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 = 1200π cos 120π t pA
i =dq/dt = − e −4t (80 cos 50 t + 1000 sin 50 t ) µ A
Chapter 1, Solution 3
(a) q(t) = ∫ i(t)dt + q(0) = (3t + 1) C
(b) q(t) = ∫ (2t + s) dt + q(v) = (t 2 + 5t) mC
(c) q(t) = ∫ 20 cos (10t + π / 6 ) + q(0) = (2sin(10t + π / 6) + 1) µ C
(d)
10e -30t
( −30 sin 40t - 40 cos t)
900 + 1600
= − e - 30t (0.16cos40 t + 0.12 sin 40t) C
q(t) = ∫ 10e -30t sin 40t + q(0) =
Chapter 1, Solution 4
q = ∫ idt = ∫
=
10
−5
5sin 6 π t dt =
cos 6π t
6π
0
5
(1 − cos 0.06π ) = 4.698 mC
6π
Chapter 1, Solution 5
q = ∫ idt = ∫
=
1
e dt mC = - e -2t
2
1
(1 − e 4 ) mC = 490 µC
2
Chapter 1, Solution 6
(a) At t = 1ms, i =
dq 80
=
= 40 mA
dt
2
(b) At t = 6ms, i =
dq
= 0 mA
dt
(c) At t = 10ms, i =
dq 80
=
= - 20 mA
4
dt
Chapter 1, Solution 7
25A,
dq 
i=
= - 25A,
dt 
 25A,
2
-2t
0<t<2
2<t<6
6<t<8
which is sketched below:
0
Chapter 1, Solution 8
q = ∫ idt =
10 × 1
+ 10 × 1 = 15 µC
2
Chapter 1, Solution 9
1
(a) q = ∫ idt = ∫ 10 dt = 10 C
0
3
5 ×1

q = ∫ idt = 10 × 1 + 10 −
 + 5 ×1
0
(b)
2 

= 15 + 10 − 25 = 22.5 C
5
(c) q = ∫ idt = 10 + 10 + 10 = 30 C
0
Chapter 1, Solution 10
q = ixt = 8 x10 3 x15 x10 − 6 = 120 µ 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
∫
t
∫
q (t ) = idt + q (0 ) = 3tdt + 0 = 1.5t 2
0
0
At t=6, q(6) = 1.5(6)2 = 54
For 6 < t < 10s,
t
t
∫
∫
q (t ) = idt + q (6 ) = 18 dt + 54 = 18 t − 54
6
6
At t=10, q(10) = 180 – 54 = 126
For 10<t<15s,
t
∫
t
∫
q (t ) = idt + q (10 ) = ( −12)dt + 126 = −12t + 246
10
10
At t=15, q(15) = -12x15 + 246 = 66
For 15<t<20s,
t
∫
q (t ) = 0 dt + q (15) =66
15
Thus,

1.5t 2 C, 0 < t < 6s

 18 t − 54 C, 6 < t < 10s
q (t ) = 
−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
2
2
w = ∫ vidt = ∫ 1200 cos 2 4 t dt
0
0
2
= 1200 ∫ ( 2 cos 8t - 1)dt (since, cos 2 x = 2 cos 2x - 1)
0
2
2

1

= 1200 sin 8t − t  = 1200 sin 16 − 2 
8
0
4

= - 2.486 kJ
Chapter 1, Solution 14
q = ∫ idt = ∫ 10(1 - e -0.5t )dt = 10(t + 2e -0.5t )
1
(a)
(b)
0
= 10(1 + 2e
-0.5
− 2 ) = 2.131 C
1
0
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 = ∫ idt = ∫
2
0
− 3 2t
3e dt =
e
2
2
-2t
= −1.5(e − 1) = 1.297 C
0
-2
(b)
5di
= −6e 2t ( 5) = −30e -2t
dt
p = vi = − 90 e − 4 t W
v=
3
(c) w = ∫ pdt = -90∫ e -4t dt =
0
3
− 90 -4t
e
= − 22.5 J
−4
0
Chapter 1, Solution 16

0<t<2
 25t mA
i(t) = 
,
100 - 25t mA 2 < t < 4

1
0< t <1
 10t V

v(t) = 10 V
1< t < 3

40 - 10t V 3 < t < 4
2
3
4
2
3
w = ∫ v(t)i(t)dt = ∫ 10 + (25t)dt + ∫ 10( 25t)dt + ∫ 10(100 − 25t)dt + ∫ ( 40 − 10t)(100 - 25t)mJ
0
1
=
1
3
2
4

250 3
250
t2 
t +
+ 250 4 t -  + ∫ 250( 4 − t) 2 dt
3
2 1
2 2 3
0

4

250 250
9
t2 


2
( 3) + 25012 − − 8 + 2  + 25016 t - 4t + 
=
+
3
2
2
3 3



= 916.7 mJ
Chapter 1, Solution 17
Σ p = 0 → -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

→
−4 I s − 2 x6 − 13 x 2 + 5 x10 = 0

→
Is = 3 A
Chapter 1, Solution 20
Since Σ p = 0
-30×6 + 6×12 + 3V0 + 28 + 28×2 - 3×10 = 0
72 + 84 + 3V0 = 210 or 3V0 = 54
V0 = 18 V
Chapter 1, Solution 21
i=
=
∆q
 photon  1  electron 
= 4 × 1011 
 ⋅ 1. 6 × 1019 ( C / electron)
 ⋅ 
∆t
 sec  8  photon 
4
× 1011 × 1. 6 × 10 −19 C/s = 0.8 × 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 =
p 1500
=
= 12.5 W
v 120
(b) w = pt = 1. 5 × 103 × 45 × 60 ⋅ J = 1.5 ×
(c) Cost = 1.125 × 10 = 11.25 cents
45
kWh = 1.125 kWh
60
Chapter 1, Solution 24
p = vi = 110 x 8 = 880 W
Chapter 1, Solution 25
4
Cost = 1.2 kW × hr × 30 × 9 cents/kWh = 21.6 cents
6
Chapter 1, Solution 26
0. 8A ⋅ h
= 80 mA
10h
(b) p = vi = 6 × 0.08 = 0.48 W
(c) w = pt = 0.48 × 10 Wh = 0.0048 kWh
(a) i =
Chapter 1, Solution 27
(a) Let T = 4h = 4 × 36005
T
q = ∫ idt = ∫ 3dt = 3T = 3 × 4 × 3600 = 43.2 kC
0
T
T
0 . 5t 

( b) W = ∫ pdt = ∫ vidt = ∫ ( 3) 10 +
dt
0
0
3600 

4×3600

0. 25t 2 

= 310t +
3600  0

= 475.2 kJ
( c)
= 3[40 × 3600 + 0. 25 × 16 × 3600]
W = 475.2 kWs, (J = Ws)
475.2
Cost =
kWh × 9 cent = 1.188 cents
3600
Chapter 1, Solution 28
(a) i =
P 30
=
= 0.25 A
V 120
( b) W = pt = 30 × 365 × 24 Wh = 262.8 kWh
Cost = $0.12 × 262.8 = $31.54
Chapter 1, Solution 29
(20 + 40 + 15 + 45)
 30 
hr + 1.8 kW  hr
60
 60 
= 2.4 + 0.9 = 3.3 kWh
Cost = 12 cents × 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 
= 2.4 + 0.9 = 3.3 kWh
Cost = 12 cents × 3.3 = 39.6 cents
w = pt = 1. 2kW
Chapter 1, Solution 33
i=
dq
→ q = ∫ idt = 2000 × 3 × 10 3 = 6 C
dt
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 = ∫ p( t ) dt = 400 × 6 + 1000 × 2 + 200 × 12 × 1200 × 2 + 400 × 2
= 7200 + 2800 = 10.4 kWh
( b)
10.4 kW
= 433.3 W/h
24 h
Chapter 1, Solution 36
160A ⋅ h
=4A
40
160Ah 160, 000h
( b) t =
=
= 6,667 days
0.001A 24h / day
(a)
i=
Chapter 1, Solution 37
q = 5 × 10 20 (− 1. 602 × 10 −19 ) = −80. 1 C
W = qv = −80. 1 × 12 = − 901.2 J
Chapter 1, Solution 38
P = 10 hp = 7460 W
W = pt = 7460 × 30 × 60 J = 13.43 × 106 J
Chapter 1, Solution 39
p = vi → i =
p 2 × 10 3
=
= 16.667 A
v
120
Chapter 2, Solution 1
v = iR
i = v/R = (16/5) mA = 3.2 mA
Chapter 2, Solution 2
p = v2/R →
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 Ω
2
3
++++ -
2A
30 Ω
60 Ω
4
40 Ω
10 Ω
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

→
I 2 = −10 A
At node 1,
I1 + I 2 = 2

→
I 1 = 2 − I 2 = 12 A
At node 4,
2 = I4 + 4

→
I 4 = 2 − 4 = −2 A
At node 3,
7 + I4 = I3

→
I3 = 7 − 2 = 5 A
Hence,
I 1 = 12 A,
I 2 = −10 A,
I 3 = 5 A,
I 4 = −2 A
Chapter 2, Solution 14
+
3V
-
+
I3
4V
+
V3 -

→
V4 = 7V
For mesh 2,
+4 + V3 + V4 = 0

→
V3 = −4 − 7 = −11V

→
V1 = V3 + 3 = −8V

→
V2 = −V1 − 2 = 6V
For mesh 3,
−3 + V1 − V3 = 0
For mesh 4,
−V1 − V2 − 2 = 0
Thus,
V1 = −8V ,
V2 = 6V ,
+
- V4
For mesh 1,
−V4 + 2 + 5 = 0
V1
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 = 0

→
v2 = 4V
For loop 2,
− v3 − 8 − 10 = 0

→
v3 = −18V

→
v1 = −6V
For loop 3,
− v1 + 12 + v3 = 0
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 3Ω = 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 Ω
-45 + 10i - 3V0 + 5i = 0
+ v0 -
But v0 = 10i,
-45 + 15i - 30i = 0
P3 = i2R = 9 x 5 = 45W
i = -3A
45V
+
+
-
5Ω
3v0
Chapter 2, Solution 22
4Ω
+ v0 6Ω
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
= −11.111
4
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
1Ω
+
6A
2Ω
vx
3Ω
Applying current division,
ix =
2
(6 A) = 2 A,
2 + 1+ 3
v x = 1i x = 2V
The current through the 1.2- Ω resistor is 0.5ix = 1A. The voltage across the 12- Ω
resistor is 1 x 4.8 = 4.8 V. Hence the power is
p=
v 2 4.8 2
=
= 1.92W
12
R
Chapter 2, Solution 24
(a)
I0 =
Vs
R1 + R2
V0 = −α I0 (R3 R4 ) = −
αV0
R1 + R 2
⋅
R3 R4
R3 + R4
V0
− αR3 R4
=
Vs (R1 + R2 )(R3 + R4 )
(b)
If R1 = R2 = R3 = R4 = R,
V0
α R α
=
⋅ = = 10
VS
2R 2 4
Chapter 2, Solution 25
V0 = 5 x 10-3 x 10 x 103 = 50V
Using current division,
I20 =
5
(0.01x50) = 0.1 A
5 + 20
V20 = 20 x 0.1 kV = 2 kV
p20 = I20 V20 = 0.2 kW
α = 40
Chapter 2, Solution 26
V0 = 5 x 10-3 x 10 x 103 = 50V
Using current division,
I20 =
5
(0.01x50) = 0.1 A
5 + 20
V20 = 20 x 0.1 kV = 2 kV
p20 = I20 V20 = 0.2 kW
Chapter 2, Solution 27
Using current division,
i1 =
4
(20) = 8 A
4+6
i2 =
6
(20) = 12 A
4+6
Chapter 2, Solution 28
We first combine the two resistors in parallel
15 10 = 6 Ω
We now apply voltage division,
v1 =
14
(40) = 20 V
14 + 6
v2 = v3 =
Hence,
6
(40) = 12 V
14 + 6
v1 = 28 V, v2 = 12 V, vs = 12 V
Chapter 2, Solution 29
The series combination of 6 Ω and 3 Ω resistors is shorted. Hence
i2 = 0 = v2
v1 = 12, i1 =
12
= 3A
4
Hence v1 = 12 V, i1 = 3 A, i2 = 0 = v2
Chapter 2, Solution 30
8Ω
i1
i
9A
By current division, i =
6Ω
+
v
-
4Ω
12
(9) = 6 A
6 + 12
i1 = 9 − 6 = 3A, v = 4i1 = 4 x 3 = 12 V
p6 = 12R = 36 x 6 = 216 W
Chapter 2, Solution 31
The 5 Ω resistor is in series with the combination of 10 (4 + 6) = 5Ω .
Hence by the voltage division principle,
v=
5
(20V) = 10 V
5+5
by ohm's law,
i=
v
10
=
= 1A
4 + 6 4+ 6
pp = i2R = (1)2(4) = 4 W
Chapter 2, Solution 32
We first combine resistors in parallel.
20 30 =
20 x30
= 12 Ω
50
10 40 =
10x 40
= 8Ω
50
Using current division principle,
8
12
i1 + i 2 =
(20) = 8A, i 3 + i 4 =
(20) = 12A
8 + 12
20
i1 =
20
(8) = 3.2 A
50
i2 =
30
(8) = 4.8 A
50
i3 =
10
(12) = 2.4A
50
i4 =
40
(12) = 9.6 A
50
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 =
8Ω
i1
10 x15
= 6Ω
10 ( 2 + 13 ) =
25
15 x15
15 (4 + 6) =
= 6Ω
25
12 (6 + 6) = 6Ω
28V
+
v1
-
+
-
6Ω
28
= 2 A and v1 = 6i1 = 12 V
8+6
We now work backward to get i2 and v2.
i1 = 2A
8Ω
6Ω
1A
1A
28V
+
12V
-
+
-
8Ω
i1 = 2A
6Ω
+
6V
-
12 Ω
6Ω
4Ω
1A
0.6A
1A
28V
Thus, v2 =
+
12V
-
+
-
12 Ω
+
6V
-
+
15 Ω
3.6V
v
13
(3 ⋅ 6) = 3 ⋅ 12, i2 = 2 = 0.24
13
15
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 Ω
50V
+
-
a
+
V1
i1 -
30 Ω
I0
+
20 Ω
i2
b
V0 5 Ω
-
-
6Ω
Combining the versions in parallel,
70 30 =
i=
70x30
= 21Ω ,
100
20 15 =
20x 5
=4 Ω
25
50
=2 A
21 + 4
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-Ω resistor is shorted. No current flows through the 1-Ω resistor. Hence v0
is the voltage across the 6Ω resistor.
I0 =
4
4
= =1 A
2 + 3 16 4
v0 = I0 (3 6 ) = 2I 0 = 2 V
Chapter 2, Solution 37
Let I = current through the 16Ω resistor. If 4 V is the voltage drop across the 6 R
combination, then 20 - 4 = 16 V in the voltage drop across the 16Ω resistor.
16
Hence, I =
= 1 A.
16
20
6R
R = 12 Ω
4= 6R=
But I =
=1
6+R
16 + 6 R
Chapter 2, Solution 38
Let I0 = current through the 6Ω resistor. Since 6Ω and 3Ω resistors are in parallel.
6I0 = 2 x 3
R0 = 1 A
The total current through the 4Ω resistor = 1 + 2 = 3 A.
Hence
vS = (2 + 4 + 2 3 ) (3 A) = 24 V
I=
vS
= 2.4 A
10
Chapter 2, Solution 39
(a)
Req = R 0 = 0
(b)
(c)
(d)
(e)
R R
+ = R
2 2
Req = (R + R ) (R + R ) = 2R 2R = R
Req = R R + R R =
1
Req = 3R (R + R R ) = 3R (R + R )
2
3
3Rx R
2 =R
=
3
3R + R
2
 R ⋅ 2R 
Req = R 2R 3R = 3R 

 3R 
2
3Rx R
2
3 = 6R
= 3R
R=
2
11
3
3R + R
3
Chapter 2, Solution 40
Req = 3 + 4 (2 + 6 3) = 3 + 2 = 5Ω
I=
10
10
=
= 2A
Re q 5
Chapter 2, Solution 41
Let R0 = combination of three 12Ω resistors in parallel
1
1
1
1
=
+ +
R o 12 12 12
Ro = 4
R eq = 30 + 60 (10 + R 0 + R ) = 30 + 60 (14 + R )
50 = 30 +
60(14 + R )
74 + R
74 + R = 42 + 3R
or R = 16 Ω
Chapter 2, Solution 42
5x 20
= 4Ω
25
(a)
Rab = 5 (8 + 20 30) = 5 (8 + 12) =
(b)
Rab = 2 + 4 (5 + 3) 8 + 5 10 (6 + 4) = 2 + 4 4 + 5 5 = 2 + 2 + 2.5 = 6.5 Ω
Chapter 2, Solution 43
5x 20 400
+
= 4 + 8 = 12 Ω
25
50
(a)
Rab = 5 20 + 10 40 =
(b)
1
1 
 1
60 20 30 =  +
+ 
 60 20 30 
Rab = 80 (10 + 10) =
−1
=
60
= 10Ω
6
80 + 20
= 16 Ω
100
Chapter 2, Solution 44
(a) Convert T to Y and obtain
20 x 20 + 20 x10 + 10 x 20 800
=
= 80 Ω
10
10
800
R2 =
= 40 Ω = R3
20
R1 =
The circuit becomes that shown below.
R1
a
R3
R2
5Ω
b
R1//0 = 0,
R3//5 = 40//5 = 4.444 Ω
Rab = R2 / /(0 + 4.444) = 40 / /4.444 = 4Ω
(b) 30//(20+50) = 30//70 = 21 Ω
Convert the T to Y and obtain
20 x10 + 10 x 40 + 40 x 20 1400
=
= 35Ω
40
40
1400
1400
R2 =
= 70 Ω , R3 =
= 140 Ω
20
10
The circuit is reduced to
that shown below.
15Ω
R1 =
R1
11 Ω
R2
R3
21 Ω
30 Ω
21 Ω
Combining the resistors in parallel
R1//15 =35//15=10.5, 30//R2=30//70 = 21
leads to the circuit below.
10.5 Ω
11 Ω
21 Ω
140 Ω
21 Ω
21 Ω
Coverting the T to Y leads to the circuit below.
10.5 Ω
11 Ω
R4
R5
R6
R4 =
21x140 + 140 x 21 + 21x 21 6321
=
= 301Ω = R6
21
21
R5 =
6321
= 45.15
140
21 Ω
10.5//301 = 10.15, 301//21 = 19.63
R5//(10.15 +19.63) = 45.15//29.78 = 17.94
Rab = 11 + 17 .94 = 28.94Ω
Chapter 2, Solution 45
(a) 10//40 = 8, 20//30 = 12, 8//12 = 4.8
Rab = 5 + 50 + 4.8 = 59.8 Ω
(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 = 32.5Ω
Chapter 2, Solution 46
(a)
30x 70
60 + 20
+ 40 +
100
80
Rab = 30 70 + 40 + 60 20 =
= 21 + 40 + 15 = 76 Ω
(b)
The 10-Ω, 50-Ω, 70-Ω, and 80-Ω resistors are shorted.
20 30 =
20x30
= 12Ω
50
40 60 =
40x 60
= 24
100
Rab = 8 + 12 + 24 + 6 + 0 + 4 = 54 Ω
Chapter 2, Solution 47
5 20 =
6 3=
5x 20
= 4Ω
25
6x3
= 2Ω
9
10 Ω
8Ω
a
b
4Ω
Rab = 10 + 4 + 2 + 8 = 24 Ω
2Ω
Chapter 2, Solution 48
R 1 R 2 + R 2 R 3 + R 3 R 1 100 + 100 + 100
=
= 30
R3
10
Ra = Rb = Rc = 30 Ω
(a)
Ra =
(b)
Ra =
30x 20 + 30x50 + 20x 50 3100
=
= 103.3Ω
30
30
3100
3100
Rb =
= 155Ω, R c =
= 62Ω
20
50
Ra = 103.3 Ω, Rb = 155 Ω, Rc = 62 Ω
Chapter 2, Solution 49
(a)
(b)
R1 =
RaRc
12 + 12
=
= 4Ω
Ra + Rb + Rc
36
R1 = R2 = R3 = 4 Ω
60x30
= 18Ω
60 + 30 + 10
60 x10
R2 =
= 6Ω
100
30x10
R3 =
= 3Ω
100
R1 =
R1 = 18Ω, R2 = 6Ω, R3 = 3Ω
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 
2
4
3R + R = R
2
800 x 10-3 = (30 x 10-3)2 R
P = I2 R
R = 889 Ω
Chapter 2, Solution 51
30 30 = 15Ω and 30 20 = 30 x 20 /(50) = 12Ω
(a)
Rab = 15 (12 + 12) = 15x 24 /(39) = 9.31 Ω
a
a
30 Ω
30 Ω
30 Ω
30 Ω
b
20 Ω
12 Ω
15 Ω
12 Ω
20 Ω
b
Converting the T-subnetwork into its equivalent ∆ network gives
(b)
Ra'b' = 10x20 + 20x5 + 5x10/(5) = 350/(5) = 70 Ω
Rb'c' = 350/(10) = 35Ω, Ra'c' = 350/(20) = 17.5 Ω
Also
30 70 = 30 x 70 /(100) = 21Ω and 35/(15) = 35x15/(50) = 10.5
Rab = 25 + 17.5 (21 + 10.5) = 25 + 17.5 31.5
Rab = 36.25 Ω
30 Ω
30 Ω
a
25 Ω
10 Ω
5Ω
b
20 Ω
a
15 Ω
25 Ω
a’
17.5 Ω
b
70 Ω
b’
35 Ω
c’
15 Ω
c’
Chapter 2, Solution 52
(a) We first convert from T to ∆ .
100 Ω
a
100 Ω
a
100 Ω
100 Ω
100 Ω
100 Ω
b
100 Ω
200 Ω
100 Ω
100 Ω
200 Ω
b
R1 =
100 Ω
100 Ω
100 Ω
100 Ω
100 Ω
100 Ω
R3
R2
100x 200 + 200x 200 + 200 x100 80000
=
= 800Ω
100
100
R2 = R3 = 80000/(200) = 400
100x 400
But
100 400 =
= 80Ω
500
We connect the ∆ to Y.
100 Ω
a
b
100 Ω
a
100 Ω
100 Ω
80 Ω
100 Ω
100 Ω
80 Ω
800 Ω
b
100 Ω
100 Ω
Rb
100 Ω
100 Ω
Rc
80 x800
64,000 400
=
=
Ω
80 + 80 + 800
960
3
80x80 20
=
Ω
Rb =
960
3
Ra = Rc =
We convert T to ∆ .
a
500/3 Ω
100 Ω
320/3 Ω
b
100 Ω
500/3 Ω
500/3 Ω
a
R2’
R1’
R3’
b
Ra
500/3 Ω
R1
R 1' =
320
320
+ 100 x
3
3 = 94,000 /(3) = 293.75Ω
320
320 /(3)
3
100 x100 + 100 x
R '2 = R 13 =
94,000 /(3)
= 313.33
100
940 /(30) 500 /(3) =
940 /(3) x500 /(3)
= 108.796
1440 /(3)
Rab = 293.75 (2 x108.796) =
293.75x 217.6
= 125 Ω
511.36
Converting the Ts to ∆ s, we have the equivalent circuit below.
(b)
100 Ω
100 Ω
a
100 Ω
300 Ω
300 Ω
100 Ω
300 Ω
100 Ω
100 Ω
100 Ω
100 Ω
a
100 Ω
300 Ω
300 Ω
b
300 Ω
100 Ω
100 Ω
100 Ω
100 Ω
b
100 Ω
300 100 = 300 x100 /(400) = 75, 300 (75 + 75) = 300 x150 /(450) = 100
Rab = 100 + 100 300 + 100 = 200 + 100 x 300 /(400)
Rab = 2.75 Ω
100 Ω
300 Ω
100 Ω
300 Ω
300 Ω
100 Ω
100 Ω
Chapter 2, Solution 53
(a)
Converting one ∆ to T yields the equivalent circuit below:
30 Ω
a’
20 Ω
a
60 Ω
b’
b
4Ω
20 Ω
c’
5Ω
80 Ω
40 x10
10 x50
40x50
= 4Ω, R b 'n =
= 5Ω, R c 'n =
= 20Ω
40 + 10 + 50
100
100
Rab = 20 + 80 + 20 + (30 + 4) (60 + 5) = 120 + 34 65
Ra'n =
Rab = 142.32 Ω
(a) We combine the resistor in series and in parallel.
30 (30 + 30) =
30x 60
= 20Ω
90
We convert the balanced ∆ s to Ts as shown below:
a
30 Ω
30 Ω
a
10 Ω
30 Ω
30 Ω
20 Ω
10 Ω
30 Ω
b
30 Ω
10 Ω
10 Ω
10 Ω
10 Ω
b
Rab = 10 + (10 + 10) (10 + 20 + 10) + 10 = 20 + 20 40
Rab = 33.33 Ω
Chapter 2, Solution 54
(a) Rab = 50 + 100 / /(150 + 100 + 150 ) = 50 + 100 / /400 = 130 Ω
(b) Rab = 60 + 100 / /(150 + 100 + 150 ) = 60 + 100 / /400 = 140 Ω
20 Ω
Chapter 2, Solution 55
We convert the T to ∆ .
I0
a
24 V
+
-
I0
20 Ω
40 Ω
60 Ω
10 Ω
20 Ω
50 Ω
a
140 Ω
60 Ω
24 V
+
35 Ω
-
70 Ω
70 Ω
b
b
Req
Req
R R + R 2 R 3 + R 3 R 1 20 x 40 + 40 x10 + 10 x 20 1400
Rab = 1 2
=
=
= 35Ω
R3
40
40
Rac = 1400/(10) = 140Ω, Rbc = 1400/(40) = 35Ω
70 70 = 35 and 140 160 = 140x60/(200) = 42
Req = 35 (35 + 42) = 24.0625Ω
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 Ω
+
100 V
-
16 Ω
15 Ω
35 Ω
12 Ω
30 Ω
16 Ω
10 Ω
20 Ω
Req
15x10 + 10x12 + 12x15 450
=
= 37.5Ω
12
12
Rac = 450/(10) = 45Ω, Rbc = 450/(15) = 30Ω
Rab =
Combining the resistors in parallel,
+
100 V
35 Ω
Req
a
37.5 Ω
30 Ω
45 Ω
c
b
20 Ω
30||20 = (600/50) = 12 Ω,
37.5||30 = (37.5x30/67.5) = 16.667 Ω
35||45 = (35x45/80) = 19.688 Ω
Req = 19.688||(12 + 16.667) = 11.672Ω
By voltage division,
v =
11.672
100 = 42.18 V
11.672 + 16
Chapter 2, Solution 57
4Ω a
2Ω
27 Ω
1Ω
18 Ω
b
d
10 Ω
36 Ω
c
e
7Ω
14 Ω
28 Ω
f
6x12 + 12x8 + 8x 6 216
=
= 18 Ω
12
12
Rac = 216/(8) = 27Ω, Rbc = 36 Ω
4x 2 + 2x8 + 8x 4 56
Rde =
=
7Ω
8
8
Ref = 56/(4) = 14Ω, Rdf = 56/(2) = 28 Ω
Rab =
Combining resistors in parallel,
280
36x 7
= 7.368Ω, 36 7 =
= 5.868Ω
38
43
27 x 3
27 3 =
= 2.7Ω
30
10 28 =
4Ω
4Ω
18 Ω
5.868 Ω
7.568 Ω
1.829 Ω
2.7 Ω
3.977 Ω
0.5964 Ω
14 Ω
7.568 Ω
14 Ω
18x 2.7
18x 2.7
=
= 1.829 Ω
18 + 2.7 + 5.867 26.567
18x5.868
=
= 3.977 Ω
26.567
5.868x 2.7
=
= 0.5904 Ω
26.567
= 4 + 1.829 + (3.977 + 7.368) (0.5964 + 14)
R an =
R bn
R cn
R eq
= 5.829 + 11.346 14.5964 = 12.21 Ω
i = 20/(Req) = 1.64 A
Chapter 2, Solution 58
The resistor of the bulb is 120/(0.75) = 160Ω
40 Ω
2.25 A
+ 90 V - 0.75 A
VS
+
-
160 Ω
1.5 A
+
80 Ω
120
Once the 160Ω and 80Ω resistors are in parallel, they have the same voltage 120V.
Hence the current through the 40Ω 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 = 42.35Ω
p = i2R
i = 1.2A + 5% = 1.2 ± 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 Ω
p = I2R = 70.52W or 57.76W, cost = $1.35
(c)
Use R2 and R3:
R = R 2 R 3 = 90 100 = 47.37Ω
p = I2R = 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 =
= 0.04Ω
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 = 10A
R=
When Rx is maximum, ix = 1A
110
= 11 Ω
10
R + Rx =
i.e., Rx = 110 - R = 99 Ω
Thus, R = 11 Ω,
Rx = 99 Ω
110
= 110 Ω
1
Chapter 2, Solution 65
Rn =
Vfs
50
− Rm =
− 1 kΩ = 4 kΩ
10mA
I fs
Chapter 2, Solution 66
20 kΩ/V = sensitivity =
1
I fs
1
kΩ / V = 50 µA
20
V
The intended resistance Rm = fs = 10(20kΩ / V) = 200kΩ
I fs
V
50 V
(a)
R n = fs − R m =
− 200 kΩ = 800 kΩ
i fs
50 µA
i.e., Ifs =
(b)
p = I fs2 R n = (50 µA) 2 (800 kΩ) = 2 mW
Chapter 2, Solution 67
(a)
By current division,
i0 = 5/(5 + 5) (2 mA) = 1 mA
V0 = (4 kΩ) i0 = 4 x 103 x 10-3 = 4 V
(b)
4k 6k = 2.4kΩ. By current division,
5
(2mA) = 1.19 mA
1 + 2.4 + 5
v '0 = (2.4 kΩ)(1.19 mA) = 2.857 V
i '0 =
v 0 − v '0
1.143
x 100% =
x100 = 28.57%
v0
4
(c)
% error =
(d)
4k 30 kΩ = 3.6 kΩ. By current division,
5
(2mA) = 1.042mA
1 + 3.6 + 5
v '0 (3.6 kΩ)(1.042 mA) = 3.75V
i '0 =
% error =
v − v '0
0.25x100
= 6.25%
x100% =
v0
4
Chapter 2, Solution 68
(a)
40 = 24 60Ω
(b)
4
= 0.1 A
16 + 24
4
i' =
= 0.09756 A
16 + 1 + 24
0.1 − 0.09756
% error =
x100% = 2.44%
0.1
i=
(c)
Chapter 2, Solution 69
With the voltmeter in place,
R2 Rm
V0 =
VS
R1 + R S + R 2 R m
where Rm = 100 kΩ without the voltmeter,
R2
VS
V0 =
R1 + R 2 + R S
100
kΩ
101
(a)
When R2 = 1 kΩ, 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 kΩ, R 2 R m =
= 9.091kΩ
110
9.091
V0 =
(40) = 9.30 V (with)
9.091 + 30
10
V0 =
(40) = 10 V (without)
10 + 30
When R2 = 100 kΩ, R 2 R m = 50kΩ
(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 Ω
8k Ω
a
b
10k Ω
12k Ω
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 Ω, IL = 1 A, find RL.
v s = i L ( R1 + R L )

→
RL =
vs
30
− R1 =
− 20 = 10Ω
iL
1
Chapter 2, Solution 72
The system can be modeled as shown.
12A
+
9V
-
R
R
R
•••
The n parallel resistors R give a combined resistance of R/n. Thus,
9 = 12 x
R
n

→
n=
12 xR 12 x15
=
= 20
9
9
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 Ω
Chapter 2, Solution 74
With the switch in high position,
6 = (0.01 + R3 + 0.02) x 5
R3 = 1.17 Ω
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 Ω
At the low position,
6 = (0.01 + R1 + R2 + R3 + 0.02) x 1
R1 = 5.97 - 1.97 = 4 Ω
R1 + R2 + R3 = 5.97
Chapter 2, Solution 75
100 Ω
R
VS
M
12 Ω
+
-
(a) When Rx = 0, then
t
Im = Ifs =
R + Rm
R2 =
E2
2
− Rm=
− 100 = 19.9kΩ
I fs
0.1x10 3
I fs
= 0.05mA
2
E
2
− (R + R m ) =
− 20kΩ = 20 kΩ
Rx =
Im
0.05x10 −3
(b) For half-scale deflection, Im =
Im =
E
R + Rm + Rx
Chapter 2, Solution 76
For series connection, R = 2 x 0.4Ω = 0.8Ω
V 2 (120) 2
p=
=
= 18 kΩ (low)
R
0.8
For parallel connection, R = 1/2 x 0.4Ω = 0.2Ω
V 2 (120) 2
p=
=
= 72 kW (high)
R
0.2
Chapter 2, Solution 77
(a)
5 Ω = 10 10 = 20 20 20 20
i.e., four 20 Ω resistors in parallel.
(b)
311.8 = 300 + 10 + 1.8 = 300 + 20 20 + 1.8
i.e., one 300Ω resistor in series with 1.8Ω resistor and
a parallel combination of two 20Ω resistors.
(c)
40kΩ = 12kΩ + 28kΩ = 24 24k + 56k 50k
i.e., Two 24kΩ resistors in parallel connected in series with two
50kΩ resistors in parallel.
(d)
42.32kΩ = 42l + 320
= 24k + 28k = 320
= 24k = 56k 56k + 300 + 20
i.e., A series combination of 20Ω resistor, 300Ω resistor, 24kΩ
resistor and a parallel combination of two 56kΩ resistors.
Chapter 2, Solution 78
The equivalent circuit is shown below:
R
VS
V0 =
+
+
V0
-
(1-α)R
-
(1 − α)R
VS = (1 − α )R 0 VS
R + (1 − α)R
V0
= (1 − α)R
VS
Chapter 2, Solution 79
Since p = v2/R, the resistance of the sharpener is
R = v2/(p) = 62/(240 x 10-3) = 150Ω
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 Ω
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 =
V 2 p2 R1
,
=
R p1 R 2
+
R2
-
Case 2
p2 =
R1
10
p1 = (12) = 30 W
4
R2
Chapter 2, Solution 81
Let R1 and R2 be in kΩ.
R eq = R 1 + R 2 5
(1)
5 R2
V0
=
VS 5 R 2 + R 1
(2)
From (1) and (2), 0.05 =
5 R1
2 = 5 R2 =
40
From (1), 40 = R1 + 2
5R 2
or R2 = 3.33 kΩ
5+ R2
R1 = 38 kΩ
Thus R1 = 38 kΩ, R2 = 3.33 kΩ
Chapter 2, Solution 82
(a)
10 Ω
40 Ω
10 Ω
80 Ω
1
2
R12
R12 = 80 + 10 (10 + 40) = 80 +
50
= 88.33 Ω
6
(b)
3
10 Ω
10 Ω
20 Ω
40 Ω
R13
80 Ω
1
R13 = 80 + 10 (10 + 40) + 20 = 100 + 10 50 = 108.33 Ω
4
(c)
20 Ω
10 Ω
R14
10 Ω
40 Ω
80 Ω
1
R14 = 80 + 0 (10 + 40 + 10) + 20 = 80 + 0 + 20 = 100 Ω
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 45mW
=
= 5mA
V1
9V
p
480mW
I2 = 2 =
= 20mA
V2
24
I1 =
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 = 40 − 5 = 35 mA
Hence, I R1 R1 = 24 - 9 = 15 V
I R 2 R 2 = 9V
R2 =
R1 =
15V
= 375 Ω
40mA
9V
= 257.14 Ω
35mA
Chapter 3, Solution 1.
40 Ω
v1
v2
8Ω
6A
2Ω
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 (9.143)2
P8Ω =
=
= 10.45 W
8
8
P4Ω =
(v 1 − v 2 )2
4
= 94.37 W
v12 (= 10.286)2
=
= 52.9 W
P2Ω =
2
2
Chapter 3, Solution 2
At node 1,
v − v2
− v1 v1
−
= 6+ 1
10
5
2
At node 2,
v2
v − v2
= 3+ 6+ 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
v
v
v
= 4 A , i2 = 0 = 2 A, i3 = 0 = 1.33 A, i4 = 0 = 67 mA
10
20
30
60
v0 = 40 V
Chapter 3, Solution 4
2A
v1
i1
4A
5Ω
i2
v2
i3
10 Ω
10 Ω
i4
5Ω
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 v 0
+
=
2k
6k
4k
v0 = 20 V
5A
Chapter 3, Solution 6
i1 + i2 + i3 = 0
v 2 − 12 v 0 v 0 − 10
+
+
=0
4
6
2
or v0 = 8.727 V
Chapter 3, Solution 7
At node a,
10 − Va Va Va − Vb
(1)
=
+

→ 10 = 6Va − 3Vb
30
15
10
At node b,
Va − Vb 12 − Vb − 9 − Vb
+
+
=0

→
24 = 2Va − 7Vb
10
20
5
Solving (1) and (2) leads to
Va = -0.556 V, Vb = -3.444V
(2)
Chapter 3, Solution 8
3Ω
i1
v1
i3
5Ω
i2
+
V0
3V
2Ω
+
–
+ 4V0
–
–
1Ω
v1 v1 − 3 v1 − 4 v 0
+
+
=0
5
1
5
2
8
v 0 = 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
3Ω
i1
v1
+ v0 –
12V
6Ω
i3
i2
+
+
v1
–
8Ω
–
+
–
2v0
At the non-reference node,
12 − v1 v1 v1 − 2 v 0
=
+
3
8
6
(1)
But
-12 + v0 + v1 = 0
v0 = 12 - v1
(2)
Substituting (2) into (1),
12 − v1 v1 3v1 − 24
=
+
3
8
6
v0 = 3.652 V
Chapter 3, Solution 10
At node 1,
v 2 − v1
v
= 4+ 1
1
8
32 = -v1 + 8v2 - 8v0
1Ω
4A
v1
8Ω
i0
2i0
v0
v2
2Ω
4Ω
(1)
At node 0,
4=
v0
v
+ 2I 0 and I 0 = 1
8
2
16 = 2v0 + v1
(2)
v2 = v1
(3)
At node 2,
2I0 =
v 2 − v1 v 2
v
+
and I 0 = 1
1
4
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
4Ω
i1 v i2
3Ω
i3
10 V
+
–
5A
6Ω
Note that i2 = -5A. At the non-reference node
10 − v
v
+5=
4
6
i1 =
v = 18
10 − v
= -2 A, i2 = -5 A
4
Chapter 3, Solution 12
10 Ω
v1
20 Ω
50 Ω
v2
i3
24 V
+
–
40 Ω
5A
At node 1,
24 − v 1
v − v 2 v1 − 0
= 1
+
10
20
40
At node 2, 5 +
v1 − v 2 v 2
=
20
50
96 = 7v1 - 2v2
500 = -5v1 + 7v2
(1)
(2)
Solving (1) and (2) gives,
v1 = 42.87 V, v2 = 102.05 V
v
v
i1 = 1 = 1.072 A, v2 = 2 = 2.041 A
40
50
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
1Ω
8Ω
2Ω
4Ω
40 V
20 V
–
+
+
–
At node 1,
40 − v 0
v1 − v 0
+5=
1
2
At node 0,
v1 − v 0
v
v + 20
+5= 0 + 0
2
4
8
Solving (1) and (2), v0 = 20 V
v1 + v0 = 70
4v1 - 7v0 = -20
(1)
(2)
Chapter 3, Solution 15
5A
v0
v1
1Ω
2Ω
4Ω
40 V
8Ω
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
v2 =
54
11
i0 = 6vi = 29.45 A
2
P65 =
v12
 54 
= v12 G =   6 = 144.6 W
R
 11 
2
 − 56 
P55 = v G = 
 5 = 129.6 W
 11 
2
2
P35 = (v L − v 3 ) G = (2) 2 3 = 12 W
2
(2)
(3)
Substituting (1) and (3) into (2),
v1 = v2 + 10 =
–
+
− 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
4Ω
2Ω
10 Ω
60 V
60 V
8Ω
+
–
3i0
60 − v1 v1 v1 − v 2
=
+
4
8
2
60 − v 2 v1 − v 2
At node 2, 3i0 +
+
=0
10
2
At node 1,
120 = 7v1 - 4v2
(1)
60 − v1
.
4
But i0 =
Hence
3(60 − v1 ) 60 − v 2 v1 − v 2
+
+
=0
4
10
2
1020 = 5v1 - 12v2
Solving (1) and (2) gives v1 = 53.08 V. Hence i0 =
60 − v1
= 1.73 A
4
(2)
Chapter 3, Solution 18
–+
v2
v1
2Ω
5A
v3
2Ω
8Ω
4Ω
10 V
+
+
v1
v3
–
(a)
At node 2, in Fig. (a), 5 =
At the supernode,
–
(b)
v 2 − v1 v 2 − v3
+
2
2
10 = - v1 + 2v2 - v3
v 2 − v1 v 2 − v 3 v1 v 3
+
=
+
2
2
4
8
From Fig. (b), - v1 - 10 + v3 = 0
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

→

→
+
 7 − 1 − 4  V1   16 

  

 − 1 7 − 2 V2  =  0 
4
2 − 7  V3   − 36 

Using MATLAB,
 10 
V = A −1 B =  4.933 
12.267 

→

→
(1)
0 = −V1 + 7V2 − 2V3
V1 − V3 V2 − V3
+
=0
8
2
4
From (1) to (3),
3+
16 = 7V1 − V2 − 4V3

→
(2)
− 36 = 4V1 + 2V2 − 7V3 (3)
AV = B
V1 = 10 V, V2 = 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

→ V1 + 4V2 + V3 = 0
(1)
4
1
4
.
V1
4Ω
.
V2
1Ω
2Ω
V3
4Ω
Between nodes 1 and 3,
− V1 + 12 + V3 = 0

→ 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 kΩ
v1
2 kΩ
v3
3v0
+
3v0
v2
+
v0
3 mA
–
1 kΩ
+
+
+
v3
v2
–
–
(b)
(a)
Let v3 be the voltage between the 2kΩ resistor and the voltage-controlled voltage source.
At node 1,
v − v 2 v1 − v 3
3x10 −3 = 1
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 v1
v − v0
=
+3+ 1
2
4
8
At node 2, 3 +
24 = 7v1 - v2
(1)
v1 − v 2 v 2 + 5v 2
=
8
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 Ω
5Ω
8V
+
+
v1
v2
–
(a)
–
(b)
Chapter 3, Solution 24
6mA
1 kΩ
2 kΩ
V1
+
30V
-
3 kΩ
V2
io
-
4 kΩ
5 kΩ
At node 1,
30 −V 1
V V − V2
=6+ 1 + 1

→ 96 = 7V1 − 2V2
1
4
2
At node 2,
(−15 −V 2) V2 V2 − V1
6+
=
+

→ 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
2Ω
10V
–
1Ω
(1)
40V
+
–
4Ω
+
–
2Ω
Using nodal analysis,
20 − v 0 40 − v 0 10 − v 0
v −0
+
+
= 0
1
2
2
4
i0 =
20 − v 0
= 0A
1
v0 = 20V
15V
+
Chapter 3, Solution 26
At node 1,
V − V3 V1 − V2
15 − V1
= 3+ 1
+

→
− 45 = 7V1 − 4V2 − 2V3
20
10
5
At node 2,
V1 − V2 4 I o − V2 V2 − V3
+
=
5
5
5
V − V3
. Hence, (2) becomes
But I o = 1
10
0 = 7V1 − 15V2 + 3V3
At node 3,
V − V3 − 10 − V3 V2 − V3
3+ 1
+
+
=0

→
− 10 = V1 + 2V2 − 5V3
10
5
5
Putting (1), (3), and (4) in matrix form produces
 7 − 4 − 2  V1   − 45 

  


→
AV = B
 7 − 15 3 V2  =  0 
1
2
− 5  V3   − 10 

Using MATLAB leads to
 − 9.835 


−1
V = A B =  − 4.982 
 − 1.96 


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,
7 11 − 4  v 1   2 
1 − 6 1   v  =  0 
 2   

4 13 − 7  v 3   − 4
7
11
∆ = 1 −6
4
13
7
2
−4
2
11
1 = 176, ∆ 1 = 0
−6
−7
−4
−4
7
∆2 = 1 0
1 = 66,
4 −4 −7
v1 =
13
11
−4
1 = 110
−7
2
∆ 3 = 1 − 6 0 = 286
4 13 − 4
∆ 1 110
∆
66
=
= 0.625V, v2 = 2 =
= 0.375V
∆
∆
176
176
v3 =
∆3
286
=
= 1.625V.
∆
176
v1 = 625 mV, v2 = 375 mV, v3 = 1.625 V.
Chapter 3, Solution 28
At node c,
Vd − Vc Vc − Vb Vc
=
+

→ 0 = −5Vb + 11Vc − 2Vd
(1)
10
4
5
At node b,
Va + 45 − Vb Vc − Vb Vb
+
=

→
− 45 = Va − 4Vb + 2Vc
(2)
8
4
8
At node a,
Va − 30 − Vd Va Va + 45 − Vb
+
+
=0

→ 30 = 7Va − 2Vb − 4Vd (3)
4
16
8
At node d,
Va − 30 − Vd Vd Vd − Vc
=
+

→ 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 

→
 7 − 2 0 − 4  V  =  30 
c

  

