQ
1.592 E––01
VM($N_0001)
1.514 E+01
VP($N_0001)
––3.421 E+01
i.e.
iX = 1.08‘33.91q A, Vx = 15.14‘––34.21q V.
Chapter 13, Solution 84.
The schematic is shown below. We set Total Pts = 1, Start Freq = 0.1592, and End
Freq = 0.1592. After simulation, the output file includes
FREQ
1.592 E––01
IM(V_PRINT1)
4.028 E+00
IP(V_PRINT1)
––5.238 E+01
FREQ
1.592 E––01
IM(V_PRINT2)
2.019 E+00
IP(V_PRINT2)
––5.211 E+01
FREQ
1.592 E––01
IM(V_PRINT3)
1.338 E+00
IP(V_PRINT3)
––5.220 E+01
i.e.
I1 = 4.028‘––52.38q A, I2 = 2.019‘––52.11q A,
I3 = 1.338‘––52.2q A.
Chapter 13, Solution 85.
Z1
VS
+
ZL/n2
For maximum power transfer,
Z1 = ZL/n2 or n2 = ZL/Z1 = 8/7200 = 1/900
n = 1/30 = N2/N1. Thus N2 = N1/30 = 3000/30 = 100 turns.
Chapter 13, Solution 86.
n = N2/N1 = 48/2400 = 1/50
ZTh = ZL/n2 = 3/(1/50)2 = 7.5 k:
Chapter 13, Solution 87.
ZTh = ZL/n2 or n =
Z L / Z Th
75 / 300 = 0.5
Chapter 13, Solution 88.
n = V2/V1 = I1/I2 or I2 = I1/n = 2.5/0.1 = 25 A
p = IV = 25x12.6 = 315 watts
Chapter 13, Solution 89.
n = V2/V1 = 120/240 = 0.5
S = I1V1 or I1 = S/V1 = 10x103/240 = 41.67 A
S = I2V2 or I2 = S/V2 = 104/120 = 83.33 A
Chapter 13, Solution 90.
(a)
n = V2/V1 = 240/2400 = 0.1
(b)
n = N2/N1 or N2 = nN1 = 0.1(250) = 25 turns
(c)
S = I1V1 or I1 = S/V1 = 4x103/2400 = 1.6667 A
S = I2V2 or I2 = S/V2 = 4x104/240 = 16.667 A
Chapter 13, Solution 91.
(a)
The kVA rating is S = VI = 25,000x75 = 1875 kVA
(b)
Since S1 = S2 = V2I2 and I2 = 1875x103/240 = 7812 A
Chapter 13, Solution 92.
(a)
V2/V1 = N2/N1 = n, V2 = (N2/N1)V1 = (28/1200)4800 = 112 V
(b)
I2 = V2/R = 112/10 = 11.2 A and I1 = nI2, n = 28/1200
I1 = (28/1200)11.2 = 261.3 mA
(c)
p = |I2|2R = (11.2)2(10) = 1254 watts.
Chapter 13, Solution 93.
(a)
For an input of 110 V, the primary winding must be connected in parallel, with
series-aiding on the secondary. The coils must be series-opposing to give 12 V. Thus the
connections are shown below.
110 V
12 V
(b)
To get 220 V on the primary side, the coils are connected in series, with seriesaiding on the secondary side. The coils must be connected series-aiding to give 50 V.
Thus, the connections are shown below.
220 V
50 V
Chapter 13, Solution 94.
V2/V1 = 110/440 = 1/4 = I1/I2
There are four ways of hooking up the transformer as an auto-transformer. However it is
clear that there are only two outcomes.
V1
V1
V1
V2
V1
V2
(1)
(2)
V2
(3)
V2
(4)
(1) and (2) produce the same results and (3) and (4) also produce the same results.
Therefore, we will only consider Figure (1) and (3).
(a)
For Figure (3), V1/V2 = 550/V2 = (440 –– 110)/440 = 330/440
Thus,
(b)
V2 = 550x440/330 = 733.4 V (not the desired result)
For Figure (1), V1/V2 = 550/V2 = (440 + 110)/440 = 550/440
Thus,
V2 = 550x440/550 = 440 V (the desired result)
Chapter 13, Solution 95.
(a)
n = Vs/Vp = 120/7200 = 1/60
(b)
Is = 10x120/144 = 1200/144
S = VpIp = VsIs
Ip = VsIs/Vp = (1/60)x1200/144 = 139 mA
Chapter 14, Solution 1.
H (Z)
Vo
Vi
H (Z)
jZ Z 0
,
1 jZ Z 0
H
H (Z)
jZRC
1 jZRC
R
R 1 jZC
where Z 0
Z Z0
1 (Z Z0 ) 2
I
1
RC
‘H (Z)
§Z·
S
tan -1 ¨ ¸
2
© Z0 ¹
This is a highpass filter. The frequency response is the same as that for P.P.14.1
except that Z0 1 RC . Thus, the sketches of H and I are shown below.
H
1
0.7071
0
Z0 = 1/RC
Z
I
90q
45q
0
Z0 = 1/RC
Z
Chapter 14, Solution 2.
H (Z)
H
R
R jZL
1
1 jZL R
1
H (Z)
1 (Z Z0 ) 2
1
,
1 jZ Z 0
where Z0
I
§Z·
- tan -1 ¨ ¸
© Z0 ¹
‘H (Z)
R
L
The frequency response is identical to the response in Example 14.1 except that
Z0 R L . Hence the response is shown below.
H
1
0.7071
Z0 = R/L
0
I
Z
Z0 = R/L
0q
Z
-45q
-90q
Chapter 14, Solution 3.
(a)
The Thevenin impedance across the second capacitor where Vo is taken is
R
Z Th R R || 1 sC R 1 sRC
VTh
1 sC
V
R 1 sC i
Vi
1 sRC
ZTh
VTh
+
1
sC
+
Vo
1 sC
˜V
Z Th 1 sC Th
Vo
(b)
Vi
(1 sRC)(1 sCZ Th )
H (s)
Vo
Vi
H (s)
1
s R C 3sRC 1
2
1
(1 sCZ Th )(1 sRC)
2
2
(40 u 10 3 )(2 u 10 -6 )
RC
1
(1 sRC)(1 sRC sRC (1 sRC))
80 u 10 -3
0.08
There are no zeros and the poles are at
- 0.383
s1
- 4.787
RC
s2
- 2.617
RC
- 32.712
Chapter 14, Solution 4.
(a)
(b)
R ||
1
jZC
R
1 jZRC
R
1 jZRC
R
jZL 1 jZRC
H (Z)
Vo
Vi
H (Z)
R
- Z RLC R jZL
H (Z)
R jZL
R jZL 1 jZC
H (Z)
- Z 2 LC jZRC
1 Z 2 LC jZRC
R
R jZL (1 jZRC)
2
jZC (R jZL)
1 jZC (R jZL)
Chapter 14, Solution 5.
(a)
(b)
1 jZC
R jZL 1 jZC
H (Z)
Vo
Vi
H (Z)
1
1 jZRC Z 2 LC
R ||
1
jZC
R
1 jZRC
jZL
jZL R (1 jZRC)
H (Z)
Vo
Vi
H (Z)
jZL Z 2 RLC
R jZL Z 2 RLC
jZL (1 jZRC)
R jZL (1 jZRC)
Chapter 14, Solution 6.
(a)
(b)
Using current division,
Io
R
H (Z)
I i R jZL 1 jZC
H (Z)
jZRC
1 jZRC Z2 LC
H (Z)
jZ5
1 jZ5 2.5 Z 2
jZ (20)(0.25)
1 jZ(20)(0.25) Z2 (10)(0.25)
We apply nodal analysis to the circuit below.
Io
Vx
Is
R
1/jZC
0.5 Vx
+ jZL
Is
Vx Vx 0.5Vx
R jZL 1 jZC
But
Is
Vx
0.5 Vx
jZL 1 jZC
Io
H (Z)
H (Z)
H (Z)
1
1
R 2 ( jZL 1 jZC)
2 ( jZL 1 jZC)
1
R
Io
Is
1
1 2 ( jZ L 1 jZ C ) R
jZ
jZ 2 (1 Z2 0.25)
jZ
2 jZ 0.5 Z 2
Chapter 14, Solution 7.
(a)
0.05
20 log10 H
2.5 u 10 -3
H 10 2.5u10
(b)
log10 H
-3
1.005773
- 6.2 20 log10 H
- 0.31 log10 H
H 10 -0.31
(c)
2 I o ( jZL 1 jZC)
1
0 .5
R jZL 1 jZC
Is
2 I o ( jZL 1 jZC)
Is
Io

o Vx
0.4898
104.7 20 log10 H
5.235 log10 H
H 10 5.235
1.718 u 10 5
jZRC
jZRC 2 (1 Z 2 LC)
Chapter 14, Solution 8.
(a)
(b)
(c)
(d)
H 0.05
H dB 20 log10 0.05
- 26.02 ,
H 125
H dB 20 log10 125
j10
2 j
H(1)
41.94 ,
I
0q
I
0q
I
63.43q
4.472‘63.43q
H dB
20 log10 4.472
13.01 ,
H(1)
3
6
1 j 2 j
3.9 j1.7
H dB
20 log10 4.254
12.577 ,
4.254‘ - 23.55q
I
- 23.55q
Chapter 14, Solution 9.
1
(1 jZ)(1 jZ 10)
H (Z)
H dB
I
-20 log10 1 jZ 20 log10 1 jZ / 10
- tan -1 (Z) tan -1 (Z / 10)
The magnitude and phase plots are shown below.
HdB
0.1
1
10
Z
100
20 log 10
-20
1
1 jZ / 10
20 log10
-40
1
1 jZ
I
0.1
-45q
1
10
Z
100
arg
1
1 jZ / 10
-90q
-135q
-180q
arg
1
1 jZ
Chapter 14, Solution 10.
H( jZ)
50
jZ(5 jZ)
10
jZ ·
§
1 jZ¨1 ¸
5 ¹
©
HdB
40
20 log1
20
10
0.1
-20
1
100
§
¨
1
20 log¨
¨
jZ
¨ 1
5
©
§ 1 ·
¸
20 log¨¨
¸
© jZ ¹
-40
I
0.1
·
¸
¸
¸
¸
¹
Z
1
10
-45q
Z
100
arg
1
1 jZ / 5
-90q
arg
-135q
1
jZ
-180q
Chapter 14, Solution 11.
H (Z)
H dB
I
5 (1 jZ 10)
jZ (1 jZ 2)
20 log10 5 20 log10 1 jZ 10 20 log10 jZ 20 log10 1 jZ 2
-90q tan -1 Z 10 tan -1 Z 2
The magnitude and phase plots are shown below.
HdB
40
34
20
14
0.1
-20
1
10
100
Z
1
10
100
Z
-40
I
90q
45q
0.1
-45q
-90q
Chapter 14, Solution 12.
T ( w)
0.1(1 jZ )
,
jZ (1 jZ / 10)
20 log 0.1 20
The plots are shown below.
|T|
(db)
20
Z
0
0.1
1
10
100
-20
-40
arg T
90o
Z
0
0.1
-90o
1
10
100
Chapter 14, Solution 13.
1 jZ
( jZ) 2 (10 jZ)
G (Z)
G dB
I
(1 10)(1 jZ)
( jZ) 2 (1 jZ 10)
-20 20 log10 1 jZ 40 log10 jZ 20 log10 1 jZ 10
-180q tan -1Z tan -1 Z 10
The magnitude and phase plots are shown below.
GdB
40
20
0.1
-20
1
10
100
Z
1
10
100
Z
-40
I
90q
0.1
-90q
-180q
Chapter 14, Solution 14.
50
25
H (Z)
H dB
1 jZ
§ jZ10 § jZ · 2 ·
¨ ¸ ¸¸
jZ¨¨1 25 © 5 ¹ ¹
©
20 log10 2 20 log10 1 jZ 20 log10 jZ
20 log10 1 jZ2 5 ( jZ 5) 2
I
§ Z10 25 ·
¸
-90q tan -1 Z tan -1 ¨
©1 Z2 5 ¹
The magnitude and phase plots are shown below.
HdB
40
26
20
6
0.1
-20
1
10
100
Z
1
10
100
Z
-40
I
90q
0.1
-90q
-180q
Chapter 14, Solution 15.
40 (1 jZ)
(2 jZ)(10 jZ)
H (Z)
H dB
I
2 (1 jZ)
(1 jZ 2)(1 jZ 10)
20 log10 2 20 log10 1 jZ 20 log10 1 jZ 2 20 log10 1 jZ 10
tan -1 Z tan -1 Z 2 tan -1 Z 10
The magnitude and phase plots are shown below.
HdB
40
20
6
0.1
-20
1
10
100
Z
1
10
100
Z
-40
I
90q
45q
0.1
-45q
-90q
Chapter 14, Solution 16.
G (Z)
jZ
jZ ·
§
100(1 jZ)¨1 ¸
© 10 ¹
2
GdB
20
0.1
20 log jZ
1
10
100
40 log
-20
-40
Z
jZ
10
20 log(1/100)
-60
I
90q
arg(jȦ)
Z
0.1
-90q
-180q
1
arg
10
100
arg
1
jZ ·
§
¨1 ¸
© 10 ¹
2
1
1 jZ
Chapter 14, Solution 17.
(1 4) jZ
(1 jZ)(1 jZ 2) 2
G (Z)
G dB
I
-20log10 4 20 log10 jZ 20 log10 1 jZ 40 log10 1 jZ 2
-90q - tan -1Z 2 tan -1 Z 2
The magnitude and phase plots are shown below.
GdB
20
0.1
-12
-20
-40
1
10
100
Z
I
90q
0.1
1
10
100
Z
-90q
-180q
Chapter 14, Solution 18.
4 (1 jZ 2) 2
50 jZ (1 jZ 5)(1 jZ 10)
G (Z)
G dB
20 log10 4 50 40 log10 1 jZ 2 20 log10 jZ
20 log10 1 jZ 5 20 log10 1 jZ 10
where 20 log10 4 50
I
-21.94
-90q 2 tan -1 Z 2 tan -1 Z 5 tan -1 Z 10
The magnitude and phase plots are shown below.
GdB
20
0.1
-20
1
10
100
Z
1
10
100
Z
-40
-60
I
180q
90q
0.1
-90q
Chapter 14, Solution 19.
jZ
100 (1 jZ 10 Z2 100)
H (Z)
H dB
I
20 log10 jZ 20 log10 100 20 log10 1 jZ 10 Z2 100
§ Z 10 ·
¸
90q tan -1 ¨
©1 Z2 100 ¹
The magnitude and phase plots are shown below.
HdB
40
20
0.1
-20
1
10
100
Z
1
10
100
Z
-40
-60
I
90q
0.1
-90q
-180q
Chapter 14, Solution 20.
10 (1 jZ Z2 )
(1 jZ)(1 jZ 10)
N(Z)
20 20 log10 1 jZ 20 log10 1 jZ 10 20 log10 1 jZ Z2
N dB
I
§ Z ·
¸ tan -1 Z tan -1 Z 10
tan -1 ¨
©1 Z2 ¹
The magnitude and phase plots are shown below.
NdB
40
20
0.1
-20
1
10
100
Z
1
10
100
Z
-40
I
180q
90q
0.1
-90q
Chapter 14, Solution 21.
T(Z)
TdB
jZ (1 jZ)
100 (1 jZ 10)(1 jZ 10 Z2 100)
20 log10 jZ 20 log10 1 jZ 20 log10 100
20 log10 1 jZ 10 20 log10 1 jZ 10 Z2 100
I
§ Z 10 ·
¸
90q tan -1 Z tan -1 Z 10 tan -1 ¨
©1 Z2 100 ¹
The magnitude and phase plots are shown below.
TdB
20
0.1
-20
1
10
100
Z
1
10
100
Z
-40
-60
I
180q
90q
0.1
-90q
-180q
Chapter 14, Solution 22.
20
20 log10 k

o k
10
A zero of slope 20 dB / dec at Z
2 
o 1 jZ 2
A pole of slope - 20 dB / dec at Z
20 
o
1
1 jZ 20
A pole of slope - 20 dB / dec at Z 100 
o
Hence,
H (Z)
10 (1 jZ 2)
(1 jZ 20)(1 jZ 100)
H (Z)
10 4 ( 2 jZ)
( 20 jZ)(100 jZ)
1
1 jZ 100
Chapter 14, Solution 23.
A zero of slope 20 dB / dec at the origin

o
A pole of slope - 20 dB / dec at Z 1 
o
1
1 jZ 1
A pole of slope - 40 dB / dec at Z 10 
o
Hence,
H (Z)
jZ
(1 jZ)(1 jZ 10) 2
H (Z)
100 jZ
(1 jZ)(10 jZ) 2
jZ
1
(1 jZ 10) 2
Chapter 14, Solution 24.
The phase plot is decomposed as shown below.
I
90q
arg (1 jZ / 10)
45q
0.1
-45q
-90q
1
arg ( jZ)
10
100
1000
Z
§
·
1
¸
arg ¨
©1 jZ / 100 ¹
k c (1 jZ 10)
jZ (1 jZ 100)
G (Z)
k c (10)(10 jZ)
jZ (100 jZ)
where k c is a constant since arg k c
G (Z)
Hence,
0.
k (10 jZ)
,
jZ (100 jZ)
where k
10k c is constant
Chapter 14, Solution 25.
Z0
1
LC
1
(40 u 10 -3 )(1 u 10 -6 )
Z(Z0 )
R
5 krad / s
2 k:
Z(Z0 4)
§ Z0
4 ·
¸
R j¨ L
Z0 C ¹
© 4
Z(Z0 4)
·
§ 5 u 10 3
4
¸
2000 j ¨
˜ 40 u 10 -3 (5 u 10 3 )(1 u 10 -6 ) ¹
© 4
Z(Z0 4)
2000 j (50 4000 5)
Z(Z0 4)
2 j0.75 k:
Z(Z0 2)
§ Z0
2 ·
¸
R j¨ L
Z0 C ¹
© 2
Z(Z0 2)
§ (5 u 10 3 )
·
2
¸
2000 j ¨
(40 u 10 -3 ) 2
(5 u 10 3 )(1 u 10 -6 ) ¹
©
Z(Z0 4)
2000 j (100 2000 5)
Z(Z0 2)
2 j0.3 k:
Z(2Z0 )
§
1 ·
¸
R j ¨ 2Z0 L 2Z0 C ¹
©
Z(2Z0 )
§
·
1
¸
2000 j ¨ (2)(5 u 10 3 )(40 u 10 -3 ) (2)(5 u 10 3 )(1 u 10 -6 ) ¹
©
Z(2Z0 )
2 j0.3 k:
Z(4Z0 )
§
1 ·
¸
R j ¨ 4Z0 L 4Z0 C ¹
©
Z(4Z0 )
§
·
1
¸
2000 j ¨ (4)(5 u 10 3 )(40 u 10 -3 ) 3
-6
(4)(5 u 10 )(1 u 10 ) ¹
©
Z(4Z0 )
2 j0.75 k:
Chapter 14, Solution 26.
(a)
fo
(b)
B
(c )
Q
1
1
2S LC
2S 5 x10 9 x10 x10 3
R
L
100
10 x10 3
Zo L
R
10 krad/s
L
LC R
1
10 6 10 x10 3
3
50 0.1x10
Chapter 14, Solution 27.
At resonance,
Z
R
B
R
L
10 : ,
1
Z0
and
Q
LC
Z0
B
Z0 L
R
Hence,
L
RQ
Z0
22.51 kHz
(10)(80)
50
16 H
14.142
C
1
Z02 L
B
R
L
Therefore,
R 10 : ,
1
(50) 2 (16)
10
16
L
25 PF
0.625 rad / s
16 H ,
C
25 PF ,
B
0.625 rad / s
Chapter 14, Solution 28.
Let R
10 : .
L
R
B
C
1
Z02 L
Q
Z0
B
10
20
0.5 H
1
(1000) 2 (0.5)
1000
20
2 PF
50
Therefore, if R 10 : then
C 2 PF ,
L 0.5 H ,
Q
50
Chapter 14, Solution 29.
jZ
Z
Z
jZ 1/jZ
1
1
jZ
j Z 1 jZ
jZ
§
1 · Z 2 jZ
j ¨Z ¸ ©
Z ¹ 1 Z2
Z
Since v( t ) and i( t ) are in phase,
1
Z
Im(Z) 0 Z Z 1 Z2
Z4 Z2 1 0
Z2
-1 r 1 4
2
Z
0.7861 rad / s
0.618
Chapter 14, Solution 30.
Select R
10 : .
L
R
Z0 Q
10
(10)(20)
C
1
Z02 L
1
(100)(0.05)
B
1
RC
0.05 H
1
(10)(0.2)
5 mH
0.2 F
0.5 rad / s
Therefore, if R 10 : then
L 5 mH , C 0.2 F ,
B
0.5 rad / s
Chapter 14, Solution 31.
ZL
XL
B
R
L
ZR
XL
o
L
XL
Z
2Sx10 x10 6 x 5.6 x10 3
40 x10
3
8.796 x10 6 rad/s
Chapter 14, Solution 32.
Since Q ! 10 ,
Z1
Z0 B
,
2
B
Z0
Q
6 u 10 6
120
Z1
6 0.025
5.975 u 10 6 rad / s
Z2
6 0.025
6.025 u 10 6 rad / s
Z2
Z0 B
2
50 krad / s
Chapter 14, Solution 33.
Q
Zo RC
Q
R
Zo L
o
o
C
L
Q
2Sf o R
80
2Sx5.6x10 6 x 40x10 3
40 x10 3
R
2Sf o Q
2Sx 5.6 x10 6 x80
14.21 PH
Chapter 14, Solution 34.
(a)
Zo
1
1
LC
8x10
3
x 60x10
1
6
1.443 krad/s
(b)
B
1
RC
(c)
Q
Zo RC 1.443x10 3 x 5x10 3 x 60x10 6
5x10 3 x 60x10 6
56.84 pF
3.33 rad/s
432.9
Chapter 14, Solution 35.
At resonance,
Y
1
R
Q
Z0 RC 
o C

o R
1
Z0
1
Y

o L
LC
1
25 u 10 -3
40 :
Q
Z0 R
80
(200 u 10 3 )(40)
1
Z02 C
1
(4 u 10 )(10 u 10 -6 )
10
B
Z0
Q
200 u 10 3
80
Z1
Z0 B
2
200 2.5
197.5 krad / s
Z1
Z0 B
2
200 2.5
202.5 krad / s
2.5 krad / s
Chapter 14, Solution 36.
Z0
1
LC
5000 rad / s
Y(Z0 )
1
R

o Z(Z0 )
R
Y(Z0 4)
§ Z0
4 ·
1
¸
j¨ C
Z0 L ¹
R
© 4
Z(Z0 4)
1
0.0005 j0.01875
Y(Z0 2)
§ Z0
2 ·
1
¸
j¨ C
Z0 L ¹
R
© 2
2 k:
0.5 j18.75 kS
1.4212 j53.3 :
0.5 j7.5 kS
10 PF
2.5 PH
1
0.0005 j0.0075
Z(Z0 2)
8.85 j132.74 :
Y(2Z0 )
§
1 ·
1
¸
j ¨ 2Z0 L 2Z0 C ¹
R
©
Z(2Z0 )
8.85 j132.74 :
Y(4Z0 )
§
1 ·
1
¸
j ¨ 4Z0 L 4Z0 C ¹
R
©
Z(4Z0 )
1.4212 j53.3 :
0.5 j7.5 kS
0.5 j18.75 kS
Chapter 14, Solution 37.
Z
jZL //( R 1
)
jZC
jZL(R R
1
)
jZ C
1
jZ L
jZ C
1 ·
L§
¨ ZL ¸
ZC ¹
C©
1 2
R 2 ( ZL )
ZC
1 ·
§L
·§
)¸
¨ jZLR ¸¨ R j(ZL ZC ¹
©C
¹©
1 2
)
R 2 ( ZL ZC
ZLR 2 Im(Z)

o
0
Thus,
Z
1
LC R 2 C 2
Chapter 14, Solution 38.
Y
1
jZC
R jZL
At resonance, Im(Y)
jZC 0 , i.e.
R jZL
R 2 Z2 L2
Z 2 ( R 2 C 2 LC) 1
Z0 C Z0 L
R 2 Z02 L2
0
L
C
R 2 Z02 L2
§ 50 ·
1
¸
¨
-3
-6 (40 u 10 )(10 u 10 ) © 40 u 10 -3 ¹
Z0
1 R2
LC L2
Z0
4841 rad / s
2
Chapter 14, Solution 39.
(a)
B
Zo
B
Z 2 Z1
2S(f 2 f1 )
1
(Z1 Z 2 )
2
1
RC
Zo
(c )
Zo
(d)
B 8S
(e)
Q
2S(88) x10 3

o
1
(b)
LC
176S
2S(90 86) x10 3
1
BR
C

o
L
8Skrad / s
176S
1
19.89nF
8Sx10 3 x 2x10 3
1
1
Z2 o C
(176S) 2 x19.89x10 9
164.4H
552.9krad / s
25.13krad / s
Zo
B
176S
8S
22
Chapter 14, Solution 40.
(a)
L = 5 + 10 = 15 mH
Z0
1
LC
1
15x10
3
x 20x10
6
1.8257 k rad/sec
(b)
Q
Z0 RC 1.8257 x10 3 x 25x10 3 x 20x10 6
B
1
RC
1
912.8
2 rad
25x10 20x10 6
3
To increase B by 100% means that B’’ = 4.
Cc
1
RBc
1
C1C 2
C1 C 2
Since Cc
10 µF
25x10 3 x 4
10PF and C1 = 20 µF, we then obtain C2 = 20 µF.
Therefore, to increase the bandwidth, we merely add another 20 µF in
series with the first one.
Chapter 14, Solution 41.
(a)
This is a series RLC circuit.
R
26
8:,
1
1
LC
0.4
Z0
(b)
Q
Z0 L
R
B
R
L
1.5811
8
L 1H,
C
1.5811 rad / s
0.1976
8 rad / s
This is a parallel RLC circuit.
3 PF and 6 PF 
o
C
2 PF ,
R
(3)(6)
3 6
2 k: ,
L
2 PF
20 mH
0.4 F
Z0
1
1
(2 u 10 -6 )(20 u 10 -3 )
LC
Q
R
Z0 L
B
1
RC
2 u 10 3
(5 u 10 3 )(20 u 10 -3 )
1
(2 u 10 )(2 u 10 -6 )
3
5 krad / s
20
250 krad / s
Chapter 14, Solution 42.
(a)
Z in
Z in
Z in
(1 jZC) || (R jZL)
R jZL
jZC
R jZL 1
jZC
R jZL
1 Z2 LC jZRC
(R jZL)(1 Z2 LC jZRC)
(1 Z2 LC) 2 Z2 R 2 C 2
At resonance, Im(Z in ) 0 , i.e.
0 ZL(1 Z2 LC) ZR 2 C
Z2 LC
Z0
(b)
L R 2C
L R 2C
LC
1 R2
C L
Z in
jZL || (R 1 jZC)
Z in
jZL (R 1 jZC)
R jZL 1 jZC
Z in
(-Z2 RLC jZL) [(1 Z2 LC) jZRC]
(1 Z2 LC) 2 Z2 R 2 C 2
jZL (1 jZRC)
(1 Z2 LC) jZRC
At resonance, Im(Z in ) 0 , i.e.
0 ZL (1 Z2 LC) Z3 R 2 C 2 L
Z2 (LC R 2 C 2 ) 1
Z0
(c)
Z in
Z in
Z in
1
LC R 2 C 2
R || ( jZL 1 jZC)
R ( jZL 1 jZC)
R (1 Z2 LC)
R jZL 1 jZC (1 Z 2 LC) jZRC
R (1 Z2 LC)[(1 Z2 LC) jZRC]
(1 Z2 LC) 2 Z2 R 2 C 2
At resonance, Im(Z in ) 0 , i.e.
0 R (1 Z2 LC) ZRC
1 Z2 LC 0
Z0
1
LC
Chapter 14, Solution 43.
Consider the circuit below.
1/jZC
Zin
(a)
R1
jZL
Z in
(R 1 || jZL) || (R 2 1 jZC)
Z in
§ R 1 jZL · §
1 ·
¨
¸ || ¨ R 2 ¸
jZC ¹
© R 1 jZL ¹ ©
R2
Z in
jZR 1 L §
1 ·
¸
˜ ¨R 2 R 1 jZL ©
jZC ¹
jR 1ZL
1
R2 jZC R 1 jZL
Z in
jZR 1 L (1 jZR 2 C)
(R 1 jZL)(1 jZR 2 C) Z2 LCR 1
Z in
- Z2 R 1 R 2 LC jZR 1 L
R 1 Z2 LCR 1 Z2 LCR 2 jZ (L R 1 R 2 C)
Z in
(-Z2 R 1 R 2 LC jZR 1 L)[R 1 Z2 LCR 1 Z2 LCR 2 jZ (L R 1 R 2 C)]
(R 1 Z2 LCR 1 Z2 LCR 2 ) 2 Z2 (L R 1 R 2 C) 2
At resonance, Im(Z in ) 0 , i.e.
0 Z3 R 1 R 2 LC (L R 1 R 2 C) ZR 1 L (R 1 Z2 LCR 1 Z2 LCR 2 )
0 Z3 R 12 R 22 LC 2 R 12 ZL Z3 R 12 L2 C
0 Z2 R 22 C 2 1 Z2 LC
Z2 (LC R 22 C 2 ) 1
Z0
Z0
Z0
(b)
At Z Z0
jZL
LC R 22 C 2
1
(0.02)(9 u 10 -6 ) (0.1) 2 (9 u 10 -6 ) 2
2.357 krad / s
2.357 krad / s ,
j(2.357 u 10 3 )(20 u 10 -3 )
R 1 || jZL
R2 1
1
jZC
Z in (Z0 )
j47.14
1 j47.14
0.1 j47.14
0.9996 j0.0212
1
j (2.357 u 10 3 )(9 u 10 -6 )
(R 1 || jZL) || (R 2 1 jZC)
0.1 j47.14
Z in (Z0 )
(0.9996 j0.0212)(0.1 j47.14)
(0.9996 j0.0212) (0.1 j47.14)
Z in (Z0 )
1:
Chapter 14, Solution 44.
We find the input impedance of the circuit shown below.
1
Z
jZ(2/3)
1/jZ
1/jZC
1
Z
3
1
jZ ,
jZ2
1 1 jZC
1
Z
-j1.5 j jC
1 jC
-j0.5 Z 1
C 2 jC
1 C2
v( t ) and i( t ) are in phase when Z is purely real, i.e.
0
-0.5 C
1 C2
1
Z
C2
1 C2
V
ZI
v( t )
1
2

o (C 1) 2

o Z
(2)(10)
20 sin( t ) V ,
1
2:
20
i.e.
Vo
20 V
or
C
1F
Chapter 14, Solution 45.
(a)
jZ
,
1 jZ
1 || jZ
1 ||
1
jZ
1 jZ
1 1 jZ
1
1 jZ
Transform the current source gives the circuit below.
jZ
I
1 jZ
jZ
1 jZ
+
1
1
1 jZ
+
Vo
Vo
(b)
1
jZ
1 jZ
˜
I
1
jZ 1 j Z
1
1 jZ 1 jZ
jZ
2 (1 jZ) 2
H (Z)
Vo
I
H (1)
1
2 (1 j) 2
H (1)
1
2 ( 2)2
0.25
Chapter 14, Solution 46.
(a) This is an RLC series circuit.
Zo
1
o
LC
(b)
(c )
C
1
1
Z2 o L
(2Sx15x10 3 ) 2 x10 x10 3
Z = R, I = V/Z = 120/20 = 6 A
Q
Zo L
R
2Sx15x10 3 x10 x10 3
20
15S
47.12
11.26nF
Chapter 14, Solution 47.
H (Z)
Vo
Vi
R
R jZL
1
1 jZL R
H(0) 1 and H(f)
0 showing that this circuit is a lowpass filter.
1
, i.e.
At the corner frequency, H(Zc )
2
1
1
2
§Zc L ·
¸
1 ¨
© R ¹
2

o 1
Zc L
R
or
Zc
R
L
Hence,
Zc
R
L
fc
1 R
˜
2S L
2Sf c
1 10 u 10 3
˜
2S 2 u 10 -3
796 kHz
Chapter 14, Solution 48.
R ||
H (Z)
1
jZC
jZL R ||
1
jZC
H (Z)
R jZC
R 1 jZC
R jZC
jZL R 1 jZC
H (Z)
R
R jZL Z 2 RLC
H(0) 1 and H(f)
0 showing that this circuit is a lowpass filter.
Chapter 14, Solution 49.
4
2
At dc, H(0)
1
H(Z)
Hence,
2
2.
H(0)
2
4
2
4 100Zc2
4 100Zc2
2
H(2)
101
0 .2
2
1 j10
0.199
In dB, 20 log10 H(2)
arg H(2)
2
8 
o Zc
4
2 j20
H(2)
2
-tan -110
- 14.023
- 84.3q
Chapter 14, Solution 50.
jZL
R jZL
H (Z)
Vo
Vi
H(0)
0 and H(f) 1 showing that this circuit is a highpass filter.
1
H (Zc )
1
2
or
Zc
fc
1 R
˜
2S L
§ R ·
¸
1 ¨
© Zc L ¹
R
L
2

o 1
2Sf c
1 200
˜
2S 0.1
318.3 Hz
R
Zc L
Chapter 14, Solution 51.
jZRC
1 jZRC
H c(Z)
jZ
jZ 1 RC
(from Eq. 14.52)
This has a unity passband gain, i.e. H(f) 1 .
1
Zc 50
RC
H ^ (Z) 10 H c(Z)
j10Z
50 jZ
j10Z
50 jZ
H (Z)
Chapter 14, Problem 52.
Design an RL lowpass filter that uses a 40-mH coil and has a cut-off frequency of
5 kHz.
Chapter 14, Solution 53.
Zc
R
R
L
2Sf c
2Sf c L
(2S)(10 5 )(40 u 10 -3 )
Chapter 14, Solution 54.
Z1
Z2
B
Z0
2Sf 1 20S u 10 3
2Sf 2 22S u 10 3
Z2 Z1
Z2 Z1
2
2S u 10 3
21S u 10 3
25.13 k:
Q
Z0
B
21S
2S
1
Z0
LC
11.5
1
Z02 C

o L
L
1
(21S u 10 ) (80 u 10 -12 )
B
R
L
R
(2S u 10 3 )(2.872)
3 2

o R
2.872 H
BL
18.045 k:
Chapter 14, Solution 55.
Zc
2Sf c
1
RC
o
R
1
2Sf c C
1
2Sx 2x10 x300x10 12
3
265.3k:
Chapter 14, Solution 56.
1
Zo
1
(25 u 10 )(0.4 u 10 6 )
3
LC
B
R
L
10
25 u 10 -3
Q
10
0.4
Z1
Zo B 2 10 0.2
Z2
Zo B 2 10 0.2 10.2 krad / s
10 krad / s
0.4 krad / s
25
Therefore,
1.56 kHz f 1.62 kHz
9.8 krad / s
or
f1
9.8
2S
or
f2
10.2
2S
1.56 kHz
1.62 kHz
Chapter 14, Solution 57.
(a)
From Eq 14.54,
R
H (s)
R sL 1
LC
,
sB
s sB Z02
H (s)
(b)
1
sC
R
and Z0
L
Since B
sRC
1 sRC s 2 LC
R
L
R
1
s2 s L LC
s
2
From Eq. 14.56,
sL H (s)
1
sC
R sL 1
sC
s2 s2 s
1
LC
R
1
L LC
s 2 Z02
s 2 sB Z02
H (s)
Chapter 14, Solution 58.
(a)
Consider the circuit below.
I
Vs
+
R
I1
1/sC
+
1/sC
R
Vo
Z(s)
R
1 §
1·
|| ¨ R ¸
sC ©
sC ¹
1 §
1·
¨R ¸
sC ©
sC ¹
R
2
R
sC
R
Z(s)
1 3sRC s 2 R 2 C 2
sC (2 sRC)
I
Vs
Z
I1
1 sC
I
2 sC R
Vo
I1R
Vs
Z (2 sRC)
R Vs
sC (2 sRC)
˜
2 sRC 1 3sRC s 2 R 2 C 2
H (s)
Vo
Vs
H (s)
ª
3
s
«
1
RC
«
3
1
3 2
«s s 2 2
¬
RC
R C
Thus, Z02
B
(b)
1 sRC
sC (2 sRC)
Z(s)
sRC
1 3sRC s 2 R 2 C 2
1
R C2
2
3
RC
º
»
»
»
¼
Z0
or
Z(s)
sL R || (R sL)
Z(s)
R 2 3sRL s 2 L2
2R sL
Vo
Vs
,
Z
I 1 ˜ sL
I1
1 rad / s
3 rad / s
Similarly,
I
1
RC
sL R
I
2R sL
R (R sL)
2R sL
R Vs
Z (2R sL)
sLR Vs
2R sL
˜ 2
2R sL R 3sRL s 2 L2
H (s)
Vo
Vs
sRL
R 3sRL s 2 L2
2
Thus, Z0
R
L
1 rad / s
B
3R
L
3 rad / s
1 § 3R ·
¨
s¸
3© L ¹
3R
R2
2
s s 2
L
L
Chapter 14, Solution 59.
(a)
(b)
1
Z0
1
LC
B
R
L
Q
Z0
B
(0.1)(40 u 10 -12 )
2 u 10 3
0 .1
2 u 10 4
0.5 u 10 6
2 u 10 4
As a high Q circuit,
B
Z1 Z0 2
Z2
(c)
Z0 0.5 u 10 6 rad / s
B
2
As seen in part (b),
250
10 4 (50 1)
490 krad / s
10 4 (50 1)
510 krad / s
Q
250
Chapter 14, Solution 60.
Consider the circuit below.
Ro
+
1/sC
Vi
+
R
Vo
sL
R (sL 1 sC)
R sL 1 sC
Z(s)
§
1 ·
R || ¨sL ¸
©
sC ¹
Z(s)
R (1 s 2 LC)
1 sRC s 2 LC
H
Vo
Vi
Z
Z Ro
R (1 s 2 LC)
R o sRR o C s 2 LCR o R s 2 LCR
R (1 s 2 LC)
Ro 1 sRC s 2 LC
Z in
Ro Z
Z in
R o sRR o C s 2 LCR o R s 2 LCR
1 sRC s 2 LC
s
jZ
Z in
R o jZRR o C Z2 LCR o R Z2 LCR
1 Z2 LC jZRC
Z in
(R o R Z2 LCR o Z2 LCR jZRR o C)(1 Z 2 LC jZRC)
(1 Z2 LC) 2 (ZRC) 2
Im(Z in )
0 implies that
- ZRC [R o R Z2 LCR o Z2 LCR ] ZRR o C (1 Z2 LC)
0
R o R Z2 LCR o Z2 LCR R o Z2 LCR o
Z LCR
2
H
R
1
Z0
LC
1
(1 u 10 )(4 u 10 -6 )
-3
H(0)
or
H max
R
Ro R
H (f )
At Z1 and Z2 , H
R
2 (R o R )
2
1
2
0
15.811 krad / s
R (1 Z2 LC)
R o jZRR o C R Z2 LCR o Z2 LCR
H max
1
0
§ 1
·
R ¨ 2 LC ¸
©Z
¹
lim
Zo f R o R
RR o C
j
LC (R R o )
Z2
Z
1
2
R
R Ro
H mzx
R (1 Z2 LC)
R o R Z 2 LC (R o R ) jZRR o C
(R o R )(1 Z2 LC)
(ZRR o C) 2 (R o R Z2 LC(R o R )) 2
10 (1 Z2 ˜ 4 u 10 -9 )
(96 u 10 -6 Z) 2 (10 Z2 ˜ 4 u 10 -8 ) 2
10 (1 Z2 ˜ 4 u 10 -9 )
(96 u 10 -6 Z) 2 (10 Z2 ˜ 4 u 10 -8 ) 2
1
2
(10 Z2 ˜ 4 u 10 -8 )( 2 ) (96 u 10 -6 Z) 2 (10 Z2 ˜ 4 u 10 -8 ) 2
(2)(10 Z2 ˜ 4 u 10 -8 ) 2
(96 u 10 -6 Z) 2 (10 Z2 ˜ 4 u 10 -8 ) 2
(96 u 10 -6 Z) 2 (10 Z2 ˜ 4 u 10 -8 ) 2
0
1.6 u 10 -15 Z4 8.092 u 10 -7 Z2 100
0
0
Z4 5.058 u 10 8 6.25 u 1016
Z2
­ 2.9109 u 10 8
®
¯ 2.1471 u 10 8
Z1
14.653 krad / s
Z2
17.061 krad / s
0
Hence,
B
Z2 Z1
17.061 14.653
2.408 krad / s
Chapter 14, Solution 61.
(a)
V
1 jZC
V,
R 1 jZC i
Since V
(b)
V
Vo
Vi
Vo
Vi
Vo
1
1 jZRC
V ,
jZRC
V
1 jZRC i
H (Z)
V
Vo
R
V,
R 1 jZC i
Since V
Vo
V ,
1
V
1 jZRC i
H (Z)
V
Vo
jZRC
1 jZRC
Chapter 14, Solution 62.
This is a highpass filter.
(a)
H (Z)
jZRC
1 jZRC
H (Z)
1
,
1 j Zc Z
H (Z)
1
1 jf c f
H (f
Vo
(b)
H (f
Vo
(c)
H (f
Vo
1
1 j ZRC
Zc
2S (1000)
1
1 j1000 f
1
1 j5
200 Hz)
120 mV
Vo
Vi
23.53 mV
1 j5
1
1 j0.5
2 kHz)
1
RC
120 mV
Vo
Vi
107.3 mV
1 j0.5
1
1 j0.1
10 kHz)
120 mV
Vo
Vi
119.4 mV
1 j0.1
Chapter 14, Solution 63.
For an active highpass filter,
H (s)
sC i R f
1 sC i R i
(1)
But
H(s)
10s
1 s / 10
(2)
Comparing (1) and (2) leads to:
Ci R f
10

o
Rf
10
Ci
Ci R i
0.1
o
Ri
0.1
100k:
Ci
10M:
Chapter 14, Solution 64.
Zf
R f ||
1
jZC f
Rf
1 jZR f C f
Zi
Ri 1
jZC i
1 jZR i C i
jZC i
Hence,
H (Z)
Vo
Vi
- Zf
Zi
- jZR f C i
(1 jZR f C f )(1 jZR i C i )
This is a bandpass filter. H(Z) is similar to the product of the transfer function
of a lowpass filter and a highpass filter.
Chapter 14, Solution 65.
V
R
V
R 1 jZC i
V
Ri
V
Ri Rf o
Since V
V ,
jZRC
V
1 jZRC i
Ri
V
Ri Rf o
H (Z)
jZRC
V
1 jZRC i
§
R · § jZRC ·
¨1 f ¸ ¨
¸
R i ¹ © 1 jZRC ¹
©
Vo
Vi
It is evident that as Z o f , the gain is 1 Rf
1
and that the corner frequency is
.
Ri
RC
Chapter 14, Solution 66.
(a)
Proof
(b)
When R 1 R 4
H (s)
(c)
R 2R 3 ,
R4
s
˜
R 3 R 4 s 1 R 2C
When R 3 o f ,
H (s)
- 1 R 1C
s 1 R 2C
Chapter 14, Solution 67.
DC gain
Rf
Ri
1
4

o R i
Corner frequency Zc
If we select R f
C
20 k: , then R i
1
(2S)(500)(20 u 10 3 )
Therefore, if R f
1
R f Cf
4R f
2S (500) rad / s
80 k: and
15.915 nF
20 k: , then R i
80 k: and C
15.915 nF
Chapter 14, Solution 68.
High frequency gain
Corner frequency Zc
If we select R i
C
Rf
Ri
5
1
R i Ci
20 k: , then R f
1
(2S)(200)(20 u 10 3 )
Therefore, if R i

o R f
5R i
2S (200) rad / s
100 k: and
39.8 nF
20 k: , then R f
100 k: and C
39.8 nF
Chapter 14, Solution 69.
This is a highpass filter with f c
1
Zc 2Sf c
RC
RC
1
2Sf c
2 kHz.
1
4S u 103
10 8 Hz may be regarded as high frequency. Hence the high-frequency gain is
R f 10
or
R f 2 .5 R
R
4
If we let R
10 k: , then R f
25 k: , and C
1
4000S u 10 4
7.96 nF .
Chapter 14, Solution 70.
(a)
Vo (s)
Vi (s)
H (s)
1
R1
where Y1
G 1 , Y2
1
R2
G 2 , Y3
sC1 , Y4
sC 2 .
G 1G 2
G 1 G 2 sC 2 (G 1 G 2 sC1 )
H (s)
(b)
Y1 Y2
Y1 Y2 Y4 (Y1 Y2 Y3 )
G 1G 2
H(f) 0
1,
G 1G 2
showing that this circuit is a lowpass filter.
H ( 0)
Chapter 14, Solution 71.
R
50 : , L
Lc
40 mH , C 1 PF
Km
L 
o 1
Kf
25K f
Cc
Km
˜ (40 u 10 -3 )
Kf
(1)
Km
C
KmKf
10 6 K f

o 1
1
Km
10 -6
KmKf
(2)
Substituting (1) into (2),
1
10 6 K f
25K f
Kf
0.2 u 10 -3
Km
25K f
5 u 10 -3
Chapter 14, Solution 72.
LcCc
K
2
f
LC
K f2
(4 u 10 -3 )(20 u 10 -6 )
(1)(2)
Kf
2 u 10 -4
Lc
Cc
L 2
K
C m
K
2
m
Km
LC
L cCc

o K f2

o K 2m
(1)(20 u 10 -6 )
(2)(4 u 10 -3 )
4 u 10 -8
Lc C
˜
Cc L
2.5 u 10 -3
5 u 10 -2
Chapter 14, Solution 73.
Rc
KmR
(12)(800 u 10 3 )
Lc
Km
L
Kf
800
(40 u 10 -6 )
1000
Cc
C
KmKf
300 u 10 -9
(800)(1000)
9.6 M:
32 PF
0.375 pF
Chapter 14, Solution 74.
300:
R '1
K m R1
3x100
R '2
KmR 2
10 x100 1 k:
L'
C'
Km
L
Kf
C
KmKf
10 2
10 6
( 2)
1
10
108
200 PH
1 nF
Chapter 14, Solution 75.
R'
KmR
L'
Km
L
Kf
C'
C
KmKf
20 x10
10
10 5
( 4)
1
10x10 5
200 :
400 PH
1 PF
Chapter 14, Solution 76.
R'
L'
50 x10 3
KmR
Km
L 10 PH
Kf
C' 40 pF
C
KmKf
o
R
50 x10 3
L 10 x10 6 x
o
o
C
50 :
10 3
10 6
10 3
10 mH
40 x10 12 x10 3 x10 6
Chapter 14, Solution 77.
L and C are needed before scaling.
B
Z0
(a)
R
L

o L
1
LC
R
B

o C
Lc
KmL
(600)(2)
Cc
C
Km
3.125 u 10 -4
600
10
5
1
Z02 L
2H
1
(1600)(2)
1200 H
0.5208 PF
312.5 PF
40 mF
(b)
(c)
Lc
L
Kf
2
10 3
Cc
C
Kf
3.125 u 10 -4
10 3
Lc
Km
L
Kf
Cc
C
KmKf
2 mH
(400)(2)
10 5
312.5 nF
8 mH
3.125 u 10 -4
(400)(10 5 )
7.81 pF
Chapter 14, Solution 78.
Rc
KmR
(1000)(1) 1 k:
Lc
Km
L
Kf
10 3
(1)
10 4
Cc
C
KmKf
0 .1 H
1
(10 )(10 4 )
3
0.1 PF
The new circuit is shown below.
1 k:
+
I
1 k:
0.1 H
0.1 PF
1 k:
Vx
Chapter 14, Solution 79.
(a)
Insert a 1-V source at the input terminals.
Ro
Io
1V
R
V1
V2
1/sC
+ +
+
sL
3Vo
Vo
There is a supernode.
1 V1
V2
R
sL 1 sC
But
V1
V2 3Vo
Also,
Vo
sL
V
sL 1 sC 2
Combining (2) and (3)
(1)

o V2

o
V1 3Vo
Vo
sL
V2
sL 1 sC
(3)
sL 1 sC
Vo
sL
V2
V1 3Vo
Vo
s 2 LC
V
1 4s 2 LC 1
(4)
Substituting (3) and (4) into (1) gives
1 V1 Vo
sC
V
R
sL 1 4s 2 LC 1
sRC
1 V1 V
1 4s 2 LC 1
V1
(2)
1 4s 2 LC
1 4s 2 LC sRC
1 4s 2 LC sRC
V1
1 4s 2 LC
Io
1 V1
R
Z in
1
Io
Z in
4sL R When R
sRC
R (1 4s 2 LC sRC)
1 sRC 4s 2 LC
sC
1
sC
5, L
Z in (s)
(5)
2, C
8s 5 0 .1 ,
10
s
At resonance,
Im(Z in )
or
(b)
Z0
0
4ZL 1
ZC
1
1
2 LC
2 (0.1)(2)
1.118 rad / s
After scaling,
Rc 
o K m R
4: 
o 40 :
5: 
o 50 :
Lc
Km
L
Kf
Cc
C
KmKf
10
( 2)
100
0 .2 H
0.1
(10)(100)
10 -4
From (5),
Z in (s)
Z0
0.8s 50 10 4
s
1
1
2 LC
2 (0.2)(10 -4 )
111.8 rad / s
Chapter 14, Solution 80.
(a)
Rc
KmR
(200)(2)
400 :
Lc
KmL
Kf
(200)(1)
10 4
20 mH
Cc
C
KmKf
0.5
(200)(10 4 )
0.25 PF
The new circuit is shown below.
20 mH
a
Ix
0.25 PF
400 :
0.5 Ix
b
(b)
Insert a 1-A source at the terminals a-b.
a
sL
V1
V2
Ix
1A
1/(sC)
R
0.5 Ix
b
At node 1,
1 sCV1 V1 V2
sL
At node 2,
V1 V2
0 .5 I x
sL
But, I x
sC V1 .
(1)
V2
R
V1 V2
0.5sC V1
sL
V2
R
(2)
Solving (1) and (2),
sL R
V1
2
s LC 0.5sCR 1
sL R
s LC 0.5sCR 1
V1
1
Z Th
2
At Z 10 4 ,
Z Th
( j10 4 )(20 u 10 -3 ) 400
( j10 4 ) 2 (20 u 10 -3 )(0.25 u 10 -6 ) 0.5( j10 4 )(0.25 u 10 -6 )(400) 1
Z Th
400 j200
0.5 j0.5
Z Th
632.5‘ - 18.435q ohms
600 j200
Chapter 14, Solution 81.
(a)
1
Z
G jZC which leads to
Z(Z)
(G jZC)(R jZL) 1
R jZ L
1
R jZ L
Z
jZL R
2
Z LC jZ(RC LG) GR 1
Z R
C LC
§ R G · GR 1
Z2 jZ¨ ¸ LC
©L C¹
j
(1)
We compare this with the given impedance:
Z(Z)
1000( jZ 1)
2
Z 2 jZ 1 2500
(2)
Comparing (1) and (2) shows that
1
C

o
1000
R G
L C

o
2
GR 1
LC
2501
C 1 mF,
G
R/L 1

o
R
C 1 mS
10 3 R 1
o
10 3 R
R
0 .4
L
Thus,
R = 0.4ȍ, L = 0.4 H, C = 1 mF, G = 1 mS
(b) By frequency-scaling, Kf =1000.
R’’ = 0.4 ȍ, G’’ = 1 mS
L
Kf
L'
0.4
10 3
0.4mH ,
C'
C
Kf
10 3
10 3
Chapter 14, Solution 82.
Cc
C
KmKf
Kf
Zcc
Z
Km
C 1
˜
Cc K f
Rc
KmR
200
1
200
1
1
-6 ˜
10 200
5 k:,
5000
thus,
R cf
2R i
10 k:
1PF
L
Chapter 14, Solution 83.
1PF

o
1
C'
C
K mKf
5PF

o
C' 0.5 pF
10 k:
o
R' K mR
20 k:
o
R ' 2 M:
10 6
100 x10 5
0.1 pF
100x10 k: 1 M:
Chapter 14, Solution 84.
The schematic is shown below. A voltage marker is inserted to measure vo. In
the AC sweep box, we select Total Points = 50, Start Frequency = 1, and End
Frequency = 1000. After saving and simulation, we obtain the magnitude and
phase plots in the probe menu as shown below.
Chapter 14, Solution 85.
We let I s 1‘0 o A so that Vo / I s
is simulated for 100 < f < 10 kHz.
Vo . The schematic is shown below. The circuit
Chapter 14, Solution 86.
The schematic is shown below. A current marker is inserted to measure I. We set
Total Points = 101, start Frequency = 1, and End Frequency = 10 kHz in the
AC sweep box. After simulation, the magnitude and phase plots are obtained in
the Probe menu as shown below.
Chapter 14, Solution 87.
The schematic is shown below. In the AC Sweep box, we set Total Points = 50,
Start Frequency = 1, and End Frequency = 100. After simulation, we obtain the
magnitude response as shown below. It is evident from the response that the
circuit represents a high-pass filter.
Chapter 14, Solution 88.
The schematic is shown below. We insert a voltage marker to measure Vo. In the
AC Sweep box, we set Total Points = 101, Start Frequency = 1, and End
Frequency = 100. After simulation, we obtain the magnitude and phase plots of
Vo as shown below.
Chapter 14, Solution 89.
The schematic is shown below. In the AC Sweep box, we type Total Points =
101, Start Frequency = 100, and End Frequency = 1 k. After simulation, the
magnitude plot of the response Vo is obtained as shown below.
Chapter 14, Solution 90.
The schematic is shown below. In the AC Sweep box, we set
Total Points = 1001, Start Frequency = 1, and End Frequency = 100k. After
simulation, we obtain the magnitude plot of the response as shown below. The
response shows that the circuit is a high-pass filter.
Chapter 14, Solution 91.
The schematic is shown below. In the AC Sweep box, we set
Total Points = 1001, Start Frequency = 1, and End Frequency = 100k. After
simulation, we obtain the magnitude plot of the response as shown below. The
response shows that the circuit is a high-pass filter.
Chapter 14, Solution 92.
The schematic is shown below. We type Total Points = 101, Start Frequency =
1, and End Frequency = 100 in the AC Sweep box. After simulating the circuit,
the magnitude plot of the frequency response is shown below.
Chapter 14, Solution 93.
L
C
R
Since
f0
1
1 R2
2S LC L2
R
L
400
240 u 10 -6
R
1
L
LC
10 7
,
6
1
LC
1
-6
(240 u 10 )(120 u 10 -12 )
1016
288
f0 #
1
10 8
2S LC
24S 2
938 kHz
R
1
.
L
LC
The result remains the same.
If R is reduced to 40 :,
Chapter 14, Solution 94.
Zc
1
RC
We make R and C as small as possible. To achieve this, we connect 1.8 k : and 3.3 k :
in parallel so that
1.8x 3.3
R
1.164 k:
1.8 3.3
We place the 10-pF and 30-pF capacitors in series so that
C = (10x30)/40 = 7.5 pF
Hence,
1
1
Zc
114.55x10 6 rad/s
RC 1.164x10 3 x 7.5x10 12
Chapter 14, Solution 95.
(a)
f0
1
2S LC
When C
f0
When C
f0
360 pF ,
1
2S (240 u 10 -6 )(360 u 10 -12 )
0.541 MHz
40 pF ,
1
2S (240 u 10 -6 )(40 u 10 -12 )
Therefore, the frequency range is
0.541 MHz f 0 1.624 MHz
1.624 MHz
(b)
Q
2SfL
R
At f 0
0.541 MHz ,
(2S )(0.541 u 10 6 )(240 u 10 -6 )
12
Q
At f 0
67.98
1.624 MHz ,
(2S )(1.624 u 10 6 )(240 u 10 -6 )
12
Q
204.1
Chapter 14, Solution 96.
Ri
Vi
L
V1
Vo
+
+
C1
C2
RL
Vo
Z2
1
sC 2
Z1
RL
1 sR 2 C 2
Z1
R L ||
Z2
1
|| (sL Z1 )
sC1
Z2
1 sL R L s 2 R L C 2 L
˜
sC1
1 sR L C 2
sL R L s 2 R L C 2 L
1
sC1
1 sR L C 2
Z2
sL R L s 2 R L LC 2
1 sR L C 2 s 2 LC1 sR L C1 s 3 R L LC1C 2
1 § sL R L s 2 R L C 2 L ·
¸
|| ¨
sC1 ©
1 sR L C 2
¹
V1
Z2
V
Z2 R i i
Vo
Z1
V
Z1 sL 1
Vo
Vi
Z2
Z1
˜
Z 2 R 2 Z1 sL
Z2
Z1
˜
V
Z 2 R 2 Z1 sL i
where
Z2
Z2 R 2
sL R L s 2 R L LC 2
sL R L s 2 R L LC 2 R i sR i R L C 2 s 2 R i LC1 sR i R L C1 s 3 R i R L LC1C 2
Z1
Z1 sL
and
RL
R L sL s 2 R L LC 2
Therefore,
Vo
Vi
R L (sL R L s 2 R L LC 2 )
(sL R L s 2 R L LC 2 R i sR i R L C 2 s 2 R i LC 1 sR i R L C 1
s 3 R i R L LC 1 C 2 )( R L sL s 2 R L LC 2 )
jZ .
where s
Chapter 14, Solution 97.
Ri
Vi
L
V1
Vo
+
+
C1
C2
RL
Vo
Z2
Z1
Z
§
1 ·
¸
sL || ¨ R L sC 2 ¹
©
sL (R L 1 sC 2 )
,
R L sL 1 sC 2
V1
Z
V
Z R i 1 sC1 i
Vo
RL
V
R L 1 sC 2 1
s
jZ
RL
Z
˜
V
R L 1 sC 2 Z R i 1 sC1 i
RL
sL (R L 1 sC 2 )
˜
R L 1 sC 2 sL (R L 1 sC 2 ) (R i 1 sC1 )(R L sL 1 sC 2 )
H (Z)
Vo
Vi
H (Z)
s 3 LR L C 1C 2
(sR i C 1 1)(s 2 LC 2 sR L C 2 1) s 2 LC 1 (sR L C 2 1)
jZ .
where s
Chapter 14, Solution 98.
B
Z2 Z1
2S (f 2 f 1 )
Z0
2Sf 0
f0
(20)(44S)
2S
2S (454 432)
44S
(20)(44S )
QB
(20)(22)
440 Hz
Chapter 14, Solution 99.
Xc
1
ZC
C
1
2S f X c
XL
ZL
1
2Sf C
1
(2S )(2 u 10 6 )(5 u 10 3 )
2S f L
10 -9
20S
L
f0
B
XL
2Sf
3 u 10 -4
4S
300
(2S )(2 u 10 6 )
1
1
2S LC
3 u 10 -4 10 -9
˜
2S
4S
20S
§ 4S ·
¸
(100) ¨
© 3 u 10 -4 ¹
R
L
1.826 MHz
4.188 u 10 6 rad / s
Chapter 14, Solution 100.
Zc
2Sf c
R
1
2Sf c C
1
RC
1
(2S )(20 u 10 3 )(0.5 u 10 -6 )
15.91 :
Chapter 14, Solution 101.
Zc
2Sf c
R
1
2Sf c C
1
RC
1
(2S )(15)(10 u 10 -6 )
1.061 k:
Chapter 14, Solution 102.
(a)
When R s
Zc
2Sf c
fc
1
2SRC
0 and R L
f , we have a low-pass filter.
1
RC
1
(2S)(4 u 10 3 )(40 u 10 -9 )
994.7 Hz
(b)
We obtain R Th across the capacitor.
R Th R L || (R R s )
R Th 5 || (4 1) 2.5 k:
fc
1
2SR Th C
fc
1.59 kHz
1
(2S )(2.5 u 10 3 )(40 u 10 -9 )
Chapter 14, Solution 103.
H (Z)
H (Z)
H (Z)
Vo
Vi
R2
,
R 2 R 1 || 1 sC
R2
R (1 sC)
R2 1
R 1 1 sC
s
jZ
R 2 (R 1 1 sC)
R 2 R 1 (1 sC)
R 2 (1 sCR 1 )
R 1 sCR 2
Chapter 14, Solution 104.
The schematic is shown below. We click Analysis/Setup/AC Sweep and enter
Total Points = 1001, Start Frequency = 100, and End Frequency = 100 k.
After simulation, we obtain the magnitude plot of the response as shown.
Chapter 15, Solution 1.
(a)
(b)
e at e - at
2
1ª 1
1 º
L > cosh(at ) @
«
2 ¬ s a s a »¼
s
s a2
e at e - at
2
1ª 1
1 º
L > sinh(at ) @
«
2 ¬ s a s a »¼
a
s a2
cosh(at )
2
sinh(at )
2
Chapter 15, Solution 2.
(a)
f ( t ) cos(Zt ) cos(T) sin(Zt ) sin(T)
F(s) cos(T) L > cos(Zt ) @ sin(T) L > sin(Zt ) @
F(s)
(b)
s cos(T) Z sin(T)
s 2 Z2
f ( t ) sin(Zt ) cos(T) cos(Zt ) sin(T)
F(s) sin(T) L > cos(Zt ) @ cos(T) L > sin(Zt ) @
F(s)
s sin(T) Z cos(T)
s 2 Z2
Chapter 15, Solution 3.
(a)
L > e -2t cos(3t ) u ( t ) @
s2
(s 2 ) 2 9
(b)
L > e -2t sin(4 t ) u ( t ) @
4
(s 2) 2 16
(c)
s
s a2
Since L > cosh(at ) @
2
s3
(s 3 ) 2 4
L > e -3t cosh(2 t ) u ( t ) @
(d)
a
s a2
Since L > sinh(at ) @
2
1
(s 4) 2 1
L > e -4t sinh( t ) u ( t ) @
(e)
2
(s 1) 2 4
L > e - t sin( 2t ) @
If
f (t) m
o F(s)
-d
F(s)
ds
t f (t) m
o
Thus, L > t e - t sin(2 t ) @
>
-d
2 (s 1) 2 4
ds
-1
@
2
˜ 2 (s 1)
((s 1) 2 4) 2
L > t e -t sin( 2t ) @
4 (s 1)
((s 1) 2 4) 2
Chapter 15, Solution 4.
(a)
G (s)
(b)
F(s)
6
s
s2 42
2
s2
5
e s
e 2s
s3
6se s
s 2 16
Chapter 15, Solution 5.
(a)
s cos(30q) 2 sin(30q)
s2 4
L > cos(2t 30q) @
L > t 2 cos(2t 30q) @
d 2 ª s cos(30q) 1 º
ds 2 «¬ s 2 4 »¼
·
-1 º
d d ª§ 3
s 1¸¸ s 2 4 »
«¨¨
ds ds «¬© 2
»¼
¹
§ 3
·
-1
-2 º
d ª 3 2
s 4 2s ¨¨
s 1¸¸ s 2 4 »
«
ds «¬ 2
»¼
© 2
¹
§ 3
·
§ 3·
§ 3
·
3
¨
¸ 2s ¨
¸ (8s 2 ) ¨
¸
2
s
1
s
1
- 2s
¨ 2
¸
¨ 2 ¸
¨ 2
¸
©
¹ ©
¹
©
¹
2
3
2
2
2
2
2
2
2
s 4
s 4
s 4
s 4
§ 3
·
(8s 2 ) ¨¨
s 1¸¸
- 3s 3s 2 3s
© 2
¹
3
2
2
2
s 4
s 4
(-3 3 s 2)(s 2 4)
s2 4
L > t 2 cos(2t 30q) @
@
L > 30 t 4 e - t
(c)
ª
º
d
L « 2t u ( t ) 4 G( t ) »
¬
¼
dt
4 3 s3 8 s 2
s2 4
3
8 12 3 s 6s 2 3s 3
s2 4
4!
(s 2) 5
(b)
30 ˜
3
3
720
(s 2 ) 5
2
4(s ˜ 1 0)
s2
2
4s
s2
(d)
2 e -(t-1) u ( t )
2 e -t u ( t )
2e
L > 2 e -(t-1) u ( t ) @
s1
(e)
Using the scaling property,
1
1
L > 5 u ( t 2) @ 5 ˜
˜
1 2 s (1 2)
(f)
L > 6 e -t 3 u ( t ) @
(g)
Let f ( t )
6
s 1 3
5˜ 2˜
1
2s
5
s
18
3s 1
G( t ) . Then, F(s) 1 .
ª dn
º
L « n G( t ) »
¬ dt
¼
ª dn
º
L « n f (t) »
¬ dt
¼
s n F(s) s n 1 f (0) s n 2 f c(0) "
ª dn
º
L « n G( t ) »
¬ dt
¼
ª dn
º
L « n f (t) »
¬ dt
¼
s n ˜ 1 s n 1 ˜ 0 s n 2 ˜ 0 "
ª dn
º
L « n G( t ) »
¬ dt
¼
sn
Chapter 15, Solution 6.
(a)
L > 2 G( t 1) @
(b)
L > 10 u ( t 2) @
10 - 2s
e
s
(c)
L > ( t 4) u ( t ) @
1 4
s2 s
(d)
L> 2e
-t
2 e -s
u ( t 4) @
L> 2e
-4
e
-(t - 4)
u ( t 4) @
2 e -4s
e 4 (s 1)
Chapter 15, Solution 7.
s
, we use the linearity and shift properties to
s 42
10 s e - s
obtain L >10 cos(4 ( t 1)) u ( t 1) @
s 2 16
(a)
Since L > cos(4t ) @
(b)
Since L > t 2 @
L > t 2 e -2 t
2
2
, L > u ( t )@
s3
1
,
s
2
, and L > u ( t 3)@
(s 2) 3
@
L > t 2 e -2 t u ( t ) u ( t 3) @
e -3s
s
2
e -3s
(s 2 ) 3
s
Chapter 15, Solution 8.
(a)
L > 2 G(3t ) 6 u (2t ) 4 e -2 t 10 e -3 t
@
1
1 1
4
10
2˜ 6˜ ˜
3
2 s 2 s2 s3
2 6
4
10
3 s s2 s3
(b)
t e -t u ( t 1)
t e -t u ( t 1)
( t 1) e -t u ( t 1) e -t u ( t 1)
( t 1) e -(t-1) e -1 u ( t 1) e -(t-1) e -1 u ( t 1)
L > t e - t u ( t 1) @
(c)
e -1 e -s
e -1 e -s
(s 1) 2
s 1
L > cos(2 ( t 1)) u ( t 1) @
s e -s
s2 4
e -(s1)
e -(s1)
(s 1) 2 s 1
(d)
Since sin(4 ( t S)) sin(4t ) cos(4S ) sin( 4S) cos(4t )
sin(4t ) u ( t S ) sin(4 ( t S )) u ( t S )
L> sin( 4 t ) > u ( t ) u ( t S )@ @
L> sin( 4 t ) u ( t ) @ L> sin( 4( t S )) u ( t S) @
4 e - Ss
4
s 2 16 s 2 16
4
˜ (1 e -Ss )
s 16
2
Chapter 15, Solution 9.
(a)
(b)
f (t)
( t 4) u ( t 2)
F(s)
e -2s 2 e -2s
2
s2
s
g( t )
2 e -4t u ( t 1)
G (s)
(c)
h(t)
( t 2) u ( t 2) 2 u ( t 2)
2 e -4 e -4(t -1) u ( t 1)
2 e -s
e 4 (s 4)
5 cos(2 t 1) u ( t )
cos(A B) cos(A) cos(B) sin(A) sin(B)
cos(2t 1) cos(2t ) cos(1) sin(2t ) sin(1)
(d)
h(t)
5 cos(1) cos(2 t ) u ( t ) 5 sin(1) sin(2t ) u ( t )
H(s)
5 cos(1) ˜
H(s)
2.702 s 8.415
s2 4 s2 4
s
2
5 sin(1) ˜ 2
s 4
s 4
2
p( t )
6u ( t 2) 6u ( t 4)
P(s)
6 - 2s 6 -4s
e e
s
s
sin(4t )
Chapter 15, Solution 10.
(a)
By taking the derivative in the time domain,
g ( t ) (-t e -t e -t ) cos( t ) t e -t sin( t )
g( t )
(b)
e -t cos( t ) t e -t cos( t ) t e -t sin( t )
G (s)
º
s 1
d ª s 1 º d ª
1
«
» «
»
2
2
2
(s 1) 1 ds ¬ (s 1) 1¼ ds ¬ (s 1) 1¼
G (s)
s 1
s 2 2s
2s 2
2
2
2 2
s 2s 2 (s 2s 2)
(s 2s 2) 2
By applying the time differentiation property,
G (s) sF(s) f (0)
where f ( t ) t e -t cos( t ) , f (0) 0
G (s)
(s) ˜
- d ª s 1 º
ds «¬ (s 1) 2 1 »¼
(s)(s 2 2s)
(s 2 2s 2) 2
s 2 (s 2)
(s 2 2s 2) 2
Chapter 15, Solution 11.
(a)
Since L > cosh(at ) @
F(s)
(b)
s 2 (s 2)
(s 2 2s 2) 2
6 (s 1)
(s 1) 2 4
Since L > sinh(at ) @
L > 3 e -2t sinh(4t ) @
s
s a2
2
6 (s 1)
s 2s 3
2
a
s a2
2
(3)(4)
(s 2) 2 16
12
s 4s 12
2
-d
> 12 (s 2 4s 12) -1 @
ds
F(s)
L > t ˜ 3 e -2t sinh(4t ) @
F(s)
(12)(2s 4)(s 2 4s 12) -2
24 (s 2)
(s 4s 12) 2
2
(c)
cosh( t )
f (t)
1
˜ (e t e - t )
2
1
8 e -3t ˜ ˜ (e t e - t ) u ( t 2)
2
4 e-2t u ( t 2) 4 e-4t u ( t 2)
4 e-4 e-2(t - 2) u ( t 2) 4 e-8 e-4(t - 2) u ( t 2)
L > 4 e -4 e -2(t -2) u ( t 2)@
4 e -4 e -2s ˜ L > e -2 u ( t )@
L > 4 e -4 e -2(t -2) u ( t 2)@
4 e -(2s 4)
s2
Similarly, L > 4 e -8 e -4(t -2) u ( t 2)@
Therefore,
4 e -(2s 4) 4 e -(2s8)
F(s)
s2
s4
4 e -(2s8)
s4
e -(2s 6) > (4 e 2 4 e -2 ) s (16 e 2 8 e -2 )@
s 2 6s 8
Chapter 15, Solution 12.
f (t)
te 2( t 1) e 2 u ( t 1)
f (s)
e s
e2
(s 2)
2
e2
( t 1)e 2 e 2( t 1) u ( t 1) e 2 e 2( t 1) u ( t 1)
e s
s2
e (s 2) §
1 ·
¨1 ¸
s 2 © s 2¹
s3
(s 2)
2
e (s 2)
Chapter 15, Solution 13.
(a) tf (t )
m
o
d
F (s)
ds
If f(t) = cost, then F ( s )
s
d
( s 2 1)(1) s (2s 1)
F
s
and
(
)
ds
s2 1
( s 2 1) 2
L (t cos t )
s2 s 1
( s 2 1) 2
(b) Let f(t) = e-t sin t.
F (s)
1
( s 1) 2 1
( s 2 2s 2)(0) (1)(2s 2)
( s 2 2s 2) 2
dF
ds
L (e t t sin t )
(c )
f (t )
t
1
s 2s 2
2
dF
ds
2( s 1)
( s 2s 2) 2
2
f
³ F (s)ds
m
o
s
sin Et , then F ( s )
Let f (t )
ª sin Et º
L«
»
¬ t ¼
f
³s
s
E
2
E
2
ds
E
1
E
E
s E2
tan 1
2
s
f
S
E
s
2
tan 1
s
E
Chapter 15, Solution 14.
f (t)
­ 5t
0 t 1
®
¯10 5t 1 t 2
We may write f ( t ) as
f ( t ) 5t > u ( t ) u ( t 1)@ (10 5t ) > u ( t 1) u ( t 2)@
5t u ( t ) 10 ( t 1) u ( t 1) 5 ( t 2) u ( t 2)
F(s)
5 10 -s 5 -2s
e 2e
s2 s2
s
F(s)
5
( 1 2 e -s e - 2 s )
s2
tan 1
E
s
Chapter 15, Solution 15.
f ( t ) 10 > u ( t ) u ( t 1) u ( t 1) u ( t 2)@
ª1 2
e -2s º
F(s) 10 « e -s s »¼
¬s s
10
(1 e -s ) 2
s
Chapter 15, Solution 16.
f (t)
5 u ( t ) 3 u ( t 1) 3 u ( t 3) 5 u ( t 4)
F(s)
1
> 5 3 e -s 3 e - 3 s 5 e - 4 s @
s
Chapter 15, Solution 17.
f (t)
f (t)
F(s)
­0
t0
°° t 2 0 t 1
®
°1 1 t 3
°¯ 0
t!3
t 2 > u ( t ) u ( t 1)@ 1> u ( t 1) u ( t 3)@
t 2 u ( t ) ( t 1) 2 u ( t 1) (-2t 1) u ( t 1) u ( t 1) u ( t 3)
t 2 u ( t ) ( t 1) 2 u ( t 1) 2 ( t 1) u ( t 1) u ( t 3)
2
2 -s e -3s
-s
(
1
e
)
e s3
s2
s
Chapter 15, Solution 18.
(a)
g( t )
G (s)
u ( t ) u ( t 1) 2 > u ( t 1) u ( t 2)@ 3 > u ( t 2) u ( t 3)@
u ( t ) u ( t 1) u ( t 2) 3 u ( t 3)
1
(1 e -s e - 2s 3 e - 3s )
s
(b)
h(t)
2 t > u ( t ) u ( t 1)@ 2 > u ( t 1) u ( t 3)@
(8 2 t ) > u ( t 3) u ( t 4)@
2t u ( t ) 2 ( t 1) u ( t 1) 2 u ( t 1) 2 u ( t 1) 2 u ( t 3)
2 ( t 3) u ( t 3) 2 u ( t 3) 2 ( t 4) u ( t 4)
2t u ( t ) 2 ( t 1) u ( t 1) 2 ( t 3) u ( t 3) 2 ( t 4) u ( t 4)
H(s)
2
2 - 3s 2 - 4 s
-s
2e
2 (1 e ) 2 e
s
s
s
2
(1 e -s e - 3s e -4s )
s2
Chapter 15, Solution 19.
Since L> G( t )@ 1 and T
2 , F(s)
1
1 e - 2s
Chapter 15, Solution 20.
Let
g1 (t)
sin(St ), 0 t 1
sin( St ) > u ( t ) u ( t 1)@
sin(St ) u ( t ) sin(St ) u ( t 1)
Note that sin(S ( t 1)) sin(St S) - sin(St ) .
So,
g1 ( t ) sin( St) u(t) sin( S( t - 1)) u(t - 1)
G 1 (s)
G (s)
S
(1 e -s )
s S2
2
G 1 (s)
1 e -2s
S (1 e -s )
(s 2 S 2 )(1 e - 2s )
Chapter 15, Solution 21.
T
Let
2S
§
t ·
f 1 ( t ) ¨1 ¸ > u ( t ) u ( t 1)@
© 2S ¹
§
t
1
1 ·
f1 ( t ) u ( t ) u(t) ( t 1) u ( t 1) ¨1 ¸ u ( t 1)
© 2S ¹
2S
2S
F1 (s)
F(s)
§
1
1
e -s
1 · -s 1
¨
¸e ˜
2
2 -1 s 2Ss
2Ss ©
2S ¹
s
F1 (s)
1 e -Ts
> 2S (-2S 1) e -s @ s > - 1 e -s @
2Ss 2
> 2S (-2S 1) e -s @ s > - 1 e -s @
2Ss 2 (1 e - 2 Ss )
Chapter 15, Solution 22.
(a)
Let
G 1 (s)
(b)
g1 (t )
2t, 0 t 1
2 t > u ( t ) u ( t 1)@
2t u ( t ) 2 ( t 1) u ( t 1) 2 u ( t 1)
2 2 e -s 2 -s
2 e
s2
s
s
G (s)
G 1 (s)
, T 1
1 e -sT
G (s)
2 (1 e -s s e -s )
s 2 (1 e -s )
Let h
h 0 u ( t ) , where h 0 is the periodic triangular wave.
Let h 1 be h 0 within its first period, i.e.
h 1 (t)
­ 2t
0 t 1
®
¯ 4 2t 1 t 2
h 1 (t)
h 1 (t)
2 t u ( t ) 2 t u ( t 1) 4u ( t 1) 2 t u ( t 1) 2 ( t 2) u ( t 2)
2 t u ( t ) 4 ( t 1) u ( t 1) 2 ( t 2) u ( t 2)
H 1 (s)
2 4 -s 2 e -2s
e 2
s2 s2
s
H 0 (s)
2 (1 e -s ) 2
s 2 (1 e -2s )
H(s)
2
(1 e -s ) 2
s2
1 2 (1 e -s ) 2
s s 2 (1 e - 2s )
Chapter 15, Solution 23.
(a)
Let
f1 (t )
f1 (t )
(b)
f1 ( t )
­1 0 t 1
®
¯- 1 1 t 2
> u (t ) u (t 1)@ > u (t 1) u (t 2)@
u ( t ) 2 u ( t 1) u ( t 2)
F1 (s)
1
(1 2 e -s e -2s )
s
F(s)
F1 (s)
, T
(1 e -sT )
F(s)
(1 e -s ) 2
s (1 e - 2s )
Let
h1 (t)
h 1 ( t ) t 2 > u ( t ) u ( t 2)@ t 2 u ( t ) t 2 u ( t 2)
t 2 u ( t ) ( t 2) 2 u ( t 2) 4 ( t 2) u ( t 2) 4 u ( t 2)
H 1 (s)
1
(1 e -s ) 2
s
2
2
4
4
- 2s
) 2 e -2s e -2s
3 (1 e
s
s
s
H(s)
H 1 (s)
, T
(1 e -Ts )
H(s)
2 (1 e -2s ) 4s e -2s (s s 2 )
s 3 (1 e - 2s )
2
Chapter 15, Solution 24.
(a)
f (0)
lim sF(s)
s of
10 lim
s of
(b)
10s 4 s
2
s of s 6s 5
lim
1
s3
1 6 5
s 2 s3 s 4
10
0
f
f (f)
lim sF(s)
lim
10s 4 s
s o 0 s 2 6s 5
0
f (0)
lim sF(s)
lim
s2 s
2
s of s 4s 6
1
s o0
s of
The complex poles are not in the left-half plane.
f (f) does not exist
(c)
f (0)
lim sF(s)
s of
2s 3 7s
2
s of (s 1)(s 2)(s 2s 5)
lim
2 7
s s3
lim
s of §
1 ·§ 2 ·§ 2 5 ·
¨1 ¸ ¨1 ¸ ¨1 2 ¸
s ¹©
s s ¹
© s ¹©
f (f)
lim sF(s)
s o0
0
1
0
2s 3 7s
lim
2
s o 0 (s 1)(s 2)(s 2s 5)
0
10
0
Chapter 15, Solution 25.
(a)
f (0)
lim sF(s)
s of
(8)(s 1)(s 3)
s of (s 2)(s 4)
lim
§ 1·§ 3 ·
(8) ¨1 ¸ ¨1 ¸
s ¹© s ¹
lim ©
s of §
2 ·§ 4 ·
¨1 ¸ ¨1 ¸
s ¹©
s¹
©
(b)
(8)(1)(3)
s o 0 ( 2)( 4)
f (f)
lim sF(s)
lim
f (0)
lim sF(s)
lim
f (0)
§1 1·
6¨ 2 4 ¸
©s
s ¹
lim
1
s of
1 4
s
s o0
s of
8
3
6s (s 1)
4
s of s 1
0
1
0
All poles are not in the left-half plane.
f (f) does not exist
Chapter 15, Solution 26.
(a)
f ( 0)
lim sF(s)
s of
s 3 3s
s of s 3 4s 2 6
lim
1
Two poles are not in the left-half plane.
f (f) does not exist
(b)
f (0)
s 3 2s 2 s
lim
2
s of (s 2)(s 2s 4)
lim sF(s)
s of
2 1
2
s
s
lim
s of §
2 ·§ 2 4 ·
¨1 ¸ ¨1 2 ¸
© s ¹© s s ¹
1
1
One pole is not in the left-half plane.
f (f) does not exist
Chapter 15, Solution 27.
u(t ) 2 e -t
(a)
f (t)
(b)
G (s)
3 (s 4) 11
s4
g( t )
3 G(t ) 11 e -4t
H(s)
4
(s 1)(s 3)
(c)
A
2,
B
H(s)
2
2
s 1 s 3
h(t)
2 e -t 2 e -3t
-2
3
11
s4
A
B
s 1 s 3
J (s)
(d)
12
(s 2) 2 (s 4)
A
B
C
2 s 2 (s 2)
s4
B
12
2
12
A (s 2) (s 4) B (s 4) C (s 2) 2
C
6,
12
(-2) 2
3
Equating coefficients :
s2 :
0 AC 
o A
-C
-3
s :
s0 :
0 6A B 4C 2A B 
o B -2A
12 8A 4B 4C -24 24 12 12
J (s)
-3
6
3
2 s 2 (s 2)
s4
j( t )
3 e -4t 3 e -2t 6 t e -2t
1
6
Chapter 15, Solution 28.
(a)
2(2) 2(4)
2 2
s5
s3
F(s)
f (t)
2
4
s3 s5
(2e 3t 4e 5t )ut ( t )
(b)
H(s)
3s 11
(s 1)(s 2 2s 5)
A
Bs C
s 1 s 2 2s 5
3s 11 A(s 2 2s 5) (Bs C)(s 1)
5A C 11; A
H(s)
B; B C
3, B
2
2s 1
o h(t)
s 1 s 2 2s 5
(A B)s 2 (2A B C)s 5A C
C3o A
2; B
2; C 1
2e t 2e t cos 2t 1.5e t sin 2t u ( t )
Chapter 15, Solution 29.
V(s)
2
As B
; 2s 2 8s 26 As 2 Bs
2
2
s (s 2) 3
V(s)
2
2(s 2)
2
3
s (s 2) 2 3 2 3 (s 2) 2 3 2
2s 26 o A
2 and B
6
2
v(t) = 2u ( t ) 2e 2 t cos 3t e 2 t sin 3t , t t 0
3
Chapter 15, Solution 30.
(a) H1 (s)
2(s 2) 2
2(s 2)
(s 2) 2 32
(s 2) 2 3 2
h1 ( t ) 2e 2 t cos 3t (b) H 2 (s)
s2 4
or
2
3
3 (s 2) 2 32
2 2 t
e
sin 3t
3
s2 4
(s 1) 2 (s 2 2s 5)
A
B
Cs D
(s 1) (s 1) 2
(s 2 2s 5)
A(s 1)(s 2 2s 5) B(s 2 2s 5) Cs(s 1) 2 D(s 1) 2
s 2 4 A(s 3 3s 2 7s 5) B(s 2 2s 5) C(s 3 2s 2 s) D(s 2 2s 1)
Equating coefficients:
s3 :
0
s2 :
1 3A B 2C D
s:
0
constant :
AC

o
7 A 2B C 2D
4
5A 5B D
C
A
ABD
6A 2B 2D
4A 2
4A 4B 1

o

o
B
A
1 / 2, C 1 / 2
5 / 4, D 1 / 4
H 2 (s)
1ª 2
5
2s 1 º
»
«
4 ¬« (s 1) (s 1) 2
(s 2 2s 5) ¼»
1ª 2
5
2(s 1) 1 º
»
«
4 ¬« (s 1) (s 1) 2
(s 1) 2 2 2 ) ¼»
Hence,
h 2 (t)
(c ) H 3 (s)
h 3 (t)
1
2e t 5te t 2e t cos 2t 0.5e t sin 2t u ( t )
4
(s 2)e s
(s 1)(s 3)
ª A
B º
es «
»
¬« (s 1) (s 3) ¼»
1 s ª 1
1 º
e «
»
2
¬« (s 1) (s 3) ¼»
1 ( t 1)
e
e 3( t 1) u ( t 1)
2
Chapter 15, Solution 31.
(a)
(b)
F(s)
10s
(s 1)(s 2)(s 3)
A
F(s) (s 1) s
-1
B
F(s) (s 2) s
-2
C
F(s) (s 3) s
-3
A
B
C
s 1 s 2 s 3
- 10
2
-5
- 20
-1
20
- 30
2
-15
F(s)
-5
20
15
s 1 s 2 s 3
f (t)
- 5 e -t 20 e -2t 15 e -3t
F(s)
2s 2 4s 1
(s 1)(s 2) 3
A
F(s) (s 1) s
-1
D
F(s) (s 2) 3
s -2
A
B
C
D
2 s 1 s 2 (s 2)
(s 2) 3
-1
-1
2s 2 4s 1 A(s 2)(s 2 4s 4) B(s 1)(s 2 4s 4)
C(s 1)(s 2) D(s 1)
Equating coefficients :
s3 :
0 AB 
o B
2
s :
s1 :
s0 :
F(s)
f(t)
(c)
2 6A 5B C A C 
o C 2 A 3
4 12A 8B 3C D 4A 3C D
4 6A D 
o D -2 A -1
1 8A 4B 2C D 4A 2C D -4 6 1 1
-1
1
3
1
2 s 1 s 2 (s 2)
(s 2) 3
-e - t e -2t 3 t e -2t t 2 -2t
e
2
f (t)
§
t2 ·
- e -t ¨1 3 t ¸ e - 2t
2¹
©
F(s)
s 1
(s 2)(s 2 2s 5)
A
-A 1
F(s) (s 2) s
-2
A
Bs C
2
s 2 s 2s 5
-1
5
s 1 A (s 2 2s 5) B (s 2 2s) C (s 2)
Equating coefficients :
AB 
o B
1
5
s2 :
0
s1 :
s0 :
1 2A 2B C 0 C 
o C 1
1 5A 2C -1 2 1
F(s)
-1 5
1 5˜ s 1
s 2 (s 1) 2 2 2
f (t)
- 0.2 e -2t 0.2 e -t cos( 2t ) 0.4 e -t sin( 2t )
-A
-1 5
1 5 (s 1)
45
2
2 s 2 (s 1) 2
(s 1) 2 2 2
Chapter 15, Solution 32.
(a)
(b)
F(s)
8 (s 1)(s 3)
s (s 2)(s 4)
A
B
C
s s2 s4
(8)(3)
(2)(4)
A
F(s) s s
B
F(s) (s 2) s
-2
C
F(s) (s 4) s
-4
0
3
(8)(-1)
(-4)
(8)(-1)(-3)
(-4)(-2)
F(s)
3
2
3
s s2 s4
f (t)
3 u(t ) 2 e -2t 3 e -4t
F(s)
s 2 2s 4
(s 1)(s 2) 2
s 2 2s 4
2
3
A
B
C
s 1 s 2 (s 2) 2
A (s 2 4s 4) B (s 2 3s 2) C (s 1)
Equating coefficients :
s2 :
1 AB 
o B 1 A
1
s :
- 2 4A 3B C 3 A C
0
s :
4 4A 2B C -B 2 
o B
A 1 B
(c)
7
C
-5 - A
F(s)
7
6
12
s 1 s 2 (s 2) 2
f (t)
7 e -t 6 (1 2 t ) e -2t
F(s)
s2 1
(s 3)(s 2 4s 5)
-12
A
Bs C
2
s 3 s 4s 5
s 2 1 A (s 2 4s 5) B (s 2 3s) C (s 3)
-6
Equating coefficients :
s2 :
1 AB 
o B 1 A
s1 :
0
4A 3B C
1 5A 3C
0
s :
B 1 A
3 A C 
o A C
-9 2A 
o A
-4
C
F(s)
5
4s 8
s 3 (s 2) 2 1
f (t)
5 e -3t 4 e -2t cos(t )
-A 3
5
-8
4 (s 2)
5
s 3 (s 2) 2 1
Chapter 15, Solution 33.
(a)
F(s)
6
6
(s 1)(s 1)
2
As B
C
2
s 1 s 1
A (s 2 s) B (s 1) C (s 2 1)
Equating coefficients :
s2 :
0 AC 
o A
-C
-A
C
2B 
o B
3
1
s :
0
AB 
o B
0
6
BC
s :
A
(b)
6 (s 1)
s4 1
-3 ,
B
3,
C
F(s)
3
- 3s 3
2
s 1 s 1
f (t)
3 e -t 3 sin( t ) 3 cos(t )
F(s)
s e - Ss
s2 1
f (t)
cos(t S ) u(t S )
3
3
- 3s
3
2
2
s 1 s 1 s 1
-3
(c)
8
s (s 1) 3
F(s)
A
B
C
D
2 s s 1 (s 1)
(s 1) 3
A
8,
8
A (s 3 3s 2 3s 1) B (s 3 2s 2 s) C (s 2 s) D s
D
-8
Equating coefficients :
s3 :
0 AB 
o B
3A 2B C
-A
AC 
o C
s2 :
0
s1 :
s0 :
0 3A B C D A D 
o D
A 8, B 8, C 8, D 8
F(s)
8
8
8
8
2 s s 1 (s 1)
(s 1) 3
f (t)
8 > 1 e -t t e -t 0.5 t 2 e -t @ u(t )
-A
B
-A
Chapter 15, Solution 34.
(a)
F(s) 10 f (t)
(b)
s2 4 3
s2 4
e -s 4 e -2s
(s 2)(s 4)
Let
1
(s 2)(s 4)
A 12
g( t )
3
s 4
2
11 G(t ) 1.5 sin( 2t )
G (s)
G (s)
11 A
B
s2 s4
B 12
§ 1
1 ·
e -s § 1
1 ·
¨
¸ 2 e -2s ¨
¸
©s 2 s 4¹
2 ©s 2 s 4¹
0.5 > e -2(t -1) e -4(t -1) @ u(t 1) 2 > e -2(t - 2) e -4(t - 2) @ u(t 2)
(c)
s 1
s (s 3)(s 4)
Let
A 1 12 ,
B
A
B
C
s s3 s4
2 3,
C
-3 4
§1 1 23
3 4 · -2s
¸e
H(s) ¨ ˜ ©12 s s 3 s 4 ¹
§ 1 2 - 3(t - 2) 3 -4(t - 2) ·
¨ e
¸ u( t 2 )
e
© 12 3
¹
4
h(t)
Chapter 15, Solution 35.
(a)
G (s)
Let
A
2,
G (s)
F(s)
f (t)
(b)
Let
s3
(s 1)(s 2)
B
2
1
s 1 s 2
A
B
s 1 s 2
-1

o g( t )
2 e - t e -2t
e -6s G (s) 
o f ( t ) g( t 6) u ( t 6)
-(t - 6)
-2(t - 6)
> 2 e e @ u( t 6)
G (s)
A 1 3,
1
(s 1)(s 4)
B
A
B
s 1 s 4
-1 3
G (s)
1
1
3 (s 1) 3 (s 4)
g( t )
1 -t
> e e -4t @
3
F(s) 4 G (s) e -2t G (s)
f ( t ) 4 g( t ) u ( t ) g ( t 2) u ( t 2)
f (t)
4 -t
> e e -4t @ u(t ) 1 > e -(t-2) e -4(t-2) @ u(t 2)
3
3
(c)
G (s)
Let
A
s
s
(s 3)(s 2 4)
A
Bs C
2
s3 s 4
- 3 13
A (s 2 4) B (s 2 3s) C (s 3)
Equating coefficients :
s2 :
0 AB 
o B
1 3B C
s1 :
0
0 4A 3C
s :
A
- 3 13 ,
B
3 13 ,
-A
C
4 13
- 3 3s 4
s 3 s2 4
13 G (s)
-3 e -3t 3 cos(2t ) 2 sin(2t )
13 g( t )
F(s)
e -s G (s)
f (t)
g( t 1) u ( t 1)
f (t)
1
> - 3 e -3(t-1) 3 cos(2 (t 1)) 2 sin( 2 (t 1))@ u(t 1)
13
Chapter 15, Solution 36.
(a)
X(s)
1
s (s 2)(s 3)
B 1 6,
2
C 1 4,
A B
C
D
2
s s
s2 s3
D
-1 9
1 A (s 3 5s 2 6s) B (s 2 5s 6) C (s 3 3s 2 ) D (s 3 2s 2 )
Equating coefficients :
0 ACD
s3 :
2
s :
0 5A B 3C 2D 3A B C
1
s :
0 6 A 5B
s0 :
1 6B 
o B 1 6
A
(b)
-5 6B
- 5 36
X(s)
- 5 36 1 6 1 4
19
2 s
s
s2 s3
x(t)
-5
1
1
1
u(t ) t e - 2t e - 3t
36
6
4
9
Y(s)
1
s (s 1) 2
A 1,
C
A
B
C
s s 1 (s 1) 2
-1
1 A (s 2 2s 1) B (s 2 s) C s
Equating coefficients :
0 AB 
o B
s2 :
(c)
-A
s :
s0 :
0 2A B C A C 
o C
1 A, B -1, C -1
Y(s)
1
1
1
s s 1 (s 1) 2
y( t )
u(t ) e -t t e -t
Z(s)
A
B
Cs D
2
s s 1 s 6s 10
1
A 1 10 ,
B
-A
-1 5
1 A (s 3 7s 2 16s 10) B (s 3 6s 2 10s) C (s 3 s 2 ) D (s 2 s)
Equating coefficients :
0 A BC
s3 :
2
s :
0 7 A 6 B C D 6 A 5B D
1
s :
0 16A 10B D 10A 5B 
o B
0
s :
-2A
1 10A 
o A 1 10
A 1 10 ,
B
-2A
-1 5 ,
C
A 1 10 ,
D
4A
4
10
10 Z(s)
1
2
s4
2
s s 1 s 6s 10
10 Z(s)
1
2
s3
1
2
s s 1 (s 3) 1 (s 3) 2 1
z( t )
0.1 > 1 2 e -t e -3t cos(t ) e -3t sin( t )@ u(t )
Chapter 15, Solution 37.
(a)
Let
12
s (s 2 4)
P(s)
A
P(s) s s
12
A (s 2 4) B s 2 C s
0
12 4
A Bs C
s s2 4
3
Equating coefficients :
s0 :
12 4A 
o A 3
1
s :
0 C
2
s :
0 AB 
o B -A
(b)
P(s)
3
3s
2
s s 4
p( t )
3 u ( t ) 3 cos(2t )
F(s)
e -2s P(s)
f (t)
3 > 1 cos( 2(t 2))@ u(t 2)
Let
G (s)
2s 1
(s 1)(s 2 9)
2
-3
As B Cs D
s2 1 s2 9
2s 1 A (s 3 9s) B (s 2 9) C (s 3 s) D (s 2 1)
Equating coefficients :
0 AC 
o C
s3 :
-A
s2 :
0
BD 
o D
s1 :
2
9A C
8A 
o A
2 8, C
8B 
o B 1 8 , D
1 9B D
G (s)
1 § 2s 1 · 1 § 2s 1 ·
¨
¸ ¨
¸
8 © s 2 1¹ 8 © s 2 9 ¹
G (s)
1
s
1
1
1
s
1
1
˜ 2
˜ 2
˜ 2
˜ 2
4 s 1 8 s 1 4 s 9 8 s 9
g( t )
1
1
1
1
cos(t ) sin( t ) cos( 3t ) sin( 3t )
4
8
4
24
36s 117
s 2 4s 13
Let
H(s)
9s2
s 2 4s 13
H(s)
9 36 ˜
s2
3
2
2 15 ˜
(s 2) 3
(s 2) 2 3 2
h(t)
9 G(t ) 36e -2 t cos( 3t ) 15e -2 t sin( 3t )
9
Chapter 15, Solution 38.
(a)
-2 8
s :
0
(c)
-B
F(s)
s 2 4s
s 2 10s 26
s 2 10s 26 6s 26
s 2 10s 26
F(s) 1 6s 26
s 10s 26
F(s) 1 6 (s 5)
4
2
2 (s 5) 1
(s 5) 2 12
f (t)
2
G(t ) 6 e -t cos(5t ) 4 e -t sin( 5t )
-1 8
(b)
F(s)
5s 2 7s 29
s (s 2 4s 29)
5s 2 7s 29
A
Bs C
2
s s 4s 29
A (s 2 4s 29) B s 2 C s
Equating coefficients :
s0 :
29 29A 
o A 1
s1 :
7
4A C 
o C
s2 :
5
AB 
o B
A 1,
B
4,
7 4A
5 A
C
3
4
3
4 (s 2)
1
5
2
2 s (s 2) 5
(s 2) 2 5 2
F(s)
1
4s 3
2
s s 4s 29
f (t)
u(t ) 4 e -2t cos(5t ) e -2t sin( 5t )
Chapter 15, Solution 39.
(a)
F(s)
2s 3 4s 2 1
(s 2 2s 17)(s 2 4s 20)
As B
Cs D
2
s 2s 17 s 4s 20
2
s 3 4s 2 1 A(s 3 4s 2 20s) B(s 2 4s 20)
C(s3 2s 2 17s) D(s 2 2s 17)
Equating coefficients :
s3 :
2 AC
2
s :
4 4 A B 2C D
1
0 20A 4B 17C 2D
s :
0
s :
1 20B 17 D
Solving these equations (Matlab works well with 4 unknowns),
D 21
A -1.6 ,
B -17.8 ,
C 3 .6 ,
F(s)
- 1.6s 17.8
3.6s 21
2
2
s 2s 17 s 4s 20
F(s)
(-1.6)(s 1)
(-4.05)(4)
(3.6)(s 2)
(3.45)(4)
2
2 2
2 2
2 (s 1) 4
(s 1) 4
(s 2) 4
(s 2) 2 4 2
f (t)
- 1.6 e -t cos(4t ) 4.05 e -t sin( 4t ) 3.6 e -2t cos(4t ) 3.45 e -2t sin( 4t )
(b)
s2 4
(s 2 9)(s 2 6s 3)
F(s)
s2 4
As B
Cs D
2
2
s 9 s 6s 3
A (s 3 6s 2 3s) B (s 2 6s 3) C (s 3 9s) D (s 2 9)
Equating coefficients :
s3 :
0 AC 
o C -A
2
s :
1 6A B D
1
s :
0 3A 6B 9C 6B 6C 
o B
4 3B 9D
s0 :
Solving these equations,
A 1 12 ,
B 1 12 ,
12 F(s)
0 
o
G (s)
E
-s5
s 5.449
F
-s5
s 0.551
G (s)
12 F(s)
f (t)
- 1 12 ,
D
A
5 12
s 1
-s5
2
2
s 9 s 6s 3
s 2 6s 3
Let
C
-C
- 6 r 36 - 12
2
-s5
s 6s 3
2
-0.551, - 5.449
E
F
s 0.551 s 5.449
s -0.551
1.133
s -5.449
- 2.133
1.133
2.133
s 0.551 s 5.449
s
1
3
1.133
2.133
˜ 2
2 2 s 3
3 s 3
s 0.551 s 5.449
2
0.08333 cos( 3t ) 0.02778 sin( 3t ) 0.0944 e -0.551t 0.1778 e -5.449t
Chapter 15, Solution 40.
ª 4s 2 7s 13 º
«
»
«¬ (s 2)(s 2 2s 5) »¼
Let H(s)
4s 2 7s 13
A
Bs C
2
s 2 s 2s 5
A(s 2 2s 5) B(s 2 2s) C(s 2)
Equating coefficients gives:
s2 :
AB
4
s:
7
constant :
H(s)
2A 2B C

o
C
13 5A 2C

o
5A 15 or A
3
s 1
2
s 2 s 2s 5
1
3, B 1
3
(s 1) 2
s 2 (s 1) 2 2 2
Hence,
h(t)
3e 2 t e t cos 2t e t sin 2t
where A cos D 1,
A sin D 1
Thus,
> 2e
h(t)
t
3e 2 t e t (A cos D cos 2t A sin D sin 2t )