 5 0 2 − 7 V   150 

 d  

We use MATLAB to invert A and obtain
AV = B
 − 10.14 


 7.847 
−1
V = A B=
− 1.736 


 − 29.17 


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

→
− 5 = 4V1 − V2 − V4
At node 2,
V1 − V2 = 2V2 + 4(V2 − V3 ) = 0

→ 0 = −V1 + 7V2 − 4V3
At node 3,
6 + 4(V2 − V3 ) = V3 − V4

→ 6 = −4V2 + 5V3 − V4
At node 4,
2 + V3 − V4 + V1 − V4 = 3V4

→
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

→
 0 − 4 5 − 1 V  =  6 
3

   
 − 1 0 − 1 5 V   2 

 4   
Using MATLAB,
 − 0.7708 


 1.209 
−1
V = A B=
2.309 


 0.7076 


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 Ω
I0
v1
10 Ω
100 V
+
120 V
20 Ω
v0
1
2
4v0
–
–+
+
–
2I0
80 Ω
At node 1,
v 1 − v 2 100 − v 1 4 v o − v 1
=
+
40
10
20
But, vo = 120 + v2
(1)
v2 = vo – 120. Hence (1) becomes
7v1 – 9vo = 280
(2)
At node 2,
Io + 2Io =
vo − 0
80
 v + 120 − v o  v o
3 1
=
40
 80

or
6v1 – 7vo = -720
from (2) and (3),
7 − 9  v 1   280 
6 − 7  v  =  − 720


 o  
∆=
∆1 =
(3)
7 −9
= −49 + 54 = 5
6 −7
280 − 9
= −8440 ,
− 720 − 7
∆2 =
7 280
= −6720
6 − 720
v1 =
∆
∆1
− 8440
− 6720
=
= −1688, vo = 2 =
− 1344 V
∆
∆
5
5
Io = -5.6 A
Chapter 3, Solution 31
1Ω
+ v0 –
v2
v1
1A
2v0
v3
2Ω
i0
4Ω
1Ω
10 V
4Ω
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
10 − v 3
v2
= v1 − v 3 +
4
2
20 = 4v1 + v2 – 2v3
At the supernode, v2 = v1 + 4io. But io =
v2 = v1 + v3
Solving (2) to (4) leads to,
v1 = 4 V, v2 = 4 V, v3 = 0 V.
(2)
(3)
v3
. Hence,
4
(4)
+
–
Chapter 3, Solution 32
5 kΩ
v1
v3
v2
+
10 kΩ
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.
4Ω
3Ω
12 V
+
5Ω
–
1Ω
2Ω
Chapter 3, Solution 34
(a)
This is a planar circuit because it can be redrawn as shown below,
7Ω
2Ω
1Ω
3Ω
6Ω
10 V
5Ω
+
–
4Ω
(b)
This is a non-planar circuit.
Chapter 3, Solution 35
30 V
20 V
+
–
+
–
i1
2 kΩ
+
i2
v0
–
5 kΩ
4 kΩ
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
4Ω
i1
12 V
+–
i2
I1
+
I2
6Ω
–
i3
2Ω
Applying mesh analysis gives,
12 = 10I1 – 6I2
-10 = -6I1 + 8I2
 6   5 − 3  I 1 
 − 5 =  − 3 4   I 
 2 
  
or
∆=
5 −3
6 −3
5
6
= 11, ∆1 =
= 9, ∆ 2 =
= −7
−3 4
−5 4
−3 −5
I1 =
∆1
9 I = ∆2 = − 7
=
, 2
∆
11
∆
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
3Ω
3V
+
v0
–
5Ω
2Ω
i1
1Ω
+
–
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
3Ω
6Ω
+ v0 –
12 V
+
–
i1
8Ω
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,
− 10 − 2 I x + 10 I 1 − 6 I 2 = 0
But I x = I 1 − I 2 . Hence,
10 = −12 I 1 + 12 I 2 + 10 I 1 − 6 I 2

→ 5 = 4 I 1 − 2 I 2
For mesh 2,
12 + 8I 2 − 6 I 1 = 0

→ 6 = 3I 1 − 4 I 2
Solving (1) and (2) leads to
I 1 = 0.8 A, I 2 = -0.9A
(1)
(2)
Chapter 3, Solution 40
2 kΩ
30V
+
i2
2 kΩ
i1
–
6 kΩ
6 kΩ
i3
4 kΩ
4 kΩ
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 Ω
i1
6V
2Ω
+–
1Ω
i2
4Ω
8V
i3
+
–
i
i2
i3
0
5Ω
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
We put (1), (2), and (3) in matrix form,
6 − 1 0  i1   3
 2 − 7 1  i  =  8 

 2   
0 − 1 6 i 3  2
6
−1 0
6 3 0
∆ = 2 − 7 1 = −234, ∆ 2 = 2 8 1 = −240
0
−1 6
0 2 6
6
−1 3
∆ 3 = 2 − 7 8 = −38
0 −1 2
At node 0, i + i2 = i3 or i = i3 – i2 =
∆3 − ∆2
− 38 − 240
=
= 1.188 A
− 234
∆
(3)
Chapter 3, Solution 42
For mesh 1,
− 12 + 50 I 1 − 30 I 2 = 0

→ 12 = 50 I 1 − 30 I 2
(1)
For mesh 2,
− 8 + 100 I 2 − 30 I 1 − 40 I 3 = 0

→ 8 = −30 I 1 + 100 I 2 − 40 I 3
For mesh 3,
(3)
− 6 + 50 I 3 − 40 I 2 = 0

→ 6 = −40 I 2 + 50 I 3
Putting eqs. (1) to (3) in matrix form, we get
0  I 1  12 
 50 − 30

   
 − 30 100 − 40  I 2  =  8 
 0
− 40 50  I 3   6 


→
(2)
AI = B
Using Matlab,
 0.48 


I = A B =  0.40 
 0.44 


−1
i.e. I1 = 0.48 A, I2 = 0.4 A, I3 = 0.44 A
Chapter 3, Solution 43
20 Ω
a
80 V
+
i1
–
30 Ω
+
i3
30 Ω
20 Ω
80 V
+
i2
–
20 Ω
30 Ω
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
+
2Ω
i3
4Ω
i2
1Ω
6V
5Ω
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,
i2 = 3 + i1
(3)
Solving (1) to (3), i1 = -3.067, i3 = -1.3333; io = i1 – i3 = -1.7333 A
+
–
Chapter 3, Solution 45
4Ω
30V
+
i3
i4
2Ω
6Ω
i1
–
8Ω
3Ω
i2
1Ω
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

→
For loop 2,
− 8i1 + 14i2 + 2vo = 0
But vo = 3i1 ,
11i1 − 8i2 = 12
(1)
− 8i1 + 14i2 + 6i1 = 0

→ i1 = 7i2
(2)
Substituting (2) into (1),
77i2 − 8i2 = 12

→ i 2 = 0.1739 A and i1 = 7i2 = 1.217 A
Chapter 3, Solution 47
First, transform the current sources as shown below.
- 6V +
2Ω
V1
8Ω
4Ω
V2
I3
4Ω
+
20V
-
I1
2Ω
V3
8Ω
I2
+
12V
-
For mesh 1,
− 20 + 14 I 1 − 2 I 2 − 8I 3 = 0

→ 10 = 7 I 1 − I 2 − 4 I 3
For mesh 2,
12 + 14 I 2 − 2 I 1 − 4 I 3 = 0

→ − 6 = − I 1 + 7 I 2 − 2 I 3
For mesh 3,
− 6 + 14 I 3 − 4 I 2 − 8I 1 = 0

→ 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 

   

→
AI = B
 − 1 7 − 2  I 2  =  − 6 
 − 4 − 2 7  I   3 

 3   
Using MATLAB,
 2 
−1
I = A B = 0.0333
1.8667 
But

→ I 1 = 2.5, I 2 = 0.0333, I 3 = 1.8667
20 − V

→ V1 = 20 − 4 I1 = 10 V
4
V2 = 2( I1 − I 2 ) = 4.933 V
Also,
V − 12
I2 = 3

→ V3 = 12 + 8I 2 = 12.267V
8
I1 =
(1)
(2)
(3)
Chapter 3, Solution 48
We apply mesh analysis and let the mesh currents be in mA.
3k Ω
I4
4k Ω
2k Ω
Io
1k Ω
+
12 V
-
I1
5k Ω
I2
+
8V
-
I3
10k Ω
6V
+
For mesh 1,
− 12 + 8 + 5I 1 − I 2 − 4 I 4 = 0

→ 4 = 5I 1 − I 2 − 4 I 4
(1)
For mesh 2,
− 8 + 13I 2 − I 1 − 10 I 3 − 2 I 4 = 0

→ 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

→ 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

→
 0 − 10 15 − 5  I  =  6 
3

   
 − 4 − 2 − 5 14  I   0 

 4   
Using MATLAB,
 7.217 


8
.
087


I = A −1 B = 
7.791 


 6 


The current through the 10k Ω resistor is Io= I2 – I3 = 0.2957 mA
Chapter 3, Solution 49
3Ω
i3
2Ω
1Ω
i1
2Ω
16 V
i2
+
–
2i0
i1
i2
0
(a)
2Ω
1Ω
2Ω
+
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
4Ω
2Ω
i3
10 Ω
8Ω
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
8Ω
2Ω
i3
1Ω
i2
40 V
+
–
4Ω
+
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 Ω
i2
–
VS
+
–
8Ω
3A
i2
i1
i3
4Ω
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 Ω
i2
–
VS
+
–
8Ω
3A
i2
i1
i3
4Ω
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

→
2 = 2I1 − I 2
(1)
For mesh 2,
− 10 + 3I 2 − I 1 − I 3 = 0

→ 10 = − I 1 + 3I 2 − I 3
For mesh 3,
(3)
− 12 + 2 I 3 − I 2 = 0

→ 12 = − I 2 + 2 I 3
Putting (1) to (3) in matrix form leads to
 2 − 1 0  I 1   2 

   
 − 1 3 − 1 I 2  = 10 
 0 − 1 2  I  12 

 3   
Using MATLAB,
 5.25 
I = A B =  8.5 
10.25
−1

→
10 V
I2
i1
4A
AI = B

→ I 1 = 5.25 mA, I 2 = 8.5 mA, I 3 = 10.25 mA
Chapter 3, Solution 55
b
c
+
1A
I2
6Ω
1A
I1
i2
I4
2Ω
i3
12 Ω
a
I3
d
I4
4A
(2)
4Ω
+–
8V
I3
0
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 –
2Ω
2Ω
12 V
+
–
i2
2Ω
2Ω
i1
2Ω
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,
 2 − 1 − 1  i1  6
 − 1 3 − 1 i  = 0

 2   
 − 1 − 1 3  i 3  0
2
−1 −1
2
6 −1
∆ = − 1 3 − 1 = 8, ∆2 = − 1 3 − 1 = 24
−1 −1 3
−1 0 3
2
−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 = 4kΩx18mA = 72V ,
V1 = 100 − V2 = 100 − 72 = 28V
Current through R is
3
3
28 =
iR =
io ,
V1 = i R R

→
(18) R
3+ R
3+ R
This leads to R = 84/26 = 3.23 k Ω
Chapter 3, Solution 58
30 Ω
i2
30 Ω
10 Ω
i1
10 Ω
i3
+
–
120 V
30 Ω
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 Ω
I0
–+
120 V
i2
10 Ω
20 Ω
i1
100V +
4v0
–
+
i3
+
–
v0
80 Ω
–
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 − 2 32 
 − 1 3 − 12


3
− 1 
 0
3
10
32
 i1  10
i  =  6 
 2  
i 3   0 
3
− 2 10
3
− 12 = 5, ∆2 = − 1
6
− 12 = −28, ∆3 = − 1
3
6 = −84
3
−1
0
−1
3
0
0
I0 = i2 = ∆2/∆ = -28/5 = -5.6 A
v0 = 8i3 = (-84/5)80 = -1344 volts
0
Chapter 3, Solution 60
0.5i0
4Ω
v1
10 V
1Ω
10 V
8Ω
v2
2Ω
+
–
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
P1Ω = (v1)2/1 = 2.041 watts, P2Ω = (v2)2/2 = 4.939 watts
P4Ω = (10 – v1)2/4 = 18.38 watts, P8Ω = (10 – v2)2/8 = 5.88 watts
Chapter 3, Solution 61
v1
is
20 Ω
v2
10 Ω
i0
+
v0
–
30 Ω
–
+ 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 Ω
(1)
Chapter 3, Solution 62
4 kΩ
100V +
8 kΩ
A
i1
–
B
i2
2 kΩ
i3
+
–
40 V
We have a supermesh. Let all R be in kΩ, 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,
i1 + 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 Ω
A
5Ω
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 Ω
i2 10 Ω
+
−
A
i0
i1
10 Ω
i2
+
–
4i0
i3
40 Ω
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 

→
AI = B
 3   

5 − 2  I 4   6 
 −1 −1 0
 0 − 1 − 1 − 2 8  I  10 
 5   

Using MATLAB leads to
 1.673 


 1.824 
I = A −1 B =  1.733 


 2.864 
 2.411 


Thus,
I 1 = 1.673 A, I 2 = 1.824 A, I 3 = 1.733 A, I 4 = 1.864 A, I 5 = 2.411 A
Chapter 3, Solution 66
Consider the circuit below.
2 kΩ
+
20V
-
2 kΩ
1 kΩ
I1
1 kΩ
+
10V
-
I2
1 kΩ
Io
2 kΩ
1 kΩ
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 kΩ
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 

 4  

Using MATLAB,

→
AI = B
 5.5 


 − 1.75 
−1
I = A B=
3.75 


 − 2.5 


Thus,
I o = − 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,
1.25 − 1
 − 1 1.5


−1
=
1.25 − 1  v 1   3 
 − 1 1.5  v  =  − 1

 2   
1 1.5 1 
, where ∆ = [(1.25)(1.5)-(-1)(-1)] = 0.875
∆  1 1.25
 v 1  1.7143 1.1429  3  3(1.7143) − 1(1.1429)  4
 v  = 1.1429 1.4286  − 1 = 3(1.1429) − 1(1.4286) = 2
  
  
 2 
Clearly v1 = 4 volts and v2 = 2 volts
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.
 8 − 1 − 5  v 1   4 
 − 1 3 − 2  v  =  2 

 2   
 − 5 − 2 7   v 3   − 1
8
∆ = −1
−1 − 5
3
−5 −2
8
4
4
− 2 = 34, ∆ 1 = 2
7
−5
−1 −5
3
−1 − 2
8
−1
− 2 = 85
7
4
∆ 2 = − 1 2 − 2 = 109, ∆ 3 = − 1 3
2 = 87
− 5 −1 7
− 5 − 2 −1
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   v 1  20
 1.75 − 0.25
 − 0.25
1
− 0.25  v 2  =  5 

   
− 0.25 1.25   v 3   5 
 − 1
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:
G 1 + G 2 + G 4

− G2

0

− G2
G1 + G 2
− G3
  v 1  − I 1 
v  =  I 
− G3
 2   2 
G 1 + G 3 + G 5   v 3   I 1 
0
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.
 6 − 2 0   i 1   12 
 − 2 12 − 2 i  =  − 8 

 2  

 0 − 2 7  i 3   − 20
Now we can solve for i2 using Cramer’s Rule.
6
∆ = −2
0
−2
12
−2
0
6
− 2 = 452, ∆ 2 = − 2
7
0
12
−8
0
− 2 = −408
− 20
7
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   i1   8 
 7 −2 0
 − 2 6 − 4 0  i   4 

 2  = 

 0 − 4 5 − 1 i 3   − 10

  

0 − 1 5  i 4   − 4 
 0
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,
 9 − 3 − 4 0   i1   6 
− 3 8
0
0  i 2   4 

=
− 4 0
6 − 1 i3   2 

   
0 − 1 2  i 4  − 3
0
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.
 V1 
− V 
2
The input voltage vector is = 
 V3 


 − V4 
R 1 + R 4 + R 6

− R4


− R6

0

− R4
− R6
R2 + R4 + R5
0
0
− R5
R6 + R7 + R8
− R8
  i 1   V1 
 i   − V 
− R5
2
 2  = 
− R8
 i 3   V3 
  

R 3 + R 5 + R 8  i 4   − V4 
0
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 = −26.88 V
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
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 kΩ
1
2 kΩ
2
+
3v0
3
6 kΩ
4
4A
4 kΩ
8 kΩ
0
This network corresponds to the Netlist.
100V +
–
Chapter 3, Solution 83
The circuit is shown below.
20 Ω
1
70 Ω
2i02
3
+ v0 –
20 V
50 Ω
+
–
2 kΩ
1
30 Ω
2A
3 kΩ
2
3v0
+
3
0
4
6 kΩ
4A
4 kΩ
8 kΩ
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 = 0.5µA and v0 = 0.5 volt.
Chapter 3, Solution 85
The amplifier acts as a source.
Rs
+
Vs
-
RL
For maximum power transfer,
R L = Rs = 9Ω
–
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 = βIB = 75IB and VCE = 2 volts, then (2) becomes 5 = 2 +2k(75IB)
which leads to IB = 20 µA.
Substituting this into (1) produces, vi = 0.7 + 0.8 = 1.5 volts.
2 kΩ
IB
40 kΩ
vi
+
VBE
+
-
–
5v
+
-
Chapter 3, Solution 90
1 kΩ
100 kΩ
vs
i1
i2
+
+
VBE
+
-
IB
500 Ω
IE
VCE
–
–
+
18V
+
-
V0
–
For loop 1, -vs + 10k(IB) + VBE + IE (500) = 0 = -vs + 0.7 + 10,000IB + 500(1 + β)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 kΩ and VTh = 2(3)/(2+6) = 0.75 volts
5 kΩ
IC
1.5 kΩ
0.75 V
IB
+
VBE
+
-
i1
i2
+
VCE
–
–
9V
+
400 Ω
+
-
V0
IE
–
For loop 1, -0.75 + 1.5kIB + VBE + 400IE = 0 = -0.75 + 0.7 + 1500IB + 400(1 + β)IB
IB = 0.05/81,900 = 0.61 µA
v0 = 400IE = 400(1 + β)IB = 49 mV
For loop 2, -400IE – VCE – 5kIC + 9 = 0, but, IC = βIB and IE = (1 + β)IB
VCE = 9 – 5kβIB – 400(1 + β)IB = 9 – 0.659 = 8.641 volts
Chapter 3, Solution 92
I1
5 kΩ
10 kΩ
VC
IB
IC
+
+
VBE
4 kΩ
IE
VCE
–
–
+
V0
–
12V
+
-
I1 = IB + IC = (1 + β)IB and IE = IB + IC = I1
Applying KVL around the outer loop,
4kIE + VBE + 10kIB + 5kI1 = 12
12 – 0.7 = 5k(1 + β)IB + 10kIB + 4k(1 + β)IB = 919kIB
IB = 11.3/919k = 12.296 µA
Also, 12 = 5kI1 + VC which leads to VC = 12 – 5k(101)IB = 5.791 volts
Chapter 3, Solution 93
1Ω
4Ω
v1
i1
24V
+
–
3v0
i
2Ω
2Ω
+
8Ω
2Ω
v2 i3
3v0
i
i2
4Ω
+
+
+
+
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.
1Ω
+
−
1V
8 (5 + 3) = 4Ω , i =
io =
5Ω
i
io
8Ω
3Ω
1
1
=
1+ 4 5
1
1
i=
= 0.1A
2
10
Chapter 4, Solution 2.
6 (4 + 2) = 3Ω, i1 = i 2 =
io =
1
A
2
1
1
i1 = , v o = 2i o = 0.5V
2
4
5Ω
4Ω
i1
io
i2
1A
8Ω
6Ω
2Ω
If is = 1µA, then vo = 0.5µV
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 ∆ .
R 3R =
3R 2 3
3
3
3
= R, R + R = R
4R
4
4
4
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 = 10Ω,
vo = 5V, io = 10/(10) = 500mA
Chapter 4, Solution 4.
If Io = 1, the voltage across the 6Ω resistor is 6V so that the current through the 3Ω
resistor is 2A.
2Ω
2A
1A
2Ω
3A
3A
i1
+
3Ω
6Ω
4Ω
Is
2Ω
4Ω
v1
−
(a)
3 6 = 2Ω , vo = 3(4) = 12V, i1 =
(b)
vo
= 3A.
4
Hence Is = 3 + 3 = 6A
If
Is = 6A
Is = 9A
Io = 1
Io = 6/(9) = 0.6667A
Is
Chapter 4, Solution 5.
2Ω
Vs
If vo = 1V,
If vs =
10
3
3Ω
v1
+
−
vo
6Ω
6Ω
1
V1 =   + 1 = 2V
3
10
2
Vs = 2  + v1 =
3
3
vo = 1
Then vs = 15
vo =
3
x15 = 4.5V
10
Chapter 4, Solution 6
Let RT = R2 // R3 =
R2 R3
RT
, then Vo =
Vs
RT +R1
R2 + R3
R2 R3
V
R2 + R3
R2 R3
RT
k= o =
=
=
R2 R3
Vs RT + R1
R1 R2 + R2 R3 + R3 R1
+ R1
R2 + R3
6Ω
Chapter 4, Solution 7
We find the Thevenin equivalent across the 10-ohm resistor. To find VTh, consider the
circuit below.
3Vx
5Ω
5Ω
+
+
15 Ω
4V
-
VTh
6Ω
+
Vx
-
From the figure,
15
(4) = 3V
15 + 5
consider the circuit below:
V x = 0,
To find RTh,
VTh =
3Vx
5Ω
5Ω
V1
V2
+
4V
-
15 Ω
+
At node 1,
V V − V2
4 − V1
= 3V x + 1 + 1
,
5
15
5
At node 2,
1A
6Ω
Vx
V x = 6 x1 = 6
-

→
258 = 3V2 − 7V1
(1)
V1 − V2
=0

→ V1 = V2 − 95
5
Solving (1) and (2) leads to V2 = 101.75 V
2
V
V
9
RTh = 2 = 101.75Ω,
p max = Th =
= 22.11 mW
1
4 RTh 4 x101.75
1 + 3V x +
(2)
Chapter 4, Solution 8.
Let i = i1 + i2,
where i1 and iL are due to current and voltage sources respectively.
6Ω
i2
i1
6Ω
4Ω 5A
20V
+
−
4Ω
(a)
i1 =
(b)
6
20
(5) = 3A, i 2 =
= 2A
6+4
6+4
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 Ω
i
ix1
15V
+
−
10 Ω
(a)
40Ω
-4A
ix2
12Ω
10Ω
(b)
40Ω
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 Ω
10 Ω
+−
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.
6Ω
io
i1
12V
+
−
2Ω
3Ω
(a)
4A
i2
6Ω
2Ω
3Ω
ix2
2Ω
4A
2Ω
(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
5Ω
6Ω
3Ω
+ vo1 −
2A
4Ω
12 Ω
io 5 Ω
+ vo1 −
5Ω
6||3 = 2 ohms, 4||12 = 3 ohms. Hence,
io = 2/2 = 1, vo1 = 5io = 5 V
For vo2, consider the circuit below.
6Ω
12V
+
−
5Ω
4Ω
6Ω
+ vo2 −
3Ω
12 Ω
12V
+
−
5Ω
+
+ vo2 −
v1
3Ω
3Ω
−
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.
5Ω
4Ω
+ vo3 −
6Ω
3Ω
12 Ω
5Ω
+
−
+ vo3 −
19V
2Ω
12 Ω
4Ω
+
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 kΩ
2 kΩ
+
30V
-
3 kΩ
i1
4 kΩ
5 kΩ
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 kΩ
0.7949 k Ω
Using current division,
i1 =
0.7949
(30mA) = 4.973 mA
4.7949
For i2, consider the circuit below.
1 kΩ
2 kΩ
3 kΩ
i2
-
4 kΩ
5 kΩ
15V
+
After successive source transformation and resistance combinations, we obtain the circuit
below:
2.42mA
i2
4 kΩ
0.7949 k Ω
Using current division,
i2 = −
0.7949
(2.42mA) = −0.4012 mA
4.7949
For i3, consider the circuit below.
6mA
1 kΩ
2 kΩ
3 kΩ
i3
4 kΩ
5 kΩ
After successive source transformation and resistance combinations, we obtain the circuit
below:
3.097mA
i3
4 kΩ
i3 = −
0.7949 k Ω
0.7949
(3.097mA) = −0.5134 mA
4.7949
Thus,
io = i1 + i2 + i3 = 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.
6Ω
4Ω
2Ω
+
+
− 20V
vo1
3Ω
−
6||(4 + 2) = 3 ohms, vo1 = (½)20 = 10 V
For vo2, consider the circuit below.
6Ω
4Ω
6Ω
4V
2Ω
2Ω
−+
+
1A
4Ω
+
3Ω
vo2
vo2
−
3Ω
−
3||6 = 2 ohms, vo2 = [2/(4 + 2 + 2)]4 = 1 V
For vo3, consider the circuit below.
6Ω
2A
4Ω
2A
2Ω
3Ω
+
3Ω
3Ω
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
+
−
1Ω
i1
2Ω
3Ω
4Ω
4||(3 + 1) = 2 ohms, Then io = [20/(2 + 2)] = 5 A, i1 = io/2 = 2.5 A
For i3, consider the circuit below.
+
2Ω
vo ’
1Ω
4Ω
i3
−
+
3Ω
16V
−
2||(1 + 3) = 4/3, vo’ = [(4/3)/((4/3) + 4)](-16) = -4
i3 = vo’/4 = -1
For i2, consider the circuit below.
2Ω
1Ω
2A
(4/3)Ω
i2
4Ω
3Ω
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
1Ω
2A
i2
3Ω
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.
4Ω
io1
12V
+
−
3Ω
10 Ω
2Ω
5Ω
10||(3 + 2 + 5) = 5 ohms, io1 = 12/(5 + 4) = (12/9) A
4A
For io2, consider the circuit below.
3Ω
io2
4Ω
2Ω
5Ω
10Ω
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.
3Ω
io3
2Ω
i2
4Ω
10 Ω
5Ω 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 Ω
10 Ω
+
90V
+
−
vx1
60 Ω
20 Ω
−
30 Ω
io 10 Ω
+
−
vx1
3A
20 Ω
12 Ω
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 Ω i ’
o
+
30 Ω
10 Ω i ’
o
vx2 −
+ vx2 −
60 Ω 6A
30 Ω
20 Ω
6A
20 Ω
12 Ω
io’ = [12/(12 + 30)]6 = 72/42, vx2 = -10io’ = -17.143 V
For vx3, consider the circuit below.
10 Ω
+
30 Ω
60 Ω
vx3
10 Ω
10 Ω
−
30 Ω
+
40V
+
−
vx3
20 Ω
io”
−
7.5Ω
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.
2Ω
1Ω
i1
10V
+
−
5i1
4Ω
i1
1Ω
+
−
10V
10i1
2Ω
+−
4Ω
-10 + 10i1 + 7i1 = 0, therefore i1 = (10/17) A
For i2, consider the circuit below.
i2
1Ω
2Ω i
o
+−
10i2
2A
4Ω
io
1Ω
+
−
2V
10i2
2Ω
+−
4Ω
-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
+
2Ω
4A
8Ω
v1
+
2Ω
6A 8Ω
−
−+
−+
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
6Ω
2A
2Ω
3Ω 4A
2Ω
(a)
From Fig. (b),
2Ω
6A
(b)
i = 6/2 = 3 A
Chapter 4, Solution 21.
To get io, transform the current sources as shown in Fig. (a).
io
6Ω
3Ω
+
− 12V
+
−
i
6V 2 A
6Ω
3Ω
+
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).
5Ω
−
+ 10V
5Ω
4Ω
10Ω
2A
(a)
i
1A
10Ω
4Ω
10Ω
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.
8Ω
10 Ω
6Ω
3Ω
5A
3A
3//6 = 2-ohm. Convert the current sources to voltages sources as shown below.
10 Ω
8Ω
+
2Ω
+
10V
-
30V
-
Applying KVL to the loop gives
− 30 + 10 + I (10 + 8 + 2) = 0

→
p = VI = I 2 R = 8 W
I = 1A
Chapter 4, Solution 24
Convert the current source to voltage source.
16 Ω
1Ω
4Ω
5Ω
+
+
48 V
10 Ω
+
12 V
-
-
Vo
-
Combine the 16-ohm and 4-ohm resistors and convert both voltages sources to current
Sources. We obtain the circuit below.
1Ω
20 Ω
2.4A
5Ω
2.4A
10 Ω
Combine the resistors and current sources.
20//5 = (20x5)/25 = 4 Ω , 2.4 + 2.4 = 4.8 A
Convert the current source to voltage source. We obtain the circuit below.
4Ω
+
19.2V
Using voltage division,
Vo =
10
(19.2) = 12.8 V
10 + 4 + 1
1Ω
+
Vo
-
10 Ω
Chapter 4, Solution 25.
Transforming only the current source gives the circuit below.
18 V
9Ω
−+
12V
+
−
5Ω
i
4Ω
vo
+
−
+
−
30 V
+−
2Ω
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 Ω
+ vx −
3A
30Ω
60Ω
30Ω
6A
20Ω
(a)
20 Ω
+
−
10 Ω
+ vx −
60V
i
(b)
12 Ω
+
−
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 Ω
+ vx −
5A
10Ω
40Ω
8A
20Ω
2A
(a)
8Ω
+
−
12 Ω
+ vx −
40V
i
(b)
20 Ω
+
−
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
+
−
4Ω
3Ω
12 V
2Ω
10 V
5Ω
+−
+−
10Ω
(a)
io
+
−
4Ω
12V
10 Ω
+
−
10Ω
22 V
(b)
io
+
−
4Ω
12V
io
10Ω
10Ω
(c)
2.2A
+
−
12V
4Ω
5Ω
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 kΩ
2 kΩ
2vo
(4/3) kΩ
−+
1.5vo
3 mA
1 kΩ
i
3 mA
+
1 kΩ
+
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 Ω
60 Ω
30 Ω
10 Ω
+
-
7ix
Combine the 60-ohm with the 10-ohm and transform the dependent source as shown
below.
24 Ω
ix
+
12V
-
30 Ω
70 Ω
0.1ix
Combining 30-ohm and 70-ohm gives 30//70 = 70x30/100 = 21-ohm. Transform the
dependent current source as shown below.
24 Ω
ix
21 Ω
+
12V
-
+
-
2.1ix
Applying KVL to the loop gives
45i x − 12 + 2.1i x = 0

→
ix =
12
= 254.8 mA
47.1
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).
3Ω
+
12V
+
−
vx
−
8Ω
vx/3
6Ω
(a)
3Ω
+
12V
+
−
vx
(24/7) Ω
−
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 Ω
10 Ω
5ix
−+
60V
+
−
50 Ω
40 Ω
(a)
15 Ω
60V
+
−
50 Ω
50 Ω
0.1ix
(b)
ix
60V
+
−
15 Ω
25 Ω
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 Ω
10 Ω
20 Ω
RTh
40 Ω
3A
+
−
v1
20 Ω
v2
+
40V
40 Ω
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
6Ω
b
3Ω
12 Ω
4Ω
(a)
2A
6Ω
+
−
v1
v2 4 Ω
+ VTh
+
12V v1
+
3Ω
12Ω
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 Ω
RTh = 5 Ω
+−
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
10Ω
a
10Ω
+
+
−
40Ω
VTh
40Ω
50V
b
b
(a)
(b)
From Fig. (a),
RTh = 10||40 = 8 ohms
From Fig. (b),
VTh = (40/(10 + 40))50 = 40V
8Ω
+
−
i
a
12 Ω
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 Ω
a
40 Ω
12 Ω
b
R N = 12 //( 20 + 40) = 10Ω
IN is found from the circuit below.
2A
20 Ω
a
40 Ω
+
120V
-
12 Ω
IN
b
Applying source transformation to the current source yields the circuit below.
20 Ω
40 Ω
+ 80 V -
+
120V
-
Applying KVL to the loop yields
− 120 + 80 + 60 I N = 0

→
I N = 40 / 60 = 0.6667 A
IN
Chapter 4, Solution 38
We find Thevenin equivalent at the terminals of the 10-ohm resistor. For RTh, consider
the circuit below.
1Ω
4Ω
5Ω
RTh
16 Ω
RTh = 1 + 5 //( 4 + 16) = 1 + 4 = 5Ω
For VTh, consider the circuit below.
V1
4Ω
1Ω
V2
5Ω
3A
+
16 Ω
VTh
+
12 V
At node 1,
V V − V2
3= 1 + 1

→
48 = 5V1 − 4V2
16
4
At node 2,
V1 − V2 12 − V2
+
=0

→
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.
5Ω
+
19.2V
-
+
Vo
-
10 Ω
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 Ω
3vab10 Ω a
+−
+
−
10 Ω
+
io
+
1V v1 −
50V
+ VTh
+
−
40 Ω4V
−
+
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
Ω
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 Ω
RTh
a
b
40Ω
(a)
20 Ω
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 Ω
+
-160V
+
−
vx
12 Ω
−
vx = [12/(12 + 28)](-160) = -48 V
Chapter 4, Solution 41
To find RTh, consider the circuit below
14 Ω
a
6Ω
5Ω
b
RTh = 5 //(14 + 6) = 4Ω = R N
Applying source transformation to the 1-A current source, we obtain the circuit below.
6Ω
14 Ω
- 14V +
VTh
a
+
6V
5Ω
3A
b
At node a,
14 + 6 − VTh
V
= 3 + Th
6 + 14
5
IN =

→
VTh = −8 V
VTh
= (−8) / 4 = −2 A
RTh
Thus,
RTh = R N = 4Ω,
VTh = −8V,
I N = −2 A
Chapter 4, Solution 42.
To find RTh, consider the circuit in Fig. (a).
20 Ω
10 Ω
30 Ω
a
20 Ω
30 Ω 30 Ω
b
a
10 Ω
10Ω
(a)
10 Ω
b
10 Ω
10 Ω
(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 Ω
+
b
−+
10 Ω
10 Ω
i1
30V
10 V
10 Ω
+
−
10 Ω
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
10Ω
b
10Ω
5Ω
(a)
10 Ω
+
−
a
+
50V va
b
+ VTh
10 Ω
+
vb
−
−
(b)
RTh = 10||10 + 5 = 10 ohms
5Ω
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
3Ω
3Ω
1Ω
a
+
−
+
−
2Ω
2Ω
i
b
5Ω
+
VTh
b
10V
5Ω
(b)
(a)
(b)
a
4Ω
24V
RTh
4Ω
1Ω
For RTh, consider the circuit in Fig. (c).
3Ω
1Ω
4Ω
3Ω
b
24V
2Ω
RTh
5Ω
1Ω
4Ω
vo
+
−
+
2Ω
5Ω
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).
6Ω
6Ω
6Ω
4Ω
RN
4A
6Ω
(a)
IN
4Ω
(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 Ω
20V
+
−
60 Ω
4Ω
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 Ω
IN 30V
+
−
Chapter 4, Solution 47
Since VTh = Vab = Vx, we apply KCL at the node a and obtain
30 − VTh VTh
=
+ 2VTh

→ VTh = 150 / 126 = 1.19 V
12
60
To find RTh, consider the circuit below.
12 Ω
Vx
a
2Vx
60 Ω
1A
At node a, KCL gives
V V
1 = 2V x + x + x

→ V x = 60 / 126 = 0.4762
60 12
V
V
RTh = x = 0.4762Ω,
I N = Th = 1.19 / 0.4762 = 2.5
RTh
1
Thus,
VTh = 1.19V ,
RTh = R N = 0.4762Ω,
I N = 2 .5 A
Chapter 4, Solution 48.
To get RTh, consider the circuit in Fig. (a).
10Io
10Io
+−
2Ω
+−
−
1A
−
2A
(a)
From Fig. (a),
VTh
4Ω
V
4Ω
+
Io
+
Io
(b)
Io = 1,
2Ω
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 Ω
40V
At the node,
+
−
vo
20 Ω
io
Isc = IN
40 Ω
(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
6Ω
6Ω
Isc = IN
4Ω
(a)
From Fig. (b),
2A
4Ω
+
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 Ω
0.4A
5Ω
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
6Ω
VTh
+
4Ω
6Ω
3Ω
2Ω
120V
4Ω
+
−
3Ω
(a)
6A
2Ω
(b)
For IN or VTh, consider the circuit in Fig. (b). After some source transformations, the
circuit becomes that shown in Fig. (c).
+ VTh
2Ω
40V
+
−
4Ω
i
2Ω
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
6Ω
4Ω
2Ω
i
+
3Ω
RN
2Ω
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 kΩ
Io
20Io
2 kΩ
RTh
b
(a)
3 kΩ
6V
+
−
a
+
Io
20Io
2 kΩ
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
2Ω
2Ω
a
+
6Ω
3Ω
+
vo
2Ω
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
6Ω
2Ω
a
+
18V
+
−
3Ω
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

→
For the right loop,
V x = −50 x 40i o = −2000io
Combining (1) and (2),
3 = 1000io − 4000io = −3000io
V x = −2000io = 2

→
3 = 1000io + 2V x
(1)
(2)

→
io = −1mA
VTh = 2
To find RTh, insert a 1-V source at terminals a-b and remove the 3-V independent
source, as shown below.
1 kΩ
ix
.
io
+
2Vx
-
40io
+
Vx
-
+
1V
-
50 Ω
2V x
= −2mA
1000
V
1
i x = 40io + x = −80mA + A = -60mA
50
50
V x = 1,
RTh =
io = −
1
= −1 / 0.060 = − 16.67Ω
ix
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 kΩ
+
−
80I
+
50 kΩ
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 kΩ
2V
+
−
a
I
vab/1000
+
80I
+
−
50 kΩ
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.
1Ω
20V
2Ω
a
2A
+
−
1Ω
4Ω
4Ω
RN
b
(a)
2Ω
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
+
−
1Ω
4Ω
a
2V
2Ω
−
+
−
+
IN
i
16V
RN
IN
(c)
(d)
3Ω
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).
2Ω
B
a
A
i
+
3Ω
vx
6Ω
0.5vx
−
(a)
10 Ω
+
−
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).
3Ω
2Ω
v1
v2
a
+
50V
+
−
+
6Ω
vx
10 Ω 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
+
−
β ib
vo
R2
Writing the node equation at node vo,
ib + βib = vo/R2 = (1 + β)ib
Isc
But
ib = (Vs – vo)/R1
vo = Vs – ibR1
Vs – ibR1 = (1 + β)R2ib, or ib = Vs/(R1 + (1 + β)R2)
Isc = IN = -βib = -βVs/(R1 + (1 + β)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 Ω
20 Ω
+ VTh
8A
50 Ω
40 Ω
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 Ω
5Ω
2A
10 Ω
a
b
3A
5Ω
2A
3A
a
3.333Ω
b
Norton Equivalent Circuit
10 V
3.333Ω
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).
2Ω
a
6Ω
6Ω
2Ω
2Ω
6Ω
b
(a)
b
2Ω
a
18 Ω
1.8 Ω
2Ω
18 Ω
a
2Ω
18 Ω
1.8 Ω
1.8 Ω
RTh
b
b
(b)
(c)
2Ω
a
6Ω
12V
6Ω
i3
+
−
12V
+
+
−
VTh
6Ω
2Ω
i1
−
+
i2
2Ω
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),
 7 − 3 − 3  i1   12 
− 3 7 − 3 i  = − 12

 2  

− 3 − 3 7  i 3   0 
−3 −3
7
7
∆ = − 3 7 − 3 = 100 ,
−3 −3 7
−3
12
∆ 2 = − 3 − 12 − 3 = −120
−3 0
7
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 Ω
10 Ω
v1
io
2vo
VS
+
−
20 Ω
+ −
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 Ω
+
vo
20 Ω
−
Applying KCL at node 1,
3Ω
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.
4Ω
1Ω
vo
io
ix
+
–
2Ω
+
−
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 = 5Ω,
VTh =
(32) = 24 V
12 + 4
Thus, the circuit can be replaced by that shown below.
5Ω
Io
+
24 V
-
+
Vo
-
Applying KVL to the loop,
− 24 + 5I o + Vo = 0

→
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).
2Ω
10V
− +
3Ω
2Ω
a
b
+
3Ω
a
VTh
5Ω
b
RTh
+
−
5Ω
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
+
4Ω
4Ω
6Ω
8Ω
6Ω
−
30V
+ −
RTh
12 Ω
i1
VTh
+
12 Ω
+
i2
8Ω
−
−
(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 kΩ v1
+
10 kΩ
vo
−
40 kΩ
3vo
30 kΩ
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 kΩ vo 22 kΩ
v1
+
100V
+
−
+
40 kΩ
vo
−
3vo
30 kΩ
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
5Ω
5Ω
+
+
15 Ω
4V
-
VTh
6Ω
+
Vx
-
From the figure,
15
(4) = 3V
15 + 5
consider the circuit below:
V x = 0,
To find RTh,
VTh =
3Vx
5Ω
5Ω
V2
V1
+
4V
-
15 Ω
+
At node 1,
V V − V2
4 − V1
= 3V x + 1 + 1
,
5
15
5
1A
6Ω
Vx
V x = 6 x1 = 6
-

→
258 = 3V2 − 7V1
(1)
At node 2,
V − V2
1 + 3V x + 1
=0

→ V1 = V2 − 95
5
Solving (1) and (2) leads to V2 = 101.75 V
2
VTh
V2
9
RTh =
= 101.75Ω,
p max =
=
= 22.11 mW
1
4 RTh 4 x101.75
(2)
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 kΩ
a
+
3 kΩ
vo
−
+
1 kΩ
−
120vo
40 kΩ
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
4Ω
6Ω
2Ω
4Ω
12V
6Ω
− +
2Ω
+
−
RTh
VTh
8V
20V
(a)
(b)
+ −
(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.
10 Ω
25 Ω
RTh
20 Ω
RTh = 10 // 20 + 25 // 5 = 325 / 30 = 10.833Ω
+
5Ω
−
10 Ω
+
60 V
-
+ VTh -
+
Va
25 Ω
20 Ω
+
5Ω
20
(60) = 40,
30
− Va + VTh + Vb = 0
Va =
Vb
-
5
(60) = 10
30

→ VTh = Va − Vb = 40 − 10 = 30 V
Vb =
2
p max
V
30 2
= Th =
= 20.77 W
4 RTh 4 x10.833
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 Ω
i
12V
i = 12/2.6 ,
+
−
2Ω
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 = ∞,

→ VTh = Voc = VL = 10.8 V
When RL = 4 ohm, VL=10.5,
IL =
VL
= 10.8 / 4 = 2.7
RL
But
VTh = VL + I L RTh

→
RTh =
VTh − V L 12 − 10.8
=
= 0.4444Ω
IL
2 .7
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

→
10
6=
VTh
10 + RTh
or
60 + 6 RTh = 10VTh
where RTh is in k-ohm.
(1)
Similarly,
30

→
VTh
30 + RTh
Solving (1) and (2) leads to
12 =
360 + 12 RTh = 30VTh
(2)
VTh = 24 V, RTh = 30kΩ
(b) Vab =
20
(24) = 9.6 V
20 + 30
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 Ω
RTh
5k Ω
B
20k Ω
10k Ω
= 30 + 10 + 20 // 5 = 44kΩ
(2)
To find VTh , consider the circuit below.
5k Ω
A
B
io
30k Ω
20k Ω
+
4mA
60 V
-
10k Ω
V A = 30 x 4 = 120,
VB =
20
(60) = 48,
25
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 99.99µA .
(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 kΩ
3 kΩ
220V
+
−
a
i1
+
6 kΩ
i2
b
vab
−
5 kΩ
10 kΩ
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 kΩ
3 kΩ
R2
6 kΩ
a
RTh
R1
b
5 kΩ
6 kΩ
a RTh
R3
18 kΩ
(b)
18 kΩ
(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
+
−
βRoix
-Vs + (Rs + Ro)ix + βRoix = 0
ix = Vs/(Rs + (1 + β)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/α = Rg + Rs + Rp
Rg + Rs = Rp((1/α) – 1) = Rp(1 - α)/α
(1a)
Combining (2) and (1a) gives,
Rs = [(1 - α)/α]Req
= (1 – 0.125)(100)/0.125 = 700 ohms
From (3) and (1a),
Rp(1 - α)/α = Rg + [(1 - α)/α]Rg = Rg/α
Rp = Rg/(1 - α) = 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 µA
For the 0 – 10 V scale,
Rm = Vfs/Ifs = 10/50 µA = 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 µA = 20 µA
VTh = 20 µA RTh + 20 µA 250 k ohms
= 4 + 20 µA RTh
(b)
(1)
A 5V reading corresponds to
I = (5/50)Ifs = 0.1 x 50 µA = 5 µA
VTh = 5 µA x RTh + 5 µA x 1 M ohm
From (1) and (2)
VTh = 5 + 5 µA RTh
(2)
0 = -1 + 15 µA 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 Ω
8Ω
10 Ω
RTh
9V
+
−
i1
40Ω
+
i2
60 Ω
8Ω
R
VTh
10 Ω
−
+
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 kΩ
12V
+
−
B
+
4 kΩ
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 Ω
30 Ω
30 Ω
R2
R1
14 Ω
a
60 Ω
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 Ω
20 Ω
I1
14 Ω
b
a
60 Ω
+
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 MΩ
Rout = 60 Ω
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 =
= −20µV
A 2 x10 5
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 =
(R 0 + R i A) v i
R 0 + (1 + A)R i
v0
R 0 + RiA
100 + 10 4 x10 5
=
⋅ 10 4
=
5
v i R 0 + (1 + A)R i 100 + (1 + 10 )
≅
100,000
10 9
⋅ 10 4 =
= 0.9999990
5
100,001
1 + 10
(
)
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 
 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 kΩ
10 kΩ
VS
+
-
Rout = 100 Ω
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 kΩ
2V
+
ia
va
2V
+
vb
1V
+
+
vo
-
-
(a)
2 kΩ
+
va
-
10 kΩ
+-
+
ia
(b)
vo
-
Since va = vb = 1V and ia = 0, no current flows through the 10 kΩ 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  v o
vs = v o 
=
 10 + 10  2
vo
=2
vs
Chapter 5, Solution 11.
8 kΩ
2 kΩ
3V
vb =
5 kΩ
+
−
a
b
io
−
+
+
10 kΩ
4 kΩ
vo
−
10
(3) = 2V
10 + 5
At node a,
3 − va va − vo
=
2
8
12 = 5va – vo
But va = vb = 2V,
vo = -2V
12 = 10 – vo
–io =
va − vo 0 − vo 2 + 2 2
+
=
+ = 1mA
8
4
8
4
i o = -1mA
Chapter 5, Solution 12.
4 kΩ
1 kΩ
1.2V
+
−
a
b
−
+
4 kΩ
2 kΩ
+
vo
−
4
2
2
vo = vo = vo
4+2
3
3
At node b,
vb =
At node a,
1 .2 − v a v a − v o
2
, but va = vb = v o
=
1
4
3
4.8 - 4 x
2
2
vo = vo − vo
3
3
va = vb =
2
9.6
vo =
3
7
is =
vo =
3x 4.8
= 2.0570V
7
1 .2 − v a − 1 .2
=
1
7
 − 1.2 
p = vsis = 1.2 
 = -205.7 mW
 7 
Chapter 5, Solution 13.
10 kΩ
a
b
1V
+
−
90 kΩ
+
−
io
100 kΩ i2
i1
4 kΩ
50 kΩ
By voltage division,
90
va =
(1) = 0.9V
100
v
50
vb =
vo = o
3
150
v0
But va = vb
= 0 .9
vo = 2.7V
3
v
v
io = i1 + i2 = o + o = 0.27mA + 0.018mA = 288 µA
10k 150k
+
vo
−
Chapter 5, Solution 14.
Transform the current source as shown below. At node 1,
10 − v1 v1 − v 2 v1 − v o
=
+
5
20
10
10 kΩ
5 kΩ
10 kΩ
20 kΩ
v1
10V
vo
v2
+
−
−
+
+
vo
−
But v2 = 0. Hence 40 - 4v1 = v1 + 2v1 - 2vo
At node 2,
v1 − v 2 v 2 − v o
=
,
20
10
40 = 7v1 - 2vo
v 2 = 0 or v1 = -2vo
From (1) and (2), 40 = -14vo - 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
v −v
v
1  vo
 −
+
(1)
i s = 1 + 1 o = v1 
R2
R3
R
R
R3
3 
 2
At the inverting terminal,
0 − v1
(2)
is =