o
A
2,
D
@
cos(2 t 45 o ) 3e 2 t u ( t )
Chapter 15, Solution 41.
Let y(t) = f(t)*h(t). For 0 < t < 1,
f ( t O)
h(O )
0
y( t )
t
2
³ (1)4OdO 2O
0
t
0
2t 2
t 1
2
O
45 o
For 1 <t<3,
f ( t O)
h (O )
0
y( t )
1
t
0
1
³ (1)4OdO ³ (1)(8 4O)dO
1 t
2
t
t
2O2 0 (8O 2O2 ) 1
8t 2 t 2 4
For 3 < t < 4
h (O)
0
y( t )
2
2
³ (8 4O)OdO 8O 2O
t 2
Thus,
y( t )
­
2t 2 , 0 t 1
°
° 8t - 2t 2 4, 1 t 3
®
2
°32 - 16t 2t , 3 t 4
°
0, otherwise
¯
2
t 2
1
t-2
32 16t 2t 2
2
f ( t O)
3
t
4
O
Chapter 15, Solution 42.
(a)
For 0 t 1 , f1 ( t O) and f 2 (O) overlap from 0 to t, as shown in Fig. (a).
y( t )
³
f1 ( t ) f 2 ( t )
f1(t - O)
t
0
(1)(O) dO
f2(O)
1
t-1
O2
2
0 t
1
t2
2
t
0
1
0
O
(a)
t-1 1
t
O
(b)
For 1 t 2 , f1 ( t O) and f 2 (O) overlap as shown in Fig. (b).
y( t )
O2
2
³
1
(1)(O) dO
t 1
t
1
t 1
t2
2
For t ! 2 , there is no overlap.
Therefore,
y( t )
(b)
­ t 2 2,
0t 1
°
2
®t t 2, 1 t 2
° 0,
otherwise
¯
For 0 t 1 , the two functions overlap as shown in Fig. (c).
y( t )
³
f1 ( t ) f 2 ( t )
f1(t - O)
(1)(1) dO
f2(O)
1
t-1
t
0
0 t
(c)
1
O
t
1
0
t-1 1
(d)
t
O
For 1 t 2 , the functions overlap as shown in Fig. (d).
y( t )
³
1
t 1
(1)(1) dO
O 1t 1
2 t
For t ! 2 , there is no overlap.
Therefore,
y( t )
(c)
0t1
­ t,
°
®2 t , 1 t 2
° 0,
otherwise
¯
For t -1 , there is no overlap. For - 1 t 0 , f1 ( t O) and f 2 (O) overlap
as shown in Fig. (e).
·
§ O2
³-1 (1)(O 1) dO ¨© 2 O¸¹ -t1
t
y( t )
f1 ( t ) f 2 ( t )
y( t )
1 2
( t 2t 1)
2
f2(t - O)
f1(O)
1
-1
t
0
1
( t 1) 2
2
1
O
1
-1
(e)
0 t
(f)
For 0 t 1 , the functions overlap as shown in Fig. (f).
y( t )
³
(1)(O 1) dO ³0 (1)(1 O) dO
-1
0
t
§
§ O2
·
O2 ·
y( t ) ¨ O ¸ 0-1 ¨ O ¸ 0t
2¹
©
©2
¹
y( t )
1
(1 2t t 2 )
2
For t ! 1 , the two functions overlap.
1
O
y( t )
³
(1)(O 1) dO ³0(1)(1 O) dO
-1
y( t )
1 §
O2 · 1
¨
O ¸ 0
2 ©
2¹
0
1
1
1
1
2
2
1
Therefore,
y( t )
0,
t -1
­
° 0.5(t 2 2t 1), - 1 t 0
°
®
2
°0.5(-t 2t 1), 0 t 1
°¯
1,
t !1
Chapter 15, Solution 43.
(a)
For 0 t 1 , x ( t O) and h (O) overlap as shown in Fig. (a).
y( t )
x(t - O)
³
x(t) h(t)
t
0
(1)(O) dO
1
O2
2
t2
2
t
0
1
h(O)
t-1
0 t
1
0
O
t-1 1
(a)
t
O
(b)
For 1 t 2 , x ( t O) and h (O) overlap as shown in Fig. (b).
y( t )
³t 1 (1)(O) dO ³1 (1)(1) dO
1
t
O2
2
1
t 1
For t ! 2 , there is a complete overlap so that
y( t )
³
t
t 1
(1)(1) dO
O tt 1
t ( t 1) 1
O 1t
-1 2
t 2t 1
2
Therefore,
y( t )
(b)
­
t 2 2,
0t1
° 2
°- (t 2) 2t 1, 1 t 2
®
1,
t!2
°
°¯
0,
otherwise
For t ! 0 , the two functions overlap as shown in Fig. (c).
y( t )
³ (1) 2 e
t
x(t) h(t)
x(t-O)
-O
0
2
dO
-2 e -O
t
0
h(O) = 2e-O
1
0
t
O
(c)
Therefore,
y( t )
(c)
2 (1 e -t ), t ! 0
For - 1 t 0 , x ( t O) and h (O) overlap as shown in Fig. (d).
y( t )
x(t) h(t)
³
t 1
0
x(t - O)
(1)(O) dO
O2
2
1
t-1 -1
t 0
t 1
0
1
( t 1) 2
2
h(O)
t+1 1
2
O
(d)
For 0 t 1 , x ( t O) and h (O) overlap as shown in Fig. (e).
y( t )
³ (1)(O) dO ³
1
0
t 1
1
(1)(2 O) dO
y( t )
O2
2
1
0
§
O2 ·
¨ 2O ¸ 1t 1
2¹
©
-1 2
1
t t
2
2
1
-1 t-1
0 t
1 t+1 2
O
(e)
For 1 t 2 , x ( t O) and h (O) overlap as shown in Fig. (f).
y( t )
³
(1)(O) dO ³1 (1)(2 O) dO
t 1
y( t )
O2
2
1
2
1
t 1
§
O2 ·
¨ 2O ¸ 12
2¹
©
-1 2
1
t t
2
2
1
0
t-1 1
t
2 t+1
O
(f)
For 2 t 3 , x ( t O) and h (O) overlap as shown in Fig. (g).
y( t )
§
O2 ·
¨
³t1 (1)(2 O) dO © 2O 2 ¸¹
2
2
t 1
9
1
3t t 2
2
2
1
0
1 t-1 2
(g)
t
t+1
O
Therefore,
y( t )
­ ( t 2 2 ) t 1 2, - 1 t 0
° 2
°- ( t 2 ) t 1 2 , 0 t 2
® 2
° ( t 2 ) 3t 9 2, 2 t 3
°¯
0,
otherwise
Chapter 15, Solution 44.
(a)
For 0 t 1 , x ( t O) and h (O) overlap as shown in Fig. (a).
y( t )
x(t) h(t)
x(t - O)
³
t
0
(1)(1) dO
h(O)
1
t-1
t
0 t
1
2
O
-1
(a)
For 1 t 2 , x ( t O) and h (O) overlap as shown in Fig. (b).
y( t )
³
(1)(1) dO ³1 (-1)(1) dO
t 1
1
t
O 1t 1 O 1t
3 2t
For 2 t 3 , x ( t O) and h (O) overlap as shown in Fig. (c).
y( t )
³
2
t 1
(1)(-1) dO
-O
2
t 1
t3
1
1
0
t-1 1
t
2
0
O
-1
1 t-1
2 t
O
-1
(b)
(c)
Therefore,
y( t )
(b)
0t 1
­ t,
° 3 2t , 1 t 2
°
®
2t3
° t 3,
°¯ 0,
otherwise
For t 2 , there is no overlap. For 2 t 3 , f1 ( t O) and f 2 (O) overlap,
as shown in Fig. (d).
y( t )
³
f1 (t ) f 2 (t )
§
O2
¨¨ Ot 2
©
t
2
·t
¸¸ 2
¹
(1)( t O) dO
t2
2t 2
2
f1(t - O)
f2(O)
1
0
1 t-1 2
t
3
4
5
O
4
5
O
(d)
1
0
1
2 t-1 3
(e)
t
For 3 t 5 , f1 ( t O) and f 2 (O) overlap as shown in Fig. (e).
§
O2 · t
¨
³t 1 (1)(t O) dO ©Ot 2 ¸¹ t1
t
y( t )
1
2
For 5 t 6 , the functions overlap as shown in Fig. (f).
§
O2 ·
¨ Ot ¸ 5t 1
(
1
)(
t
)
d
O
O
t 1
2¹
©
³
5
y( t )
-1 2
t 5t 12
2
1
0
1
2
3
4 t-1 5
(f)
Therefore,
­ ( t 2 2 ) 2t 2,
2t3
°
1 2,
3t5
°
® 2
°- (t 2) 5t 12, 5 t 6
°¯
0,
otherwise
y( t )
Chapter 15, Solution 45.
(a)
(b)
f ( t ) G( t )
³
f (t ) G(t )
f (t )
f (t) u(t)
³
t
0
t
0
Since u ( t O)
f ( t ) u( t )
³
t
o
f (O) G( t O) dO
f (O) u ( t O) dO
­1 O t
®
¯0 O! t
f ( O ) dO
f (O) O
t
t
O
Alternatively,
F(s)
s
L^ f ( t ) u ( t )`
­ F(s) ½
¾
L1 ®
¯ s ¿
f (t) u(t)
³
t
o
f (O) dO
Chapter 15, Solution 46.
(a)
Let y( t )
x1 (t) x 2 (t)
³
t
0
x 2 ( t ) x 1 ( t O) dO
For 0 t 3 , x 1 ( t O) and x 2 (O) overlap as shown in Fig. (a).
³ 4e
t
y( t )
-2O
0
4 e - t ³0 e -O dO
t
e -(t -O ) dO
4
4 (e - t e - 2 t )
x2(O)
x1(t-O)
0
O
3
t
(a)
For t ! 3 , the two functions overlap as shown in Fig. (b).
³ 4e
3
y( t )
0
-2 O
e -(t -O ) dO
4 e - t - e -O
3
0
4 e - t (1 e -3 )
4
3
0
(b)
t
O
Therefore,
y( t )
(b)
­ 4 (e - t e -2t ), 0 t 3
® -t
-3
t!3
¯4e (1 e ),
For 1 t 2 , x 1 (O) and x 2 ( t O) overlap as shown in Fig. (c).
y( t )
x1 (t) x 2 (t)
³
t
1
(1)(1) dO
O 1t
x2(t - O)
t 1
x1(O)
1
0
t-1 1
t
2
O
3
(c)
For 2 t 3 , the two functions overlap completely.
y( t )
³
t
t 1
(1)(1) dO
O tt 1
t ( t 1) 1
For 3 t 4 , the two functions overlap as shown in Fig. (d).
y( t )
³
3
t 1
(1)(1) dO
O 3t 1
4 t
1
0
1
2 t-1 3
t
O
(d)
Therefore,
y( t )
(c)
­ t 1, 1 t 2
° 1,
2t3
°
®
°4 t , 3 t 4
°¯ 0,
otherwise
For - 1 t 0 , x 1 ( t O) and x 2 (O) overlap as shown in Fig. (e).
y( t )
x1 ( t ) x 2 ( t )
³
t
-1
(1) 4 e -(t-O) dO
4 e -t ³-1 e O dO
t
y( t )
4 > 1 e -(t1) @
x1(t - O)
x2(O)
1
-1
t
-1
1
O
(e)
For 0 t 1 ,
y( t )
³
y( t )
4 e -t e O
0
-1
(1) 4 e -(t -O) dO ³0 (-1) 4 e -(t -O) dO
t
4 e -t e O
0
-1
t
0
8 e -t 4 e -(t1) 4
For t ! 1 , the two functions overlap completely.
y( t )
³
(1) 4 e -(t-O ) dO ³0 (-1) 4 e -(t-O) dO
-1
y( t )
4 e -t e O
0
Therefore,
1
0
-1
4 e -t e O
>
1
0
8 e -t 4 e -(t1) 4 e -(t1)
@
­
4 1 e -(t 1) ,
-1 t 0
°
-t
-(t 1)
4,
0t1
® 8e 4e
t
(t
1)
(t
1)
°8 e 4 e
4e
,
t !1
¯
y( t )
Chapter 15, Solution 47.
f1 ( t )
f 2 (t)
cos( t )
L -1 > F1 (s) F2 (s)@
³
cos(A) cos(B)
1
> cos(A B) cos(A B)@
2
t
0
cos(O) cos(t O) dO
L -1 > F1 (s) F2 (s)@
1 t
[cos( t ) cos( t 2O )] dO
2 ³0
L -1 > F1 (s) F2 (s)@
1
1 sin( t 2O)
cos(t ) ˜ O 0t ˜
2
2
-2
L-1 > F1 (s) F2 (s)@
0.5 t cos(t ) 0.5 sin( t )
t
0
Chapter 15, Solution 48.
(a)
2
s 2s 5
Let G (s)
g( t )
e -t sin(2 t )
F(s)
G (s) G (s)
f (t)
L -1 > G (s) G (s)@
f (t)
³
t
0
2
(s 1) 2 2 2
2
³
t
0
g (O) g ( t O) dO
e -O sin( 2O) e -( t O) sin( 2( t O)) dO
sin(A) sin(B)
1
> cos(A B) cos(A B)@
2
f (t)
1 - t t -O
e ³ e > cos(2t ) cos(2( t 2O))@ dO
2 0
f (t)
t
e -t
e -t
-2 O
cos(2 t ) ³0 e dO 2
2
f (t)
e -t
e -2O
cos(2 t ) ˜
2
-2
f (t)
t
1 -t
e -t
-2 t
e cos(2 t ) (-e 1) cos(2 t ) ³0 e -2 O cos(4O) dO
4
2
t
0
e -t
2
³
³
t
0
t
0
e -2O cos(2 t 4O) dO
e -2O > cos(2 t ) cos(4O) sin( 2 t ) sin( 4O)@ dO
t
e -t
sin( 2 t ) ³0 e -2O sin( 4O) dO
2
f (t)
1 -t
e cos(2t ) (1 e -2 t )
4
ª e -2O
º
e -t
cos(2t ) «
- 2cos(4O) 4 sin(4O) » 0t
2
¬ 4 16
¼
f (t)
(b)
ª e -2O
º
e -t
sin(2t ) «
- 2sin(4O) 4 cos(4O) » 0t
2
¬ 4 16
¼
e -t
e -3t
e -t
e -3t
cos( 2t ) cos( 2t ) cos( 2t ) cos( 2t ) cos(4t )
2
4
20
20
e -3t
e -t
cos( 2t ) sin( 4t ) sin( 2t )
10
10
e -t
e -t
sin( 2t ) sin( 4t ) sin( 2t ) cos(4t )
20
10
X(s)
Let
2
,
s 1
Y(s)
s
s4
y( t )
cos(2t ) u ( t )
x(t)
2 e -t u ( t ) ,
F(s)
X(s) Y(s)
f (t)
L -1 > X(s) Y(s)@
f (t)
³
f (t)
2 e -t ˜
f (t)
2 -t t
e > e cos(2t ) 2 sin(2t ) 1
5
f (t)
2
4
2
cos( 2t ) sin( 2t ) e -t
5
5
5
t
0
cos(2O) ˜ 2 e
³ y(O) x (t O) dO
f
0
-(t O )
dO
eO
cos(2O) 2 sin( 2O)
1 4
t
0
@
Chapter 15, Solution 49.
Let x(t) = u(t) –– u(t-1) and y(t) = h(t)*x(t).
y( t )
ª 4 1 e s º
( )»
L1 >H(s)X(s)@ L1 «
s ¼»
¬« s 2 s
ª 4(1 e s ) º
L1 «
»
¬« s(s 2) ¼»
But
1
s(s 2)
Y(s)
A
B
s s2
1 ª1
1 º
«
2 ¬ s s 2 »¼
ª1
1
es es º
2« »
s
s 2 »¼
«¬ s s 2
y( t ) 2[1 e 2 t ]u ( t ) 4[1 e 2( t 1) ]u ( t 1)
Chapter 15, Solution 50.
Take the Laplace transform of each term.
>s
2
V(s) s v(0) v c(0)@ 2 > s V(s) v(0)@ 10 V(s)
s 2 V(s) s 2 2s V(s) 2 10 V(s)
(s 2s 10) V(s)
2
V(s)
s 3 7s
3s
s 2
s 4
s 3 7s
(s 2 4)(s 2 2s 10)
3s
s 4
2
3s
s 4
2
s 3 7s
s2 4
As B
Cs D
2
2
s 4 s 2s 10
A (s 3 2s 2 10s) B (s 2 2s 10) C (s 3 4s) D (s 2 4)
Equating coefficients :
1 AC 
o C 1 A
s3 :
2
s :
0 2A B D
1
7 10A 2B 4C 6A 2B 4
s :
0 10B 4D 
o D
s0 :
Solving these equations yields
9
12
A
,
B
,
26
26
-2.5 B
17
,
26
C
D
- 30
26
V(s)
1 ª 9s 12
17s 30 º
2
2
26 «¬ s 4 s 2s 10 »¼
V(s)
º
s 1
47
2
1 ª 9s
6˜ 2
17 ˜
«
2
2
2 2
2 »
(s 1) 3
(s 1) 3 ¼
s 4
26 ¬ s 4
v( t )
47
17
6
9
cos( 2t ) sin( 2t ) e -t cos( 3t ) e -t sin( 3t )
78
26
26
26
Chapter 15, Solution 51.
Taking the Laplace transform of the differential equation yields
>s V(s) sv(0) v' (0)@ 5>sV(s) v(0)]@ 6V(s)
10
s 1
2
or s 2 5s 6 V(s) 2s 4 10
Let V(s)
A
B
C
,
s 1 s 2 s 3
10
s 1
A

o
5,
B
V(s)
0,
Hence,
v( t ) (5e t 3e 3t )u ( t )
C
2s 2 16s 24
(s 1)(s 2)(s 3)
3
Chapter 15, Solution 52.
Take the Laplace transform of each term.
>s
2
I(s) s i(0) i c(0)@ 3 > s I(s) i(0)@ 2 I(s) 1 0
(s 2 3s 2) I(s) s 3 3 1 0
s5
(s 1)(s 2)
I(s)
A
4,
B
A
B
s 1 s 2
-3
4
3
s 1 s 2
I(s)
i( t )
(4 e -t 3 e -2t ) u(t )
Chapter 15, Solution 53.
Take the Laplace transform of each term.
>s
2
(s 2 5s 6) Y(s) s 4 5
(s 2 5s 6) Y(s)
Y(s)
A
s
s 4
Y(s) s y(0) y c(0)@ 5 > s Y(s) y(0)@ 6 V(s)
s9
(s 2) Y(s) s
-2
27
,
4
s
s 4
2
s
2
s 4
s 3 9s 2 5s 36
(s 2)(s 3)(s 2 4)
2
s (s 9)(s 2 4)
s2 4
A
B
Cs D
2
s2 s3 s 4
B
(s 3) Y(s) s
-3
- 75
13
When s
0,
36
(2)(3)(4)
A B D
2 3 4

o D
5
26
When s 1 ,
46 5
(12)(5)
A B C D
3 4 5 5

o C
1
52
27 4 75 13 1 52 ˜ s 5 26
s2 s3
s2 4
Thus, Y(s)
27 - 2t 75 - 3t 1
5
e e cos( 2t ) sin( 2t )
4
13
52
52
y( t )
Chapter 15, Solution 54.
Taking the Laplace transform of the differential equation gives
> s 2 V(s) s v(0) vc(0)@ 3> s V(s) v(0)@ 2 V(s)
(s 2 3s 2) V(s)
5
1
s3
V(s)
2s
(s 3)(s 2 3s 2)
V(s)
A
B
C
s 1 s 2 s 3
A
3 2,
V(s)
B -4 ,
2s
s3
2s
(s 1)(s 2)(s 3)
C
52
32
52
4
s 1 s 2 s 3
v( t )
(1.5 e -t 4 e -2t 2.5 e -3t ) u(t )
5
s3
Chapter 15, Solution 55.
Take the Laplace transform of each term.
>s
3
Y(s) s 2 y(0) s yc(0) ycc(0)@ 6 > s 2 Y(s) s y(0) yc(0)@
8 > s Y(s) y(0)@
s 1
(s 1) 2 2 2
Setting the initial conditions to zero gives
s 1
s 2s 5
(s 3 6 s 2 8s) Y(s)
Y(s)
A
2
(s 1)
s (s 2)(s 4)(s 2 2s 5)
1
,
40
1
,
20
B
C
A
B
C
Ds E
2
s s 2 s 4 s 2s 5
-3
,
104
D
-3
,
65
E
-7
65
Y(s)
3s 7
1 1 1
1
3
1
1
˜ ˜
˜
˜
40 s 20 s 2 104 s 4 65 (s 1) 2 2 2
Y(s)
3 (s 1)
1 1 1
1
3
1
1
1
4
˜ ˜
˜
˜
˜
2
2 40 s 20 s 2 104 s 4 65 (s 1) 2
65 (s 1) 2 2 2
y( t )
1
1
3 -4t 3 -t
2
u(t ) e - 2t e e cos( 2t ) e -t sin( 2t )
40
20
104
65
65
Chapter 15, Solution 56.
Taking the Laplace transform of each term we get:
12
4 > s V(s) v(0)@ V(s) 0
s
ª
12 º
«¬ 4 s s »¼ V(s)
V(s)
8s
4s 12
2
v( t )
8
2s
s 3
2 cos
2
3t
Chapter 15, Solution 57.
Take the Laplace transform of each term.
> s Y(s) y(0)@ 9 Y(s)
s
s 4
2
s
§s2 9 ·
s
¨
¸ Y(s) 1 2
s 4
© s ¹
Y(s)
s 3 s 2 4s
(s 2 4)(s 2 9)
s 3 s 2 4s
s2 s 4
s2 4
As B Cs D
s2 4 s2 9
A (s 3 9s) B (s 2 9) C (s 3 4s) D (s 2 4)
Equating coefficients :
s0 :
0 9B 4D
1
s :
4 9 A 4C
2
1 B D
s :
3
s :
1 AC
Solving these equations gives
A
0,
Y(s)
B
- 4 5,
-4 5 s9 5
s2 4 s2 9
y( t )
C 1,
D
95
-4 5
95
s
2
2
2
s 4 s 9 s 9
- 0.4 sin( 2t ) cos( 3t ) 0.6 sin( 3t )
Chapter 15, Solution 58.
We take the Laplace transform of each term and obtain
10
6V(s) [sV (s) v(0)] V(s)
s
V(s)
e
2s
(s 3)e 2s 3e 2s
(s 3) 2 1

o
V(s)
se 2s
s 2 6s 10
Hence,
v(t )
ª¬e 3( t 2) cos(t 2) 3e 3( t 2) sin(t 2) º¼ u (t 2)
Chapter 15, Solution 59.
Take the Laplace transform of each term of the integrodifferential equation.
> s Y(s) y(0)@ 4 Y(s) 3 Y(s)
s
(s 2 4s 3) Y(s)
§ 6
·
s¨
1¸
©s 2 ¹
Y(s)
s ( 4 s)
(s 2)(s 2 4s 3)
Y(s)
A
B
C
s 1 s 2 s 3
A
2 .5 ,
Y(s)
B
6
s2
( 4 s) s
(s 1)(s 2)(s 3)
6,
C
-10.5
2.5
6
10.5
s 1 s 2 s 3
y( t )
2.5 e -t 6 e -2t 10.5 e -3t
Chapter 15, Solution 60.
Take the Laplace transform of each term of the integrodifferential equation.
3
4
2 > s X(s) x (0)@ 5 X(s) X(s) s
s
(2s 2 5s 3) X(s)
X(s)
2s 4 4s
2
s 16
2s 3 4s 2 36s 64
(2s 2 5s 3)(s 2 16)
4
s 2 16
2s 3 4s 2 36s 64
s 2 16
s 3 2s 2 18s 32
(s 1)(s 1.5)(s 2 16)
X(s)
A
A
B
Cs D
2
s 1 s 1.5 s 16
(s 1) X(s) s
B (s 1.5) X(s) s
When s
-6.235
-1
-1.5
7.329
0,
- 32
(1.5)(16)
A
B D
1.5 16

o D
0.2579
s3 2s 2 18s 32 A (s3 1.5s 2 16s 24) B (s3 s 2 16s 16)
C (s3 2.5s 2 1.5s) D (s 2 2.5s 1.5)
Equating coefficients of the s3 terms,
1 A BC 
o C
X(s)
-0.0935
- 6.235 7.329 - 0.0935s 0.2579
s 1
s 1.5
s 2 16
x(t)
- 6.235 e -t 7.329 e -1.5t 0.0935 cos(4t ) 0.0645 sin( 4t )
Chapter 16, Solution 1.
Consider the s-domain form of the circuit which is shown below.
1
1/s
I(s)
+
1/s
s
I(s)
i( t )
i( t )
1s
1 s 1 s
1
s s 1
2
1
(s 1 2) 2 ( 3 2) 2
§ 3 ·
e - t 2 sin ¨¨
t ¸¸
2
3
©
¹
2
1.155 e -0.5t sin (0.866t ) A
Chapter 16, Solution 2.
8/s
s
4
s
+
+
Vx
2
4
4
s Vx 0 Vx 0
8
s
2
4
s
Vx Vx (4s 8) 0
(16s 32)
(2s 2 4s)Vx s 2 Vx
s
Vx (3s 2 8s 8)
0
16s 32
s
s2
·
§
¸
¨
0.25
0.125 ¸
0.125
¨
16
¨ s
8¸
4
8
4
s j
s j
¸
¨
3 ¹
3
3
3
©
Vx
16
vx
(4 2e (1.3333 j0.9428) t 2e (1.3333 j0.9428) t )u ( t ) V
s(3s 2 8s 8)
§2 2
vx = 4u ( t ) e 4 t / 3 cos¨¨
© 3
· 6 4t / 3 § 2 2
t ¸¸ e
sin ¨¨
2
¹
© 3
Chapter 16, Solution 3.
s
+
1/2
5/s
1/8
Vo
Current division leads to:
Vo
1
§
·
¨
¸
1 5¨
2
¸
8s¨1 1
¸
¨ s¸
©2 8
¹
5
10 16s
5
16(s 0.625)
vo(t) = 0.3125 1 e 0.625t u ( t ) V
·
t ¸¸ V
¹
Chapter 16, Solution 4.
The s-domain form of the circuit is shown below.
6
1/(s + 1)
s
+
+
10/s
Vo(s)
Using voltage division,
§ 1 ·
¨
¸
s © s 1¹
Vo (s)
10 s
s 6 10
Vo (s)
10
(s 1)(s 6s 10)
10
§ 1 ·
10
¨
¸
s 6s 10 © s 1 ¹
2
2
A
Bs C
2
s 1 s 6s 10
A (s 2 6s 10) B (s 2 s) C (s 1)
Equating coefficients :
s2 :
0 AB 
o B
s1 :
0
s :
0
6A B C
10 10A C
-A
5A C 
o C
5A 
o A
2, B
-5A
-2, C
-10
2 (s 3)
2
4
2
2 s 1 (s 3) 1
(s 3) 2 12
Vo (s)
2
2s 10
2
s 1 s 6s 10
v o (t)
2 e -t 2 e -3t cos(t ) 4 e -3t sin( t ) V
Chapter 16, Solution 5.
Io
1
s2
s
2
2
s
V
§
·
¸
1 ¨¨
1
¸
s2¨1 1 s ¸
¨ ¸
©s 2 2¹
Io
Vs
2
·
1 §
2s
¸
¨¨
s 2 © s 2 s 2 ¸¹
2s
(s 2)(s 0.5 j1.3229)(s 0.5 j1.3229)
s2
(s 2)(s 0.5 j1.3229)(s 0.5 j1.3229)
(0.5 j1.3229) 2
(0.5 j1.3229) 2
1
(1.5 j1.3229)( j2.646) (1.5 j1.3229)( j2.646)
s2
s 0.5 j1.3229
s 0.5 j1.3229
i o (t)
e 2 t 0.3779e 90q e t / 2 e j1.3229 t 0.3779e 90q e t / 2 e j1.3229 t u ( t ) A
or
e 2 t 0.7559 sin 1.3229 t u ( t ) A
Chapter 16, Solution 6.
2
Io
10/s
5
s2
s
Use current division.
Io
i o (t)
5
s2
10 s 2
s2
s
5s
5(s 1)
s 2s 10
2
2
5
5e t cos 3t e t sin 3t
3
(s 1) 3
2
5
(s 1) 2 3 2
Chapter 16, Solution 7.
The s-domain version of the circuit is shown below.
1/s
1
+
Ix
2s
2
s 1
––
Z
1
(2s)
1
Z 1 // 2s 1 s
1
s
2s
s
2
1 2s 2
x
s 1 2s 2 2s 1
V
Z
Ix
1
2s
2s 2 2s 1
1 2s 2
1 2s 2
2s 2 1
2
(s 1)(s s 0.5)
A
Bs C
2
(s 1) (s s 0.5)
2s 2 1 A(s 2 s 0.5) B(s 2 s) C(s 1)
s2 :
2
s:
0
constant :
Ix
i x (t)
AB
ABC
2C
1 0.5A C or 0.5A 3
0.5t
C

o
2
A
6
4(s 0.5)
s 1 (s 0.5) 2 0.866 2
6
4s 2
s 1 (s 0.5) 2 0.75
>6 4e

o
@
cos 0.866 t u ( t ) A
6, B
-4
Chapter 16, Solution 8.
(a) Z
1
(b)
Z
1
1 //(1 2s)
s
1 (1 2s)
s 2 2s
1 1
1
1
2 s
1
s
3s 2 3s 2
2s(s 1)
Z
s 2 1.5s 1
s(s 1)
2s(s 1)
3s 2 3s 2
Chapter 16, Solution 9.
The s-domain form of the circuit is shown in Fig. (a).
(a)
Z in
2 || (s 1 s)
2 (s 1 s)
2 s 1 s
2 (s 2 1)
s 2 2s 1
1
s
2
s
1/s
(b)
2
2/s
1
(a)
(b)
The s-domain equivalent circuit is shown in Fig. (b).
2 || (1 2 s)
2 (1 2 s)
3 2 s
1 2 || (1 2 s)
Z in
§ 5s 6 ·
¸
s || ¨
© 3s 2 ¹
2 (s 2)
3s 2
5s 6
3s 2
§ 5s 6 ·
¸
s ˜¨
© 3s 2 ¹
§ 5s 6 ·
¸
s ¨
© 3s 2 ¹
s (5s 6)
3s 2 7s 6
Chapter 16, Solution 10.
To find ZTh, consider the circuit below.
1/s
Vx
+
1V
2
-
Vo
2Vo
Applying KCL gives
1 2Vo
Vx
2 1/ s
But Vo
2
Vx . Hence
2 1/ s
1
4Vx
2 1/ s
ZTh
Vx
1
Vx
2 1/ s

o
Vx
(2s 1)
3s
(2s 1)
3s
To find VTh, consider the circuit below.
1/s
Vy
+
2
s 1
2
Vo
-
Applying KCL gives
2
2Vo
s 1
Vo
2

o
Vo
4
3(s 1)
2Vo
1
But Vy 2Vo x Vo
s
VTh
2
Vo (1 )
s
Vy
0
4 §s 2·
¨
¸
3(s 1) © s ¹
4(s 2)
3s(s 1)
Chapter 16, Solution 11.
The s-domain form of the circuit is shown below.
4/s
1/s
+
s
I1
2
+
I2
4/(s + 2)
Write the mesh equations.
1
s
§
4·
¨ 2 ¸ I1 2 I 2
©
s¹
-4
s2
(1)
-2 I1 (s 2) I 2
(2)
Put equations (1) and (2) into matrix form.
ª 1 s º ª 2 4 s - 2 º ª I1 º
« - 4 (s 2) » « - 2
s 2»¼ «¬ I 2 »¼
¬
¼ ¬
'
2 2
(s 2s 4) ,
s
I1
'1
'
'1
s 2 4s 4
,
s (s 2)
1 2 ˜ (s 2 4s 4)
(s 2)(s 2 2s 4)
1 2 ˜ (s 2 4s 4)
'2
-6
s
A
Bs C
2
s 2 s 2s 4
A (s 2 2s 4) B (s 2 2s) C (s 2)
Equating coefficients :
s2 :
1 2 AB
1
s :
- 2 2A 2B C
s0 :
4 A 2C
2
Solving these equations leads to
A
2,
-3 2,
B
I1
- 3 2s 3
2
s 2 (s 1) 2 ( 3 ) 2
I1
2
-3
(s 1)
-3
3
˜
˜
2
2
2
s 2 2 (s 1) ( 3 )
2 3 (s 1) ( 3 ) 2
i1 ( t )
i 2 (t)
-3
> 2 e -2t 1.5 e -t cos(1.732t ) 0.866 sin(1.732t )@ u(t ) A
'2
'
I2
C
-6
s
˜
2
s 2 (s 2s 4)
-3
3
-3
(s 1) ( 3 ) 2
2
- 1.732 e -t sin(1.732t ) u(t ) A
e - t sin( 3t )
Chapter 16, Solution 12.
We apply nodal analysis to the s-domain form of the circuit below.
s
10/(s + 1)
10
Vo 3
s 1
s
s
+
1/(2s)
4
Vo
2sVo
4
(1 0.25s s 2 ) Vo
Vo
Vo
10
15
s 1
15s 25
(s 1)(s 2 0.25s 1)
10 15s 15
s 1
A
Bs C
2
s 1 s 0.25s 1
3/s
A
(s 1) Vo
15s 25
40
7
s -1
A (s 2 0.25s 1) B (s 2 s) C (s 1)
Equating coefficients :
s2 :
0 AB 
o B -A
1
s :
15 0.25A B C -0.75A C
0
25 A C
s :
A
40 7 ,
Vo
- 40 7 ,
B
- 40 135
40
s
7
7
7
s 1 § 1 ·2 3
¨s ¸ © 2¹
4
C 135 7
3
§ 155 2 ·
40 1
40
2
¸
¨
˜
2
2
7
7 s 1 7 § 1·
¹
©
3 § 1·
3
3
¨s ¸ ¨s ¸ © 2¹
© 2¹
4
4
s
1
2
v o (t)
§ 3 · (155)(2)
§ 3 ·
40 - t 40 - t 2
e e cos ¨¨
t ¸¸ e - t 2 sin ¨¨
t ¸¸
7
7
© 2 ¹ (7)( 3 )
© 2 ¹
v o (t)
5.714 e -t 5.714 e -t 2 cos(0.866t ) 25.57 e -t 2 sin( 0.866t ) V
Chapter 16, Solution 13.
Consider the following circuit.
1/s
2s
Vo
Io
2
1/(s + 2)
Applying KCL at node o,
1
s2
Vo
Vo
2s 1 2 1 s
s 1
V
2s 1 o
1
Vo
2s 1
(s 1)(s 2)
Io
Vo
2s 1
1
(s 1)(s 2)
A 1,
B
A
B
s 1 s 2
-1
1
1
s 1 s 2
Io
i o (t)
e -t e -2t u(t ) A
Chapter 16, Solution 14.
We first find the initial conditions from the circuit in Fig. (a).
1:
4:
+
5V
+
vc(0)
io
(a)
i o (0 )
5 A , v c (0 )
0V
We now incorporate these conditions in the s-domain circuit as shown in Fig.(b).
1
4
Vo
Io
15/s
+
2s
(b)
At node o,
Vo 15 s Vo 5 Vo 0
1
2s s 4 4 s
0
5/s
4/s
§
1
s ·
¸ Vo
¨1 © 2s 4 (s 1) ¹
15 5
s s
10
s
4s 2 4s 2s 2 s 2
Vo
4s (s 1)
Vo
40 (s 1)
5s 2 6s 2
Io
Io
5s 2 6s 2
Vo
4s (s 1)
Vo 5
4 (s 1)
5
2
2s s s (s 1.2s 0.4) s
5 A
Bs C
2
s s s 1.2s 0.4
4 (s 1)
A (s 2 1.2s 0.4) B s s C s
Equating coefficients :
s0 :
4 0.4A 
o A 10
s1 :
2
s :
4 1.2A C 
o C
-1.2A 4
AB 
o B
-10
0
-A
Io
5 10
10s 8
2
s s s 1.2s 0.4
Io
10 (s 0.6)
10 (0.2)
15
2
2 s (s 0.6) 0.2
(s 0.6) 2 0.2 2
i o (t)
> 15 10 e -0.6t
-8
cos(0.2 t ) sin( 0.2 t ) @ u(t ) A
Chapter 16, Solution 15.
First we need to transform the circuit into the s-domain.
s/4
10
Vo
+
3Vx
+
5/s
Vx
+
5
s2
5
Vo Vo 3Vx Vo 0
s2
s/4
5/s
10
5s
s2
0
Vx 5
s2
40Vo 120Vx 2s 2 Vo sVo 5
o Vo
s2
Vo But, Vx
0
(2s 2 s 40)Vo 120Vx We can now solve for Vx.
5 ·
5s
§
(2s 2 s 40)¨ Vx ¸ 120Vx s 2¹
s2
©
2(s 2 0.5s 40)Vx
Vx
5
10
0
(s 2 20)
s2
(s 2 20)
(s 2)(s 2 0.5s 40)
Chapter 16, Solution 16.
We first need to find the initial conditions. For t 0 , the circuit is shown in Fig. (a).
To dc, the capacitor acts like an open circuit and the inductor acts like a short circuit.
2:
+
1:
1F
Vo/2
Vo
+
(a)
+
1H
io
3V
5s
s2
Hence,
-3
3
i L (0)
io
-1 A ,
v c (0)
§ - 1·
-(2)(-1) ¨ ¸
©2¹
vo
-1 V
2.5 V
We now incorporate the initial conditions for t ! 0 as shown in Fig. (b).
2
+
Vo
1
1/s
s
+
5/(s + 2)
I1
2.5/s
+
Vo/2
+
I2
+
-1 V
Io
(b)
For mesh 1,
- 5 § 1·
1
2.5 Vo
¨ 2 ¸ I1 I 2 s2 ©
s¹
s
s
2
But,
Vo
Io
0
I2
§ 1·
§ 1 1·
¨ 2 ¸ I1 ¨ ¸ I 2
©
©2 s¹
s¹
5
2.5
s2 s
For mesh 2,
V
§
1·
1
2.5
¨1 s ¸ I 2 I1 1 o ©
s¹
s
2
s
§1
1
1·
- I1 ¨ s ¸ I 2
©2
s
s¹
2.5
1
s
(1)
0
(2)
Put (1) and (2) in matrix form.
ª 1
1 1 º
ª I1 º
«2 s
2 s »« »
«
»
1
1
1
«
»« I »
¬ 2¼
s
«¬ s
2
s »¼
'
3
2s 2 ,
s
Io
I2
'2
'
- 2s 2 13
ª 5
2.5 º
«s 2 s »
«
»
2
.
5
«
»
«¬ s 1 »¼
'2
-2 4
5
s s (s 2)
- 2s 2 13
(s 2)(2s 2 2s 3)
A
Bs C
2
s 2 2s 2s 3
A (2s 2 2s 3) B (s 2 2s) C (s 2)
Equating coefficients :
s2 :
- 2 2A B
1
0 2A 2 B C
s :
0
s :
13 3A 2C
Solving these equations leads to
A 0.7143 , B -3.429 , C
Io
Io
i o (t)
5.429
0.7143 3.429 s 5.429 0.7143 1.7145 s 2.714
s2
2s 2 2s 3
s2
s 2 s 1.5
0.7143 1.7145 (s 0.5) (3.194)( 1.25 )
s2
(s 0.5) 2 1.25 (s 0.5) 2 1.25
> 0.7143 e
-2t
@
1.7145 e -0.5t cos(1.25t ) 3.194 e -0.5t sin(1.25t ) u(t ) A
Chapter 16, Solution 17.
We apply mesh analysis to the s-domain form of the circuit as shown below.
2/(s+1)
+ I3
1/s
s
I1
1
I2
1
4
For mesh 3,
§ 1·
2
1
¨ s ¸ I 3 I1 s I 2
s 1 © s¹
s
0
For the supermesh,
§ 1·
§1 ·
¨1 ¸ I1 (1 s) I 2 ¨ s ¸ I 3
© s¹
©s ¹
But
I1
(1)
0
I2 4
(3)
Substituting (3) into (1) and (2) leads to
§
§ 1·
§ 1·
1·
¨ 2 s ¸ I 2 ¨s ¸ I 3 4 ¨1 ¸
©
© s¹
© s¹
s¹
§ 1·
§ 1·
- ¨s ¸ I 2 ¨s ¸ I 3
© s¹
© s¹
Adding (4) and (5) gives
2
2 I2 4 s 1
I2
2
1
s 1
(2)
-4
2
s s 1
(4)
(5)
i o (t)
( 2 e -t ) u(t ) A
i 2 (t)
Chapter 16, Solution 18.
3 e s
vs(t) = 3u(t) –– 3u(t––1) or Vs = s
s
3
(1 e s )
s
1:
Vs
+
+
1/s
2:
Vo
V
Vo Vs
sVo o
2
1
Vo
0 o (s 1.5)Vo
3
(1 e s )
s(s 1.5)
Vs
2 ·
§2
s
¨ ¸(1 e )
s
s
1
.
5
©
¹
v o ( t ) [(2 2e 1.5t )u ( t ) (2 2e 1.5( t 1) )u ( t 1)] V
Chapter 16, Solution 19.
We incorporate the initial conditions in the s-domain circuit as shown below.
2
V1
2I
Vo
+
I
4/(s + 2)
+
1/s
1/s
s
2
At the supernode,
4 (s 2) V1
2
2
V1 1
sVo
s s
§ 1 1·
1
2
2 ¨ ¸ V1 s Vo
©2 s¹
s
s2
But
Vo
V1 2 I and
Vo
V1 I
2 (V1 1)
s
(1)
V1 1
s

o V1
Vo 2 s
(s 2) s
s Vo 2
s2
Substituting (2) into (1)
2
1 § 2s 1 ·ª§ s ·
2 º
¨
¸«¨
¸ Vo 2
s Vo
s2
s © s ¹¬© s 2 ¹
s 2 »¼
2
1 2 (2s 1)
2 s2
s s (s 2)
2s 2 9s
s (s 2)
Vo
A
Vo
2s 9
s2
2s 9
s 4s 1
2
2.443 ,
B
ª§ 2s 1 · º
«¬¨© s 2 ¸¹ s »¼ Vo
s 2 4s 1
Vo
s2
A
B
s 0.2679 s 3.732
-0.4434
2.443
0.4434
s 0.2679 s 3.732
Therefore,
v o ( t ) ( 2.443 e -0.2679t 0.4434 e -3.732t ) u(t ) V
(2)
Chapter 16, Solution 20.
We incorporate the initial conditions and transform the current source to a voltage source
as shown.
2/s
1
1/s
Vo
+ 1/(s + 1)
+
1
At the main non-reference node, KCL gives
1 (s 1) 2 s Vo Vo Vo 1
11 s
1
s s
s
2 s Vo
s 1
(s 1)(s 1 s) Vo s
s 1
2
s 1
s
(2s 2 1 s) Vo
Vo
- 2s 2 4s 1
(s 1)(2s 2 2s 1)
Vo
- s 2s 0.5
(s 1)(s 2 s 0.5)
A
(s 1) Vo
- s 2 2s 0.5
s -1
A
Bs C
2
s 1 s s 0.5
1
A (s 2 s 0.5) B (s 2 s) C (s 1)
Equating coefficients :
s2 :
-1 A B 
o B
1
s :
s0 :
Vo
v o (t)
s 1
s
-2
-2 A BC 
o C -1
- 0.5 0.5A C 0.5 1 -0.5
1
2s 1
2
s 1 s s 0.5
2 (s 0.5)
1
s 1 (s 0.5) 2 (0.5) 2
> e -t 2 e -t 2 cos(t 2)@ u(t ) V
s
1/s
Chapter 16, Solution 21.
The s-domain version of the circuit is shown below.
1
s
V1
+
Vo
2/s
2
1/s
10/s
At node 1,
10
V1
V1 Vo s
s
Vo
1
s
2
At node 2,
V1 Vo Vo
sVo
s
2

o

o
10
( s 1)V1 (
s2
1)Vo
2
(1)
s
V1 Vo ( s 2 1)
2
(2)
Substituting (2) into (1) gives
10
s2
( s 1)( s s / 2 1)Vo ( 1)Vo
2
Vo
10
s ( s 2s 1.5)
s ( s 2 2s 1.5)Vo
2
2
A
Bs C
2
s s 2s 1.5
A( s 2 2 s 1.5) Bs 2 Cs
s2 :
0 A B
s:
0 2A C
constant :
10 1.5 A

o
10
Vo
20 ª 1
s2
º
2
«
3 ¬ s s 2 s 1.5 »¼
A
20 / 3, B
-20/3, C
º
0.7071
20 ª 1
s 1
1.414
« 2
2
2
2 »
3 ¬ s ( s 1) 0.7071
( s 1) 0.7071 ¼
Taking the inverse Laplace tranform finally yields
v o (t)
>
-40/3
@
20
1 e t cos 0.7071t 1.414e t sin 0.7071t u ( t ) V
3
Chapter 16, Solution 22.
The s-domain version of the circuit is shown below.
4s
V1
12
s 1
At node 1,
12
s 1
1
V1 V1 V2
1
4s
At node 2,
V1 V2
4s
V2 s
V2
2 3
V2
2
o
3/s
12
s 1
1· V
§
V1 ¨1 ¸ 2
© 4s ¹ 4s
(1)
V1
§4
·
V2 ¨ s 2 2s 1¸
©3
¹
(2)
o
Substituting (2) into (1),
ª§ 4
1· 1º
·§
V2 «¨ s 2 2s 1¸¨1 ¸ »
¹© 4s ¹ 4s ¼
© 3
12
s 1
V2
9
9
7
9
(s 1)(s 2 s )
4
8
A(s 2 3·
§4 2 7
¨ s s ¸V2
3
2¹
©3
A
Bs C
(s 1) (s 2 7 s 9 )
4
8
7
9
s ) B(s 2 s) C(s 1)
4
8
Equating coefficients:
s2 :
0
AB
s:
0
7
ABC
4
3
AC
4
constant :
9
9
3
AC
A 
o
8
8

o
A
24, B
C
3
A
4
-24, C
-18
24s 18
7
9
(s 2 s )
4
8
24
(s 1)
V2
3
24
24(s 7 / 8)
7
23
7
23
(s 1)
(s ) 2 (s ) 2 8
64
8
64
Taking the inverse of this produces:
v 2 (t)
>24e
t
@
24e 0.875t cos(0.5995t ) 5.004e 0.875t sin(0.5995t ) u ( t )
Similarly,
§4
·
9¨ s 2 2s 1¸
©3
¹
7
9
(s 1)(s 2 s )
4
8
V1
§4
·
9¨ s 2 2s 1¸
©3
¹
D
(s 1)
Es F
7
9
(s 2 s )
4
8
7
9
D(s 2 s ) E(s 2 s) F(s 1)
4
8
Equating coefficients:
s2 :
s:
9
v1 ( t )
>8e
t
3
DF
4
9
3
D F or 3
D 
o
8
8
8
(s 1)
V1
DE
7
D E F or 6
4
18
constant :
Thus,
12
4s
7
9
(s 2 s )
4
8

o
D
8, E
F
3
6 D
4
4, F
0
8
4(s 7 / 8)
7/2
7
23
7
23
(s 1)
(s ) 2 (s ) 2 8
64
8
64
@
4e 0.875t cos(0.5995t ) 5.838e 0.875t sin(0.5995t ) u ( t )
Chapter 16, Solution 23.
The s-domain form of the circuit with the initial conditions is shown below.
V
I
4/s
R
sL
-2/s
1/sC
5C
At the non-reference node,
4 2
5C
s s
6 5 sC
s
s
1 ·
CV § 2
¨s ¸
RC LC ¹
s ©
5s 6 C
s s RC 1 LC
V
But
V V
sCV
R sL
2
1
RC
1
10 80
5s 480
s 8s 20
V
2
V
sL
I
1.25s 120
s (s 2 8s 20)
i( t )
20
5 (s 4)
(230)(2)
2
2 (s 4) 2
(s 4) 2 22
5s 480
4s (s 2 8s 20)
I
I
1
4 80
5 e -4t cos( 2t ) 230 e -4t sin( 2t ) V
v( t )
A
1
LC
8,
6,
B
6 6s 46.75
s s 2 8s 20
A
Bs C
2
s s 8s 20
-6 ,
C
-46.75
6
6 (s 4)
(11.375)(2)
2
2 s (s 4) 2
(s 4) 2 22
6 u(t ) 6 e -4t cos( 2t ) 11.375 e -4t sin( 2t ), t ! 0
Chapter 16, Solution 24.
At t = 0-, the circuit is equivalent to that shown below.
+
4:
9A
5:
vo
-
v o (0)
5x
4
(9)
45
20
For t > 0, we have the Laplace transform of the circuit as shown below after
transforming the current source to a voltage source.
4:
16 :
Vo
+
36V
10A
2/s
5:
-
Applying KCL gives
36 Vo
10
20
sVo Vo
2
5

o
Thus,
v o (t)
Vo
3.6 20s
s(s 0.5)
A
B
,
s s 0.5
>7.2 12.8e @ u(t)
0.5 t
A
7.2, B
12.8
Chapter 16, Solution 25.
For t ! 0 , the circuit in the s-domain is shown below.
6
s
I
+
9/s
(2s)/(s2 + 16)
+
V
+
2/s
Applying KVL,
2s §
9· 2
¨6 s ¸ I s¹ s
s 16 ©
2
I
V
0
4s 2 32
(s 2 6s 9)(s 2 16)
9
2
I
s
s
2
36s 2 288
s s (s 3) 2 (s 2 16)
2 A
B
C
Ds E
2
2
s s s 3 (s 3)
s 16
36s 2 288 A (s 4 6s 3 25s 2 96s 144) B (s 4 3s 3 16s 2 48s)
C (s 3 16s) D (s 4 6s 3 9s 2 ) E (s 3 6s 2 9s)
Equating coefficients :
288 144A
s0 :
1
s :
0 96A 48B 16C 9E
2
36 25A 16B 9D 6E
s :
3
0 6A 3B C 6D E
s :
4
s :
0 A B D
Solving equations (1), (2), (3), (4) and (5) gives
A 2 , B -1.7984 , C -8.16 , D
(1)
(2)
(3)
(4)
(5)
-0.2016 ,
E
2.765
V(s)
4 1.7984
8.16
0.2016 s (0.6912)(4)
2 s
s 3 (s 3)
s 2 16
s 2 16
v( t )
4 u(t ) 1.7984 e -3t 8.16 t e -3t 0.2016 cos(4t ) 0.6912 sin( 4t ) V
Chapter 16, Solution 26.
Consider the op-amp circuit below.
R2
1/sC
R1
0
+
Vs
+
At node 0,
Vs 0
R1
0 Vo
(0 Vo ) sC
R2
Vs
§ 1
·
sC ¸ - Vo
R1 ¨
©R2
¹
Vo
Vs
-1
sR 1C R 1 R 2
But
R1
R2
20
10
So,
Vo
Vs
-1
s2
Vs
3 e -5t
2,
R 1C (20 u 103 )(50 u 10-6 ) 1

o Vs
3 (s 5)
+
Vo
-3
(s 2)(s 5)
Vo
3
(s 2)(s 5)
- Vo
A 1,
A
B
s2 s5
B -1
1
1
s5 s2
Vo
vo (t)
e -5t e -2t u(t )
Chapter 16, Solution 27.
Consider the following circuit.
2s
10/(s + 3)
For mesh 1,
10
s3
10
s3
+
I1
s
2s
1
1
(1 2s) I1 I 2 s I 2
(1 2s) I1 (1 s) I 2
For mesh 2,
0 (2 2s) I 2 I1 s I1
0 -(1 s) I1 2 (s 1) I 2
(1) and (2) in matrix form,
ª10 (s 3) º ª 2s 1 - (s 1) ºª I1 º
» « - (s 1) 2 (s 1) »« I »
«
0
¼¬ 2 ¼
¼ ¬
¬
'
I2
3s 2 4s 1
(1)
(2)
'1
20 (s 1)
s3
'2
10 (s 1)
s3
Thus
I1
'1
'
20 (s 1)
(s 3)( 3s 2 4s 1)
I2
'2
'
10 (s 1)
(s 3)( 3s 2 4s 1)
I1
2
Chapter 16, Solution 28.
Consider the circuit shown below.
s
1
+
+
6/s
I1
2s
s
I2
For mesh 1,
6
(1 2s) I1 s I 2
s
Vo
2
(1)
For mesh 2,
0 s I1 (2 s) I 2
I1
§ 2·
- ¨1 ¸ I 2
© s¹
Substituting (2) into (1) gives
§ 2·
6
-(1 2s)¨1 ¸ I 2 s I 2
© s¹
s
or
I2
-6
s 5s 2
2
(2)
- (s 2 5s 2)
I2
s
Vo
- 12
s 5s 2
2 I2
2
- 12
(s 0.438)(s 4.561)
Since the roots of s 2 5s 2 0 are -0.438 and -4.561,
Vo
A
B
s 0.438 s 4.561
A
- 12
4.123
-2.91 ,
B
- 12
- 4.123
2.91
- 2.91
2.91
s 0.438 s 4.561
Vo (s)
2.91 > e -4.561t e 0.438t @ u(t ) V
vo (t)
Chapter 16, Solution 29.
Consider the following circuit.
1
1:2
+
10/(s + 1)
Let
Io
ZL
4/s
8 ||
4
s
(8)(4 s)
8 4 s
8
2s 1
When this is reflected to the primary side,
Zin
1
ZL
, n
n2
Zin
1
2
2s 1
Io
10 1
˜
s 1 Zin
2
2s 3
2s 1
10 2s 1
˜
s 1 2s 3
8
Io
10s 5
(s 1)(s 1.5)
A
-10 ,
I o (s)
A
B
s 1 s 1.5
B 20
- 10
20
s 1 s 1.5
io (t)
>
@
10 2 e -1.5t e t u(t ) A
Chapter 16, Solution 30.
4
s 1 3
Y(s)
H(s) X(s) ,
Y(s)
12 s 2
(3s 1) 2
Y(s)
4 8
s
4
1
˜
˜
2 3 9 (s 1 3)
27 (s 1 3) 2
Let G (s)
X(s)
12
3s 1
4 8s 4 3
3 (3s 1) 2
-8
s
˜
9 (s 1 3) 2
Using the time differentiation property,
·
-8 d
- 8 § - 1 -t 3
¨ t e e -t 3 ¸
g( t )
˜ (t e -t 3 )
¹
9 dt
9©3
g( t )
8 -t 3 8 -t 3
te e
27
9
y( t )
4
8 -t 3 8 -t 3 4 -t 3
u(t) te e te
3
27
9
27
y( t )
4 -t 3
8
4
te
u( t ) e - t 3 27
9
3
Hence,
Chapter 16, Solution 31.
x(t)
u(t) 
o X(s)
1
s
10s
s2 4
y( t ) 10 cos(2t ) 
o Y(s)
H(s)
Y(s)
X(s)
10s 2
s2 4
Chapter 16, Solution 32.
(a)
Y(s)
H(s) X(s)
s3
1
˜
s 4s 5 s
2
s3
s (s 4s 5)
2
A
Bs C
2
s s 4s 5
s 3 A (s 2 4s 5) Bs 2 Cs
Equating coefficients :
3 5A 
o A 3 5
s0 :
s1 :
1 4A C 
o C 1 4A - 7 5
s2 :
0 AB 
o B -A - 3 5
Y(s)
35 1
3s 7
˜ 2
s 5 s 4s 5
Y(s)
0.6 1 3 (s 2) 1
˜
s 5 (s 2) 2 1
y( t )
> 0.6 0.6 e -2t cos(t ) 0.2 e -2t sin(t )@ u(t )
(b)
x ( t ) 6 t e -2t

o X(s)
6
(s 2) 2
s3
6
˜
s 4s 5 (s 2) 2
Y(s)
H(s) X(s)
Y(s)
6 (s 3)
(s 2) 2 (s 2 4s 5)
2
A
B
Cs D
2 2
s 2 (s 2) s 4s 5
Equating coefficients :
s3 :
0 AC 
o C -A
2
0 6 A B 4C D 2 A B D
s :
1
s :
6 13A 4B 4C 4D 9A 4B 4D
0
18 10A 5B 4D 2A B
s :
Solving (1), (2), (3), and (4) gives
A 6,
B 6,
C -6 ,
D
Y(s)
6
6
6s 18
2 s 2 (s 2)
(s 2) 2 1
Y(s)
6
6
6 (s 2)
6
2 2
s 2 (s 2)
(s 2) 1 (s 2) 2 1
y( t )
1
s
H(s)
Y(s)
,
X(s)
Y(s)
4
1
2s
(3)(4)
2
s 2 (s 3) (s 2) 16 (s 2) 2 16
X(s)
4
-18
> 6 e -2t 6 t e -2t 6 e -2t cos(t ) 6 e -2t sin(t )@ u(t )
Chapter 16, Solution 33.
H(s) s Y(s)
(1)
(2)
(3)
(4)
s
2s2
12 s
2
2
2 (s 3) s 4s 20 s 4s 20
Chapter 16, Solution 34.
Consider the following circuit.
2
s
Vo
+
+
Vs
4
10/s
Vo(s)
Using nodal analysis,
Vs Vo
s2
Vo Vo
4 10 s
Vs
§ 1
1 s·
(s 2) ¨
¸ Vo
© s 2 4 10 ¹
Vs
1
2s 2 9s 30 Vo
20
Vo
Vs
20
2s 9s 30
§ 1
·
1
¨1 (s 2) (s 2 2s) ¸ Vo
© 4
¹
10
2
Chapter 16, Solution 35.
Consider the following circuit.
I
Vs
2/s
s
V1
+
+
2I
Vo
At node 1,
2I I
V1
,
s3
where I
Vs V1
2s
3
3˜
Vs V1
2s
V1
s3
3s
3s
Vs V1
2
2
V1
s3
§ 1
3s ·
¨
¸ V1
©s 3 2 ¹
3s
V
2 s
V1
3s (s 3)
V
3s 2 9s 2 s
Vo
3
V
s3 1
Vo
Vs
H(s)
9s
V
3s 9s 2 s
2
9s
3s 9s 2
2
Chapter 16, Solution 36.
From the previous problem,
V1
s3
3I
I
But
I
3s
V
3s 9s 2 s
2
s
V
3s 9s 2 s
Vs
2
3s 2 9s 2
Vo
9s
s
3s 2 9 s 2
˜
Vo
3s 2 9 s 2
9s
H(s)
Vo
I
9
Vo
9
Chapter 16, Solution 37.
(a)
Consider the circuit shown below.
3
+
Vs
2s
+
I1
I2
2/s
Vx
+
4Vx
For loop 1,
Vs
§ 2·
2
¨3 ¸ I1 I 2
©
s
s¹
(1)
For loop 2,
§
2
2·
4Vx ¨ 2s ¸ I 2 I1
©
s
s¹
0
§2·
(I1 I 2 ) ¨ ¸
©s¹
But,
Vx
So,
§
2
2·
8
(I1 I 2 ) ¨ 2s ¸ I 2 I1
©
s
s¹
s
0
0
·
§6
-6
I1 ¨ 2s ¸ I 2
¹
©s
s
(2)
In matrix form, (1) and (2) become
ª Vs º ª3 2 s
- 2 s ºª I1 º
« 0 » « - 6 s 6 s 2s »« I »
¼¬ 2 ¼
¬ ¼ ¬
'
§
· § 6 ·§ 2 ·
2 ·§ 6
¨3 ¸¨ 2s ¸ ¨ ¸¨ ¸
©
¹ © s ¹© s ¹
s ¹© s
'
18
6s 4
s
'1
§6
·
¨ 2s ¸ Vs ,
©s
¹
'2
6
V
s s
I1
I1
Vs
(b)
I2
'1
'
(6 s 2s)
V
18 s 4 6s s
3 ss
9 s23
s2 3
3s 2 2s 9
'2
'
2 § '1 ' 2 ·
¨
¸
s© ' ¹
Vx
2
(I I )
s 1 2
Vx
2 s Vs (6 s 2s 6 s)
'
I2
Vx
6 s Vs
- 4Vs
- 4Vs
'
-3
2s
Chapter 16, Solution 38.
(a)
Consider the following circuit.
1
Is
V1
s
Vo
Io
+
Vs
+
1/s
1/s
1
Vo
At node 1,
Vs V1
1
Vs
s V1 V1 Vo
s
§
1·
1
¨1 s ¸ V1 Vo
©
s¹
s
(1)
At node o,
V1 Vo
s
V1
s Vo Vo
(s 1) Vo
(s 2 s 1) Vo
(2)
Substituting (2) into (1)
Vs (s 1 1 s)(s 2 s 1)Vo 1 s Vo
Vs
H 1 (s)
(b)
(d).
1
s 2s 3s 2
3
2
Is
Vs V1
Is
(s 3 s 2 2s 1)Vo
H 2 (s)
(c)
Vo
Vs
(s 3 2s 2 3s 2)Vo
Io
Vo
Is
(s 3 2s 2 3s 2)Vo (s 2 s 1)Vo
1
s s 2s 1
3
2
Vo
1
H 3 (s)
Io
Is
Vo
Is
H 2 (s)
1
s s 2s 1
H 4 (s)
Io
Vs
Vo
Vs
H 1 (s)
1
s 2s 3s 2
3
2
3
2
Chapter 16, Solution 39.
Consider the circuit below.
Va
Vb
Vs
+
1/sC
+
+
R
Io
Vo
Since no current enters the op amp, I o flows through both R and C.
Vo
§
1·
-I o ¨ R ¸
©
sC ¹
Va
Vb
Vo
Vs
H(s)
- Io
sC
Vs
R 1 sC
1 sC
sRC 1
Chapter 16, Solution 40.
(a)
(b)
H(s)
Vo
Vs
h(t)
R - Rt L
e
u( t )
L
v s (t)
u(t) 
o Vs (s) 1 s
Vo
R L
V
sR L s
A 1,
Vo
vo (t)
R
R sL
B
R L
sR L
R L
s (s R L)
A
B
s sR L
-1
1
1
s sR L
u ( t ) e -Rt L u ( t )
(1 e -Rt L ) u(t )
Chapter 16, Solution 41.
Y(s)
H(s) X(s)
h(t)
2 e -t u ( t ) 
o
H(s)
2
s 1
v i (t)
5 u(t) 
o Vi (s)
Y(s)
10
s (s 1)
A 10 ,
X(s)
5s
A
B
s s 1
B
-10
Y(s)
10 10
s s 1
y( t )
10 (1 e -t ) u(t )
Chapter 16, Solution 42.
2s Y(s) Y(s)
(2s 1) Y(s)
X(s)
X(s)
H(s)
Y(s)
X(s)
1
2s 1
h (t )
0.5 e -t 2 u(t )
1
2 (s 1 2)
Chapter 16, Solution 43.
1ȍ
u(t)
+
i(t)
1F
1H
First select the inductor current iL and the capacitor voltage vC to be the state
variables.
Applying KVL we get:
u ( t ) i v C i' 0; i
v 'C
Thus,
v 'C
i
i'
v C i u(t)
Finally we get,
ª v c º ª 0 1 º ª v C º ª0 º
« C » «
» « » « » u ( t ) ; i( t )
¬« i c ¼» ¬ 1 1¼ ¬ i ¼ ¬1¼
>0 1@ª«
vC º
» >0@u ( t )
¬ i ¼
Chapter 16, Solution 44.
1/8 F
1H
+
+
4u ( t )
2ȍ
vx
4ȍ
First select the inductor current iL and the capacitor voltage vC to be the state
variables.
Applying KCL we get:
v 'C
8
v
iL x 2
i 'L
4u ( t ) v x
vx
v 'C
vC 4
8
0; or v 'C
8i L 4v x
v 'C
vC 2
v C 4i L 2v x ; or v x
v 'C
8i L 1.3333v C 5.333i L
i 'L
4u ( t ) 0.3333v C 1.3333i L
0.3333v C 1.3333i L
1.3333v C 2.666i L
Now we can write the state equations.
ª v 'C º
« ' »
«¬ i L »¼
2.666 º ª v C º ª0º
ª 1.3333
« 0.3333 1.3333» « i » «4» u ( t ); v x
¬
¼¬ L ¼ ¬ ¼
ª0.3333º ª v C º
«1.3333» « i »
¬
¼¬ L ¼
Chapter 16, Solution 45.
First select the inductor current iL (current flowing left to right) and the capacitor voltage
vC (voltage positive on the left and negative on the right) to be the state variables.
Applying KCL we get:
v 'C v o
iL
4
2
0 or v 'C
i 'L
vo v 2
vo
v C v1
v 'C
4i L 2 v C 2 v1
i 'L
v C v1 v 2
4i L 2 v o
ª i c º ª0 1º ª i L º ª1 1º ª v1 ( t ) º
« L c» «
» ; v o (t)
»«
»« »«
«¬ v C »¼ ¬4 2¼ ¬ v C ¼ ¬2 0 ¼ ¬ v 2 ( t )¼
Chapter 16, Solution 46.
>0
ªi º
ª v (t) º
1@ « L » >1 0@ « 1 »
¬ v 2 ( t )¼
¬v C ¼
First select the inductor current iL (left to right) and the capacitor voltage vC to be
the state variables.
Letting vo = vC and applying KCL we get:
v
i L v 'C C i s
4
i 'L
Thus,
ªv ' º
« 'C »
¬« i L ¼»
0 or v 'C
0.25v C i L i s
v C v s
ª 0.25 1º ª v 'C º ª0 1º ª v s º
; v o (t)
« »
« 1
0»¼ «¬ i 'L »¼ «¬1 0»¼ «¬ i s »¼
¬
ª1º ª v C º ª0 0º ª v s º
«0 » « i » «0 0 » « i »
¬ ¼¬ L ¼ ¬
¼¬ s ¼
Chapter 16, Solution 47.
First select the inductor current iL (left to right) and the capacitor voltage vC (+ on the
left) to be the state variables.
Letting i1 =
v 'C
and i2 = iL and applying KVL we get:
4
Loop 1:
§ v'
·
v1 v C 2¨ C i L ¸
¨ 4
¸
©
¹
0 or v 'C
4i L 2 v C 2 v1
Loop 2:
§
v 'C ·¸ '
¨
2 iL i v 2 0 or
¸ L
¨
4
¹
©
4i 2v C 2v1
v2
i 'L 2i L L
2
v C v1 v 2
i1
ªi c º
« L c»
«¬ v C »¼
4i L 2 v C 2 v1
4
i L 0.5v C 0.5v1
ª0 1º ª i L º ª1 1º ª v1 ( t ) º
« 4 2» « v » « 2 0 » « v ( t ) » ;
¬
¼¬ C¼ ¬
¼¬ 2 ¼
ª i1 ( t ) º
«i ( t )»
¬2 ¼
ª1 0.5º ª i L º ª0.5 0º ª v1 ( t ) º
«1
0 »¼ «¬ v C »¼ «¬ 0 0»¼ «¬ v 2 ( t )»¼
¬
Chapter 16, Solution 48.
Let x1 = y(t). Thus, x1'
y'
x 2 and x '2
ycc
3x1 4 x 2 z( t )
This gives our state equations.
ª x1' º
« ' »
¬« x 2 ¼»
1 º ª x 1 º ª0 º
ª0
« 3 4» « x » «1» z( t ); y( t )
¬
¼¬ 2 ¼ ¬ ¼
>1 0@ª«
x1 º
» >0@z( t )
¬x 2 ¼
Chapter 16, Solution 49.
Let x1
y( t ) and x 2
x1' z
y ' z or y '
x2 z
Thus,
x '2
y cc z '
6x1 5( x 2 z) z ' 2z z '
6x1 5x 2 3z
This now leads to our state equations,
ª x1' º
« ' »
«¬ x 2 »¼
1 º ª x1 º ª 1 º
ª0
« 6 5» « x » « 3» z( t ); y( t )
¬
¼¬ 2 ¼ ¬ ¼
>1 0@ ª«
x1 º
» >0@z( t )
¬x 2 ¼
Chapter 16, Solution 50.
Let x1 = y(t), x2 = x1' , and x 3
Thus,
x "3
x '2 .
6x1 11x 2 6x 3 z( t )
We can now write our state equations.
ª x1' º
« ' »
«x 2 »
«x ' »
¬« 3 ¼»
1
0 º ª x 1 º ª0 º
ª0
«0
0
1 »» «« x 2 »» ««0»» z( t ); y( t )
«
«¬ 6 11 6»¼ «¬ x 3 »¼ «¬1»¼
ª x1 º
>1 0 0@««x 2 »» >0@z( t )
«¬ x 3 »¼
Chapter 16, Solution 51.
We transform the state equations into the s-domain and solve using Laplace
transforms.
sX(s) x (0)
§1·
AX(s) B¨ ¸
©s¹
Assume the initial conditions are zero.
(sI A)X(s)
§1·
B¨ ¸
©s¹
X(s)
ªs 4 4º
« 2
s »¼
¬
Y(s)
X1 (s)
1
ª0 º § 1 ·
« 2» ¨ s ¸
¬ ¼© ¹
4 ºª 0 º
ªs
«
»«
»
s 2 4s 8 ¬2 s 4¼ ¬(2 / s)¼
8
1
s4
s s 2 4s 8
1
s(s 2 4s 8)
1
1
2
(s 2)
s4
2
2
2
2
s (s 2) 2
s (s 2) 2
(s 2) 2 2 2
y(t) = 1 e 2 t cos 2t sin 2t u ( t )
Chapter 16, Solution 52.
Assume that the initial conditions are zero. Using Laplace transforms we get,
X(s)
X1
1 º
ªs 2
« 2 s 4»
¬
¼
3s 8
2
2
s((s 3) 1 )
1
ª1 1 º ª1 / s º
« 4 0» « 2 / s »
¬
¼¬
¼
ªs 4 1 º ª3 / s º
«
s 2»¼ «¬4 / s »¼
s 6s 10 ¬ 2
1
2
0.8 0.8s 1.8
s
(s 3) 2 12
0.8
s3
1
0.8
.6
s
(s 3) 2 12
(s 3) 2 12
x1 ( t )
(0.8 0.8e 3t cos t 0.6e 3t sin t )u ( t )
4s 14
X2
1.4 1.4s 4.4
s
(s 3) 2 12
s((s 3) 2 12
1.4
s3
1
1.4
0.2
2
2
s
(s 3) 1
(s 3) 2 12
x 2 (t)
(1.4 1.4e 3t cos t 0.2e 3t sin t )u ( t )
y1 ( t )
2 x 1 ( t ) 2 x 2 ( t ) 2 u ( t )
(2.4 4.4e 3t cos t 0.8e 3t sin t )u ( t )
x1 ( t ) 2u ( t )
y 2 (t)
(1.2 0.8e 3t cos t 0.6e 3t sin t )u ( t )
Chapter 16, Solution 53.
If Vo is the voltage across R, applying KCL at the non-reference node gives
Is
Vo
V
sC Vo o
R
sL
Vo
Io
H(s)
Is
1
1
sC R
sL
Vo
R
§1
1·
¨ sC ¸ Vo
©R
sL ¹
sRL Is
sL R s 2 RLC
sL Is
s RLC sL R
2
Io
Is
sL
s RLC sL R
2
s RC
s s RC 1 LC
2
The roots
s1, 2
-1
1
1
r
2 2RC
(2RC)
LC
both lie in the left half plane since R, L, and C are positive quantities.
Thus, the circuit is stable.
Chapter 16, Solution 54.
(a)
H1 (s)
3
,
s 1
H 2 (s)
H(s)
H1 (s) H 2 (s)
h (t )
L-1 > H(s)@
1
s4
3
(s 1)(s 4)
ª A
B º
L-1 «
¬ s 1 s 4 »¼
A 1,
B -1
-t
-4t
h ( t ) ( e e ) u( t )
(b)
Since the poles of H(s) all lie in the left half s-plane, the system is stable.
Chapter 16, Solution 55.
Let
Vo1 be the voltage at the output of the first op amp.
Vo1
Vs
1 sC
R
H(s)
Vo
Vs
h(t)
t
R C2
lim h ( t )
t of
1
,
sRC
Vo
Vo1
1
sRC
1
s R 2C 2
2
2
f , i.e. the output is unbounded.
Hence, the circuit is unstable.
Chapter 16, Solution 56.
1
sC
1
sL sC
sL ˜
1
sL ||
sC
sL
1 s 2 LC
V2
V1
sL
1 s 2 LC
sL
R
1 s 2 LC
V2
V1
1
RC
1
1
s2 s ˜
RC LC
sL
s RLC sL R
2
s˜
Comparing this with the given transfer function,
1
1
2
,
6
RC
LC
If R 1 k: ,
C
1
2R
500 PF
L
1
6C
333.3 H
Chapter 16, Solution 57.
The circuit in the s-domain is shown below.
R1
Vi
L
V1
+
+
C
R2
Vx
Z
Z
1
|| (R 2 sL)
sC
(1 sC) ˜ (R 2 sL)
R 2 sL 1 sC
R 2 sL
1 s 2 LC sR 2 C
V1
Z
V
R1 Z i
Vo
R2
V
R 2 sL 1
R2
Z
˜
V
R 2 sL R 1 Z i
R 2 sL
R2
1 s 2 LC sR 2 C
˜
R 2 sL
R 2 sL
R1 1 s 2 LC sR 2 C
Vo
Vi
R2
Z
˜
R 2 sL R 1 Z
Vo
Vi
R2
s R 1 LC sR 1 R 2 C R 1 R 2 sL
Vo
Vi
R2
R 1 LC
§R2
1 · R1 R 2
¸
s 2 s¨
© L R 1C ¹ R 1 LC
2
Comparing this with the given transfer function,
R2
R2
R1 R 2
1
5
6
25
R 1 LC
L R 1C
R 1 LC
Since R 1
5
4 : and R 2 1 : ,
1
1

o LC
4 LC
20
6
1
1
L 4C
25
5
4 LC
(2)

o LC
1
20
Substituting (1) into (2),
1
6 20 C 
o 80 C 2 24 C 1 0
4C
Thus, C
1
,
4
1
20
(1)
When C
1
,
4
L
1
20 C
1
.
5
When C
1
,
20
L
1
20 C
1.
Therefore, there are two possible solutions.
C 0.25 F
L 0.2 H
or
C 0.05 F
L
1H
Chapter 16, Solution 58.
We apply KCL at the noninverting terminal at the op amp.
(Vs 0) Y3 (0 Vo )(Y1 Y2 )
Y3 Vs - (Y1 Y2 )Vo
Vo
Vs
Let
Y1
- Y3
Y1 Y2
sC1 ,
Vo
Vs
Y2
- sC 2
sC1 1 R 1
1 R1 ,
Y3
sC 2
- sC 2 C1
s 1 R 1 C1
Comparing this with the given transfer function,
C2
1
1,
10
R 1 C1
C1
If R 1
1 k: ,
C1
C2
1
10 4
100 PF
Chapter 16, Solution 59.
Consider the circuit shown below. We notice that V3
Vo and V2
V3
Vo .
Y4
Y1
Y2
V1
+
Vin
V2
+
Vo
Y3
At node 1,
(Vin V1 ) Y1 (V1 Vo ) Y2 (V1 Vo ) Y4
Vin Y1 V1 (Y1 Y2 Y4 ) Vo (Y2 Y4 )
At node 2,
(V1 Vo ) Y2 (Vo 0) Y3
V1 Y2 (Y2 Y3 ) Vo
V1
(1)
Y2 Y3
Vo
Y2
(2)
Substituting (2) into (1),
Y2 Y3
Vin Y1
˜ (Y1 Y2 Y4 ) Vo Vo (Y2 Y4 )
Y2
Vin Y1 Y2
Vo
Vin
Vo (Y1 Y2 Y22 Y2 Y4 Y1 Y3 Y2 Y3 Y3 Y4 Y22 Y2 Y4 )
Y1 Y2
Y1 Y2 Y1 Y3 Y2 Y3 Y3 Y4
Y1 and Y2 must be resistive, while Y3 and Y4 must be capacitive.
Let
Y1
1
,
R1
Y2
1
,
R2
Y3
sC1 ,
Y4
sC 2
Vo
Vin
1
R 1R 2
sC1 sC1
1
s 2 C1 C 2
R 1R 2 R 1 R 2
Vo
Vin
1
R 1 R 2 C1C 2
§ R1 R 2 ·
1
¸
s2 s ˜¨
© R 1 R 2 C 2 ¹ R 1 R 2 C1 C 2
Choose R 1
1 k: , then
1
R 1 R 2 C1 C 2
10 6
and
R1 R 2
R 1R 2 C 2
100
We have three equations and four unknowns. Thus, there is a family of solutions. One
such solution is
R 2 1 k: , C1 50 nF , C 2 20 PF
Chapter 16, Solution 60.
With the following MATLAB codes, the Bode plots are generated as shown below.
num=[1 1];
den= [1 5 6];
bode(num,den);
Chapter 16, Solution 61.
We use the following codes to obtain the Bode plots below.
num=[1 4];
den= [1 6 11 6];
bode(num,den);
Chapter 16, Solution 62.
The following codes are used to obtain the Bode plots below.
num=[1 1];
den= [1 0.5 1];
bode(num,den);
Chapter 16, Solution 63.
We use the following commands to obtain the unit step as shown below.
num=[1 2];
den= [1 4 3];
step(num,den);
Chapter 16, Solution 64.
With the following commands, we obtain the response as shown below.
t=0:0.01:5;
x=10*exp(-t);
num=4;
den= [1 5 6];
y=lsim(num,den,x,t);
plot(t,y)
Chapter 16, Solution 65.
We obtain the response below using the following commands.
t=0:0.01:5;
x=1 + 3*exp(-2*t);
num=[1 0];
den= [1 6 11 6];
y=lsim(num,den,x,t);
plot(t,y)
Chapter 16, Solution 66.
We obtain the response below using the following MATLAB commands.
t=0:0.01:5;
x=5*exp(-3*t);
num=1;
den= [1 1 4];
y=lsim(num,den,x,t);
plot(t,y)
Chapter 16, Solution 67.
Using the result of Practice Problem 16.14,
Vo
- Y1 Y2
Vi Y2 Y3 Y4 (Y1 Y2 Y3 )
When Y1
sC1 ,
C1
0.5 PF
Y2
1
,
R1
R1
10 k:
Y3
Y2 ,
Y4
sC 2 ,
Vo
Vi
- sC1 R 1
1 R sC 2 (sC1 2 R 1 )
Vo
Vi
- sC1 R 1
s C1C 2 R 12 s ˜ 2C 2 R 1 1
Vo
Vi
- s (0.5 u 10 -6 )(10 u 10 3 )
s 2 (0.5 u 10 -6 )(1 u 10 -6 )(10 u 10 3 ) 2 s (2)(1 u 10 -6 )(10 u 10 3 ) 1
Vo
Vi
- 100 s
s 400 s 2 u 10 4
1 PF
C2
- sC1 R 1
1 sC 2 R 1 (2 sC1 R 1 )
2
1
2
2
Therefore,
a - 100 ,
400 ,
b
c
2 u 10 4
Chapter 16, Solution 68.
(a)
Let
Y(s)
Y(f)
lim
i.e.
0.25
K (s 1)
s of
s3
Hence, Y(s)
(b)
K (s 1)
s3
K (1 1 s)
s of 1 3 s
lim
K
K.
s1
4 (s 3)
Consider the circuit shown below.
t=0
Vs = 8 V
+
I
YS
Vs
8 u(t) 
o Vs
I
Vs
Z
I
A
B
s s3
A
i( t )
Y(s) Vs (s)
2 3,
B
8s
8 s 1
˜
4s s 3
2 (s 1)
s (s 3)
-4 3
1
> 2 4 e -3t @ u(t ) A
3
Chapter 16, Solution 69.
The gyrator is equivalent to two cascaded inverting amplifiers. Let V1 be the
voltage at the output of the first op amp.
V1
-R
V
R i
Vo
- 1 sC
V1
R
Io
Vo
R
-Vi
1
V
sCR i
Vo
sR 2 C
Vo
Io
sR 2 C
Vo
Io
sL, when L
R 2C
Chapter 17, Solution 1.
(a)
This is periodic with Z = S which leads to T = 2S/Z = 2.
(b)
y(t) is not periodic although sin t and 4 cos 2St are independently
periodic.
(c)
Since sin A cos B = 0.5[sin(A + B) + sin(A –– B)],
g(t) = sin 3t cos 4t = 0.5[sin 7t + sin(––t)] = ––0.5 sin t + 0.5 sin7t
which is harmonic or periodic with the fundamental frequency
Z = 1 or T = 2S/Z = 2S.
(d)
h(t) = cos 2 t = 0.5(1 + cos 2t). Since the sum of a periodic function and
a constant is also periodic, h(t) is periodic. Z = 2 or T = 2S/Z = S.
(e)
The frequency ratio 0.6|0.4 = 1.5 makes z(t) periodic.
Z = 0.2S or T = 2S/Z = 10.
(f)
p(t) = 10 is not periodic.
(g)
g(t) is not periodic.
Chapter 17, Solution 2.
(a)
The frequency ratio is 6/5 = 1.2. The highest common factor is 1.
Z = 1 = 2S/T or T = 2S.
(b)
Z = 2 or T = 2S/Z = S.
(c)
f3(t) = 4 sin2 600S t = (4/2)(1 –– cos 1200S t)
Z = 1200S or T = 2S/Z = 2S/(1200S) = 1/600.
(d)
f4(t) = ej10t = cos 10t + jsin 10t. Z = 10 or T = 2S/Z = 0.2S.
Chapter 17, Solution 3.
T = 4, Zo = 2S/T = S/2
g(t) = 5,
10,
0,
0<t<1
1<t<2
2<t<4
T
1
0
0
2
ao = (1/T) ³ g( t )dt = 0.25[ ³ 5dt + ³ 10dt ] = 3.75
an = (2/T)
1
T
³ g( t ) cos(nZo t )dt = (2/4)[
0
1
³ 5 cos(
0
2
nS
nS
t )dt + ³ 10 cos( t )dt ]
1
2
2
2
1
nS
2
2
nS
t + 10
sin
t ] = (––1/(nS))5 sin(nS/2)
= 0.5[ 5
sin
2 0
nS
nS
2 1
an =
bn = (2/T)
(5/(nS))(––1)(n+1)/2,
0,
T
³ g( t ) sin(nZ t )dt =
o
0
(2/4)[
0
1
= 0.5[
nS
1
³ 5 sin( 2
n = odd
n = even
2
t )dt + ³ 10 sin(
1
nS
t )dt ]
2
2
nS
nS
2x5
2 x10
t ––
t ] = (5/(nS))[3 –– 2 cos nS + cos(nS/2)]
cos
cos
2 0
2 1
nS
nS
Chapter 17, Solution 4.
f(t) = 10 –– 5t, 0 < t < 2, T = 2, Zo = 2S/T = S
ao = (1/T)
an = (2/T)
=
T
2
2
0
0
0
2
³ f ( t )dt = (1/2) ³ (10 5t )dt = 0.5[10t (5t / 2)] = 5
T
³ f ( t ) cos(nZo t )dt = (2/2)
0
2
³ (10) cos(nSt )dt ––
0
2
2
³ (10 5t ) cos(nSt )dt
0
2
³ (5t ) cos(nSt )dt
0
2
5
5t
sin nSt = [––5/(n2S2)](cos 2nS –– 1) = 0
= 2 2 cos nSt +
n S
nS
0
0
bn = (2/2)
2
³ (10 5t ) sin(nSt )dt
0
2
2
0
0
³ (10) sin(nSt )dt –– ³ (5t ) sin(nSt )dt
=
2
2
5
5t
cos nSt = 0 + [10/(nS)](cos 2nS) = 10/(nS)
= 2 2 sin nSt +
n S
nS
0
0
f(t) = 5 Hence
10 f 1
¦ sin(nSt )
S n 1n
Chapter 17, Solution 5.
T
2S,
Z
2S / T 1
T
ao
1
z( t )dt
T³
1
[1xS 2 xS]
2S
0
T
an
2
z( t ) cos nZo dt
T³
0
T
bn
2
z( t ) cos nZo dt
T³
0
S
1
1
1 cos ntdt ³
S
S
0.5 f
n odd
³ 2 cos ntdt
S
S
2S
0
S
1
1
1sin ntdt ³
S
S
6
sin nt
n 1 nS
¦
2S
0
Thus,
z( t )
0.5
³ 2 sin ntdt
1
2
S
2S
sin ..nt 0 sin nt S
nS
nS
1
2
2S
S
cos nt 0 cos nt S
nS
nS
0
­ 6
° , n odd
® nS
°̄0, n even
Chapter 17, Solution 6.
2S
2
2, Zo
T
ao
1 2
y( t )dt
2 ³0
S
1
(4 x1 2 x1)
2
Since this is an odd function, a n
bn
2 2
y( t ) sin( nZo t )dt
2 ³0
6
2
3
0.
1
2
³0 4 sin(nSt )dt ³1 2 sin(nSt )dt
4
2
(cos(2nS) cos(nS))
(cos(nS) 1) nS
nS
4
2
1
2
cos(nSt ) 1
cos(nSt ) 0 nS
nS
2
4
(1 cos(nS))
(1 cos(nS)) nS
nS
y( t )
2
(1 cos(nS))
nS
0,
n even
4
, n odd
nS
4 f 1
3
¦ sin(nSt )
Sn 1 n
n odd
Chapter 17, Solution 7.
Z
T 12,
2S / T
T
an
2
f ( t ) cos nZo dt
T³
0
S
,
6
a0
0
4
10
2
4
1
[ 10 cos nSt / 6dt ³ (10) cos nSt / 6dt ]
6 ³
10
10
4
10
sin nSt / 6 2 sin nSt / 6 4
nS
nS
T
bn
2
f ( t ) sin nZo dt
T³
0
10
>2 sin 2nS / 3 sin nS / 3 sin 5nS / 3@
nS
4
10
2
4
1
[ 10 sin nSt / 6dt ³ (10) sin nSt / 6dt ]
6 ³
10
10
4
10
cos nSt / 6 2 cos nSnt / 6 4
nS
nS
10
>cos 5nS / 3 cos nS / 3 2 sin 2nS / 3@
nS
f
¦ a n cos nSt / 6 b n sin nSt / 6
f (t)
n 1
where an and bn are defined above.
Chapter 17, Solution 8.
2(1 t ), - 1 t 1,
f (t)
T
ao
1
f ( t )dt
T³
0
T
1
1
2( t 1)dt
2³
T
2
f ( t ) sin nZo dt
T³
0
2
f (t)
2S / T
S
1
2
t t
2
1
1
2
f ( t ) cos nZo dt
T³
2
2( t 1) cos nStdt
2³
1
0
bn
Zo
1
T
an
2,
1
2
2( t 1) sin nStdt
2³
1
1
§ 1
·
t
1
2¨¨
cos nSt sin nSt sin nSt ¸¸
nS
nS
© n 2S2
¹ 1
1
§
·
t
1
1
sin nSt cos nSt cos nSt ¸¸
2¨¨ 2 2
nS
nS
© n S
¹ 1
4 f (1) n
cos nSt
¦
Sn 1 n
Chapter 17, Solution 9.
f(t) is an even function, bn=0.
T
Z
8,
T
ao
2S / T
1
f (t )dt
T ³0
S /4
2
º
2ª
« ³ 10 cos St / 4dt 0»
8 ¬0
¼
10 4
( ) sin St / 4
4 S
2
10
0
S
3.183
0
4
cos nS
nS
an
4
T
T /2
³
2
40
[ 10 cos St / 4 cos nSt / 4dt 0]
8 ³0
f (t ) cos nZ o dt
0
2
5³ >cos St (n 1) / 4 cos St (n 1) / 4@dt
0
For n = 1,
2
2
a1
ª2
º
5« sin St / 2dt t »
¬S
¼0
5³ [cos St / 2 1]dt
0
10
For n>1,
2
an
a2
20
20
S (n 1)t
S (n 1)
sin
sin
S (n 1)
S (n 1)
4
4
0
10
S
sin S 20
S
sin S / 2
20
20
S (n 1)
S (n 1)
sin
sin
S (n 1)
S (n 1)
2
2
20
10
sin 2S sin S
4S
S
6.3662,
a3
10,
6.362,
0
Thus,
a0
3.183,
a1
a2
a3
0,
b1
0
b2
b3
Chapter 17, Solution 10.
T
Zo
2,
2S / T
T
cn
1
h ( t )e jnZo t dt
³
T
0
cn
S
2
1 ª 1 jnSt
4
e
dt
(2)e jnSt dt º»
³
³
«
0
1
2¬
¼
>
j
4e jSn 4 2e j2nS 2e jnS
2nS
Thus,
f
f (t )
§ j6 · jnSt
¨
¸e
n f © nS ¹
¦
n odd
@
1 ª 4e jnSt 1 2e jnSt 2 º
»
«
2 «¬ jnS 0
jnS 1 »¼
j
[6 cos nS 6]
2nS
­ 6j
°
, n odd
,
® nS
°̄ 0, n even
Chapter 17, Solution 11.
T
Zo
4,
2S / T
T
cn
1
y( t )e jnZo t dt
T³
0
cn
S/ 2
1
1ª 0
( t 1)e jnSt / 2 dt ³ (1)e jnSt / 2 dt º»
³
«
0
4 ¬ 1
¼
1 ª e jnSt / 2
2 jnSt / 2 0
2 jnSt / 2 1 º
e
e
« 2 2 ( jnSt / 2 1) 1 jnS
0»
4 «¬ n S / 4
jnS
»¼
1ª 4
2
4
2 jnS / 2
2 jnS / 2
2 º
e jnS / 2 ( jnS / 2 1) e
e
«
4 ¬ n 2 S 2 jnS n 2 S 2
jnS
jnS
jnS »¼
But
e jnS / 2
cn
cos nS / 2 j sin nS / 2
1
n 2S2
j sin nS / 2,
e jnS / 2
cos nS / 2 j sin nS / 2
j sin nS / 2
>1 j( jnS / 2 1) sin nS / 2 nS sin nS / 2@
f
y( t )
¦
n f
1
2 2
n S
>1 j( jnS / 2 1) sin nS / 2 nS sin nS / 2@e jnSt / 2
Chapter 17, Solution 12.
A voltage source has a periodic waveform defined over its period as
v(t) = t(2S - t) V,
for all 0 < t < 2S
Find the Fourier series for this voltage.
v(t) = 2S t –– t2, 0 < t < 2S, T = 2S, Zo = 2S/T = 1
ao =
T
(1/T) ³ f ( t )dt
0
1 2S
(2St t 2 )dt
³
0
2S
1
(St 2 t 3 / 3)
2S
2S
0
4S 3
(1 2 / 3)
2S
2S 2
3
2 T
an = ³ (2St t 2 ) cos(nt )dt
T 0
2S
1 ª 2S
2St
º
cos(nt ) sin(nt )»
2
«
n
S ¬n
¼0
>
1
2nt cos(nt ) 2 sin(nt ) n 2 t 2 sin( nt )
3
Sn
2
1
(1 1) 3 4nS cos(2Sn )
2
Sn
n
bn =
2 T
(2nt t 2 ) sin( nt )dt
T ³0
@
2S
0
4
n2
1
(2nt t 2 ) sin(nt )dt
S³
2S
2n 1
1
S
(sin(nt ) nt cos(nt )) 0 3 (2nt sin(nt ) 2 cos(nt ) n 2 t 2 cos(nt ))
2
0
S n
Sn
4 S 4S
n
n
Hence,
f(t) =
0
2S 2 f 4
¦ 2 cos(nt )
3
n 1 n
Chapter 17, Solution 13.
T = 2S, Zo = 1
T
ao = (1/T) ³ h( t )dt
0
2S
1 S
[ ³ 10 sin t dt + ³ 20 sin( t S) dt ]
S
2S 0
>
1
S
2S
10 cos t 0 20 cos( t S) S
2S
an = (2/T)
@
30
S
T
³ h( t ) cos(nZ t )dt
0
o
S
= [2/(2S)] ª ³ 10 sin t cos( nt )dt «¬ 0
³
2S
S
20 sin( t S) cos( nt )dt º
»¼
Since sin A cos B = 0.5[sin(A + B) + sin(A –– B)]
sin t cos nt = 0.5[sin((n + 1)t) + sin((1 –– n))t]
sin(t –– S) = sin t cos S –– cost sin S = ––sin t
sin(t –– S)cos(nt) = ––sin(t)cos(nt)
an =
2S
1 ª S
10³ [sin([1 n ]t ) sin([1 n ]t )]dt 20³ [sin([1 n ]t ) sin([1 n ]t )]dt º
»¼
S
2S «¬ 0
5
=
S
ª§ cos([1 n ]t ) cos([1 n ]t ) · S § 2 cos([1 n ]t ) 2 cos([1 n ]t ) · 2 S º
«¨ ¸ ¨
¸ »
1 n
1 n
1 n
1 n
¹0 ©
¹ S »¼
«¬©
But,
3
3 cos([1 n ]S) 3 cos([1 n ]S) º
ª 3
«¬1 n 1 n »¼
1 n
1 n
5
S
an =
[1/(1+n)] + [1/(1-n)] = 1/(1––n2)
cos([n––1]S) = cos([n+1]S) = cos S cos nS –– sin S sin nS = ––cos nS
an = (5/S)[(6/(1––n2)) + (6 cos(nS)/(1––n2))]
= [30/(S(1––n2))](1 + cos nS) = [––60/(S(n––1))], n = even
= 0,
n = odd
T
bn = (2/T) ³ h ( t ) sin nZo t dt
0
S
2S
0
S
= [2/(2S)][ ³ 10 sin t sin nt dt + ³ 20( sin t ) sin nt dt
But,
sin A sin B = 0.5[cos(A––B) –– cos(A+B)]
sin t sin nt = 0.5[cos([1––n]t) –– cos([1+n]t)]
S
bn = (5/S){[(sin([1––n]t)/(1––n)) –– (sin([1+n]t)/ (1 n )] 0
2S
+ [(2sin([1-n]t)/(1-n)) –– (2sin([1+n]t)/ (1 n )] S }
=
Thus,
5
S
ª sin([1 n ]S) sin([1 n ]S) º
«¬ »¼ = 0
1 n
1 n
h(t) =
30 60 f cos( 2kt )
¦
S
S k 1 ( 4k 2 1)
Chapter 17, Solution 14.
Since cos(A + B) = cos A cos B –– sin A sin B.
f
10
·
§ 10
cos(nS / 4) cos( 2nt ) 3
sin(nS / 4) sin( 2nt ) ¸
f(t) = 2 ¦ ¨ 3
n 1
¹
n 1©n 1
Chapter 17, Solution 15.
(a)
Dcos Zt + Esin Zt = A cos(Zt - T)
where
f(t) = 10 A =
D 2 E 2 , T = tan-1(E/D)
A =
16
1
6 , T = tan-1((n2+1)/(4n3))
2
( n 1)
n
f
¦
n 1
(b)
2
2
§
16
1
1 n 1 ·
¸
cos¨¨ 10nt tan
( n 2 1) 2 n 6
4n 3 ¸¹
©
Dcos Zt + Esin Zt = A sin(Zt + T)
where
D 2 E 2 , T = tan-1(D/E)
A =
f(t) = 10 f
¦
n 1
§
16
1
4n 3 ·
1
¸
sin¨¨ 10nt tan
( n 2 1) 2 n 6
n 2 1 ¸¹
©
Chapter 17, Solution 16.
If v2(t) is shifted by 1 along the vertical axis, we obtain v2*(t) shown below, i.e.
v2*(t) = v2(t) + 1.
v2*(t)
2
1
-2 -1
0
1
2
3
4
5
t
Comparing v2*(t) with v1(t) shows that
v2*(t) = 2v1((t + to)/2)
where (t + to)/2 = 0 at t = -1 or to = 1
Hence
v2*(t) = 2v1((t + 1)/2)
But
v2*(t) = v2(t) + 1
v2(t) + 1 = 2v1((t+1)/2)
v2(t) = -1 + 2v1((t+1)/2)
= -1 + 1 v2(t) = 8
S2
8
S2
ª
º
§ t 1·
§ t 1· 1
§ t 1· 1
«cos S¨ 2 ¸ 9 cos 3S¨ 2 ¸ 25 cos 5S¨ 2 ¸ "»
¹
©
¹
©
¹
©
¬
¼
ª § St S · 1
º
§ 5St 5S ·
§ 3St 3S · 1
«cos¨ 2 2 ¸ 9 cos¨ 2 2 ¸ 25 cos¨ 2 2 ¸ "»
¹
©
¹
©
¹
¬ ©
¼
v2(t) = 8
S2
ª § S t · 1 § 3 St ·
º
1
§ 5 St ·
«sin¨ 2 ¸ 9 sin¨ 2 ¸ 25 sin¨ 2 ¸ "»
©
¹
©
¹
¬ © ¹
¼
Chapter 17, Solution 17.
We replace t by ––t in each case and see if the function remains unchanged.
(a)
1 –– t,
neither odd nor even.
(b)
t2 –– 1,
even
(c)
cos nS(-t) sin nS(-t) = - cos nSt sin nSt,
odd
(d)
sin2 n(-t) = (-sin St)2 = sin2 St,
even
(e)
e t,
neither odd nor even.
Chapter 17, Solution 18.
(a)
T = 2 leads to Zo = 2S/T = S
f1(-t) = -f1(t), showing that f1(t) is odd and half-wave symmetric.
(b)
T = 3 leads to Zo = 2S/3
f2(t) = f2(-t), showing that f2(t) is even.
(c)
T = 4 leads to Zo = S/2
f3(t) is even and half-wave symmetric.
Chapter 17, Solution 19.
This is a half-wave even symmetric function.
ao = 0 = bn, Zo = 2S/T S/2
an =
4
T
³
T/2
0
4t º
ª
«¬1 T »¼ cos(nZo t )dt
= [4/(nS)2](1 cos nS)
f (t) =
8
S2
f
¦
n odd
= 8/(n2S2),
=
0,
1
§ nSt ·
cos¨
¸
2
n
© 2 ¹
Chapter 17, Solution 20.
This is an even function.
bn = 0, T = 6, Z = 2S/6 = S/3
ao =
2
T
³
T/2
0
f ( t )dt
3
2ª 2
(
4
t
4
)
dt
4 dt º
³
³
«
»¼
1
2
6¬
n = odd
n = even
2
1ª 2
( 2 t 4 t ) 4(3 2)º = 2
1
¼»
3 «¬
=
4
T
an =
³
T/4
0
f ( t ) cos( nSt / 3)dt
2
= (4/6)[ ³ ( 4 t 4) cos( nSt / 3)dt +
1
³
3
2
4 cos( nSt / 3)dt ]
2
3
16 ª 3
3
16 ª 9
§ nSt ·º
§ nSt ·º
§ nSt ·
§ nSt · 3t
=
sin¨
sin¨
sin¨
cos¨
¸
¸» ¸
¸
2 2
«
«
6 ¬ nS © 3 ¹»¼ 2
6 ¬n S
© 3 ¹ nS © 3 ¹ nS © 3 ¹¼1
= [24/(n2S2)][cos(2nS/3) cos(nS/3)]
f(t) = 2 Thus
24 f 1
¦
S2 n 1 n2
ª § 2Sn ·
§ nSt ·
§ Sn · º
«cos¨ 3 ¸ cos¨ 3 ¸ » cos¨ 3 ¸
©
¹
¹
©
¹¼
¬ ©
At t = 2,
f(2) = 2 + (24/S2)[(cos(2S/3) cos(S/3))cos(2S/3)
+ (1/4)(cos(4S/3) cos(2S/3))cos(4S/3)
+ (1/9)(cos(2S) cos(S))cos(2S) + -----]
= 2 + 2.432(0.5 + 0 + 0.2222 + -----)
f(2) = 3.756
Chapter 17, Solution 21.
This is an even function.
bn = 0, T = 4, Zo = 2S/T = S/2.
f(t) = 2 2t,
= 0,
0<t<1
1<t<2
2 1
ao =
2(1 t )dt
4 ³0
ª
t2 º
« t 2 » = 0.5
¼0
¬
an =
4
T
³
T/2
0
f ( t ) cos( nZo t )dt
1
4 1
§ nSt ·
2(1 t ) cos¨
¸dt
³
4 0
© 2 ¹
= [8/(S2n2)][1 cos(nS/2)]
f
1
2
f(t) =
8
¦n S
n 1
2
2
ª
§ nSt ·
§ nS · º
«1 cos¨ 2 ¸ » cos¨ 2 ¸
©
¹
© ¹¼
¬
Chapter 17, Solution 22.
Calculate the Fourier coefficients for the function in Fig. 16.54.
f(t)
4
-5 -4 -3 -2 -1
0
1
Figure 17.61
2
3
4
5
t
For Prob. 17.22
This is an even function, therefore bn = 0. In addition, T=4 and Zo = S/2.
ao =
an =
2
T
³
4
T
T2
0
³
f ( t )dt
T2
0
2 1
4 tdt
4 ³0
f ( t ) cos(Zo nt )dt
t2
1
0
1
4 1
4 t cos( nSt / 2)dt
4 ³0
1
2t
ª 4
º
sin( nSt / 2)»
4 « 2 2 cos( nSt / 2) nS
¬n S
¼0
an =
16
8
sin( nS / 2)
(cos( nS / 2) 1) 2 2
nS
n S
Chapter 17, Solution 23.
f(t) is an odd function.
f(t) = t, 1< t < 1
ao = 0 = an, T = 2, Zo = 2S/T = S
bn =
=
4
T
³
T/2
0
4 1
t sin( nSt )dt
2 ³0
f ( t ) sin( nZo t )dt
2
>sin(nSt ) nSt cos(nSt )@ 10
2
n S
2
= [2/(nS)]cos(nS) = 2(1)n+1/(nS)
f(t) =
2
S
f
( 1) n 1
sin( nSt )
n
1
¦
n
Chapter 17, Solution 24.
(a)
This is an odd function.
ao = 0 = an, T = 2S, Zo = 2S/T = 1
bn =
4
T
³
T/2
0
f ( t ) sin(Zo nt )dt
f(t) = 1 + t/S,
bn =
0<t<S
4 S
(1 t / S) sin( nt )dt
2S ³0
S
=
2ª 1
1
t
º
cos( nt ) 2 sin( nt ) cos( nt )»
«
S¬ n
n S
nS
¼0
= [2/(nS)][1 2cos(nS)] = [2/(nS)][1 + 2(1)n+1]
a2 = 0, b2 = [2/(2S)][1 + 2(1)] = 1/S = 0.3183
(b)
Zn = nZo = 10 or n = 10
a10 = 0, b10 = [2/(10S)][1 cos(10S)] = 1/(5S)
Thus the magnitude is A10 =
and the phase is
2
a 210 b10
= 1/(5S) = 0.06366
I10 = tan1(bn/an) = 90q
f
(c)
f(t) =
2
¦ nS [1 2 cos(nS)] sin(nt ) S
n 1
f(S/2) =
f
2
¦ nS [1 2 cos(nS)] sin(nS / 2) S
n 1
For n = 1,
f1 = (2/S)(1 + 2) = 6/S
For n = 2,
f2 = 0
For n = 3,
f3 = [2/(3S)][1 2cos(3S)]sin(3S/2) = 6/(3S)
For n = 4,
f4 = 0
For n = 5,
f5 = 6/(5S), ----
Thus, f(S/2) = 6/S 6/(3S) + 6/(5S) 6/(7S) --------= (6/S)[1 1/3 + 1/5 1/7 + --------]
f(S/2) # 1.3824
which is within 8% of the exact value of 1.5.
(d)
From part (c)
f(S/2) = 1.5 = (6/S)[1 1/3 + 1/5 1/7 + - - -]
(3/2)(S/6) = [1 1/3 + 1/5 1/7 + - - -]
or S/4 = 1 1/3 + 1/5 1/7 + - - -
Chapter 17, Solution 25.
This is an odd function since f(t) = f(t).
ao = 0 = an, T = 3, Zo = 2S/3.
bn =
4
T
³
T/2
0
f ( t ) sin( nZo t )dt
4 1
t sin(2Snt / 3)dt
3 ³0
1
3t
4ª 9
§ 2Snt ·
§ 2Snt ·º
=
cos¨
sin¨
¸
¸»
2 2
«
3 ¬ 4S n
© 3 ¹ 2nS
© 3 ¹¼ 0
=
f(t) =
3
4ª 9
§ 2Sn ·º
§ 2Sn ·
cos¨
sin¨
¸»
¸
2 2
«
3 ¬ 4S n
© 3 ¹¼
© 3 ¹ 2nS
f
ª 3
¦ «S n
n 1
2
¬
2
2
§ 2 Sn ·
§ 2S n · º § 2S t ·
cos¨
sin¨
¸
¸
¸ » sin¨
© 3 ¹ nS
© 3 ¹¼ © 3 ¹
Chapter 17, Solution 26.
T = 4, Zo = 2S/T = S/2
1ª 1
1 dt 4 «¬ ³0
ao =
1 T
f ( t )dt
T ³0
an =
2 T
f ( t ) cos( nZo t )dt
T ³0
an =
2ª 2
1 cos( nSt / 2)dt 4 «¬ ³1
³
3
1
³
3
2
2 dt ³ 1 dt º = 1
»¼
3
4
2 cos( nSt / 2)dt ³ 1 cos( nSt / 2)dt º
»¼
3
4
2
3
4
ª2
nSt º
nSt
2
nSt
4
sin
= 2 « sin
sin
»
2 3 ¼»
2 2 nS
2 1 nS
¬« nS
=
4
nS
nS º
ª 3nS
«¬sin 2 sin 2 »¼
bn =
2 T
f ( t ) sin( nZo t )dt
T ³0
=
nSt
2ª 2
dt 1 sin
³
«
1
2
4¬
³
3
2
2 sin
nSt
dt 2
³
4
3
1 sin
nSt º
dt »
2
¼
2
3
4
ª 2
nSt º
nSt
2
nSt
4
cos
cos
cos
= 2«
»
2 3 ¼»
2 2 nS
2 1 nS
¬« nS
=
4
>cos(nS) 1@
nS
Hence
f(t) =
1
f
4
¦ nS >(sin( 3nS / 2) sin(nS / 2)) cos( nSt / 2) (cos( nS) 1) sin(nSt / 2)@
n 1
Chapter 17, Solution 27.
(a)
(b)
odd symmetry.
ao = 0 = an, T = 4, Zo = 2S/T = S/2
f(t)
= t, 0 < t < 1
= 0,
nSt
4 1
bn =
dt
t sin
³
2
4 0
=
1<t<2
1
nSt º
nSt 2 t
ª 4
cos
sin
2
2
«¬ n S
2 »¼ 0
2
nS
nS
nS
2
4
0
cos
sin
2
2
2
nS
n S
2
= 4(1)(n1)/2/(n2S2),
n = odd
2(1)n/2/(nS),
n = even
a3 = 0, b3 = 4(1)/(9S2) = 0.045
(c)
b1 = 4/S2, b2 = 1/S, b3 = 4/(9S2), b4 = 1/(2S), b5 = (25S2)
Frms =
a 2o 1
¦ a 2n b 2n
2
Frms2 = 0.56bn2 = [1/(2S2)][(16/S2) + 1 + (16/(8S2)) + (1/4) + (16/(625S2))]
= (1/19.729)(2.6211 + 0.27 + 0.00259)
Frms =
0.14659 = 0.3829
Compare this with the exact value of Frms =
2
T
1
³ t dt
0
2
1 / 6 = 0.4082
Chapter 17, Solution 28.
This is half-wave symmetric since f(t T/2) = f(t).
ao = 0, T = 2, Zo = 2S/2 = S
an =
4
T
³
T/2
0
4 1
( 2 2 t ) cos( nSt )dt
2 ³0
f ( t ) cos( nZo t )dt
1
t
1
ª1
º
= 4 « sin( nSt ) 2 2 cos( nSt ) sin( nSt )»
nS
n S
¬ nS
¼0
= [4/(n2S2)][1 cos(nS)] =
8/(n2S2),
0,
n = odd
n = even
1
bn = 4 ³ (1 t ) sin( nSt )dt
0
1
t
1
ª 1
º
= 4 «
cos( nSt )»
cos( nSt ) 2 2 sin( nSt ) nS
n S
¬ nS
¼0
= 4/(nS), n = odd
f(t) =
f
§
¦ ¨© n
k 1
8
4
·
cos( nSt ) sin(nSt ) ¸ , n = 2k 1
2
nS
S
¹
2
Chapter 17, Solution 29.
This function is half-wave symmetric.
T = 2S, Zo = 2S/T = 1, f(t) = t, 0 < t < S
For odd n,
an =
2
T
bn =
2
S
³
S
³
S
0
0
( t ) cos( nt )dt = ( t ) sin( nt )dt
2
>cos(nt ) nt sin(nt )@ 0S = 4/(n2S)
2
n S
2
>sin(nt ) nt cos(nt )@ 0S = 2/n
2
n S
Thus,
f
1
ª 2
º
f(t) = 2¦ « 2 cos( nt ) sin(nt )» ,
n
¼
k 1 ¬n S
n = 2k 1
Chapter 17, Solution 30.
cn
1
T
T/2
³
T/2
1 ª T/2
f ( t ) cos nZ o tdt j³
f ( t ) sin nZ o tdt º»
³
«
T / 2
T ¬ T / 2
¼
f ( t )e jnZo t dt
T / 2
(a)
(1)
The second term on the right hand side vanishes if f(t) is even. Hence
cn
(b)
2
T
T/2
³ f (t ) cos nZo tdt
0
The first term on the right hand side of (1) vanishes if f(t) is odd. Hence,
cn
j2
T
T/2
³ f (t ) sin nZo tdt
0
Chapter 17, Solution 31.
If h ( t )
f (Dt ),
T' T / D