→ v1 = −i s R1
R1
Combining (1) and (2) leads to


v
vo
RR 
R
R 

→
= − R1 + R3 + 1 3 
i s 1 + 1 + 1  = − o
R3
is
R2 

 R2 R3 
(b) For this case,
vo
20 x 40 

= − 20 + 40 +
 kΩ = - 92 kΩ
is
25 

Chapter 5, Solution 16
10k Ω
5k Ω
ix
va
vb
iy
+
vo
2k Ω
+
0.5V
-
8k Ω
Let currents be in mA and resistances be in k Ω . At node a,
0 .5 − v a v a − v o
=

→ 1 = 3v a − vo
5
10
(1)
But
8
10
vo

→ vo = v a
(2)
8+2
8
Substituting (2) into (1) gives
10
8
1 = 3v a − v a

→ v a =
8
14
Thus,
0 .5 − v a
ix =
= −1 / 70 mA = − 14.28 µA
5
v − vb v o − v a
10
0 .6 8
iy = o
+
= 0 .6 ( v o − v a ) = 0 .6 ( v a − v a ) =
x mA = 85.71 µA
2
10
8
4 14
v a = vb =
Chapter 5, Solution 17.
(a)
(b)
(c)
G=
vo
R
12
= − 2 = − = -2.4
vi
R1
5
vo
80
=−
= -16
vi
5
vo
2000
=−
= -400
vi
5
Chapter 5, Solution 18.
Converting the voltage source to current source and back to a voltage source, we have the
circuit shown below:
10 20 =
20
kΩ
3
1 MΩ
(20/3) kΩ 50 kΩ
−
+
+
−
2vi/3
+
vo
−
vo = −
1000 2v i
⋅
20 3
50 +
3
vo
200
=−
= -11.764
v1
17
Chapter 5, Solution 19.
We convert the current source and back to a voltage source.
24=
(4/3) kΩ
(2/3)V
+
−
4 kΩ
0V
4
3
10 kΩ
−
+
vo
5 kΩ
10k  2 
  = -1.25V
 4 3
 4x k
 3
v
v −0
= -0.375mA
io = o + o
5k
10k
vo = −
Chapter 5, Solution 20.
8 kΩ
4 kΩ
9V
a
+
−
4 kΩ
vs
2 kΩ
b
−
+
+
−
+
vo
−
At node a,
9 − va va − vo va − vb
=
+
4
8
4
18 = 5va – vo - 2vb
(1)
At node b,
va − vb vb − vo
=
4
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 = 10kΩ, then Rf = 150 kΩ.
Chapter 5, Solution 23
At the inverting terminal, v=0 so that KCL gives
vs − 0
0 0 − vo
=
+
R1
R2
Rf
→
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
+
+
=0
R1
R2
Rf

→
 1

 + 1 + 1 v1 − v s = vo
R R
R f 
R2 R f
2
 1
(1)
Applying KCL at node 2 gives
R3
v1 v1 − v s
+
=0

→ v1 =
vs
R3 + R4
R3
R4
Substituting (2) into (1) yields
(2)
 R
R
R  R3  1 
 −  v s
vo = R f  3 + 3 − 4 
 R1 R f R2  R3 + R4  R2 
i.e.
 R
R
R  R3  1 
 − 
k = R f  3 + 3 − 4 
 R1 R f R2  R3 + R4  R2 
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
-
8k Ω
+
2k Ω
vo
-
vb = 0.4 =
8
vo = 0.8vo
8+ 2

→
Hence,
io =
v o 0 .5
=
= 0.1 mA
5k 5k
vo = 0.4 / 0.8 = 0.5 V
io
5k Ω
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 

G = v0/(vi) = 10.2
(b)
vi = v0/(G) = 15/(10.2) cos 120πt = 1.471 cos 120πt V
Chapter 5, Solution 28.
−
+
+
−
At node 1,
0 − v1 v1 − v o
=
10k
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 µA
Chapter 5, Solution 29
R1
va
vb
+
vi
-
va =
R2
R2
vi ,
R1 + R2
But v a = vb
vb =

→
+
-
R1
R1
vo
R1 + R2
R2
R1
vi =
vo
R1 + R2
R1 + R2
Or
v o R2
=
vi
R1
Chapter 5, Solution 30.
The output of the voltage becomes
vo = vi = 12
30 20 = 12kΩ
By voltage division,
vx =
12
(1.2) = 0.2V
12 + 60
ix =
p=
v 2x 0.04
=
= 2µW
R
20k
vx
0 .2
=
= 10µA
20k 20k
+
R2
vo
-
Chapter 5, Solution 31.
After converting the current source to a voltage source, the circuit is as shown below:
12 kΩ
3 kΩ
1
6 kΩ v
o
v1
12 V
+
−
+
−
2
vo
6 kΩ
At node 1,
12 − v1 v1 − v o v1 − v o
=
+
3
6
12
48 = 7v1 - 3vo
(1)
At node 2,
v1 − v o v o − 0
=
= ix
6
6
v1 = 2vo
(2)
From (1) and (2),
48
11
vo
ix =
= 0.7272mA
6k
vo =
Chapter 5, Solution 32.
Let vx = the voltage at the output of the op amp. The given circuit is a non-inverting
amplifier.
 50 
v x = 1 +  (4 mV) = 24 mV
 10 
60 30 = 20kΩ
By voltage division,
vo =
v
20
v o = o = 12mV
20 + 20
2
ix =
vx
24mV
=
= 600nA
(20 + 20)k 40k
p=
v o2 144x10 −6
=
= 204nW
R
60x10 3
Chapter 5, Solution 33.
After transforming the current source, the current is as shown below:
1 kΩ
4 kΩ
4V
vi
+
−
va
+
−
2 kΩ
vo
3 kΩ
This is a noninverting amplifier.
3
 1
v o = 1 +  v i = v i
2
 2
Since the current entering the op amp is 0, the source resistor has a OV potential drop.
Hence vi = 4V.
vo =
3
(4) = 6V
2
Power dissipated by the 3kΩ resistor is
v o2 36
=
= 12mW
R 3k
ix =
va − vo 4 − 6
=
= -2mA
R
1k
Chapter 5, Solution 34
v1 − vin v1 − vin
+
=0
R1
R2
(1)
R3
vo
R3 + R 4
(2)
but
va =
Combining (1) and (2),
v1 − va +
R1
R
v 2 − 1 va = 0
R2
R2

R 
R
v a 1 + 1  = v1 + 1 v 2
R2
 R2 
R
R 
R 3v o 
1 + 1  = v1 + 1 v 2
R2
R3 + R 4  R 2 
vo =
vO =

R3 + R 4 
R
 v1 + 1 v 2 
R2 

R 
R 3 1 + 1  
 R2 
R3 + R 4
( v1R 2 + v 2 )
R 3 ( R1 + R 2 )
Chapter 5, Solution 35.
Av =
vo
R
= 1 + f = 10
Ri
vi
If Ri = 10kΩ, Rf = 90kΩ
Rf = 9Ri
Chapter 5, Solution 36
VTh = Vab
But
VTh
R1
Vab . Thus,
R1 + R2
R
R + R2
= Vab = 1
v s = (1 + 2 )v s
R1
R1
vs =
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
v
RTh = o = 0
io
Chapter 5, Solution 37.

R
R
R
v o = −  f v1 + f v 2 + f v 3 
R3 
R2
 R1
30
30
 30

= −  (1) + (2) + (−3)
20
30
 10

vo = -3V
Chapter 5, Solution 38.

R
R
R
R
v o = −  f v1 + f v 2 + f v 3 + f v 4 
R4 
R3
R2
 R1
50
50
50
 50

= −  (10) + (−20) + (50) + (−100)
20
10
50
 25

= -120mV
Chapter 5, Solution 39
This is a summing amplifier.
Rf
Rf 
 Rf
50
50
 50

vo = −
v1 +
v2 +
v3  = − (2) + v 2 + (−1)  = −9 − 2.5v 2
R2
R3 
20
50
 10

 R1
Thus,
vo = −16.5 = −9 − 2.5v 2

→ v 2 = 3 V
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
+
+
=0
R1
R2
R3

→
v1 v 2 v3
1
1
1
+
+
= va ( +
+ ) (1)
R1 R2 R3
R1 R2 R3
But
v a = vb =
R
vo
R + Rf
(2)
Substituting (2) into (1)gives
Rvo
v1 v 2 v3
1
1
1
+
+
=
( +
+ )
R1 R2 R3 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 = 40kΩ
The averaging amplifier is as shown below:
v1
v2
v3
v4
Chapter 5, Solution 42
1
R f = R1 = 10 kΩ
3
R1 = 40 kΩ
10 kΩ
R2 = 40 kΩ
R3 = 40 kΩ
R4 = 40 kΩ
−
+
vo
Chapter 5, Solution 43.
In order for

R
R
R
R
v o =  f v1 + f v 2 + f v 3 + f v 4 
R2
R3
R4 
 R1
to become
1
(v 1 + v 2 + v 3 + v 4 )
4
R
Rf 1
12
=
Rf = i =
= 3kΩ
Ri 4
4
4
vo = −
Chapter 5, Solution 44.
R4
R3
v1
v2
a
R1
−
+
b
R2
At node b,
v b − v1 v b − v 2
+
=0
R1
R2
At node a,
0 − va va − vo
=
R3
R4
v1 v 2
+
R1 R 2
vb =
1
1
+
R1 R 2
(1)
vo
1+ R4 / R3
(2)
va =
But va = vb. We set (1) and (2) equal.
vo
R v + R 1 v1
= 2 1
1+ R4 / R3
R1 + R 2
or
vo =
vo
(R 3 + R 4 )
(R 2 v1 + R 1 v1 )
R 3 (R 1 + R 2 )
Chapter 5, Solution 45.
This can be achieved as follows:
 R
(− v1 ) + R v 2 
v o = −
R/2 
R / 3

R
R
= −  f (− v1 ) + f v 2 
R2 
 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<100kΩ).
R
v1
R
R
−
+
R/3
-v1
v2
R/2
−
+
vo
Chapter 5, Solution 46.
v1 1
R
R
R
1
+ ( − v 2 ) + v 3 = f 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 = 10kΩ,
a solution is given below.
− vo =
10 kΩ
v2
v1
10 kΩ
−
+
30 kΩ
30 kΩ
10 kΩ
-v2
v3
20 kΩ
−
+
vo
Chapter 5, Solution 47.
If a is the inverting terminal at the op amp and b is the noninverting terminal,
then,
vb =
10 − v a v a − v o
3
(8) = 6V, v a = v b = 6V and at node a,
=
2
4
3 +1
which leads to vo = –2 V and io =
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) = 0.125µA
For loop 2, (40 + 60) i2 = -5
i2 = -0.05µA
But, 10i + v1 - 5 = 0
60i + v2 + 5 = 0
v1 = 5 - 10i = 3.75mV
v2 = -5 - 60i = -2mV
As a difference amplifier,
R
80
[3.75 − (−2)]mV
v o = 2 (v 2 − v 1 ) =
20
R1
= 23mV
Chapter 5, Solution 49.
R1 = R3 = 10kΩ, R2/(R1) = 2
i.e.
R2 = 2R1 = 20kΩ = R4
vo =
Verify:
=2
R 2 1 + R1 / R 2
R
v 2 − 2 v1
R1 1 + R 3 / R 4
R1
(1 + 0.5)
v 2 − 2v1 = 2(v 2 − v1 )
1 + 0.5
Thus, R1 = R3 = 10kΩ, R2 = R4 = 20kΩ
Chapter 5, Solution 50.
(a)
We use a difference amplifier, as shown below:
v1
R1
R2
−
+
v2
vo =
(b)
R1
vo
R2
R2
(v 2 − v1 ) = 2(v 2 − v1 ), i.e. R2/R1 = 2
R1
If R1 = 10 kΩ then R2 = 20kΩ
We may apply the idea in Prob. 5.35.
v 0 = 2 v1 − 2 v 2
 R
(− v1 ) + R v 2 
= −
R/2 
R / 2
R

R
= −  f (− v1 ) + f v 2 
R2 
 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 = 10kΩ.
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 Ω .
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,
v1 − v a v a − v o
=
R1
R2
R2
vb =
v2
R1 + R 2
va =
But va = vb. Setting (1) and (2) equal gives
R v + R 1v o
R2
v2 = 2 1
R1 + R 2
R1 + R 2
R 2 v1 + R 1 v o
R1 + R 2
(1)
(2)
v 2 − v1 =
R1
vo = vi
R2
vo R 2
=
vi
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 v A − v a
+
=
R1 / 2
Rg
R1 / 2
or
v1 − v A +
At node B,
v2 − vB vB − vA vB − vb
=
+
R1 / 2
R1 / 2
Rg
or
v2 − vB −
R1
(v B − v A ) = v A − v a
2R g
R1
(v B − v A ) = v B − v b
2R g
Subtracting (1) from (2),
v 2 − v1 − v B + v A −
2R 1
(v B − v A ) = v B − v A − v b + v a
2R g
Since, va = vb,
v 2 − v1 
R 
v
= 1 + 1 (v B − v A ) = i


2
2
 2R g 
(1)
(2)
vB − vA =
or
vi
⋅
2
1
R
1+ 1
2R g
(3)
But for the difference amplifier,
R2
(v B − v A )
R1 / 2
R
vB − vA = 1 vo
2R 2
vo =
or
Equating (3) and (4),
R1
v
vo = i ⋅
2R 2
2
vo R 2
=
⋅
vi
R1
(c)
At node a,
At node b,
(4)
1
R
1+ 1
2R g
1
R
1+ 1
2R g
v1 − v a v a − v A
=
R1
R2 /2
2R 1
2R 1
vA
va −
v1 − v a =
R2
R2
2R 1
2R 1
vB
vb −
v2 − vb =
R2
R2
(1)
(2)
Since va = vb, we subtract (1) from (2),
− 2R 1
v
(v B − v A ) = i
2
R2
− R2
vi
vB − vA =
2R 1
v 2 − v1 =
or
(3)
At node A,
va − vA vB − vA vA − vo
+
=
R2 /2
Rg
R/2
va − vA +
R2
(v B − v A ) = v A − v o
2R g
(4)
At node B,
vb − vB vB − vA vB − 0
−
=
R/2
Rg
R/2
vb − vB −
R2
(v B − v A ) = v B
2R g
(5)
Subtracting (5) from (4),
v B −v A +
R2
(v B − v A ) = v A − v B − v o
Rg

R 
2(v B − v A )1 + 2  = − v o
 2R 
g 

Combining (3) and (6),
− R 2 
R 
v i 1 + 2  = −v o
 2R 
R1
g 

v o R 2 
R 
=
1+ 2 
vi
R 1  2R g 
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
(6)
Chapter 5, Solution 56.
There is a cascading system of two inverting amplifiers.
− 12  − 12 

 v s = 6v s
4  6 
v
i o = s = 3v s mA
2k
vo =
(a)
(b)
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 o′ =
R2
100
( v 2 − v1 ) =
(1 + 4) = 10 mA
R1
50
The second stage is a non-inverter.
R
R


v o = 1 +  v o ′ = 1 + 10 mA = 40 mV(given)
 40 
 40 
Which leads to,
R = 120 kΩ
Chapter 5, Solution 58.
By voltage division, the input to the voltage follower is:
v1 =
3
(0.6) = 0.45V
3 +1
vo =
10
− 10
v1 − v1 = −7 v1 = −3.15
2
5
io =
0 − vo
= 0.7875mA
4k
Thus
Chapter 5, Solution 59.
Let a be the node between the two op amps.
va = vo
The first stage is a summer
va =
or
− 10
10
vs −
vo = vo
5
20
1.5vs = -2vs
vo − 2
=
= -1.333
v s 1 .5
Chapter 5, Solution 60.
Transform the current source as shown below:
4 kΩ
10 kΩ
5 kΩ
−
+
5is
v1
+
−
io
3 kΩ
+
−
3Ω
2 kΩ
Assume all currents are in mA. The first stage is a summer
v1 =
By voltage division,
− 10
(5i s ) − 10 v o = −10i s − 2.5v o
5
4
(1)
v1 =
3
1
vo = vo
3+3
2
(2)
Alternatively, we notice that the second stage is a non-inverter.
 1 
vo = 
 v1 = 2 v1
3+ 3
From (1) and (2),
0.5v o = −10i s − 2.5v o
v o = −2i o = −
10i s
3
3vo = 10is
io 5
= = 1.667
is 3
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.
R2
v1
R1
R
R
v 0 = − 4 v 01 − 4 v 2
R5
R3
v 01 = −
v1 =
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
vi – 2 vo
R1
Rf
The second stage is a follower. By voltage division
(1)
vo = v2 =
R4
v1
R3 + R4
v1 =
R3 + R4
vo
R4
(2)
From (1) and (2),
 R3 
R
R
 v o = − 2 v i − 2 v o
1 +
Rf
R1
 R4 
 R3 R2 
R
 v o = − 2 v i
1 +
+
R1
 R4 Rf 
vo
R
1
=− 2 ⋅
R
R
vi
R1
1+ 3 + 2
R4 R4
− R 2R 4
=
R 1 (R 2 + R 3 + R 4 )
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 
R6
 R1 
 R R  R R
R 
v o 1 − 2 4  =  2 4 − 4  v i
 R 3 R 5   R 1R 5 R 6 
vo =
R4
R5
R 2R 4 R 4
−
vo
R 1R 3 R 6
=
R R
vi
1− 2 4
R 3R 5
Chapter 5, Solution 64
G4
G
G1
+
G3
1
G
+
0V
vs
v
2
0V +
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)
→

(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

→
vo =
48
v o ' = − 21.6 mV
40
Chapter 5, Solution 66.
100  40 
100
− 110
(6) −
(2)
 − (4) −
25
20  20 
10
= −24 + 40 − 20 = -4V
vo =
Chapter 5, Solution 67.
80  80 
80
 − (0.5) − (0.2)
40  20 
20
= 3.2 − 0.8 = 2.4V
vo = −
Chapter 5, Solution 68.
If Rq = ∞, the first stage is an inverter.
Va = −
15
(10) = −30mV
5
when Va is the output of the first op amp.
The second stage is a noninverting amplifier.
 6
v o = 1 +  v a = (1 + 3)(−30) = -120mV
 2
Chapter 5, Solution 69.
In this case, the first stage is a summer
va = −
15
15
(10) − v o = −30 − 1.5v o
5
10
For the second stage,
 6
v o = 1 +  v a = 4v a = 4(− 30 − 1.5v o )
 2
120
vo = −
= -17.143mV
7 v o = −120
7
Chapter 5, Solution 70.
The output of amplifier A is
vA = −
30
30
(10) − (2) = −9
10
10
The output of amplifier B is
vB = −
20
20
(3) − (4) = −14
10
10
40 kΩ
vA
vB
20 kΩ
a
60 kΩ
−
+
b
vo
10 kΩ
vb =
60
(−14) = −2V
60 + 10
At node a,
vA − va va − vo
=
20
40
But va = vb = -2V, 2(-9+2) = -2-vo
Therefore,
vo = 12V
Chapter 5, Solution 71
20k Ω
5k Ω
100k Ω
40k Ω
+
+
2V
-
v2
10k Ω
80k Ω
+
20k Ω
+
vo
-
+
+
3V
-
10k Ω
v1
+
-
30k Ω
v3
50k Ω
20
50
(2) = −8, v3 = (1 + )v1 = 8
5
30
100 
 100
vo = −
v2 +
v3  = −(−20 + 10) = 10 V
80 
 40
v1 = 3,
v2 = −
Chapter 5, Solution 72.
Since no current flows into the input terminals of ideal op amp, there is no voltage
drop across the 20 kΩ 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) = -1V
v2 = −
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
100
(0.6) = −6V
10
32
v2 = −
(0.4) = −8V
1.6
v − v2
−6+8
io = 1
=−
= 100 µA
20k
20k
v1 = −
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 µA
(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 µA
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 µA
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 µA
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 kΩ
20 kΩ
20 kΩ
vi
vi
30 kΩ
+−
40 kΩ
80 kΩ
40 kΩ
b
80 kΩ
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 kΩ
2 kΩ
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 kΩ
5 kΩ a
5is
+
−
b
−
+
4 kΩ
io
2 kΩ
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 

 R 2R 3

R 2RL
 R 4 +
R2 + RL




Chapter 6, Solution 1.
i=C
(
)
dv
= 5 2e −3t − 6 + e −3 t = 10(1 - 3t)e-3t A
dt
p = vi = 10(1-3t)e-3t ⋅ 2t e-3t = 20t(1 - 3t)e-6t W
Chapter 6, Solution 2.
1 2 1
Cv1 = (40)(120) 2
2
2
1
1
w2 = Cv12 = (40)(80) 2
2
2
w1 =
∆w = w 1 − w 2 = 20(120 2 − 80 2 ) = 160 kW
Chapter 6, Solution 3.
i=C
280 − 160
dv
= 40x10 −3
= 480 mA
dt
5
Chapter 6, Solution 4.
v=
=
1 t
idt + v(0)
C ∫o
1
6 sin 4 tdt + 1
2∫
= 1 - 0.75 cos 4t
Chapter 6, Solution 5.
1 t
idt + v(0)
C ∫o
For 0 < t < 1, i = 4t,
t
1
v=
4t dt + 0 = 100t2 kV
− 6 ∫o
20x10
v(1) = 100 kV
v=
For 1 < t < 2, i = 8 - 4t,
t
1
v=
(8 − 4t )dt + v(1)
20x10 −6 ∫1
= 100 (4t - t2 - 3) + 100 kV
100t 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
dv
=
dt 2x10 −3
10
dv
= 30x10 −6 x
= 150mA
i= C
dt
2x10 −3
Thus the current i is sketched below.
i(t) (mA)
150
4
8
2
6
t (msec)
10
-150
Chapter 6, Solution 7.
v=
1
1
idt + v( t o ) =
∫
C
50x10 −3
=
t
∫ 4tx10
o
2t 2
+ 10 = 0.04k2 + 10 V
50
−3
dt + 10
12
Chapter 6, Solution 8.
(a) i = C
dv
= −100 ACe −100t − 600 BCe −600t
dt
i (0) = 2 = −100 AC − 600 BC

→
(1)
5 = − A − 6B
v (0 + ) = v (0 − )

→ 50 = A + B
Solving (2) and (3) leads to
A=61, B=-11
(b) Energy =
(2)
(3)
1 2
1
Cv (0) = x 4 x10 −3 x 2500 = 5 J
2
2
(c ) From (1),
i = −100 x61x 4 x10 −3 e −100t − 600 x11x 4 x10 −3 e −600t = − 24.4e −100t − 26.4e −600t A
Chapter 6, Solution 9.
v(t) =
(
)
(
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
i=C
)
1 t
6 1 − e − t dt + 0 = 12 t + e − t V
∫
o
12
dv
dv
= 2 x10 −3
dt
dt
 16t , 0 < t < 1µs

v =  16, 1 < t < 3 µs
64 - 16t, 3 < t < 4 µs

 16 x10 6 , 0 < t < 1µs
dv 
=  0, 1 < t < 3 µs
dt 
6
- 16x10 , 3 < t < 4 µs
 32 kA, 0 < t < 1µs

i (t ) =  0, 1 < t < 3 µs
- 32 kA, 3 < t < 4µs

Chapter 6, Solution 11.
1 t
idt + v(0)
C ∫o
For 0 < t < 1,
v=
t
1
40 x10 −3 dt = 10t kV
− 6 ∫o
4x10
v(1) = 10 kV
v=
For 1 < t < 2,
v=
1 t
vdt + v(1) = 10kV
C ∫1
For 2 < t < 3,
1
4x10 −6
= -10t + 30kV
v=
t
∫ (−40x10
2
−3
)dt + v(2)
Thus
0 < t <1
10 t ⋅ kV,

v(t) = 10kV,
1< t < 2
− 10 t + 30kV, 2 < t < 3
Chapter 6, Solution 12.
dv
= 3x10 −3 x 60(4π)(− sin 4π t)
dt
= - 0.7e π sin 4πt A
i=C
P = vi = 60(-0.72)π cos 4π t sin 4π t = -21.6π sin 8π t W
W=
∫
t
o
1
pdt = − ∫ 8 21.6π sin 8π t dt
o
21.6π
cos 8π
=
8π
1/ 8
o
= -5.4J
Chapter 6, Solution 13.
Under dc conditions, the circuit becomes that shown below:
i1
10 Ω
50 Ω
i2
20 Ω
+
+
v1
30 Ω
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
4
4
= =
C eq C 30
(b)
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 = 0.1J
2
2
1
w30 = x30x10 −6 x100 2 = 0.15J
2
w20 =
(b)
When they are connected in series as in Fig. (b):
v1 =
w20 =
C2
30
x100 = 60, v2 = 40
V=
50
C1 + C 2
1
x30x10 −6 x 60 2 = 36 mJ
2
w30 =
1
x30x10 −6 x 40 2 = 24 mJ
2
Chapter 6, Solution 16
C eq = 14 +
Cx80
= 30
C + 80

→
C = 20 µF
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 µF
Hence
1
1
1
1
1
=
+
+
=
C eq 20 30 60 10
Ceq = 10 µF
Chapter 6, Solution 19.
We combine 10-, 20-, and 30- µ F capacitors in parallel to get 60 µ F. The 60 - µ F
capacitor in series with another 60- µ F capacitor gives 30 µ F.
30 + 50 = 80 µ F, 80 + 40 = 120 µ F
The circuit is reduced to that shown below.
12
120
12
80
120- µ F capacitor in series with 80 µ F gives (80x120)/200 = 48
48 + 12 = 60
60- µ F capacitor in series with 12 µ F gives (60x12)/72 = 10 µ 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.
4µF in series with 12µF = (4x12)/16 = 3µF
3µF in parallel with 3µF = 6µF
6µF in series with 6µF = 3µF
3µF in parallel with 2µF = 5µF
5µF in series with 5µF = 2.5µF
Hence Ceq = 2.5µF
Chapter 6, Solution 22.
Combining the capacitors in parallel, we obtain the equivalent circuit shown below:
a
b
40 µF
60 µF
30 µF
20 µF
Combining the capacitors in series gives C1eq , where
1
1
1
1
1
=
+
+
=
1
C eq 60 20 30 10
Thus
Ceq = 10 + 40 = 50 µF
C1eq = 10µF
Chapter 6, Solution 23.
(a)
(b)
3µF is in series with 6µF
v4µF = 1/2 x 120 = 60V
v2µF = 60V
3
v6µF =
(60) = 20V
6+3
v3µF = 60 - 20 = 40V
3x6/(9) = 2µF
Hence w = 1/2 Cv2
w4µF = 1/2 x 4 x 10-6 x 3600 = 7.2mJ
w2µF = 1/2 x 2 x 10-6 x 3600 = 3.6mJ
w6µF = 1/2 x 6 x 10-6 x 400 = 1.2mJ
w3µF = 1/2 x 3 x 10-6 x 1600 = 2.4mJ
Chapter 6, Solution 24.
20µF is series with 80µF = 20x80/(100) = 16µF
14µF is parallel with 16µF = 30µF
(a)
v30µF = 90V
v60µF = 30V
v14µF = 60V
80
v20µF =
x 60 = 48V
20 + 80
v80µF = 60 - 48 = 12V
(b)
1 2
Cv
2
w30µF = 1/2 x 30 x 10-6 x 8100 = 121.5mJ
w60µF = 1/2 x 60 x 10-6 x 900 = 27mJ
w14µF = 1/2 x 14 x 10-6 x 3600 = 25.2mJ
w20µF = 1/2 x 20 x 10-6 x (48)2 = 23.04mJ
w80µF = 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 C 2
=
v 2 C1
vs = v1 + v2 =
Similarly, v1 =
C + C2
C2
v2
v2 + v2 = 1
C1
C1
v2 =
C2
vs
C1 + C 2
(b) For capacitors in parallel
Q1 Q 2
=
C1 C 2
C
C + C2
Q2
Qs = Q1 + Q2 = 1 Q 2 + Q 2 = 1
C2
C2
v1 = v2 =
or
C2
C1 + C 2
C1
Q1 =
Qs
C1 + C 2
Q2 =
i=
dQ
dt
i1 =
C1
is ,
C1 + C 2
i2 =
C2
is
C1 + C 2
Chapter 6, Solution 26.
(a)
Ceq = C1 + C2 + C3 = 35µF
(b)
Q1 = C1v = 5 x 150µC = 0.75mC
Q2 = C2v = 10 x 150µC = 1.5mC
Q3 = C3v = 20 x 150 = 3mC
(c)
w=
1
1
C eq v 2 = x35x150 2 µJ = 393.8mJ
2
2
C1
vs
C1 + C 2
Chapter 6, Solution 27.
1
1
1
1
1 1
1
7
=
+
+
= + +
=
C eq C1 C 2 C 3 5 10 20 20
(a)
Ceq =
(b)
(c)
20
µF = 2.857µF
7
Since the capacitors are in series,
20
Q1 = Q2 = Q3 = Q = Ceqv =
x 200µV = 0.5714mV
7
1
1 20
w = C eq v 2 = x x 200 2 µJ = 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 µF
Cc
20 µF
Ca
 1  1   1  1   1  1 
   +    +   
1
10 40
10 30
30 40
=         
1
Ca
30
3
1
1
2
=
+ +
=
40 10 40 10
Ca = 5µF
1
1
1
+
+
2
1
= 400 300 1200 =
1
30
C6
10
Cb = 15µF
1
1
1
+
+
1
4
= 400 300 1200 =
1
Cc
15
40
Cc = 3.75µF
Cb in parallel with 50µF = 50 + 15 = 65µF
Cc in series with 20µF = 23.75µF
65x 23.75
= 17.39µF
65µF in series with 23.75µF =
88.75
17.39µF in parallel with Ca = 17.39 + 5 = 22.39µF
Hence Ceq = 22.39µF
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 = 3C
C
5
5
2
Cx
C
C
in parallel with C = C + 3 = 1.6 C
5
5
(b)
2C
Ceq
2C
1
1
1
1
=
+
=
C eq 2C 2C C
Ceq = C
Chapter 6, Solution 30.
1 t
idt + i(0)
C ∫o
For 0 < t < 1, i = 60t mA,
10 −3 t
vo =
60tdt + 0 = 10 t 2 kV
3x10 −6 ∫o
vo(1) = 10kV
vo =
For 1< t < 2, i = 120 - 60t mA,
10 −3 t
vo =
(120 − 60t )dt + v o (1)
3x10t −6 ∫1
= [40t – 10t2 ] 1 + 10kV
= 40t – 10t2 - 20
10t 2 kV,
0 < t <1
v o (t) = 
2
40t − 10 t − 20kV, 1 < t < 2
Chapter 6, Solution 31.
20 tmA,
i s ( t ) = 20mA,
− 50 + 10 t ,
0 < t <1
1< t < 3
3< t <5
Ceq = 4 + 6 = 10µF
1 t
v=
idt + v(0)
C eq ∫o
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
∫ 20t dt + 0 = t
o
t
2
kV
∫ 20dt + v(1) = 2(t − 1) + 1kV
1
t
∫ 10(t − 5)dt + v(3)
3
= t 2 − 5 + 3t +5kV = t 2 − 5t + 11kV
 t 2 kV,
0 < t <1

v( t ) = 2t − 1kV,
1< t < 3
2
 t − 5t + 11kV, 3 < t < 5

dv
dv
= 6x10 −6
dt
dt
0 < t <1
12 tmA,

= 12mA,
1< t < 3
12 − 30mA, 3 < t < 5
i 1 = C1
dv
dv
= 4x10 −6
dt
dt
0 < t <1
8tmA,

= 8mA,
1< t < 3
8t − 20mA, 3 < t < 5
i1 = C 2
Chapter 6, Solution 32.
(a)
Ceq = (12x60)/72 = 10 µ F
t
10 −3
v1 =
30e − 2t dt + v1 (0) = − 1250e − 2t
−6 ∫
12 x10 0
t
10 −3
v2 =
30e − 2t dt + v 2 (0) = 250e − 2t
−6 ∫
60 x10 0
(b)
t
0
t
0
+ 50 = − 1250e − 2t + 1300
+ 20 = 250e − 2t − 230
At t=0.5s,
v1 = −1250e −1 + 1300 = 840.15,
w12 µF =
1
x12 x10 −6 x(840.15) 2 = 4.235 J
2
1
x 20 x10 −6 x(−138.03) 2 = 0.1905 J
2
1
= x 40 x10 −6 x(−138.03) 2 = 0.381 J
2
w20 µF =
w40 µF
v 2 = 250e −1 − 230 = −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
2
-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
60 x10 −3
=
= 200 mH
∆i / ∆t 0.6 /(2)
Chapter 6, Solution 36.
di 1
= x10 −3 (12)(2)(− sin 2 t )V
dt 4
= - 6 sin 2t mV
v=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.
di
= 12 x10 −3 x 4(100) cos100t
dt
= 4.8 cos 100t V
v=L
p = vi = 4.8 x 4 sin 100t cos 100t = 9.6 sin 200t
w=
t
11 / 200
o
o
∫ pdt = ∫
9.6 sin 200 t
9.6
/ 200
cos 200t 11
J
o
200
= −48(cos π − 1)mJ = 96 mJ
=−
Chapter 6, Solution 38.
v=L
di
= 40x10 −3 (e − 2 t − 2te − 2 t )dt
dt
= 40(1 − 2t )e −2 t mV, t > 0
Chapter 6, Solution 39
di
1

→ i = ∫ 0t idt + i(0)
L
dt
v=L
i=
1
200x10
t
(3t 2
−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
v=L
di
di
= 20 x10 −3
dt
dt
 10t , 0 < t < 1 ms

i =  20 - 10t, 1 < t < 3 ms
- 40 + 10t, 3 < t < 4 ms

 10 x10 3 ,
di 
= - 10x10 3 ,
dt 
3
10x10 ,
 200 V,

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
1 t
vdt + i(0) =   ∫ 20 1 − 2 − 2 t dt + 0.3
∫
L 0
2 o
 1

= 10 t + e − 2t  to +0. 3 = 10t + 5e − 2t − 4. 7 A
 2

At t = ls, i = 10 - 4.7 + 5e-2 = 5.977 A
w=
1 2
L i = 35.72J
2
Chapter 6, Solution 42.
1 t
1 t
vdt
i
(
0
)
+
=
v( t )dt − 1
L ∫o
5 ∫o
10 t
For 0 < t < 1, i = ∫ dt − 1 = 2t − 1 A
5 0
i=
For 1 < t < 2, i = 0 + i(1) = 1A
1
10dt + i(2) = 2t 2t +1
∫
5
= 2t - 3 A
For 2 < t < 3, i =
For 3 < t < 4, i = 0 + i(3) = 3 A
1 t
10dt + i(4) = 2 t 4t +3
5 ∫4
= 2t - 5 A
For 4 < t < 5, i =
 2t − 1A,

1A,
Thus, i (t ) =  2t − 3 A,

3 A,
 2t − 5,
0 < t <1
1< t < 2
2<t<3
3<t < 4
4<t <5
Chapter 6, Solution 43.
2
1
1
w = L ∫ idt = Li( t ) − Li (−∞)
−∞
2
2
1
= x80 x10 −3 x 60x10 −3 − 0
2
= 144 µJ
t
(
)
Chapter 6, Solution 44.
i=
1 t
1 t
vdt + i(t o ) = ∫ (4 + 10 cos 2t )dt − 1
∫
L to
5 o
= 0.8t + sin 2t -1
Chapter 6, Solution 45.
i(t) =
1 t
v( t ) + i(0)
L ∫o
For 0 < t < 1, v = 5t
i=
1
10x10 −3
t
∫ 5t dt + 0
o
= 0.25t2 kA
For 1 < t < 2, v = -10 + 5t
i=
1
10x10 −3
t
∫ (−10 + 5t )dt + i(1)
1
t
= ∫ (0.5t − 1)dt + 0.25kA
1
= 1 - t + 0.25t2 kA
0.25t 2 kA,
0 < t <1
i( t ) = 
2
1 − t + 0.25t kA, 1 < t < 2
Chapter 6, Solution 46.
Under dc conditions, the circuit is as shown below:
2Ω
iL
+
3A
vC
4Ω
−
By current division,
iL =
4
(3) = 2A, vc = 0V
4+2
wL =
1 2 11 2
L i L =  (2) = 1J
2
22
wc =
1
1
C v c2 = (2)( v) = 0J
2
2
Chapter 6, Solution 47.
Under dc conditions, the circuit is equivalent to that shown below:
R
+
5A
iL =
2Ω
2
10
10R
(5) =
, v c = Ri L =
R+2
R+2
R+2
vC
−
iL
1 2
100R 2
Cv c = 80x10 −6 x
2
(R + 2) 2
1
100
w L = Li12 = 2x10 −3 x
2
(R + 2) 2
If wc = wL,
wc =
100R 2
2x10 −3 x100
80x10 x
=
(Rx 2) 2
(R + 2) 2
80 x 10-3R2 = 2
−6
R = 5Ω
Chapter 6, Solution 48.
Under dc conditions, the circuit is as shown below:
4Ω
iL1
+
+
30V
+
−
vC1
−
i L1 = i L 2 =
30
= 3A
4+6
v C1 = 6i L1 = 18V
v C 2 = 0V
iL2
vC2
−
6Ω
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 = 3H
(c)
L eq = 4 (2 + 3 6) = 4 4 = 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
1
1
1
1
=
+
+
=
L 60 20 30 10
L eq = 10 (25 + 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) = 4 + 3
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
= 1.4 L
2 L + 0.5L
(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
5
Lx L
2 

3 = 5L
= L L + L =
5
8
3 

L+ L
3
Chapter 6, Solution 57.
Let v = L eq
di
dt
di
+ v2
dt
i2 = i – i1
i = i1 + i2
di
di
v
v 2 = 3 1 or 1 = 2
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
 3
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
= 8.375H
8
(1)
(2)
(3)
(4)
(5)
Chapter 6, Solution 58.
v=L
di
di
= 3 = 3 x slope of i(t).
dt
dt
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 = (L1 + L 2 )
di
dt
vs
di
=
dt L1 + L 2
di
di
v 1 = L1 , v 2 = L 2
dt
dt
L1
L2
v1 =
vs , vL =
vs
L1 + L 2
L1 + L 2
(b)
v i = v 2 = L1
di1
di
= L2 2
dt
dt
i s = i1 + i 2
di s di1 di 2
(L + L 2 )
v
v
=
+
=
+
=v 1
L1 L 2
dt
dt
dt
L1 L 2
L1 L 2 di s
L2
1
1
dt =
is
vdt =
i1 =
∫
∫
L1 L1 + L 2 dt
L1 + L 2
L1
i2 =
1
1
vdt =
∫
L2
L2
L1 L 2 di s
L1
dt =
is
+
L
dt
L
+
L
1
2
1
2
∫L
Chapter 6, Solution 60
Leq = 3 // 5 =
vo = Leq
(
15
8
)
di 15 d
=
4e − 2t = − 15e − 2t
dt 8 dt
t
io =
t
I
1
vo (t )dt + io (0) = 2 + ∫ (−15)e − 2t = 2 + 1.5e − 2t
∫
L0
50
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
i1 =
i s = 0.4 6e − 2 t = 2.4e-2t mA
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 µV
dt
dt
di
d
v 2 = L = 12x10 −3
6e −2 t x10 −3 = -144e-2t µV
dt
dt
(
(d)
w 10 mH =
= 0.8e − 4 t
)
(
)
(
)
(
1
x30x10 −3 36e − 4 t x10 −6
2
t=
1
2
)
µJ
= 24.36nJ
1
w 30 mH = x30 x10 −3 5.76e − 4 t x10 −6 t =1 / 2
2
= 11.693nJ
1
w 20 mH = x 20x10 −3 12.96e − 4 t x10 −6 t =1 / 2
2
= 17.54 nJ
(
(
)
)
Chapter 6, Solution 62.
(a)
Leq = 25 + 20 // 60 = 25 +
v = Leq
di
dt
20 x60
= 40 mH
80
t

→
i=
1
10 −3
(
)
(
0
)
v
t
dt
+
i
=
12e −3t dt + i (0) = −0.1(e −3t − 1) + i (0)
Leq ∫
40 x10 −3 ∫0
Using current division,
60
3
1
i1 =
i = i, i 2 = i
80
4
4
3
i1 (0) = i (0)

→ 0.75i (0) = −0.01
4

→
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 = - 75e -3t + 65 mA
4
i2 = - 25e -3t + 21.67 mA
(b) i1 =
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.
v1 = L
di1
di  2,
=2 1 =
dt
dt − 2,
 4,
di2
di2 
v2 = L
=2
=  0,
dt
dt 
− 4,
0<t <3
3<t <6
0<t<2
2<t<4
4<t<6
i (0) = −0.01333
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 (∞) = 12 / 4 = 3,
τ = L / R = 1/ 8
i (t ) = i (∞) + [i (0) − i (∞)]e − t / ι = 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
dv
= 20(−200)(50e − 200 t x10 −3 )
dt
= -200e-200tA
(
dv
= 20(−200) 50e − 200 t x10 −3
dt
= -200e-200t A
i2 = L2
(d)
)
i = i1 + i2 = -250e-200t A
Chapter 6, Solution 66.
L eq = 20 + 16 + 60 40 = 36 + 24 = 60mH
v=L
di
dt
1 t
vdt + i(0)
L ∫o
t
1
=
12 sin 4t dt + 0 mA
−3 ∫o
60x10
i = −50 cos 4t ot = 50(1 - cos 4t) mA
i=
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 ∫
− 10 3
vo =
10 sin 50t dt
2 ∫
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
∫
RC
1 t
vo = − ∫ 10dt + 0 = −2t
5 o
The op amp will saturate at vo = ± 12
vo = −
-12 = -2t
t = 6s
Chapter 6, Solution 69.
RC = 4 x 106 x 1 x 10-6 = 4
vo = −
1
1
v i dt = − ∫ v i dt
∫
RC
4
For 0 < t < 1, vi = 20, v o = −
1 t
20dt = -5t mV
4 ∫o
1 t
10dt + v(1) = −2.5( t − 1) − 5
4 ∫1
= -2.5t - 2.5mV
For 1 < t < 2, vi = 10, v o = −
1 t
20dt + v(2) = 5( t − 2) − 7.5
4 ∫2
= 5t - 17.5 mV
For 2 < t < 4, vi = - 20, v o = +
1 t
10dt + v(4) = 2.5( t − 4) + 2.5
4 ∫4
= 2.5t - 7.5 mV
For 4 < t < 5m, vi = -10, v o =
1 t
20dt + v(5) = −5( t − 5) + 5
4 ∫5
= - 5t + 30 mV
For 5 < t < 6, vi = 20, v o = −
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 kΩ, C =
= 0.2µF
50 x100 x10 3
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
∫
∫
R 1C
R 2C
R 2C ∫
For the given problem, C = 2µF,
vo = −
R1C = 1
R2C = 1/(4)
R3C = 1/(10)
R1 = 1/(C) = 1006/(2) = 500 kΩ
R2 = 1/(4C) = 500kΩ/(4) = 125 kΩ
R3 = 1/(10C) = 50 kΩ
Chapter 6, Solution 72.
The output of the first op amp is
v1 = −
1
1
v i dt = −
3
∫
RC
10x10 x 2 x10 −6
t
∫ idt = −
o
100 t
2
= - 50t
vo = −
1
1
v i dt = −
3
∫
RC
20x10 x 0.5x10 −6
t
∫ (−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,
0 − v v − vo
=
R
R
2v - vo = 0
(1)
R
R
a
R
v
−
+
R
v
vo
b
vi
At node b,
+
−
vi − v v − vo
dv
=
+C
R
R
dt
+
C
−
v i = 2v − v o + RC
dv
dt
(2)
Combining (1) and (2),
v i =v o −v o +
RC dv o
2 dt
or
vo =
2
v i dt
RC ∫
showing that the circuit is a noninverting integrator.
Chapter 6, Solution 74.
RC = 0.01 x 20 x 10-3 sec
v o = − RC
dv i
dv
= −0.2 m sec
dt
dt
 − 2V,
v o = 2V,
− 2V,
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.
v 0 = − RC
dv i
, RC = 250 x10 3 x10x10 −6 = 2.5
dt
v o = −2.5
d
(12t ) = -30 mV
dt
Chapter 6, Solution 76.
dv i
, RC = 50 x 103 x 10 x 10-6 = 0.5
dt
− 10, 0 < t < 5
dv
v o = 0.5 i = 
5<t <5
dt 5,
v o = − RC
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 0 − v0
d
=
+ C (0 − v o )
dt
R
RF
R F C = 10 6 x10 −6 = 1
15
dv 

Hence v i = − v o + o 
dt 

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
2dv o
= 10 sin 2 t −
− vo
dt
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
1000kΩ
1000kΩ dv o
= f (t) −
vo −
2
5000kΩ
200kΩ dt
dt
or
d 2 vo
dv
+ 5 o + 2v o = f ( t )
2
dt
dt
Chapter 6, Solution 81
We can write the equation as
d 2v
= −5v − 2 f (t )
dt 2
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 10µF capacitors in series gives 5µF, 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
= 2.4µC
∆q 2.4 x10−6
=
= 150nF
C=
∆v (36 − 20)
Chapter 6, Solution 85.
It is evident that differentiating i will give a waveform similar to v. Hence,
di
v=L
dt
4 t ,0 < t < 1
i=
8 − 4 t ,1 < t < 2
v=L
But,
di 4L,0 < t < 1
=
dt − 4L,1 < t < 2
5mV,0 < t < 1
v=
− 5mV,1 < t < 2
Thus, 4L = 5 x 10-3
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
1000
= 10 x
µF = 1250µF
3
8
(b)
vT = 8 x 100V = 800V
w=
(
)
1
1
C eq v T2 = 1250 x10 −6 (800) 2
2
2
= 400J
Chapter 7, Solution 1.
Applying KVL to Fig. 7.1.
1 t
∫ i dt + Ri = 0
C -∞
Taking the derivative of each term,
i
di
+R =0
C
dt
di
dt
or
=−
i
RC
Integrating,
 i( t )  - t
=
ln
 I 0  RC
i( t ) = I 0 e - t RC
v( t ) = Ri( t ) = RI 0 e - t RC
or
v(t ) = V0e- t RC
Chapter 7, Solution 2.
τ = R th C
where R th is the Thevenin equivalent at the capacitor terminals.
R th = 120 || 80 + 12 = 60 Ω
τ = 60 × 0.5 × 10 -3 = 30 ms
Chapter 7, Solution 3.
(a) RTh = 10 // 10 = 5kΩ,
τ = RTh C = 5 x10 3 x 2 x10 −6 = 10 ms
(b) RTh = 20 //(5 + 25) + 8 = 20Ω,
τ = RTh C = 20 x0.3 = 6s
Chapter 7, Solution 4.
τ = R eq C eq
where C eq =
C1C 2
,
C1 + C 2
τ=
R eq =
R 1R 2
R1 + R 2
R 1 R 2 C1C 2
( R 1 + R 2 )(C 1 + C 2 )
Chapter 7, Solution 5.
v( t ) = v(4) e -(t -4) τ
where v(4) = 24 ,
τ = RC = (20)(0.1) = 2
-(t - 4) 2
v( t ) = 24 e
v(10) = 24 e -6 2 = 1.195 V
Chapter 7, Solution 6.
v o = v ( 0) =
2
(24) = 4 V
10 + 2
v( t ) = voe − t / τ , τ = RC = 40 x10−6 x 2 x103 =
2
25
v( t ) = 4e −12.5t V
Chapter 7, Solution 7.
v( t ) = v(0) e - t τ ,
τ = 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.
8Ω
i2
i
i1
0.5 V
+
−
10 Ω
+
v=1
−
i1 =
1
= 0.1 ,
10
i = i1 + i 2 = 0.1 +
i2 =
1 13
=
16 80
1 80
R th = =
i 13
80
8
τ = R th C =
× 0.1 =
13
13
-13t 8
v( t ) = 20 e
V
1 − 0.5 1
=
8
16
+
−
1V
Chapter 7, Solution 8.
(a)
τ = RC =
1
4
dv
dt
-4t
- 0.2 e = C (10)(-4) e-4t
-i = C

→ C = 5 mF
1
= 50 Ω
4C
1
τ = RC = = 0.25 s
4
1
1
w C (0) = CV02 = (5 × 10 -3 )(100) = 250 mJ
2
2
1 1
1
w R = × CV02 = CV02 (1 − e -2t 0 τ )
2 2
2
1
→ e -8t 0 =
0.5 = 1 − e -8t 0 
2
8t 0
or
e =2
1
t 0 = ln (2) = 86.6 ms
8
R=
(b)
(c)
(d)
Chapter 7, Solution 9.
v( t ) = v(0) e- t τ ,
τ = R eq C
R eq = 2 + 8 || 8 + 6 || 3 = 2 + 4 + 2 = 8 Ω
τ = R eq C = (0.25)(8) = 2
v( t ) = 20 e -t 2 V
Chapter 7, Solution 10.
io
15 Ω
i
10 Ω
iT
+
10 mF
4Ω
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 
→ i o =
R th = 4 + 10 || 15 = 4 + 6 = 10 Ω
τ = R th C = (10)(10 × 10 -3 ) = 0.1
v( t ) = v(0) e - t τ = 50 e -10t
dv
iC = C
= (10 × 10 -3 )(-500 e -10t )
dt
i C = - 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
∫
L
R
Differentiating both sides,
v 1 dv
+
=0 
→
L R dt
v = A e -Rt L
dv R
+ v=0
dt L
If the initial current is I 0 , then
v(0) = I 0 R = A
v = I 0 R e -t τ ,
τ=
L
R
1 t
∫ v(t ) dt
L -∞
- τ I 0 R -t τ t
i=
e -∞
L
i = - I 0 R e -t τ
i=
i( t ) = I 0 e - t τ
Chapter 7, Solution 12.
When t < 0, the switch is closed and the inductor acts like a short circuit to dc. The 4 Ω
resistor is short-circuited so that the resulting circuit is as shown in Fig. (a).
3Ω
12 V
i(0-)
+
−
4Ω
(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
τ = = = 0.5
R 4
Hence,
i( t ) = i(0) e - t τ = 4 e -2t A
Chapter 7, Solution 13.
L
R th
where R th is the Thevenin resistance at the terminals of the inductor.
τ=
R th = 70 || 30 + 80 || 20 = 21 + 16 = 37 Ω
2 × 10 -3
τ=
= 81.08 µs
37
Chapter 7, Solution 14
Converting the wye-subnetwork to delta gives
16 Ω
R2
80mH
R1
R3
30 Ω
R1 =
10 x 20 + 20 x50 + 50 x10
= 1700 / 20 = 85Ω,
20
R2 =
1700
= 34Ω ,
50
R3 =
1700
= 170Ω
10
30//170 = (30x170)/200 = 25.5 Ω , 34//16=(34x16)/50 =10.88 Ω
RTh = 85 //( 25.5 + 10.88) =
85 x36.38
= 25.476Ω,
121.38
τ=
80 x10 −3
L
=
= 3.14 ms
RTh
25.476
Chapter 7, Solution 15
(a) RTh = 12 + 10 // 40 = 20Ω,
(b) RTh = 40 // 160 + 8 = 40Ω,
L
= 5 / 20 = 0.25s
RTh
L
τ=
= (20 x10 −3 ) / 40 = 0.5 ms
RTh
τ=
Chapter 7, Solution 16.
τ=
(a)
L eq
R eq
L eq = L and R eq = R 2 +
τ=
(b)
R 1R 3
R 2 (R 1 + R 3 ) + R 1 R 3
=
R1 + R 3
R1 + R 3
L( R 1 + R 3 )
R 2 (R 1 + R 3 ) + R 1 R 3
R 3 (R 1 + R 2 ) + R 1 R 2
L1 L 2
R 1R 2
=
and R eq = R 3 +
L1 + L 2
R1 + R 2
R1 + R 2
L1L 2 (R 1 + R 2 )
τ=
(L 1 + L 2 ) ( R 3 ( R 1 + R 2 ) + R 1 R 2 )
where L eq =
Chapter 7, Solution 17.
i( t ) = i(0) e - t τ ,
τ=
14 1
L
=
=
R eq
4 16
i( t ) = 2 e -16t
v o ( t ) = 3i + L
di
= 6 e-16t + (1 4)(-16) 2 e-16t
dt
v o ( t ) = - 2 e -16t V
Chapter 7, Solution 18.
If v( t ) = 0 , the circuit can be redrawn as shown below.
+
0.4 H
Req
vo(t)
−
i(t)
6
L 2 5 1
τ= = × =
,
5
R 5 6 3
-t τ
-3t
i( t ) = i(0) e = e
di - 2
v o ( t ) = -L =
(-3) e -3t = 1.2 e -3t V
dt
5
R eq = 2 || 3 =
Chapter 7, Solution 19.
i
1V
− +
10 Ω
i1
i1
i2
i/2
i2
40 Ω
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 
→ i =
30
1
R th = = 30 Ω
i
L
6
τ=
=
= 0.2 s
R th 30
i( t ) = 2 e -5t A
Chapter 7, Solution 20.
(a).
L
1
=

→ R = 50L
R 50
di
-v= L
dt
-50t
- 150 e = L(30)(-50) e -50t 
→ L = 0.1 H
τ=
R = 50L = 5 Ω
(b).
(c).
(d).
i.e.
L
1
=
= 20 ms
R 50
1
1
w = L i 2 (0) = (0.1)(30) 2 = 45 J
2
2
Let p be the fraction
1
1
L I 0 ⋅ p = L I 0 ( 1 − e -2t 0 τ )
2
2
-(2)(10) 50
p = 1− e
= 1 − e -0.4 = 0.3296
p = 33%
τ=
Chapter 7, Solution 21.
The circuit can be replaced by its Thevenin equivalent shown below.
Rth
Vth
Vth =
+
−
2H
80
(60) = 40 V
80 + 40
80
+R
3
Vth
40
I = i(0) = i(∞) =
=
R th 80 3 + R
R th = 40 || 80 + R =
2
1
1  40 
 =1
w = L I 2 = (2)
2
2  R + 80 3 
40
40
=1 
→ R =
R + 80 3
3
R = 13.33 Ω
Chapter 7, Solution 22.
i( t ) = i(0) e - t τ ,
τ=
L
R eq
R eq = 5 || 20 + 1 = 5 Ω ,
τ=
2
5
i( t ) = 10 e -2.5t A
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 Ω resistor, 1/3 H inductor, and the (3+1) Ω resistor are in parallel,
they always have the same voltage.
2
2
+
= 1.5 
→ 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) = ,
τ=
=
=
3
R th 4 3 4
-i =
i( t ) = i(0) e - t τ = -1.5 e -4t , t > 0
di
v L = v o = L = -1.5(-4)(1/3) e -4t
dt
-4t
v o = 2 e V, t > 0
vx =
1
v = 0.5 e -4t V , t > 0
3 +1 L
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)
(c) x ( t ) = ( 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)
(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.
v1 ( t ) = u ( t + 1) − u ( t ) + [ u ( t − 1) − u ( t )]
v1 ( t ) = u(t + 1) − 2 u(t ) + u(t − 1)
(a)
v 2 ( t ) = ( 4 − t ) [ u ( t − 2) − u ( t − 4) ]
v 2 ( t ) = -( t − 4) u ( t − 2) + ( t − 4) u ( t − 4)
v 2 ( t ) = 2 u(t − 2) − r(t − 2) + r(t − 4)
(b)
v 3 ( t ) = 2 [ u(t − 2) − u(t − 4)] + 4 [ u(t − 4) − u(t − 6)]
v 3 ( t ) = 2 u(t − 2) + 2 u(t − 4) − 4 u(t − 6)
(c)
v 4 ( t ) = -t [ u ( t − 1) − u ( t − 2)] = -t u(t − 1) + t u ( t − 2)
v 4 ( t ) = (-t + 1 − 1) u ( t − 1) + ( t − 2 + 2) u ( t − 2)
v 4 ( t ) = - r(t − 1) − u(t − 1) + r(t − 2) + 2 u(t − 2)
(d)
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 4tδ (t − 1) = cos 4δ (t − 1) = −0.6536δ (t − 1) , which is sketched below.
z(t)
0
1
t
-0.653 δ (t )
Chapter 7, Solution 30.
(a)
∫
4 t 2 δ( t − 1) dt = 4 t 2
(b)
∫
cos(2πt ) δ( t − 0.5) dt = cos(2πt )
10
0
∞
-∞
t =1
=4
t = 0.5
= cos π = - 1
Chapter 7, Solution 31.
(a)
(b)
= e = 112 × 10
∫ [ e δ(t − 2)] dt = e
∫ [ 5 δ(t ) + e δ(t ) + cos 2πt δ(t )] dt = ( 5 + e + cos(2πt ))
∞
- 4t 2
- 4t 2
-∞
∞
-t
-t
-∞
Chapter 7, Solution 32.
(a)
(b)
t
t
t
1
4
1
1
∫ u (λ )dλ = ∫ 1dλ = λ
5
(c )
1
4
0
1
= t −1
∫ r (t − 1)dt = ∫ 0dt + ∫ (t − 1)dt =
0
∫ (t − 6)
1
2
-9
-16
t=2
δ (t − 2)dt = (t − 6) 2
t2
− t 14 = 4.5
2
t =2
= 16
t =0
= 5 +1+1 = 7
Chapter 7, Solution 33.
i( t ) =
1 t
∫ v(t ) dt + i(0)
L 0
i( t ) =
10 -3
10 × 10 -3
∫ 20 δ(t − 2) dt + 0
t
0
i ( t ) = 2 u( t − 2 ) A
Chapter 7, Solution 34.
(a)
d
[u ( t − 1) u ( t + 1)] = δ( t − 1)u ( t + 1) +
dt
u ( t − 1)δ( t + 1) = δ( t − 1) • 1 + 0 • δ( t + 1) = δ( t − 1)
(b)
d
[r ( t − 6) u ( t − 2)] = u ( t − 6)u ( t − 2) +
dt
r ( t − 6)δ( t − 2) = u ( t − 6) • 1 + 0 • δ( t − 2) = u ( t − 6)
d
[sin 4t u (t − 3)] = 4 cos 4t u ( t − 3) + sin 4tδ( t − 3)
dt
= 4 cos 4t u ( t − 3) + sin 4x3δ( t − 3)
(c)
= 4 cos 4t u ( t − 3) − 0.5366δ( t − 3)
Chapter 7, Solution 35.
(a)
v( t ) = A e -5t 3 , v(0) = A = -2
v( t ) = - 2 e -5t 3 V
(b)
v( t ) = A e 2t 3 , v(0) = A = 5
v( t ) = 5 e 2t 3 V
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.
•
1
vh + 4 v
h
=0
v h = Ae −t / 4

→
v = 10 + Ae −0.25t
v(0) = 2 = 10 + A
v = 10 − 8e −0.25t

→
A = −8
(a) τ = 4 s
(b) v(∞) = 10 V
(c ) v = 10 − 8e −0.25t
Chapter 7, Solution 38
Let i = ip +ih
•
i h + 3ih = 0
Let i p = ku (t ),
•
ip = 0,

→
3ku (t ) = 2u (t )
ih = Ae −3t u (t )

→
k=
2
3
ip =
2
u (t )
3
2
i = ( Ae −3t + )u (t )
3
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 V
4 +1
After t = 0,
v( t ) = v(∞) + [ v(0) − v(∞)] e - t τ
τ = RC = (4)(2) = 8 , v(0) = 4 ,
v( t ) = 20 + (8 − 20) e - t 8
v( t ) = 20 − 12 e -t 8 V
v(∞) = 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
4Ω
−
+
−
8V
12 V
+
−
3Ω
3Ω
(a)
(b)
v( t ) = v(∞) + [ v(0) − v(∞)] e - t τ
v(∞) = 12 ,
v(0) = 4 ,
τ = RC = (2)(3) = 6
-t 6
v( t ) = 12 + (4 − 12) e
v( t ) = 12 − 8 e -t 6 V
Chapter 7, Solution 40.
(a)
(b)
Before t = 0, v = 12 V .
After t = 0, v( t ) = v(∞) + [ v(0) − v(∞)] e - t τ
v(∞) = 4 ,
v(0) = 12 ,
τ = RC = (2)(3) = 6
-t 6
v( t ) = 4 + (12 − 4) e
v( t ) = 4 + 8 e - t 6 V
Before t = 0, v = 12 V .
After t = 0, v( t ) = v(∞) + [ v(0) − v(∞)] e - t τ
After transforming the current source, the circuit is shown below.
t=0
2Ω
+
−
12 V
v(∞) = 12 ,
v(0) = 12 ,
v = 12 V
Chapter 7, Solution 41.
v(0) = 0 ,
v(∞) =
R eq C = (6 || 30)(1) =
30
(12) = 10
16
(6)(30)
=5
36
v( t ) = v(∞) + [ v(0) − v(∞)] e - t τ
v( t ) = 10 + (0 − 10) e - t 5
v( t ) = 10 ( 1 − e -0.2t ) V
4Ω
5F
τ = RC = (2)(5) = 10
Chapter 7, Solution 42.
(a)
v o ( t ) = v o (∞) + [ v o (0) − v o (∞)] e - t τ
4
v o (∞) =
(12) = 8
v o (0) = 0 ,
4+2
4
τ = R eq C eq , R eq = 2 || 4 =
3
4
τ = (3) = 4
3
v o ( t ) = 8 − 8 e -t 4
v o ( t ) = 8 ( 1 − e -0.25t ) V
(b)
For this case, v o (∞) = 0 so that
v o ( t ) = v o (0) e -t τ
4
v o (0) =
(12) = 8 ,
4+2
v o ( t ) = 8 e -t 12 V
τ = 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−

→ 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 τ ,
τ = 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 Ω
vC
1
0.5
=
,
i o = 0.5 i =
80 80
80
1 80
R th = =
= 160 Ω ,
τ = R th C = 480
i o 0.5
v C (0) = 64 V
i=
v C ( t ) = 64 e - t 480
dv C
 1 
 64 e - t 480
0.5 i = -i C = -C
= -3 

dt
480 
i( t ) = 0.8 e -t 480 A
Chapter 7, Solution 44.
R eq = 6 || 3 = 2 Ω ,
τ = RC = 4
v( t ) = v(∞) + [ v(0) − v(∞)] e - t τ
Using voltage division,
3
3
v(0) =
(30) = 10 V ,
v(∞) =
(12) = 4 V
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
4
dt
Chapter 7, Solution 45.
For t < 0, v s = 5 u ( t ) = 0 
→ v(0) = 0
For t > 0, v s = 5 ,
v(∞) =
4
5
(5) =
4 + 12
4
R eq = 7 + 4 || 12 = 10 ,
τ = R eq C = (10)(1 2) = 5
v( t ) = v(∞) + [ v(0) − v(∞)] e - t τ
v( t ) = 1.25 ( 1 − e -t 5 ) V
i( t ) = C
dv  1  - 5  - 1  - t 5
=     e
dt  2  4  5 
i( t ) = 0.125 e -t 5 A
Chapter 7, Solution 46.
τ = RTh C = (2 + 6) x0.25 = 2s,
v(0) = 0,
v(∞) = 6i s = 6 x5 = 30
v(t ) = v(∞) + [v(0) − v(∞)]e − t / τ = 30(1 − e − t / 2 ) V
Chapter 7, Solution 47.
For t < 0, u ( t ) = 0 ,
u ( t − 1) = 0 ,
v(0) = 0
For 0 < t < 1, τ = RC = (2 + 8)(0.1) = 1
v(0) = 0 ,
v(∞) = (8)(3) = 24
v( t ) = v(∞) + [ v(0) − v(∞)] e - t τ
v( t ) = 24( 1 − e - t )
For t > 1,
v(1) = 24( 1 − e -1 ) = 15.17
- 6 + v(∞) - 24 = 0 
→ v(∞) = 30
v( t ) = 30 + (15.17 − 30) e -(t-1)
v( t ) = 30 − 14.83 e -(t-1)
Thus,
(
)
 24 1 − e - t V ,
0<t<1
v( t ) = 
-(t -1)
V,
t >1
 30 − 14.83 e
Chapter 7, Solution 48.
For t < 0,
u (-t) = 1 ,
For t > 0,
u (-t) = 0 ,
R th = 20 + 10 = 30 ,
v(0) = 10 V
v(∞) = 0
τ = R th C = (30)(0.1) = 3
v( t ) = v(∞) + [ v(0) − v(∞)] e - t τ
v( t ) = 10 e -t 3 V
 - 1
dv
= (0.1)  10 e - t 3
3
dt
- 1 -t 3
e A
i( t ) =
3
i( t ) = C
Chapter 7, Solution 49.
For 0 < t < 1, v(0) = 0 ,
R eq = 4 + 6 = 10 ,
v(∞) = (2)(4) = 8
τ = R eq C = (10)(0.5) = 5
v( t ) = v(∞) + [ v(0) − v(∞)] e - t τ
v( t ) = 8 ( 1 − e - t 5 ) V
For t > 1,
v(1) = 8 ( 1 − e -0.2 ) = 1.45 ,
v( t ) = v(∞) + [ v(1) − v(∞)] e -( t −1) τ
v( t ) = 1.45 e -( t −1) 5 V
Thus,
(
v(∞) = 0
)
 8 1 − e -t 5 V , 0 < t < 1
v( t ) = 
- ( t −1 ) 5
V,
t >1
 1.45 e
Chapter 7, Solution 50.
For the capacitor voltage,
v( t ) = v(∞) + [ v(0) − v(∞)] e- t τ
v(0) = 0
For t < 0, we transform the current source to a voltage source as shown in Fig. (a).
1 kΩ
1 kΩ
+
30 V
+
−
v
−
(a)
2
(30) = 15 V
2 +1+1
R th = (1 + 1) || 2 = 1 kΩ
1
1
τ = R th C = 10 3 × × 10 -3 =
4
4
-4t
v( t ) = 15 ( 1 − e ) , t > 0
v(∞) =
2 kΩ
We now obtain i x from v(t). Consider Fig. (b).
iT 1 kΩ
v
ix
1 kΩ
30 mA
1/4 mF
(b)
But
i x = 30 mA − i T
v
dv
iT =
+C
R3
dt
i T ( t ) = 7.5 ( 1 − e -4t ) mA +
i T ( t ) = 7.5 ( 1 + e -4t ) mA
1
× 10 -3 (-15)(-4) e -4t A
4
Thus,
i x ( t ) = 30 − 7.5 − 7.5 e -4t mA
i x ( t ) = 7.5 ( 3 − e -4t ) mA , t > 0
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
R
di
-R
=
dt
i − VS R
L
Integrating both sides,
2 kΩ

V  i(t ) - R
ln  i − S  I 0 =
t

R
L
 i − VS R  - t
=
ln 
 I0 − VS R  τ
or
i − VS R
= e- t τ
I0 − VS R
i( t ) =
VS 
VS  -t τ
e
+  I0 −
R 
R
which is the same as Eq. (7.60).
Chapter 7, Solution 52.
20
= 2 A,
i(∞) = 2 A
10
i( t ) = i(∞) + [ i(0) − i(∞)] e- t τ
i(0) =
i( t ) = 2 A
Chapter 7, Solution 53.
25
=5A
3+ 2
After t = 0,
i( t ) = i(0) e - t τ
L 4
τ = = = 2,
i(0) = 5
R 2
i( t ) = 5 e - t 2 A
i=
(a)
Before t = 0,
(b)
Before t = 0, the inductor acts as a short circuit so that the 2 Ω and 4 Ω
resistors are short-circuited.
i( t ) = 6 A
After t = 0, we have an RL circuit.
i( t ) = i(0) e - t τ ,
i( t ) = 6 e - 2 t 3 A
τ=
L 3
=
R 2
Chapter 7, Solution 54.
(a)
Before t = 0, i is obtained by current division or
4
i( t ) =
(2) = 1 A
4+4
After t = 0,
i( t ) = i(∞) + [ i(0) − i(∞)] e- t τ
L
τ=
,
R eq = 4 + 4 || 12 = 7 Ω
R eq
τ=
3.5 1
=
7
2
i(0) = 1 ,
i(∞) =
6
3
4 || 12
(2) =
(2) =
7
4+3
4 + 4 || 12
6 
6
+  1 −  e -2 t
7 
7
1
i( t ) = ( 6 − e - 2t ) A
7
10
=2A
Before t = 0, i( t ) =
2+3
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(∞) , consider the circuit below, at t = when the inductor
becomes a short circuit,
v
τ=
i
10 V
2Ω
+
−
24 V
+
−
6Ω
10 − v 24 − v v
+
=

→ v = 9
2
6
3
v
i(∞) = = 3 A
3
i( t ) = 3 + (2 − 3) e -9 t 4
i( t ) = 3 − e - 9 t 4 A
2H
3Ω
Chapter 7, Solution 55.
For t < 0, consider the circuit shown in Fig. (a).
0.5 H
io
3Ω
24 V
io
+
−
+
−
0.5 H
i
+
4io
2Ω
v
−
(a)
8Ω
20 V
+
v
+
−
−
(b)
3i o + 24 − 4i o = 0 
→ i o = 24
v
v( t ) = 4i o = 96 V
i = = 48 A
2
For t > 0, consider the circuit in Fig. (b).
i( t ) = i(∞) + [ i(0) − i(∞)] e- t τ
20
i(∞) =
=2A
i(0) = 48 ,
8+ 2
L
0.5 1
R th = 2 + 8 = 10 Ω , τ =
=
=
R th 10 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 Ω ,
τ=
L
= 0.05
R
i( t ) = i(∞) + [ i(0) − i(∞)] e- t τ
i(0) is found by applying nodal analysis to the following circuit.
2Ω
5Ω
i
vx
12 Ω
2A
6Ω
+
20 Ω
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+

→ v x = 12
Since 20 || 5 = 4 ,
4
i(∞) =
(4) = 1.6
4+6
i( t ) = 1.6 + (2 − 1.6) e- t 0.05 = 1.6 + 0.4 e-20t
di 1
v( t ) = L = (0.4) (-20) e -20t
dt 2
v( t ) = - 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.
6Ω
30 V
+
−
i
i1
i2
5Ω
20 Ω
20
30
30
=
= 3,
i1 =
(3) = 2.4 ,
6 + 5 || 20 10
25
i 1 ( 0 ) = 2 .4 A ,
i 2 ( 0 ) = 0 .6 A
i=
i 2 = 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 Ω and 20 Ω resistors.
L
2.5 1
i1 ( t ) = i1 (0) e - t τ1 ,
τ1 = 1 =
=
R1
5
2
i1 ( t ) = 2.4 e -2t A
i 2 ( t ) = i 2 (0) e - t τ 2 ,
τ2 =
L2
4 1
=
=
R 2 20 5
i 2 ( t ) = 0.6 e -5t A
Chapter 7, Solution 58.
For t < 0,
v o (t) = 0
For t > 0,
i(0) = 10 ,
R th = 1 + 3 = 4 Ω ,
20
=5
1+ 3
L 14 1
τ=
=
=
R th
4 16
i(∞) =
i( t ) = i(∞) + [ i(0) − i(∞)] e- t τ
i( t ) = 5 ( 1 + e-16t ) A
di
1
= 15 ( 1 + e -16t ) + (-16)(5) e-16t
dt
4
-16t
v o ( t ) = 15 − 5 e V
vo (t ) = 3i + L
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
2+ 4
i( t ) = i(∞) + [ i(0) − i(∞)] e- t τ
i( t ) = 1 − e-4t
i(∞) =
di
= (1.5)(-4)(-e- 4t )
dt
v o ( t ) = 6 e -4t V
vo (t ) = L
τ=
L 1 .5
=
= 0.25
R eq
6
Chapter 7, Solution 60.
Let I be the inductor current.
For t < 0,
u(t) = 0 
→ i(0) = 0
For t > 0,
R eq = 5 || 20 = 4 Ω ,
τ=
L
8
= =2
R eq 4
i(∞) = 4
i( t ) = i(∞) + [ i(0) − i(∞)] e- t τ
i( t ) = 4 ( 1 − e - t 2 )
 - 1
di
= (8)(-4)  e - t 2
2
dt
v( t ) = 16 e -0.5t V
v( t ) = L
Chapter 7, Solution 61.
The current source is transformed as shown below.
4Ω
20u(-t) + 40u(t)
+
−
0.5 H
L 12 1
=
= ,
i(0) = 5 ,
R
4
8
i( t ) = i(∞) + [ i(0) − i(∞)] e - t τ
i( t ) = 10 − 5 e -8t A
τ=
di  1 
=  (-5)(-8) e -8t
dt  2 
v( t ) = 20 e -8t V
v( t ) = L
Chapter 7, Solution 62.
L
2
=
=1
R eq 3 || 6
For 0 < t < 1, u ( t − 1) = 0 so that
τ=
i(∞) = 10
i(0) = 0 ,
i( t ) =
i(∞) =
1
6
1
( 1 − e -t )
6
1
( 1 − e -1 ) = 0.1054
6
1 1 1
i(∞) = + =
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
0<t<1
i( t ) = 
6
 0.5 − 0.3946 e -(t -1) A
t>1
Chapter 7, Solution 63.
10
=2
5
For t < 0,
u (- t ) = 1 ,
i(0) =
For t > 0,
u (-t) = 0 ,
i(∞) = 0
L
0.5 1
τ=
=
=
R th
4 8
R th = 5 || 20 = 4 Ω ,
i( t ) = i(∞) + [ i(0) − i(∞)] e - t τ
i( t ) = 2 e -8t A
di  1 
=  (-8)(2) e-8t
dt  2 
v( t ) = - 8 e -8t V
v( t ) = L
Chapter 7, Solution 64.
Let i be the inductor current.
For t < 0, the inductor acts like a short circuit and the 3 Ω resistor is shortcircuited so that the equivalent circuit is shown in Fig. (a).
6Ω
10 Ω
+
−
3Ω
(a)
6Ω
i
10 Ω
+
−
io
v
i
3Ω
2Ω
(b)
i = i(0) =
For t > 0,
10
= 1.667 A
6
R th = 2 + 3 || 6 = 4 Ω ,
τ=
L
4
= =1
R th 4
To find i(∞) , consider the circuit in Fig. (b).
10 − v v v
10
= +

→ v =
6
3 2
6
v 5
i = i(∞) = =
2 6
i( t ) = i(∞) + [ i(0) − i(∞)] e - t τ
5  10 5 
5
i( t ) = +  −  e - t = ( 1 − e - t ) A
6  6 6
6
v o is the voltage across the 4 H inductor and the 2 Ω resistor
5
di 10 10 - t
10 10 - t
− e
=
+ e + (4) (-1) e - t =
6
dt 6 6
6 6
v o ( t ) = 1.667 ( 1 − e -t ) V
v o (t) = 2 i + 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) = 0 ,
R th = 5 || 20 = 4 ,
10
=2
5
L
2 1
τ=
= =
R th 4 2
i(∞) =
i( t ) = i(∞) + [ i(0) − i(∞)] e- t τ
i( t ) = 2 ( 1 − e -2t ) A
i(1) = 2 ( 1 − e-2 ) = 1.729
For t > 1,
i(∞) = 0
since vs = 0
i( t ) = i(1) e- ( t −1) τ
i( t ) = 1.729 e-2( t −1) A
Thus,
 2 ( 1 − e - 2t ) A 0 < t < 1
i( t ) = 
t>1
 1.729 e - 2( t −1) A
Chapter 7, Solution 66.
Following Practice Problem 7.14,
v( t ) = VT e - t τ
τ = R f C = (10 × 103 )(2 × 10- 6 ) =
VT = v(0) = -4 ,
1
50
v( t ) = -4 e -50t
v o ( t ) = -v( t ) = 4 e -50t , t > 0
i o (t) =
v o (t)
4
=
e -50t = 0.4 e -50t mA , t > 0
Ro
10 × 10 3
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 v1 − 0 v1 − v o
=
+
R
R
R

→ v1 =
2
v
3 o
At the noninverting terminal,
dv
v − v1
C o + o
=0
dt
R
dv
1
2
− RC o = v o − v1 = v o − v o = v o
3
3
dt
dv o
v
=− o
dt
3RC
v o ( t ) = VT e - t 3RC
VT = vo (0) = 5 V ,
τ = 3RC = (3)(10 × 103 )(1 × 10- 6 ) =
3
100
v o ( t ) = 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 kΩ and 100 kΩ resistors are in series since no current enters the
op amp terminals. As t → ∞ , the capacitor acts like an open circuit so that
20 + 100
48
v x (∞) =
(4) =
20 + 100 + 10
13
R th = 20 + 100 = 120 kΩ ,
τ = R th C = (120 × 103 )(25 × 10-3 ) = 3000
v x ( t ) = v x (∞) + [ v x (0) − v x (∞)] e- t τ
48
v x ( t ) = ( 1 − e - t 3000 )
13
vo (t ) =
40
100
( 1 − e -t 3000 ) V
vx (t) =
120
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 
→ v o = v s − v
From (1),
dv v s
=
=0
dt RC
- t vs
-1
v=
v s dt + v(0) =
∫
RC
RC
Since v is constant,
(1)
(2)
(3)
RC = (20 × 10 3 )(5 × 10 -6 ) = 0.1
- 20 t
mV = -200 t mV
v=
0.1
From (3),
v o = v s − v = 20 + 200 t
v o = 20 ( 1 + 10t ) mV
Chapter 7, Solution 71.
Let v = voltage across the capacitor.
Let v o = voltage across the 8 kΩ resistor.
For t < 2, v = 0 so that v(2) = 0 .
For t > 2, we have the circuit shown below.
10 kΩ
10 kΩ
20 kΩ
−
+
+
4V
+
−
100 mF
v
−
+
io
8 kΩ
vo
−
Since no current enters the op amp, the input circuit forms an RC circuit.
τ = RC = (10 × 10 3 )(100 × 10 -3 ) = 1000
v( t ) = v(∞) + [ v(2) − v(∞)] e -( t − 2 ) τ
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(∞) + [ v(0) − v(∞)] e - t τ
v(0) = -2 V , v(∞) = 3 V , τ = RC = (10 × 10 3 )(10 × 10 -6 ) = 0.1
v( t ) = 3 + (-2 - 3) e -10t = 3 − 5 e -10t
dv
= (10 × 10 -6 )(-5)(-10) e -10t
dt
i o = 0.5 e -10t mA , t > 0
io = C
Chapter 7, Solution 73.
Consider the circuit below.
Rf
v1
R1
v2
+
v1
+
−
C
v
v3
−
−
+
+
vo
−
At node 2,
v1 − v 2
dv
=C
dt
R1
At node 3,
(1)
C
dv v 3 − v o
=
dt
Rf
(2)
But v 3 = 0 and v = v 2 − v 3 = v 2 . Hence, (1) becomes
v1 − v
dv
=C
R1
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
t<0
v( t ) = 
-t τ
t>0
 v1 + ( v T − v1 ) e
where v T = v(0) = 1 and v1 = 4
τ = R 1C = (10 × 10 3 )(20 × 10 -6 ) = 0.2
 1
t<0
v( t ) = 
-5t
t>0
4 − 3 e
From (2),
dv
= (20 × 10 3 )(20 × 10 -6 )(15 e -5t )
dt
v o = -6 e -5t , t > 0
v o = -R f C
v o = - 6 e -5t u(t ) V
Chapter 7, Solution 74.
Let v = capacitor voltage.
Rf
v1
R1
v2
+
v1
+
−
C
v
v3
−
−
+
+
vo
−
v(0) = 0
i s = 10 µA . Consider the circuit below.
For t < 0,
For t > 0,
dv v
+
dt R
v( t ) = v(∞) + [ v(0) − v(∞)] e - t τ
is = C
(1)
(2)
It is evident from the circuit that
τ = RC = (2 × 10 −6 )(50 × 10 3 ) = 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(∞) = i s R = (10 × 10 −6 )(50 × 10 3 ) = 0.5 V
Then,
But
v( t ) = 0.5 ( 1 − e -10t ) V
is =
0 − vo
Rf

→ v o = -i s R f
Combining (1), (3), and (4), we obtain
- Rf
dv
vo =
v − RfC
R
dt
-1
dv
v o = v − (10 × 10 3 )(2 × 10 -6 )
5
dt
-10t
-2
v o = -0.1 + 0.1e − (2 × 10 )(0.5)( - 10 e -10t )
v o = 0.2 e -10t − 0.1
v o = 0.1 ( 2 e -10t − 1) V
(3)
(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 kΩ
R1
vo = -ioR
Also, i o =
i.e.
v
dv
+C ,
R2
dt
(1)
R 2 = 10 kΩ , C = 2 µF
- vs
dv
v
=
+C
dt
R1 R 2
(2)
This is a step response.
v( t ) = v(∞) + [ v(0) − v(∞)] e - t τ ,
where τ = R 2 C = (10 × 10 3 )(2 × 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 → ∞
- vs
v(∞)
= io =
R2
R1
- R2
- 10
(4) = -2
v(∞) =
vs =
i.e.
20
R1
v( t ) = -2 + (1 + 2) e -50t
v( t ) = -2 + 3 e -50t
But
v = vs − vo
or
v o = v s − v = 4 + 2 − 3 e -50 t
v o = 6 − 3 e -50 t V
- vs
-4
=
= -0.2 mA
R 1 20k
dv
v
+C
= - 0.2 mA
io =
dt
R2
io =
or
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(∞) = -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(∞) = 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,
R
4
i o (∞ ) = −
= −0.5 A,
RTh = (3 + 5) // 4 = 8 / 3, τ = Th = 80 / 3
5+3
L
io (t ) = io (∞) + [io (0) − io (∞)]e −t / τ = − 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) = i2 (0) = vo (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 = 3 // 6 = 2Ω,
τ=
RTh
= 2 / 4 = 0 .5
L
i L (t ) = i L (∞) + [i L (0) − i L (∞)]e −t / τ = 0 + 3e −t / 0.5 = 3e −2t A
(c) i1 (∞) =
30
= 2 A,
10 + 5
vo (t ) = L
3
i 2 (∞ ) = − i L (∞ ) = 0 A
9
di L
dt

→
v o (∞ ) = 0 V
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.
τ = RC 
→ R =
3 × 10 -3
τ
=
= 30 Ω
C 100 × 10 -6
Chapter 7, Solution 83.
v(∞) = 120,
v(0) = 0,
τ = RC = 34 x10 6 x15 x10 −6 = 510s
v(t ) = v(∞) + [v(0) − v(∞)]e − t / τ
85.6 = 120(1 − e − t / 510 )

→
Solving for t gives
t = 510 ln 3.488 = 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 (t ) = I o (1 − e − t / τ ),
0.6 I o = I o (1 − e −50t )
τ = R / L = 0.16 / 8 = 1 / 50

→
t=
1
1
ln
= 18.33 ms.
50 0.4
Chapter 7, Solution 85.
(a)
τ = RC = (4 × 106 )(6 × 10-6 ) = 24 s
Since v( t ) = v(∞) + [ v(0) − v(∞)] e - t τ
v( t 1 ) − v(∞) = [ v(0) − v(∞)] e - t1 τ
v( t 2 ) − v(∞) = [ v(0) − v(∞)] e- t 2 τ
Dividing (1) by (2),
v( t1 ) − v(∞)
= e( t 2 − t1 ) τ
v( t 2 ) − v(∞)
 v( t ) − v(∞) 

t 0 = t 2 − t1 = τ ln  1
 v( t 2 ) − v(∞) 
(b)
 75 − 120 
 = 24 ln (2) = 16.63 s
t 0 = 24 ln 
 30 − 120 
Since t 0 < t , the light flashes repeatedly every
τ = RC = 24 s
(1)
(2)
Chapter 7, Solution 86.
v( t ) = v(∞) + [ v(0) − v(∞)] e- t τ
v(∞) = 12 ,
v(0) = 0
-t τ
v( t ) = 12 ( 1 − e )
v( t 0 ) = 8 = 12 ( 1 − e- t 0 τ )
8
1
= 1 − e- t 0 τ 
→ e- t 0 τ =
12
3
t 0 = τ ln (3)
For R = 100 kΩ ,
τ = RC = (100 × 103 )(2 × 10-6 ) = 0.2 s
t 0 = 0.2 ln (3) = 0.2197 s
For R = 1 MΩ ,
τ = RC = (1 × 106 )(2 × 10-6 ) = 2 s
t 0 = 2 ln (3) = 2.197 s
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
= 1.2 A
100
For t > 0, we have an RL circuit
L
50
τ= =
= 0.1 ,
i(∞) = 0
R 100 + 400
i( t ) = i(∞) + [ i(0) − i(∞)] e - t τ
i( t ) = 1.2 e -10t
At t = 100 ms = 0.1 s,
i(0.1) = 1.2 e -1 = 0.441 A
which is the same as the current through the resistor.
Chapter 7, Solution 88.
(a)
τ = RC = (300 × 10 3 )(200 × 10 -12 ) = 60 µs
As a differentiator,
T > 10 τ = 600 µs = 0.6 ms
Tmin = 0.6 ms
i.e.
(b)
τ = RC = 60 µs
As an integrator,
T < 0.1τ = 6 µs
Tmax = 6 µs
i.e.
Chapter 7, Solution 89.
Since τ < 0.1 T = 1 µs
L
< 1 µs
R
L < R × 10 -6 = (200 × 10 3 )(1 × 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 = v i 
→
=
10
10 R s + R p
Rs =
Also,
1
6 2
R p = = MΩ
9
9 3
τ = R th C s = 15 µs
6 (2 3)
where R th = R p || R s =
= 0.6 MΩ
6+2 3
τ
15 × 10 -6
=
= 25 pF
Cs =
R th 0.6 × 10 6
Chapter 7, Solution 91.
12
= 240 mA ,
i(∞) = 0
50
i( t ) = i(∞) + [ i(0) − i(∞)] e - t τ
i( t ) = 240 e - t τ
L 2
τ= =
R R
i( t 0 ) = 10 = 240 e - t 0 τ
i o (0) =
e t 0 τ = 24 
→ t 0 = τ ln (24)
t0
2
5
τ=
=
= 1.573 =
R
ln (24) ln (24)
2
R=
= 1.271 Ω
1.573
Chapter 7, Solution 92.
 10
 10 -3
dv
= 4 × 10 -9 ⋅  2 ×- 10
i=C
dt

 5 × 10 -6
0 < t < tR
tR < t < tD
 20 µA
0 < t < 2 ms
i( t ) = 
- 8 mA 2 ms < t < 2 ms + 5 µs
which is sketched below.
i(t)
5 µs
20 µA
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).
6Ω
VS
+
−
6Ω
6Ω
+
+
vL
10 H
−
(a)
v
10 µF
−
(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(∞) = 0 A, v(∞) = 0 V
Chapter 8, Solution 2.
(a)
At t = 0-, the equivalent circuit is shown in Figure (a).
25 kΩ
20 kΩ
iR
+
−
80V
iL
+
60 kΩ v
−
(a)
25 kΩ
20 kΩ
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 + iL
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 µF = 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(∞) = iL(∞) = 80/45k = 1.778 mA
iC(∞) = Cdv(∞)/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 Ω
40 Ω
+
vC
+
vR
+
+
−
10 Ω
−
vR
+
−
−
10V
2A
iL
vC
−
10 Ω
+
−
(a)
10V
(b)
(c)
As t approaches infinity, we end up with the equivalent circuit shown in
Figure (b).
iL(∞) = 10(2)/(40 + 10) = 400 mA
vC(∞) = 2[10||40] –10 = 16 – 10 = 6V
vR(∞) = 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
3Ω
i
40V
+
−
+
v
5Ω
−
(a)
3Ω
0.25 H
+ vL − iC
i
+
−
40V
iR
0.1F
4A
5Ω
(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 Ω vC
1H
iC +
0.25F
vL −
+
6Ω
−
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(∞) = 6(4)/10 = 2.4 A
v(∞) = 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(∞) = [R/(R + Rs)]Vs
vL(∞) = 0 V
Chapter 8, Solution 7.
s2 + 4s + 4 = 0, thus s1,2 =
− 4 ± 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 ± 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 ± 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 ± 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 =
−2± 4−4
= -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,
ωo =
1
LC
=
1
0.01x 4
= 5
For critical damping, ωo = α = 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
α = R/(2L) = 20/(2x2) = 5 and ωo =
1
LC
=
1
0.04
= 5
ωo = α 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
α = R/(2L) = 20/(2x2) = 5 and ωo =
1
LC
ωo = α 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.
α = R/(2L) = (40 + 60)/5 = 20 and ωo =
1
LC
=
1
−3
10 x 2.5
ωo = α 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]
Hence,
B = -9.6 or i(t) = [-9.6te-20t] A
Chapter 8, Solution 17.
i(0) = I0 = 0, v(0) = V0 = 4 x15 = 60
di(0)
1
= − (RI0 + V0 ) = −4(0 + 60) = −240
dt
L
1
1
ωo =
=
= 10
LC
1 1
4 25
R
10
α=
=
= 20, which is > ωo .
2L 2 1
4
s = −α ± α 2 − ωo2 = −20 ± 300 = −20 ± 10 3 = −2.68, − 37.32
i( t ) = A1e − 2.68t + A 2e −37.32 t
di(0)
i(0) = 0 = A1 + A 2 ,
= −2.68A1 − 37.32A 2 = −240
dt
This leads to A1 = −6.928 = −A 2
(
i( t ) = 6.928 e −37.32 t − e − 268t
Since, v( t ) =
)
1 t
∫ i( t )dt + 60, we get
C 0
v(t) = (60 + 64.53e-2.68t – 4.6412e-37.32t) V
= 20
Chapter 8, Solution 18.
When the switch is off, we have a source-free parallel RLC circuit.
ωo =
α < ωo
1
LC
=
1
0.25 x1

→
α=
= 2,
1
= 0.5
2 RC
underdamped case ω d = ω o − α 2 = 4 − 0.25 = 1.936
2
Io(0) = i(0) = initial inductor current = 20/5 = 4A
Vo(0) = v(0) = initial capacitor voltage = 0 V
v(t ) = e −αt ( A1 cos ω d t + A2 sin ω d t ) = e −0.5αt ( A1 cos1.936t + A2 sin 1.936t )
v(0) =0 = A1
dv
= e −0.5αt (−0.5)( A1 cos1.936t + A2 sin 1.936t ) + e −0.5αt (−1.936 A1 sin 1.936t + 1.936 A2 cos1.936t )
dt
(V + RI o )
dv(0)
( 0 + 4)
=− o
=−
= −4 = −0.5 A1 + 1.936 A2
dt
RC
1
Thus,