o
T'
an '
2
h ( t ) cos nZo ' tdt
T' ³
0
Let Dt
O, ,
dt
dO / D ,
Zo '
2S
T'
T'
0
DT ' T
2D
f (O) cos nZo OdO / D
T ³
0
Similarly,
bn ' bn
DZo
2
f (Dt ) cos nZo ' tdt
T' ³
T
an '
2S
T/D
an
Chapter 17, Solution 32.
When is = 1 (DC component)
i = 1/(1 + 2) = 1/3
Zn = 3n, Is = 1/n2‘0q
For n t 1,
I = [1/(1 + 2 + jZn2)]Is = Is/(3 + j6n)
1
‘0q
2
n
=
3 1 4n 2 ‘ tan 1 (6n / 3)
1
3n 2 1 4n 2
‘ tan(2n )
Thus,
i(t) =
1
3
f
¦
n 1
1
3n
1 4n
2
2
cos( 3n tan 1 ( 2n ))
Chapter 17, Solution 33.
For the DC case, the inductor acts like a short, Vo = 0.
For the AC case, we obtain the following:
Vo Vs
V
jnSVo
o 10
j2nS
4
§
5 ··
§
¨¨1 j¨ 2.5nS ¸ ¸¸Vo
nS ¹ ¹
©
©
Vo
Vs
Vs
5 ·
§
1 j¨ 2.5nS ¸
nS ¹
©
A n ‘4 n
An
0
4
nS
1
4
5 ·
§
1 j¨ 2.5nS ¸
nS ¹
©
nS j(2.5n 2 S 2 5)
4
n 2 S 2 (2.5n 2 S 2 5) 2
; 4n
§ 2.5n 2 S 2 5 ·
¸
tan 1 ¨
¨
¸
n
S
©
¹
f
¦ A n sin(nSt 4 n ) V
v o (t)
n 1
Chapter 17, Solution 34.
For any n, V = [10/n2]‘(nS/4), Z = n.
1 H becomes jZnL = jn and 0.5 F becomes 1/(jZnC) = j2/n
2:
jn
+
+
V
j2/n
Vo
Vo = {j(2/n)/[2 + jn j(2/n)]}V = {j2/[2n + j(n2 2)]}[(10/n2)‘(nS/4)]
20‘((nS / 4) S / 2)
n
2
4n (n 2 2) 2 ‘ tan 1 ((n 2 2) / 2n )
2
20
n
2
n 4
2
vo(t) =
‘[(nS / 4) (S / 2) tan 1 ((n 2 2) / 2n )]
f
¦
n 1
§
nS S
n2 cos¨¨ nt tan 1
4
2
2n
n2 4
©
20
n2
2·
¸¸
¹
Chapter 17, Solution 35.
If vs in the circuit of Fig. 17.72 is the same as function f2(t) in Fig. 17.57(b),
determine the dc component and the first three nonzero harmonics of vo(t).
1:
1H
+
vS
+
1F
1:
vo
Figure 17.72
For Prob. 17.35
f2(t)
2
1
-2 -1
0
1
2
Figure 17.57(b)
3
4
t
5
For Prob. 17.35
The signal is even, hence, bn = 0. In addition, T = 3, Zo = 2S/3.
vs(t)
ao =
an =
=
= 1 for all 0 < t < 1
= 2 for all 1 < t < 1.5
2ª 1
1dt 3 «¬ ³0
2dt º
»¼
1.5
³
1
4ª 1
cos(2nSt / 3)dt 3 «¬ ³0
1.5
³
1
4
3
2 cos(2nSt / 3)dt º
»¼
4ª 3
6
2
1
1.5 º
sin(
2
n
t
/
3
)
S
S
sin(
2
n
t
/
3
)
sin(2nS / 3)
0
1
»¼
3 «¬ 2nS
2nS
nS
4 2 f 1
vs(t) =
¦ sin(2nS / 3) cos(2nSt / 3)
3 Sn 1n
Now consider this circuit,
1:
j2nS/3
+
vS
+
-j3/(2nS)
1:
vo
Let Z = [-j3/(2nS)](1)/(1 –– j3/(2nS)) = -j3/(2nS - j3)
Therefore, vo = Zvs/(Z + 1 + j2nS/3). Simplifying, we get
vo =
j9 v s
12nS j( 4n 2 S 2 18)
For the dc case, n = 0 and vs = ¾ V and vo = vs/2 = 3/8 V.
We can now solve for vo(t)
ª3 f
§ 2nSt
·º
vo(t) = « ¦ A n cos¨
4 n ¸ » volts
© 3
¹¼
¬8 n 1
where A n
6
sin( 2nS / 3)
nS
·
§ 4n 2 S 2
6 ¸¸
16n S ¨¨
¹
© 3
2
2
and 4 n
2
3 ·
§ nS
90 o tan 1 ¨
¸
© 3 2nS ¹
where we can further simplify An to this, A n
9 sin( 2nS / 3)
nS 4n 4 S 4 81
Chapter 17, Solution 36.
f
vs(t) =
¦A
n
cos( nt T n )
n 1
n odd
where Tn = tan1[(3/(nS))/(1/(nS))] = tan1(3) = 100.5q
An =
Sn
9
1
2 2 sin 2
2
2
n S
n S
2
nS
1
9 sin 2
2
nS
Zn = n and 2 H becomes jZnL = j2n
Let
Z = 1||j2n = j2n/(1 + j2n)
If Vo is the voltage at the non-reference node or across the 2-H inductor.
Vo = ZVs/(1 + Z) = [j2n/(1 + j2n)]Vs/{1 + [j2n/(1 + j2n)]}
= j2nVs/(1 + j4n)
But
Vs = An‘Tn
Vo = j2n An‘Tn/(1 + j4n)
Io = Vo/j = [2n An‘Tn]/ 1 16n 2 ‘tan14n
1 §
nS ·
¨ 9 sin 2
¸2n
nS ¨©
2 ¸¹
‘100.5q tan14n
=
2
1 16n
Since sin(nS/2) = (1)(n1)/2 for n = odd, sin2(nS/2) = 1
2 10
‘ 100.5q tan 1 4n
S
Io =
1 16n 2
io(t) =
2 10 f
¦
S n1
1
n odd
1 16n
2
cos(nt 100.5q tan 1 4n )
Chapter 17, Solution 37.
From Example 15.1,
vs(t) = 5 20 f 1
¦ sin(nSt ),
S k 1n
n = 2k 1
For the DC component, the capacitor acts like an open circuit.
Vo = 5
For the nth harmonic,
Vs = [20/(nS)]‘0q
10 mF becomes 1/(jZnC) = j/(nSx10x103) = j100/(nS)
100
Vs
nS
vo =
100
j
20
nS
j
vo(t) =
100
S
¦
j100
20
20nS j100 nS
1
n 25 n 2 S 2
100‘ 90q tan 1
nS 25 n 2 S 2
sin(nSt 90q tan 1
5
)
nS
5
nS
Chapter 17, Solution 38.
v s (t)
Vo
For dc, Z n
1 2 f 1
¦ sin nSt ,
2 S k 1n
jZ n
Vs ,
1 jZ n
0,
Vo
0.5,
2k 1
nS
Vo
0
2
‘ 90 o
nS
nS‘90 o
2
‘90 o
1 n 2 S 2 ‘ tan 1 nS nS
f
v o (t)
Zn
Vs
For nth harmonic, Vs
n
¦
k 1
2
2 2
1 n S
x
2‘ tan 1 nS
1 n 2S2
cos(nSt tan 1 nS),
n
2k 1
Chapter 17, Solution 39.
Comparing vs(t) with f(t) in Figure 15.1, vs is shifted by 2.5 and the magnitude is
5 times that of f(t).
Hence
10 f 1
vs(t) = 5 n = 2k 1
¦ sin(nSt ),
S k 1n
T = 2, Zo = 2S//T = S, Zn = nZo = nS
For the DC component,
For the kth harmonic,
io = 5/(20 + 40) = 1/12
Vs = (10/(nS))‘0q
100 mH becomes jZnL = jnSx0.1 = j0.1nS
50 mF becomes 1/(jZnC) = j20/(nS)
I 20 :
VS
40 :
Io
j20/(nS)
+
j0.1nS
Z
j20
( 40 j0.1nS)
n
S
Let Z = j20/(nS)||(40 + j0.1nS) =
j20
40 j0.1nS
nS
j20( 40 j0.1nS
j20 40nS j0.1n 2 S 2
=
802nS j( 2n 2 S 2 1200)
40nS j(0.1n 2 S 2 20)
Zin = 20 + Z =
I =
Vs
Z in
400nS j( n 2 S 2 200)
nS[802nS j( 2n 2 S 2 1200)]
j20
I
nS
Io =
=
=
2nS j800
40nS j(0.1n 2 S 2 20)
j20I
40nS j(0.1n 2 S 2 20)
j20
( 40 j0.1nS)
nS
j200
nS[802nS j( 2n 2 S 2 1200)]
200‘ 90q tan 1{(2n 2 S 2 1200) /(802nS)}
nS (802) 2 ( 2n 2 S 2 1200) 2
Thus
io(t) =
where
1
200
S
20
Tn
In =
f
¦I
n
sin(nSt T n ) ,
k 1
90q tan 1
2n 2 S 2 1200
802nS
1
n (804nS) (2n 2 S 2 1200)
2
n = 2k 1
Chapter 17, Solution 40.
T = 2, Zo = 2S/T = S
1
ao =
T
an =
2
T
1 1
( 2 2 t )dt
2 ³0
T
³ v( t )dt
0
T
³ v( t ) cos(nSt )dt
0
1
ª
t2 º
t
«
2 »¼ 0
¬
1/ 2
1
³ 2(1 t ) cos(nSt )dt
0
1
1
t
ª1
º
sin( nSt )»
= 2 « sin( nSt ) 2 2 cos( nSt ) n S
nS
¬ nS
¼0
0,
2
= 2 2 (1 cos nS)
n S
bn =
n
4
, n
n S2
2
2 T
v ( t ) sin( nSt )dt
T ³0
even
odd
4
S ( 2n 1) 2
2
1
2 ³ (1 t ) sin( nSt )dt
0
1
1
t
ª 1
º
cos( nSt ) 2 2 sin( nSt ) cos( nSt )»
= 2«
n S
nS
¬ nS
¼0
vs(t) =
1
2
where In
¦A
tan 1
n
2
nS
cos( nSt M n )
S( 2n 1) 2
, An
2n
4
16
4
2
n S
S ( 2n 1) 4
2
For the DC component, vs = 1/2. As shown in Figure (a), the capacitor acts
like an open circuit.
1:
0.5V
+
Vx
i
+
2Vx
+
Vx
(a)
Vo
+
3:
Vo
1:
Vx
+
2Vx
Vo
+
VS
+
Vo
3:
(1/4)F
(b)
Applying KVL to the circuit in Figure (a) gives
But
Adding (1) and (2),
––0.5 –– 2Vx + 4i = 0
(1)
––0.5 + i + Vx = 0 or ––1 + 2Vx + 2i = 0
(2)
––1.5 + 6i = 0 or i = 0.25
Vo = 3i = 0.75
For the nth harmonic, we consider the circuit in Figure (b).
Zn = nS, Vs = An‘––I, 1/(jZnC) = ––j4/(nS)
At the supernode,
(Vs –– Vx)/1 = ––[nS/(j4)]Vx + Vo/3
Vs = [1 + jnS/4]Vx + Vo/3
But
(3)
––Vx –– 2Vx + Vo = 0 or Vo = 3Vx
Substituting this into (3),
Vs = [1 + jnS/4]Vx + Vx = [2 + jnS/4]Vx
= (1/3)[2 + jnS/4]Vo = (1/12)[8 + jnS]Vo
Vo = 12Vs/(8 + jnS) =
Vo =
12
64 n S
2
2
12A n ‘ I
64 n 2 S 2 ‘ tan 1 (nS / 8)
4
16
4
‘[tan 1 (nS / 8) tan 1 (S(2n 1) /(2n ))]
2
4
n S
S (2n 1)
2
Thus
vo(t) =
where
Vn =
3
4
f
¦V
n
cos( nSt T n )
n 1
12
64 n 2 S 2
4
16
4
2
n S
S ( 2n 1) 4
2
Tn = tan––1(nS/8) –– tan––1(S(2n –– 1)/(2n))
Chapter 17, Solution 41.
For the full wave rectifier,
T = S, Zo = 2S/T = 2, Zn = nZo = 2n
Hence
vin(t) =
2 4 f
1
¦ 2
cos 2nt
S S n 1 4n 1
For the DC component,
Vin = 2/S
The inductor acts like a short-circuit, while the capacitor acts like an open circuit.
Vo = Vin = 2/S
For the nth harmonic,
Vin = [––4/(S(4n2 –– 1))]‘0q
2 H becomes jZnL = j4n
0.1 F becomes 1/(jZnC) = ––j5/n
Z = 10||(––j5/n) = ––j10/(2n –– j)
Vo = [Z/(Z + j4n)]Vin = ––j10Vin/(4 + j(8n –– 10))
= §
j10
4‘0q ·
¨¨ ¸
4 j(8n 10) © S(4n 2 1) ¸¹
40‘{90q tan 1 (2n 2.5)}
=
S(4n 2 1) 16 (8n 10) 2
Hence
vo(t) =
2
S
f
¦A
n
cos( 2nt T n )
n 1
where
An =
20
S( 4n 2 1) 16n 2 40n 29
Tn = 90q –– tan––1(2n –– 2.5)
Chapter 17, Solution 42.
vs
20 f 1
5
¦ sin nSt, n
S k 1n
Vs 0
R
jZ n C(0 Vo )
2k - 1
o
Vo
j
Vs , Z n
Z n RC
nZ o
For n = 0 (dc component), Vo=0.
For the nth harmonic,
Vo
1‘90 o 20
‘ 90 o
nSRC nS
20
10 5
n 2 S 2 x10 4 x 40 x10 9
2n 2 S 2
Hence,
v o (t)
10 5 f 1
¦
2S 2 k 1 n 2
cos nSt , n
2k - 1
Alternatively, we notice that this is an integrator so that
v o (t)
1
v s dt
RC ³
10 5 f 1
¦
2S 2 k 1n 2
cos nSt , n
2k - 1
nS
Chapter 17, Solution 43.
a 02 (a)
Vrms =
(b)
Irms =
(c)
P = VdcIdc +
1 f 2
(a n b 2n )
¦
2n1
1
30 2 (20 2 10 2 ) = 33.91 V
2
1
6 2 (4 2 2 2 ) = 6.782 A
2
1
¦ Vn I n cos(4 n ) n )
2
= 30x6 + 0.5[20x4cos(45o-10o) –– 10x2cos(-45o+60o)]
= 180 + 32.76 –– 9.659 = 203.1 W
Chapter 17, Solution 44.
>
1
60 cos 25 o 10 cos 45 o 0
2
(a)
p
(b)
The power spectrum is shown below.
vi
p
@
27.19 3.535 0
27.19
3.535
0
1
2
3
Z
Chapter 17, Solution 45.
Zn = 1000n
jZnL = j1000nx2x10––3 = j2n
1/(jZnC) = ––j/(1000nx40x10––6) = ––j25/n
30.73 W
Z = R + jZnL + 1/(jZnC) = 10 + j2n –– j25/n
I = V/Z
For n = 1, V1 = 100, Z = 10 + j2 –– j25 = 10 –– j23
I1 = 100/(10 –– j23) = 3.987‘73.89q
For n = 2, V2 = 50, Z = 10 + j4 –– j12.5 = 10 –– j8.5
I2 = 50/(10 –– j8.5) = 3.81‘40.36q
For n = 3, V3 = 25, Z = 10 + j6 –– j25/3 = 10 –– j2.333
I3 = 25/(10 –– j2.333) = 2.435‘13.13q
Irms = 0.5 3.987 2 3.812 2.435 2 = 3.014 A
1 3
p = VDCIDC + ¦ Vn I n cos(T n I n )
2n1
= 0 + 0.5[100x3.987cos(73.89q) + 50x3.81cos(40.36q)
+ 25x2.435cos(13.13q)]
= 0.5[110.632 + 145.16 + 59.28] = 157.54 watts
Chapter 17, Solution 46.
(a)
This is an even function
Irms =
f(t) =
2 T/2 2
f ( t )dt
T ³0
1 T 2
f ( t )dt
T ³0
2 2t,
0,
0 t 1
1 t 2
T = 4, Zo = 2S/T = S/2
Irms2 =
2 1
4(1 t ) 2 dt
³
0
4
2( t t 2 t 3 / 3)
1
0
= 2(1 –– 1 + 1/3) = 2/3 or
Irms
(b)
= 0.8165 A
From Problem 16.14,
an = [8/(n2S2)][1 –– cos(nS/2)], ao = 0.5
a1 = 8/S2, a2 = 4/S2, a3 = 8/(9S2), a4 = 0, a5 = 9/(25S2), a6 = 4/(9S2)
Irms =
ao 1 f 2
¦ An #
2n1
64 64 16 ·
1
1 §
4 ¨ 64 16 ¸
81 625 81 ¹
4 2S ©
0.66623
Irms = 0.8162 A
Chapter 17, Solution 47.
Let I = IDC + I1 + I2
For the DC component
IDC = [5/(5 + 10)](3) = 1 A
I
j8
5:
10 :
Is
For AC, Z = 100
For Is = 0.5‘––60q
jZL = j100x80x10––3 = j8
In = 5Is/(5 + 10 + j8)
I1 = 10‘––60q/(15 + j8) or |I1| = 10/ 15 2 8 2
For Is = 0.5‘––120q
I2 = 2.5‘––120q/(15 + j8) or |I2| = 2.5/ 15 2 8 2
p10 = (IDC2 + |I1|2/2 + |I2|2/2)10 = (1 + [100/(2x289)] + [6.25/(2x289)])x10
p10 = 11.838 watts
Chapter 17, Solution 48.
(a) For the DC component, i(t) = 20 mA. The capacitor acts like an open circuit so that
v = Ri(t) = 2x103x20x10––3 = 40
For the AC component,
Zn = 10n, n = 1,2
1/(jZnC) = ––j/(10nx100x10––6) = (––j/n) k:
Z = 2||(––j/n) = 2(––j/n)/(2 –– j/n) = ––j2/(2n –– j)
V = ZI = [––j2/(2n –– j)]I
For n = 1,
V1 = [––j2/(2 –– j)]16‘45q = 14.311‘––18.43q mV
For n = 2,
V2 = [––j2/(4 –– j)]12‘––60q = 5.821‘––135.96q mV
v(t) = 40 + 0.014311cos(10t –– 18.43q) + 0.005821cos(20t –– 135.96q) V
(b)
p = VDCIDC +
1 f
¦ Vn I n cos(T n I n )
2n1
= 20x40 + 0.5x10x0.014311cos(45q + 18.43q)
+0.5x12x0.005821cos(––60q + 135.96q)
= 800.1 mW
Chapter 17, Solution 49.
(a)
2
Z rms
T
1 2
z ( t )dt
T³
0
Z rms
2S º
S
1 ª
« 1dt ³ 4dt »
2S « ³
»¼
S
¬0
1
(5S)
2S
2.5
1.581
(b)
Z 2 rms
a 2o 1 f 2
(a n b 2 n )
¦
2n 1
Z rms
(c )
1 1 f 36
¦
4 2 n 1n 2 S 2
1
1
4 18S 2
1.7086
§ 1.7086 ·
%error ¨
1¸ x100
¹
© 1.581
8.071
1
·
§ 1 1 1
...¸
¨1 ¹
© 4 9 16 25
2.9193
Chapter 17, Solution 50.
1
T
cn =
=
³
T
0
f ( t )e jZo nt dt,
Zo
2n
1
S
1 1 jnSt
te
dt
2 ³
Using integration by parts,
u = t and du = dt
dv = e––jnStdt which leads to v = ––[1/(2jnS)]e––jnSt
t
e jnSt
cn = 2 jnS
1
1
>
1 1 jnSt
e
dt
2 jnS ³1
@
j jnS
1
e
e jnSt
=
e jnSt 2 2
2
nS
2n S ( j)
1
1
= [j/(nS)]cos(nS) + [1/(2n2S2)](e––jnS –– ejnS)
j( 1) n
2j
cn =
sin( nS)
nS
2n 2 S 2
j( 1) n
nS
Thus
f
f(t) =
¦ c n e jnZot =
n f
f
¦ ( 1)
n f
n
j jnSt
e
nS
Chapter 17, Solution 51.
T
Zo
2,
2S / T
T
cn
1
f ( t )e jnZo t dt
T³
0
cn
S
1
j2n 3 S 3
f (t)
2
1 2 jnSt
t e
dt
2³
0
(4n 2 S 2 j4nS)
f
¦
n f n
2
2 2
S
2
n 2S2
(1 jnS)e jnSt
1 e jnSt
2
n 2 S 2 t 2 2 jnSt 2 0
2 ( jnS) 3
(1 jnS)
Chapter 17, Solution 52.
cn =
=
1
T
³
T
0
f ( t )e jZo nt dt,
2n
1
Zo
S
1 1 jnSt
te
dt
2 ³
Using integration by parts,
u = t and du = dt
dv = e––jnStdt which leads to v = ––[1/(2jnS)]e––jnSt
t
cn = e jnSt
2 jnS
1
1 1 jnSt
e
dt
2 jnS ³1
1
>
@
j jnS
1
=
e jnSt e
e jnSt
2 2
2
nS
2n S ( j)
1
1
= [j/(nS)]cos(nS) + [1/(2n2S2)](e––jnS –– ejnS)
cn =
j( 1) n
2j
sin( nS)
nS
2n 2 S 2
j( 1) n
nS
Thus
f(t) =
f
¦c e
n
jnZo t
=
n f
f
¦ ( 1)
n f
n
j jnSt
e
nS
Chapter 17, Solution 53.
Zo = 2S/T = 2S
cn =
³
T
0
e t e jnZo t dt
1
³e
(1 jnZo ) t
0
1
e (1 j2 nS ) t
=
1 j2Sn
1
0
dt
>
= [1/(j2nS)][1 –– e––1(cos(2Sn) –– jsin(2nS))]
= (1 –– e––1)/(1 + j2nS) = 0.6321/(1 + j2nS
0.6321e j2 nSt
f(t) = ¦
n f 1 j2nS
f
@
1
e (1 j 2 n S ) 1
1 j2nS
Chapter 17, Solution 54.
T = 4, Zo = 2S/T = S/2
cn =
1
T
³
T
0
f ( t )e jZo nt dt
=
1 ª 1 jnSt / 2
2e
dt 4 «¬ ³0
=
j
2e jnS / 2 2 e jnS e jnS / 2 e j2 nS e jnS
2nS
=
j
3e jnS / 2 3 2e jnS
2nS
f(t) =
³
2
1
1e jnSt / 2 dt ³
4
2
1e jnSt / 2 dt º
»¼
>
>
f
¦c e
n
@
@
jnZo t
n f
Chapter 17, Solution 55.
T = 2S, Zo = 2S/T = 1
1
T
cn =
But
i(t) =
cn =
1
2S
³
S
0
³
T
0
i( t )e jnZo t dt
sin( t ),
0,
sin( t )e jnSt dt
0tS
S t 2S
1
2S
³
S
0
1 jt
(e e jt )e jnt dt
2j
S
1 ª e jt (1 n )
e jt (1 n ) º
4Sj «¬ j(1 n ) j(1 n ) »¼ 0
1 ª e jS (1 n ) 1 e jS( n 1) 1º
4 «¬ 1 n
1 n »¼
>
1
e jS (1 n ) 1 ne jS (1 n ) n e jS (1 n ) 1 ne jS (1 n ) n
2
4S( n 1)
@
But ejS = cos(S) + jsin(S) = ––1 = e––jS
>
1
cn =
e jnS e jnS ne jnS ne jnS 2
2
4 S( n 1)
Thus
i(t) =
f
1 e jnS
¦ 2S(1 n
n f
2
)
e jnSt
Chapter 17, Solution 56.
co = ao = 10, Zo = S
co = (an –– jbn)/2 = (1 –– jn)/[2(n2 + 1)]
f
f(t) = 10 (1 jn) jnSt
e
2
1)
¦ 2(n
n f
nz0
Chapter 17, Solution 57.
ao = (6/––2) = ––3 = co
cn = 0.5(an ––jbn) = an/2 = 3/(n3 –– 2)
f
f(t) = 3 ¦n
n f
nz0
3
3
e j50nt
2
Chapter 17, Solution 58.
cn = (an –– jbn)/2, (––1) = cos(nS), Zo = 2S/T = 1
cn = [(cos(nS) –– 1)/(2Sn2)] –– j cos(nS)/(2n)
Thus
f(t) =
S
cos(nS) · jnt
§ cos(nS) 1
j
¦¨
¸e
2
4
2n ¹
© 2Sn
@
1 e jnS
2 S(1 n 2 )
Chapter 17, Solution 59.
For f(t), T = 2S, Zo = 2S/T = 1.
ao = DC component = (1xS + 0)/2S = 0.5
For
h(t), T = 2, Zo = 2S/T = S.
ao = (3x1 –– 2x1)/2 = 0.5
Thus by replacing Zo = 1 with Zo = S and multiplying the magnitude by five,
we obtain
f
1
j5e j( 2n 1) St
h(t) =
¦
2 n f ( 2n 1)S
nz0
Chapter 17, Solution 60.
From Problem 16.17,
ao = 0 = an, bn = [2/(nS)][1 –– 2 cos(nS)], co = 0
cn = (an –– jbn)/2 = [j/(nS)][2 cos(nS) –– 1], n z 0.
Chapter 17, Solution 61.
Zo = 1.
(a)
f(t) = ao +
¦A
n
cos(nZ o t I n )
= 6 + 4cos(t + 50q) + 2cos(2t + 35q)
+ cos(3t + 25q) + 0.5cos(4t + 20q)
= 6 + 4cos(t)cos(50q) –– 4sin(t)sin(50q) + 2cos(2t)cos(35q)
–– 2sin(2t)sin(35q) + cos(3t)cos(25q) –– sin(3t)sin(25q)
+ 0.5cos(4t)cos(20q) –– 0.5sin(4t)sin(20q)
= 6 + 2.571cos(t) –– 3.73sin(t) + 1.635cos(2t)
–– 1.147sin(2t) + 0.906cos(3t) –– 0.423sin(3t)
+ 0.47cos(4t) –– 0.171sin(4t)
(b)
frms =
1 f 2
¦ An
2n1
a o2 frms2 = 62 + 0.5[42 + 22 + 12 + (0.5)2] = 46.625
frms = 6.828
Chapter 17, Solution 62.
(a) Zo
20
2S / T
(b) f ( t )
ao 
o
T
f
¦ A n cos(nZo t I n )
2S
20
0.3141s
3 4 cos(20t 90 o ) 5.1cos(40t 90 o ) ...
n 1
3 4 sin 20 t 5.1sin 40t 2.7 sin 60 t 1.8 sin 80t ....
f (t)
Chapter 17, Solution 63.
This is an even function.
T = 3, Zo = 2S/3, bn = 0.
f(t) =
ao =
an =
1, 0 t 1
2, 1 t 1.5
2 T/2
f ( t )dt
T ³0
4 T/2
f ( t ) cos(nZo t )dt
T ³0
1.5
2ª 1
1dt ³ 2 dt º = (2/3)[1 + 1] = 4/3
³
»¼
1
3 «¬ 0
1.5
4ª 1
1cos(2nSt / 3)dt ³ 2 cos(2nSt / 3)dt º
³
»¼
1
3 «¬ 0
4ª 3
6
§ 2nSt ·
§ 2nSt ·
= «
sin ¨
sin ¨
¸ ¸
3 «¬ 2nS © 3 ¹ 0 2nS © 3 ¹ 1
1
= [––2/(nS)]sin(2nS/3)
1.5
º
»
»¼
4 2 f 1 § 3nS · § 2nSt ·
f2(t) = ¦ sin ¨
¸ cos¨
¸
3 Sn 1n © 3 ¹ © 3 ¹
ao = 4/3 = 1.3333, Zo = 2S/3, an = ––[2/(nS)]sin(2nSt/3)
An =
2
§ 2nS ·
sin ¨
¸
nS © 3 ¹
a 2n b 2n
A1 = 0.5513, A2 = 0.2757, A3 = 0, A4 = 0.1375, A5 = 0.1103
The amplitude spectra are shown below.
1.333
An
0.551
0.275
0.1378 0.1103
0
0
1
2
3
4
5
n
Chapter 17, Solution 64.
The amplitude and phase spectra are shown below.
An
3.183
2.122
1.591
0.4244
0
2S
4S
2S
4S
6S
Z
In
0
6S
Z
-180o
Chapter 17, Solution 65.
an = 20/(n2S2), bn = ––3/(nS), Zn = 2n
An = a 2n b 2n
=
400
9
2 2
4 4
n S
n S
3
44.44
1 2 2 , n = 1, 3, 5, 7, 9, etc.
nS
n S
n
1
3
5
7
9
An
2.24
0.39
0.208
0.143
0.109
In = tan––1(bn/an) = tan––1{[––3/(nS)][n2S2/20]} = tan––1(––nx0.4712)
n
1
3
5
7
9
f
In
––25.23q
––54.73q
––67q
––73.14q
––76.74q
––90q
2.24
0
2
6
10
14
18
Zn
––30q
––25.23q
An
0.39
0.208
0
2
6
10
––60q
––54.73q
0.0.143 0.109
14
18
Zn
––90q
In
––67q
––73.14q
––76.74q
Chapter 17, Solution 66.
The schematic is shown below. The waveform is inputted using the attributes of
VPULSE. In the Transient dialog box, we enter Print Step = 0.05, Final Time = 12,
Center Frequency = 0.5, Output Vars = V(1) and click enable Fourier. After simulation,
the output plot is shown below. The output file includes the following Fourier
components.
FOURIER COMPONENTS OF TRANSIENT RESPONSE V(1)
DC COMPONENT = 5.099510E+00
HARMONIC
NO
1
2
3
4
5
6
7
8
9
FREQUENCY FOURIER
NORMALIZED
(HZ)
COMPONENT COMPONENT
5.000E-01
1.000E+00
1.500E+00
2.000E+00
2.500E+00
3.000E+00
3.500E+00
4.000E+00
4.500E+00
3.184E+00
1.593E+00
1.063E+00
7.978E-01
6.392E-01
5.336E-01
4.583E-01
4.020E-01
3.583E-01
1.000E+00
5.002E-01
3.338E-01
2.506E-01
2.008E-01
1.676E-01
1.440E-01
1.263E-01
1.126E-01
PHASE
(DEG)
1.782E+00
3.564E+00
5.347E+00
7.129E+00
8.911E+00
1.069E+01
1.248E+01
1.426E+01
1.604E+01
NORMALIZED
PHASE (DEG)
0.000E+00
1.782E+00
3.564E+00
5.347E+00
7.129E+00
8.911E+00
1.069E+01
1.248E+01
1.426E+01
TOTAL HARMONIC DISTORTION = 7.363360E+01 PERCENT
Chapter 17, Solution 67.
The Schematic is shown below. In the Transient dialog box, we type ““Print step = 0.01s,
Final time = 36s, Center frequency = 0.1667, Output vars = v(1),”” and click Enable
Fourier. After simulation, the output file includes the following Fourier components,
FOURIER COMPONENTS OF TRANSIENT RESPONSE V(1)
DC COMPONENT = 2.000396E+00
HARMONIC FREQUENCY FOURIER NORMALIZED
NO
(HZ) COMPONENT COMPONENT (DEG)
1
2
3
4
5
6
7
8
9
1.667E-01
3.334E-01
5.001E-01
6.668E-01
8.335E-01
1.000E+00
1.167E+00
1.334E+00
1.500E+00
2.432E+00
6.576E-04
5.403E-01
3.343E-04
9.716E-02
7.481E-06
4.968E-02
1.613E-04
6.002E-02
1.000E+00
2.705E-04
2.222E-01
1.375E-04
3.996E-02
3.076E-06
2.043E-02
6.634E-05
2.468E-02
PHASE
NORMALIZED
PHASE (DEG)
-8.996E+01
-8.932E+01
9.011E+01
9.134E+01
-8.982E+01
-9.000E+01
-8.975E+01
-8.722E+01
9.032E+01
0.000E+00
6.467E-01
1.801E+02
1.813E+02
1.433E-01
-3.581E-02
2.173E-01
2.748E+00
1.803E+02
TOTAL HARMONIC DISTORTION = 2.280065E+01 PERCENT
Chapter 17, Solution 68.
The schematic is shown below. We set the final time = 6T=12s and the center frequency
= 1/T = 0.5. When the schematic is saved and run, we obtain the Fourier series from the
output file as shown below.
FOURIER COMPONENTS OF TRANSIENT RESPONSE V(1)
DC COMPONENT =
HARMONIC
NORMALIZED
NO
(DEG)
1
2
3
4
5
6
7
8
9
1.990000E+00
FREQUENCY
FOURIER
NORMALIZED
PHASE
(HZ)
COMPONENT
COMPONENT
(DEG)
1.273E+00
6.367E-01
4.246E-01
3.185E-01
2.549E-01
2.125E-01
1.823E-01
1.596E-01
1.419E-01
1.000E+00
5.001E-01
3.334E-01
2.502E-01
2.002E-01
1.669E-01
1.431E-01
1.253E-01
1.115E-01
5.000E-01
1.000E+00
1.500E+00
2.000E+00
2.500E+00
3.000E+00
3.500E+00
4.000E+00
4.500E+00
Chapter 17, Solution 69.
9.000E-01
-1.782E+02
2.700E+00
-1.764E+02
4.500E+00
-1.746E+0
6.300E+00
-1.728E+02
8.100E+00
PHASE
0.000E+00
1.791E+02
1.800E+00
-1.773E+02
3.600E+00
-1.755E+02
5.400E+00
-1.737E+02
7.200E+00
The schematic is shown below. In the Transient dialog box, set Print Step = 0.05 s, Final
Time = 120, Center Frequency = 0.5, Output Vars = V(1) and click enable Fourier. After
simulation, we obtain V(1) as shown below. We also obtain an output file which
includes the following Fourier components.
FOURIER COMPONENTS OF TRANSIENT RESPONSE V(1)
DC COMPONENT = 5.048510E-01
HARMONIC FREQUENCY FOURIER NORMALIZED
NO
(HZ) COMPONENT COMPONENT (DEG)
1
2
3
4
5
6
7
8
9
PHASE
NORMALIZED
PHASE (DEG)
5.000E-01 4.056E-01 1.000E+00 -9.090E+01 0.000E+00
1.000E+00 2.977E-04 7.341E-04 -8.707E+01 3.833E+00
1.500E+00 4.531E-02 1.117E-01 -9.266E+01 -1.761E+00
2.000E+00 2.969E-04 7.320E-04 -8.414E+01 6.757E+00
2.500E+00 1.648E-02 4.064E-02 -9.432E+01 -3.417E+00
3.000E+00 2.955E-04 7.285E-04 -8.124E+01 9.659E+00
3.500E+00 8.535E-03 2.104E-02 -9.581E+01 -4.911E+00
4.000E+00 2.935E-04 7.238E-04 -7.836E+01 1.254E+01
4.500E+00 5.258E-03 1.296E-02 -9.710E+01 -6.197E+00
TOTAL HARMONIC DISTORTION = 1.214285E+01 PERCENT
Chapter 17, Solution 70.
The schematic is shown below. In the Transient dialog box, we set Print Step = 0.02 s,
Final Step = 12 s, Center Frequency = 0.5, Output Vars = V(1) and V(2), and click
enable Fourier. After simulation, we compare the output and output waveforms as
shown. The output includes the following Fourier components.
FOURIER COMPONENTS OF TRANSIENT RESPONSE V(1)
DC COMPONENT = 7.658051E-01
HARMONIC FREQUENCY FOURIER NORMALIZED
NO
(HZ) COMPONENT COMPONENT (DEG)
1
2
3
4
5
6
7
8
9
5.000E-01 1.070E+00
1.000E+00 3.758E-01
1.500E+00 2.111E-01
2.000E+00 1.247E-01
2.500E+00 8.538E-02
3.000E+00 6.139E-02
3.500E+00 4.743E-02
4.000E+00 3.711E-02
4.500E+00 2.997E-02
1.000E+00
3.512E-01
1.973E-01
1.166E-01
7.980E-02
5.738E-02
4.433E-02
3.469E-02
2.802E-02
1.004E+01
-3.924E+01
-3.985E+01
-5.870E+01
-5.680E+01
-6.563E+01
-6.520E+01
-7.222E+01
-7.088E+01
PHASE
NORMALIZED
PHASE (DEG)
0.000E+00
-4.928E+01
-4.990E+01
-6.874E+01
-6.685E+01
-7.567E+01
-7.524E+01
-8.226E+01
-8.092E+01
TOTAL HARMONIC DISTORTION = 4.352895E+01 PERCENT
Chapter 17, Solution 71.
The schematic is shown below. We set Print Step = 0.05, Final Time = 12 s, Center
Frequency = 0.5, Output Vars = I(1), and click enable Fourier in the Transient dialog box.
After simulation, the output waveform is as shown. The output file includes the
following Fourier components.
FOURIER COMPONENTS OF TRANSIENT RESPONSE I(L_L1)
DC COMPONENT = 8.374999E-02
HARMONIC FREQUENCY FOURIER NORMALIZED
NO
(HZ) COMPONENT COMPONENT (DEG)
1
2
3
4
5
6
7
8
9
PHASE
NORMALIZED
PHASE (DEG)
5.000E-01 2.287E-02 1.000E+00 -6.749E+01 0.000E+00
1.000E+00 1.891E-04 8.268E-03 8.174E+00 7.566E+01
1.500E+00 2.748E-03 1.201E-01 -8.770E+01 -2.021E+01
2.000E+00 9.583E-05 4.190E-03 -1.844E+00 6.565E+01
2.500E+00 1.017E-03 4.446E-02 -9.455E+01 -2.706E+01
3.000E+00 6.366E-05 2.783E-03 -7.308E+00 6.018E+01
3.500E+00 5.937E-04 2.596E-02 -9.572E+01 -2.823E+01
4.000E+00 6.059E-05 2.649E-03 -2.808E+01 3.941E+01
4.500E+00 2.113E-04 9.240E-03 -1.214E+02 -5.387E+01
TOTAL HARMONIC DISTORTION = 1.314238E+01 PERCENT
Chapter 17, Solution 72.
T = 5, Zo = 2S/T = 2S/5
f(t) is an odd function. ao = 0 = an
4 T/2
f ( t ) sin(nZo t )dt
T ³0
bn =
4 10
10 sin(0.4nSt )dt
5 ³0
1
20
8x 5
[1 cos(0.4nS)]
= cos(0.4Snt ) =
2 nS
nS
0
f(t) =
20 f 1
¦ [1 cos(0.4nS)]sin(0.4nSt )
S n 1n
Chapter 17, Solution 73.
p =
2
VDC
1 V2
¦ n
R
2
R
= 0 + 0.5[(22 + 12 + 12)/10] = 300 mW
Chapter 17, Solution 74.
(a)
An =
a 2n b 2n ,
I = tan––1(bn/an)
A1 =
6 2 8 2 = 10,
I1 = tan––1(6/8) = 36.87q
A2 =
3 2 4 2 = 5,
I2 = tan––1(3/4) = 36.87q
i(t) = {4 + 10cos(100St –– 36.87q) –– 5cos(200St –– 36.87q)} A
(b)
p = I 2DC R 0.5¦ I 2n R
= 2[42 +0.5(102 + 52)] = 157 W
Chapter 17, Solution 75.
The lowpass filter is shown below.
R
+
+
C
vs
-
vo
-
vs
AW 2A f 1
nSW
sin
cos nZo t
¦
T
T n 1n
T
Vo
1
jZ n C
Vs
1
R
jZ n C
For n=0, (dc component), Vo
Vs
1
Vs ,
1 jZ n RC
AW
T
Zn
nZo
2nS / T
(1)
For the nth harmonic,
2A
nSW
sin
‘ 90 o
2
2 2
1
nT
T
1 Z n R C ‘ tan Z n RC
1
Vo
When n=1, | Vo |
2A
nSW
sin
x
T
T
x
1
1
(2)
2
4S
R 2C 2
T
From (1) and (2),
AW
T
1
50 x
2A
S
sin
T
10
4S 2 2 2
R C
T
1
1
1
T
10 5
2SR
10 2 x 3.09 x10 4
2
4S
R 2C 2
T
1010
o
C
4S 2 2 2
R C
T