→
A2 = −2.066
v(t ) = −2.066e −0.5t sin 1.936t
Chapter 8, Solution 19.
For t < 0, the equivalent circuit is shown in Figure (a).
10 Ω
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 = α.
1
ωo =
LC
=
1
4
= 0.5 = ωd
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.
2Ω
i
12
+ −
−
vC
+
v(0) = -12V and i(0) = 12/2 = 6A
For t > 0, we have a series RLC circuit.
α = R/(2L) = 2/(2x0.5) = 2
ωo = 1/ LC = 1 / 0.5x 1 4 = 2 2
Since α is less than ωo, we have an under-damped response.
ωd = ωo2 − α 2 = 8 − 4 = 2
i(t) = (Acos2t + Bsin2t)e-2t
i(0) = 6 = A
di/dt = -2(6cos2t + Bsin2t)e-2t + (-2x6sin2t + 2Bcos2t)e-αt
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 Ω
24V
6Ω
t=0
i
3H
+
−
24 Ω
+
(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
α = R/(2L) = 30/6 = 5
ωo = 1 / LC = 1 / 3x1 / 27 = 3, clearly α > ωo (overdamped response)
s1,2 = − α ± α 2 − ωo2 = −5 ± 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.
α = 20 = 1/(2RC) or RC = 1/40
(1)
ωd = 50 = ωo2 − α 2 which leads to 2500 + 400 = ωo2 = 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 µF
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,
α = 1/(2RCo), ωo = 1/ LC o
α = 1 = 1/(2RCo), we then have Co = 1/(2R) = 1/20 = 50 mF
ωo = 1/ 0.5x 0.5 = 6.32 > α (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.
α = 1/(2RC) = 1/(2x5x10-3) = 100
ωo = 1/ LC = 1 / 0.1x10 −3 = 100
ωo = α (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
2Ω
30V
+
−
io(t)
+
t=0, note this is a
make before break
switch so the
inductor current is
not interrupted.
Figure 8.78
8Ω
For Problem 8.25.
At t = 0-, vo(0) = (8/(2 + 8)(30) = 24
For t > 0, we have a source-free parallel RLC circuit.
α = 1/(2RC) = ¼
ωo = 1/ LC = 1 / 1x 1 4 = 2
Since α is less than ωo, we have an under-damped response.
ωd = ωo2 − α 2 = 4 − (1 / 16) = 1.9843
vo(t) = (A1cosωdt + A2sinωdt)e-αt
(1/4)F
vo(t)
−
vo(0) = 24 = A1 and io(t) = C(dvo/dt) = 0 when t = 0.
dvo/dt = -α(A1cosωdt + A2sinωdt)e-αt + (-ωdA1sinωdt + ωdA2cosωdt)e-αt
at t = 0, we get dvo(0)/dt = 0 = -αA1 + ωdA2
Thus, A2 = (α/ωd)A1 = (1/4)(24)/1.9843 = 3.024
vo(t) = (24cosωdt + 3.024sinωdt)e-t/4 volts
Chapter 8, Solution 26.
s2 + 2s + 5 = 0, which leads to s1,2 =
− 2 ± 4 − 20
= -1±j4
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 ± 16 − 32
= −2 ± j2
2
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 ± 36 − 32
s1, 2 =
= −4,−2
2
Hence,
i (t ) = I s + Ae −2t + Be −4t
8I s = 12

→
i (0) = 0

→
I s = 1.5
0 = 1.5 + A + B
(1)
di
= −2 Ae − 2t − 4 Be − 4t
dt
di(0)
= 2 = −2 A − 4 B

→ 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 = ±j2 (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
(c)
s2 + 2s + 1 = 0, s1,2 = -1, -1
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 = −α + α 2 − ω o ,
s 2 = −800 = −α − α 2 − ω o
2
s1 + s 2 = −1300 = −2α

→
α = 650 =
2
R
2L
Hence,
L=
s1 − s 2 = 300 = 2 α 2 − ω o
R
200
=
= 153.8 mH
2α 2 x650
2
C=

→
ω o = 623.45 =
1
= 16.25µF
(632.45) 2 L
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 Ω
i
vC(0+) = 40 V
10 Ω
i1 40 Ω
+
+
+
v
50V
−
+
−
v
vL
−
10 Ω
−
0.5H
(a)
50V
(b)
Chapter 8, Solution 32.
For t = 0-, the equivalent circuit is shown below.
2A
i
+
v
−
6Ω
i(0-) = 0, v(0-) = -2x6 = -12V
For t > 0, we have a series RLC circuit with a step input.
α = R/(2L) = 6/2 = 3, ωo = 1/ LC = 1 / 0.04
s = − 3 ± 9 − 25 = −3 ± 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 Ω
i
i
+
30V
+
−
1H
+
v
5Ω
−
v
10 Ω
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.
α = R/(2L) = 5/2 = 2.5
ω o = 1 / LC = 1 / 4 = 0.25, clearly α > ωo (overdamped response)
s1,2 = − α ± α 2 − ω 2o = −2.5 ± 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.
α = R/(2L) = 0, ωo = 1/ LC = 1/
1
1
+
= 8, s = ±j8
16 4
Since α is less than ωo, 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.
α = R/(2L) = 2/2 = 1, ωo = 1/ LC = 1/ 1 / 5 =
5
s1,2 = − α ± α 2 − ω 2o = −1 ± 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 Ω
i
10 Ω
5H
+
15V
+
−
2Ω
20 V
0.2 F
+ −
v
−
α = R/(2L) = (2 + 5 + 1)/(2x5) = 0.8
ωo = 1/ LC = 1/ 5x 0.2 = 1
s1,2 = − α ± α 2 − ω2o = −0.8 ± 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
6Ω
6Ω
6Ω
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
ωo = 1/ LC = 1/ (1 / 2)(1 / 8) = 4
α = R/(2L) = (4)/(2x(1/2)) = 4
α = ωo, 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 Ω
v
i1
5Ω
−
10 Ω
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
ωo = 1/ LC = 1/ (1 / 3)(3 / 4) = 2
α = R/(2L) = (4)/(2x(3/4)) = 8/3
s1,2 = − α ± α 2 − ω 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 Ω
60V
+
−
+ v −
20 Ω
(a)
30 Ω
0.5F
0.25H
20 Ω 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
ωo = 1/ LC = 1/ (1 / 2)(1 / 4) =
8
α = R/(2L) = (12)/(0.5) = 24
Since α > ωo, we have an overdamped response.
s1,2 = − α ± α 2 − ω 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
+
6Ω
v
14 Ω
−
24V
12V
+ −
ωo = 1/ LC = 1/ 2 x 0.02 = 5
α = R/(2L) = (6 + 14)/(2x2) = 5
+
−
Since α = ωo, 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
4Ω
5A
20 Ω
5Ω
1H
i
+
10 µF
20V
+
−
0.04F
v
−
(a)
(b)
ωo = 1/ LC = 1/ 1x1 / 25 = 5
α = R/(2L) = (4)/(2x1) = 2
s1,2 = − α ± α 2 − ω 2o = -2 ± j4.583
Thus,
v(t) = Vs + [(Acosωdt + Bsinωdt)e-2t],
where ωd = 4.583 and Vs = 20
v(0) = 50/3 = 20 + A or A = -10/3
i(t) = Cdv/dt = C(-2) [(Acosωdt + Bsinωdt)e-2t] + Cωd[(-Asinωdt + Bcosωdt)e-2t]
i(0) = 0 = -2A + ωdB
B = 2A/ωd = -20/(3x4.583) = -1.455
i(t) = C{[(0cosωdt + (-2B - ωdA)sinωdt)]e-2t}
= (1/25){[(2.91 + 15.2767) sinωdt)]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
− +
1Ω
5Ω
6Ω
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.
ωo = 1/ LC = 1/ 1x1 / 25 = 5
α = R/(2L) = (6)/(2) = 3
s1,2 = − α ± α 2 − ω 2o = -3 ± j4
Thus,
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.
α=
R
2L
R = 2αL = 2 x8 x0.5 = 8Ω

→
ω d = ω o 2 − α 2 = 30
ωo =
1
LC
ω o = 900 − 64 = 836

→

→
C=
1
ω oL
2
=
1
= 2.392 mF
836 x0.5
Chapter 8, Solution 44.
α=
R 1000
=
= 500,
2L
2 x1
ωo > α

→
ωo =
1
LC
=
1
100 x10
−9
= 10 4
underdamped.
Chapter 8, Solution 45.
ωo = 1/ LC = 1/ 1x 0.5 =
2
α = R/(2L) = (1)/(2x2x0.5) = 0.5
Since α < ωo, we have an underdamped response.
s1,2 = − α ± α 2 − ω 2o = -0.5 ± j1.323
Thus,
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
5µF
v
2 kΩ
−
6mA
α = 1/(2RC) = (1)/(2x2x103 x5x10-6) = 50
ωo = 1/ LC = 1/ 8x10 3 x 5x10 −6 = 5,000
Since α < ωo, we have an underdamped response.
s1,2 = − α ± α 2 − ωo2 ≅ -50 ± 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.
α = 1/(2RC) = (1)/(2x5x0.01) = 10
ωo = 1/ LC = 1/ 1x 0.01 = 10
Since α = ωo, 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.
α = 1/(2RC) = (1)/(2x1x0.25) = 2
ωo = 1/ LC = 1/ 1x 0.25 = 2
Since α = ωo, 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.
α = 1/(2RC) = (1)/(2x5x0.05) = 2
ωo = 1/ LC = 1/ 5x 0.05 = 2
Since α = ωo, 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 Ω
10 mF
6A
40 Ω
v
−
Is = 3 + 6 = 9A and R = 10||40 = 8 ohms
α = 1/(2RC) = (1)/(2x8x0.01) = 25/4 = 6.25
ωo = 1/ LC = 1/ 4x 0.01 = 5
Since α > ωo, we have a overdamped response.
s1,2 = − α ± α 2 − ω 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 = ∞).
α = 1/(2RC) = 0 and ωo = 1/ LC which leads to s1,2 = ± jωo
v = Acosωot + Bsinωot, v(0) = 0 A
iC = Cdv/dt = -i
dv/dt = ωoBsinωot = -i/C
dv(0)/dt = ωoB = -io/C therefore B = io/(ωoC)
v(t) = -(io/(ωoC))sinωot V where ωo =
LC
Chapter 8, Solution 52.
α = 300 =
1
2 RC
ω d = ω o 2 − α 2 = 400
(1)

→
ω o = 400 2 − 300 2 = 264.575 =
From (2),
C=
1
= 285.71µF
(264.575) 2 x50 x10 −3
From (1),
R=
1
1
=
(3500) = 5.833Ω
2αC 2 x300
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 =
v dv 1 dv 1 d 2 v
+
+
+
2 dt 4 dt 2 dt 2
s 2 + 2.5s + 3 = 0

→
d 2v
dv
+ 2.5 + 3v = 0
2
dt
dt

→
s = −1.25 ± j1.199
v = Ae −1.25t cos1.199t + Be −1.25t sin 1.199t
v(0) = 2=A. Let w=1.199
dv
= −1.25( Ae −1.25t cos wt + Be −1.25t sin wt ) + w(− Ae −1.25t sin wt + Be −1.25t cos wt )
dt
dv(0)
= 0 = −1.25 A + Bw
dt

→
B=
1.25 X 2
= 2.085
1.199
v = 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 ) ∫ 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.
4Ω
i
6Ω
i
0.04F
+
−
20
io
0.25H
Applying KVL to the larger loop,
-20 +6io +0.25dio/dt + 25 ∫ (i o + i)dt = 0
Taking the derivative,
6dio/dt + 0.25d2io/dt2 + 25(io + i) = 0
For the smaller loop,
4 + 25 ∫ (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(∞) = 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.
α = R/(2L) = (20)/(2x1) = 10
ωo = 1/ LC = 1/ 1x (1 / 36) = 6
Since α > ωo, we have a overdamped response.
s1,2 = − α ± α 2 − ωo2 = -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 Ω
+
+
(1/36)F
v
−
8Ω
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
[v(0) + Ri(0)]
(4 + 0)
dv(0)
=−
=−
= − 8 V/s
dt
RC
0.5
(b) For t ≥ 0 , the circuit is a source-free RLC parallel circuit.
α=
1
1
=
= 1,
2 RC 2 x0.5 x1
ωo =
1
LC
=
1
0.25 x1
=2
ω d = ω 2 o − α 2 = 4 − 1 = 1.732
Thus,
v(t ) = e − t ( A1 cos1.732t + A2 sin 1.732t )
v(0) = 4 = A1
dv
= −e −t A1 cos1.732t − 1.732e −t A1 sin 1.732t − e −t A2 sin 1.732t + 1.732e −t A2 cos1.732t
dt
dv(0)
= −8 = − A1 + 1.732 A2

→
A2 = −2.309
dt
v(t ) = e − t (4 cos 1.732t − 2.309 sin 1.732t ) V
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, ω o =
=
= 2,

→ α = ω 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)]

→ − 2 A + B = − (32 + 0),
dt
L
4
α=
B = −4
i (t ) = 2e −2t − 4te −2t
t
v=
t
t
1
idt + v(0) = 32 ∫ e − 2t dt − 64 ∫ te − 2t dt = −16e − 2t
∫
C0
0
0
t
−
0
64 − 2t
e (−2t − 1)
4
t
0
v = 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)
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
+
4Ω
vC
0.25F
6Ω
−
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(∞) + [Ae-2t + Be-5t],
where iL(∞) 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.
α = 1/(2RC) = (1)/(2x3x(1/18)) = 3
ωo = 1/ LC = 1/ 2x1 / 18 = 3
Since α = ωo, 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
dv
=C 1
R
dt
(1)
At node 2,
dv
v2 − vo
=C o
(2)
dt
R
As a voltage follower, v1 = v 2 = v . Hence (2) becomes
dv
(3)
v = v o + RC o
dt
and (1) becomes
dv
v s = v + RC
(4)
dt
Substituting (3) into (4) gives
v s = vo + RC
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 = v s
2
dt
dt
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
Combining (1), (2), and (3), eliminating v1 and v2 we get,
d 2 vo  1 
d 2vo
− 100 v o = 0
−
v o =
dt 2  R 2 C 2 
dt 2
(3)
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 µF, 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 = ±j
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,
d ( v1 − v o )
d ( v 1 − 0)
v in − v1
= C1
+ C2
dt
dt
R1
At node 2,
C2
(1)
− vo
d ( v 1 − 0) 0 − v o
dv1
, or
=
=
dt
R2
dt
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
vin
1
v1
C2
2
0V
−
+
vo
From (2) and (3),
−
vo
d 2 v o R 1 dv o
dv
dv
R C dv o
= 1 = in + 1 1
+ R 1 C1
+
C2R 2
dt
dt
C 2 R 2 dt
R 2 dt
dt 2
d 2 vo
vo
1  1
1  dv o
1 dv in


+
=−
+
+
2
R 2  C1 C 2  dt
C1 C 2 R 2 R 1
R 1C1 dt
dt
But C1C2R1R2 = 10-4 x10-4 x104 x104 = 1
1
R2
 1
1 
2
2
 =

= 4
=2
+
−4
 C1 C 2  R 2 C1 10 x10
d 2 vo
dv
dv
+ 2 o + v o = − in
2
dt
dt
dt
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(∞) = 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
=0
dt
C2R 2
From (2),
From (1) at t = 0+,
dv (0+)
dv o (0+ )
1
1− 0
= −C1 o
which leads to
=−
= −1
dt
dt
C1 R 1
R1
Substituting this into (4) gives B = –1
Thus,
v(t) = –te-tu(t) V
(4)
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 Ω
5Ω
+
20 V
-
2F
4H
Hence the dual circuit is shown below.
2H
20A
0.1 Ω
4F
0.2 Ω
Chapter 8, Solution 75.
The dual circuit is connected as shown in Figure (a). It is redrawn in Figure (b).
0.1 Ω
12V
+
−
10 Ω
12A
24A
0.5 F
24V
0.25 Ω
4Ω
10 H
10 H
10 µF
(a)
0.1 Ω
2F
0.5 H
24A
12A
0.25 Ω
(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 Ω
0.05 Ω
1/3 Ω
10 Ω
20 Ω
60 A
30 Ω
120 A
+ −
– +
60 V
2V
120 V
+ −
4H
1F
1H
2A
4F
(a)
0.05 Ω
60 A
120 A
1H
0.1 Ω
1/30 Ω
1/4 F 2V
(b)
+
−
Chapter 8, Solution 77.
The dual is constructed in Figure (a) and redrawn in Figure (b).
– +
5A
5V
2Ω
1/3 Ω
1/2 Ω
1F
1Ω
1/4 H
1H
3Ω
12V
1Ω
1/4 F
+
−
12 A
(a)
1Ω
2Ω
1/4 F
1/3 Ω
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.
α = 1/(2RC) = 1/(2x3x1/30) = 5
ωo = 1 / LC = 1 / 60 x10 −3 x1 / 30 = 22.36
α < ωo produces an underdamped response.
s1, 2 = −α ± α 2 − ω o2 = –5 ± 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,
α = ωo

→
1
=
2 RC
1
LC
Hence,
C=
0.25
L
=
= 434 µF
2
4 x144
4R
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 = −α − α 2 − ωo2
s 2 = −α + α 2 − ω o2
s1 + s2 = –2α = –12,000, therefore α = 6,000 = R/(2L)
L = R/12,000 = 60,000/12,000 = 5H
s 2 = −α + α 2 − ωo2 = –2,000
α − α 2 − ωo2 = 2,000
6,000 − α 2 − ωo2 = 2,000
α 2 − ωo2 = 4,000
α2 – ωo2 = 16x106
ωo2 = α2 – 16x106 = 36x106 – 16x106
ωo = 103 20 = 1 / LC
C = 1/(20x106x5) = 10 nF
Chapter 8, Solution 81.
t = 1/α = 0.25 leads to α = 4
But,
α 1/(2RC) or,
C = 1/(2αR) = 1/(2x4x200) = 625 µF
ωd = ωo2 − α 2
ωo2 = ωd2 + α 2 = (2π4x10 3 ) 2 + 16 ≅ (2π4 x10 3 0 2 = 1/(LC)
This results in L = 1/(64π2x106x625x10-6) = 2.533 µH
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.
ω = 103 rad/s
(a)
angular frequency
(b)
frequency
f =
ω
= 159.2 Hz
2π
(c)
period
T =
1
= 6.283 ms
f
(d)
Since sin(A) = cos(A – 90°),
vs = 12 sin(103t + 24°) = 12 cos(103t + 24° – 90°)
vs in cosine form is
vs = 12 cos(103t – 66°) V
(e)
vs(2.5 ms) = 12 sin((10 3 )(2.5 × 10 -3 ) + 24°)
= 12 sin(2.5 + 24°) = 12 sin(143.24° + 24°)
= 2.65 V
Chapter 9, Solution 2.
(a)
amplitude = 8 A
(b)
ω = 500π = 1570.8 rad/s
(c)
f =
(d)
Is = 8∠-25° A
Is(2 ms) = 8 cos((500π )(2 × 10 -3 ) − 25°)
= 8 cos(π − 25°) = 8 cos(155°)
= -7.25 A
ω
= 250 Hz
2π
Chapter 9, Solution 3.
(a)
4 sin(ωt – 30°) = 4 cos(ωt – 30° – 90°) = 4 cos(ωt – 120°)
(b)
-2 sin(6t) = 2 cos(6t + 90°)
(c)
-10 sin(ωt + 20°) = 10 cos(ωt + 20° + 90°) = 10 cos(ωt + 110°)
Chapter 9, Solution 4.
(a)
v = 8 cos(7t + 15°) = 8 sin(7t + 15° + 90°) = 8 sin(7t + 105°)
(b)
i = -10 sin(3t – 85°) = 10 cos(3t – 85° + 90°) = 10 cos(3t + 5°)
Chapter 9, Solution 5.
v1 = 20 sin(ωt + 60°) = 20 cos(ωt + 60° − 90°) = 20 cos(ωt − 30°)
v2 = 60 cos(ωt − 10°)
This indicates that the phase angle between the two signals is 20° and that v1 lags
v2.
Chapter 9, Solution 6.
(a)
v(t) = 10 cos(4t – 60°)
i(t) = 4 sin(4t + 50°) = 4 cos(4t + 50° – 90°) = 4 cos(4t – 40°)
Thus, i(t) leads v(t) by 20°.
(b)
v1(t) = 4 cos(377t + 10°)
v2(t) = -20 cos(377t) = 20 cos(377t + 180°)
Thus, v2(t) leads v1(t) by 170°.
(c)
x(t) = 13 cos(2t) + 5 sin(2t) = 13 cos(2t) + 5 cos(2t – 90°)
X = 13∠0° + 5∠-90° = 13 – j5 = 13.928∠-21.04°
x(t) = 13.928 cos(2t – 21.04°)
y(t) = 15 cos(2t – 11.8°)
phase difference = -11.8° + 21.04° = 9.24°
Thus, y(t) leads x(t) by 9.24°.
Chapter 9, Solution 7.
If f(φ) = cosφ + j sinφ,
df
= -sinφ + j cos φ = j (cos φ + j sin φ) = j f (φ )
dφ
df
= j dφ
f
Integrating both sides
ln f = jφ + ln A
f = Aejφ = cosφ + j sinφ
f(0) = A = 1
i.e. f(φ) = ejφ = cosφ + j sinφ
Chapter 9, Solution 8.
(a)
(b)
(c)
15∠45°
15∠45°
+ j2 =
+ j2
5∠ - 53.13°
3 − j4
= 3∠98.13° + j2
= -0.4245 + j2.97 + j2
= -0.4243 + j4.97
(2 + j)(3 – j4) = 6 – j8 + j3 + 4 = 10 – j5 = 11.18∠-26.57°
8∠ - 20°
(-5 − j12)(10)
8∠ - 20°
10
+
+
=
11.18∠ - 26.57°
25 + 144
(2 + j)(3 - j4) - 5 + j12
= 0.7156∠6.57° − 0.2958
− j0.71
= 0.7109 + j0.08188 −
0.2958 − j0.71
= 0.4151 − j0.6281
10 + (8∠50°)(13∠-68.38°) = 10+104∠-17.38°
= 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∠-10° +
1 − j2
2.236 ∠ - 63.43°
= 4∠-10° +
3∠6°
3∠6°
= 4∠-10° + 0.7453∠-69.43°
= 3.939 – j0.6946 + 0.2619 – j0.6978
= 4.201 – j1.392
(c)
8∠10° + 6 ∠ - 20° 7.879 + j1.3892 + 5.638 − j2.052
=
9∠80° − 4∠50°
1.5628 + j8.863 − 2.571 − j3.064
13.533∠ - 2.81°
13.517 − j0.6629
=
=
5.886∠99.86°
− 1.0083 + j5.799
= 2.299∠-102.67°
= -0.5043 – j2.243
Chapter 9, Solution 10.
(a) z1 = 6 − j8, z 2 = 8.66 − j 5, and z 3 = −4 − j 6.9282
z1 + z 2 + z 3 = 10.66 − j19.93
(b)
z1 z 2
= 9.999 + j 7.499
z3
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) = − 1.2749 + j 0.1520
(b)
50∠ − 30 o
= − 2.0833
24∠150 o
(c) (2+j3)(8-j5) –(-4) = 35 +j14
Chapter 9, Solution 14.
(a)
3 − j14
= − 0.5751 + j 0.5116
− 15 + j11
(b)
(62.116 + j 231.82 + 138.56 − j80)(60 − j80)
24186 − 6944.9
=
= − 1.922 − j11.55
(67 + j84)(16.96 + j10.5983)
246.06 + j 2134.7
(c) (− 2 + j 4 )
2
(260 − j120) = − 256.4 − j 200.89
Chapter 9, Solution 15.
(a)
(b)
(c)
10 + j6 2 − j3
= -10 – j6 + j10 – 6 + 10 – j15
-5
-1 + j
= -6 – j11
20∠ − 30° - 4∠ - 10°
= 60∠15° + 64∠-10°
16∠0°
3∠45°
= 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 + 75°) = 10 cos(4t + 75° − 180°)
= 10 cos(4t − 105°)
The phasor form is 10∠-105°
(b)
5 sin(20t – 10°) = 5 cos(20t – 10° – 90°)
= 5 cos(20t – 100°)
The phasor form is 5∠-100°
(c)
4 cos(2t) + 3 sin(2t) = 4 cos(2t) + 3 cos(2t – 90°)
The phasor form is 4∠0° + 3∠-90° = 4 – j3 = 5∠-36.87°
Chapter 9, Solution 17.
(a)
Let A = 8∠-30° + 6∠0°
= 12.928 – j4
= 13.533∠-17.19°
a(t) = 13.533 cos(5t + 342.81°)
(b)
We know that -sinα = cos(α + 90°).
Let B = 20∠45° + 30∠(20° + 90°)
= 14.142 + j14.142 – 10.261 + j28.19
= 3.881 + j42.33
= 42.51∠84.76°
b(t) = 42.51 cos(120πt + 84.76°)
(c)
Let C = 4∠-90° + 3∠(-10° – 90°)
= -j4 – 0.5209 – j2.954
= 6.974∠265.72°
c(t) = 6.974 cos(8t + 265.72°)
Chapter 9, Solution 18.
(a)
v1 ( t ) = 60 cos(t + 15°)
(b)
V2 = 6 + j8 = 10∠53.13°
v 2 ( t ) = 10 cos(40t + 53.13°)
(c)
i1 ( t ) = 2.8 cos(377t – π/3)
(d)
I 2 = -0.5 – j1.2 = 1.3∠247.4°
i 2 ( t ) = 1.3 cos(103t + 247.4°)
Chapter 9, Solution 19.
(a)
3∠10° − 5∠-30° = 2.954 + j0.5209 – 4.33 + j2.5
= -1.376 + j3.021
= 3.32∠114.49°
Therefore,
3 cos(20t + 10°) – 5 cos(20t – 30°) = 3.32 cos(20t +
114.49°)
(b)
4∠-90° + 3∠-45° = -j40 + 21.21 – j21.21
= 21.21 – j61.21
= 64.78∠-70.89°
Therefore,
40 sin(50t) + 30 cos(50t – 45°) = 64.78 cos(50t – 70.89°)
(c)
Using sinα = cos(α − 90°),
20∠-90° + 10∠60° − 5∠-110° = -j20 + 5 + j8.66 + 1.7101 + j4.699
= 6.7101 – j6.641
= 9.44∠-44.7°
Therefore,
20 sin(400t) + 10 cos(400t + 60°) – 5 sin(400t – 20°)
= 9.44 cos(400t – 44.7°)
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,
v = 8.966 cos(377t − 4.399 o )
(b) I = 10∠0 o + jω 8∠20 o − 90 o ,
ω = 5 , i.e. I = 10 + 40∠20 o = 49.51∠16.04 o
i = 49.51 cos(5t + 16.04 o )
Chapter 9, Solution 21.
(a) F = 5∠15 o − 4∠− 30 o − 90 o = 6.8296 + j 4.758 = 8.3236∠34.86 o
f (t ) = 8.324 cos(30t + 34.86 o )
(b) G = 8∠ − 90 o + 4∠50 o = 2.571 − j 4.9358 = 5.565∠ − 62.49 o
g (t ) = 5.565 cos(t − 62.49 o )
1
(10∠0 o + 5∠ − 90 o ), ω = 40
jω
i.e. H = 0.25∠ − 90 o + 0.125∠ − 180 o = − j 0.25 − 0.125 = 0.2795∠ − 116.6 o
(c) H =
h(t ) = 0.2795 cos(40t − 116.6 o )
Chapter 9, Solution 22.
t
dv
Let f(t) = 10v(t ) + 4 − 2 ∫ v(t )dt
dt
−∞
2V
F = 10V + jω 4V −
, ω = 5, V = 20∠ − 30 o
jω
F = 10V + j 20V − j 0.4V = (10 − j19.6)(17.32 − j10) = 440.1∠ − 92.97 o
f (t ) = 440.1 cos(5t − 92.97 o )
Chapter 9, Solution 23.
(a)
v(t) = 40 cos(ωt – 60°)
(b)
V = -30∠10° + 50∠60°
= -4.54 + j38.09
= 38.36∠96.8°
v(t) = 38.36 cos(ωt + 96.8°)
(c)
I = j6∠-10° = 6∠(90° − 10°) = 6∠80°
i(t) = 6 cos(ωt + 80°)
(d)
2
+ 10∠-45° = -j2 + 7.071 – j7.071
j
= 11.5∠-52.06°
i(t) = 11.5 cos(ωt – 52.06°)
I =
Chapter 9, Solution 24.
(a)
V
= 10∠0°, ω = 1
jω
V (1 − j) = 10
10
V=
= 5 + j5 = 7.071∠45°
1− j
Therefore,
v(t) = 7.071 cos(t + 45°)
V+
(b)
4V
= 20∠(10° − 90°), ω = 4
jω

4
V  j4 + 5 +  = 20 ∠ - 80°
j4 

20∠ - 80°
= 3.43∠ - 110.96°
V=
5 + j3
Therefore,
v(t) = 3.43 cos(4t – 110.96°)
jωV + 5V +
Chapter 9, Solution 25.
(a)
2jωI + 3I = 4∠ - 45°, ω = 2
I (3 + j4) = 4∠ - 45°
4∠ - 45° 4∠ - 45°
=
= 0.8∠ - 98.13°
I=
3 + j4
5∠53.13°
Therefore,
i(t) = 0.8 cos(2t – 98.13°)
(b)
I
+ jωI + 6I = 5∠22°, ω = 5
jω
(- j2 + j5 + 6) I = 5∠22°
5∠22°
5∠22°
I=
=
= 0.745∠ - 4.56°
6 + j3 6.708∠26.56°
Therefore,
i(t) = 0.745 cos(5t – 4.56°)
10
Chapter 9, Solution 26.
I
= 1∠0°, ω = 2
jω

1
I  j2 + 2 +  = 1
j2 

1
= 0.4∠ - 36.87°
I=
2 + j1.5
Therefore,
i(t) = 0.4 cos(2t – 36.87°)
jωI + 2I +
Chapter 9, Solution 27.
V
= 110∠ - 10°, ω = 377
jω

j100 
 = 110∠ - 10°
V  j377 + 50 −

377 
V (380.6∠82.45°) = 110∠ - 10°
V = 0.289 ∠ - 92.45°
jωV + 50V + 100
Therefore, v(t) = 0.289 cos(377t – 92.45°).
Chapter 9, Solution 28.
i( t ) =
v s ( t ) 110 cos(377 t )
=
= 13.75 cos(377t) A.
R
8
Chapter 9, Solution 29.
Z=
1
1
=
= - j 0.5
6
jωC j (10 )(2 × 10 -6 )
V = IZ = (4∠25°)(0.5∠ - 90°) = 2 ∠ - 65°
Therefore
v(t) = 2 sin(106t – 65°) V.
Chapter 9, Solution 30.
Z = jωL = j (500)(4 × 10 -3 ) = j2
V 60 ∠ - 65°
I= =
= 30∠ - 155°
Z
2∠90°
Therefore,
i(t) = 30 cos(500t – 155°) A.
Chapter 9, Solution 31.
Thus,
i(t) = 10 sin(ωt + 30°) = 10 cos(ωt + 30° − 90°) = 10 cos(ωt − 60°)
I = 10∠-60°
v(t) = -65 cos(ωt + 120°) = 65 cos(ωt + 120° − 180°) = 65 cos(ωt − 60°)
Thus,
V = 65∠-60°
Z=
V 65∠ - 60°
=
= 6.5 Ω
I 10∠ - 60°
Since V and I are in phase, the element is a resistor with R = 6.5 Ω.
Chapter 9, Solution 32.
V = 180∠10°,
Z=
I = 12∠-30°,
ω = 2
V 180∠10°
=
= 15∠40° = 11.49 + j 9.642 Ω
I 12∠ - 30°
One element is a resistor with R = 11.49 Ω.
The other element is an inductor with ωL = 9.642 or
L = 4.821 H.
Chapter 9, Solution 33.
110 = v 2R + v 2L
v L = 110 2 − v 2R
v L = 110 2 − 85 2 = 69.82 V
Chapter 9, Solution 34.
v o = 0 if ωL =
ω=
1
ωC

→ ω =
1
(5 × 10 −3 )(2 × 10 − 3 )
1
LC
= 100 rad/s
Chapter 9, Solution 35.
Vs = 5∠0°
jωL = j (2)(1) = j2
1
1
=
= - j2
jωC j (2)(0.25)
j2
j2
Vs = 5∠0° = (1∠90°)(5∠0°) = 5∠90°
2 − j2 + j2
2
Thus, v o ( t ) = 5 cos(2t + 90°) = -5 sin(2t) V
Vo =
Chapter 9, Solution 36.
Let Z be the input impedance at the source.
10 µF
jωL = j 200 x100 x10 −3 = j 20

→
100 mH
1
1
=
= − j 500
jωC j10 x10 −6 x 200

→
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
= 26.61∠ − 3.896 o mA
o
2255∠ − 6.104
i = 266.1 cos(200t − 3.896 o )
Chapter 9, Solution 37.
jωL = j (5)(1) = j5
1
1
=
= -j
jωC j (5)(0.2)
Let Z1 = - j ,
Z 2 = 2 || j5 =
Then,
Ix =
(2)( j5)
j10
=
2 + j5 2 + j5
Z2
I ,
Z1 + Z 2 s
where I s = 2∠0°
j10
j20
2 + j5
Ix =
(2) =
= 2.12 ∠32°
j10
5 + j8
- j+
2 + j5
Therefore,
i x ( t ) = 2.12 sin(5t + 32°) A
Chapter 9, Solution 38.
1
F 
→
6
(a)
1
1
=
= - j2
jωC j (3)(1 / 6)
- j2
(10 ∠45°) = 4.472∠ - 18.43°
4 − j2
Hence, i(t) = 4.472 cos(3t – 18.43°) A
I=
V = 4I = (4)(4.472∠ - 18.43°) = 17.89∠ - 18.43°
Hence, v(t) = 17.89 cos(3t – 18.43°) V
1
F 
→
12
(b)
3H 
→
1
1
=
= - j3
jωC j (4)(1 / 12)
jωL = j (4)(3) = j12
V 50∠0°
= 10∠36.87°
=
Z 4 − j3
Hence, i(t) = 10 cos(4t + 36.87°) A
I=
j12
(50∠0°) = 41.6 ∠33.69°
8 + j12
Hence, v(t) = 41.6 cos(4t + 33.69°) V
V=
Chapter 9, Solution 39.
Z = 8 + j5 || (- j10) = 8 +
I=
( j5)(- j10)
= 8 + j10
j5 − j10
V 40 ∠0°
20
=
=
= 3.124∠ - 51.34°
Z 8 + j10 6.403∠51.34°
I1 =
- j10
I = 2 I = 6.248∠ - 51.34°
j5 − j10
I2 =
j5
I = - I = 3.124∠128.66°
- j5
Therefore,
i1 ( t ) = 6.248 cos(120πt – 51.34°) A
i 2 ( t ) = 3.124 cos(120πt + 128.66°) A
Chapter 9, Solution 40.
(a)
For ω = 1 ,
1H 
→
jωL = j (1)(1) = j
1
1
0.05 F 
→
=
= - j20
jωC j (1)(0.05)
- j40
Z = j + 2 || (- j20) = j +
= 1.98 + j0.802
2 − j20
V
4 ∠0°
4∠0°
=
=
= 1.872 ∠ - 22.05°
Z 1.98 + j0.802 2.136∠22.05°
Hence, i o ( t ) = 1.872 cos(t – 22.05°) A
Io =
(b)
For ω = 5 ,
1H 
→
jωL = j (5)(1) = j5
1
1
0.05 F 
→
=
= - j4
jωC j (5)(0.05)
- j4
Z = j5 + 2 || (- j4) = j5 +
= 1.6 + j4.2
1 − j2
4∠0°
4∠0°
V
=
=
= 0.89∠ - 69.14°
Z 1.6 + j4 4.494∠69.14°
Hence, i o ( t ) = 0.89 cos(5t – 69.14°) A
Io =
(c)
For ω = 10 ,
1H 
→ jωL = j (10)(1) = j10
1
1
0.05 F 
→
=
= - j2
jωC j (10)(0.05)
- j4
Z = j10 + 2 || (- j2) = j10 +
= 1 + j9
2 − j2
V 4∠0°
4 ∠0°
=
= 0.4417 ∠ - 83.66°
=
Z 1 + j9 9.055∠83.66°
Hence, i o ( t ) = 0.4417 cos(10t – 83.66°) A
Io =
Chapter 9, Solution 41.
ω = 1,
1H 
→
jωL = j (1)(1) = j
1
1
=
= -j
jωC j (1)(1)
1F 
→
Z = 1 + (1 + j) || (- j) = 1 +
I=
Vs
10
=
,
Z 2− j
- j+1
= 2− j
1
I c = (1 + j) I
V = (- j)(1 + j) I = (1 − j) I =
Thus,
(1 − j)(10)
= 6.325∠ - 18.43°
2− j
v(t) = 6.325 cos(t – 18.43°) V
Chapter 9, Solution 42.
ω = 200
50 µF 
→
1
1
=
= - j100
jωC j (200)(50 × 10 -6 )
0.1 H 
→
jωL = j (200)(0.1) = j20
50 || -j100 =
Vo =
(50)(-j100) - j100
=
= 40 − j20
50 − j100
1 - j2
j20
j20
(60∠0°) =
(60∠0°) = 17.14 ∠90°
j20 + 30 + 40 − j20
70
Thus, v o ( t ) = 17.14 sin(200t + 90°) V
or
v o ( t ) = 17.14 cos(200t) V
Chapter 9, Solution 43.
ω= 2
1H 
→ jωL = j (2)(1) = j2
1F 
→
Io =
1
1
=
= - j0.5
jωC j (2)(1)
j2 − j0.5
j1.5
I=
4∠0° = 3.328∠33.69°
j2 − j0.5 + 1
1 + j1.5
Thus, i o ( t ) = 3.328 cos(2t + 33.69°) A
Chapter 9, Solution 44.
ω = 200
10 mH 
→ jωL = j (200)(10 × 10 -3 ) = j2
5 mF 
→
1
1
=
= -j
jωC j (200)(5 × 10 -3 )
Y=
1 1
1
3+ j
+ +
= 0.25 − j0.5 +
= 0.55 − j0.4
4 j2 3 − j
10
Z=
1
1
=
= 1.1892 + j0.865
Y 0.55 − j0.4
I=
6∠0°
6∠0°
=
= 0.96 ∠ - 7.956°
5 + Z 6.1892 + j0.865
Thus, i(t) = 0.96 cos(200t – 7.956°) A
Chapter 9, Solution 45.
We obtain I o by applying the principle of current division twice.
I
I2
Z1
I2
Io
-j2 Ω
Z2
(a)
Z 1 = - j2 ,
(b)
Z 2 = j4 + (-j2) || 2 = j4 +
I2 =
Z1
- j2
- j10
I=
(5∠0°) =
Z1 + Z 2
- j2 + 1 + j3
1+ j
Io =
 - j  - j10  - 10
- j2

=
I2 = 
= -5 A
2 - j2
1 - j  1 + j  1 + 1
- j4
= 1 + j3
2 - j2
Chapter 9, Solution 46.
i s = 5 cos(10 t + 40°) 
→ I s = 5∠40°
Let
0.1 F 
→
1
1
=
= -j
jωC j (10)(0.1)
0.2 H 
→
jωL = j (10)(0.2) = j2
Z1 = 4 || j2 =
2Ω
j8
= 0.8 + j1.6 ,
4 + j2
Z2 = 3 − j
Io =
Z1
0.8 + j1.6
(5∠40°)
Is =
3.8 + j0.6
Z1 + Z 2
Io =
(1.789∠63.43°)(5∠40°)
= 2.325∠94.46°
3.847 ∠8.97°
Thus, i o ( t ) = 2.325 cos(10t + 94.46°) A
Chapter 9, Solution 47.
First, we convert the circuit into the frequency domain.
Ix
5∠0˚
Ix =
2Ω
j4
+
−
-j10
20 Ω
5
5
5
=
=
= 0.4607∠52.63°
− j10(20 + j4) 2 + 4.588 − j8.626 10.854∠ − 52.63°
2+
− j10 + 20 + j4
is(t) = 0.4607cos(2000t +52.63˚) A
Chapter 9, Solution 48.
Converting the circuit to the frequency domain, we get:
10 Ω
V1 30 Ω
Ix
20∠-40˚
+
−
We can solve this using nodal analysis.
j20
-j20
V1 − 20∠ − 40° V1 − 0
V −0
=0
+
+ 1
10
j20
30 − j20
V1(0.1 − j0.05 + 0.02307 + j0.01538) = 2∠ − 40°
2∠40°
= 15.643∠ − 24.29°
0.12307 − j0.03462
15.643∠ − 24.29°
=
= 0.4338∠9.4°
30 − j20
= 0.4338 sin(100 t + 9.4°) A
V1 =
Ix
ix
Chapter 9, Solution 49.
Z T = 2 + j2 || (1 − j) = 2 +
I
( j2)(1 − j)
=4
1+ j
Ix
1Ω
j2 Ω
-j Ω
j2
j2
I=
I,
j2 + 1 − j
1+ j
1+ j
1+ j
I=
Ix =
j2
j4
where I x = 0.5∠0° =
Ix =
1
2
1+ j
1+ j
(4) =
= 1 − j = 1.414∠ - 45°
j4
j
v s ( t ) = 1.414 sin(200t – 45°) V
Vs = I Z T =
Chapter 9, Solution 50.
Since ω = 100, the inductor = j100x0.1 = j10 Ω and the capacitor = 1/(j100x10-3)
= -j10Ω.
j10
5∠40˚
Ix
+
-j10
20 Ω
vx
−
Using the current dividing rule:
− j10
5∠40° = − j2.5∠40° = 2.5∠ − 50°
− j10 + 20 + j10
Vx = 20I x = 50∠ − 50°
Ix =
v x = 50 cos(100t − 50°) V
Chapter 9, Solution 51.
0.1 F 
→
1
1
=
= - j5
jωC j (2)(0.1)
0.5 H 
→
jωL = j (2)(0.5) = j
The current I through the 2-Ω resistor is
Is
1
I=
Is =
,
1 − j5 + j + 2
3 − j4
I s = (10)(3 − j4) = 50∠ - 53.13°
where I = 10 ∠0°
Therefore,
i s ( t ) = 50 cos(2t – 53.13°) A
Chapter 9, Solution 52.
5 || j5 =
j25
j5
=
= 2.5 + j2.5
5 + j5 1 + j
Z1 = 10 ,
Z 2 = - j5 + 2.5 + j2.5 = 2.5 − j2.5
I2
IS
Z1
Z2
I2 =
Z1
10
4
Is =
Is =
I
12.5 − j2.5
5− j s
Z1 + Z 2
Vo = I 2 (2.5 + j2.5)
 4
8∠30° = 
5 −
Is =

10 (1 + j)
 I s (2.5)(1 + j) =
I
j
5− j s
(8∠30°)(5 − j)
= 2.884∠-26.31° A
10 (1 + j)
Chapter 9, Solution 53.
Convert the delta to wye subnetwork as shown below.
Z1
Io
Z2
2Ω
Z3
+
10 Ω
8Ω
60∠ − 30 V
o
-
Z
Z1 =
− j 2 x4
= 0.1532 − j 0.7692,
10 − j 2
Z3 =
12
= 1.1538 + j 0.2308
10 − j 2
Z2 =
j6 x4
= −0.4615 + j 2.3077,
10 − j 2
( Z 3 + 8) //( Z 2 + 10) = (9.1538 + j 0.2308) //(9.5385 + j 2.3077) = 4.726 + j 0.6062
Z = 2 + Z 1 + 4.726 + j 0.6062 = 6.878 − j 0.163
Io =
60∠ − 30 o
60∠ − 30 o
=
= 8.721∠ − 28.64 o A
o
Z
6.88∠ − 1.3575
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
Z
V1
V2
+
−
V1 = I o (1 − j) = 2 (1 − j)
V2 = 2V1 = 4 (1 − j)
Vs = V1 + V2 = 6 (1 − j)
Vs = 8.485∠-45° V
Chapter 9, Solution 55.
12 Ω
-j20 V
I
I1
Z
I2
+
−
-j4 Ω
+
Vo
−
I1 =
Vo
4
= = -j0.5
j 8 j8
I2 =
I 1 (Z + j8) (-j0.5)(Z + j8) Z
=
= +j
- j4
- j4
8
I = I 1 + I 2 = -j0.5 +
Z
Z
+ j = + j0.5
8
8
- j20 = 12 I + I 1 (Z + j8)
Z j - j
- j20 = 12  +  + (Z + j8)
 8 2 2
j8 Ω
3
1
- 4 - j26 = Z  − j 
2
2
Z=
- 4 - j26 26.31∠261.25°
=
= 16.64∠279.68°
3
1 1.5811∠ - 18.43°
−j
2
2
Z = 2.798 – j16.403 Ω
Chapter 9, Solution 56.
3H

→
jωL = j 30
3F

→
1
= − j / 30
jω C
1.5F
1
= − j / 15
jω C

→
−j
15 = − j 0.06681
j 30 //( − j / 15) =
j
j 30 −
15
j 30 x
Z=
−j
− j 0.033(2 − j 0.06681)
//(2 − j 0.06681) =
= 6 − j 333 mΩ
30
− j 0.033 + 2 − j 0.06681
Chapter 9, Solution 57.
2H
1F

→

→
jωL = j 2
1
=−j
jω C
Z = 1 + j2 //( 2 − j) = 1 +
j2(2 − j)
= 2.6 + j1.2
j2 + 2 − j
Y = 1 = 0.3171 − j0.1463 S
Z
Chapter 9, Solution 58.
(a)
10 mF 
→
1
1
=
= - j2
jωC j (50)(10 × 10 -3 )
10 mH 
→
jωL = j (50)(10 × 10 -3 ) = j0.5
Z in = j0.5 + 1 || (1 − j2)
1 − j2
Z in = j0.5 +
2 − j2
Z in = j0.5 + 0.25 (3 − j)
Z in = 0.75 + j0.25 Ω
(b)
0.4 H 
→
jωL = j (50)(0.4) = j20
0.2 H 
→
jωL = j (50)(0.2) = j10
1
1
=
= - j20
jωC j (50)(1 × 10 -3 )
1 mF 
→
For the parallel elements,
1
1
1
1
=
+
+
Z p 20 j10 - j20
Z p = 10 + j10
Then,
Z in = 10 + j20 + Z p = 20 + j30 Ω
Chapter 9, Solution 59.
Z eq = 6 + (1 − j2) || (2 + j4)
Z eq = 6 +
(1 − j2)(2 + j4)
(1 − j2) + (2 + j4)
Z eq = 6 + 2.308 − j1.5385
Z eq = 8.308 – j1.5385 Ω
Chapter 9, Solution 60.
Z = (25 + j15) + (20 − j 50) //(30 + j10) = 25 + j15 + 26.097 − j 5.122 = 51.1 + j 9.878Ω
Chapter 9, Solution 61.
All of the impedances are in parallel.
1
1
1
1
1
=
+
+ +
Z eq 1 − j 1 + j2 j5 1 + j3
1
= (0.5 + j0.5) + (0.2 − j0.4) + (- j0.2) + (0.1 − j0.3) = 0.8 − j0.4
Z eq
Z eq =
1
= 1 + j0.5 Ω
0.8 − j0.4
Chapter 9, Solution 62.
2 mH 
→
jωL = j (10 × 10 3 )(2 × 10 -3 ) = j20
1
1
1 µF 
→
=
= - j100
3
jωC j (10 × 10 )(1 × 10 -6 )
50 Ω
+
1∠0° A
+
V
j20 Ω
−
Vin
+
−
-j100 Ω
V = (1∠0°)(50) = 50
Vin = (1∠0°)(50 + j20 − j100) + (2)(50)
Vin = 50 − j80 + 100 = 150 − j80
Z in =
Vin
= 150 – j80 Ω
1∠0°
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
200 + j450
z2 =
= 30 − j13.333, z3 =
= 10 + j22.5
j15
20
z1 =
8Ω
–j12 Ω
–j16 Ω
z2
10 Ω
z1
ZT
z3
–j16 Ω
10 Ω
Now all we need to do is to combine impedances.
z 2 (10 − j16) =
(30 − j13.333)(10 − j16)
= 8.721 − j8.938
40 − j29.33
z3 (10 − j16) = 21.70 − j3.821
ZT = 8 − j12 + z1 (8.721 − j8.938 + 21.7 − j3.821) = 34.69 − j6.93Ω
Chapter 9, Solution 64.
− j10(6 + j8)
= 19 − j5Ω
6 − j2
30∠90°
I=
= −0.3866 + j1.4767 = 1.527∠104.7° A
ZT
ZT = 4 +
Chapter 9, Solution 65.
Z T = 2 + (4 − j6) || (3 + j4)
ZT = 2 +
(4 − j6)(3 + j4)
7 − j2
Z T = 6.83 + j1.094 Ω = 6.917∠9.1° Ω
I=
V
120 ∠10°
=
= 17.35∠0.9° A
Z T 6.917 ∠9.1°
Chapter 9, Solution 66.
Z T = (20 − j5) || (40 + j10) =
(20 − j5)(40 + j10) 170
=
(12 − j)
60 + j5
145
Z T = 14.069 – j1.172 Ω = 14.118∠-4.76°
I=
V
60∠90°
=
= 4.25∠94.76°
Z T 14.118∠ - 4.76°
I
I1
I2
20 Ω
j10 Ω
+
I1 =
40 + j10
8 + j2
I=
I
60 + j5
12 + j
I2 =
20 − j5
4− j
I=
I
60 + j5
12 + j
Vab = -20 I 1 + j10 I 2
Vab
−
Vab =
- (160 + j40)
10 + j40
I+
I
12 + j
12 + j
Vab =
- 150
(-12 + j)(150)
I=
I
12 + j
145
Vab = (12.457 ∠175.24°)(4.25∠97.76°)
Vab = 52.94∠273° V
Chapter 9, Solution 67.
(a)
20 mH 
→
jωL = j (10 3 )(20 × 10 -3 ) = j20
1
1
12.5 µF 
→
=
= - j80
3
jωC j (10 )(12.5 × 10 -6 )
Z in = 60 + j20 || (60 − j80)
( j20)(60 − j80)
Z in = 60 +
60 − j60
Z in = 63.33 + j23.33 = 67.494 ∠20.22°
Yin =
(b)
1
= 0.0148∠-20.22° S
Z in
10 mH 
→
20 µF 
→
jωL = j (10 3 )(10 × 10 -3 ) = j10
1
1
=
= - j50
jωC j (10 3 )(20 × 10 -6 )
30 || 60 = 20
Z in = - j50 + 20 || (40 + j10)
(20)(40 + j10)
Z in = - j50 +
60 + j10
Z in = 13.5 − j48.92 = 50.75∠ - 74.56°
Yin =
1
= 0.0197∠74.56° S = 5.24 + j18.99 mS
Z in
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
1
1
1
= +
= (1 + j2)
Yo 4 - j2 4
Yo =
4
(4)(1 − j2)
=
= 0.8 − j1.6
1 + j2
5
Yo + j = 0.8 − j0.6
1
1 1
1
=
+
+
= (1) + ( j0.333) + (0.8 + j0.6)
Yo ′ 1 - j3 0.8 − j0.6
1
Yo ′
= 1.8 + j0.933 = 2.028∠27.41°
Yo ′ = 0.4932∠ - 27.41° = 0.4378 − j0.2271
Yo ′ + j5 = 0.4378 + j4.773
1
1
1
0.4378 − j4.773
= +
= 0.5 +
Yeq 2 0.4378 + j4.773
22.97
1
= 0.5191 − j0.2078
Yeq
Yeq =
0.5191 − j0.2078
= 1.661 + j0.6647 S
0.3126
Chapter 9, Solution 70.
Make a delta-to-wye transformation as shown in the figure below.
a
Zan
Zbn
Zeq
n
Zcn
b
c
8Ω
2Ω
-j5 Ω
Z an =
(- j10)(10 + j15)
(10)(15 − j10)
=
= 7 − j9
5 − j10 + 10 + j15
15 + j5
Z bn =
(5)(10 + j15)
= 4.5 + j3.5
15 + j5
Z cn =
(5)(- j10)
= -1 − j3
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 +
(6.5 + j3.5)(7 − j8)
13.5 − j4.5
Z eq = 7 − j9 + 5.511 − j0.2
Z eq = 12.51 − j9.2 = 15.53∠-36.33° Ω
Chapter 9, Solution 71.
We apply a wye-to-delta transformation.
j4 Ω
Zab
b
a
Zac
Zbc
Zeq
-j2 Ω
1Ω
c
Z ab =
2 − j2 + j4 2 + j2
=
= 1− j
j2
j2
Z ac =
2 + j2
= 1+ j
2
Z bc =
2 + j2
= -2 + j2
-j
j4 || Z ab = j4 || (1 − j) =
1 || Z ac = 1 || (1 + j) =
( j4)(1 − j)
= 1.6 − j0.8
1 + j3
(1)(1 + j)
= 0.6 + j0.2
2+ j
j4 || Z ab + 1 || Z ac = 2.2 − j0.6
1
1
1
1
=
+
+
Z eq - j2 - 2 + j2 2.2 − j0.6
= j0.5 − 0.25 − j0.25 + 0.4231 + j0.1154
= 0.173 + j0.3654 = 0.4043∠64.66°
Z eq = 2.473∠-64.66° Ω = 1.058 – j2.235 Ω
Chapter 9, Solution 72.
Transform the delta connections to wye connections as shown below.
a
j2 Ω
j2 Ω
-j18 Ω
-j9 Ω
j2 Ω
R1
R2
R3
b
- j9 || - j18 = - j6 ,
R1 =
(20)(20)
= 8 Ω,
20 + 20 + 10
R2 =
Z ab = j2 + ( j2 + 8) || (j2 − j6 + 4) + 4
Z ab = 4 + j2 + (8 + j2) || (4 − j4)
Z ab = 4 + j2 +
(8 + j2)(4 − j4)
12 - j2
Z ab = 4 + j2 + 3.567 − j1.4054
Z ab = 7.567 + j0.5946 Ω
(20)(10)
= 4Ω,
50
R3 =
(20)(10)
= 4Ω
50
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 Ω
j2 Ω
-j18 Ω
-j9 Ω
j2 Ω
R1
R2
R3
b
( j8)(- j6)
48
=
= - j4.8
j8 + j8 − j6 j10
Z 2 = Z1 = -j4.8
( j8)( j8) - 64
Z3 =
=
= j6.4
j10
j10
Z1 =
(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) = 46.4 + j9.6
46.4 + j9.6
= 1.5 − j7.25
j6.4
46.4 + j9.6
= 3.574 + j6.688
Zb =
4 − j4.8
46.4 + j9.6
= 1.727 + j8.945
Zc =
2 − j4.8
Za =
(6∠90°)(7.583∠61.88°)
= 07407 + j3.3716
3.574 + j12.688
(-j4)(1.5 − j7.25)
- j4 || Z a =
= 0.186 − j2.602
1.5 − j11.25
j6 || Z b =
j12 || Z c =
(12∠90°)(9.11∠79.07°)
= 0.5634 + j5.1693
1.727 + j20.945
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∠75.42° Ω = 0.3796 + j1.46 Ω
Chapter 9, Solution 74.
One such RL circuit is shown below.
20 Ω
V
20 Ω
+
+
j20 Ω
Vi = 1∠0°
j20 Ω
Vo
−
Z
We now want to show that this circuit will produce a 90° phase shift.
Z = j20 || (20 + j20) =
V=
( j20)(20 + j20) - 20 + j20
=
= 4 (1 + j3)
20 + j40
1 + j2
Z
4 + j12
1 + j3 1
Vi =
(1∠0°) =
= (1 + j)
Z + 20
24 + j12
6 + j3 3
Vo =
 j  1
j20
 (1 +
V =
20 + j20
1 + j  3
 j
j)  = = 0.3333∠90°
 3
This shows that the output leads the input by 90°.
Chapter 9, Solution 75.
Since cos(ωt ) = sin(ωt + 90°) , we need a phase shift circuit that will cause the
output to lead the input by 90°. This is achieved by the RL circuit shown
below, as explained in the previous problem.
10 Ω
10 Ω
+
+
j10 Ω
Vi
j10 Ω
−
Vo
−
This can also be obtained by an RC circuit.
Chapter 9, Solution 76.
Let Z = R – jX, where X =
| Z |= R 2 + X 2
X = | Z |2 − R 2 = 1162 = 662 = 95.394

→
C=
1
1
=
ωC 2πfC
1
1
= 27.81µF
=
2πfX 2πx 60x95.394
Chapter 9, Solution 77.
(a)
- jX c
V
R − jX c i
1
1
where X c =
=
= 3.979
ωC (2π)(2 × 10 6 )(20 × 10 -9 )
Vo =
Vo
- j3.979
=
=
Vi 5 - j3.979
Vo
=
Vi
3.979
25 + 15.83
3.979
5 + 3.979
2
2
∠(-90° + tan -1 (3.979 5))
∠(-90° − 38.51°)
Vo
= 0.6227 ∠ - 51.49°
Vi
Therefore, the phase shift is 51.49° lagging
(b)
θ = -45° = -90° + tan -1 (X c R )
45° = tan -1 (X c R ) 
→ R = X c =
ω = 2πf =
f=
1
ωC
1
RC
1
1
=
= 1.5915 MHz
2πRC (2π )(5)(20 × 10 -9 )
Chapter 9, Solution 78.
8+j6
R
Z
-jX
Z = R //[8 + j (6 − X )] =
R[8 + j (6 − X )]
=5
R + 8 + j (6 − X )
i.e 8R + j6R – jXR = 5R + 40 + j30 –j5X
Equating real and imaginary parts:
8R = 5R + 40 which leads to R=13.33Ω
6R-XR =30-5 which leads to X=4.125Ω.
Chapter 9, Solution 79.
(a)
Consider the circuit as shown.
20 Ω
V2
40 Ω
V1
30 Ω
+
Vi
+
j10 Ω
j30 Ω
−
j60 Ω
Vo
−
Z2
Z1
( j30)(30 + j60)
= 3 + j21
30 + j90
( j10)(43 + j21)
Z 2 = j10 || (40 + Z1 ) =
= 1.535 + j8.896 = 9.028∠80.21°
43 + j31
Z1 = j30 || (30 + j60) =
Let Vi = 1∠0° .
Z2
(9.028∠80.21°)(1∠0°)
Vi =
21.535 + j8.896
Z 2 + 20
V2 = 0.3875∠57.77°
V2 =
Z1
3 + j21
(21.213∠81.87°)(0.3875∠57.77°)
V2 =
V2 =
Z1 + 40
43 + j21
47.85∠26.03°
V1 = 0.1718∠113.61°
V1 =
j60
j2
2
V1 =
V1 = (2 + j)V1
30 + j60
1 + j2
5
Vo = (0.8944∠26.56°)(0.1718∠113.6°)
Vo = 0.1536∠140.2°
Vo =
Therefore, the phase shift is 140.2°
(b)
The phase shift is leading.
(c)
If Vi = 120 V , then
Vo = (120)(0.1536∠140.2°) = 18.43∠140.2° V
and the magnitude is 18.43 V.
Chapter 9, Solution 80.
200 mH 
→
Vo =
(a)
jωL = j (2π )(60)(200 × 10 -3 ) = j75.4 Ω
j75.4
j75.4
Vi =
(120∠0°)
R + 50 + j75.4
R + 50 + j75.4
When R = 100 Ω ,
j75.4
(75.4∠90°)(120∠0°)
(120 ∠0°) =
Vo =
150 + j75.4
167.88∠26.69°
Vo = 53.89∠63.31° V
(b)
When R = 0 Ω ,
j75.4
(75.4∠90°)(120 ∠0°)
(120∠0°) =
Vo =
50 + j75.4
90.47 ∠56.45°
Vo = 100∠33.55° V
(c)
To produce a phase shift of 45°, the phase of Vo = 90° + 0° − α = 45°.
Hence, α = phase of (R + 50 + j75.4) = 45°.
For α to be 45°,
R + 50 = 75.4
Therefore,
R = 25.4 Ω
Chapter 9, Solution 81.
Let
Z1 = R 1 ,
Z2 = R 2 +
1
,
jωC 2
Zx =
Z3
Z
Z1 2
Rx +
R3 
1
1 
R 2 +

=
jωC x R 1 
jωC 2 
Rx =
R3
1200
R2 =
(600) = 1.8 kΩ
R1
400
Z 3 = R 3 , and Z x = R x +
R1
 400 
1 R3   1 
(0.3 × 10 -6 ) = 0.1 µF
→ C x =
C2 = 
=   
 1200 
Cx  R1   C2 
R3
Chapter 9, Solution 82.
Cx =
R1
 100 
(40 × 10 -6 ) = 2 µF
Cs = 
 2000 
R2
Chapter 9, Solution 83.
Lx =
R2
 500 
(250 × 10 -3 ) = 104.17 mH
Ls = 
1200 
R1
1
.
jωC x
Chapter 9, Solution 84.
Let
1
Z2 = R 2 ,
,
jωC s
R1
jωC s
R1
Z1 =
=
1
jωR 1C s + 1
R1 +
jωC s
Z1 = R 1 ||
Since Z x =
Z 3 = R 3 , and Z x = R x + jωL x .
Z3
Z ,
Z1 2
R x + jωL x = R 2 R 3
jωR 1C s + 1 R 2 R 3
=
(1 + jωR 1C s )
R1
R1
Equating the real and imaginary components,
R 2R 3
Rx =
R1
ωL x =
R 2R 3
(ωR 1C s ) implies that
R1
L x = R 2 R 3Cs
Given that R 1 = 40 kΩ , R 2 = 1.6 kΩ , R 3 = 4 kΩ , and C s = 0.45 µF
R 2 R 3 (1.6)(4)
=
kΩ = 0.16 kΩ = 160 Ω
R1
40
L x = R 2 R 3 C s = (1.6)(4)(0.45) = 2.88 H
Rx =
Chapter 9, Solution 85.
Let
1
,
jωC 2
R4
- jR 4
Z4 =
=
jωR 4 C 4 + 1 ωR 4 C 4 − j
Z1 = R 1 ,
Since Z 4 =
Z3
Z
Z1 2
Z2 = R 2 +

→ Z1 Z 4 = Z 2 Z 3 ,
Z 3 = R 3 , and Z 4 = R 4 ||
1
.
jωC 4

- jR 4 R 1
j 

= R 3 R 2 −
ωR 4 C 4 − j
ωC 2 

jR 3
- jR 4 R 1 (ωR 4 C 4 + j)
= R 3R 2 −
2
2 2
ω R 4C4 + 1
ωC 2
Equating the real and imaginary components,
R 1R 4
= R 2R 3
2
ω R 24 C 24 + 1
(1)
2
R3
ωR 1 R 4 C 4
=
2
2 2
ω R 4 C 4 + 1 ωC 2
(2)
Dividing (1) by (2),
1
= ωR 2 C 2
ωR 4 C 4
1
ω2 =
R 2C2R 4C4
1
ω = 2πf =
R 2C2 R 4C4
f=
1
2π R 2 R 4 C 2 C 4
Chapter 9, Solution 86.
Y=
1
1
1
+
+
240 j95 - j84
Y = 4.1667 × 10 -3 − j0.01053 + j0.0119
Z=
1
1000
1000
=
=
Y 4.1667 + j1.37 4.3861∠18.2°
Z = 228∠-18.2° Ω
Chapter 9, Solution 87.
Z1 = 50 +
-j
1
= 50 +
(2π)(2 × 10 3 )(2 × 10 -6 )
jωC
Z1 = 50 − j39.79
Z 2 = 80 + jωL = 80 + j (2π)(2 × 10 3 )(10 × 10 -3 )
Z 2 = 80 + j125.66
Z 3 = 100
1
1
1
1
=
+
+
Z Z1 Z 2 Z 3
1
1
1
1
=
+
+
Z 100 50 − j39.79 80 + j125.66
1
= 10 -3 (10 + 12.24 + j9.745 + 3.605 − j5.663)
Z
= (25.85 + j4.082) × 10 -3
= 26.17 × 10 -3 ∠8.97°
Z = 38.21∠-8.97° Ω
Chapter 9, Solution 88.
(a)
(b)
Z = - j20 + j30 + 120 − j20
Z = 120 – j10 Ω
1
1
=
would cause the capacitive
ωC 2πf C
impedance to double, while ωL = 2πf L would cause the inductive
impedance to halve. Thus,
Z = - j40 + j15 + 120 − j40
Z = 120 – j65 Ω
If the frequency were halved,
Chapter 9, Solution 89.

1 

Z in = jωL ||  R +
jωC 


1 
L

jωL  R +
+ jωL R
jωC 

C
=
Z in =
1

1 
R + jωL +

R + jωL −
jωC

ωC 
Z in =


L
1 

 + jωL R  R − jωL −


C
ωC 

1 

R + ωL −

ωC 
2
2
To have a resistive impedance, Im(Z in ) = 0 . Hence,
 L 
1 
=0
ωL R 2 −  ωL −
 C 
ωC 
ωR 2 C = ωL −
1
ωC
ω2 R 2 C 2 = ω2 LC − 1
ω2 R 2 C 2 + 1
L=
ω2 C
(1)
Ignoring the +1 in the numerator in (1),
L = R 2 C = (200) 2 (50 × 10 -9 ) = 2 mH
Chapter 9, Solution 90.
Let
Vs = 145∠0° ,
I=
X = jωL = j (2π)(60) L = j377 L
Vs
145∠0°
=
80 + R + jX 80 + R + jX
V1 = 80 I =
50 =
(80)(145)
80 + R + jX
(80)(145)
80 + R + jX
Vo = (R + jX) I =
110 =
(1)
(R + jX)(145∠0°)
80 + R + jX
(R + jX)(145)
80 + R + jX
(2)
From (1) and (2),
50
80
=
110
R + jX
11 
R + jX = (80)  
5
R 2 + X 2 = 30976
From (1),
(80)(145)
80 + R + jX =
= 232
50
(3)
6400 + 160R + R 2 + X 2 = 53824
160R + R 2 + X 2 = 47424
(4)
Subtracting (3) from (4),
160R = 16448 
→ R = 102.8 Ω
From (3),
X 2 = 30976 − 10568 = 20408
X = 142.86 = 377 L 
→ L = 0.3789 H
Chapter 9, Solution 91.
Z in =
1
+ R || jωL
jωC
Z in =
-j
jωLR
+
ωC R + jωL
- j ω 2 L2 R + jωLR 2
=
+
ωC
R 2 + ω 2 L2
To have a resistive impedance, Im(Z in ) = 0 .
Hence,
-1
ωLR 2
=0
+ 2
ωC R + ω2 L2
1
ωLR 2
= 2
ωC R + ω2 L2
R 2 + ω2 L2
C=
ω2 LR 2
where ω = 2π f = 2π × 10 7
C=
9 × 10 4 + (4π 2 × 1014 )(400 × 10 −12 )
(4π 2 × 1014 )(20 × 10 − 6 )(9 × 10 4 )
C=
9 + 16π 2
nF
72π 2
C = 235 pF
Chapter 9, Solution 92.
(a) Z o =
Z
=
Y
100∠75 o
= 471.4∠13.5 o Ω
o
−6
450∠48 x10
(b) γ = ZY = 100∠75 o x 450∠48 o x10 −6 = 0.2121∠61.5 o
Chapter 9, Solution 93.
Z = Zs + 2 ZA + ZL
Z = (1 + 0.8 + 23.2) + j(0.5 + 0.6 + 18.9)
Z = 25 + j20
IL =
VS
115∠0°
=
Z 32.02 ∠38.66°
I L = 3.592∠-38.66° A
Chapter 10, Solution 1.
ω=1
10 cos( t − 45°) 
→ 10∠ - 45°
5 sin( t + 30°) 
→ 5∠ - 60°
1H 
→
1F 
→
jωL = j
1
= -j
jωC
The circuit becomes as shown below.
3Ω
10∠-45° V
+
−
Vo
jΩ
2 Io
+
−
5∠-60° V
Applying nodal analysis,
(10∠ - 45°) − Vo (5∠ - 60°) − Vo Vo
+
=
3
j
-j
j10∠ - 45° + 15∠ - 60° = j Vo
Vo = 10 ∠ - 45° + 15∠ - 150° = 15.73∠247.9°
Therefore,
v o ( t ) = 15.73 cos(t + 247.9°) V
Chapter 10, Solution 2.
ω = 10
4 cos(10t − π 4) 
→ 4∠ - 45°
20 sin(10 t + π 3) 
→ 20 ∠ - 150°
1H 
→
jωL = j10
1
1
0.02 F 
→
=
= - j5
jωC j 0.2
The circuit becomes that shown below.
10 Ω
Vo
Io
20∠-150° V
+
−
j10 Ω
4∠-45° A
-j5 Ω
Applying nodal analysis,
Vo Vo
(20∠ - 150°) − Vo
+ 4∠ - 45° =
+
10
j10 - j5
20 ∠ - 150° + 4∠ - 45° = 0.1(1 + j) Vo
Io =
Therefore,
Vo 2 ∠ - 150° + 4 ∠ - 45°
= 2.816 ∠150.98°
=
j10
j (1 + j)
i o ( t ) = 2.816 cos(10t + 150.98°) A
Chapter 10, Solution 3.
ω= 4
2 cos(4t ) 
→ 2∠0°
16 sin(4 t ) 
→ 16∠ - 90° = -j16
2H 
→
jωL = j8
1
1
1 12 F 
→
=
= - j3
jωC j (4)(1 12)
The circuit is shown below.
4Ω
-j16 V
+
−
-j3 Ω
Vo
1Ω
j8 Ω
2∠0° A
6Ω
Applying nodal analysis,
Vo
Vo
- j16 − Vo
+2=
+
4 − j3
1 6 + j8

- j16
1
1 
V
+ 2 = 1 +
+
4 − j3
 4 − j3 6 + j8  o
Vo =
3.92 − j2.56 4.682∠ - 33.15°
=
= 3.835∠ - 35.02°
1.22 + j0.04
1.2207 ∠1.88°
v o ( t ) = 3.835 cos(4t – 35.02°) V
Therefore,
Chapter 10, Solution 4.
16 sin(4 t − 10°) 
→ 16∠ - 10°, ω = 4
1H 
→
jωL = j4
0.25 F 
→
Ix
16∠-10° V
+
−
1
1
=
= -j
jωC j (4)(1 4)
j4 Ω
V1
-j Ω
+
0.5 Ix
1Ω
Vo
−
(16∠ - 10°) − V1 1
V
+ Ix = 1
j4
2
1− j
But
Ix =
So,
(16∠ - 10°) − V1
j4
3 ((16∠ - 10°) − V1 )
V
= 1
j8
1− j
V1 =
48∠ - 10°
- 1 + j4
Using voltage division,
1
48∠ - 10°
Vo =
V1 =
= 8.232∠ - 69.04°
1− j
(1 - j)(-1 + j4)
v o ( t ) = 8.232 sin(4t – 69.04°) V
Therefore,
Chapter 10, Solution 5.
Let the voltage across the capacitor and the inductor be Vx and we get:
Vx − 0.5I x − 10∠30° Vx Vx
+
+
=0
4
− j2 j3
(3 + j6 − j4)Vx − 1.5I x = 30∠30° but I x =
Vx
= j0.5Vx
− j2
Combining these equations we get:
(3 + j2 − j0.75)Vx = 30∠30° or Vx =
I x = j0.5
30∠30°
3 + j1.25
30∠30°
= 4.615∠97.38° A
3 + j1.25
Chapter 10, Solution 6.
Let Vo be the voltage across the current source. Using nodal analysis we get:
Vo − 4Vx
Vo
20
−3+
= 0 where Vx =
Vo
20 + j10
20
20 + j10
Combining these we get:
Vo
4Vo
Vo
−
−3+
= 0 → (1 + j0.5 − 3)Vo = 60 + j30
20 20 + j10
20 + j10
Vo =
60 + j30
20(3)
or Vx =
= 29.11∠–166˚ V.
− 2 + j0.5
− 2 + j0.5
Chapter 10, Solution 7.
At the main node,
120∠ − 15 o − V
V
V
= 6∠30 o +
+
− j30 50
40 + j20

→
115.91 − j31.058
− 5.196 − j3 =
40 + j20

1
j
1
+
+ 
V
 40 + j20 30 50 
V=
− 3.1885 − j4.7805
= 124.08∠ − 154 o V
0.04 + j0.0233
Chapter 10, Solution 8.
ω = 200,
100mH
50µF

→

→
jωL = j200x 0.1 = j20
1
1
=
= − j100
jωC j200x 50x10 − 6
The frequency-domain version of the circuit is shown below.
0.1 Vo
40 Ω
V1
6∠15
o
20 Ω
+
Vo
-
Io
V2
-j100 Ω
j20 Ω
At node 1,
or
V
V1
V − V2
6∠15 o + 0.1V1 = 1 +
+ 1
20 − j100
40
5.7955 + j1.5529 = (−0.025 + j 0.01)V1 − 0.025V2
(1)
At node 2,
V
V1 − V2
= 0.1V1 + 2
j20
40
From (1) and (2),
→
0 = 3V1 + (1 − j2)V2
(−0.025 + j0.01) − 0.025 V1   (5.7955 + j1.5529) 
 =


3
(1 − j2)  V2  
0


or
(2)
AV = B
Using MATLAB,
V = inv(A)*B
V2 = −110.3 + j161.09
leads to V1 = −70.63 − j127.23,
V − V2
Io = 1
= 7.276∠ − 82.17 o
40
Thus,
i o ( t ) = 7.276 cos(200 t − 82.17 o ) A
Chapter 10, Solution 9.
10 cos(10 3 t ) 
→ 10 ∠0°, ω = 10 3
10 mH 
→
jωL = j10
50 µF 
→
1
1
=
= - j20
3
jωC j (10 )(50 × 10 -6 )
Consider the circuit shown below.
20 Ω
V1
-j20 Ω
V2
j10 Ω
Io
10∠0° V
+
−
20 Ω
+
4 Io
30 Ω
Vo
−
At node 1,
10 − V1 V1 V1 − V2
=
+
20
20
- j20
10 = (2 + j) V1 − jV2
(1)
At node 2,
V1 − V2
V
V2
V
, where I o = 1 has been substituted.
= (4) 1 +
20
- j20
20 30 + j10
(-4 + j) V1 = (0.6 + j0.8) V2
V1 =
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
3
170
V2 =
⋅
= 6.154 ∠70.26°
30 + j10
3 + j 0.6 − j26.2
v o ( t ) = 6.154 cos(103 t + 70.26°) V
Therefore,
Chapter 10, Solution 10.
50 mH
2µF

→

→
jωL = j2000x50 x10 − 3 = j100,
ω = 2000
1
1
=
= − j250
jωC j2000 x 2x10 − 6
Consider the frequency-domain equivalent circuit below.
V1
36<0o
2k Ω
-j250
j100
V2
0.1V1
4k Ω
At node 1,
36 =
V1
V
V − V2
+ 1 + 1
2000 j100 − j250

→
36 = (0.0005 − j0.006)V1 − j0.004V2
(1)
At node 2,
V
V1 − V2
= 0.1V1 + 2
4000
− j250
→
0 = (0.1 − j0.004)V1 + (0.00025 + j0.004)V2 (2)
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 ) 
→ 1∠0°, ω = 2
8 sin( 2t + 30°) 
→ 8∠ - 60°
1H 
→
jωL = j2
12F 
→
1
1
=
= -j
jωC j (2)(1 2)
2H 
→
jωL = j4
14F 
→
1
1
=
= - j2
jωC j (2)(1 4)
Consider the circuit below.
2 Io
2 Io
2 Io
2 -j Ω
2
2 Io
2
I
-j Ω
2
2 Io
At node 1,
(8∠ - 60°) − V1 V1 V1 − V2
=
+
2
-j
j2
8∠ - 60° = (1 + j) V1 + j V2
(1)
At node 2,
1+
V1 − V2 (8∠ - 60°) − V2
=0
+
j2
j4 − j2
V2 = 4 ∠ - 60° + j + 0.5 V1
Substituting (2) into (1),
1 + 8∠ - 60° − 4 ∠30° = (1 + j1.5) V1
Therefore,
(2)
V1 =
1 + 8∠ - 60° − 4∠30°
1 + j1.5
Io =
V1 1 + 8∠ - 60° − 4 ∠30°
= 5.024∠ - 46.55°
=
-j
1.5 − j
i o ( t ) = 5.024 cos(2t – 46.55°)
Chapter 10, Solution 12.
20 sin(1000t ) 
→ 20 ∠0°, ω = 1000
10 mH 
→
jωL = j10
50 µF 
→
1
1
=
= - j20
3
jωC j (10 )(50 × 10 -6 )
The frequency-domain equivalent circuit is shown below.
2 Io
V1
10 Ω
V2
Io
20∠0° A
20 Ω
-j20 Ω
j10 Ω
At node 1,
20 = 2 I o +
V1 V1 − V2
+
,
20
10
where
Io =
V2
j10
20 =
2V2 V1 V1 − V2
+
+
j10 20
10
400 = 3V1 − (2 + j4) V2
(1)
At node 2,
2V2 V1 − V2
V
V
+
= 2 + 2
j10
10
- 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 = (1 + j0.5) V2
Therefore,
V2 =
400
1 + j0.5
Io =
V2
40
=
= 35.74 ∠ - 116.6°
j10 j (1 + j0.5)
(2)
i o ( t ) = 35.74 sin(1000t – 116.6°) A
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 Ω
40∠30º V
+
−
18 Ω
j6 Ω
+
Vx
−
3Ω
50∠0º V
+
−
Vx − 40∠30° Vx Vx − 50
+
+
=0
− j2
3
18 + j6
which leads to Vx = 29.36∠62.88˚ A.
Chapter 10, Solution 14.
At node 1,
0 − V1 0 − V1 V2 − V1
+
+
= 20∠30°
- j2
10
j4
- (1 + j2.5) V1 − j2.5 V2 = 173.2 + j100
(1)
At node 2,
V2 V2 V2 − V1
+
+
= 20∠30°
j2 - j5
j4
- j5.5 V2 + j2.5 V1 = 173.2 + j100
Equations (1) and (2) can be cast into matrix form as
1 + j2.5 j2.5  V1   - 200 ∠30°
=
 j2.5
- j5.5 V2   200 ∠30° 

∆=
1 + j2.5 j2.5
= 20 − j5.5 = 20.74∠ - 15.38°
j2.5
- j5.5
∆1 =
∆2 =
- 200 ∠30° j2.5
= j3 (200∠30°) = 600∠120°
200 ∠30° - j5.5
1 + j2.5 - 200∠30°
j2.5
200∠30°
V1 =
∆1
= 28.93∠135.38°
∆
V2 =
∆2
= 49.18∠124.08°
∆
= (200 ∠30°)(1 + j5) = 1020∠108.7°
(2)
Chapter 10, Solution 15.
We apply nodal analysis to the circuit shown below.
5A
2Ω
V2
I
+
−
-j20 V
jΩ
V1
-j2 Ω
2I
4Ω
At node 1,
V
V − V2
- j20 − V1
= 5+ 1 + 1
2
- j2
j
- 5 − j10 = (0.5 − j0.5) V1 + j V2
At node 2,
5 + 2I +
V1 − V2 V2
,
=
j
4
where I =
V2 =
V1
- j2
5
V1
0.25 − j
(2)
Substituting (2) into (1),
- 5 − j10 −
j5
= 0.5 (1 − j) V1
0.25 − j
(1 − j) V1 = -10 − j20 −
j40
1 − j4
( 2 ∠ - 45°) V1 = -10 − j20 +
V1 = 15.81∠313.5°
160 j40
−
17 17
(1)
I=
V1
= (0.5∠90°)(15.81∠313.5°)
- j2
I = 7.906∠43.49° A
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
V
+
+1+ j = 2
10
- j5
j10
10 (1 + j) = - (1 + j2) V1 + (1 + j) V2
Thus,
 j40
10 (1 +

  3 + j4 - 2 (1 + j2)  V1 
=
j)   - (1 + j2)
1 + j  V2 
∆=
3 + j4 - 2 (1 + j2)
= 5 − j = 5.099 ∠ - 11.31°
- (1 + j2)
1+ j
∆1 =
j40
- 2 (1 + j2)
= −60 + j100 = 116.62 ∠120.96°
10 (1 + j)
1+ j
∆2 =
3 + j4
j40
- (1 + j2) 10 (1 + j)
= -90 + j110 = 142.13∠129.29°
∆1
= 22.87∠132.27° V
∆
∆2
V2 =
= 27.87∠140.6° V
∆
V1 =
Chapter 10, Solution 17.
Consider the circuit below.
j4 Ω
100∠20° V
1Ω
Io
+
−
2Ω
V1
V2
3Ω
-j2 Ω
At node 1,
100∠20° − V1 V1 V1 − V2
=
+
j4
3
2
100 ∠20° =
V1
(3 + j10) − j2 V2
3
(1)
At node 2,
100∠20° − V2 V1 − V2 V2
+
=
1
2
- j2
100 ∠20° = -0.5 V1 + (1.5 + j0.5) V2
From (1) and (2),
100∠20°  - 0.5
0.5 (3 + j)  V1 
=
100∠20° 1 + j10 3
- j2  V2 

 
∆=
- 0.5
1.5 + j0.5
= 0.1667 − j4.5
1 + j10 3
- j2
∆1 =
∆2 =
100∠20° 1.5 + j0.5
= -55.45 − j286.2
100∠20°
- j2
- 0.5
100∠20°
1 + j10 3 100∠20°
= -26.95 − j364.5
(2)
V1 =
∆1
= 64.74 ∠ - 13.08°
∆
V2 =
∆2
= 81.17 ∠ - 6.35°
∆
Io =
V1 − V2 ∆ 1 − ∆ 2 - 28.5 + j78.31
=
=
2
2∆
0.3333 − j 9
I o = 9.25∠-162.12°
Chapter 10, Solution 18.
Consider the circuit shown below.
8Ω
V1
j6 Ω V
2
4Ω
j5 Ω
+
4∠45° A
2Ω
+
2 Vx
Vx
-j Ω
-j2 Ω
−
−
At node 1,
4∠45° =
V1 V1 − V2
+
2
8 + j6
200 ∠45° = (29 − j3) V1 − (4 − j3) V2
(1)
At node 2,
V1 − V2
V
V2
,
+ 2Vx = 2 +
8 + j6
- j 4 + j5 − j2
where Vx = V1
(104 − j3) V1 = (12 + j41) V2
12 + j41
V
104 − j3 2
Substituting (2) into (1),
V1 =
200∠45° = (29 − j3)
Vo
(2)
(12 + j41)
V − (4 − j3) V2
104 − j3 2
200 ∠45° = (14.21∠89.17°) V2
V2 =
200∠45°
14.21∠89.17°
Vo =
- j2
- j2
- 6 − j8
V2 =
V2 =
V2
4 + j5 − j2
4 + j3
25
Vo =
10∠233.13°
200∠45°
⋅
25
14.21∠89.17°
Vo = 5.63∠189° V
Chapter 10, Solution 19.
We have a supernode as shown in the circuit below.
j2 Ω
V1
V2
4Ω
V3
+
2Ω
Vo
-j4 Ω
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 V3 − V2
=
j2
4
(0.8 − j2) V1 + V2 + (-1 + j2) V3 = 0
Subtracting (2) from (1),
(2)
0 = 1.2V1 + j V2
But at the supernode,
V1 = 12 ∠0° + V2
V2 = V1 − 12
or
Substituting (4) into (3),
0 = 1.2V1 + j (V1 − 12)
V1 =
j12
= Vo
1.2 + j
Vo =
12∠90°
1.562∠39.81°
(3)
(4)
Vo = 7.682∠50.19° V
Chapter 10, Solution 20.
The circuit is converted to its frequency-domain equivalent circuit as shown below.
R
+
Vm∠0°
+
−
jωL
Vo
1
jωC
−
Let
Z = jωL ||
1
=
jωC
L
C
1
jωL +
jωC
=
jωL
1 − ω2 LC
jωL
Z
jωL
1 − ω2 LC
Vo =
Vm =
Vm =
V
jωL
R+Z
R (1 − ω2 LC) + jωL m
R+
1 − ω2 LC
Vo =


ωL

90° − tan -1
∠
2
R (1 − ω LC) 
R 2 (1 − ω2 LC) 2 + ω2 L2 
ωL Vm
If
Vo = A∠φ , then
A=
and
ωL Vm
R 2 (1 − ω 2 LC) 2 + ω 2 L2
φ = 90° − tan -1
ωL
R (1 − ω 2 LC)
Chapter 10, Solution 21.
(a)
Vo
=
Vi
1
jωC
R + jωL +
1
jωC
As ω → ∞ ,
(b)
Vo
=
Vi
1
LC
Vo
=
Vi
,
jωL
R + jωL +
As ω → ∞ ,
1
LC
1
jωC
1
jRC ⋅
1
=
-j L
R C
LC
− ω2 LC
=
1 − ω2 LC + jωRC
Vo
= 0
Vi
Vo 1
= = 1
Vi 1
At ω = 0 ,
At ω =
1
1 − ω LC + jωRC
2
Vo 1
= = 1
Vi 1
Vo
= 0
Vi
At ω = 0 ,
At ω =
=
,
Vo
=
Vi
−1
jRC ⋅
1
LC
=
j L
R C
Chapter 10, Solution 22.
Consider the circuit in the frequency domain as shown below.
R1
R2
Vs
+
−
1
jωC
jωL
Let
Z = (R 2 + jωL) ||
+
Vo
−
1
jωC
1
(R + jωL)
R 2 + jωL
jωC 2
Z=
=
1
1 + jωR 2 − ω2 LC
R 2 + jωL +
jωC
R 2 + jωL
Vo
1 − ω2 LC + jωR 2 C
Z
=
=
R 2 + jωL
Vs Z + R 1
R1 +
1 − ω2 LC + jωR 2 C
Vo
R 2 + jωL
=
2
Vs R 1 + R 2 − ω LCR 1 + jω (L + R 1 R 2 C)
Chapter 10, Solution 23.
V − Vs
V
+
+ jωCV = 0
1
R
jωL +
jω C
V+
jωRCV
− ω2LC + 1
+ jωRCV = Vs
 1 − ω2LC + jωRC + jωRC − jω3RLC2 

 V = Vs
2


−
ω
1
LC


V=
(1 − ω2 LC)Vs
1 − ω2LC + jωRC(2 − ω2LC)
Chapter 10, Solution 24.
For mesh 1,
 1
1 
1
 I1 −
+
Vs = 
I
jωC 2 2
 jωC1 jωC 2 
(1)
For mesh 2,

1 
−1
I
I 1 +  R + jωL +
jωC 2
jωC 2  2

Putting (1) and (2) into matrix form,
−1
1
 1

+

  I1 
Vs 
jωC1 jωC 2
jωC 2


=
0
−1
1  I 2 
  
R + jωL +

jω C 2
jωC 2 
0=
(2)
 1
1 
1 
1
+ 2
 R + jωL +
∆ =
+
jωC 2  ω C1C 2
 jωC1 jωC 2 

1 

∆ 1 = Vs  R + jωL +
jωC 2 

and

1 

Vs  R + jωL +
jωC 2 

∆1
I1 =
=
∆  1
1 
1
1 

 R + jωL +
+ 2
+
jωC 2  ω C1 C 2
 jωC 1 jωC 2 
I2 =
Vs
jωC 2
∆2
=
∆  1
1 
1 
1
+ 2

 R + jωL +
+
jωC 2  ω C1 C 2
 jωC 1 jωC 2 
Chapter 10, Solution 25.
ω= 2
10 cos(2t ) 
→ 10∠0°
∆2 =
Vs
jωC 2
6 sin(2t ) 
→ 6 ∠ - 90° = -j6
2H 
→
jωL = j4
1
1
0.25 F 
→
=
= - j2
jωC j (2)(1 4)
The circuit is shown below.
4Ω
j4 Ω
Io
10∠0° V
+
−
I1
-j2 Ω
I2
+
−
6∠-90° V
For loop 1,
- 10 + (4 − j2) I 1 + j2 I 2 = 0
5 = (2 − j) I 1 + j I 2
(1)
For loop 2,
j2 I 1 + ( j4 − j2) I 2 + (- j6) = 0
I1 + I 2 = 3
In matrix form (1) and (2) become
 2 − j j   I 1   5
 1 1   I  =  3
 2   

∆ = 2 (1 − j) ,
∆ 1 = 5 − j3 ,
(2)
∆ 2 = 1 − j3
∆1 − ∆ 2
4
=
= 1 + j = 1.414 ∠45°
2 (1 − j)
∆
i o ( t ) = 1.414 cos(2t + 45°) A
I o = I1 − I 2 =
Therefore,
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 = 0
- 2 I 1 + (2 − j2) I 2 + (3 + j) I 3 = 0
(1)
(2)
At node A,
I o = I1 − I 2
(3)
At node B,
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),
1  
4
- 2  I 1 
 0 =  - (14 + j4) 17 + j3 I 
  
 2 
(4)
(5)
(6)
∆ = 40 + j4
∆1 =
1
-2
0 17 + j3
I 3 = 5 I 2 − 4 I1 =
Vo = 30 I 3 =
Therefore,
= 17 + j3 ,
∆2 =
4
1
- (14 + j4) 0
= 14 + j4
5 ∆ 2 − 4 ∆1
2 + j8
=
∆
40 + j4
15 (1 + j4)
= 6.154∠70.25°
10 + j
v o ( t ) = 6.154 cos(103 t + 70.25°) V
Chapter 10, Solution 27.
For mesh 1,
- 40 ∠30° + ( j10 − j20) I 1 + j20 I 2 = 0
4 ∠30° = - j I 1 + j2 I 2
(1)
For mesh 2,
50 ∠0° + (40 − j20) I 2 + j20 I 1 = 0
5 = - j2 I 1 − (4 − j2) I 2
From (1) and (2),
 4∠30°  - j
j2  I 1 
=
 5   - j2 - (4 − j2)  I 

 
 2 
∆ = -2 + 4 j = 4.472∠116.56°
(2)
∆ 1 = -(4 ∠30°)(4 − j2) − j10 = 21.01∠211.8°
∆ 2 = - j5 + 8∠120° = 4.44 ∠154.27°
I1 =
∆1
= 4.698∠95.24° A
∆
I2 =
∆2
= 0.9928∠37.71° A
∆
Chapter 10, Solution 28.
1H

→
jωL = j4,

→
1F
1
1
=
= − j0.25
jωC j1x 4
The frequency-domain version of the circuit is shown below, where
V1 = 10∠0 o ,
V2 = 20∠ − 30 o .
1
j4
j4
1
-j0.25
+
+
V1
-
I1
V1 = 10∠0 o ,
1
I2
V2
-
V2 = 20∠ − 30 o
Applying mesh analysis,
10 = (2 + j3.75)I1 − (1 − j0.25)I 2
(1)
− 20∠ − 30 o = −(1 − j0.025)I1 + (2 + j3.75)I 2
(2)
From (1) and (2), we obtain
10
  2 + j3.75 − 1 + j0.25  I1 

 
 = 

 − 17.32 + j10   − 1 + j0.25 2 + j3.75  I 2 
Solving this leads to
I1 = 1.3602 − j0.9769 = 1.6747∠ − 35.69 o ,
I 2 = −4.1438 + j2.111 = 4.6505∠153o
Hence,
i1 = 1.675 cos(4t − 35.69 o ) A,
i 2 = 4.651cos(46 + 153o ) A
Chapter 10, Solution 29.
For mesh 1,
(5 + j5) I 1 − (2 + j) I 2 − 30 ∠20° = 0
30 ∠20° = (5 + j5) I 1 − (2 + j) I 2
(1)
For mesh 2,
(5 + j3 − j6) I 2 − (2 + j) I 1 = 0
0 = - (2 + j) I 1 + (5 − j3) I 2
From (1) and (2),
30∠20°  5 + j5 - (2 + j)  I 1 
 0  =  - (2 + j) 5 - j3  I 

 
 2 
∆ = 37 + j6 = 37.48∠9.21°
∆ 1 = (30 ∠20°)(5.831∠ - 30.96°) = 175∠ - 10.96°
∆ 2 = (30 ∠20°)(2.356 ∠26.56°) = 67.08∠46.56°
I1 =
∆1
= 4.67∠-20.17° A
∆
I2 =
∆2
= 1.79∠37.35° A
∆
(2)
Chapter 10, Solution 30.
Consider the circuit shown below.
I2
j4 Ω
10∠0° V
+
−
1Ω
Io
I1
3Ω
2Ω
I3
-j2 Ω
For mesh 1,
100 ∠20° = (3 + j4) I 1 − j4 I 2 − 3 I 3
(1)
0 = - j4 I 1 + (3 + j4) I 2 − j2 I 3
(2)
For mesh 2,
For mesh 3,
0 = -3 I 1 − 2 I 2 + (5 − j2) I 3
Put (1), (2), and (3) into matrix form.
3 + j4 - j4
- 3  I 1  100∠20°

  

j4
3
+
j4
j2
0
=
I
2






 - 3

-2
5 - j2  I 3  
0
3 + j4 - j4
-3
∆ = - j4 3 + j4 - j2 = 106 + j30
-3
-2
5 - j2
3 + j4 100∠20° - 3
∆ 2 = - j4
0
- j2 = (100∠20°)(8 + j26)
-3
0
5 - j2
3 + j4 - j4 100∠20°
∆ 3 = - j4 3 + j4
0
= (100∠20°)(9 + j20)
-3
-2
0
Io = I3 − I2 =
∆ 3 − ∆ 2 (100∠20°)(1 − j6)
=
∆
106 + j30
I o = 5.521∠-76.34° A
(3)
Chapter 10, Solution 31.
Consider the network shown below.
80 Ω
100∠120° V
+
−
I1
Io
-j40 Ω
j60 Ω
I2
-j40 Ω
20 Ω
I3
+
−
60∠-30° V
For loop 1,
- 100 ∠20° + (80 − j40) I 1 + j40 I 2 = 0
10 ∠20° = 4 (2 − j) I 1 + j4 I 2
(1)
j40 I 1 + ( j60 − j80) I 2 + j40 I 3 = 0
0 = 2 I1 − I 2 + 2 I 3
(2)
60 ∠ - 30° + (20 − j40) I 3 + j40 I 2 = 0
- 6 ∠ - 30° = j4 I 2 + 2 (1 − j2) I 3
(3)
For loop 2,
For loop 3,
From (2),
2 I 3 = I 2 − 2 I1
Substituting this equation into (3),
- 6 ∠ - 30° = -2 (1 − j2) I 1 + (1 + j2) I 2
(4)
From (1) and (4),
 10∠120°   4 (2 − j)
j4  I 1 
 - 6∠ - 30° =  - 2 (1 − j2) 1 + j2 I 
 2 
 