o
30.9
W
4Sx10 3
3.09 x10 4
24.59 mF
Chapter 17, Solution 76.
vs(t) is the same as f(t) in Figure 16.1 except that the magnitude is multiplied by
10. Hence
20 f 1
vo(t) = 5 ¦ sin( nSt ) , n = 2k –– 1
S k 1n
T = 2, Zo = 2S/T = 2S, Zn = nZo = 2nS
jZnL = j2nS; Z = R||10 = 10R/(10 + R)
Vo = ZVs/(Z + j2nS) = [10R/(10R + j2nS(10 + R))]Vs
Vo =
10R‘ tan 1{(nS / 5R )(10 R )}
100R 2 4n 2 S 2 (10 R ) 2
Vs
Vs = [20/(nS)]‘0q
The source current Is is
Vs
Is =
Z j2nS
Vs
10R
j2nS
10 R
20
nS
10R j2nS(10 R )
(10 R )
20
‘ tan 1{( nS / 3)(10 R )}
nS
100R 2 4n 2 S 2 (10 R ) 2
(10 R )
=
1
¦ Vsn I sn cos(T n I n )
2
ps = VDCIDC +
For the DC case, L acts like a short-circuit.
Is =
5
10R
10 R
5(10 R )
, Vs = 5 = Vo
10R
ª
§ 1 § S
··
tan ¨ (10 R ) ¸ ¸¸
«
2 (10 R ) cos¨
¨
25(10 R ) 1 «§ 20 ·
©5
¹¹
©
¨ ¸
10R
2 «© S ¹
100R 2 4S 2 (10 R ) 2
«
¬
ps
º
§
§ 2S
··
(10 R ) 2 cos¨¨ tan 1 ¨ (10 R ) ¸ ¸¸
»
© 5
¹¹
§ 10 ·
©
¨ ¸
"»
2
2
2
»
S
© ¹
100R 16S (10 R )
»
¼
2
ps =
=
VDC 1 f Von
¦
R
2n1 R
º
25 1 ª
100R
100R
«
"»
2
2
2
2
2
2
R 2 ¬100R 4S (10 R )
100R 10S (10 R )
¼
We want po = (70/100)ps = 0.7ps. Due to the complexity of the terms, we
consider only the DC component as an approximation. In fact the DC component
has the latgest share of the power for both input and output signals.
25
R
7 25(10 R )
x
10
10R
100 = 70 + 7R which leads to R = 30/7 = 4.286 :
Chapter 17, Solution 77.
(a) For the first two AC terms, the frequency ratio is 6/4 = 1.5 so that the highest
common factor is 2. Hence Zo = 2.
T = 2S/Zo = 2S/2 = S
(b) The average value is the DC component = ––2
Vrms =
(c)
2
Vrms
ao 1 f 2
(a n b 2n )
¦
2n1
1
(2) 2 (10 2 8 2 6 2 3 2 12 ) = 121.5
2
Vrms = 11.02 V
Chapter 17, Solution 78.
(a)
2
VDC
Vn2
1
p =
¦
R
2
R
2
2
Vn ,rms
VDC
¦
R
R
= 0 + (402/5) + (202/5) + (102/5) = 420 W
(b)
5% increase = (5/100)420 = 21
pDC = 21 W =
2
VDC
2
which leads to VDC
R
21R
105
VDC = 10.25 V
Chapter 17, Solution 79.
From Table 17.3, it is evident that an = 0,
bn = 4A/[S(2n –– 1)], A = 10.
A Fortran program to calculate bn is shown below. The result is also shown.
C
FOR PROBLEM 17.79
DIMENSION B(20)
10
A = 10
PIE = 3.142
C = 4.*A/PIE
DO 10 N = 1, 10
B(N) = C/(2.*FLOAT(N) –– 1.)
PRINT *, N, B(N)
CONTINUE
STOP
END
n
1
2
3
4
5
6
7
8
9
10
bn
12.731
4.243
2.546
1.8187
1.414
1.1573
0.9793
0.8487
0.7498
0.67
Chapter 17, Solution 80.
From Problem 17.55,
cn = [1 + e––jnS]/[2S(1 –– n2)]
This is calculated using the Fortran program shown below. The results are also
shown.
C
FOR PROBLEM 17.80
COMPLEX X, C(0:20)
10
PIE = 3.1415927
A = 2.0*PIE
DO 10 N = 0, 10
IF(N.EQ.1) GO TO 10
X = CMPLX(0, PIE*FLOAT(N))
C(N) = (1.0 + CEXP(––X))/(A*(1 –– FLOAT(N*N)))
PRINT *, N, C(N)
CONTINUE
STOP
END
n
0
1
2
3
4
5
6
7
8
9
10
cn
0.3188 + j0
0
––0.1061 + j0
0
––0.2121x10––1 + j0
0
––0.9095x10––2 + j0
0
––0.5052x10––2 + j0
0
––0.3215x10––2 + j0
Chapter 17, Solution 81.
(a)
A
0
f(t) =
2A 4A f
1
cos(nZo t )
¦
2
S
S n 1 4n 1
The total average power is
pavg = Frms2R = Frms2 since R = 1 ohm.
Pavg = Frms2 =
(b)
2T
T
1 T 2
f ( t )dt = 0.5A2
³
0
T
From the Fourier series above
|co| = 2A/2, |cn| = 4A/[S(4n2 –– 1)]
3T
Zo
0
2Zo
4Zo
6Zo
8Zo
n
0
1
2
3
4
2|cn|2
4A2/(S2)
8A2/(9S2)
2A2/(225S2)
8A2/(1225S2)
8A2/(3969S2)
|cn|
2A/S
2A/(3S)
2A/(15S)
2A/(35S)
2A/(63S)
(c)
81.1%
(d)
0.72%
Chapter 17, Solution 82.
2
VDC
1 f Vn2
P =
¦
R
2n1 R
Assuming V is an amplitude-phase form of Fourier series. But
|An| = 2|Cn|, co = ao
|An|2 = 4|Cn|2
Hence,
f
c o2
c 2n
2¦
P =
R
n 1 R
Alternatively,
2
Vrms
P =
R
where
2
Vrms
a o2 1 f 2
¦ An
2n1
f
c o2 2¦ c 2n
n 1
f
¦c
n f
2
n
= 102 + 2(8.52 + 4.22 + 2.12 + 0.52 + 0.22)
= 100 + 2x94.57 = 289.14
P = 289.14/4 = 72.3 W
% power
81.1%
18.01%
0.72%
0.13%
0.04%
Chapter 18, Solution 1.
f ' ( t ) G( t 2) G( t 1) G( t 1) G( t 2)
jZF(Z) e j2 Z e jZ e jZ e jZ2
2 cos 2Z 2 cos Z
F(Z) =
2[cos 2Z cos Z]
jZ
Chapter 18, Solution 2.
f (t)
ªt,
«0,
¬
0 t 1
otherwise
f ””(t)
f ‘‘(t)
1
G(t)
0
t
t
1
––G’’(t-1)
-G(t-1)
-G(t-1)
f"(t) = G(t) - G(t - 1) - G'(t - 1)
Taking the Fourier transform gives
-Z2F(Z) = 1 - e-jZ - jZe-jZ
F(Z) =
1
³ te
or F(Z)
But
F(Z)
0
³xe
ax
j Zt
dt
eax
(ax 1) c
a2
dx
e jZ
jZ
(1 jZ)e jZ 1
Z2
2
( jZt 1) 10
>
@
1
1 jZ e jZ 1
2
Z
Chapter 18, Solution 3.
f (t)
F(Z)
F(Z) =
1
t , 2 t 2,
2
1 jZt
³2 2 t e dt
2
f ' (t)
1
, 2t 2
2
e jZt
( jZt 1) 2 2
2
2( jZ)
>
@
>
@
1
e jZ2 ( jZ2 1) e jZ2 ( jZ2 1)
2
2Z
1
jZ2 e jZ2 e jZ2 e jZ2 e jZ2
2
2Z
1
jZ4 cos 2Z j2 sin 2Z
2Z 2
j
(sin 2Z 2Z cos 2Z)
Z2
Chapter 18, Solution 4.
2į(t+1)
g’’
2
––1
0
1
t
––2
––2į(t––1)
4į(t)
2į’’(t+1)
g””
––1
0
––2į(t+1)
1
t
––2
––2į(t––1)
––2į’’(t––1)
g cc
2G( t 1) 2Gc( t 1) 4G( t ) 2G( t 1) 2Gc( t 1)
( jZ) 2 G (Z)
2e jZ 2 jZe jZ 4 2e jZ 2 jZe jZ
4 cos Z 4Z sin Z 4
4
G (Z)
Z2
(cos Z Z sin Z 1)
Chapter 18, Solution 5.
h’’(t)
1
0
––1
t
1
––2į(t)
h””(t)
1
į(t+1)
1
––1
––2į’’(t)
h cc( t )
t
0
––į(t––1)
G( t 1) G( t 1) 2Gc( t )
( jZ) 2 H(Z)
e jZ e jZ 2 j Z
H(Ȧ) =
2j 2j
sin Z
Z Z2
2 j sin Z 2 jZ
Chapter 18, Solution 6.
0
F(Z)
³ (1)e
jZt
1
Re F(Z)
1
dt ³ te jZt dt
0
0
1
1
0
³ cos Ztdt ³ t cos Ztdt
· 1
§ 1
1
t
0
sin Zt 1 ¨¨
cos Zt sin Zt ¸¸ 0
2
Z
Z
¹
©Z
1
Z2
Chapter 18, Solution 7.
(a)
f1 is similar to the function f(t) in Fig. 17.6.
f ( t 1)
f1 ( t )
Since F(Z)
F1 (Z)
jZ
2(cos Z 1)
jZ
e F(Z)
2e jZ (cos Z 1)
jZ
Alternatively,
f 1' ( t ) G( t ) 2G( t 1) G( t 2)
jZF1 (Z) 1 2e jZ e j2 Z e jZ (e jZ 2 e jZ )
e jZ (2 cos Z 2)
F1(Z) =
(b)
2e jZ (cos Z 1)
jZ
f2 is similar to f(t) in Fig. 17.14.
f2(t) = 2f(t)
F2(Z) =
4(1 cos Z)
Z2
(cos Z 1)
Chapter 18, Solution 8.
1
³ 2e
F(Z)
(a)
2
jZt
dt ³ (4 2 t )e jZt dt
0
1
2 jZt 1
4 jZt 2
2 jZt
2
e
e
e
( jZt 1) 1
1
0
2
jZ
jZ
Z
2
F(Z)
(b)
Z
2
2 jZ 2
4 j2Z 2
e
e
(1 j2Z)e j2Z
2
jZ
jZ jZ
Z
g(t) = 2[ u(t+2) –– u(t-2) ] - [ u(t+1) –– u(t-1) ]
G (Z)
4 sin 2Z 2 sin Z
Z
Z
Chapter 18, Solution 9.
(a)
y(t) = u(t+2) –– u(t-2) + 2[ u(t+1) –– u(t-1) ]
Y(Z)
1
³ ( 2 t ) e
(b) Z(Z)
2
4
sin 2Z sin Z
Z
Z
jZt
dt
2e jZt
0
Z2
1
( jZt 1) 0
2
Z2
Chapter 18, Solution 10.
(a)
x(t) = e2tu(t)
X(Z) = 1/(2 + jZ)
(b)
e ( t )
ªe t , t ! 0
« t
«¬e , t 0
Y(Z)
1
³
1
y( t )e jZt dt
³
0
1
1
e t e jZt dt ³ e t e jZt dt
0
2e j Z
Z2
(1 jZ)
e (1 jZ) t
1 jZ
0
1
e (1 jZ) t
(1 jZ)
1
0
ª cos Z jsin Z cos Z jsin Z º
2
e 1 «
»
2
1 jZ
1 jZ
1 Z
¬
¼
>
2
1 e 1 (cos Z Z sin Z)
2
1 Z
Y(Z) =
Chapter 18, Solution 11.
f(t) = sin S t [u(t) - u(t - 2)]
F(Z)
³
2
0
sin St e jZt dt
1 2 j St
e e j St e jZt dt
³
0
2j
1 ª 2 j( Z S ) t
e j( Z S) t )dt º
(e
»¼
2 j «¬ ³0
1 ª
1
e j( Z S ) t 2 º
e j( Z S ) t 02 «
0»
j(Z S) ¼
2 j ¬ j(Z S)
1 § 1 e j2 Z 1 e j2 Z ·
¸
¨
2 ¨© S Z
S Z ¸¹
1
2S 2Se j2 Z
2
2(S Z )
2
S
e jZ 2 1
2
Z S
F(Z) =
2
Chapter 18, Solution 12.
(a)
F(Z)
³
f
0
e t e jZt dt
1
e (1 jZ) t
1 jZ
2
0
³
2
0
e (1 jZ) t dt
e 2 jZ 2 1
1 jZ
@
³
H(Z)
(b)
0
1
1
e jZt dt ³ (1)e jZt dt
0
1
1 jZ
1 e jZ e 1
jZ
jZ
4 sin 2 Z / 2
jZ
1
(2 2 cos Z)
jZ
§ sin Z / 2 ·
jZ¨
¸
© Z/ 2 ¹
2
Chapter 18, Solution 13.
(a) We know that F[cos at ] S[G(Z a ) G(Z a )] .
Using the time shifting property,
F[cos a ( t S / 3a )]
(b) sin S( t 1)
Se jZS / 3a [G(Z a ) G(Z a )]
sin St cos S cos St sin S
Se jS / 3G(Z a ) Se jS / 3G(Z a )
sin St
g(t) = -u(t+1) sin (t+1)
Let x(t) = u(t)sin t, then X(Z)
1
1
( jZ) 2 1
1 Z2
Using the time shifting property,
G (Z)
1
1 Z2
e jZ
e jZ
Z2 1
(c ) Let y(t) = 1 + Asin at, then Y(Z)
h(t) = y(t) cos bt
2SG(Z) jSA[G(Z a ) G(Z a )]
Using the modulation property,
1
H(Z)
[Y(Z b) Y(Z b)]
2
H(Z)
S>G(Z b) G(Z b)@ jSA
>G(Z a b) G(Z a b) G(Z a b) G(Z a b)@
2
4
(d) I(Z)
jZ t
³ (1 t )e dt
0
e j Zt e j Zt
4
( jZt 1) 0
2
jZ Z
1
Z2
e j4Z e j4Z
( j4Z 1)
jZ
Z2
Chapter 18, Solution 14.
(a)
cos(3t S) cos 3t cos S sin 3t sin S
f ( t ) e t cos 3t u ( t )
F(Z) =
cos 3t (1) sin 3t (0)
1 jZ
2
1 jZ 9
(b)
g(t)
1
-1
1
t
-1
g’’(t)
S
-1
1
t
-S
g' ( t )
S cos St>u ( t 1) u ( t 1)@
g" ( t ) S 2 g( t ) SG( t 1) SG( t 1)
Z 2 G (Z) S 2 G (Z) Se jZ Se jZ
S 2 Z 2 G (Z) S(e jZ e jZ ) 2 jS sin Z
G(Z) =
2 jS sin Z
Z2 S 2
Alternatively, we compare this with Prob. 17.7
f(t) = g(t - 1)
F(Z) = G(Z)e-jZ
cos(3t )
GZ
S
e jZ e jZ
2
Z S
F(Z)e jZ
2
j2S sin Z
Z2 S 2
2 jS sin Z
S 2 Z2
G(Z) =
(c)
cos S( t 1) cos St cos S sin St sin S cos St (1) sin St (0)
Let x ( t ) e 2( t 1) cos S( t 1)u ( t 1) e 2 h ( t )
and
y( t ) e 2 t cos(St )u ( t )
Y(Z)
2 jZ
(2 jZ) 2 S 2
y( t ) x ( t 1)
Y(Z) X(Z)e jZ
X(Z)
2
2 jZ S 2
X(Z)
e 2 H(Z)
H(Z)
e 2 X(Z)
=
(d)
2 jZ e j Z
2 jZ e jZ 2
2
2 jZ S 2
e 2 t sin( 4t )u ( t ) y( t )
p( t ) x ( t )
where y( t ) e 2 t sin 4t u ( t )
Let x ( t )
Y (Z)
X(Z)
2 jZ
2
2 jZ 4 2
Y(Z)
2 jZ
2
2 jZ 16
cos St
p(Z)
(e)
X(Z)
jZ 2
2
jZ 2 16
§
8 jZ 2
1 ·
e
3 2¨¨ SG(Z) ¸¸e jZ2
jZ
jZ ¹
©
Q(Z)
6 jZ 2
e 3 2SG(Z)e jZ 2
jZ
Q(Z) =
Chapter 18, Solution 15.
e j3Z e jZ3
(a)
F(Z)
(b)
Let g( t )
F(Z)
2 j sin 3Z
2G( t 1), G (Z)
2e jZ
t
F §¨ ³ g ( t ) dt ·¸
© f
¹
G (Z)
SF(0)G(Z)
jZ
2e j Z
2SG(1)G(Z)
jZ
=
(c)
F >G(2t )@
F(Z)
2e jZ
jZ
1
˜1
2
1
1
˜ 1 jZ
3
2
1 jZ
3 2
Chapter 18, Solution 16.
(a) Using duality properly
t o
2
Z2
2
o 2S Z
t2
4
o 4S Z
t2
or
§4·
F(Z) = F ¨ 2 ¸
©t ¹
(b)
e
4S Z
2a
a Z2
at
2
2a
a t2
2S e
2
8
a t2
2
4S e
§ 8 ·
G(Z) = F ¨
2 ¸
©4t ¹
a Z
2 Z
4S e
2 Z
Chapter 18, Solution 17.
Since H(Z) = F cos Z0 t f ( t )
(a)
where F(Z) = F >u t
HZ
@
SG Z 1
>F Z Z0 F Z Z0
2
1
, Z0
jZ
2
1ª
1
1 º
SG Z 2 «SG Z 2 »
2¬
j Z 2
j Z2 ¼
@
S
>G Z 2 G Z 2 @ j ª« Z 2 Z 2 º»
2
2 ¬ Z 2 Z2 ¼
H(Z) =
(b)
S
>G Z 2 G Z 2 @ 2jZ
2
Z 4
G(Z) = F >sin Z0 t f ( t )@
where F(Z) = F >u t
GZ
@
j
>F Z Z0 F Z Z0
2
SG Z @
1
jZ
º
jª
1
1
SG Z 10 SG Z 10 «
2¬
j Z 10
j Z 10 »¼
jS
>G Z 10 G Z 10 @ j ª« j j º»
2
2 ¬ Z 10 Z 10 ¼
=
jS
>G Z 10 G Z 10 @ 2 10
2
Z 100
Chapter 18, Solution 18.
Let f t
e t u t
f t cos t
Hence Y Z
FZ
1
j jZ
1
>F Z 1 F Z 1 @
2
º
1ª
1
1
«
»
2 ¬1 j Z 1 1 j Z 1 ¼
1 ª 1 jZ j 1 jZ j º
«
»
2 ¬ >1 j Z 1 @>1 j Z 1 @¼
1 jZ
1 jZ j jZ j Z 2 1
=
1 jZ
2 jZ Z 2 2
Chapter 18, Solution 19.
f
1 1 j2 St
e e j2 St e jZt dt
³
0
2
FZ
³
FZ
1 1 j Z 2 S t
e
e j Z 2 S t dt
³
0
2
f
f ( t )e jZt dt
>
@
1
º
1
1ª
1
e j Z 2S t »
e j Z 2 S t «
2 ¬ j Z 2S
j Z 2S
¼0
1 ª e j Z 2 S 1 e j Z 2 S 1 º
«
»
2 ¬ j Z 2S
j Z 2S ¼
But
e j2 S
cos 2S j sin 2S 1 e j2 S
FZ
1 § e jZ 1 ·§ 1
1 ·
¸¸¨
¨¨
¸
2©
j ¹© Z 2S Z 2S ¹
=
jZ
e jZ 1
2
Z 4S
2
Chapter 18, Solution 20.
(a)
F (cn) = cnG(Z)
F c n e jnZo t
§ f
·
F ¨ ¦ c n e jnZo t ¸
© n f
¹
(b)
T
cn
2S
c n G Z nZo
f
¦c
n f
Zo
1 T
f t e jnZo t dt
³
0
T
n
G Z nZ o
2S
1
T
1 § S
jnt
¨ ³0 1˜ e dt 0 ·¸
©
¹
2S
1 § 1 jnt
¨ e
2S ¨© jn
But e jnS
cn
S
0
·
¸¸
¹
j
e jnS 1
2Sn
cos nS j sin nS
>
@
cos nS
ª 0,
« j ,
¬ nS
j
n
1 1
2nS
(1) n
n even
n odd , n z 0
for n = 0
cn
1 S
1 dt
2S ³0
1
2
Hence
f (t)
f
1
j jnt
¦
e
2 n f nS
n z0
n odd
F(Z) =
f
1
j
GZ ¦
G Zn
2
n f nS
nz0
n odd
Chapter 18, Solution 21.
Using Parseval’’s theorem,
f
³ f f
2
1 f
| F(Z) | 2 dZ
³
f
2S
( t )dt
If f(t) = u(t+a) –– u(t+a), then
f
³f
f 2 ( t )dt
a
³a
(1) 2 dt
2
2a
1 f
§ sin aZ ·
4a 2 ¨
¸ dZ
³
2 S f
© aZ ¹
or
2
§ sin aZ ·
³ f ¨© aZ ¸¹ dZ
f
4Sa
4a 2
S
as required.
a
Chapter 18, Solution 22.
F >f ( t ) sin Zo t @
³
f
f
f (t)
e j Z o t e j Z o t j Zt
e dt
2j
f
1ª f
f ( t )e j Z Zo t dt ³ e j Z Zo t dt º
³
«
»¼
f
f
2j ¬
1
>F Z Z o F Z Zo
2j
=
@
Chapter 18, Solution 23.
(a) f(3t) leads to
F >f 3t
1
10
˜
3 2 jZ / 3 5 jZ / 3
30
6 jZ 15 jZ
@
1
10
˜
2 2 jZ / 2 15 jZ / 2
(b) f(2t)
1
1
F Z 2 F Z 2
2
2
(c) f(t) cos 2t
5
>2 j Z 2 @>5 j Z 2 @
(d) F >f ' t
(e)
³
t
f
@
f t dt
20
4 jZ 10 jZ
20e jZ / 2
4 jZ 10 jZ
f(2t-1) = f [2(t-1/2)]
=
30
6 jZ 15 jZ
jZ F Z
5
>2 j Z 2 >5 j Z 2
jZ10
2 jZ 5 jZ
FZ
SF 0 G Z
jZ
@@
10
x10
SG Z
jZ 2 jZ 5 jZ
2x5
=
10
SG Z
jZ 2 jZ 5 jZ
Chapter 18, Solution 24.
(a) X Z
F Z + F [3]
= 6SG Z (b) y t
YZ
j jZ
e 1
Z
f t2
e jZ 2 F Z
je j2Z jZ
e 1
Z
(c) If h(t) = f '(t)
HZ
(d) g t
jZ F Z
jZ
j jZ
e 1
Z
§2 ·
§5 ·
4f ¨ t ¸ 10f ¨ t ¸, G (Z)
©3 ¹
©3 ¹
6˜
=
j
3
Z
2
e j3Z / 2 1 1 e jZ
3 §3 ·
3 §3 ·
4 x F¨ Z ¸ 10x F¨ Z ¸
2 ©2 ¹
5 ©5 ¹
6 j j3Z / 5
e
1
3
Z
5
j4 j3Z / 2
j10 j3Z / 5
e
1 e
1
Z
Z
Chapter 18, Solution 25.
10
s s2
(a) F s
A
10
2
5, B
A
B
, s
s s2
10
2
5
FZ
5
5
jZ jZ 2
f(t) =
5
sgn t 5e 2 t u t
2
(b) F Z
Fs
jZ 4
jZ 1 j Z 2
s4
s 1 s 2
jZ
A
B
j Z 1 jZ 2
A
B
, s
s 1 s 2
jZ
A = 5, B = 6
FZ
6
5
1 jZ 2 jZ
f(t) = 5e t 6e 2 t u t
Chapter 18, Solution 26.
(a) f ( t )
e ( t 2) u ( t )
(b) h ( t )
te 4 t u ( t )
(c) If x ( t )
u ( t 1) u ( t 1)
By using duality property,

o
X(Z)
2
sin Z
Z
G (Z)
2u (Z 1) 2u (Z 1)

o
2 sin t
St
g( t )
Chapter 18, Solution 27.
100
s s 10
(a) Let F s
100
10
A
A
B
, s
s s 10
100
10
10, B
jZ
10
10
10
jZ jZ 10
FZ
f(t) = 5 sgn t 10e 10 t u t
10s
2s 3s
(b) G s
A
GZ
20
5
4, B
A
B
, s
2s s3
30
5
jZ
6
4
6
jZ 2 jZ 3
g(t) = 4e 2 t u t 6e 3 t u t
(c) H Z
60
60
jZ j40Z 1300
jZ 20 900
2
2
h(t) = 2e 20 t sin( 30t ) u t
yt
1 f G Z e jZt dZ
2S ³f 2 jZ jZ 1
1 1
S˜
2 2
1
S
4
Chapter 18, Solution 28.
(a)
f (t)
1 1
2 (5)(2)
(b)
f (t)
f (t)
Let
0.05
10
e j2 t
2S ( j2)( j2 1)
( 2 j)e j2 t
2S
1 f 20G(Z 1)e jZt
dZ
2S ³f (2 jZ)(3 5Z)
20e jt
2S(5 5 j)
(d)
1
20
1 f 10G(Z 2) jZt
e dZ
2S ³f jZ( jZ 1)
j5 e j2 t
2S 1 j2
(c)
SG(Z) e jZt
1 f
dZ
2S ³f (5 jZ)(2 jZ)
1 f
F(Z)e jZt dZ
2S ³f
20
e jt
2S (2 j)(3 j)
(1 j)e jt
S
F(Z)
5SG(Z)
5
(5 jZ) jZ(5 jZ)
F1 (Z) F2 (Z)
f1 ( t )
1 f 5SG(Z) jZt
e dZ
2S ³f 5 jZ
5S 1
˜
2S 5
F2 (s)
5
s(5 s)
A
B
, A 1, B
s s5
F2 (Z)
1
1
jZ jZ 5
f 2 (t)
1
sgn( t ) e 5 t
2
f (t)
f1 (t ) f 2 (t )
1
u(t) e 5t
2
u( t ) e 5 t
0.5
1
Chapter 18, Solution 29.
(a)
f(t) = F -1 [G(Z)] F -1 >4G(Z 3) 4G(Z 3)@
1 4 cos 3t
1
1 8 cos 3t
=
S
2S
2S
(b)
(c)
If h ( t )
u ( t 2) u ( t 2)
H(Z)
2 sin 2Z
Z
G (Z)
4H(Z)
g(t) =
4 sin 2t
St
1 8 sin 2 t
˜
2S
t
g( t )
Since
cos(at) ↔SG(Z a ) SG(Z a )
Using the reversal property,
2S cos 2Z l SG( t 2) SG( t 2)
or F -1 >6 cos 2Z@
3G(t 2) 3G(t 2)
Chapter 18, Solution 30.
(a)
y( t )
sgn( t )
H(Z)
(b) X(Z)
Y(Z)
X(Z)
1
,
1 jZ
H(Z)

o
2(a jZ)
jZ
Y(Z)
2
,
jZ
Y(Z)
X(Z)
2a
jZ

o

o
h(t)
2
1
a jZ
h(t)
2G( t ) a[u ( t ) u ( t )]
1
2 jZ
1 jZ
1
1
2 jZ
2 jZ
(c ) In this case, by definition, h ( t )
y( t )
G( t ) e 2 t u ( t )
e at sin bt u ( t )
Chapter 18, Solution 31.
(a)
Y(Z)
X(Z)
1
H(Z)
,
(a jZ)
2
Y(Z)
H(Z)
1
a jZ
(b)
By definition, x ( t )
(c )
Y(Z)
X(Z)
Y(Z)
H(Z)
1
(a jZ)
jZ
2(a jZ)

o
y( t )
H(Z)
,
1
a jZ
x(t)
e at u ( t )
u ( t 1) u ( t 1)
2
jZ
1
a
2 2(a jZ)

o
Chapter 18, Solution 32.
(a)
e jZ
jZ 1
and F Z
e t 1 u ( t 1)
Since
F1 Z
f(-t)
e jZ
jZ 1
f1 t
f1(t) = e t 1 u t 1
(b)
From Section 17.3,
2
t 1
2Se
2
Z
If F2 Z
2e
f2(t) =
2
S t 1
2
Z
, then
e t 1 u t 1
x(t)
a
1
G( t ) e at u ( t )
2
2
(b)
By partial fractions
1
F3 Z
jZ 1
Hence f 3 t
=
(d)
2
jZ 1
2
1
1
1
1
4
4
4 4
2
2
jZ 1
jZ 1
jZ 1
jZ 1
1 t
te e t te t e t u t
4
1
1
t 1 e t u t t 1 e t u t
4
4
1 f
F1 Z e jZt dZ
2S ³ f
f4 t
1
1 f G Z e jZt
=
³
2S f 1 j2Z
2S
Chapter 18, Solution 33.
(a)
2 sin St>u t 1 u t 1 @
Let x t
From Problem 17.9(b),
4 jS sin Z
S 2 Z2
Applying duality property,
XZ
f t
(b)
1
X t
2S
2 j sin t
S2 t 2
f(t) =
2 j sin t
t 2 S2
FZ
j
j
cos 2Z j sin 2Z cos Z j sin Z
Z
Z
j j2 Z
e e jZ
Z
f t
e jZ e j 2 Z
jZ
jZ
1
1
sgn t 1 sgn t 2
2
2
But sgn( t )
2u ( t ) 1
f t
u t 1 1
1
u t2 2
2
= u t 1 u t 2
Chapter 18, Solution 34.
First, we find G(Z) for g(t) shown below.
gt
g' t
10>u t 2 u t 2 @ 10>u t 1 u t 1 @
10>G t 2 G t 2 @ 10>G t 1 G t 1 @
The Fourier transform of each term gives
g(t)
20
10
––2
––1
0
1
t
2
g ‘‘(t)
10G(t+2)
10G(t+1)
––2
––1
0
1
––10G(t-1)
2
––10G(t-2)
jZG Z 10 e jZ2 e jZ2 10 e jZ e jZ
20 j sin 2Z 20 j sin Z
GZ
20 sin 2Z 20 sin Z
Z
Z
Note that G(Z) = G(-Z).
40 sinc(2Z) + 20 sinc(Z)
t
FZ
2SG Z
f t
1
Gt
2S
= (20/S)sinc(2t) + (10/S)sinc(t)
Chapter 18, Solution 35.
(a)
x(t) = f[3(t-1/3)]. Using the scaling and time shifting properties,’’’’
X(Z)
(b)
1
1
e jZ / 3
3 2 jZ / 3
e jZ / 3
(6 jZ)
Using the modulation property,
Y(Z)
1
[F(Z 5) F(Z 5)]
2
(c )
Z(Z)
jZF(Z)
(d)
H(Z)
F(Z)F(Z)
(e)
I(Z)
j
º
1ª
1
1
«
2 ¬ 2 j(Z 5) 2 j(Z 5) »¼
jZ
2 jZ
d
F(Z)
dZ
1
(2 jZ) 2
j
(0 j)
1
(2 jZ) 2
(2 jZ) 2
Chapter 18, Solution 36.
HZ
Vo Z
Vo Z
Vi Z
H Z Vi Z
10Vi Z
2 jZ
1ª 1
1 º
«
2 ¬ jZ 7 jZ 3 »¼
(a)
vi = 4G(t)
Vi(Z) = 4
40
2 jZ
Vo Z
v o ( t ) 40e 2 t u t
vo(2) = 40e––4 = 0.7326 V
(b)
vi
6e t u t
Vo Z
60
2 jZ 1 jZ
Vo s
60
s 2 s 1
A
60
1
60, B
60
1
v o (t)
60 e t e 2 t u ( t )
2
e 4
@
@
jZ
60
60
60
1 jZ 2 jZ
v o (2)
= 7.021 V
(c)
A
B
, s
s 1 s 2
Vo Z
>
60>e
6
1 jZ
Vi Z
60 0.13533 0.01831
vi(t) = 3 cos 2t
Vi(Z) = S[G(Z + 2) + G(Z- 2)]
Vo
v o (t)
10S>G Z 2 G Z 2
2 jZ
@
1 f
Vo Z e jZt dZ
2S ³f
jZt
f G Z2 e
G Z 2 jZt
5³
e dZ 5 ³
dZ
f
f 2 jZ
2 jZ
f
5e j2 t 5e j2 t
2 j2 2 j2
5
2 2
>e
j 2 t 45q
e j 2 t 45q
@
5
cos 2 t 45q
2
5
cos 4 45q
2
vo 2
5
cos 229.18q 45q
2
vo(2) = ––3.526 V
Chapter 18, Solution 37.
2 jZ
j2Z
2 jZ
By current division,
HZ
H(Z) =
Io Z
Is Z
j2Z
2 jZ
j2Z
4
2 jZ
j2Z
j2Z 8 j4Z
jZ
4 j3Z
Chapter 18, Solution 38.
Vi Z
Vo Z
SG Z 1
jZ
10
Vi Z
10 jZ2
Let Vo Z
V1 Z V2 Z
5 §
1 ·
¨¨ SG Z ¸¸
5 jZ ©
jZ ¹
5SG Z
5
5 j Z j Z 5 jZ
V2 Z
5
s s5
V2 Z
1
1
jZ 5 j Z
V1
5SG Z
5 jZ
5S 1
˜
2S 5
v1(t)
But
A
B
s s5
v0 t
sgn t
A = 1, B = -1, s = jZ
v2 t
1
sgn( t ) e 5 t
2
v1 t
1 f 5SG Z jZt
e dZ
2S ³f 5 jZ
0.5
v1 t v 2 t
1 2u t
0.5 0.5 sgn t e 5 t
0.5 0.5 u t e 5 t u t
vo t
u t e 5 t u t
Chapter 18, Solution 39.
f
Vs (Z)
³ (1 t )e
jZt
dt
f
I(Z)
Vs (Z)
3
10 jZx10
1
1
1 jZ
e
2
jZ Z
Z2
10 3
3
§ 1
1
1 jZ ·
¨¨ ¸¸
e
2
10 jZ © jZ Z
Z2
¹
6
Chapter 18, Solution 40.
v( t ) G( t ) 2G( t 1) G( t 2)
Z 2 V(Z) 1 2e jZ e jZ2
Now
VZ
1 2e jZ e jZ2
Z2
ZZ
2
1
jZ
1 j2Z
jZ
2e jZ e jZ2 1
jZ
˜
2
1 j2Z
Z
VZ
ZZ
I
1
0.5 0.5e jZ2 e jZ
jZ 0.5 jZ
1
s(s 0.5)
But
IZ
i(t) =
A
B
s s 0.5
A = 2, B = -2
2
2
0.5 0.5e jZ2 e jZ 0.5 0.5e jZ2 e jZ
jZ
0.5 jZ
1
1
sgn( t ) sgn(t 2) sgn( t 1) e 0.5t u(t ) e 0.5( t 2 ) u(t 2) 2e 0.5( t 1) u(t 1)
2
2
Chapter 18, Solution 41.
2
1
2 jZ
+
+
V
1/s
0.5s
V
1
2V
2 jZ
jZ V 2
2
jZ
2
j Z 2Z 4 V
V(Ȧ) =
0
jZ
j4Z 2 jZ
2 jZ(4.5 j2Z)
(2 jZ)(4 2Z 2 jZ)
4Z 2 j9Z
2 jZ
2
Chapter 18, Solution 42.
By current division, I o
(a)
2
˜I Z
2 jZ
For i(t) = 5 sgn (t),
10
jZ
2
10
˜
2 jZ jZ
IZ
Io
20
s(s 2)
Let I o
20
jZ(2 jZ)
A
B
, A 10, B
s s2
10
10
10
j Z 2 jZ
Io Z
io(t) = 5 sgn( t ) 10e 2 t u(t )A
i(t)
(b)
i’’(t)
4
4G(t)
1
1
t
i' ( t ) 4G( t ) 4G( t 1)
jZ I Z 4 4e jZ
IZ
Io
4 1 e jZ
jZ
8 1 e jZ
jZ(2 jZ)
§ 1
1 ·
¸¸ 1 e jZ
4¨¨ Z
Z
j
2
j
©
¹
4
4
4e j Z 4e j Z
jZ 2 jZ
jZ
2 jZ
t
––4G(t––1)
io(t) = 2 sgn( t ) 2 sgn( t 1) 4e 2 t u(t ) 4e 2( t 1) u t 1 A
Chapter 18, Solution 43.
1
jZ C

o
20 mF
Vo
Vo
1
j20x10
3
Z
50
,
jZ
5e t
is
50
40
Is x
50
jZ
40 jZ
50
,
(s 1.25)(s 5)
A
B
s 1.25 s 5
40 ª
1
1 º
«
3 ¬ jZ 1.25 jZ 5 »¼
v o (t)
s

o
Is
1
5 jZ
jZ
40 1.25t
(e
e 5 t ) u ( t )
3
Chapter 18, Solution 44.
1H
jZ
We transform the voltage source to a current source as shown in Fig. (a) and then
combine the two parallel 2: resistors, as shown in Fig. (b).
Io
+
Vs/2
2
2 Vo
Io
+
Vs/2
1 Vo
jZ
(a)
2 2 1:, I o
Vo
jZ I o
V
1
˜ s
1 jZ 2
jZ Vs
2(1 jZ)
(b)
jZ
v s ( t ) 10G t 10G( t 2)
10 10e j2 Z
jZ Vs Z
Vs Z
10 1 e j2Z
jZ
5 1 e j2 Z
1 jZ
Hence Vo
v o (t )
5
5
e j2 Z
1 jZ 1 jZ
5e t u ( t ) 5e ( t 2) u ( t 2)
v o (1)
5e 1 1 0
1.839 V
Chapter 18, Solution 45.
Vo
1
jZ
1
2 jZ jZ
(2)
2
( jZ 1)
2

o
v o (t)
2te t u ( t )
Chapter 18, Solution 46.
1
F
4
1
jZ
2H
3G( t )
1
4
j4
Z
jZ2
3
1
1 jZ
e t u(t)
The circuit in the frequency domain is shown below:
2:
Vo
Io(Z)
––j4/Z
1/(1+jZ)
+
3
+
j2Z
At node Vo, KCL gives
1
Vo
3 Vo
1 jZ
j4
2
Z
Vo
j2Z
2
2Vo jZ3 jZVo
1 jZ
j2Vo
Z
2
jZ3
1 jZ
Vo
j2
2 jZ Z
2 jZ3 3Z 2
Vo
1 jZ
Io Z
j2 ·
j2Z
§
j2Z¨ 2 jZ ¸
Z¹
©
Io(Z) =
2 jZ 2 3Z 2
4 6Z 2 j(8Z 2Z 3 )
Chapter 18, Solution 47.
Transferring the current source to a voltage source gives the circuit below:
1/(jZ)
2:
+
8V
+
Vo
1:
jZ/2
Let Z in
2 1
jZ
2
By voltage division,
jZ
2 2
jZ
1
2
1
jZ
Vo Z
1
Z in
jZ
4 j3Z
2 jZ
8
1 jZZ in
˜8
=
8 2 jZ
2 jZ jZ4 3Z 2
=
8 2 jZ
2 jZ5 3Z 2
8
jZ 4 j3Z
1
2 jZ
Chapter 18, Solution 48.
1
jZC
0.2F
j5
Z
As an integrator,
RC
20 x 10 3 x 20 x 10 6
0.4
vo
1 t
v i dt
RC ³o
Vo
º
1 ª Vi
SVi (0)G Z »
«
RC ¬ jZ
¼
Io
Vo
mA
20
º
1 ª
2
SG Z »
«
0 .4 ¬ j Z 2 j Z
¼
ª
º
2
0.125 «
SG Z »
¬ jZ 2 jZ
¼
0.125 0.125
0.125SG Z
jZ
2 jZ
0.125 sgn( t ) 0.125e 2 t u t i o (t)
0.125
SG Z e jZt dt
³
2S
0.125 0.25u ( t ) 0.125e 2 t u ( t ) 0.125
2
io(t) = 0.625 0.25u(t ) 0.125e 2t u(t ) mA
Chapter 18, Solution 49.
Consider the circuit shown below:
jZ
j2Z
jZ
+
VS
Vs
+
i1
i2
1 : vo
2:
S>G Z 1 G Z 2 @
For mesh 1, Vs 2 j2Z I1 2I 2 jZI 2
Vs 2 1 jZ I1 2 jZ I 2
For mesh 2,
I1
0
0
(1)
3 j Z I 2 2 I 1 jZ I 1
3 Z I2
2Z
(2)
Substituting (2) into (1) gives
Vs
2
2 1 jZ 3 jZ I 2
2 jZ I 2
2 jZ
Vs 2 Z
>2 3 j4Z Z
I 2 2 j4Z Z
2
2
@
4 j4Z Z 2 I 2
I2
s 2 Vs
,s
s 4s 2
Vo
I2
2
jZ
jZ 2 S >G Z 1 G Z 1 @
2
jZ jZ4 2
1 f
v o Z e jZt dZ
³
f
2S
v o (t)
³
f
f
1
1
jZ 2 e jZt G Z 1 dZ
jZ 2 e jZt G Z 1 dZ
2
2
2
2
jZ jZ4 2
jZ jZ4 2
1
1
j 2 e jt
j 2 e jt
2
2
1 j4 2 1 j4 2
v o (t)
1
1
2 j 1 j4
2 j 1 j4 e jt
2
e jt 2
17
17
1
1
6 j7 e jt 6 j7 e jt
34
34
0.271e j t 13.64q 0.271e j t 13.64q
vo(t) = 0.542 cos(t 13.64q)V
Chapter 18, Solution 50.
Consider the circuit shown below:
j0.5Z
1:
VS
+
i1
jZ
i2
jZ
+
1:
vo
For loop 1,
2 1 jZ I1 j0.5ZI 2
(1)
0
For loop 2,
1 jZ I 2 j0.5ZI1
(2)
0
From (2),
1 jZ I 2
j0.5Z
I1
2
1 jZ I 2
jZ
Substituting this into (1),
2 1 jZ I 2 jZ
2
I2
jZ
2
2 jZ
3 ·
§
¨ 4 j4Z Z 2 ¸I 2
2 ¹
©
I2
2 jZ
4 j4Z 1.5Z 2
Vo
I2
Vo
4
jZ
3
8
8Z
j
jZ
3
3
2 jZ
4 j4Z 1.5 jZ
2
·
§4
4¨ jZ ¸
¹
©3
· § 8 ·¸
§4
¨ jZ ¸ ¨¨
¹ © 3 ¸¹
©3
2
Vo ( t )
2
2
16
3
2
· §¨ 8 ·¸
§4
¨ jZ ¸ ¨
¹ © 3 ¸¹
©3
2
§ 8 ·
§ 8 ·
4e 4t / 3 cos¨¨
t ¸¸ u(t ) 5.657e 4t / 3 sin¨¨
t ¸¸u(t ) V
3
3
©
¹
©
¹
Chapter 18, Solution 51.
1
jZ
Z 1 //
Vo
1
jZ
1
1
jZ
Z
Vo
Z2
1
1 jZ
1
2
1 jZ
1
1 jZ
2
1 jZ
1
,
(s 1)(s 1.5)
Vo
A
B
s 1 s 1.5
f
W
f 2 ( t )dt
³
f
f
4 ³ (e
s
1
2
3 2 jZ 1 jZ
jZ
2
2
s 1 s 1.5
o
v o (t)
f
4 ³ (e t e 1.5t ) 2 dt
0
f
2t
2e
2.5 t
e
3t
)dt
0
W
§ e 2t
e 2.5t e 3t ·¸
2
4¨
¨ 2
2
.
5
3 ¸¹
©
0
1 2 1
4( )
2 2.5 3
0.1332 J
Chapter 18, Solution 52.
f
J = 2 ³ f 2 ( t ) dt
0
2(e t e 1.5t )u ( t )
1 f
2
F(Z) dZ
³
0
S
1 f
1
dZ
=
2
³
0
S 9 Z2
f
1
tan 1 (Z / 3)
3S
0
1 S
= (1/6)
3S 2
Chapter 18, Solution 53.
J =
³
f
0
2
F(Z) dZ
f
2S³ f 2 ( t ) dt
f
f(t) =
e 2t ,
t0
e 2 t ,
t!0
0
f
J = 2S ª ³ e 4 t dt ³ e 4 t dt º
«¬ f
»¼
0
ª e 4t
2S «
«¬ 4
0
f
e 4 t
4
º
» = 2S[(1/4) + (1/4)] = S
0 »
¼
f
Chapter 18, Solution 54.
W1: =
³
f
f
f 2 ( t ) dt
f
16 ³ e 2 t dt
8e 2 t
0
f
0
= 8J
Chapter 18, Solution 55.
f(t) = 5e2e––tu(t)
F(Z) = 5e2/(1 + jZ), |F(Z)|2 = 25e4/(1 + Z2)
W1:
1 f
2
=
F(Z) dZ
³
S 0
25e 4
S
³
f
0
1
dZ
1 Z2
= 12.5e4 = 682.5 J
or
W1: =
³
f
f
f 2 ( t ) dt
f
25e 4 ³ e 2 t dt
0
Chapter 18, Solution 56.
W1: =
But,
³
f
f
f 2 ( t ) dt
³
f
0
e 2 t sin 2 (2 t ) dt
sin2(A) = 0.5(1 –– cos(2A))
f
25e 4
tan 1 (Z)
S
0
12.5e4 = 682.5 J
W1: =
³
f
0
e
2 t
0.5[1 cos(4 t )]dt
f
1 e 2 t
2 2
0
f
e 2 t
[2 cos(4 t ) 4 sin(4t )]
4 16
0
= (1/4) + (1/20)(––2) = 0.15 J
Chapter 18, Solution 57.
W1: =
or
W1: =
³
f
f
i 2 ( t ) dt
³
0
4e 2 t dt
f
2e 2 t
0
f
= 2J
I(Z) = 2/(1 –– jZ), |I(Z)|2 = 4/(1 + Z2)
1 f
2
I(Z) dZ
³
f
2S
4 f
1
dZ
³
f
2S
(1 Z 2 )
f
4
tan 1 (Z)
S
0
4S
= 2J
S2
In the frequency range, ––5 < Z < 5,
5
4
W =
tan 1 Z
S
0
4
tan 1 (5)
S
4
(1.373) = 1.7487
S
W/ W1: = 1.7487/2 = 0.8743 or 87.43%
Chapter 18, Solution 58.
Zm = 200S = 2Sfm which leads to fm = 100 Hz
(a)
Zc = Sx104 = 2Sfc which leads to fc = 104/2 = 5 kHz
(b)
lsb = fc –– fm = 5,000 –– 100 = 4,900 Hz
(c)
usb = fc + fm = 5,000 + 100 = 5,100 Hz
Chapter 18, Solution 59.
H(Z)
Vo (Z)
Vi (Z)
10
6
2 jZ 4 j Z
2
5
3
2 jZ 4 jZ
§ 5
3 · 4
¸¸
¨¨
© 2 jZ 4 jZ ¹ 1 jZ
20
12
, s jZ
(s 1)(s 2) (s 1)(s 4)
Vo (Z)
H(Z)Vi (Z)
Using partial fraction,
Vo (Z)
A
B
C
D
s 1 s 2 s 1 s 4
16
20
4
1 jZ 2 jZ 4 jZ
v o (t)
16e t 20e 2 t 4e 4 t u ( t ) V
Thus,
Chapter 18, Solution 60.
2
+
Is
1/jȦ
jȦ
V
V
jZI s
1
jZ
1
2 jZ
jZ
jZI s
1 Z 2 j2Z
Since the voltage appears across the inductor, there is no DC component.
V1
2S‘90q8
50.27‘90q
38.48 j12.566
1 4S 2 j4S
V2
4S‘90q5
2
1 16S j8S
1.2418‘ 71.92q
62.83‘90q
156.91 j25.13
0.3954‘ 80.9q
v( t ) 1.2418 cos( 2St 41.92q) 0.3954 cos( 4St 129.1q) mV
Chapter 18, Solution 61.
lsb = 8,000,000 –– 5,000 = 7,995,000 Hz
usb = 8,000,000 + 5,000 = 8,005,000 Hz
Chapter 18, Solution 62.
For the lower sideband, the frequencies range from
10,000 –– 3,500 Hz = 6,500 Hz to 10,000 –– 400 Hz = 9,600 Hz
For the upper sideband, the frequencies range from
10,000 + 400 Hz = 10,400 Hz to 10,000 + 3,500 Hz = 13,500 Hz
Chapter 18, Solution 63.
Since fn = 5 kHz, 2fn = 10 kHz
i.e. the stations must be spaced 10 kHz apart to avoid interference.
'f = 1600 –– 540 = 1060 kHz
The number of stations = 'f /10 kHz = 106 stations
Chapter 18, Solution 64.
'f = 108 –– 88 MHz = 20 MHz
The number of stations = 20 MHz/0.2 MHz = 100 stations
Chapter 18, Solution 65.
Z = 3.4 kHz
fs = 2Z = 6.8 kHz
Chapter 18, Solution 66.
Z = 4.5 MHz
fc = 2Z = 9 MHz
Ts = 1/fc = 1/(9x106) = 1.11x10––7 = 111 ns
Chapter 18, Solution 67.
We first find the Fourier transform of g(t). We use the results of Example 17.2 in
conjunction with the duality property. Let Arect(t) be a rectangular pulse of height A and
width T as shown below.
Arect(t) transforms to Atsinc(Z2/2)
f(t)
F(Z)
A
Z
t
––T/2
T/2
G(Z)
Z
––Zm/2
According to the duality property,
AWsinc(Wt/2)
becomes 2SArect(W)
g(t) = sinc(200St) becomes 2SArect(W)
where AW = 1 and W/2 = 200S or T = 400S
i.e. the upper frequency Zu = 400S = 2Sfu or fu = 200 Hz
The Nyquist rate = fs = 200 Hz
The Nyquist interval = 1/fs = 1/200 = 5 ms
Zm/2
Chapter 18, Solution 68.
The total energy is
WT =
³
f
f
v 2 ( t ) dt
Since v(t) is an even function,
WT =
³
f
0
2500e
4 t
dt
e 4 t
5000
4
f
= 1250 J
0
V(Z) = 50x4/(4 + Z2)
1 5
| V(Z) | 2 dZ
2S ³1
W =
But
³ (a
2
1
dx
x 2 )2
1 5 (200) 2
dZ
2S ³1 (4 Z 2 ) 2
1 ª x
1
º
tan 1 ( x / a )» C
2 « 2
2
a
2a ¬ x a
¼
5
2x10 4 1 ª Z
1
º
W =
tan 1 (Z / 2)»
2
«
S 8 ¬4 Z
2
¼1
= (2500/S)[(5/29) + 0.5tan-1(5/2) –– (1/5) –– 0.5tan––1(1/2) = 267.19
W/WT = 267.19/1250 = 0.2137 or 21.37%
Chapter 18, Solution 69.
The total energy is
WT =
=
W =
1 f
2
F(Z) dZ
³
f
2S
>
1 f 400
dZ
2S ³f 4 2 Z 2
400
(1 / 4) tan 1 (Z / 4)
S
@
1 2
2
F(Z) dZ
³
0
2S
>
f
0
100 S
= 50
S 2
400
(1 / 4) tan 1 (Z / 4)
2S
@
2
0
= [100/(2S)]tan––1(2) = (50/S)(1.107) = 17.6187
W/WT = 17.6187/50 = 0.3524 or 35.24%
Chapter 19, Solution 1.
To get z 11 and z 21 , consider the circuit in Fig. (a).
1:
4:
I2 = 0
+
I1
6:
V1
Io
+
2:
V2
(a)
z 11
V1
I1
Io
1
I ,
2 1
z 21
V2
I1
1 6 || (4 2)
4:
V2
2Io
I1
1:
To get z 22 and z 12 , consider the circuit in Fig. (b).
I1 = 0
1:
4:
Io '
+
+
6:
V1
2:
V2
(b)
z 22
V2
I2
Io'
2
I
2 10 2
z 12
V1
I2
2 || (4 6) 1.667 :
1:
1
I ,
6 2
V1
6Io ' I2
I2
Hence,
ª4
1 º
[z ] «
»:
¬ 1 1.667 ¼
Chapter 19, Solution 2.
Consider the circuit in Fig. (a) to get z 11 and z 21 .
1:
Io '
1:
1:
1:
+
I1
1:
V1
1:
I2 = 0
Io
+
1:
V2
1:
1:
1:
1:
(a)
z 11
V1
I1
z 11
§
3·
2 1 || ¨ 2 ¸
©
4¹
Io
1
Io'
1 3
Io'
1
I
1 11 4 1
Io
1 4
˜ I
4 15 1
V2
Io
z 21
V2
I1
2 1 || [ 2 1 || (2 1) ]
2
(1)(11 4)
1 11 4
1 '
I
4 o
4
I
15 1
1
I
15 1
1
I
15 1
1
15
z 12
0.06667
2
11
15
2.733
To get z 22 , consider the circuit in Fig. (b).
I1 = 0
1:
1:
1:
1:
+
+
1:
V1
1:
1:
V2
1:
1:
1:
1:
(b)
z 22
V2
I2
2 1 || (2 1 || 3)
z 11
2.733
Thus,
ª 2.733 0.06667 º
[z ] «
»:
¬ 0.06667 2.733 ¼
Chapter 19, Solution 3.
(a)
To find z 11 and z 21 , consider the circuit in Fig. (a).
-j :
Io
+
I1
I2 = 0
j:
V1
1:
V1
I1
V2
(a)
z 11
+
j || (1 j)
By current division,
j
Io
I
j 1 j 1
j I1
j (1 j)
1 j
j 1 j
I2
V2
Io
z 21
V2
I1
jI 1
j
To get z 22 and z 12 , consider the circuit in Fig. (b).
-j :
I1 = 0
+
+
j:
V1
1:
I2
V2
(b)
z 22
V2
I2
V1
jI 2
z 12
V1
I2
1 || ( j j)
0
j
Thus,
ª 1 j j º
[z ] «
:
0 »¼
¬ j
(b)
To find z 11 and z 21 , consider the circuit in Fig. (c).
j:
-j :
+
I1
+
1:
V1
I2 = 0
V2
-j :
1:
(c)
z 11
V1
I1
V2
(1.5 j0.5) I 1
z 21
V2
I1
j 1 1 || (-j) 1 j -j
1.5 j0.5
1 j
1.5 j0.5
To get z 22 and z 12 , consider the circuit in Fig. (d).
I1 = 0
j:
-j :
+
+
1:
V1
V2
-j :
1:
(d)
z 22
V2
I2
-j 1 1 || (-j) 1.5 - j1.5
V1
(1.5 j0.5) I 2
z 12
V1
I2
1.5 j0.5
Thus,
ª 1.5 j0.5 1.5 j0.5 º
[z ] «
»:
¬ 1.5 j0.5 1.5 j1.5 ¼
Chapter 19, Solution 4.
Transform the 3 network to a T network.
Z1
Z3
Z2
I2
Z1
(12)( j10)
12 j10 j5
Z2
- j60
12 j5
Z3
50
12 j5
j120
12 j5
The z parameters are
Z2
(-j60)(12 - j5)
144 25
z 12
z 21
z 11
Z1 z 12
( j120)(12 j5)
z 12
169
1.775 j4.26
z 22
Z 3 z 21
(50)(12 j5)
z 21
169
1.7758 j5.739
-1.775 - j4.26
Thus,
ª 1.775 j4.26 - 1.775 j4.26 º
[z ] «
»:
¬ - 1.775 j4.26 1.775 j5.739 ¼
Chapter 19, Solution 5.
Consider the circuit in Fig. (a).
1
s
I2 = 0
Io
+
I1
V1
1
1/s
1/s
(a)
1 §
1·
1 || || ¨1 s ¸
s ©
s¹
V2
z 11
+
1
s
§
1·
|| ¨1 s ¸
1 ©
s¹
1
s
§ 1 ·§
1·
¨
¸¨1 s ¸
© s 1 ¹©
s¹
§ 1 ·
1
¨
¸ 1 s © s 1¹
s
z 11
s2 s 1
s 3 2s 2 3s 1
1 ||
Io
Io
1
s
1
1
1 || 1 s s
s
I1
s
s 2s 3s 1
3
2
1
s 1
1
1
1 s s 1
s
I1
s
s 1
s
s2 s 1
s 1
I1
I1
V2
1
Io
s
I1
s 2s 2 3s 1
z 21
V2
I1
s 2s 3s 1
3
1
3
2
Consider the circuit in Fig. (b).
s
1
I1 = 0
+
+
1
V1
1/s
1/s
(b)
Hence,
V2
1 §
1·
|| ¨1 s 1 || ¸
s ©
s¹
z 22
V2
I2
z 22
§ 1 ·§
1 ·
¨ ¸¨1 s ¸
© s ¹©
s 1¹
1
1
1 s s
s 1
z 22
s 2 2s 2
s 3 2s 2 3s 1
z 12
z 21
1 §
1 ·
¸
|| ¨1 s s ©
s 1¹
1
s 1
s
1 s s2 s 1
1 s I2
ª
º
s2 s 1
1
« s 3 2s 2 3s 1 s 3 2s 2 3s 1 »
»
[z ] «
1
s 2 2s 2
«
»
¬ s 3 2s 2 3s 1 s 3 2s 2 3s 1 ¼
Chapter 19, Solution 6.
To find z 11 and z 21 , connect a voltage source V1 to the input and leave the output
open as in Fig. (a).
I1
10 :
Vo
20 :
+
+
V1
30 :
0.5 V2
V2
(a)
V1 Vo
10
0.5 V2 Vo
,
50
V1
§3 · V
Vo 5 ¨ Vo ¸ o
©5 ¹ 5
I1
V1 Vo
10
3 .2
V
10 o
30
V
20 30 o
where V2
4.2 Vo
0.32 Vo
z 11
V1
I1
4.2 Vo
0.32 Vo
13.125 :
z 21
V2
I1
0.6 Vo
0.32 Vo
1.875 :
To obtain z 22 and z 12 , use the circuit in Fig. (b).
10 :
20 :
I2
+
V1
0.5 V2
(b)
30 :
+
V2
3
V
5 o
I2
V2
0.5333 V2
30
1
1.875 :
0.5333
0.5 V2 V2
I2
z 22
V1
V2 (20)(0.5 V2 )
z 12
V1
I2
- 9 V2
0.5333 V2
-9 V2
-16.875 :
Thus,
ª 13.125 - 16.875 º
[z ] «
:
1.875 »¼
¬ 1.875
Chapter 19, Solution 7.
To get z11 and z21, we consider the circuit below.
20 :
I1
I2=0
100 :
+
+
vx
V1
50 :
60 :
-
-
12vx
+
V1 Vx
20
I1
Vx Vx 12Vx
50
160
V1 Vx
20
81 V1
( )
121 20