∆=
∆2 =
8 − j4
- j4
= 32 + j20 = 37.74∠32°
- 2 + j4 1 + j2
8 − j4 10∠120°
= -4.928 + j82.11 = 82.25∠93.44°
- 2 + j4 - 6∠ - 30°
Io = I2 =
∆2
= 2.179∠61.44° A
∆
Chapter 10, Solution 32.
Consider the circuit below.
j4 Ω
Io
+
2Ω
4∠-30° V
Vo
I1
+
−
3 Vo
−
For mesh 1,
where
(2 + j4) I 1 − 2 (4∠ - 30°) + 3 Vo = 0
Vo = 2 (4∠ - 30° − I 1 )
Hence,
(2 + j4) I 1 − 8∠ - 30° + 6 (4 ∠ - 30° − I 1 ) = 0
4 ∠ - 30° = (1 − j) I 1
I 1 = 2 2 ∠15°
or
Io =
3 Vo
3
=
(2)(4∠ - 30° − I 1 )
- j2 - j2
I o = j3 (4 ∠ - 30° − 2 2 ∠15°)
I o = 8.485∠15° A
Vo =
- j2 I o
= 5.657∠-75° V
3
Chapter 10, Solution 33.
Consider the circuit shown below.
I2
-j2 Ω
5A
I4
2Ω
jΩ
I
-j20 V
+
−
I1
I2
-j2 Ω
2I
I3
4Ω
For mesh 1,
j20 + (2 − j2) I 1 + j2 I 2 = 0
(1 − j) I 1 + j I 2 = - j10
(1)
For the supermesh,
( j − j2) I 2 + j2 I 1 + 4 I 3 − j I 4 = 0
(2)
Also,
I 3 − I 2 = 2 I = 2 (I 1 − I 2 )
I 3 = 2 I1 − I 2
(3)
I4 = 5
(4)
For mesh 4,
Substituting (3) and (4) into (2),
(8 + j2) I 1 − (- 4 + j) I 2 = j5
(5)
Putting (1) and (5) in matrix form,
 1− j
j  I 1   - j10 
8 + j2 4 − j I  =  j5 


 2  
∆ = -3 − j5 ,
I = I1 − I 2 =
∆ 1 = -5 + j40 ,
∆ 2 = -15 + j85
∆ 1 − ∆ 2 10 − j45
=
= 7.906∠43.49° A
∆
- 3 − j5
Chapter 10, Solution 34.
The circuit is shown below.
Io
I2
5Ω
3A
20 Ω
8Ω
40∠90° V
+
−
-j2 Ω
I3
10 Ω
I1
j15 Ω
j4 Ω
For mesh 1,
- j40 + (18 + j2) I 1 − (8 − j2) I 2 − (10 + j4) I 3 = 0
For the supermesh,
(13 − j2) I 2 + (30 + j19) I 3 − (18 + j2) I 1 = 0
(1)
(2)
Also,
I2 = I3 − 3
Adding (1) and (2) and incorporating (3),
- j40 + 5 (I 3 − 3) + (20 + j15) I 3 = 0
3 + j8
I3 =
= 1.465∠38.48°
5 + j3
I o = I 3 = 1.465∠38.48° A
(3)
Chapter 10, Solution 35.
Consider the circuit shown below.
4Ω
j2 Ω
I3
8Ω
1Ω
-j3 Ω
10 Ω
20 V
+
−
I1
-j4 A
I2
-j5 Ω
For the supermesh,
- 20 + 8 I 1 + (11 − j8) I 2 − (9 − j3) I 3 = 0
(1)
Also,
I 1 = I 2 + j4
(2)
(13 − j) I 3 − 8 I 1 − (1 − j3) I 2 = 0
(3)
For mesh 3,
Substituting (2) into (1),
(19 − j8) I 2 − (9 − j3) I 3 = 20 − j32
(4)
Substituting (2) into (3),
- (9 − j3) I 2 + (13 − j) I 3 = j32
(5)
From (4) and (5),
 19 − j8 - (9 − j3)  I 2   20 − j32 
 - (9 − j3) 13 − j  I  =  j32 

 3  

∆ = 167 − j69 ,
∆ 2 = 324 − j148
∆ 2 324 − j148 356.2∠ - 24.55°
=
=
∆
167 − j69 180.69∠ - 22.45°
I 2 = 1.971∠-2.1° A
I2 =
Chapter 10, Solution 36.
Consider the circuit below.
j4 Ω
-j3 Ω
+
I1
4∠90° A
2Ω
Vo
I2
+
−
−
2Ω
2Ω
I3
2∠0° A
Clearly,
I 1 = 4 ∠90° = j4
and
I 3 = -2
For mesh 2,
(4 − j3) I 2 − 2 I 1 − 2 I 3 + 12 = 0
(4 − j3) I 2 − j8 + 4 + 12 = 0
- 16 + j8
= -3.52 − j0.64
I2 =
4 − j3
Thus,
Vo = 2 (I 1 − I 2 ) = (2)(3.52 + j4.64) = 7.04 + j9.28
Vo = 11.648∠52.82° V
Chapter 10, Solution 37.
I1
+
120∠ − 90 o V
-
Ix
Z
Z=80-j35 Ω
I2
Iy
Iz
12∠0° V
120∠ − 30 o V
+
Z
I3
For mesh x,
ZI x − ZI z = − j120
(1)
ZI y − ZI z = −120∠30 o = −103.92 + j60
(2)
− ZI x − ZI y + 3ZI z = 0
(3)
For mesh y,
For mesh z,
Putting (1) to (3) together leads to the following matrix equation:
0
(−80 + j35)  I x  
− j120
 (80 − j35)


  

0
(80 − j35) (−80 + j35)  I y  =  − 103.92 + j60 

 (−80 + j35) (−80 + j35) (240 − j105)  I  

0

 z  


→
Using MATLAB, we obtain
 - 1.9165 + j1.4115 


I = inv(A) * B =  - 2.1806 - j0.954 
 - 1.3657 + j0.1525 


I1 = I x = −1.9165 + j1.4115 = 2.3802∠143.6 o A
I 2 = I y − I x = −0.2641 − j2.3655 = 2.3802∠ − 96.37 o A
I 3 = − I y = 2.1806 + j0.954 = 2.3802∠23.63o A
Chapter 10, Solution 38.
Consider the circuit below.
AI = B
Io
I1
2∠0° A
2Ω
j2 Ω
1Ω
I2
+
−
-j4 Ω
4∠0° A
I3
I4
10∠90° V
1Ω
A
Clearly,
I1 = 2
(1)
For mesh 2,
(2 − j4) I 2 − 2 I 1 + j4 I 4 + 10 ∠90° = 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),
1 − j2 j2  I 2   2 − j5
 j2 1 − j I  =  2 + j6

 4  

∆ = 3 − j3 ,
∆ 1 = 9 − j11
- ∆ 1 - (9 − j11) 1
=
= (-10 + j)
∆
3 − j3
3
I o = 3.35∠174.3° A
Io = -I2 =
(4)
(5)
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  12∠64 o 
−8
 (28 − j15)
 

(8 − j9) − j16  I 2  =  0 
 −8



j15
− j16 (10 + j)  I 3   0 



Using MATLAB,
I = inv(A)*B
I1 = −0.128 + j0.3593 = 0.3814∠109.6 o A
I 2 = −0.1946 + j0.2841 = 0.3443∠124.4 o A
I 3 = 0.0718 − j0.1265 = 0.1455∠ − 60.42 o A
or
AI = B
I x = 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).
4Ω
2Ω
iO1
+
−
8V
(a)
Clearly,
i O1 = 8 2 = 4 A
For i O 2 , consider the circuit in Fig. (b).
4Ω
2Ω
IO2
10∠0° V
+
−
j4 Ω
(b)
If we transform the voltage source, we have the circuit in Fig. (c), where 4 || 2 = 4 3 Ω .
IO2
2.5∠0° A
4Ω
(c)
By the current division principle,
43
I O2 =
(2.5∠0°)
4 3 + j4
I O 2 = 0.25 − j0.75 = 0.79∠ - 71.56°
2Ω
j4 Ω
i O 2 = 0.79 cos(4t − 71.56°) A
Thus,
Therefore,
i O = i O1 + i O 2 = 4 + 0.79 cos(4t – 71.56°) A
Chapter 10, Solution 41.
Let vx = v1 + v2.
For v1 we let the DC source equal zero.
5Ω
+
+
–
20∠0˚
1Ω
–j
V1
−
V1 − 20 V1 V1
+
+
= 0 which simplifies to (1j − 5 + 5 j)V1 = 100 j
5
−j 1
V1 = 2.56∠–39.8˚ or v1 = 2.56sin(500t – 39.8˚) V
Setting the AC signal to zero produces:
5Ω
1Ω
+
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 ω =2.
2 − j4
1 − j2
6 = 12
= 3.724 − j3.31 = 4.983∠ − 41.63°
3 + j2 + 2 − j4
5 − j2
i1= 4.983sin(2t – 41.63˚) A
I1 =
–j4
2Ω
i1
3Ω
is
j2
For i2, we transform vs = 12cos(4t – 30˚) into the frequency domain and get
Vs = 12∠–30˚.
Thus, I 2 =
12∠ − 30°
= 5.385∠8.2° or i2 = 5.385cos(4t + 8.2˚) A
2 − j2 + 3 + j4
–j2
2Ω
i2
3Ω
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).
4Ω
2Ω
iO1
+
−
8V
(a)
Clearly,
i O1 = 8 2 = 4 A
For i O 2 , consider the circuit in Fig. (b).
4Ω
2Ω
IO2
10∠0° V
+
−
j4 Ω
(b)
If we transform the voltage source, we have the circuit in Fig. (c), where 4 || 2 = 4 3 Ω .
IO2
2.5∠0° A
4Ω
2Ω
(c)
By the current division principle,
43
I O2 =
(2.5∠0°)
4 3 + j4
I O 2 = 0.25 − j0.75 = 0.79∠ - 71.56°
Thus,
i O 2 = 0.79 cos(4t − 71.56°) A
Therefore,
i O = i O1 + i O 2 = 4 + 0.79 cos (89)(4t – 71.56°) A
j4 Ω
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 , ω = 6 , 5 H

→
jωL = j30
The frequency-domain circuit is shown below.
20 Ω
16 Ω
Is
Let Z = 16 //( 20 + j30) =
+
V1
-
16(20 + j30)
= 11.8 + j3.497 = 12.31∠16.5 o
36 + j30
V1 = I s Z = (12∠10 o )(12.31∠16.5 o ) = 147.7∠26.5 o
For v2 , ω = 2 , 5 H
j30

→

→
v1 = 147.7 cos(6 t + 26.5 o ) V
jωL = j10
The frequency-domain circuit is shown below.
20 Ω
16 Ω
-
j10
+
V2
-
+
Vs
-
Using voltage division,
16
16(50∠0 o )
V2 =
Vs =
= 21.41∠ − 15.52 o
16 + 20 + j10
36 + j10

→
v 2 = 21.41sin(2t − 15.52 o ) V
Thus,
v x = 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 Ω
IT
I1
20∠-150° V
+
−
j10 Ω
-j5 Ω
(a)
j10 || - j5 = - j10
20 ∠ - 150° 2∠ - 150°
IT =
=
10 − j10
1− j
Using current division,
- j5
- j5 2 ∠ - 150°
I1 =
IT =
⋅
= - (1 + j) ∠ - 150°
j10 − j5
j5
1− j
For I 2 , consider the circuit in Fig. (b).
I2
10 Ω
j10 Ω
(b)
10 || - j5 =
Using current division,
- j10
2− j
-j5 Ω
4∠-45° A
I2 =
- j10 (2 − j)
(4∠ - 45°) = -2 (1 + j) ∠ - 45°
- j10 (2 − j) + j10
I o = I 1 + I 2 = - 2 ∠ - 105° − 2 2 ∠0°
I o = -2.462 + j1.366 = 2.816∠150.98°
i o = 2.816 cos(10t + 150.98°) A
Therefore,
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).
6Ω
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).
ω= 2
2H 
→ jωL = j4
1
1
1
F 
→
=
= - j6
12
jωC j (2)(1 / 12)
+
6Ω
-j6 Ω
4∠0° A
V2
−
(b)
Applying nodal analysis,
V
V
V 1 j j 
4 = 2 + 2 + 2 =  + −  V2
6 - j6 j4  6 6 4 
V2 =
24
= 21.45∠26.56°
1 − j0.5
j4 Ω
Hence,
v 2 = 21.45 sin( 2 t + 26.56°) V
For v 3 , consider the circuit in Fig. (c).
ω=3
2H 
→ jωL = j6
1
1
1
F 
→
=
= - j4
12
jωC j (3)(1 / 12)
6Ω
12∠0° V
j6 Ω
+
+
−
-j4 Ω
V3
−
(c)
At the non-reference node,
12 − V3 V3 V3
=
+
6
- j4 j6
12
V3 =
= 10.73∠ - 26.56°
1 + j0.5
Hence,
v 3 = 10.73 cos(3t − 26.56°) V
Therefore,
v o = 10 + 21.45 sin(2t + 26.56°) + 10.73 cos(3t – 26.56°) V
Chapter 10, Solution 47.
Let i o = i1 + i 2 + i 3 , where i1 , i 2 , and i 3 are respectively due to the 24-V dc source, the
ac voltage source, and the ac current source. For i1 , consider the circuit in Fig. (a).
1Ω
24 V
1/6 F
− +
2Ω
(a)
Since the capacitor is an open circuit to dc,
2H
i1
4Ω
i1 =
24
=4A
4+2
For i 2 , consider the circuit in Fig. (b).
ω=1
2H 
→ jωL = j2
1
1
F 
→
= - j6
6
jωC
1Ω
j2 Ω
-j6 Ω
I2
10∠-30° V
+
−
I1
2Ω
I2
4Ω
(b)
For mesh 1,
- 10 ∠ - 30° + (3 − j6) I 1 − 2 I 2 = 0
10 ∠ - 30° = 3 (1 − 2 j) I 1 − 2 I 2
(1)
For mesh 2,
0 = -2 I 1 + (6 + j2) I 2
I 1 = (3 + j) I 2
(2)
Substituting (2) into (1)
10 ∠ - 30° = 13 − j15 I 2
I 2 = 0.504 ∠19.1°
Hence,
i 2 = 0.504 sin( t + 19.1°) A
For i 3 , consider the circuit in Fig. (c).
ω=3
2H 
→ jωL = j6
1
1
1
F 
→
=
= - j2
jωC j (3)(1 / 6)
6
1Ω
j6 Ω
-j2 Ω
I3
2Ω
(c)
2∠0° A
4Ω
2 || (1 − j2) =
2 (1 − j2)
3 − j2
Using current division,
2 (1 − j2)
⋅ (2∠0°)
2 (1 − j2)
3 − j2
=
I3 =
2 (1 − j2)
13 + j3
4 + j6 +
3 − j2
I 3 = 0.3352 ∠ - 76.43°
Hence
i 3 = 0.3352 cos(3t − 76.43°) A
Therefore,
i o = 4 + 0.504 sin(t + 19.1°) + 0.3352 cos(3t – 76.43°) A
Chapter 10, Solution 48.
Let i O = i O1 + i O 2 + i O 3 , where i O1 is due to the ac voltage source, i O 2 is due to the dc
voltage source, and i O3 is due to the ac current source. For i O1 , consider the circuit in
Fig. (a).
ω = 2000
50 cos(2000t ) 
→ 50∠0°
→
40 mH 
jωL = j (2000)(40 × 10 -3 ) = j80
1
1
=
= - j25
jωC j (2000)(20 × 10 -6 )
→
20 µF 
I
50∠0° V
-j25 Ω
IO1
+
−
80 Ω
j80 Ω
(a)
80 || (60 + 100) = 160 3
50
30
=
I=
160 3 + j80 − j25 32 + j33
Using current division,
60 Ω
100 Ω
- 80 I
-1
10∠180°
= I=
80 + 160 3
46∠45.9°
= 0.217 ∠134.1°
i O1 = 0.217 cos(2000 t + 134.1°) A
I O1 =
I O1
Hence,
For i O 2 , consider the circuit in Fig. (b).
iO2
80 Ω
100 Ω
60 Ω
+
−
24 V
(b)
i O2 =
24
= 0.1 A
80 + 60 + 100
For i O3 , consider the circuit in Fig. (c).
ω = 4000
2 cos(4000t ) 
→ 2∠0°
→
40 mH 
20 µF 
→
jωL = j (4000)(40 × 10 -3 ) = j160
1
1
=
= - j12.5
jωC j (4000)(20 × 10 -6 )
-j12.5 Ω
I2
IO3
80 Ω
j160 Ω
I3
2∠0° A
I1
100 Ω
60 Ω
(c)
For mesh 1,
I1 = 2
For mesh 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
I3 =
= 1.1782∠7.38°
16 + j44.25
(2)
(3)
Hence,
I O 3 = - I 3 = -1.1782∠7.38°
i O 3 = -1.1782 sin( 4000t + 7.38°) A
Therefore,
i O = 0.1 + 0.217 cos(2000t + 134.1°) – 1.1782 sin(4000t + 7.38°) A
Chapter 10, Solution 49.
8 sin( 200t + 30°) 
→ 8∠30°, ω = 200
5 mH 
→
1 mF 
→
jωL = j (200)(5 × 10 -3 ) = j
1
1
=
= - j5
jωC j (200)(1 × 10 -3 )
After transforming the current source, the circuit becomes that shown in the figure below.
5Ω
40∠30° V
3Ω
I
+
−
40 ∠30°
40 ∠30°
=
= 4.472∠56.56°
5 + 3 + j − j5
8 − j4
i = 4.472 sin(200t + 56.56°) A
I=
jΩ
-j5 Ω
Chapter 10, Solution 50.
50 cos(10 5 t ) 
→ 50 ∠0°, ω = 10 5
0.4 mH 
→
0.2 µF 
→
jωL = j (10 5 )(0.4 × 10 -3 ) = j40
1
1
=
= - j50
5
jωC j (10 )(0.2 × 10 -6 )
After transforming the voltage source, we get the circuit in Fig. (a).
j40 Ω
+
20 Ω
2.5∠0° A
-j50 Ω
80 Ω
Vo
−
(a)
Let
Z = 20 || - j 50 =
and
- j100
2 − j5
Vs = (2.5∠0°) Z =
- j250
2 − j5
With these, the current source is transformed to obtain the circuit in Fig.(b).
j40 Ω
Z
Vs
+
+
−
80 Ω
Vo
−
(b)
By voltage division,
80
80
- j250
Vs =
⋅
- j100
Z + 80 + j40
2 − j5
+ 80 + j40
2 − j5
8 (- j250)
Vo =
= 36.15∠ - 40.6°
36 − j42
v o = 36.15 cos(105 t – 40.6°) V
Vo =
Therefore,
Chapter 10, Solution 51.
The original circuit with mesh currents and a node voltage labeled is shown below.
Io
j10 Ω
4∠-60° V
-j20 Ω
40 Ω
1.25∠0° A
The following circuit is obtained by transforming the voltage sources.
Io
4∠-60° V
j10 Ω
-j20 Ω
40 Ω
Use nodal analysis to find Vx .
 1
1
1
4 ∠ - 60° + 1.25∠0° = 
+
+  Vx
 j10 - j20 40 
3.25 − j3.464 = (0.025 − j0.05) Vx
3.25 − j3.464
Vx =
= 81.42 + j24.29 = 84.97 ∠16.61°
0.025 − j0.05
Thus, from the original circuit,
40 ∠30° − Vx (34.64 + j20) − (81.42 + j24.29)
I1 =
=
j10
j10
- 46.78 − j4.29
I1 =
= -0.429 + j4.678 = 4.698∠95.24° A
j10
Vx − 50 ∠0° 31.42 + j24.29
=
40
40
I 2 = 0.7855 + j0.6072 = 0.9928∠37.7° = 0.9928∠37.7° A
I2 =
Chapter 10, Solution 52.
We transform the voltage source to a current source.
60∠0°
= 6 − j12
Is =
2 + j4
1.25∠0° A
The new circuit is shown in Fig. (a).
-j2 Ω
Ix
2Ω
Is = 6 – j12 A
4Ω
6Ω
j4 Ω
5∠90° A
-j3 Ω
(a)
Let
6 (2 + j4)
= 2.4 + j1.8
8 + j4
Vs = I s Z s = (6 − j12)(2.4 + j1.8) = 36 − j18 = 18 (2 − j)
Z s = 6 || (2 + j4) =
With these, we transform the current source on the left hand side of the circuit to a
voltage source. We obtain the circuit in Fig. (b).
Zs
-j2 Ω
Ix
Vs
4Ω
+
−
j5 A
-j3 Ω
(b)
Let
Z o = Z s − j2 = 2.4 − j0.2 = 0.2 (12 − j)
Vs
18 (2 − j)
Io =
=
= 15.517 − j6.207
Z o 0.2 (12 − j)
With these, we transform the voltage source in Fig. (b) to a current source. We obtain the
circuit in Fig. (c).
Ix
Io
4Ω
Zo
-j3 Ω
(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∠17.35° A
Chapter 10, Solution 53.
We transform the voltage source to a current source to obtain the circuit in Fig. (a).
-j3 Ω
j4 Ω
+
4Ω
5∠0° A
j2 Ω
2Ω
Vo
-j2 Ω
−
(a)
Let
j8
= 0.8 + j1.6
4 + j2
Vs = (5∠0°) Z s = (5)(0.8 + j1.6) = 4 + j8
Z s = 4 || j2 =
With these, the current source is transformed so that the circuit becomes that shown in
Fig. (b).
-j3 Ω
Zs
Vs
j4 Ω
+
+
−
2Ω
-j2 Ω
Vo
−
(b)
Let
Z x = Z s − j3 = 0.8 − j1.4
V
4 + j8
= −3.0769 + j4.6154
Ix = s =
Z s 0.8 − j1.4
With these, we transform the voltage source in Fig. (b) to obtain the circuit in Fig. (c).
j4 Ω
+
Ix
Zx
2Ω
-j2 Ω
Vo
−
(c)
1.6 − j2.8
= 0.8571 − j0.5714
2.8 − j1.4
Vy = I x Z y = (−3.0769 + j4.6154) ⋅ (0.8571 − j0.5714) = j5.7143
Z y = 2 || Z x =
Let
With these, we transform the current source to obtain the circuit in Fig. (d).
j4 Ω
Zy
Vy
+
+
−
-j2 Ω
Vo
−
(d)
Using current division,
Vo =
- j2 ( j5.7143)
- j2
Vy =
= (3.529 – j5.883) V
Z y + j4 − j2
0.8571 − j0.5714 + j4 − j2
Chapter 10, Solution 54.
50 x(− j 30)
= 13.24 − j 22.059
50 − j 30
We convert the current source to voltage source and obtain the circuit below.
50 //(− j 30) =
40 Ω
+
115.91 –j31.06V
13.24 – j22.059 Ω
j20 Ω
+
-
I
134.95-j74.912 V
V
-
+
-
Applying KVL gives
-115.91 + j31.058 + (53.24-j2.059)I -134.95 + j74.912 = 0
or I =
− 250.86 + j105.97
= −4.7817 + j1.8055
53.24 − j 2.059
But − V + (40 + j20)I + V = 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 Ω
10 Ω
Zth
-j10 Ω
(a)
( j20)(- j10)
j20 − j10
= 10 − j20 = 22.36∠-63.43° Ω
Z N = Z th = 10 + j20 || (- j10) = 10 +
To find Vth , consider the circuit in Fig. (b).
j20 Ω
10 Ω
+
50∠30° V
+
−
-j10 Ω
Vth
−
(b)
Vth =
IN =
- j10
(50∠30°) = -50∠30° V
j20 − j10
Vth
- 50 ∠30°
=
= 2.236∠273.4° A
Z th 22.36 ∠ - 63.43°
(b)
To find Z th , consider the circuit in Fig. (c).
-j5 Ω
8Ω
Zth
j10 Ω
(c)
Z N = Z th = j10 || (8 − j5) =
( j10)(8 − j5)
= 10∠26° Ω
j10 + 8 − j5
To obtain Vth , consider the circuit in Fig. (d).
-j5 Ω
Io
4∠0° A
8Ω
j10 Ω
+
Vth
−
(d)
By current division,
8
32
Io =
(4∠0°) =
8 + j10 − j5
8 + j5
Vth = j10 I o =
IN =
j320
= 33.92∠58° V
8 + j5
Vth 33.92 ∠58°
=
= 3.392∠32° A
10 ∠26°
Z th
Chapter 10, Solution 56.
(a)
To find Z th , consider the circuit in Fig. (a).
j4 Ω
6Ω
-j2 Ω
Zth
(a)
( j4)(- j2)
= 6 − j4
j4 − j2
= 7.211∠-33.69° Ω
Z N = Z th = 6 + j4 || (- j2) = 6 +
By placing short circuit at terminals a-b, we obtain,
I N = 2∠0° A
Vth = Z th I th = (7.211∠ - 33.69°) (2∠0°) = 14.422∠-33.69° V
(b)
To find Z th , consider the circuit in Fig. (b).
j10 Ω
30 Ω
60 Ω
-j5 Ω
(b)
30 || 60 = 20
(- j5)(20 + j10)
20 + j5
= 5.423∠-77.47° Ω
Z N = Z th = - j5 || (20 + j10) =
Zth
To find Vth and I N , we transform the voltage source and combine the 30 Ω
and 60 Ω resistors. The result is shown in Fig. (c).
j10 Ω
4∠45° A
20 Ω
a
IN
-j5 Ω
b
(c)
20
2
(4∠45°) = (2 − j)(4∠45°)
20 + j10
5
= 3.578∠18.43° A
IN =
Vth = Z th I N = (5.423∠ - 77.47°) (3.578∠18.43°)
= 19.4∠-59° V
Chapter 10, Solution 57.
To find Z th , consider the circuit in Fig. (a).
5Ω
-j10 Ω
2Ω
Zth
j20 Ω
(a)
( j20)(5 − j10)
5 + j10
= 18 − j12 = 21.633∠-33.7° Ω
Z N = Z th = 2 + j20 || (5 − j10) = 2 +
To find Vth , consider the circuit in Fig. (b).
5Ω
-j10 Ω
2Ω
+
60∠120° V
+
−
j20 Ω
Vth
−
(b)
j20
j4
(60 ∠120°) =
(60∠120°)
5 − j10 + j20
1 + j2
= 107.3∠146.56° V
Vth =
IN =
Vth 107.3∠146.56°
=
= 4.961∠-179.7° A
Z th 21.633∠ - 33.7°
Chapter 10, Solution 58.
Consider the circuit in Fig. (a) to find Z th .
8Ω
Zth
j10 Ω
-j6 Ω
(a)
( j10)(8 − j6)
= 5 (2 + j)
8 + j4
= 11.18∠26.56° Ω
Z th = j10 || (8 − j6) =
Consider the circuit in Fig. (b) to find Vth .
Io
+
8Ω
5∠45° A
j10 Ω
-j6 Ω
(b)
Io =
8 − j6
4 − j3
(5∠45°) =
(5∠45°)
8 − j6 + j10
4 + j2
Vth = j10 I o =
( j10)(4 − j3)(5∠45°)
= 55.9∠71.56° V
(2)(2 + j)
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).
3Ω
10∠-45° V
jΩ
jΩ
a
+
+
−
3Ω
+
−
-j Ω
Vo
−
Zth
5∠-60° A
b
(b)
(a)
3Ω
jΩ
Zth
a
+
10∠-45° V
+
−
Vth
+
−
5∠-60° A
Vth
+
+
−
Vo
−
(c)
From Fig. (b),
Z th = 3 || j =
j3
3
= (1 + j3)
3 + j 10
From Fig.(c),
10∠ - 45° − Vth 5∠ - 60° − Vth
=0
+
3
j
10 ∠ - 45° − 15∠30°
Vth =
1 − j3
From Fig. (d),
−
(d)
-j Ω
b
-j
V = 10∠ - 45° − 15∠30°
Z th − j th
Vo = 15.73∠247.9° V
Vo =
Therefore,
v o = 15.73 cos(t + 247.9°) V
Chapter 10, Solution 60.
(a)
To find Z th , consider the circuit in Fig. (a).
10 Ω
-j4 Ω
a
j5 Ω
Zth
4Ω
b
(a)
Z th = 4 || (- j4 + 10 || j5) = 4 || (- j4 + 2 + j4)
Z th = 4 || 2 = 1.333 Ω
To find Vth , consider the circuit in Fig. (b).
10 Ω
V1
-j4 Ω
V2
+
20∠0° V
+
−
j5 Ω
4∠0° A
4Ω
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 = V2 = 9.615∠33.69° V
(b)
To find Z th , consider the circuit in Fig. (c).
Zth
c
d
10 Ω
-j4 Ω
j5 Ω
4Ω
(c)

j10 

Z th = - j4 || (4 + 10 || j5) = - j4 ||  4 +
2 + j

- j4
Z th = - j4 || (6 + j4) =
(6 + j4) = 2.667 – j4 Ω
6
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∠56.31° V
Chapter 10, Solution 61.
First, we need to find Vth and Z th across the 1 Ω resistor.
4Ω
-j3 Ω
j8 Ω
6Ω
Zth
(a)
From Fig. (a),
Z th = (4 − j3) || (6 + j8) =
(4 − j3)(6 + j8)
= 4.4 − j0.8
10 + j5
Z th = 4.472∠-10.3° Ω
4Ω
-j16 V
+
−
-j3 Ω
j8 Ω
+
2A
Vth
−
(b)
From Fig. (b),
- j16 − Vth
Vth
+2=
4 − j3
6 + j8
3.92 − j2.56
Vth =
= 20.93∠ - 43.45°
0.22 + j0.4
Vth
20.93∠ - 43.45°
=
1 + Z th
5.46 ∠ - 8.43°
Vo = 3.835∠ - 35.02°
Vo =
Therefore,
v o = 3.835 cos(4t – 35.02°) V
6Ω
Chapter 10, Solution 62.
First, we transform the circuit to the frequency domain.
12 cos( t ) 
→ 12∠0°, ω = 1
2H 
→
1
F 
→
4
1
F 
→
8
jωL = j2
1
= - j4
jωC
1
= - j8
jωC
To find Z th , consider the circuit in Fig. (a).
3 Io
Io
4Ω
Vx
j2 Ω
1
Ix
2
-j4 Ω
-j8 Ω
+
−
1V
(a)
At node 1,
Vx Vx
1 − Vx
,
+
+ 3Io =
4 - j4
j2
Thus,
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
I x = (0.75 + j0.5) Vx − j
3
8
I x = -0.1 + j0.425
Z th =
1
= -0.5246 − j2.229 = 2.29∠ - 103.24° Ω
Ix
- Vx
4
To find Vth , consider the circuit in Fig. (b).
3 Io
Io
4Ω
j2 Ω
V1
V2
1
12∠0° V
+
−
2
-j4 Ω
-j8 Ω
+
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
V
+ 3Io = 2
j2
- j8
72 = (6 + j4) V1 − j3 V2
(2)
From (1) and (2),
 24  2 + j - j2  V1 
 72  =  6 + j4 - j3  V 
 2
  
∆ = -5 + j6 ,
Vth = V2 =
Thus,
∆ 2 = - j24
∆2
= 3.073∠ - 219.8°
∆
2
(2)(3.073∠ - 219.8°)
Vth =
2 + Z th
1.4754 − j2.229
6.146∠ - 219.8°
Vo =
= 2.3∠ - 163.3°
2.673∠ - 56.5°
Vo =
Therefore,
v o = 2.3 cos(t – 163.3°) V
12 − V1
4
Chapter 10, Solution 63.
Transform the circuit to the frequency domain.
4 cos(200t + 30°) 
→ 4∠30°, ω = 200
10 H 
→
5 µF 
→
jωL = j (200)(10) = j2 kΩ
1
1
=
= - j kΩ
jωC j (200)(5 × 10 -6 )
Z N is found using the circuit in Fig. (a).
-j kΩ
j2 kΩ
ZN
2 kΩ
(a)
Z N = - j + 2 || j2 = - j + 1 + j = 1 kΩ
We find I N using the circuit in Fig. (b).
-j kΩ
4∠30° A
j2 kΩ
(b)
j2 || 2 = 1 + j
By the current division principle,
1+ j
IN =
(4 ∠30°) = 5.657 ∠75°
1+ j − j
Therefore,
i N = 5.657 cos(200t + 75°) A
Z N = 1 kΩ
2 kΩ
IN
Chapter 10, Solution 64.
Z N is obtained from the circuit in Fig. (a).
60 Ω
40 Ω
ZN
-j30 Ω
j80 Ω
(a)
Z N = (60 + 40) || ( j80 − j30) = 100 || j50 =
(100)( j50)
100 + j50
Z N = 20 + j40 = 44.72∠63.43° Ω
To find I N , consider the circuit in Fig. (b).
60 Ω
3∠60° A
j80 Ω
(b)
For mesh 1,
100 I 1 − 60 I s = 0
I 1 = 1.8∠60°
For mesh 2,
( j80 − j30) I 2 − j80 I s = 0
I 2 = 4.8∠60°
I N = I 1 − I 2 = 3∠60° A
40 Ω
I2
-j30 Ω
IN
Is
I s = 3∠60°
I1
Chapter 10, Solution 65.
5 cos(2 t ) 
→ 5∠0°, ω = 2
4H 
→
1
F 
→
4
1
F 
→
2
jωL = j (2)(4) = j8
1
1
=
= - j2
jωC j (2)(1 / 4)
1
1
=
= -j
jωC j (2)(1 / 2)
To find Z N , consider the circuit in Fig. (a).
2Ω
ZN
-j2 Ω
-j Ω
(a)
Z N = - j || (2 − j2) =
- j (2 − j2) 1
= (2 − j10)
2 − j3
13
To find I N , consider the circuit in Fig. (b).
5∠0° V
2Ω
+ −
-j2 Ω
-j Ω
IN
(b)
IN =
5∠0°
= j5
-j
The Norton equivalent of the circuit is shown in Fig. (c).
Io
IN
ZN
(c)
j8 Ω
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∠ - 77.47°
Therefore, i o = 0.542 cos(2t – 77.47°) A
Chapter 10, Solution 66.
ω = 10
0.5 H 
→
jωL = j (10)(0.5) = j5
1
1
10 mF 
→
=
= - j10
jωC j (10)(10 × 10 -3 )
-j10 Ω
Vx
+
10 Ω
j5 Ω
Vo
2 Vo
1A
−
(a)
To find Z th , consider the circuit in Fig. (a).
Vx
Vx
,
+
j5 10 − j10
V
19 Vx
- 10 + j10
1+
= x 
→ Vx =
10 − j10
j5
21 + j2
1 + 2 Vo =
Z N = Z th =
where Vo =
10Vx
10 − j10
Vx
14.142 ∠135°
= 0.67∠129.56° Ω
=
1
21.095∠5.44°
To find Vth and I N , consider the circuit in Fig. (b).
12∠0° V
-j10 Ω
− +
+
+
-j2 A
10 Ω
Vo
j5 Ω
I
2 Vo
Vth
−
−
(b)
where
Thus,
(10 − j10 + j5) I − (10)(- j2) + j5 (2 Vo ) − 12 = 0
Vo = (10)(- j2 − I )
(10 − j105) I = -188 − j20
188 + j20
I=
- 10 + j105
Vth = j5 (I + 2 Vo ) = j5 (21I + j40) = j105 I − 200
j105 (188 + j20)
Vth =
− 200 = -11.802 + j2.076
- 10 + j105
Vth = 11.97∠170° V
IN =
Vth
11.97 ∠170°
= 17.86∠40.44° A
=
Z th 0.67 ∠129.56°
Chapter 10, Solution 67.
10(13 − j5) 12(8 + j6)
+
= 11.243 + j1.079Ω
23 − j5
20 + j6
10
(8 + j6)
Va =
(60∠45 o ) = 13.78 + j21.44,
Vb =
(60∠45 o ) = 25.93 + j454.37Ω
23 − j5
20 + j6
V
VTh = Va − Vb = 433.1∠ − 1.599 o V,
I N = Th = 38.34∠ − 97.09 o A
Z Th
Z N = Z Th = 10 //(13 − j5) + 12 //(8 + j6) =
Chapter 10, Solution 68.
1H

→
jωL = j10x1 = j10
1
1
1
F

→
=
= − j2
1
20
jω C
j10 x
20
We obtain VTh using the circuit below.
Io
4Ω
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 = Vo = − j10I o =
− j60
= 11.52∠ − 50.19 o
4 − j10 / 3
v Th = 11.52 sin(10 t − 50.19 o )
To find RTh, we insert a 1-A source at terminals a-b, as shown below.
Io
4Ω
a
+
Vo/3
1
4I o + Vo = 0
3
1 + 4I o =
→
Vo Vo
+
− j2 j10
+
-
V
Io = − o
12
-j2 j10
4Io
Vo
-
1<0o
Combining the two equations leads to
Vo =
1
= 1.2293 − j1.4766
0.333 + j0.4
V
Z Th = o = 1.2293 − 1.477Ω
1
Chapter 10, Solution 69.
This is an inverting op amp so that
Vo - Z f
-R
=
=
= -jωRC
Vs
Zi
1 jωC
When Vs = Vm and ω = 1 RC ,
1
Vo = - j ⋅
⋅ RC ⋅ Vm = - j Vm = Vm ∠ - 90°
RC
Therefore,
v o ( t ) = Vm sin(ωt − 90°) = - Vm cos(ωt)
Chapter 10, Solution 70.
This may also be regarded as an inverting amplifier.
2 cos(4 × 10 4 t ) 
→ 2 ∠0°, ω = 4 × 10 4
1
1
10 nF 
→
=
= - j2.5 kΩ
4
jωC j (4 × 10 )(10 × 10 -9 )
Vo - Z f
=
Vs
Zi
where Z i = 50 kΩ and Z f = 100k || (- j2.5k ) =
- j100
kΩ .
40 − j
Vo
- j2
=
Vs 40 − j
Thus,
If Vs = 2 ∠0° ,
Vo =
Therefore,
- j4
4 ∠ - 90°
=
= 0.1∠ - 88.57°
40 − j 40.01∠ - 1.43°
v o ( t ) = 0.1 cos(4x104 t – 88.57°) V
Chapter 10, Solution 71.
8 cos(2t + 30 o )
→ 8∠30 o
1
1
0. 5µF

→
=
= − j1kΩ
jωC j2x 0.5x10 − 6
At the inverting terminal,
Vo − 8∠30 o Vo − 8∠30 o 8∠30 o
+
=
− j1k
10k
2k
Vo =

→
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 ) 
→ 4 ∠0°, ω = 10 4
1
1
1 nF 
→
=
= - j100 kΩ
4
jωC j (10 )(10 -9 )
Consider the circuit as shown below.
50 kΩ
4∠0° V
+
−
Therefore,
+
−
-j100 kΩ
At the noninverting node,
4 − Vo
Vo
=
50
- j100
Io =
Vo

→ Vo =
Vo
Io
100 kΩ
4
1 + j0.5
Vo
4
=
mA = 35.78∠ - 26.56° µA
100k (100)(1 + j0.5)
i o ( t ) = 35.78 cos(104 t – 26.56°) µA
Chapter 10, Solution 73.
As a voltage follower, V2 = Vo
1
1
=
= -j20 kΩ
3
jωC1 j (5 × 10 )(10 × 10 -9 )
1
1
C 2 = 20 nF 
→
=
= -j10 kΩ
3
jωC 2 j (5 × 10 )(20 × 10 -9 )
C1 = 10 nF 
→
Consider the circuit in the frequency domain as shown below.
-j20 kΩ
Is 10 kΩ
20 kΩ
V1
VS
+
−
V2
+
−
Io
Vo
-j10 kΩ
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
(2)
1
Vo = -j Vs
3
2
1
V1 = (1 + j2)Vo =  − j  Vs
3
3
Is =
Vs − V1 (1 3)(1 − j)
=
Vs
10k
10k
Is 1− j
=
Vs 30k
Vs 30k
=
= 15 (1 + j) k
Is 1− j
Z in = 21.21∠45° kΩ
Z in =
Chapter 10, Solution 74.
Zi = R1 +
1
,
jωC1
Zf = R 2 +
1
jωC 2
1
Vo - Z f
jωC 2  C 1   1 + jωR 2 C 2 

=
=
Av =
=  
1
Vs
Zi
 C 2   1 + jωR 1C 1 
R1 +
jωC1
R2 +
Av =
At ω = 0 ,
As ω → ∞ ,
Av =
C1
C2
R2
R1
At ω =
1
,
R 1 C1
 C   1 + j R 2 C 2 R 1C1 

Av =  1  
1+ j

 C2  
At ω =
1
,
R 2C2
C 

1+ j

Av =  1  
 C 2   1 + j R 1C1 R 2 C 2 
Chapter 10, Solution 75.
ω = 2 × 10 3
→
C1 = C 2 = 1 nF 
1
1
=
= -j500 kΩ
3
jωC1 j (2 × 10 )(1 × 10 -9 )
Consider the circuit shown below.
100 kΩ
-j500 kΩ
-j500 kΩ
V2
V1
VS
+
−
20 kΩ
+
−
100 kΩ
+
Vo
20 kΩ
−
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
R3
Vo =
R3 + R4
2
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
2
=
= 0.0554∠43.88°
Vs 26 − j25
Chapter 10, Solution 76.
(3)
(4)
Let the voltage between the -jk Ω capacitor and the 10k Ω resistor be V1.
2∠30 o − V1 V1 − Vo V1 − Vo
=
+
− j4k
10k
20k

→
(1)
2∠30 o = (1 − j0.6)V1 + j0.6Vo
Also,
V1 − Vo
Vo
=
− j2k
10k
→
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
Vs − V1
= jωC V1
R1
Vs = (1 + jωR 1C1 ) V1
At node 2,
0 − V1 V1 − Vo
=
+ jωC 2 (V1 − Vo )
R3
R2
 R3

V1 = (Vo − V1 ) 
+ jωC 2 R 3 
R2

R2
+
Vo
−
(1)


1
 V1
Vo = 1 +
 (R 3 R 2 ) + jωC 2 R 3 
From (1) and (2),


Vs
R2
1 +

Vo =
1 + jωR 1C1  R 3 + jωC 2 R 2 R 3 
(2)
Vo
R 2 + R 3 + jωC 2 R 2 R 3
=
Vs (1 + jωR 1C 1 ) ( R 3 + jωC 2 R 2 R 3 )
Chapter 10, Solution 78.
2 sin(400t ) 
→ 2∠0°, ω = 400
1
1
0.5 µF 
→
=
= - j5 kΩ
jωC j (400)(0.5 × 10 -6 )
1
1
0.25 µF 
→
=
= - j10 kΩ
jωC j (400)(0.25 × 10 -6 )
Consider the circuit as shown below.
20 kΩ
10 kΩ V
1
2∠0° V
+
−
-j5 kΩ
V2
+
−
Vo
40 kΩ
-j10 kΩ
10 kΩ
20 kΩ
At node 1,
2 − V1
V
V − V2 V1 − Vo
= 1 + 1
+
10
- j10
- j5
20
4 = (3 + j6) V1 − j4 V2 − Vo
(1)
V1 − V2 V2
=
− j5
10
V1 = (1 − j0.5) V2
(2)
At node 2,
But
V2 =
20
1
Vo = Vo
20 + 40
3
(3)
From (2) and (3),
1
V1 = ⋅ (1 − j0.5) Vo
3
Substituting (3) and (4) into (1) gives
(4)

1
4
1
4 = (3 + j6) ⋅ ⋅ (1 − j0.5) Vo − j Vo − Vo = 1 − j  Vo

3
3
6
Vo =
24
= 3.945∠9.46°
6− j
Therefore,
v o ( t ) = 3.945 sin(400t + 9.46°) V
Chapter 10, Solution 79.
5 cos(1000t ) 
→ 5∠0°, ω = 1000
0.1 µF 
→
1
1
=
= - j10 kΩ
jωC j (1000)(0.1 × 10 -6 )
0.2 µF 
→
1
1
=
= - j5 kΩ
jωC j (1000)(0.2 × 10 -6 )
Consider the circuit shown below.
20 kΩ
-j10 kΩ
10 kΩ
Vs = 5∠0° V
+
−
−
+
40 kΩ
V1
-j5 kΩ
Since each stage is an inverter, we apply Vo =
−
+
- 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∠0°

Vo = 
 10  20 − j10 
Vo = 16 (2 + j) = 35.78∠26.56°
Therefore,
v o ( t ) = 35.78 cos(1000t + 26.56°) V
Chapter 10, Solution 80.
4 cos(1000t − 60°) 
→ 4∠ - 60°, ω = 1000
0.1 µF 
→
1
1
=
= - j10 kΩ
jωC j (1000)(0.1 × 10 -6 )
0.2 µF 
→
1
1
=
= - j5 kΩ
jωC j (1000)(0.2 × 10 -6 )
The two stages are inverters so that
 20
20  - j5 