o
o
Vx
z11
V1
I1
40
V1
121
29.88
+
V2
-
V2
13Vx
) 12Vx
160
57
57 40
Vx (
)V1
8
8 121
V2
70.37 I1 
o z 21
70.37
I1
60(
To get z12 and z22, we consider the circuit below.
20 :
I1=0
57 40 20x121
(
)
I1
8 121
81
I2
100 :
+
+
50 :
vx
V1
+
60 :
-
-
V2
12vx
-
+
Vx
50
V2
100 50
1
V2 ,
3
z 22
V2
I2
11.11
V1
Vx
1 / 0.09
1
V2
3
11.11
I2
3
I2
V2 V2 12Vx
150
60
3.704I 2

o
Thus,
[z]
ª 29.88 3.704º
« 70.37 11.11» :
¬
¼
z12
0.09V2
V1
I2
3.704
Chapter 19, Solution 8.
To get z11 and z21, consider the circuit below.
j4 :
I1 -j2 :
x
5:
I2 =0
x
j6 :
+
j8 :
+
V2
V1
10 :
-
-
V1
(10 j2 j6)I1
V2
10I1 j4I1

o

o
V1
I1
z11
z 21
V2
I1
10 j4
(10 j4)
To get z22 and z12, consider the circuit below.
j4 :
I1=0 -j2 :
x
j6 :
5:
x
j8 :
I2
+
+
V2
V1
10 :
-
-
V2
(5 10 j8)I 2
V1
(10 j4)I 2

o

o
V2
I2
z 22
z12
V1
I2
15 j8
(10 j4)
Thus,
[z]
ª (10 j4) (10 j4)º
« (10 j4) (15 j8) » :
¬
¼
Chapter 19, Solution 9.
It is evident from Fig. 19.5 that a T network is appropriate for realizing the z
parameters.
6:
R2
R1
2:
4:
R3
R1
z 11 z 12
10 4
6:
R2
z 22 z 12
64
2:
R3
z 12
4:
z 21
Chapter 19, Solution 10.
(a)
This is a non-reciprocal circuit so that the two-port looks like the one
shown in Figs. (a) and (b).
I1
z11
z22
+
+
V1
I2
z12 I2
+
+
z21 I1
V2
(a)
(b)
This is a reciprocal network and the two-port look like the one shown in
Figs. (c) and (d).
z11 –– z12
I1
z22 –– z12
I2
+
+
z12
V1
V2
(c)
25 :
I1
10 :
I2
+
+
+
20 I2
V1
+
5 I1
V2
(b)
z 11 z 12
1
z 22 z 12
2s
2
s
1
1
0.5 s
1
s
z 12
I1
1:
0.5 F
2H
+
I2
+
1F
V1
V2
(d)
Chapter 19, Solution 11.
This is a reciprocal network, as shown below.
1+j5
3+j
1:
j5 :
3:
j1 :
5:
5-j2
-j2 :
Chapter 19, Solution 12.
This is a reciprocal two-port so that it can be represented by the circuit in Figs. (a)
and (b).
I1
z11 –– z12
z22 –– z12
I2
+
+
z12
V1
V2
2:
(a)
I1
8:
2:
+
+
4:
V1
V2
(b)
From Fig. (b),
V1
I2
(8 4 || 4) I 1
10 I 1
Io
2:
By current division,
Io
V2
V1
1
I ,
2 1
I1
10 I 1
V2
2Io
I1
0 .1
Chapter 19, Solution 13.
This is a reciprocal two-port so that the circuit can be represented by the circuit below.
40 :
120‘0q V
rms
+
I1
50 :
10 I2
20 :
+
+
30 I1
We apply mesh analysis.
For mesh 1,
- 120 90 I 1 10 I 2 0 
o 12 9 I 1 I 2
For mesh 2,
30 I 1 120 I 2 0 
o I 1 -4 I 2
Substituting (2) into (1),
- 12
12 -36 I 2 I 2 -35 I 2 
o I 2
35
P
1
2
I2 R
2
2
1 § 12 ·
¨ ¸ (100)
2 © 35 ¹
5.877 W
I2
100 :
(1)
(2)
Chapter 19, Solution 14.
To find Z Th , consider the circuit in Fig. (a).
I2
I1
+
ZS
+
V1
Vo = 1
(a)
V1
V2
z 11 I 1 z 12 I 2
z 21 I 1 z 22 I 2
V2
1,
(1)
(2)
But
Hence,
V1
- Z s I1
(z 11 Z s ) I 1 z 12 I 2
0

o I 1
- z 12
I
z 11 Z s 2
§ - z 21 z 12
·
1 ¨
z 22 ¸ I 2
© z 11 Z s
¹
Z Th
V2
I2
1
I2
z 22 z 21 z 12
z 11 Z s
To find VTh , consider the circuit in Fig. (b).
ZS
+
VS
I1
I2 = 0
+
+
V1
V2 = VTh
(b)
I2
0,
V1
Vs I 1 Z s
Substituting these into (1) and (2),
Vs I 1 Z s
V2
z 21 I 1
VTh
V2
Vs
z 11 Z s

o I 1
z 11 I 1
z 21 Vs
z 11 Z s
z 21 Vs
z 11 Z s
Chapter 19, Solution 15.
(a) From Prob. 18.12,
z 22 ZTh
(b) VTh
z12z 21
z11 Zs
z 21
Vs
z11 Zs
120 80x 60
40 10
ZL
ZTh
80
(120)
40 10
Pmax
24
24:
192
V 2Th
8R Th
192 2
8x 24
192 W
Chapter 19, Solution 16.
As a reciprocal two-port, the given circuit can be represented as shown in Fig. (a).
5:
15‘0q V
10 –– j6 :
+
4 –– j6 :
a
j6 :
(a)
j4 :
b
At terminals a-b,
Z Th
(4 j6) j6 || (5 10 j6)
j6 (15 j6)
15
Z Th
4 j6 4 j6 2.4 j6
Z Th
6.4 :
VTh
j6
(15‘0q)
j6 5 10 j6
j6
6‘90q V
The Thevenin equivalent circuit is shown in Fig. (b).
6.4 :
+
+
6‘90q V
Vo
j4 :
(b)
From this,
Vo
v o (t)
j4
( j6)
6.4 j4
3.18‘148q
3.18 cos( 2t 148q) V
Chapter 19, Solution 17.
To obtain z 11 and z 21 , consider the circuit in Fig. (a).
4:
+
I1
V1
Io '
2:
Io
I2 = 0
+
V2
8:
6:
(a)
In this case, the 4-: and 8-: resistors are in series, since the same current, I o , passes
through them. Similarly, the 2-: and 6-: resistors are in series, since the same current,
I o ' , passes through them.
V1
I1
z 11
Io
But
(12)(8)
20
(4 8) || (2 6) 12 || 8
8
I
8 12 1
2
I
5 1
- V2 4 I o 2 I o '
V2
-4 I o 2 I o '
z 21
V2
I1
-2
5
4 .8 :
3
I
5 1
Io'
0
-8
6
I1 I1
5
5
-2
I
5 1
-0.4 :
To get z 22 and z 12 , consider the circuit in Fig. (b).
4:
I1 = 0
+
+
2:
V1
V2
8:
6:
(b)
z 22
V2
I2
(4 2) || (8 6)
z12
z 21
-0.4 :
6 || 14
(6)(14)
20
Thus,
ª 4.8 - 0.4 º
[z ] «
»:
¬ - 0.4 4.2 ¼
We may take advantage of Table 18.1 to get [y] from [z].
' z (4.8)(4.2) (0.4) 2 20
4 .2 :
I2
y 11
z 22
'z
y 21
- z 21
'z
[y ]
ª 0.21 0.02 º
« 0.02 0.24 » S
¼
¬
4 .2
20
0 .4
20
0.21
0.02
y 12
- z 12
'z
y 22
z 11
'z
0 .4
20
4 .8
20
0.02
0.24
Thus,
Chapter 19, Solution 18.
To get y 11 and y 21 , consider the circuit in Fig.(a).
6:
I1
3:
I2
+
V1
+
6:
3:
V2 = 0
(a)
V1
(6 6 || 3) I 1
y 11
I1
V1
I2
y 21
1
8
-6
I
63 1
I2
V1
8 I1
- 2 V1
3 8
- V1
12
-1
12
To get y 22 and y 12 , consider the circuit in Fig.(b).
I1
6:
Io
3:
I2
+
V1 = 0
6:
3:
(b)
+
V2
y 22
I2
V2
I1
- Io
,
2
I1
- I2
6
1
3 || (3 6 || 6)
1
3 || 6
Io
§ - 1 ·§ 1 ·
¨ ¸¨ V2 ¸
© 6 ¹© 2 ¹
y 12
I1
V2
[y ]
ª 1 -1 º
« 8 12 »
«
»S
« -1 1 »
¬ 12 2 ¼
-1
12
1
2
3
I
3 6 2
1
I
3 2
- V2
12
y 21
Thus,
Chapter 19, Solution 19.
Consider the circuit in Fig.(a) for calculating y 11 and y 21 .
1
I1
I2
+
V1
+
1/s
s
1
(a)
V1
y 11
I2
§1 ·
¨ || 2 ¸ I 1
©s ¹
I1
V1
2s
I
2 (1 s) 1
2s 1
2
(- 1 s )
I
(1 s) 2 1
2
I
2s 1 1
s 0 .5
- I1
2s 1
- V1
2
V2 = 0
y 21
I2
V1
-0.5
To get y 22 and y 12 , refer to the circuit in Fig.(b).
1
I1
I2
+
1/s
V1 = 0
s
1
(b)
V2
(s || 2) I 2
y 22
I2
V2
I1
2s
I
s2 2
s2
2s
-s
I
s2 2
0 .5 1
s
-s s 2
˜
V
s 2 2s 2
y 12
I1
V2
[y ]
ª s 0.5
- 0.5 º
« - 0.5 0.5 1 s » S
¼
¬
-0.5
Thus,
Chapter 19, Solution 20.
To get y11 and y21, consider the circuit below.
- V2
2
+
V2
3ix
2:
I1
I2
+
V1
I1
+
ix
4:
6:
V2 =0
-
-
Since 6-ohm resistor is short-circuited, ix = 0
V1
I1(4 // 2)
I2
8
I1
6
4
I1
42
o
2 6
( V1)
3 8
y11
1
V1
2
I1
V1
0.75
o
y 21
I2
V1
0.5
To get y22 and y12, consider the circuit below.
3ix
2:
I1
+
V1=0
ix
4:
-
+
6 : V2
-
I2
ix
V2
,
6
I1
V
3i x 2
2
V
i x 3i x 2
2
I2
0

o
V2
6
o
I1
V2
y12
y 22
I2
V2
1
6
0.1667
0
Thus,
[ y]
0 º
ª 0.75
« 0.5 0.1667 » S
¬
¼
Chapter 19, Solution 21.
To get y 11 and y 21 , refer to Fig. (a).
I1
0.2 V1
V1
I2
+
+
V1
5:
10 :
V2 = 0
(a)
At node 1,
I1
V1
0.2 V1
5
I2
-0.2 V1
0.4 V1

o y 11
I2
V1

o y 21
I1
V1
0 .4
-0.2
To get y 22 and y 12 , refer to the circuit in Fig. (b).
I1
0.2 V1
V1
I2
+
V1 = 0
5:
10 :
+
V2
(b)
Since V1
0 , the dependent current source can be replaced with an open circuit.
I2
V2

o y 22
V2
10 I 2
y 12
I1
V2
[y ]
ª 0.4
0º
« - 0.2 0.1» S
¬
¼
1
10
0 .1
0
Thus,
Consequently, the y parameter equivalent circuit is shown in Fig. (c).
I1
I2
+
+
0.2 V1
0.4 S
V1
0.1 S
V2
(c)
Chapter 19, Solution 22.
(a)
To get y 11 and y 21 refer to the circuit in Fig. (a).
I1
2:
V1
Vo
3:
+
+
V1
+
I2
1:
Vx
Vx/2
V2 = 0
(a)
At node 1,
I1
V1 V1 Vo
1
2
At node 2,
V1 Vo V1
2
2
Vo
3

o I 1
1.5 V1 0.5 Vo

o 1.2 V1
Substituting (2) into (1) gives,
Vo
(1)
(2)
I1
1.5 V1 0.6 V1
0.9 V1
I2
- Vo
3

o y 21
-0.4 V1
I1
V1

o y 11
I2
V1
0 .9
-0.4
To get y 22 and y 12 refer to the circuit in Fig. (b).
I1
2:
V1
+
+
V1 = 0
Vx
3:
1:
Vx/2
I2
+
V2
(b)
Vx V1 0 so that the dependent current source can be replaced by an
open circuit.
I2 1
V2 (3 2 0) I 2 5 I 2 
o y 22
0 .2
V2 5
I1
I 1 - I 2 -0.2 V2 
o y 12
-0.2
V2
Thus,
ª 0.9 - 0.2 º
[y ] «
»S
¬ - 0.4 0.2 ¼
(b)
To get y 11 and y 21 refer to Fig. (c).
j:
Io '
I1
Io
1:
1:
Io''
I2
+
V1
+
-j :
V2 = 0
Zin
Z in
j || (1 1 || -j)
(c)
§
-j ·
¸
j || ¨1 © 1 j¹
j || (1.5 j0.5)
j (1.5 j0.5)
1.5 j0.5
0.6 j0.8
I1
V1

o y 11
1
Z in
V1
Z in I 1
Io
j
I ,
1.5 j0.5 1
I o ''
-j
I
1 j o
- I2
Io Io
- I2
(1.2 j0.4)(0.6 j0.8) V1
y 21
I2
V1
'
I1
2 j
(1.5 j0.5) 2
2 j
I1 I
2 .5
5 1
-0.4 j1.2
0.6 j0.8
1.5 j0.5
I
1.5 j0.5 1
Io'
I1
(1 j)(1.5 j0.5)
''
1
0.6 j0.8
(1.2 j0.4) I 1
(0.4 j1.2) V1
y 12
To get y 22 refer to the circuit in Fig.(d).
j:
1:
I1
1:
I2
+
+
-j :
V1 = 0
(d)
Z out
y 22
j || (1 1 || - j)
1
Z out
0.6 j0.8
0.6 j0.8
Thus,
[y ]
ª 0.6 j0.8 - 0.4 j1.2 º
« - 0.4 j1.2 0.6 j0.8 » S
¬
¼
Zout
V2
Chapter 19, Solution 23.
(a)
y12
1 //
y11 y12
1
s
1
s 1

o
1
y 22 y12

o
y11

o
s
[ y]
y12
1 y12
y 22
1
s y12
ªs 2
« s 1
«
« 1
«¬ s 1
1
s 1
1
s 1
s
s2
s 1
1
s 1
s 2 s 1
s 1
1 º
s 1 »
»
s 2 s 1»
s 1 »¼
(b) Consider the network below.
I1
1
+
+
Vs
-
V1
-
I2
+
[y]
V2
-
2
Vs
I1 V1
(1)
V2
2I 2
(2)
I1
y11V1 y12 V2
(3)
I2
y 21V1 y 22 V2
(4)
From (1) and (3)
Vs V1
y11V1 y12 V2

o
(1 y11 )V1 y12 V2
Vs
(5)
From (2) and (4),
0.5V2
y 21V1 y 22 V2
o
V1
1
(0.5 y 22 )V2
y 21
(6)
Substituting (6) into (5),
Vs
2
s
(1 y11)(0.5 y 22 )
V2 y12V2
y 21

o
V2
2/s
ª
º
1
(1 y11)(0.5 y 22 )»
« y12 y 21
¬
¼
2(s 1)
2/s
V2
2
1
§ 2s 3 ·§¨ 1 s s 1 ·¸
(s 1)¨
¸ s 1
s 1 ¸¹
© s 1 ¹¨© 2
s(2s3 6s 2 7.5s 3.5)
Chapter 19, Solution 24.
Since this is a reciprocal network, a 3 network is appropriate, as shown below.
Y2
Y1
Y3
(a)
4:
1/4 S
1/4 S
1/8 S
(b)
4:
8:
(c)
Y1
y 11 y 12
Y2
- y 12
Y3
y 22 y 21
1 1
2 4
1
S,
4
1
S,
4
3 1
8 4
1
S,
8
Z1
4:
Z2
4:
Z3
8:
Chapter 19, Solution 25.
This is a reciprocal network and is shown below.
0.5 S
0.5S
1S
Chapter 19, Solution 26.
To get y 11 and y 21 , consider the circuit in Fig. (a).
4:
2:
V1
+
1
2
+
Vx
1:
2 Vx
I2
+
V2 = 0
(a)
At node 1,
V1 Vx
2 Vx
2
But
I1
V1 Vx
2
Also,
I2 Vx
4
y 21
Vx Vx
1
4
V1 2 V1
2
2 Vx
I2
V1

o 2 V1
1.5 V1

o I 2

o y 11
1.75 Vx
(1)
-Vx
I1
V1
1 .5
-3.5 V1
-3.5
To get y 22 and y 12 , consider the circuit in Fig.(b).
4:
2:
1
2
+
I1
1:
Vx
2 Vx
I2
+
V2
(b)
At node 2,
I2
2 Vx V2 Vx
4
(2)
At node 1,
2 Vx V2 Vx
4
Substituting (3) into (2) gives
1
I 2 2 Vx Vx
2
y 22
I2
V2
-1.5
I1
- Vx
2
V2
2
Vx Vx
2
1
1.5 Vx
3
V
2 x

o V2
-1.5 V2

o y 12
I1
V2
0 .5
-Vx
(3)
Thus,
ª 1.5 0.5 º
« - 3.5 - 1.5 » S
¬
¼
[y ]
Chapter 19, Solution 27.
Consider the circuit in Fig. (a).
4:
I1
I2
+
+
V1
0.1 V2
+
10 :
20 I1
V2 = 0
(a)
V1
4 I1
I2
20 I 1

o y 11
5 V1
I1
V1
I1
4 I1

o y 21
0.25
I2
V1
5
Consider the circuit in Fig. (b).
I1
4:
I2
+
0.1 V2
V1 = 0
+
10 :
+
V2

o y 22
I2
V2
0 .6
20 I1
(b)
4 I1
I2
0.1 V2
20 I 1 
o y 12
V2
10
I1
V2
0.5 V2 0.1 V2
0 .1
4
0.025
0.6 V2
Thus,
[y ]
ª 0.25 0.025 º
S
« 5
0.6 »¼
¬
Alternatively, from the given circuit,
V1 4 I 1 0.1 V2
I 2 20 I 1 0.1 V2
Comparing these with the equations for the h parameters show that
h 11 4 ,
h 12 -0.1,
h 21 20 ,
h 22 0.1
Using Table 18.1,
y 11
1
h11
1
4
y 21
h 21
h 11
20
4
0.25 ,
y 12
- h 12
h 11
0 .1
4
5,
y 22
'h
h 11
0 .4 2
4
0.025
0 .6
as above.
Chapter 19, Solution 28.
We obtain y 11 and y 21 by considering the circuit in Fig.(a).
1:
4:
I2
+
I1
+
6:
V1
(a)
1 6 || 4
y 11
I1
V1
I2
-6
I
10 1
1
Z in
V2 = 0
Z in
2:
3 .4
0.2941
§ - 6 ·§ V1 ·
¨ ¸¨ ¸
© 10 ¹© 3.4 ¹
-6
V
34 1
y 21
I2
V1
-6
34
-0.1765
To get y 22 and y 12 , consider the circuit in Fig. (b).
I1
1:
4:
Io
+
+
6:
V1 = 0
2:
I2
V2
(b)
6·
§
2 || ¨ 4 ¸
7¹
©
1
y 22
2 || (4 6 || 1)
y 22
24
34
I1
-6
I
7 o
Io
I1
-6
V
34 2

o y 12
[y ]
ª 0.2941 - 0.1765º
« - 0.1765 0.7059 » S
¬
¼
(2)(34 7)
2 (34 7)
34
24
V2
I2
0.7059
2
I
2 (34 7) 2
I1
V2
-6
34
14
I
48 2
7
V
34 2
-0.1765
Thus,
The equivalent circuit is shown in Fig. (c). After transforming the current source to a
voltage source, we have the circuit in Fig. (d).
6/34 S
1A
4/34 S
(c)
18/34 S
2:
8.5 :
5.667 :
+
+
8.5 V
1.889 :
V
2:
(d)
V
(2 || 1.889)(8.5)
2 || 1.889 8.5 5.667
P
V2
R
(0.5454) 2
2
(0.9714)(8.5)
0.9714 14.167
0.5454
0.1487 W
Chapter 19, Solution 29.
(a)
Transforming the ' subnetwork to Y gives the circuit in Fig. (a).
1:
1:
Vo
+
10 A
+
2:
V1
-4 A
V2
(a)
It is easy to get the z parameters
z 12 z 21 2 , z 11 1 2
'z
z 11 z 22 z 12 z 21
y 11
z 22
'z
3
5
3,
94
3
y 21
- z 12
'z
5
y 12
y 22 ,
z 22
-2
5
Thus, the equivalent circuit is as shown in Fig. (b).
2/5 S
I1
I2
+
+
10 A
V1
1/5 S
1/5 S
V2
(b)
-4 A
I1
10
3
2
V1 V2
5
5
I2
-4
-2
3
V1 V2
5
5
10
V1 1.5 V2

o 50

o - 20

o V1
Substituting (2) into (1),
50 30 4.5 V2 2 V2
V1
(b)
10 1.5 V2
3 V1 2 V2
(1)
-2 V1 3 V2
10 1.5 V2

o V2
(2)
8V
22 V
For direct circuit analysis, consider the circuit in Fig. (a).
For the main non-reference node,
Vo
10 4

o Vo 12
2
10
V1 Vo
1

o V1
10 Vo
-4
V2 Vo
1

o V2
Vo 4
22 V
8V
Chapter 19, Solution 30.
(a)
Convert to z parameters; then, convert to h parameters using Table 18.1.
z 11 z 12 z 21 60 : ,
z 22 100 :
'z
z 11 z 22 z 12 z 21
h 11
'z
z 22
h 21
- z 21
z 22
2400
100
-0.6 ,
6000 3600
24 ,
2400
h 12
z 12
z 22
60
100
h 22
1
z 22
0.01
0 .6
Thus,
[h]
(b)
Similarly,
z 11
ª 24 :
0.6 º
« - 0.6 0.01 S »
¼
¬
30 :
'z
600 400
h11
200
20
h 21
-1
[h]
ª 10 :
1 º
« - 1 0.05 S »
¼
¬
z 12
z 21
z 22
h12
20
20
1
h 22
1
20
0.05
20 :
200
10
Thus,
Chapter 19, Solution 31.
We get h11 and h 21 by considering the circuit in Fig. (a).
1:
2:
V3
V4
1:
I2
+
I1
2:
V1
4 I1
(a)
At node 1,
I1
V3 V3 V4
2
2

o 2 I 1
2 V3 V4
(1)
-2 V3 6 V4
(2)
At node 2,
V3 V4
4 I1
2
8 I1
-V3 3 V4
V4
1

o 16 I 1
Adding (1) and (2),
18 I 1 5 V4 
o V4
V3 3 V4 8 I 1 2.8 I 1
V1 V3 I 1 3.8 I 1
h11
V1
I1
I2
- V4
1
3.6 I 1
3 .8 :
-3.6 I 1
I2
I1

o h 21
-3.6
To get h 22 and h12 , refer to the circuit in Fig. (b). The dependent current source can be
replaced by an open circuit since 4 I 1 0 .
1:
I1
1:
2:
I2
+
2:
V1
+
4 I1 = 0
V2
(b)
V1
2
V
2 2 1 2
2
V
5 2
I2
V2
2 2 1
[h]
ª 38 : 0.4 º
« - 3.6 0.2 S »
¼
¬
V2
5

o h12

o h 22
I2
V2
V1
V2
0 .4
1
5
0 .2 S
Thus,
Chapter 19, Solution 32.
(a)
We obtain h11 and h 21 by referring to the circuit in Fig. (a).
1
s
s
+
I1
I2
+
1/s
V1
V2 = 0
(a)
§
1·
¨1 s s || ¸ I 1
©
s¹
V1
V1
I1
h11
s 1
By current division,
-1 s
I2
I
s 1 s 1
§
s ·
¨1 s 2
¸I
©
s 1¹ 1
s
s 1
2
- I1

o h 21
s 1
I2
I1
-1
s 1
2
To get h 22 and h12 , refer to Fig. (b).
I1 = 0
1
s
s
I2
+
+
1/s
V1
V2
(b)
V1
1s
V
s 1 s 2
V2

o h12
s 1
V2
§ 1·
¨s ¸ I 2
© s¹

o h 22
[h]
ª
s
« s 1 s2 1
«
-1
«¬
2
s 1
V1
V2
2
I2
V2
1
s 1
2
1
s 1 s
s
s 1
s
I2 = 0
2
Thus,
(b)
1 º
s 1 »
s »
»
s2 1 ¼
2
To get g11 and g 21 , refer to Fig. (c).
I1
1
s
+
V1
+
1/s
V2
(c)
V1
§
1·
¨1 s ¸ I 1
©
s¹
V2
1s
V
1 s 1 s 1
I1
V1

o g 11
1
1 s 1 s
V1

o g 21
s s 1
2
s
s s 1
2
V2
V1
1
s s 1
2
To get g 22 and g 12 , refer to Fig. (d).
I1
1
s
s
+
I2
+
1/s
V1 = 0
I2
V2
(d)
V2
g 22
I1
§ 1
·
¨s || (s 1) ¸ I 2
© s
¹
V2
I2
s
§
(s 1) s ·
¸I 2
¨s © 1 s 1 s ¹
s 1
s s 1
-1 s
I
1 s 1 s 2
2
- I2

o g 12
s s 1
2
Thus,
[g ]
ª
« s2
«
«¬ 2
s
º
s
-1
2
s1
s s1 »
s1 »
1
s 2
»
s1
s s1 ¼
Chapter 19, Solution 33.
To get h11 and h21, consider the circuit below.
I1
I2
-1
s s 1
2
4:
j6 :
+
I1
V1
Also, I 2
5(4 j6)I1
9 j6
5 //( 4 j6)I1
-j3 :
5:
V1
-
5
I1
9 j6
I2

o
h 21
V1
I1
h11
I2
I1
+
V2=0
-
3.0769 j1.2821
0.3846 j0.2564
To get h22 and h12, consider the circuit below.
4:
j6 :
I2
I1
+
V1
V2
-j3 :
5:
V1
-
5
V2
9 j6
+
-

o
j3 //(9 j6)I 2
h12

o
V1
V2
5
9 j6
h 22
I2
V2
0.3846 j0.2564
1
j3 //(9 j6)
0.0769 j0.2821
Thus,
[h ]
+
V2
ª 3.0769 j1.2821 0.3846 j0.2564º
»
«
«¬ 0.3846 j0.2564 0.0769 j0.2821»¼
9 j3
j3(9 j6)
Chapter 19, Solution 34.
Refer to Fig. (a) to get h11 and h 21 .
300 :
10 :
50 :
2
1
+
I1
V1
+
100 :
Vx
I2
+
+
10 Vx
V2 = 0
(a)
At node 1,
I1
Vx
But
Vx Vx 0
100
300
300
I
4 1

o 300 I 1
(1)
4 Vx
75 I 1
V1
10 I 1 Vx
I2
0 10 Vx Vx
50
300
85 I 1

o h11
V1
I1
85 :
At node 2,
h 21
I2
I1
Vx Vx
5 300
75
75
I1 I
5
300 1
14.75 I 1
14.75
To get h 22 and h 12 , refer to Fig. (b).
300 :
I1 = 0 10 :
50 :
1
+
V1
+
Vx
+
100 :
(b)
10 Vx
2
I2
+
V2
At node 2,
I2
V2 V2 10 Vx
400
50
But
Vx
100
V
400 2
Hence,
400 I 2
h 22

o 400 I 2
V2
4
9 V2 20 V2
I2
V2
9 V2 80 Vx
29
400
V2
4
29 V2
0.0725 S
V1
V2

o h 12
V1
Vx
[h]
ª 85 :
0.25 º
« 14.75 0.0725 S »
¬
¼
1
4
0.25
To get g 11 and g 21 , refer to Fig. (c).
300 :
I1
10 :
50 :
1
Vx
I2 = 0
+
+
+
V1
2
+
100 :
10 Vx
V2
(c)
At node 1,
I1
Vx Vx 10 Vx
100
350
But
I1
V1 Vx
10
or
Vx
V1 10 I 1

o 350 I 1

o 10 I 1
14.5 Vx
(2)
V1 Vx
(3)
Substituting (3) into (2) gives
350 I 1 14.5 V1 145 I 1

o 495 I 1
14.5 V1
g 11
I1
V1
V2
§ 11
·
(50) ¨
Vx ¸ 10 Vx
© 350 ¹
-8.4286 Vx
-8.4286 V1 84.286 I 1
§ 14.5 ·
-8.4286 V1 (84.286) ¨
¸ V1
© 495 ¹
14.5
495
0.02929 S
At node 2,
V2
-5.96 V1

o g 21
V2
V1
-5.96
To get g 22 and g 12 , refer to Fig. (d).
300 :
Io
10 :
I1
+
+
+
V1 = 0
Vx
Io
50 :
+
100 :
10 Vx
I2
V2
(d)
10 || 100
I2
V2 10 Vx
V2
50
300 9.091
309.091 I 2
But
Vx
9.091
7.1818 V2 61.818 Vx
9.091
V
309.091 2
Substituting (5) into (4) gives
309.091 I 2 9 V2
0.02941 V2
(4)
(5)
g 22
V2
I2
34.34 :
Io
V2
309.091
I1
- 100
I
110 o
34.34 I 2
309.091
- 34.34 I 2
(1.1)(309.091)
g 12
I1
I2
[g ]
ª 0.02929 S - 0.101 º
« - 5.96
34.34 : »¼
¬
-0.101
Thus,
Chapter 19, Solution 35.
To get h11 and h 21 consider the circuit in Fig. (a).
1:
I1
4:
1:2
+
+
V1
V2 = 0
(a)
ZR
4
n2
4
4
V1
(1 1) I 1
I1
I2
- N2
N1
I2
1
2 I1

o h 11
-2 
o h 21
I2
I1
V1
I1
-1
2
2:
-0.5
To get h 22 and h 12 , refer to Fig. (b).
1:
I1 = 0
4:
1:2
I2
+
+
V1
V2
(b)
Since I 1
Hence,
0 , I2 0 .
h 22 0 .
At the terminals of the transformer, we have V1 and V2 which are related as
V2
V1
N2
N1
2 
o h12
n
V1
V2
1
2
0 .5
Thus,
[h]
ª 2 : 0.5 º
« - 0.5 0 »
¼
¬
Chapter 19, Solution 36.
We replace the two-port by its equivalent circuit as shown below.
4:
I1
16 :
2 I1
+
10 V
+
+
3 V2
V1
+
100 || 25
V2
I2
-2 I1
100 : V2
20 :
(20)(2 I 1 )
40 I 1
- 10 20 I 1 3 V2 0
10 20 I 1 (3)(40 I 1 ) 140 I 1
(1)
25 :
I1
1
,
14
V1
16 I 1 3 V2
136
14
I2
§ 100 ·
¸ (2 I 1 )
¨
© 125 ¹
-8
70
V2
(a)
V2
V1
40
136
(b)
I2
I1
- 1.6
(c)
I1
V1
1
136
7.353 u 10 -3 S
(d)
V2
I1
40
1
40 :
40
14
0.2941
Chapter 19, Solution 37.
(a)
We first obtain the h parameters. To get h11 and h 21 refer to Fig. (a).
6:
3:
I2
+
I1
+
6:
V1
3:
(a)
3 || 6
V1
2
( 6 2) I 1
V2 = 0
8 I1

o h11
V1
I1
8:
-6
I
3 6 1
I2
-2
I
3 1

o h 21
I2
I1
-2
3
To get h 22 and h12 , refer to the circuit in Fig. (b).
I1 = 0
6:
3:
I2
+
6:
V1
+
3:
V2
(b)
3 || 9
9
4
V2
9
I
4 2
V1
6
V
63 2
[h]
ª
2 º
«8 : 3 »
«-2 4 »
S»
«¬
3 9 ¼

o h 22
2
V
3 2
I2
V2
4
9

o h12
V1
V2
2
3
The equivalent circuit of the given circuit is shown in Fig. (c).
I1
8:
I2
+
10 V
+
2/3 V2
+
-2/3 I1
9/4 : V2
(c)
2
8 I 1 V2
3
10
(1)
5:
2 § 9·
I ¨5 || ¸
3 1© 4¹
V2
2 § 45 ·
I ¨ ¸
3 1 © 29 ¹
30
I
29 1
29
V
30 2
I1
(2)
Substituting (2) into (1),
§ 29 ·
2
(8) ¨ ¸ V2 V2
© 30 ¹
3
300
252
V2
(b)
10
1.19 V
By direct analysis, refer to Fig.(d).
6:
3:
+
10 V
+
6:
3:
V2
5:
(d)
10
-A current source. Since
6
6 || 6 3 : , we combine the two 6-: resistors in parallel and transform
10
u 3 5 V voltage source shown in Fig. (e).
the current source back to
6
Transform the 10-V voltage source to a
3:
3:
+
5V
+
V2
(e)
3 || 5
(3)(5)
8
15
8
3 || 5 :
15 8
(5)
6 15 8
V2
75
63
1.19 V
Chapter 19, Solution 38.
We replace the two-port by its equivalent circuit as shown below.
200 :
I1 800 :
I2
+
10 V
+
V1
+
10-4 V2
+
50 I1
200 k: V2
Z in
V2
Vs
,
I1
200 || 50
-50 I 1 (40 u 10 3 )
For the left loop,
Vs 10 -4 V2
1000
40 k:
(-2 u 10 6 ) I 1
I1
Vs 10 -4 (-2 u 10 6 I 1 ) 1000 I 1
Vs 1000 I 1 200 I 1 800I 1
Z in
Vs
I1
Z in
Z s h11 Z in
(10 -4 )(50)(50 u 10 3 )
200 800 1 (0.5 u 10 -5 )(50 u 10 3 )
800 :
Alternatively,
h12 h 21 Z L
1 h 22 Z L
800 :
50 k:
Chapter 19, Solution 39.
To get g11 and g21, consider the circuit below which is partly obtained by converting the
delta to wye subnetwork.
I1
R1
R2
I2
+
+
R3
V2
V1
10 :
-
R1
4 x8
88 4
V2
13.2
V1
13.2 1.6
V1
I1(1.6 3.2 10)
1.6
R2,
8x8
20
R3
0.8919V1
14.8I1
o

o
3.2
V2
V1
g 21
g11
I1
V1
0.8919
1
14.8
0.06757
To get g22 and g12, consider the circuit below.
1.6 :
1.6 :
I1
+
V1=0
V2
13.2 :
-
I2
I1
V2
13.2
I2
13.2 1.6
0.8919I 2
I 2 (1.6 13.2 // 1.6)

o

o
3.027I 2
[g ]
I1
I2
g12
0.8919
V2
I2
g 22
3.027
ª0.06757 0.8919º
« 0.8919
3.027 »¼
¬
Chapter 19, Solution 40.
To get g 11 and g 21 , consider the circuit in Fig. (a).
-j6 :
I1
j10 :
I2 = 0
+
+
V1
12 :
V2
(a)
V1
g 21
(12 j6) I 1
12 I 1
(12 j6) I 1
V2
V1
I1
V1

o g 11
2
2 j
1
12 j6
0.0667 j0.0333 S
0.8 j0.4
To get g 12 and g 22 , consider the circuit in Fig. (b).
I1
-j6 :
j10 :
I2
+
V1 = 0
12 :
(b)
I2
I1
- 12
I
12 - j6 2

o g 12
I1
I2
- 12
12 - j6
V2
( j10 12 || -j6) I 2
g 22
V2
I2
[g ]
ª 0.0667 j0.0333 S - 0.8 j0.4 º
«
0.8 j0.4
2.4 j5.2 : »¼
¬
j10 (12)(-j6)
12 - j6
- g 21
-0.8 j0.4
2.4 j5.2 :
Chapter 19, Solution 41.
For the g parameters
I 1 g 11 V1 g 12 I 2
V2 g 21 V1 g 22 I 2
V1 Vs I 1 Z s
and
But
V2 - I 2 Z L g 21 V1 g 22 I 2
0 g 21 V1 (g 22 Z L ) I 2
or
V1
- (g 22 Z L )
I2
g 21
Substituting this into (1),
(g 22 g 11 Z L g 11 g 21 g 12 )
I1
I2
- g 21
or
I2
I1
- g 21
g 11 Z L ' g
Also,
V2
g 21 (Vs I 1 Z s ) g 22 I 2
g 21 Vs g 21 Z s I 1 g 22 I 2
g 21 Vs Z s (g 11 Z L ' g ) I 2 g 22 I 2
But
I2
- V2
ZL
(1)
(2)
ª V2 º
g 21 Vs [ g 11 Z s Z L ' g Z s g 22 ]«
»
¬ ZL ¼
V2
V2 [ Z L g 11 Z s Z L ' g Z s g 22 ]
g 21 Vs
ZL
V2
Vs
g 21 Z L
Z L g 11 Z s Z L ' g Z s g 22
V2
Vs
g 21 Z L
Z L g 11 Z s Z L g 11 g 22 Z s g 21 g 12 Z s g 22
V2
Vs
g 21 Z L
(1 g 11 Z s )(g 22 Z L ) g 12 g 21 Z s
Chapter 19, Solution 42.
(a)
The network is shown in Fig. (a).
20 :
I1
I2
+
+
100 :
V1
+
-0.5 I2
0.5 I1
V2
(a)
(b)
The network is shown in Fig. (b).
2:
I1
+
V1
s
I2
+
10 :
+
12 V1
V2
(b)
Chapter 19, Solution 43.
(a)
To find A and C , consider the network in Fig. (a).
Z
I1
I2
+
V1
+
V2
(a)
V1
V2

o A
V1
V2
I1
0 
o C
I1
V2
1
0
To get B and D , consider the circuit in Fig. (b).
Z
I1
I2
+
V1
+
V2 = 0
(b)
V1
Z I1 ,
B
- V1
I2
D
- I1
I2
I2
- Z I1
- I1
1
Hence,
ª1 Zº
[T] «
»
¬0 1 ¼
Z
- I1
(b)
To find A and C , consider the circuit in Fig. (c).
I1
I2
+
V1
+
Z
V2
(c)
V1
V2
V1
Z I1

o A
V1
V2

o C
V2
1
I1
V2
1
Z
Y
To get B and D , refer to the circuit in Fig.(d).
I2
+
I1
+
Y
V1
V2 = 0
(d)
V1
V2
0
I2
- I1
B
- V1
I2
0,
D
- I1
I2
Thus,
ª 1 0º
[T] «
»
¬Y 1¼
Chapter 19, Solution 44.
To determine A and C , consider the circuit in Fig.(a).
1
j15 :
Io
-j10 :
I1
-j20 :
Io '
V1
+
I2 = 0
Io
20 :
+
V2
(a)
V1
[ 20 (- j10) || ( j15 j20) ] I 1
V1
ª
(-j10)(-j5) º
« 20 - j15 » I 1
¼
¬
Io
'
ª
10 º
20
j
I
«¬
3 »¼ 1
I1
Io
§ - j10 ·
¨¨
¸¸ I 1
j
10
j
5
©
¹
V2
(-j20) I o 20 I o ' = j
A
V1
V2
C
I1
V2
§2·
¨ ¸ I1
©3¹
(20 j10 3) I 1
40 ·
§
¨ 20 j ¸I1
3 ¹
©
1
40
20 j
3
40 ·
40
§
I1 20I1 = ¨ 20 j ¸I1
3 ¹
3
©
0.7692 + j0.3461
0.03461 + j0.023
To find B and D , consider the circuit in Fig. (b).
j15 :
-j10 :
I1
-j20 :
I2
+
+
V1
20 :
V2 = 0
(b)
We may transform the ' subnetwork to a T as shown in Fig. (c).
Z1
( j15)(-j10)
j15 j10 j20
Z2
(-j10)(-j20)
- j15
-j
Z3
( j15)(-j20)
- j15
j20
I1
j10
40
3
j10 :
j20 :
I2
+
+
V1
20 –– j40/3 :
V2 = 0
(c)
- I2
20 j40 3
I
20 j40 3 j20 1
3 j
3 j2
3 j2
I
3 j 1
D
- I1
I2
V1
ª
( j20)(20 j40 3) º
« j10 20 j40 3 j20 » I 1
¬
¼
0.5385 j0.6923
V1
[ j10 2 (9 j7) ] I 1
B
- V1
I2
B
-6.923 j25.385 :
[T]
- j I 1 (24 j18)
- (3 - j2)
I
3 j 1
j I 1 (24 j18)
6
(-15 j55)
13
ª 0.7692 j0.3461 - 6.923 j25.385 : º
« 0.03461 j0.023 S 0.5385 j0.6923 »
¬
¼
Chapter 19, Solution 45.
To obtain A and C, consider the circuit below.
I1
sL
1/sC
I2 =0
+
+
V1
R1
V2
-
R2
V2
R1
V1
R1 R 2 sL
V2
I1R1
o
o
C
I1
V2
A
1
R1
To obtain B and D, consider the circuit below.
V1
V2
R1 R 2 sL
R1
I1
sL
1/sC
I2
+
+
V1
R1
V2=0
-
R2
R1
V1
R1 ·
§
¨
¸
¨ R 2 sL sC ¸I1
1 ¸
¨
R1 ¨
¸
sC ¹
©
R1 1
sC
I1
B
sR1C
I1
1 sR1C
I2
V
1
I2

o
1 sR1C
sR1C
I
1
I2
D
>(1 sR1C)(R 2 sL) R1@ (1 sR1C) I
1 sR1C
sR1C
1
>R1 (1 sR1C)(R 2 sL)@
sR1C
Chapter 19, Solution 46.
To get A and C , refer to the circuit in Fig.(a).
I1
V1
1:
+
1:
1
+
Vo
I2 = 0
2
Ix
2:
+
4 Ix
V2
(a)
At node 1,
I1
Vo Vo V2
2
1

o 2 I 1
3 Vo 2 V2
(1)
2
At node 2,
Vo V2
1
4 Vo
2
4Ix

o Vo
2 Vo
-V2
(2)
From (1) and (2),
2 I1
But
I1
-5 V2

o C
V1 Vo
1
V1 V2
V1 V2
- 2.5 V2
V1
V2
A
I1
V2
-5
2

o V1
-2.5 S
-3.5 V2
-3.5
To get B and D , consider the circuit in Fig. (b).
I1
1:
+
+
V1
1:
1
I2
2
Ix
+
2:
Vo
4 Ix
V2 = 0
(b)
At node 1,
I1
Vo Vo
2
1
I2 Vo
4Ix
1

o 2 I 1
(3)
3 Vo
At node 2,
–– I 2
Vo 2 Vo
Adding (3) and (4),
2 I1 I 2
D
- I1
I2
0
0 
o I 2
0 
o I 1
0 .5
-0.5 I 2
-3 Vo
(4)
(5)
But
I1
V1 Vo
1

o V1
Substituting (5) and (4) into (6),
-1
-1
V1
I2 I2
2
3
B
- V1
I2
5
6
I 1 Vo
(6)
-5
I
6 2
0.8333 :
Thus,
ª - 3.5 0.8333 : º
[T] «
- 0.5 »¼
¬ - 2.5 S
Chapter 19, Solution 47.
To get A and C, consider the circuit below.
6:
I1
1:
+
+
Vx
V1
-
V1 Vx
1
V2
I1
4:
2:
+
5Vx
+
V2
-
-
-
Vx Vx 5Vx
2
10
4(0.4Vx ) 5Vx
V1 Vx
1

o
3.4Vx
1.1Vx Vx
0.1Vx
I2=0
V1 1.1Vx

o
o
A
V1
V2
C
1.1 / 3.4
I1
V2
0.3235
0.1 / 3.4
0.02941
Chapter 19, Solution 48.
(a)
Refer to the circuit below.
I2
I1
+
V1
+
[T]
ZL
V2
V1 4 V2 30 I 2
I 1 0.1 V2 I 2
When the output terminals are shorted, V2
So, (1) and (2) become
V1 -30 I 2
and
Hence,
V1
30 :
Z in
I1
(b)
0.
I1
-I2
When the output terminals are open-circuited, I 2 0 .
So, (1) and (2) become
V1 4 V2
I 1 0.1 V2
or
V2 10 I 1
V1 40 I 1
Z in
(c)
(1)
(2)
V1
I1
40 :
When the output port is terminated by a 10-: load, V2
So, (1) and (2) become
V1 -40 I 2 30 I 2 -70 I 2
I 1 - I 2 I 2 -2 I 2
V1 35 I 1
Z in
V1
I1
35 :
-10 I 2 .
A ZL B
CZL D
Alternatively, we may use Z in
Chapter 19, Solution 49.
To get A and C , refer to the circuit in Fig.(a).
1/s
I1
I2 = 0
+
V1
+
1:
1/s
1/s
1:
V2
(a)
1 ||
But
1
s
1s
11 s
1
s 1
V2
1 || 1 s
V
1 s 1 || 1 s 1
A
V2
V1
1
s 1
1
1
s s 1
V1
§ 1 · §1
1 ·
¸ || ¨ ¸
I1 ¨
© s 1¹ © s s 1¹
V1
I1
§ 1 · § 2s 1 ·
¸
¸˜¨
¨
© s 1 ¹ © s (s 1) ¹
1
2s 1
s 1 s (s 1)
V1
V2 ˜
2s 1
s
s
2s 1
§ 1 · § 2s 1 ·
¸
¸ || ¨
I1¨
© s 1 ¹ © s (s 1) ¹
2s 1
(s 1)(3s 1)
V2 2s 1
˜
s
I1
Hence,
C
V2
I1
2s 1
(s 1)(3s 1)
(s 1)(3s 1)
s
To get B and D , consider the circuit in Fig. (b).
1/s
I1
I2
+
V1
+
1:
1/s
1:
1/s
V2 = 0
(b)
V1
§ 1 1·
I 1 ¨1 || || ¸
© s s¹
§ 1·
I 1 ¨1 || ¸
© 2s ¹
I2
-1
I
s 1 1
1
1
s 1 s
D
- I1
I2
V1
§ 1 ·§ 2s 1 ·
¨
¸¨
¸I
© 2s 1 ¹© - s ¹ 2
I1
2s 1
-s
I
2s 1 1
2s 1
s
2
1
s
I2
-s
Thus,
ª
2
«
2s 1
[T] «
(s 1)( 3s 1)
«¬
s
º
»
1»
2 »
s¼
1
s
Chapter 19, Solution 50.
To get a and c, consider the circuit below.

o B
- V1
I2
1
s
I1=0
2
s
I2
+
+
4/s
V1
V2
-
-
V1
4/s
V2
s 4/s
4
V2
(s 4 / s)I 2 or
I2
V2
s 4/s

o
V2
2
s 4
(1 0.25s 2 )V1
s 4/s

o
V2
a
c
V1
I2
V1
1 0.25s 2
s 0.25s3
s2 4
To get b and d, consider the circuit below.
2
I1
s
I2
+
+
V1=0
4/s
V2
-
I1
V2
4/s
I2
2 4/s
-
4
(s 2 // )I2
s
2I 2
s2

o
d
I
2
I1
1 0.5s
(s 2 2s 4)
I2
s2
(s 2 2s 4)( s 2)
I1
s2
2
[t]

o
b
V
2
I1
0.5s 2 s 2
ª 0.25s 2 1 0.5s 2 s 2º
»
«
2
« 0.25s s
0.5s 1 »
»
« s2 4
¼
¬
Chapter 19, Solution 51.
To get a and c , consider the circuit in Fig. (a).
j:
1:
I1 = 0
-j3 :
I2
+
j2 :
V1
+
j:
V2
(a)
I 2 ( j j3)
-jI 2
V2
V1
-j2 I 2
a
V2
V1
- j2 I 2
- jI 2
c
I2
V1
1
-j
2
j
To get b and d , consider the circuit in Fig. (b).
j:
I1
1:
-j3 :
I2
+
j2 :
V1 = 0
j:
(b)
For mesh 1,
0
or
I2
I1
(1 j2) I1 j I 2
1 j2
j
2 j
+
V2
d
- I2
I1
-2 j
For mesh 2,
I 2 ( j j3) j I 1
I 1 (2 j)(- j2) j I 1
V2
V2
b
- V2
I1
(-2 j5) I 1
2 j5
Thus,
[t ]
ª 2 2 j5 º
« j -2 j»
¬
¼
Chapter 19, Solution 52.
It is easy to find the z parameters and then transform these to h parameters and T
parameters.
ª R1 R 2
[z ] «
¬ R2
'z
(a)
º
R 2 R 3 »¼
R2
(R 1 R 2 )(R 2 R 3 ) R 22
R 1R 2 R 2 R 3 R 3 R 1
ª 'z
«z
[h] « 22
-z
« 21
¬ z 22
z 12 º
z 22 »
»
1
»
z 22 ¼
ª R 1R 2 R 2 R 3 R 3 R 1
«
R2 R3
«
- R2
«
R2 R3
¬
R2 º
R2 R3 »
»
1
»
R2 R3 ¼
Thus,
h 11
R1 R 2R 3
,
R2 R3
h 12
'z º
z 21 »
»
z 22
»
z 21 ¼
ª R1 R 2
« R
2
«
1
«
¬ R2
R 1R 2 R 2 R 3 R 3 R 1 º
»
R2
»
R2 R3
»
R2
¼
R2
R2 R3
- h 21 ,
as required.
(b)
ª z 11
«z
[T] « 21
1
«
¬ z 21
h 22
1
R2 R3
Hence,
A
1
R1
, B
R2
R3 R1
(R 2 R 3 ) , C
R2
1
, D
R2
1
R3
R2
as required.
Chapter 19, Solution 53.
For the z parameters,
V1 z11 I1 z12 I 2
V2 z12 I1 z 22 I 2
(1)
(2)
For ABCD parameters,
V1 A V2 B I 2
I1 C V2 D I 2
From (4),
I1 D
V2
I
C C 2
Comparing (2) and (5),
1
,
z 21
C
(3)
(4)
(5)
z 22
D
C
Substituting (5) into (3),
§ AD
·
A
V1
B¸ I 2
I1 ¨
© C
¹
C
A
AD BC
I1 I2
C
C
Comparing (6) and (1),
A
z11
C
Thus,
ªA
«
[Z] = « C
1
«
¬C
(6)
z 12
'T º
C»
D»
»
C¼
AD BC
C
'T
C
Chapter 19, Solution 54.
For the y parameters
I 1 y 11 V1 y 12 V2
I 2 y 21 V1 y 22 V2
From (2),
I 2 y 22
V1
V
y 21 y 21 2
or
V1
(1)
(2)
- y 22
1
V2 I
y 12
y 21 2
(3)
Substituting (3) into (1) gives
- y 11 y 22
y 11
I1
V2 y 12 V2 I
y 21
y 21 2
or
I1
- 'y
y 21
V2 y 11
I
y 21 2
(4)
Comparing (3) and (4) with the following equations
V1 A V2 B I 2
I 1 C V2 D I 2
clearly shows that
A
- y 22
,
y 21
B
-1
,
y 21
C
- 'y
y 21
,
D
- y 11
y 21
as required.
Chapter 19, Solution 55.
For the z parameters
V1 z11 I1 z12 I 2
V2 z 21 I1 z 22 I 2
From (1),
z
1
I1
V1 12 I 2
z11
z11
Substituting (3) into (2) gives
(1)
(2)
(3)
or
V2
§
z z ·
z 21
V1 ¨ z 22 21 12 ¸ I 2
z11 ¹
z11
©
V2
'
z 21
V1 z I 2
z11
z11
(4)
Comparing (3) and (4) with the following equations
I1 g11 V1 g12 I 2
V2 g 21 V1 g 22 I 2
indicates that
- z 12
z 21
1
, g 12
, g 21
,
g 11
z 11
z 11
z 11
g 22
'z
z 11
as required.
Chapter 19, Solution 56.
(a) ' y
(2 j)(3 j) j4
7 j5
ª y 22 / ' y
« y / '
¬ 21 y
y12 / ' y º
y11 / ' y »¼
[z]
(b) [h ]
y12 / y11 º
ª 1 / y11
«y / y
»
¬ 21 11 ' y / y11 ¼
(c ) [ t ]
ª y11 / y12
« ' / y
y 12
¬
ª0.2162 j0.2973 0.2703 j0.3784º
«0.0946 j0.0676 0.2568 j0.0405 »:
¬
¼
ª 0.4 j0.2 0.8 j1.6º
« 0.4 j0.2 3.8 j0.6 »
¬
¼
1 / y12 º
y 22 / y12 »¼
j0.25 º
ª 0.25 j0.5
« 1.25 j1.75 0.25 j0.75»
¬
¼
Chapter 19, Solution 57.
'T
(3)(7) (20)(1) 1
ªA
«
[z ] « C
1
«¬
C
'T º
C»
D»
»
C¼
ª3 1º
«1 7»:
¼
¬
ªD
«
[y ] « B
-1
«¬
B
- 'T º
B »
A »
»
B ¼
ªB
«
[h] « -D1
«¬
D
'T º
D»
C»
»
D¼
ªC
«
[g ] « A
1
«¬
A
- 'T º
A »
B »
»
A ¼
ªD
«'
[t ] « CT
«
¬ 'T
B º
'T »
A»
»
'T ¼
ª 7
« 20
« -1
«¬
20
-1
20
3
20
º
»
»S
»¼
ª 20
1 º
« 7 : 7 »
« -1
1 »
S»
«¬
7
7 ¼
ª1
«3S
« 1
«¬
3
-1 º
3 »
20 »
:»
¼
3
ª 7 20 : º
«1S
3 »¼
¬
Chapter 19, Solution 58.
The given set of equations is for the h parameters.
ª1 :
2 º
' h (1)(0.4) (2)(-2)
[h] «
»
¬ - 2 0.4 S¼
(a)
ª 1
«h
[y ] « 11
h
« 21
¬ h11
(b)
ª
«
[T] «
«
¬
- 'h
h 21
- h 22
h 21
- h12
h11
'h
h11
º
» ª 1 -2 º
» «
»S
» ¬ - 2 4.4 ¼
¼
- h11
h 21
-1
h 21
º
»
»
»
¼
ª 2.2 0.5 : º
« 0.2 S 0.5 »
¬
¼
4 .4
Chapter 19, Solution 59.
'g
(0.06)(2) (-0.4)(2)
0.12 0.08
0.2
(a)
ª
«
[z] «
«
¬«
1
g 11
g 21
g 11
'g
(b)
ª
«
[y ] «
«
«¬
(c)
ª g 22
« '
g
[h] « - g
« 21
¬« ' g
- g 12 º
'g »
»
g 11 »
' g »¼
(d)
ª
«
[T] «
«
¬«
g 22 º
g 21 »
»
'g
»
g 21 ¼»
g 22
- g 21
g 22
1
g 21
g 11
g 21
- g 12 º
g 11 » ª 16.667 6.667 º
» «
»:
'g
» ¬ 3.333 3.333 ¼
g 11 ¼»
g 12
g 22
1
g 22
º
» ª 0.1 - 0.2 º
» «
»S
» ¬ - 0.1 0.5 ¼
»¼
ª 10 :
2 º
« - 1 0.3 S »
¬
¼
ª 5
10 : º
« 0.3 S
1 »¼
¬
Chapter 19, Solution 60.
'y
y 11 y 22 y 12 y 21
(a)
0.3 0.02
ª
«
[z ] «
«
«¬
0.28
y 22
'y
- y 21
'y
- y 12
'y
y 11
'y
º
»
»
»
»¼
ª 1.786 0.7143 º
« 0.3571 2.143 » :
¬
¼
- y 12 º
y 11 » ª 1.667 : 0.3333 º
» «
»
'y
» ¬ - 0.1667 0.4667 S ¼
y 11 »¼
(b)
ª
«
[h] «
«
«¬
1
y 11
y 21
y 11
(c)
ª
«
[t ] «
«
¬«
- y 11
y 12
- 'y
y 12
-1 º
y 12 »
»
- y 22 »
y 12 ¼»
ª 3
5:º
« 1.4 S 2.5 »
¬
¼
Chapter 19, Solution 61.
(a)
To obtain z 11 and z 21 , consider the circuit in Fig. (a).
1:
Io
1:
1:
+
I1
+
1:
V1
V2
(a)
V1
I 1 [1 1 || (1 1) ]
z 11
V1
I1
Io
5
3
1
I
1 2 1
- V2 I o I 1
V2
1
I I
3 1 1
I2 = 0
1
I
3 1
0
4
I
3 1
§ 2·
I 1 ¨1 ¸
© 3¹
5
I
3 1
z 21
V2
I1
4
3
To obtain z 22 and z 12 , consider the circuit in Fig. (b).
1:
1:
I1
1:
+
+
1:
V1
V2
I2
(b)
Due to symmetry, this is similar to the circuit in Fig. (a).
5
4
z 22 z 11
,
z 21 z 12
3
3
ª
«
[z ] «
«
¬
5
3
4
3
4º
3»
»:
5»
3¼
(b)
ª
«
[h] «
«
¬
'z
z 22
- z 21
z 22
z 12
z 22
1
z 22
º
»
»
»
¼
(c)
ª
«
[T] «
«
¬
z 11
z 21
1
z 21
'z
z 21
z 22
z 21
º
»
»
»
¼
Chapter 19, Solution 62.
Consider the circuit shown below.
ª3
4 º
«5: 5 »
« -4 3 »
S»
«¬
5
5 ¼
ª 5
« 4
«3
«¬ S
4
3 º
:
4 »
5 »
»
4 ¼
I1
10 k:
a
40 k:
+
+
I2
+
50 k:
b
30 k:
V1
Ib
20 k:
Since no current enters the input terminals of the op amp,
V1 (10 30) u 10 3 I 1
But
Va
Vb
30
V
40 1
V2
(1)
3
V
4 1
Vb
3
V
3
20 u 10
80 u 10 3 1
which is the same current that flows through the 50-k: resistor.
Ib
Thus,
V2
40 u 10 3 I 2 (50 20) u 10 3 I b
V2
40 u 10 3 I 2 70 u 10 3 ˜
V2
21
V 40 u 10 3 I 2
8 1
V2
105 u 10 3 I 1 40 u 10 3 I 2
3
V
80 u 10 3 1
From (1) and (2),
ª 40 0 º
[z ] «
» k:
¬ 105 40 ¼
'z
z 11 z 22 z 12 z 21
16 u 10 8
(2)
ª
ªA Bº «
[T] «
» «
¬C D¼ «
¬
'z
z 21
z 22
z 21
z 11
z 21
1
z 21
º
» ª 0.381 15.24 k: º
» «
0.381 »¼
» ¬ 9.52 PS
¼
Chapter 19, Solution 63.
To get z11 and z21, consider the circuit below.
1:3
I1
x
+
4:
V1
I2=0
x +
+
+
V’’1
V’’2
-
-
9:
-
V2
-
9
ZR
n2
1,
n
4
I1
5
V1
(4 // ZR )I1
V2
V2 ' nV1' nV1
3

o
V1
I1
z11
3(4 / 5)I1

o
0.8
z 21
V2
I1
2.4
To get z21 and z22, consider the circuit below.
I1=0
1:3
x
+
V1
4:
x +
+
V’’1
V’’2
-
-
-
ZR '
I2
+
9:
V2
-
n 2 (4)
36 ,
n
3
V2
(9 // ZR ' )I 2
V1
V2
n
V2
3
9x36
I2
45
o
o
2.4I 2
V2
I2
z 22
V1
I2
z 21
7.2
2.4
Thus,
ª0.8 2.4º
«2.4 7.2» :
¬
¼
[z]
Chapter 19, Solution 64.
1
jZC
1 PF 
o
-j
(10 )(10 -6 )
3
- j k:
Consider the op amp circuit below.
40 k:
I1
20 k:
Vx
10 k:
1
+
+
2
-j k:
V1
I2
+
V2
At node 1,
V1 Vx
20
V1
Vx Vx 0
-j
10
(3 j20) Vx
(1)
At node 2,
Vx 0
10
But
I1
0 V2
40
V1 Vx
20 u 10 3
Substituting (2) into (3) gives

o Vx
-1
V
4 2
(2)
(3)
I1
V1 0.25 V2
20 u 10 3
50 u 10 -6 V1 12.5 u 10 -6 V2
(4)
Substituting (2) into (1) yields
-1
V1
(3 j20) V2
4
or
0
V1 (0.75 j5) V2
(5)
Comparing (4) and (5) with the following equations
I 1 y 11 V1 y 12 V2
I 2 y 21 V1 y 22 V2
indicates that I 2
0 and that
12.5 u 10 -6 º
»S
0.75 j5 ¼
[y ]
ª 50 u 10 -6
«
1
¬
'y
(77.5 j25. 12.5) u 10 -6
ª
«
[h] «
«
¬«
1
y 11
y 21
y 11
(65 j250) u 10 -6
- y 12 º
- 0.25 º
y 11 » ª 2 u 10 4 :
» «
»
4
'y
1.3 j5 S ¼
» ¬ 2 u 10
y 11 ¼»
Chapter 19, Solution 65.
The network consists of two two-ports in series. It is better to work with z parameters
and then convert to y parameters.
ª4 2º
For N a ,
[z a ] «
»
¬2 2¼
For N b ,
ª 2 1º
[z b ] «
»
¬ 1 1¼
ª6 3º
[z ] [z a ] [z b ] «
»
¬3 3¼
'z
18 9
9
- z 12 º
'z »
»
z 11
»
'z ¼
ª z 22
« '
[y ] « z
-z
« 21
¬ 'z
ª 1
« 3
« -1
«¬
3
-1 º
3 »S
2 »
»
3 ¼
Chapter 19, Solution 66.
Since we have two two-ports in series, it is better to convert the given y parameters to z
parameters.
' y y 11 y 22 y 12 y 21 (2 u 10 -3 )(10 u 10 -3 ) 0 20 u 10 -6
ª
«
[z a ] «
«
¬«
y 22
'y
- y 21
'y
- y 12
'y
y 11
'y
º
» ª 500 :
0 º
» «
100 : »¼
» ¬ 0
¼»
ª 500 0 º ª 100 100 º ª 600 100 º
[z ] «
»«
» «
»
¬ 0 100 ¼ ¬ 100 100 ¼ ¬ 100 200 ¼
i.e.
V1
V2
z 11 I 1 z 12 I 2
z 21 I 1 z 22 I 2
or
V1
V2
600 I 1 100 I 2
100 I 1 200 I 2
(1)
(2)
But, at the input port,
Vs V1 60 I 1
(3)
and at the output port,
V2 Vo
(4)
-300 I 2
From (2) and (4),
100 I 1 200 I 2
I 1 -5I 2
-300 I 2
Substituting (1) and (5) into (3),
Vs 600 I 1 100 I 2 60 I 1
(5)
(660)(-5) I 2 100 I 2
-3200I 2
From (4) and (6),
Vo
V2
- 300 I 2
- 3200 I 2
(6)
0.09375
Chapter 19, Solution 67.
The y parameters for the upper network is
ª 2 -1 º
[y ] «
»,
¬ -1 2 ¼
ª
«
[z a ] «
«
«¬
- y 12
'y
y 11
'y
y 22
'y
- y 21
'y
º
»
»
»
»¼
ª2
«3
«
«1
¬3
'y
4 1 3
1º
3»
»
2»
3¼
ª1 1º
[z b ] «
»
¬1 1¼
ª5 3 4 3º
[z ] [z a ] [z b ] «
»
¬4 3 5 3¼
'z
25 16
9
9
ª
«
[T] «
«
¬
z 11
z 21
1
z 21
1
'z
z 21
z 22
z 21
º
»
»
»
¼
ª 1.25 0.75 : º
« 0.75 S 1.25 »
¬
¼
Chapter 19, Solution 68.
ª 4 -2º
For the upper network N a , [y a ] «
»
¬-2 4 ¼
ª 2 -1 º
and for the lower network N b , [y b ] «
»
¬1 2 ¼
For the overall network,
ª 6 -3º
[y ] [y a ] [y b ] «
»
¬ -3 6 ¼
'y
36 9
ª
«
[h] «
«
«¬
27
- y 12 º ª 1
-1 º
y 11 » « 6 : 2 »
» «
»
'y
» « -1 9 S »
2 ¼
y 11 »¼ ¬ 2
1
y 11
y 21
y 11
Chapter 19, Solution 69.
We first determine the y parameters for the upper network N a .
To get y 11 and y 21 , consider the circuit in Fig. (a).
n
1
,
2
ZR
V1
(2 Z R ) I 1
y 11
I1
V1
I2
- I1
n
-2 I 1
y 21
I2
V1
-s
s2
§
4·
¨ 2 ¸ I1
©
s¹
1s
n2
4
s
§ 2s 4 ·
¨
¸I
© s ¹ 1
s
2 (s 2)
- s V1
s2
To get y 22 and y 12 , consider the circuit in Fig. (b).
I1
2:
1/s
2:1
I2
+
+
V1 =0
V2
I2
(b)
ZR'
V2
§1·
(n 2 )(2) ¨ ¸ (2)
©4¹
§1
·
¨ ZR ' ¸I2
©s
¹
y 22
I2
V2
I1
-n I2
y 12
I1
V2
1
2
§s 2·
¨
¸I
© 2s ¹ 2
§ 1 1·
¨ ¸I2
© s 2¹
2s
s2
§ - 1 ·§ 2s ·
¨ ¸¨
¸V
© 2 ¹© s 2 ¹ 2
§ -s ·
¨
¸V
©s 2¹ 2
-s
s2
ª
s
« 2 (s 2)
[y a ] «
-s
«
¬ s2
-s º
s2 »
2s »
»
s2 ¼
For the lower network N b , we obtain y 11 and y 21 by referring to the network in Fig. (c).
I1
2
I2
+
V1
+
s
V2 = 0
(c)
V1
2 I1

o y 11
I1
V1
1
2
I2
- V1
2
- I1
I2
V1

o y 21
-1
2
To get y 22 and y 12 , refer to the circuit in Fig. (d).
I1
2
I2
+
+
s
V1 = 0
I2
V2
(d)
V2
(s || 2) I 2
I1
- I2 ˜
-s
s2
y 12
I1
V2
-1
2
2s
I
s2 2

o y 22
§ - s ·§ s 2 ·
¨
¸¨
¸V
© s 2 ¹© 2s ¹ 2
s2
2s
I2
V2
- V2
2
ª12
-1 2 º
[y b ] «
»
¬ - 1 2 (s 2) 2s ¼
[y ] [y a ] [y b ]
ª s1
« s2
«
« - (3s 2)
¬ 2 (s 2)
- (3s 2)
2 (s 2)
5s 2 4s 4
2s (s 2)
º
»
»
»
¼
Chapter 19, Solution 70.
We may obtain the g parameters from the given z parameters.
ª 25 20 º
[z a ] «
' z a 250 100 150
»,
¬ 5 10 ¼
ª 50 25 º
[z b ] «
»,
¬ 25 30 ¼
' zb
1500 625 875
ª
«
[g ] «
«
¬
1
z 11
z 21
z 11
- z 12
z 11
'z
z 11
º
»
»
»
¼
ª 0.04 - 0.8 º
[g a ] «
,
6 »¼
¬ 0.2
[g ] [g a ] [ g b ]
ª 0.02 - 0.5 º
[g b ] «
»
¬ 0.5 17.5 ¼
ª 0.06 S - 1.3 º
« 0.7
23.5 : »¼
¬
Chapter 19, Solution 71.
This is a parallel-series connection of two two-ports. We need to add their g parameters
together and obtain z parameters from there.
For the transformer,
V1
1
V2 , I1
2
2 I 2
Comparing this with
V1
AV2 BI2 ,
I1
CV2 DI 2
shows that
ª0.5 0º
« 0 2»
¬
¼
[Tb1]
To get A and C for Tb2 , consider the circuit below.
I1
+
V1
-
4:
I2 =0
5:
2:
+
V2
-
V1
9I1,
A
V1
V2
V2
5I1
9 / 5 1.8,
C
I1
V2
1/ 5
0.2
Chapter 19, Solution 72.
Consider the network shown below.
I1
Ia1
+
Va1
+
V1
Ia2
Na
Ib1
+
Vb1
Va1
Ia2
Vb1
I b2
V1
V2
I2
I1
+
Va2
+
V2
Ib2
Nb
25 I a1 4 Va 2
- 4 I a1 Va 2
16 I b1 Vb 2
- I b1 0.5 Vb 2
Va1 Vb1
Va 2 Vb 2
I a 2 I b2
I a1
Now, rewrite (1) to (4) in terms of I 1 and V2
Va1 25 I 1 4 V2
I a 2 - 4 I 1 V2
Vb1 16 I b1 V2
I b 2 - I b1 0.5 V2
Adding (5) and (7),
I2
+
Vb2
(1)
(2)
(3)
(4)
(5)
(6)
(7)
(8)
V1
25 I 1 16 I b1 5 V2
(9)
Adding (6) and (8),
I 2 - 4 I 1 I b1 1.5 V2
I b1
I a1
(10)
I1
(11)
Because the two networks N a and N b are independent,
I 2 - 5 I 1 1.5 V2
V2 3.333 I 1 0.6667 I 2
or
(12)
Substituting (11) and (12) into (9),
25
5
V1 41I 1 I1 I
1.5
1.5 2
V1
57.67 I 1 3.333 I 2
(13)
Comparing (12) and (13) with the following equations
V1 z 11 I 1 z 12 I 2
V2 z 21 I 1 z 22 I 2
indicates that
ª 57.67 3.333 º
[z ] «
»:
¬ 3.333 0.6667 ¼
Alternatively,
ª 25 4 º
[h a ] «
»,
¬-4 1¼
[h] [h a ] [h b ]
ª 'h
« h
[z ] « 22
-h
« 21
¬ h 22
as obtained previously.
h12
h 22
1
h 22
ª 16 1 º
[h b ] «
»
¬ - 1 0.5 ¼
ª 41 5 º
« - 5 1.5 »
¬
¼
'h
º
» ª 57.67 3.333 º
» «
»:
3
.
333
0
.
6667
¬
¼
»
¼
61.5 25
86.5
Chapter 19, Solution 73.
From Example 18.14 and the cascade two-ports,
ª2 3º
[Ta ] [Tb ] «
»
¬1 2¼
ª 2 3 ºª 2 3 º
[T] [Ta ][Tb ] «
»
»«
¬ 1 2 ¼¬ 1 2 ¼
When the output is short-circuited, V2
V1 - B I 2 ,
Hence,
V1 B 12
Z in
:
I1 D 7
ª 7 12 : º
«4S
7 »¼
¬
0 and by definition
I1 - D I 2
Chapter 19, Solution 74.
From Prob. 18.35, the transmission parameters for the circuit in Figs. (a) and (b) are
ª1 Zº
[Ta ] «
»,
¬0 1 ¼
ª 1 0º
[Tb ] «
»
¬1 Z 1 ¼
Z
Z
(a)
(b)
We partition the given circuit into six subcircuits similar to those in Figs. (a) and (b) as
shown in Fig. (c) and obtain [T] for each.
s
s
1
1/s
T1
T2
T3
1
T4
T5
1/s
T6
ª1 0 º
[T1 ] «
»,
¬1 1 ¼
ª1 sº
[T2 ] «
»,
¬ 0 1¼
ª1 0 º
[T3 ] «
»
¬s 1¼
[T4 ] [T2 ] ,
[T5 ] [T1 ] ,
[T6 ] [T3 ]
ª 1 0 ºª 1 0 º
[T] [T1 ][T2 ][T3 ][T4 ][T5 ][T6 ] [T1 ][T2 ][T3 ][T4 ]«
»«
»
¬ 1 1 ¼¬ s 1 ¼
0º
ª 1
[T1 ][T2 ][T3 ][T4 ] «
»
¬ s 1 1 ¼
0º
ª1 sºª 1
[T1 ][T2 ][T3 ] «
»
«
»
¬ 0 1 ¼ ¬ s 1 1 ¼
ª 1 0 º ª s2 s 1 s º
[T1 ][T2 ] «
»
»«
1¼
¬ s 1 ¼ ¬ s 1
s º
ª 1 s º ª s2 s 1
[T1 ] «
« 3
»
»
2
2
¬ 0 1 ¼ ¬ s s 2s 1 s 1 ¼
ª 1 0 º ª s 4 s 3 3s 2 2s 1 s 3 2s º
»
«1 1 » «
3
2
s2 1 ¼
¬
¼ ¬ s s 2s 1
ª s 4 s 3 3s 2 2s 1
º
s 3 2s
[T] « 4
»
3
2
3
2
¬ s 2s 4s 4s 2 s s 2s 1 ¼
Note that AB CD 1 as expected.
Chapter 19, Solution 75.
(a) We convert [za] and [zb] to T-parameters. For Na, ' z
[Ta ]
For Nb, ' y
[Tb ]
ªz11 / z 21 ' z / z 21 º
« 1/ z
z 22 / z 21 »¼
21
¬
80 8
4 º
ª 2
«0.25 1.25»
¬
¼
88 .
ª y 22 / y 21 1 / y 21 º
«' /y
»
y 21 y11 / y 21 ¼
¬
ª 5 0.5º
« 44 4 »
¬
¼
40 24 16 .
[T] [Ta ][Tb ]
17 º
ª 186
« 56.25 5.125»
¬
¼
We convert this to y-parameters. ' T
ª D / B ' T / Bº
« 1 / B
A / B »¼
¬
[ y]
(b)
AD BC
3.
ª0.3015 0.1765º
«
»
«¬0.0588 10.94 »¼
The equivalent z-parameters are
[z]
ªA / C ' T / Cº
« 1/ C D / C »
¬
¼
ª 3.3067 0.0533º
« 0.0178 0.0911»
¬
¼
Consider the equivalent circuit below.
I1
z11
z22
+
I2
+
+
+
Vi
z12 I2
-
-
z21 I1
ZL
Vo
-
-
Vi
z11I1 z12 I 2
(1)
Vo
z 21I1 z 22 I 2
(2)
But Vo
I 2 ZL
o
I2
Vo / ZL
(3)
From (2) and (3) ,
Vo
V
z 21I1 z 22 o
ZL
o
I1
·
§ 1
z
Vo ¨¨
22 ¸¸
© z 21 ZL z 21 ¹
(4)
Vo.
Vi
0.0051
Substituting (3) and (4) into (1) gives
Vi
Vo
§ z11 z11z 22 · z12
¨¨
¸¸ © z 21 z 21ZL ¹ ZL
194.3
o
Chapter 19, Solution 76.
To get z11 and z21, we open circuit the output port and let I1 = 1A so that
V1
V2
z11
V1, z 21
V2
I1
I1
The schematic is shown below. After it is saved and run, we obtain
z11
V1
3.849,
z 21
V2
1.122
Similarly, to get z22 and z12, we open circuit the input port and let I2 = 1A so that
z12
V1
I2
V1,
z 22
V2
I2
V2
The schematic is shown below. After it is saved and run, we obtain
z12
V1
1.122,
z 22
V2
3.849
Thus,
[z]
ª3.949 1.122 º
«1.122 3.849» :
¬
¼
Chapter 19, Solution 77.
We follow Example 19.15 except that this is an AC circuit.
(a)
We set V2 = 0 and I1 = 1 A. The schematic is shown below. In the AC Sweep
Box, set Total Pts = 1, Start Freq = 0.1592, and End Freq = 0.1592. After simulation,
the output file includes
FREQ
IM(V_PRINT2)
IP(V_PRINT2)
1.592 E––01
3.163 E––.01
––1.616 E+02
FREQ
VM($N_0001)
VP($N_0001)
1.592 E––01
9.488 E––01
––1.616 E+02
From this we obtain
h11 = V1/1 = 0.9488‘––161.6q
h21 = I2/1 = 0.3163‘––161.6q.
(b)
In this case, we set I1 = 0 and V2 = 1V. The schematic is shown below. In the
AC Sweep box, we set Total Pts = 1, Start Freq = 0.1592, and End Freq = 0.1592.
After simulation, we obtain an output file which includes
FREQ
VM($N_0001)
VP($N_0001)
1.592 E––01
3.163 E––.01
1.842 E+01
FREQ
IM(V_PRINT2)
IP(V_PRINT2)
1.592 E––01
9.488 E––01
––1.616 E+02
From this,
h12 = V1/1 = 0.3163‘18.42q
h21 = I2/1 = 0.9488‘––161.6q.
Thus,
ª0.9488‘ 161.6q 0.3163‘18.42q º
[h] = «
»
¬0.3163‘ 161.6q 0.9488‘ 161.6q¼
Chapter 19, Solution 78
For h11 and h21, short-circuit the output port and let I1 = 1A. f Z / 2S 0.6366 . The
schematic is shown below. When it is saved and run, the output file contains the
following:
FREQ
IM(V_PRINT1)IP(V_PRINT1)
6.366E-01
FREQ
1.202E+00
1.463E+02
VM($N_0003) VP($N_0003)
6.366E-01
3.771E+00
-1.350E+02
From the output file, we obtain
I2
1.202‘146.3o ,
V1
3.771‘ 135o
so that
h11
V1
1
3.771‘ 135o ,
h 21
I2
1
1.202‘146.3o
For h12 and h22, open-circuit the input port and let V2 = 1V. The schematic is shown
below. When it is saved and run, the output file includes:
FREQ
VM($N_0003) VP($N_0003)
6.366E-01
FREQ
1.202E+00
-3.369E+01
IM(V_PRINT1)IP(V_PRINT1)
6.366E-01
3.727E-01
-1.534E+02
From the output file, we obtain
I2
0.3727‘ 153.4o ,
V1
1.202‘ 33.69o
so that
h12
V1
1
1.202‘ 33.69o ,
h 22
I2
1
0.3727‘ 153.4o
Thus,
[h ]
ª3.771‘ 135o
«
¬« 1.202‘146.3
1.202‘ 33.69o º
»
0.3727‘ 153.4o ¼»
Chapter 19, Solution 79
We follow Example 19.16.
(a)
We set I1 = 1 A and open-circuit the output-port so that I2 = 0. The schematic
is shown below with two VPRINT1s to measure V1 and V2. In the AC Sweep box, we
enter Total Pts = 1, Start Freq = 0.3183, and End Freq = 0.3183. After simulation, the
output file includes
FREQ
VM(1)
VP(1)
3.183 E––01
4.669 E+00
––1.367 E+02
FREQ
VM(4)
VP(4)
3.183 E––01
2.530 E+00
––1.084 E+02
From this,
z11 = V1/I1 = 4.669‘––136.7q/1 = 4.669‘––136.7q
z21 = V2/I1 = 2.53‘––108.4q/1 = 2.53‘––108.4q.
(b)
In this case, we let I2 = 1 A and open-circuit the input port. The schematic is
shown below. In the AC Sweep box, we type Total Pts = 1, Start Freq = 0.3183, and
End Freq = 0.3183. After simulation, the output file includes
FREQ
VM(1)
VP(1)
3.183 E––01
2.530 E+00
––1.084 E+02
FREQ
VM(2)
VP(2)
3.183 E––01
1.789 E+00
––1.534 E+02
From this,
z12 = V1/I2 = 2.53‘––108.4q/1 = 2.53‘––108..4q
Thus,
z22 = V2/I2 = 1.789‘––153.4q/1 = 1.789‘––153.4q.
ª4.669‘ 136.7q 2.53‘ 108.4q º
[z] = «
»
¬ 2.53‘ 108.4q 1.789‘ 153.4q¼
Chapter 19, Solution 80
To get z11 and z21, we open circuit the output port and let I1 = 1A so that
z11
V1
I1
V1,
z 21
V2
I1
V2
The schematic is shown below. After it is saved and run, we obtain
z11
V1
29.88,
z 21
V2
70.37
Similarly, to get z22 and z12, we open circuit the input port and let I2 = 1A so that
z12
V1
I2
V1,
z 22
V2
I2
V2
The schematic is shown below. After it is saved and run, we obtain
z12
V1
3.704,
z 22
[z]
ª 29.88 3.704º
« 70.37 11.11» :
¼
¬
Thus,
V2
11.11
Chapter 19, Solution 81
(a)
We set V1 = 1 and short circuit the output port. The schematic is shown below.
After simulation we obtain
y11 = I1 = 1.5, y21 = I2 = 3.5
(b)
We set V2 = 1 and short-circuit the input port. The schematic is shown below.
Upon simulating the circuit, we obtain
y12 = I1 = ––0.5, y22 = I2 = 1.5
ª1.5 0.5º
[Y] = «
»
¬ 3.5 1.5 ¼
Chapter 19, Solution 82
We follow Example 19.15.
(a)
Set V2 = 0 and I1 = 1A. The schematic is shown below. After simulation, we
obtain
h11 = V1/1 = 3.8, h21 = I2/1 = 3.6
(b)
Set V1 = 1 V and I1 = 0. The schematic is shown below. After simulation, we
obtain
h12 = V1/1 = 0.4, h22 = I2/1 = 0.25
Hence,
ª 3.8 0.4 º
[h] = «
»
¬ 3.6 0.25¼
Chapter 19, Solution 83
To get A and C, we open-circuit the output and let I1 = 1A. The schematic is shown
below. When the circuit is saved and simulated, we obtain V1 = 11 and V2 = 34.
A
V1
V2
0.3235,
C
I1
V2
1
34
0.02941
Similarly, to get B and D, we open-circuit the output and let I1 = 1A. The schematic
is shown below. When the circuit is saved and simulated, we obtain V1 = 2.5 and I2
= -2.125.
B
V
1
I2
2.5
2.125
1.1765,
D
I
1
I2
1
2.125
Thus,
[T ]
ª 0.3235 1.1765 º
«0.02941 0.4706»
¬
¼
0.4706
Chapter 19, Solution 84
(a)
Since A =
V1
V2
and C =
I2 0
I1
V2
, we open-circuit the output port and let V1
I2 0
= 1 V. The schematic is as shown below. After simulation, we obtain
A = 1/V2 = 1/0.7143 = 1.4
C = I2/V2 = 1.0/0.7143 = 1.4
(b)
To get B and D, we short-circuit the output port and let V1 = 1. The schematic is
shown below. After simulating the circuit, we obtain
B = ––V1/I2 = ––1/1.25 = ––0.8
D = ––I1/I2 = ––2.25/1.25 = ––1.8
ªA B º
ª1.4 0.8º
« C D» = «1.4 1.8 »
¬
¼
¬
¼
Thus
Chapter 19, Solution 85
(a)
Since A =
V1
V2
and C =
I2 0
I1
V2
, we let V1 = 1 V and openI2 0
circuit the output port. The schematic is shown below. In the AC Sweep box, we set
Total Pts = 1, Start Freq = 0.1592, and End Freq = 0.1592. After simulation, we obtain
an output file which includes
FREQ
1.592 E––01
IM(V_PRINT1)
6.325 E––01
IP(V_PRINT1)
1.843 E+01
FREQ
1.592 E––01
VM($N_0002)
6.325 E––01
VP($N_0002)
––7.159 E+01
From this, we obtain
A =
1
V2
1
1.581‘71.59q
0.6325‘ 71.59q
C =
(b)
I1
V2
0.6325‘18.43q
0.6325‘ 71.59q
V1
I2
Similarly, since B =
1‘90q = j
and D = V2 0
I1
I2
, we let V1 = 1 V and shortV2 0
circuit the output port. The schematic is shown below. Again, we set Total Pts = 1, Start
Freq = 0.1592, and End Freq = 0.1592 in the AC Sweep box. After simulation, we get
an output file which includes the following results:
FREQ
1.592 E––01
IM(V_PRINT1)
5.661 E––04
IP(V_PRINT1)
8.997 E+01
FREQ
1.592 E––01
IM(V_PRINT3)
9.997 E––01
IP(V_PRINT3)
––9.003 E+01
From this,
B = 1
I2
1
0.9997‘ 90q
D = I1
I2
5.661x10 4 ‘89.97q
= 5.561x10––4
0.9997‘ 90q
1‘90q
j
j
ªA B º
ª1.581‘71.59q
º
« C D» = «
4 »
j
5.661x10 ¼
¬
¼
¬
Chapter 19, Solution 86
(a)
By definition, g11 =
I1
V1
, g21 =
I2 0
V1
V2
.
I2 0
We let V1 = 1 V and open-circuit the output port. The schematic is shown below. After
simulation, we obtain
g11 = I1 = 2.7
g21 = V2 = 0.0
(b)
Similarly,
g12 =
I1
I2
, g22 =
V1 0
V2
I2
V1 0
We let I2 = 1 A and short-circuit the input port. The schematic is shown below. After
simulation,
g12 = I1 = 0
g22 = V2 = 0
ª 2.727S 0º
[g] = «
0»¼
¬ 0
Thus
Chapter 19, Solution 87
(a)
Since
a =
V2
V1
and c =
I1 0
I2
V1
,
I1 0
we open-circuit the input port and let V2 = 1 V. The schematic is shown below. In the
AC Sweep box, set Total Pts = 1, Start Freq = 0.1592, and End Freq = 0.1592. After
simulation, we obtain an output file which includes
FREQ
1.592 E––01
IM(V_PRINT2)
5.000 E––01
IP(V_PRINT2)
1.800 E+02
FREQ
1.592 E––01
VM($N_0001)
5.664 E––04
VP($N_0001)
8.997 E+01
From this,
a =
1
5.664x10 4 ‘89.97q
1765‘ 89.97q
c =
0.5‘180q
5.664x10 4 ‘89.97q
882.28‘ 89.97q
(b)
Similarly,
b = V2
I1
and d = V1 0
I2
I1
V1 0
We short-circuit the input port and let V2 = 1 V. The schematic is shown below. After
simulation, we obtain an output file which includes
FREQ
1.592 E––01
IM(V_PRINT2)
5.000 E––01
IP(V_PRINT2)
1.800 E+02
FREQ
1.592 E––01
IM(V_PRINT3)
5.664 E––04
IP(V_PRINT3)
––9.010 E+01
From this, we get
b = d = Thus
1
4
5.664x10 ‘ 90.1q
= ––j1765
0.5‘180q
= j888.28
5.664x10 4 ‘ 90.1q
ª j1765 j1765º
[t] = «
»
¬ j888.2 j888.2 ¼
Chapter 19, Solution 88
To get Z in , consider the network in Fig. (a).
Rs
I1
I2
+
+
Vs
+
Two-Port
V1
Zin
But
RL
V2
I1
I2
y 11 V1 y 12 V2
y 21 V1 y 22 V2
I2
- V2
RL
V2
- y 21 V1
y 22 1 R L
(a)
(1)
(2)
y 21 V1 y 22 V2
(3)
Substituting (3) into (1) yields
or
I1
§ - y 21 V1 ·
¸,
y 11 V1 y 12 ˜ ¨
© y 22 1 R L ¹
I1
§ ' y y 11 YL ·
¸ V1 ,
¨
© y 22 YL ¹
Z in
V1
I1
y 22 YL
' y y 11 YL
Ai
I2
I1
y 21 V1 y 22 V2
I1
y 21 Z in y 22 y 21 Z in
y 22 YL
Ai
y 21 YL
' y y 11 YL
Av
V2
V1
From (3),
- y 21
y 22 YL
YL
'y
y 11 y 22 y 12 y 21
§y
y 21 Z in ¨¨ 22
© I1
§ y 22 YL
¨
¨' y Y
11
L
© y
1
RL
· § - y 21 V1
¸¸ ¨¨
¹ © y 22 YL
·
¸¸
¹
·§
¸¨ y 21 y 22 y 21
¸¨
y 22 YL
¹©
·
¸¸
¹
To get Z out , consider the circuit in Fig. (b).
I1
I2
+
Rs
+
Two-Port
V1
V2
(b)
But
Z out
V2
I2
V2
y 21 V1 y 22 V2
V1
- R s I1
Substituting this into (1) yields
I 1 - y 11 R s I 1 y 12 V2
(1 y 11 R s ) I 1 y 12 V2
I1
or
V1
V2
y 12 V2
1 y 11 R s
- V1
Rs
- y 12 R s
1 y 11 R s
Substituting this into (4) gives
1
Z out
y 12 y 21 R s
y 22 1 y 11 R s
y 22
Z out
1 y 11 R s
y 11 y 22 R s y 21 y 22 R s
y 11 Ys
' y y 22 Ys
Zout
(4)
Chapter 19, Solution 89
Av
- h fe R L
h ie (h ie h oe h re h fe ) R L
Av
- 72 ˜ 10 5
2640 (2640 u 16 u 10 -6 2.6 u 10 -4 u 72) ˜ 10 5
Av
- 72 ˜ 10 5
2640 1824
dc gain
- 1613
20 log A v
20 log (1613)
64.15
Chapter 19, Solution 90
(a)
Z in
1500
500
h ie h re h fe R L
1 h oe R L
10 -4 u 120 R L
2000 1 20 u 10 -6 R L
12 u 10 -3
1 2 u 10 -5 R L
500 10 -2 R L 12 u 10 -3 R L
500 u 10 2 0.2 R L
R L 250 k:
(b)
Av
- h fe R L
h ie (h ie h oe h re h fe ) R L
Av
- 120 u 250 u 10 3
2000 (2000 u 20 u 10 -6 120 u 10 -4 ) u 250 u 10 3
Av
- 30 u 10 6
2 u 10 3 7 u 10 3
- 3333
Ai
(c)
h fe
1 h oe R L
120
1 20 u 10 -6 u 250 u 10 3
Z out
R s h ie
(R s h ie ) h oe h re h fe
Z out
2600
k:
40
Av
Vc
Vb
20
600 2000
(600 2000) u 20 u 10 -6 10 -4 u 120
65 k:
Vc
Vs

o Vc
A v Vs
-3333 u 4 u 10 -3
- 13.33 V
Chapter 19, Solution 91
Rs
1.2 k: ,
RL
4 k:
Av
- h fe R L
h ie (h ie h oe h re h fe ) R L
Av
- 80 u 4 u 10 3
1200 (1200 u 20 u 10 -6 1.5 u 10 -4 u 80) u 4 u 10 3
Av
- 32000
1248
(b)
Ai
h fe
1 h oe R L
(c)
Z in
h ie h re A i
Z in
1200 1.5 u 10 -4 u 74.074 # 1.2 k:
(a)
(d)
- 25.64
80
1 20 u 10 -6 u 4 u 10 3
Z out
R s h ie
(R s h ie ) h oe h re h fe
Z out
1200 1200
2400 u 20 u 10 -6 1.5 u 10 -4 u 80
74.074
2400
0.0468
51.282 k:
Chapter 19, Solution 92
Due to the resistor R E 240 : , we cannot use the formulas in section 18.9.1. We will
need to derive our own. Consider the circuit in Fig. (a).
Rs
Ib
hie
Ic
+
+
hre Vc
Vs
+
+
hfe Ib
Vb
Vc
IE
RE
(a)
Zin
But
hoe
IE
Ib Ic
(1)
Vb
h ie I b h re Vc (I b I c ) R E
(2)
Ic
h fe I b Vc
Vc
RE 1
(3)
h oe
-Ic R L
(4)
Substituting (4) into (3),
Ic
or
h fe I b RL
RE 1
Ic
h oe
h fe (1 R E h oe )
1 h oe (R L
Ai
Ic
Ib
Ai
100(1 240x30 x10 6 )
1 30 u 10 -6 (4,000 240)
Ai
79.18
From (3) and (5),
(5)
RL
Ic
h fe (1 R E )h oe
Ib
1 h oe (R L R E )
h fe I b Vc
RE 1
(6)
h oe
Substituting (4) and (6) into (2),
Vb (h ie R E ) I b h re Vc I c R E
Vb
1
Av
Vb
Vc
1
Av
Vc (h ie R E )
V
h re Vc c R E
RL
º
§
1 · ª h fe (1 R E h oe )
¸¸ «
¨¨ R E h fe »
h oe ¹ ¬1 h oe (R L R E )
¼
©
(h ie R E )
º
§
1 · ª h fe (1 R E h oe )
¸¸ «
¨¨ R E h fe »
h oe ¹ ¬1 h oe (R L R E )
¼
©
1
§
¨ 240 30 x10 6
©
h re RE
RL
(7)
(4000 240)
240
10 -4 6
4000
º
· ª100(1 240 x 30 x10 )
100»
¸«
-6
¹ ¬ 1 30 u 10 u 4240
¼
1
Av
6.06x10 3 10 -4 0.06
Av
––15.15
-0.066
From (5),
Ic
h fe
I
1 h oe R L b
We substitute this with (4) into (2) to get
Vb (h ie R E ) I b (R E h re R L ) I c
Vb
§ h (1 R E h oe )
·
(h ie R E ) I b (R E h re R L ) ¨¨ fe
I b ¸¸
© 1 h oe (R L R E ) ¹
Z in
Vb
Ib
Z in
(100)(240 u 10 -4 u 4 u 10 3 )(1 240x30x10 6 )
4000 240 1 30 u 10 -6 u 4240
Z in
12.818 k:
h ie R E h fe (R E h re R L )(1 R E h oe )
1 h oe (R L R E )
(8)
To obtain Z out , which is the same as the Thevenin impedance at the output, we introduce
a 1-V source as shown in Fig. (b).
Rs
hie
Ib
Ic
+
+
+
hre Vc
Vb
hfe Ib
hoe
IE
RE
Zout
From the input loop,
I b (R s h ie ) h re Vc R E (I b I c )
Vc
1V
(b)
But
So,
+
Vc
0
1
I b (R s h ie R E ) h re R E I c
(9)
0
From the output loop,
Ic
or
Ib
Vc
RE 1
h oe
h fe I b
h oe
h fe I b
R E h oe 1
h oe
h fe
Ic
h fe 1 R E h oe
(10)
Substituting (10) into (9) gives
§I
(R s R E h ie ) ¨¨ c
© h fe
·
¸¸ h re R E I c ¹
R s R E h ie
Ic R E Ic
h fe
h
·
(R s R E h ie )§¨ oe
h fe ¸¹
©
1 R E h oe
R s R E h ie
1 R E h oe
§ h oe
¨¨
© h fe
·
¸¸ h re
¹
0
Ic
Z out
Z out
Z out
ª R R E h ie º
(h oe h fe ) « s
» h re
¬ 1 R E h oe ¼
R E (R s R E h ie ) h fe
1
Ic
R E h fe R s R E h ie
ª R s R E h ie º
«
» h oe h re h fe
¬ 1 R E h oe ¼
240 u 100 (1200 240 4000)
ª1200 240 4000 º
-6
-4
« 1 240 x 30 x10 6 » u 30 u 10 10 u 100
¬
¼
24000 5440
0.152
193.7 k:
Chapter 19, Solution 93
We apply the same formulas derived in the previous problem.
1
Av
1
Av
(h ie R E )
§
º
1 · ª h fe (1 R E h oe )
¨¨ R E ¸¸ «
h fe »
h oe ¹ ¬1 h oe (R L R E )
©
¼
h re RE
RL
200
(2000 200)
2.5 u 10 -4 3800
ª150(1 0.002)
º
150»
(200 10 5 ) «
¬ 1 0.04
¼
1
Av
0.004 2.5 u 10- 4 0.05263
Av
––17.74
Ai
h fe (1 R E h oe )
1 h oe (R L R E )
Z in
h ie R E -0.05638
150(1 200x10 5 )
1 10 -5 u (200 3800)
h fe (R E h re R L )(1 R E h oe )
1 h oe (R L R E )
144.5
Z in
2000 200 Z in
Z in
2200 28966
31.17 k:
R E h fe R s R E h ie
ª R s R E h ie º
«
» h oe h re h fe
¬ 1 R E h oe ¼
Z out
200 u 150 1000 200 2000
ª 3200 u 10 -5 º
-4
» 2.5 u 10 u 150
«
¬ 1.002 ¼
Z out
Z out
(150)(200 2.5 u 10 -4 u 3.8 u 10 3 )(1.002)
1.04
33200
- 0.0055
––6.148 M:
Chapter 19, Solution 94
We first obtain the ABCD parameters.
ª 200 0 º
[h] «
Given
»,
¬ 100 10 -6 ¼
ª
«
[T] «
«
¬
'h
h 21
- h 22
h 21
- h11
h 21
-1
h 21
'h
º
» ª - 2 u 10 -6
» «
-8
» ¬ - 10
¼
h11 h 22 h12 h 21
-2 º
»
- 10 -2 ¼
The overall ABCD parameters for the amplifier are
ª - 2 u 10 -6
- 2 ºª - 2 u 10 -6
-2
[T] «
-8
-2 »«
-8
- 10 ¼¬ - 10
- 10 -2
¬ - 10
'T
2 u 10 -12 2 u 10 -12
ªB
«
[h] « D
-1
«¬
D
'T
D
C
D
0
º
0 º
» ª 200
» «¬ - 10 -4 10 -6 »¼
»¼
2 u 10 -4
º ª 2 u 10 -8
»#«
¼ ¬ 10 -10
2 u 10 -2 º
»
10 -4 ¼
Thus,
h re
0,
-10 -4 ,
h ie
200 ,
Av
(10 4 )(4 u 10 3 )
200 (2 u 10 -4 0) u 4 u 10 3
Z in
h ie h re h fe R L
1 h oe R L
h fe
200 0
10 -6
h oe
2 u 10 5
200 :
Chapter 19, Solution 95
Let Z A
1
y 22
s 4 10s 2 8
s 3 5s
Using long division,
5s 2 8
s 3
s 5s
ZA
i.e.
L1
1H
s L1 Z B
and
ZB
5s 2 8
s 3 5s
as shown in Fig (a).
L1
ZB
y22 = 1/ZA
(a)
1
ZB
s 3 5s
5s 2 8
YB
0.2s 3.4s
5s 2 8
C2
0 .2 F
and
YB
Using long division,
where
sC 2 YC
YC
3.4s
5s 2 8
as shown in Fig. (b).
L1
C2
Yc = 1/ZC
(b)
ZC
1
YC
5s 2 8
3.4s
5s
8
3.4 3.4s
i.e. an inductor in series with a capacitor
5
L3
1.471 H and
3.4
C4
s L3 3.4
8
1
s C4
0.425 F
Thus, the LC network is shown in Fig. (c).
0.425 F
1.471 H
1H
0.2 F
(c)
Chapter 19, Solution 96
This is a fourth order network which can be realized with the network shown in Fig. (a).
L1
L3
C2
C4
(a)
' (s)
(s 4 3.414s 2 1) (2.613s 3 2.613s)
1:
1
2.613s 2.613s
s 4 3.414s 2 1
1
2.613s 3 2.613s
3
H(s)
which indicates that
-1
2.613s 2.613s
s 4 3.414s 1
2.613s 3 2.613s
y 21
3
y 22
We seek to realize y 22 . By long division,
2.414s 2 1
y 22 0.383s 2.613s 3 2.613s
i.e.
C4
and
0.383 F
YA
s C 4 YA
2.414s 2 1
2.613s 3 2.613s
as shown in Fig. (b).
L1
YA
L3
C2
C4
y22
(b)
ZA
1
YA
2.613s 3 2.613s
2.414s 2 1
By long division,
i.e.
ZA
1.082s L3
1.082 H
1.531s
2.414s 2 1
and
s L3 Z B
ZB
1.531s
2.414s 2 1
as shown in Fig.(c).
L1
ZB
L3
C2
C4
(c)
i.e.
YB
1
ZB
C2
1.577 F
1.577s 1
1.531s
s C2 and
L1
1
s L1
1.531 H
Thus, the network is shown in Fig. (d).
1.531 H
1.577 F
1.082 H
1:
0.383 F
(d)
Chapter 19, Solution 97
H(s)
s3
(s 3 12s) (6s 2 24)
Hence,
y 22
6s 2 24
s 3 12s
1
ZA
s C3
where Z A is shown in the figure below.
s3
s 3 12s
6s 2 24
1 3
s 12s
(1)
C1
C3
L2
ZA
y22
We now obtain C 3 and Z A using partial fraction expansion.
Let
6s 2 24
s (s 2 12)
6s 2 24
A Bs C
s s 2 12
A (s 2 12) Bs 2 Cs
Equating coefficients :
s0 :
24 12A 
o A 2
1
s :
0 C
2
s :
6 AB 
o B 4
Thus,
6s 2 24
2
4s
2
2
s (s 12) s s 12
Comparing (1) and (2),
1
C3
A
But
(2)
1
F
2
1
ZA
s 2 12
4s
1
ZA
sC1 1
3
s
4
s
(3)
1
s L2
Comparing (3) and (4),
1
1
C1
F
and
L2
H
4
3
Therefore,
C1 0.25 F ,
L 2 0.3333 H ,
(4)
C3
0.5 F
Chapter 19, Solution 98
'h
1 0 .8
[Ta ]
[Tb ]
[T ]
0 .2
ª ' h / h 21 h11 / h 21 º
« h / h
»
¬ 22 21 1 / h 21 ¼
[Ta ][Tb ]
ª 0.001
« 2.5x10 6
¬
10 º
0.005»¼
ª2.6x105
0.06 º
«
»
8
5x105 ¼»
¬«1.5x10
We now convert this to z-parameters
[z]
ª1.733x103
«
7
¬«6.667 x10
ªA / C ' T / Cº
« 1/ C D / C »
¬
¼
1000
I1
0.0267 º
»
3.33x103 ¼»
z11
+
z22
+
I2
+
+
Vs
z12 I2
-
-
z21 I1
Vo
ZL
-
Vs
(1000 z11)I1 z12 I 2
(1)
Vo
z 22 I 2 z 21I1
(2)
But Vo
I 2 ZL
o
I2
Vo / ZL
(3)
Substituting (3) into (2) gives
I1
·
§ 1
z
Vo ¨¨
22 ¸¸
© z 21 z 21ZL ¹
We substitute (3) and (4) into (1)
(4)
Vs
·
§ 1
z
z
(1000 z11)¨¨
22 ¸¸ Vo 12 Vo
ZL
© z11 z 21ZL ¹
7.653x10 4 2.136 x105
744PV
Chapter 19, Solution 99
Z ab
Z1 Z 3
Z1 Z 3
Z cd
Z ac
Z c (Z a Z b )
Za Zb Zc
Z2 Z3
Z2 Z3
(1)
Z a || (Z b Z c )
Z a (Z b Z c )
Za Zb Zc
Z1 Z 2
Z1 Z 2
Z c || (Z b Z a )
(2)
Z b || (Z a Z c )
Z b (Z a Z c )
Za Zb Zc
(3)
Z b (Z c Z a )
Za Zb Zc
(4)
Subtracting (2) from (1),
Z1 Z 2
Adding (3) and (4),
Z1
ZbZc
Za Zb Zc
(5)
Subtracting (5) from (3),
Z2
ZaZb
Za Zb Zc
(6)
Subtracting (5) from (1),
Z3
ZcZa
Za Zb Zc
(7)
Using (5) to (7)
Z1Z 2 Z 2 Z 3 Z 3 Z1
Z a Z b Z c (Z a Z b Z c )
(Z a Z b Z c ) 2
Z1Z 2 Z 2 Z 3 Z 3 Z1
Za ZbZc
Za Zb Zc
(8)
Dividing (8) by each of (5), (6), and (7),
Z1Z 2 Z 2 Z 3 Z 3 Z1
Za
Z1
Zb
Z1Z 2 Z 2 Z 3 Z 3 Z1
Z3
Zc
Z1Z 2 Z 2 Z 3 Z 3 Z1
Z2
as required. Note that the formulas above are not exactly the same as those in Chapter 9
because the locations of Z b and Z c are interchanged in Fig. 18.122.