⋅ (4∠ - 60°) +
Vo = 
V 
50 o  10 
 - j10
=
-j
-j 2
⋅ ( j2) ⋅ (4∠ - 60°) + ⋅ Vo
2
2 5
(1 + j 5) Vo = 4∠ - 60°
Vo =
Therefore,
4∠ - 60°
= 3.922 ∠ - 71.31°
1+ j 5
v o ( t ) = 3.922 cos(1000t – 71.31°) 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($N_0001)
5.019 E+01
which means that
Vo = 7.684∠50.19o V
Chapter 10, Solution 83.
The schematic is shown below. The frequency is f = ω / 2π =
1000
= 159.15
2π
When the circuit is saved and simulated, we obtain from the output file
FREQ
1.592E+02
VM(1)
6.611E+00
Thus,
VP(1)
-1.592E+02
vo = 6.611cos(1000t – 159.2o) V
Chapter 10, Solution 84.
The schematic is shown below. We set PRINT to print Vo in the output file. In AC
Sweep box, we set Total Pts = 1, Start Freq = 0.1592, and End Freq = 0.1592. After
simulation, we obtain the output file which includes:
FREQ
VM($N_0003)
1.592 E-01
1.664 E+00
VP($N_0003)
E+02
Vo = 1.664∠-146.4o V
Namely,
Chapter 10, Solution 85.
The schematic is shown below. We let ω = 1 rad/s so that L=1H and C=1F.
When the circuit is saved and simulated, we obtain from the output file
FREQ
1.591E-01
VM(1)
2.228E+00
VP(1)
-1.675E+02
From this, we conclude that
Vo = 2.228∠ − 167.5 V
-1.646
Chapter 10, Solution 86.
We insert three pseudocomponent PRINTs at nodes 1, 2, and 3 to print V1, V2, and V3,
into the output file. Assume that w = 1, we set Total Pts = 1, Start Freq = 0.1592, and
End Freq = 0.1592. After saving and simulating the circuit, we obtain the output file
which includes:
FREQ
VM($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($N_0003)
-1.524
E+02
Therefore,
V1 = 15.91∠169.6o V V2 = 5.172∠-138.6o V V3 = 2.27∠-152.4o V
Chapter 10, Solution 88.
The schematic is shown below. We insert IPRINT and PRINT to print Io and Vo in the
output file. Since w = 4, f = w/2π = 0.6366, we set Total Pts = 1, Start Freq = 0.6366,
and End Freq = 0.6366 in the AC Sweep box. After simulation, the output file includes:
FREQ
VM($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 Vin − V2
=
R1
R2
- Vin + V2 =
R2
V
R 1 in
At node 3,
V2 − Vin Vin − V4
=
R3
1 jωC
(1)
- Vin + V4 =
Vin − V2
jωCR 3
(2)
From (1) and (2),
- R2
V
jωCR 3 R 1 in
- Vin + V4 =
Thus,
I in =
R2
Vin − V4
=
V
R4
jωCR 3 R 1 R 4 in
Z in =
Vin jωCR 1R 3 R 4
=
= jωL eq
I in
R2
L eq =
where
R 1R 3 R 4C
R2
Chapter 10, Solution 90.
Let
Z 4 = R ||
1
R
=
jωC 1 + jωRC
Z3 = R +
1
1 + jωRC
=
jωC
jωC
Consider the circuit shown below.
Z3
Vi
+
−
R1
+
Z4
Vo
−
R2
Vo =
Z4
R2
Vi −
V
Z3 + Z 4
R1 + R 2 i
R
Vo
R2
1 + jωC
=
−
R
1 + jωRC R 1 + R 2
Vi
+
1 + jωC
jωC
=
jωRC
R2
−
2
jωRC + (1 + jωRC)
R1 + R 2
Vo
R2
jωRC
=
−
2
2 2
Vi 1 − ω R C + j3ωRC R 1 + R 2
For Vo and Vi to be in phase,
Vo
must be purely real. This happens when
Vi
1 − ω2 R 2 C 2 = 0
ω=
1
= 2πf
RC
f=
or
1
2πRC
At this frequency,
Av =
Vo 1
R2
= −
Vi 3 R 1 + R 2
Chapter 10, Solution 91.
(a)
Let
V2 = voltage at the noninverting terminal of the op amp
Vo = output voltage of the op amp
Z p = 10 kΩ = R o
Z s = R + jωL +
As in Section 10.9,
1
jωC
Zp
V2
=
=
Vo Z s + Z p
Ro
R + R o + jωL −
j
ωC
ωCR o
V2
=
Vo ωC (R + R o ) + j (ω2 LC − 1)
For this to be purely real,
1
ωo2 LC − 1 = 0 
→ ωo =
fo =
1
2π LC
=
LC
1
2π (0.4 × 10 -3 )(2 × 10 -9 )
f o = 180 kHz
(b)
At oscillation,
Ro
ωo CR o
V2
=
=
Vo ωo C (R + R o ) R + R o
This must be compensated for by
Vo
80
= 1+
=5
Av =
V2
20
Ro
1
=
R + Ro 5

→ R = 4R o = 40 kΩ
Chapter 10, Solution 92.
Let
V2 = voltage at the noninverting terminal of the op amp
Vo = output voltage of the op amp
Zs = R o
Z p = jωL ||
As in Section 10.9,
ωRL
1
1
=
|| R =
1
1
jωC
ωL + jR (ω2 LC − 1)
+ jωC +
R
jωL
ωRL
Zp
V2
ωL + jR (ω2 LC − 1)
=
=
ωRL
Vo Z s + Z p
Ro +
ωL + jR (ω2 LC − 1)
V2
ωRL
=
Vo ωRL + ωR o L + jR o R (ω2 LC − 1)
For this to be purely real,
ωo2 LC = 1 
→ f o =
(a)
1
2π LC
At ω = ωo ,
ωo RL
V2
R
=
=
Vo ωo RL + ωo R o L R + R o
This must be compensated for by
Vo
Rf
1000k
Av =
= 1+
= 1+
= 11
V2
Ro
100k
Hence,
R
1
=

→ R o = 10R = 100 kΩ
R + R o 11
(b)
1
fo =
2π (10 × 10 -6 )(2 × 10 -9 )
f o = 1.125 MHz
Chapter 10, Solution 93.
As shown below, the impedance of the feedback is
jωL
1
jωC2
ZT =
1
jωC1
1
jωC1

1 

||  jωL +
jωC 2 

ZT
-j 
-j 
1
 jωL +

− ωLC 2
ωC1 
ωC 2 
ω
ZT =
=
-j
-j
j (C1 + C 2 − ω2 LC1C 2 )
+ jωL +
ωC1
ωC 2
In order for Z T to be real, the imaginary term must be zero; i.e.
C1 + C 2 − ωo2 LC1C 2 = 0
C1 + C 2
1
ωo2 =
=
LC1C 2
LC T
1
fo =
2π LC T
Chapter 10, Solution 94.
If we select C1 = C 2 = 20 nF
CT =
Since f o =
1
2π LC T
L=
C1 C 2
C1
=
= 10 nF
C1 + C 2
2
,
1
1
=
= 10.13 mH
2
2
(2πf ) C T (4π )(2500 × 10 6 )(10 × 10 -9 )
Xc =
1
1
=
= 159 Ω
ωC 2 (2π )(50 × 10 3 )(20 × 10 -9 )
We may select R i = 20 kΩ and R f ≥ R i , say R f = 20 kΩ .
Thus,
C1 = C 2 = 20 nF,
R f = R i = 20 kΩ
L = 10.13 mH
Chapter 10, Solution 95.
First, we find the feedback impedance.
C
ZT
L2
L1

1 

Z T = jωL1 ||  jωL 2 +
jωC 


j 

jωL1  jωL 2 −

ω2 L1C (1 − ωL 2 )
ωC 
ZT =
=
j
j (ω2 C (L1 + L 2 ) − 1)
jωL1 + jωL 2 −
ωC
In order for Z T to be real, the imaginary term must be zero; i.e.
ωo2 C (L1 + L 2 ) − 1 = 0
ωo = 2πf o =
fo =
1
C ( L1 + L 2 )
1
2π C (L 1 + L 2 )
Chapter 10, Solution 96.
(a)
Consider the feedback portion of the circuit, as shown below.
jωL
Vo
V2 =
+
−
jωL
V
R + jωL 1
V1
R
R

→ V1 =
Applying KCL at node 1,
Vo − V1 V1
V1
=
+
jωL
R R + jωL

1
1

Vo − V1 = jωL V1  +
 R R + jωL 
V2
jωL
R + jωL
V2
jωL
(1)
 j2ωRL − ω2 L2 

Vo = V1 1 +
R (R + jωL) 

(2)
From (1) and (2),
 R + jωL  j2ωRL − ω2 L2 
V
1 +
Vo = 
R (R + jωL)  2
 jωL 
Vo R 2 + jωRL + j2ωRL − ω2 L2
=
jωRL
V2
V2
=
Vo
1
R − ω2 L2
3+
jωRL
2
V2
1
=
Vo 3 + j (ωL R − R ωL )
(b)
V2
must be real,
Vo
Since the ratio
ωo L
R
−
=0
R
ωo L
ωo L =
R2
ωo L
ωo = 2πf o =
fo =
(c)
R
L
R
2π L
When ω = ωo
V2 1
=
Vo 3
This must be compensated for by A v = 3 . But
R2
Av = 1+
=3
R1
R 2 = 2 R1
Chapter 11, Solution 1.
v( t ) = 160 cos(50t )
i( t ) = -20 sin(50t − 30°) = 2 cos(50t − 30° + 180° − 90°)
i( t ) = 20 cos(50t + 60°)
p( t ) = v( t ) i( t ) = (160)(20) cos(50t ) cos(50t + 60°)
p( t ) = 1600 [ cos(100 t + 60°) + cos(60°) ] W
p( t ) = 800 + 1600 cos(100t + 60°) W
P=
1
1
Vm I m cos(θ v − θi ) = (160)(20) cos(60°)
2
2
P = 800 W
Chapter 11, Solution 2.
First, transform the circuit to the frequency domain.
30 cos(500t ) 
→ 30 ∠0° ,
ω = 500
0.3 H 
→
→
20µF 
I
jωL = j150
1
-j
=
= - j100
jωC (500)(20)(10 -6 )
I2
-j100 Ω
I1
30∠0° V
I1 =
+
−
j150 Ω
200 Ω
30∠0°
= 0.2∠ − 90° = - j0.2
j150
i1 ( t ) = 0.2 cos(500 t − 90°) = 0.2 sin(500 t )
I2 =
30∠0°
0.3
=
= 0.1342∠26.56° = 0.12 + j0.06
200 − j100 2 − j
i 2 ( t ) = 0.1342 cos(500 t + 25.56°)
I = I 1 + I 2 = 0.12 − j0.14 = 0.1844 ∠ - 49.4°
i( t ) = 0.1844 cos(500t − 35°)
For the voltage source,
p( t ) = v( t ) i( t ) = [ 30 cos(500t ) ] × [ 0.1844 cos(500t − 35°) ]
At t = 2 s ,
p = 5.532 cos(1000) cos(1000 − 35°)
p = (5.532)(0.5624)(0.935)
p = 2.91 W
For the inductor,
p( t ) = v( t ) i( t ) = [ 30 cos(500t ) ] × [ 0.2 sin(500t ) ]
At t = 2 s ,
p = 6 cos(1000) sin(1000)
p = (6)(0.5624)(0.8269)
p = 2.79 W
For the capacitor,
Vc = I 2 (- j100) = 13.42∠ - 63.44°
p( t ) = v( t ) i( t ) = [13.42 cos(500 − 63.44°) ] × [ 0.1342 cos(500t + 25.56°)
At t = 2 s ,
p = 18 cos(1000 − 63.44°) cos(1000 + 26.56°)
p = (18)(0.991)(0.1329)
p = 2.37 W
For the resistor,
VR = 200 I 2 = 26.84 ∠25.56°
p( t ) = v( t ) i( t ) = [ 26.84 cos(500t + 26.56°) ] × [ 0.1342 cos(500t + 26.56°) ]
At t = 2 s ,
p = 3.602 cos 2 (1000 + 25.56°)
p = (3.602)(0.1329 2
p = 0.0636 W
Chapter 11, Solution 3.
ω= 2
10 cos(2t + 30°) 
→ 10∠30° ,
1H 
→
jωL = j2
1
= -j2
jωC
0.25 F 
→
I
4Ω
I1
2Ω
I2
10∠30° V
+
−
j2 || (2 − j2) =
I=
j2 Ω
( j2)(2 − j2)
= 2 + j2
2
10 ∠30°
= 1.581∠11.565°
4 + 2 + j2
I1 =
j2
I = j I = 1.581∠101.565°
2
I2 =
2 − j2
I = 2.236 ∠56.565°
2
For the source,
S = V I* =
1
(10∠30°)(1.581∠ - 11.565°)
2
S = 7.905∠18.43° = 7.5 + j2.5
The average power supplied by the source = 7.5 W
For the 4-Ω 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 = I 2 Z L = (2.236) 2 ( j2) = j5
2
2
The average power absorbed by the inductor = 0 W
-j2 Ω
For the 2-Ω resistor, the average power absorbed is
1
1
2
P = I 1 R = (1.581) 2 (2) = 2.5 W
2
2
For the capacitor,
S=
1
1
2
I 1 Z c = (1.581) 2 (- j2) = - j2.5
2
2
The average power absorbed by the capacitor = 0 W
Chapter 11, Solution 4.
20 Ω
50 V
+
−
10 Ω
I1
-j10 Ω
I2
j5 Ω
For mesh 1,
50 = (20 − j10) I 1 + j10 I 2
5 = (2 − j) I 1 + j I 2
(1)
0 = (10 + j5 − j10) I 2 + j10 I 1
0 = (2 − j) I 2 + j2 I 1
(2)
For mesh 2,
In matrix form,
5  2 − j
j  I 1 
 0 =  j2 2 − j I 
 2 
  
∆ = 5 − j4 ,
∆ 1 = 5 (2 − j) ,
I1 =
∆ 1 5 (2 − j)
=
= 1.746∠12.1°
∆
5 − j4
I2 =
∆ 2 - j10
=
= 1.562 ∠128.66°
∆
5 - j4
For the source,
S=
1
V I 1* = 43.65∠ - 12.1°
2
∆ 2 = -j10
The average power supplied = 43.65 cos(12.1°) = 42.68 W
For the 20-Ω resistor,
1
2
P = I 1 R = 30.48 W
2
For the inductor and capacitor,
P=0W
For the 10-Ω resistor,
1
2
P = I 2 R = 12.2 W
2
Chapter 11, Solution 5.
Converting the circuit into the frequency domain, we get:
1Ω
8∠–40˚
I1Ω =
+
−
2Ω
j6
8∠ − 40°
= 1.6828∠ − 25.38°
j6(2 − j2)
1+
j6 + 2 − j2
1.6828 2
P1Ω =
1 = 1.4159 W
2
P3H = P0.25F = 0
I 2Ω =
j6
1.6828∠ − 25.38° = 2.258
j6 + 2 − j2
2.258 2
P2Ω =
2 = 5.097 W
2
–j2
Chapter 11, Solution 6.
20 Ω
50 V
+
−
I1
10 Ω
I2
-j10 Ω
j5 Ω
For mesh 1,
(4 + j2) I 1 − j2 (4 ∠60°) + 4 Vo = 0
Vo = 2 (4 ∠60° − I 2 )
(1)
(2)
For mesh 2,
(2 − j) I 2 − 2 (4∠60°) − 4Vo = 0
Substituting (2) into (3),
(2 − j) I 2 − 8∠60° − 8 (4 ∠60° − I 2 ) = 0
I2 =
(3)
40∠60°
10 − j
Hence,

40∠60°  - j8∠60°
=
Vo = 2  4 ∠60° −
10 − j 
10 − j

Substituting this into (1),
14 − j 
j32 ∠60°

(4 + j2) I 1 = j8∠60° +
= ( j8∠60°) 
10 − j
10 − j 
I1 =
(4∠60°)(1 + j14)
= 2.498∠125.06°
21 + j8
P4 =
1
1
2
I 1 R = (2.498) 2 (4) = 12.48 W
2
2
Chapter 11, Solution 7.
20 Ω
50 V
+
−
I1
10 Ω
-j10 Ω
I2
j5 Ω
Applying KVL to the left-hand side of the circuit,
8∠20° = 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 +

→ V1 =
(1)
10 − j5
Vo
10
Vo
10 − j5
Vo +
=0
j50
10
I o = j0.025 Vo
Substituting (2) into (1),
8∠20° = 0.1 Vo (1 + j)
(2)
Vo =
80∠20°
1+ j
I1 =
Vo 10
=
∠ - 25°
10
2
P=
 1 100 
1
2
(10) = 250 W
I 1 R =  
 2  2 
2
Chapter 11, Solution 8.
We apply nodal analysis to the following circuit.
V1 Io -j20 Ω
V2
I2
j10 Ω
6∠0° A
0.5 Io
40 Ω
At node 1,
6=
At node 2,
V1 V1 − V2
V1 = j120 − V2
+
j10
- j20
(1)
0 .5 I o + I o =
V2
40
V1 − V2
- j20
But,
Io =
Hence,
1.5 (V1 − V2 ) V2
=
- j20
40
3V1 = (3 − j) V2
(2)
Substituting (1) into (2),
j360 − 3V2 − 3V2 + j V2 = 0
V2 =
j360 360
=
(-1 + j6)
6 − j 37
I2 =
V2
9
=
(-1 + j6)
40 37
1
2
P = I2 R =
2
2
1 9 

 (40) = 43.78 W
2  37 
Chapter 11, Solution 9.
 6
Vo = 1 +  Vs = (4)(2) = 8 V rms
 2
P10 =
Vo2 64
=
mW = 6.4 mW
R 10
The current through the 2 -kΩ resistor is
Vs
= 1 mA
2k
P2 = I 2 R = 2 mW
Similarly,
P6 = I 2 R = 6 mW
Chapter 11, Solution 10.
No current flows through each of the resistors. Hence, for each resistor,
P = 0 W.
Chapter 11, Solution 11.
ω = 377 ,
R = 10 4 ,
C = 200 × 10 -9
ωRC = (377)(10 4 )(200 × 10 -9 ) = 0.754
tan -1 (ωRC) = 37.02°
Z ab =
10k
1 + (0.754) 2
∠ - 37.02° = 6.375∠ - 37.02° kΩ
i( t ) = 2 sin(377 t + 22°) = 2 cos(377 t − 68°) mA
I = 2 ∠ - 68°
2
S= I
2
rms
 2 × 10 -3 
 (6.375∠ - 37.02°) × 10 3
Z ab = 

2 
S = 12.751∠ - 37.02° 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
8Ω
Z Th
-j2 Ω
(a)
(8)(-j2) 8
= 8 || -j2 =
= (1 − j4) = 0.471 − j1.882
8 − j2 17
Z L = Z *Th = 0.471 + j1.882 Ω
We find VTh using the circuit in Fig. (b).
Io
+
8Ω
-j2 Ω
Vth
4∠0° A
−
(b)
Io =
- j2
(4∠0°)
8 − j2
VTh = 8 I o =
- j64
8 − j2
2
Pmax =
(b)
VTh
8RL
2
 64 


 68 
=
= 15.99 W
(8)(0.471)
We obtain Z Th from the circuit in Fig. (c).
5Ω
-j3 Ω
j2 Ω
4Ω
Zth
(c)
Z Th = j2 + 5 || (4 − j3) = j2 +
Z L = Z *Th = 2.5 − j1.167 Ω
(5)(4 − j3)
= 2.5 + j1.167
9 − j3
Chapter 11, Solution 13.
(a)
We find Z Th at the load terminals using the circuit in Fig. (a).
j100 Ω
80 Ω
Zth
-j40 Ω
(a)
(-j40)(80 + j100)
= 51.2 − j1.6
80 + j60
Z Th = -j40 || (80 + j100) =
Z L = Z *Th = 51.2 + j1.6 Ω
(b)
We find VTh at the load terminals using Fig. (b).
Io
j100 Ω
+
3∠20° A
80 Ω
-j40 Ω
Vth
−
(b)
Io =
80
(8)(3∠20°)
(3∠20°) =
80 + j100 − j40
8 + j6
VTh = - j40 I o =
(- j40)(24∠20°)
8 + j6
2
Pmax =
VTh
8RL
2
 40

 ⋅ 24
 10

=
= 22.5 W
(8)(51.2)
From Fig.(d), we obtain VTh using the voltage division principle.
5Ω
-j3 Ω
j2 Ω
10∠30° V
+
−
4Ω
+
Vth
−
(d)
 4 − j3 
 4 − j3  10

(10∠30°) = 
 ∠30°
VTh = 

 9 − j3 
 3 − j  3
2
Pmax =
VTh
8RL
2
 5 10 

⋅ 
 10 3 
=
= 1.389 W
(8)(2.5)
Chapter 11, Solution 14.
I
j24 Ω
–j10 Ω
16 Ω
40∠90º A
VTh
10 Ω
j8 Ω
Z Th = − j10 +
+
ZTh
_
(10 + j24)(16 + j8)
= − j10 + 8.245 + j7.7 = 8.245 − j2.3Ω
10 + j24 + 16 + j8
Z = Z∗Th = 8.245 + j2.3Ω
10
j40(16 + j8)
10 + j24 + 16 + j8
= 173.55∠65.66° = 71.53 + j158.12 V
VTh = I(16 + j8) =
2
VTh
Pmax =
I 2rms 8.245 =
2
2
8.245 = 456.6 W
(2x8.245) 2
Chapter 11, Solution 15.
To find Z Th , insert a 1-A current source at the load terminals as shown in Fig. (a).
1Ω
1
-j Ω
2
+
2 Vo
jΩ
Vo
1A
−
(a)
At node 1,
Vo Vo V2 − Vo
+
=
1
j
-j
At node 2,
1 + 2 Vo =
V2 − Vo
-j

→ Vo = j V2

→ 1 = j V2 − (2 + j) Vo
Substituting (1) into (2),
1 = j V2 − (2 + j)( j) V2 = (1 − j) V2
V2 =
1
1− j
VTh =
V2 1 + j
=
= 0.5 + j0.5
1
2
Z L = Z *Th = 0.5 − j0.5 Ω
(1)
(2)
We now obtain VTh from Fig. (b).
1Ω
12∠0° V
-j Ω
+
+
+
−
Vo
jΩ
2 Vo
Vth
−
−
(b)
12 − Vo Vo
=
1
j
- 12
Vo =
1+ j
2 Vo +
Vo − (- j × 2 Vo ) + VTh = 0
VTh = -(1 + j2)Vo =
(12)(1 + j2)
1+ j
2
Pmax =
VTh
2
8RL
12 5 


 2 
= 90 W
=
(8)(0.5)
Chapter 11, Solution 16.
ω = 4,
1H

→
jωL = j 4,

→
1 / 20F
1
1
=
= − j5
jωC j 4 x1 / 20
We find the Thevenin equivalent at the terminals of ZL. To find VTh, we use the circuit
shown below.
0.5Vo
2Ω
4Ω
V1
V2
+
+
10<0o
-
+
Vo
-
-j5
j4
VTh
-
At node 1,
V
V − V2
10 − V1
= 1 + 0.25V1 + 1
2
− j5
4
At node 2,
V1 − V2
V
+ 0.25V1 = 2
4
j4

→

→
5 = V1 (1 + j 0.2) − 0.25V2
0 = 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 Ω
30 Ω
a
b
40 Ω
j20 Ω
(a)
Z Th = 30 || j20 + 40 || (- j10) =
(30)( j20) (40)(-j10)
+
30 + j20
40 − j10
Z Th = 9.23 + j13.85 + 2.353 − j9.41
Z Th = 11.583 + j4.44 Ω
Z L = Z *Th = 11.583 − j4.44 Ω
We obtain VTh from Fig. (b).
I1
I2
-j10 Ω
30 Ω
j5 A
+ VTh −
40 Ω
j20 Ω
(b)
(1)
(2)
Using current division,
30 + j20
I1 =
( j5) = -1.1 + j2.3
70 + j10
I2 =
40 − j10
( j5) = 1.1 + j2.7
70 + j10
VTh = 30 I 2 + j10 I 1 = 10 + j70
Pmax =
VTh
2
8RL
=
5000
= 53.96 W
(8)(11.583)
Chapter 11, Solution 18.
We find Z Th at terminals a-b as shown in the figure below.
40 Ω
40 Ω
-j10 Ω
80 Ω
a
j20 Ω
Zth
b
Z Th = j20 + 40 || 40 + 80 || (-j10) = j20 + 20 +
Z Th = 21.23 + j10.154
Z L = Z *Th = 21.23 − j10.15 Ω
Chapter 11, Solution 19.
At the load terminals,
Z Th = - j2 + 6 || (3 + j) = -j2 +
Z Th = 2.049 − j1.561
R L = Z Th = 2.576 Ω
(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 ∠0°) Z = 8.196 + j1.756
Pmax =
VTh
2
8RL
=
70.258
= 3.409 W
20.608
Chapter 11, Solution 20.
Combine j20 Ω and -j10 Ω 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).
Io
40 Ω
V1
4 Io
V2
+ −
-j20 Ω
-j10 Ω
1A
(a)
At the supernode,
1=
V1
V
V
+ 1 + 2
40 - j20 - j10
40 = (1 + j2) V1 + j4 V2
Also,
V1 = V2 + 4 I o ,
1.1 V1 = V2

→ V1 =
Substituting (2) into (1),
V 
40 = (1 + j2)  2  + j4 V2
 1 .1 
(1)
where I o =
V2
1 .1
- V1
40
(2)
V2 =
44
1 + j6.4
Z Th =
V2
= 1.05 − j6.71 Ω
1
R L = Z Th = 6.792 Ω
To find VTh , consider the circuit in Fig. (b).
40 Ω
Io
V1
4 Io
V2
+ −
+
120∠0° V
+
−
-j20 Ω
-j10 Ω
Vth
−
(b)
At the supernode,
V
V
120 − V1
= 1 + 2
40
- j20 - j10
120 = (1 + j2) V1 + j4 V2
Also,
V1 = V2 + 4 I o ,
V1 =
(3)
where I o =
120 − V1
40
V2 + 12
1 .1
(4)
Substituting (4) into (3),
109.09 − j21.82 = (0.9091 + j5.818) V2
VTh = V2 =
Pmax =
109.09 − j21.82
= 18.893∠ - 92.43°
0.9091 + j5.818
VTh
8RL
2
(18.893) 2
=
= 6.569 W
(8)(6.792)
Chapter 11, Solution 21.
We find Z Th at terminals a-b, as shown in the figure below.
100 Ω
-j10 Ω
a
40 Ω
Zth
50 Ω
j30 Ω
b
Z Th = 50 || [ - j10 + 100 || (40 + j30) ]
where 100 || (40 + j30) =
(100)(40 + j30)
= 31.707 + j14.634
140 + j30
Z Th = 50 || (31.707 + j4.634) =
(50)(31.707 + j4.634)
81.707 + j4.634
Z Th = 19.5 + j1.73
R L = Z Th = 19.58 Ω
Chapter 11, Solution 22.
i (t ) = 4 sin t ,
I
2
rms
=
1
0<t <π
π
16 sin
π∫
2
tdt =
0
I rms = 8 = 2.828 A
16  t sin 2t 
 −

π 2
4 
π
0
=
16 π
( − 0) = 8
π 2
Chapter 11, Solution 23.
15, 0 < t < 2
v( t ) = 
 5, 2 < t < 6
2
Vrms
=
1
6
[ ∫ 15
2
2
0
]
dt + ∫2 5 2 dt =
6
550
6
Vrms = 9.574 V
Chapter 11, Solution 24.
 5, 0 < t < 1
v( t ) = 
- 5, 1 < t < 2
T = 2,
2
=
Vrms
1
2
[∫ 5
1
0
2
]
dt + ∫1 (-5) 2 dt =
2
25
[1 + 1] = 25
2
Vrms = 5 V
Chapter 11, Solution 25.
[
1 T 2
1 1
2
3
f
(
t
)
dt
(−4) 2 dt + ∫ 1 0dt + ∫2 4 2 dt
=
∫
∫
0
0
T
3
1
32
= [16 + 0 + 16] =
3
3
2
f rms
=
f rms =
32
= 3.266
3
]
Chapter 11, Solution 26.
5 0< t<2
v( t ) = 
10 2 < t < 4
T = 4,
2
Vrms
=
1
4
[∫ 5
2
2
0
]
4
1
dt + ∫2 (10) 2 dt = [50 + 200 ] = 62.5
4
Vrms = 7.906 V
Chapter 11, Solution 27.
T = 5,
I
2
rms
i( t ) = t , 0 < t < 5
1 t3
1 5 2
= ∫0 t dt = ⋅
5 3
5
5
0
=
125
= 8.333
15
I rms = 2.887 A
Chapter 11, Solution 28.
2
Vrms
=
2
rms
V
1
5
[ ∫ (4t )
2
2
dt + ∫2 0 2 dt
2
0
=
0
1 16 t 3
= ⋅
5 3
5
]
16
(8) = 8.533
15
Vrms = 2.92 V
2
Vrms
8.533
P=
=
= 4.267 W
R
2
Chapter 11, Solution 29.
 20 − 2t 5 < t < 15
i( t ) = 
- 40 + 2t 15 < t < 25
T = 20 ,
[∫
]
2
I eff
=
1
20
2
I eff
=
25
1  15
2
(
100
20
t
t
)
dt
( t 2 − 40 t + 400) dt 
−
+
+
∫
∫


5
15
5
I
2
eff
15
5
(20 − 2 t ) 2 dt + ∫15 (-40 + 2t) 2 dt
25
1
t 3  15
2

=  100 t − 10 t +  5
5 
3
 25 
 t3
+  − 20 t 2 + 400 t  15

 
3
1
2
I eff
= [83.33 + 83.33 ] = 33.332
5
I eff = 5.773 A
2
P = I eff
R = 400 W
Chapter 11, Solution 30.
t 0<t<2
v( t ) = 
- 1 2 < t < 4
2
Vrms
=
1
4
[∫ t
2
2
0
]
4

1 8
dt + ∫2 (-1) 2 dt =  + 2  = 1.1667

43
Vrms = 1.08 V
Chapter 11, Solution 31.
V
2
rms
2
1
2
 1 4
1
1

2
= ∫ v(t )dt =  ∫ (2t ) dt + ∫ (−4) 2 dt  =  + 16 = 8.6667
20
2 0

1
 2 3
Vrms = 2.944 V
Chapter 11, Solution 32.
I 2rms =
I
2
rms
2
1 1
(10t 2 ) 2 dt + ∫ 0 dt 
∫

1
2  0
t5
= 50 ∫0 t dt = 50 ⋅
5
1
4
1
0
= 10
I rms = 3.162 A
Chapter 11, Solution 33.
 10
0 < t <1

i( t ) =  20 − 10t 1 < t < 2
 0
2<t<3

I 2rms =
1
3
[ ∫ 10
1
0
2
dt + ∫1 (20 − 10t ) 2 dt + 0
2
]
3 I 2rms = 100 + 100∫1 (4 − 4t + t 2 ) dt = 100 + (100)(1 3) = 133.33
2
I rms =
133.33
= 6.667 A
3
Chapter 11, Solution 34.
[
1 T 2
1 2
3
f ( t )dt = ∫ 0 (3t ) 2 dt + ∫ 2 6 2 dt
∫
0
T
3
2


1  9t 3
=
+ 36 = 20

3 3
0


2
f rms
=
f rms = 20 = 4.472
]
Chapter 11, Solution 35.
2
Vrms
=
1
6
[ ∫ 10
1
0
2
dt + ∫1 20 2 dt + ∫2 30 2 dt + ∫4 20 2 dt + ∫5 10 2 dt
6
5
4
2
]
1
2
Vrms
= [100 + 400 + 1800 + 400 + 100 ] = 466.67
6
Vrms = 21.6 V
Chapter 11, Solution 36.
(a) Irms = 10 A
2
 3 

→
(b) V rms = 4 + 

 2
36
(c)
= 9.055 A
I rms = 64 +
2
2
(d)
2
Vrms =
Vrms = 16 +
9
= 4.528 V (checked)
2
25 16
+
= 4.528 V
2
2
Chapter 11, Solution 37.
i = i1 + i2 + i3 = 8 + 4 sin(t + 10 o ) + 6 cos(2t + 30 o )
I rms = I 21rms + I 2 2 rms + I 2 3rms = 64 +
16 36
+
= 90 = 9.487 A
2
2
Chapter 11, Solution 38.
0.5 H 
→
jωL = j (2π )(50)(0.5) = j157.08
Z = R + jX L = 30 + j157.08
S=
V
2
Z*
Apparent power = S =
=
(210) 2
30 − j157.08
(210) 2
= 275.6 VA
160

157.08 
 = cos(79.19°)
pf = cos θ = cos  tan -1 
 36 

pf = 0.1876 (lagging)
Chapter 11, Solution 39.
Z T = j4 || (12 − j8) =
( j4)(12 − j8)
12 − j4
Z T = 0.4 (3 + j11) = 4.56 ∠74.74°
pf = cos(74.74°) = 0.2631
Chapter 11, Solution 40.
At node 1,
120∠30 o − V1 V1 V1 − V2
=
+
20
j30
50

→
103.92 + j60 = V1 (1.4 − j0.6667) − 0.4V2
(1)
At node 2,
V1 − V2 V2
V
=
+ 2
50
10 − j 40

→
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 Ω = 0 = P− j 40 Ω
P10 Ω =
P50 Ω
1 | V1 − V2 | 2
=
= 4603.1 / 100 = 46.03 W
R
2
P20 Ω =
(b) I =
V 2 rms 1 | V2 | 2
=
= 173.3 / 20 = 8.665 W
R
2 R
1 | 120∠30 o − V1 | 2
= 3514 / 40 = 87.86 W
R
2
120∠30 o − V1
= 2.944 − j 0.3467,
20
1
S = Vs I • = 142.5 − j106.3,
2
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) = -j2 || -j3 =
(-j2)(-j3)
= -j6
j
Z T = 4 − j6 = 7.211∠ - 56.31°
pf = cos(-56.31°) = 0.5547 (leading)
(b)
j2 || (4 + j) =
( j2)(4 + j)
= 0.64 + j1.52
4 + j3
Z = 1 || (0.64 + j1.52 − j) =
0.64 + j0.44
= 0.4793∠21.5°
1.64 + j0.44
pf = cos(21.5°) = 0.9304 (lagging)
Chapter 11, Solution 42.
pf = 0.86 = cos θ 
→ θ = 30.683°
Q = S sin θ 
→ S =
S = V I*

→ I * =
Q
5
=
= 9.798 kVA
sin θ sin(30.683°)
S 9.798 × 10 3 ∠30.683°
=
= 44.536 ∠30.683°
220
V
Peak current = 2 × 44.536 = 62.98 A
Apparent power = S = 9.798 kVA
Chapter 11, Solution 43.
(a) Vrms = V 21rms + V 2 2 rms + V 2 3rms = 25 +
(b) P =
9 1
+ = 30 = 5.477 V
2 2
V 2 rms
= 30 / 10 = 3 W
R
Chapter 11, Solution 44.
pf = 0.65 = cosθ

→
θ = 49.46 o
S = S (cosθ + j sin θ ) = 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 +
60 2
= 2200
2
I rms = 12 +

→
Vrms = 46.9 V
0.5 2
= 1.125 = 1.061A
2
(b) P = Vrms I rms = 49.74 W
Chapter 11, Solution 46.
(a)
S = V I * = (220∠30°)(0.5∠ - 60°) = 110∠ - 30°
S = 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 = V I * = (250∠ - 10°)(6.2 ∠25°) = 1550∠15°
S = 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 = V I * = (120∠0°)(2.4∠15°) = 288∠15°
S = 278.2 + j74.54 VA
Apparent power = 288 VA
Real power = 278.2 W
Reactive power = 74.54 VAR
pf is lagging because current lags voltage
(d)
S = V I * = (160 ∠45°)(8.5∠ - 180°) = 1360∠ - 135°
S = - 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)
V = 112 ∠10° ,
I = 4∠ - 50°
1
S = V I * = 224∠60° = 112 + j194 VA
2
Average power = 112 W
Reactive power =194 VAR
(b)
V = 160 ∠0° ,
I = 25∠45°
1
S = V I * = 200∠ - 45° = 141.42 − j141.42 VA
2
Average power = 141.42 W
Reactive power = - 141.42 VAR
(c)
S=
2
V
Z*
=
(80) 2
= 128∠30° = 90.51 + j64 VA
50∠ - 30°
Average power = 90.51 W
Reactive power = 64 VAR
(d)
2
S = I Z = (100)(100∠45°) = 7.071 + j7.071 kVA
Average power = 7.071 kW
Reactive power = 7.071 kVAR
Chapter 11, Solution 48.
(a)
S = P − jQ = 269 − j150 VA
(b)
pf = cos θ = 0.9 
→ θ = 25.84°
Q = S sin θ 
→ S =
Q
2000
=
= 4588.31
sin θ sin(25.84°)
P = S cos θ = 4129.48
S = 4129 − j2000 VA
(c)
Q 450
=
= 0.75
S 600
pf = 0.6614
Q = S sin θ 
→ sin θ =
θ = 48.59 ,
P = S cos θ = (600)(0.6614) = 396.86
S = 396.9 + j450 VA
(d)
S=
V
2
Z
=
(220) 2
= 1210
40
P = S cos θ 
→ cos θ =
P 1000
=
= 0.8264
S 1210
θ = 34.26°
Q = S sin θ = 681.25
S = 1000 + j681.2 VA
Chapter 11, Solution 49.
(a)
4
sin(cos -1 (0.86)) kVA
0.86
S = 4 + j2.373 kVA
S = 4+ j
(b)
pf =
P 1.6
=
0.8 = cos θ 
→ sin θ = 0.6
S
2
S = 1.6 − j2 sin θ = 1.6 − j1.2 kVA
(c)
S = Vrms I *rms = (208∠20°)(6.5∠50°) VA
S = 1.352 ∠70° = 0.4624 + j1.2705 kVA
2
(d)
V
(120) 2
14400
S= * =
=
Z
40 − j60 72.11∠ - 56.31°
S = 199.7 ∠56.31° = 110.77 + j166.16 VA
Chapter 11, Solution 50.
(a)
S = P − jQ = 1000 − j
1000
sin(cos -1 (0.8))
0.8
S = 1000 − j750
But,
Z =
*
Vrms
S=
Vrms
2
Z*
2
S
(220) 2
=
= 30.98 + j23.23
1000 − j750
Z = 30.98 − j23.23 Ω
(b)
2
S = I rms Z
Z=
(c)
Z =
*
S
I rms
2
Vrms
S
=
2
1500 + j2000
= 10.42 + j13.89 Ω
(12) 2
V
2
(120) 2
=
=
= 1.6 ∠ - 60°
2S
(2)(4500 ∠60°)
Z = 1.6 ∠60° = 0.8 + j1.386 Ω
Chapter 11, Solution 51.
(a)
Z T = 2 + (10 − j5) || (8 + j6)
ZT = 2 +
(10 − j5)(8 + j6)
110 + j20
= 2+
18 + j
18 + j
Z T = 8.152 + j0.768 = 8.188∠5.382°
pf = cos(5.382°) = 0.9956 (lagging)
2
(b)
V
1
(16) 2
S = V I* =
=
2
2 Z * (2)(8.188∠ - 5.382°)
S = 15.63∠5.382°
P = S cos θ = 15.56 W
(c)
Q = S sin θ = 1.466 VAR
(d)
S = S = 15.63 VA
(e)
S = 15.63∠5.382° = 15.56 + j1.466 VA
Chapter 11, Solution 52.
2000
0.6 = 2000 + j1500
0 .8
S B = 3000 x 0.4 − j3000 x 0.9165 = 1200 − j2749
S A = 2000 + j
SC = 1000 + j500
S = S A + S B + SC = 4200 − j749
4200
(a)
pf =
(b)
S = Vrms I ∗rms 
→ I ∗rms =
4200 2 + 749 2
= 0.9845 leading.
Irms = 35.55∠–55.11˚ A.
4200 − j749
= 35.55∠ − 55.11°
120∠45°
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)
SB
S + SC
S
5640∠0.2°
= 66.46∠ − 29.8°
+ A
=
=
120∠30°
Vrms
Vrms
Vrms
2
I = 2 x 66.46∠29.88° = 93.97∠29.8° A
(b)
pf = cos(0.2˚) ≈ 1.0 lagging.
Chapter 11, Solution 54.
(a)
S = P − jQ = 1000 − j
1000
sin(cos -1 (0.8))
0.8
S = 1000 − j750
But,
Z =
*
Vrms
S=
Vrms
2
Z*
2
=
S
(220) 2
= 30.98 + j23.23
1000 − j750
Z = 30.98 − j23.23 Ω
(b)
2
S = I rms Z
Z=
(c)
Z =
*
S
I rms
2
Vrms
S
=
1500 + j2000
= 10.42 + j13.89 Ω
(12) 2
2
=
V
2
2S
=
(120) 2
= 1.6 ∠ - 60°
(2)(4500 ∠60°)
Z = 1.6 ∠60° = 0.8 + j1.386 Ω
Chapter 11, Solution 55.
We apply mesh analysis to the following circuit.
-j20 Ω
j10 Ω
I3
+
−
40∠0° V rms
I1
20 Ω
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,
 2  1 − j - 1  I1 
 - j5 =  - 2 2 + j I 
  
 2 
50∠90° V rms
For mesh 1,
∆ = 1− j ,
∆ 1 = 4 − j3 ,
I1 =
∆ 1 4 − j3 1
=
= (7 − j) = 3.535∠8.13°
∆
1− j
2
I2 =
∆ 2 - 1 − j5
=
= 2 − j3 = 3.605∠ - 56.31°
∆
1− j
(1)
(2)
∆ 2 = -1 − j5
I 3 = I1 − I 2 = (3.5 + j0.5) − (2 − j3) = 1.5 + j3.5 = 3.808∠66.8°
For the 40-V source,
1

S = -V I 1* = -(40)  ⋅ (7 − j)  = - 140 + j20 VA
2

For the capacitor,
S = I1
2
Z c = - j250 VA
2
R = 290 VA
2
Z L = j130 VA
For the resistor,
S = I3
For the inductor,
S = I2
For the j50-V source,
S = V I *2 = ( j50)(2 + j3) = - 150 + j100 VA
Chapter 11, Solution 56.
- j2 || 6 =
(6)(- j2)
= 0.6 − j1.8
6 − j2
3 + j4 + (-j2) || 6 = 3.6 + j2.2
The circuit is reduced to that shown below.
Io
+
2∠30° A
3.6 + j2.2 Ω
5Ω
Vo
−
Io =
3.6 + j2.2
(2∠30°) = 0.95∠47.08°
8.6 + j2.2
Vo = 5 I o = 4.75∠47.08°
S=
1
1
Vo I *s = ⋅ (4.75∠47.08°)(2∠ - 30°)
2
2
S = 4.75∠17.08° = 4.543 + j1.396 VA
Chapter 11, Solution 57.
Consider the circuit as shown below.
4Ω
24∠0° V
+
−
Vo
-j1 Ω
V1
2Ω
+
1Ω
j2 Ω
V2
−
At node o,
24 − Vo Vo Vo − V1
=
+
4
1
-j
2 Vo
24 = (5 + j4) Vo − j4 V1
At node 1,
(1)
Vo − V1
V
+ 2 Vo = 1
-j
j2
V1 = (2 − j4) Vo
(2)
Substituting (2) into (1),
24 = (5 + j4 − j8 − 16) Vo
Vo =
- 24
,
11 + j4
V1 =
(-24)(2 - j4)
11 + j4
The voltage across the dependent source is
V2 = V1 + (2)(2 Vo ) = V1 + 4 Vo
V2 =
(-24)(6 − j4)
- 24
⋅ (2 − j4 + 4) =
11 + j4
11 + j4
S=
1
1
V2 I * = V2 (2 Vo* )
2
2
S=
(-24)(6 − j4) - 24  576 
(6 − j4)
⋅
=
11 + j4
11 - j4  137 
S = 25.23 − j16.82 VA
Chapter 11, Solution 58.
Ix -j3 kΩ
8 mA
From the left portion of the circuit,
0.2
Io =
= 0.4 mA
500
20 I o = 8 mA
4 kΩ
j1 kΩ
10 kΩ
From the right portion of the circuit,
16
4
Ix =
mA
(8 mA) =
7− j
4 + 10 + j − j3
S = Ix
2
R=
(16 × 10 -3 ) 2
⋅ (10 × 10 3 )
50
S = 51.2 mVA
Chapter 11, Solution 59.
Consider the circuit below.
Ix -j3 kΩ
8 mA
4+
j1 kΩ
4 kΩ
10 kΩ
Vo
Vo
240 − Vo
=
+
50
- j20 40 + j30
88 = (0.36 + j0.38) Vo
Vo =
88
= 168.13∠ - 46.55°
0.36 + j0.38
I1 =
Vo
= 8.41∠43.45°
- j20
I2 =
Vo
= 3.363∠ - 83.42°
40 + j30
Reactive power in the inductor is
1
1
2
S = I 2 Z L = ⋅ (3.363) 2 ( j30) = j169.65 VAR
2
2
Reactive power in the capacitor is
1
1
2
S = I 1 Z c = ⋅ (8.41) 2 (- j20) = - j707.3 VAR
2
2
Chapter 11, Solution 60.
S1 = 20 + j
20
sin(cos -1 (0.8)) = 20 + j15
0.8
S 2 = 16 + j
16
sin(cos -1 (0.9)) = 16 + j7.749
0.9
S = S1 + S 2 = 36 + j22.749 = 42.585∠32.29°
S = Vo I * = 6 Vo
But
Vo =
S
= 7.098 ∠ 32.29°
6
pf = cos(32.29°) = 0.8454 (lagging)
Chapter 11, Solution 61.
Consider the network shown below.
I2
Io
S2
I1
+
Vo
So
S1
S3
−
S 2 = 1.2 − j0.8 kVA
S3 = 4 + j
4
sin(cos -1 (0.9)) = 4 + j1.937 kVA
0.9
Let
S 4 = S 2 + S 3 = 5.2 + j1.137 kVA
But
S4 =
1
V I*
2 o 2
2 S 4 (2)(5.2 + j1.137) × 10 3
I =
=
= 22.74 − j104
Vo
100 ∠90°
*
2
I 2 = 22.74 + j104
2
sin(cos -1 (0.707)) = 2 − j2 kVA
0.707
Similarly,
S1 = 2 − j
But
S1 =
1
Vo I 1*
2
I 1* =
2 S 1 (4 − j4) × 10 3
=
= -40 − j40
Vo
j100
I 1 = -40 + j40
I o = I 1 + I 2 = -17.26 + j144 = 145∠96.83°
So =
1
Vo I *o
2
So =
1
⋅ (100∠90°)(145∠ - 96.83°) VA
2
S o = 7.2 − j0.862 kVA
Chapter 11, Solution 62.
Consider the circuit below
0.2 + j0.04 Ω
I
I2
0.3 + j0.15 Ω
I1
Vs
+
−
+
+
V1
V2
−
−
S 2 = 15 − j
15
sin(cos -1 (0.8)) = 15 − j11.25
0.8
S 2 = V2 I *2
But
I *2 =
S 2 15 − j11.25
=
V2
120
I 2 = 0.125 + j0.09375
V1 = V2 + I 2 (0.3 + j0.15)
V1 = 120 + (0.125 + j0.09375)(0.3 + j0.15)
V1 = 120.02 + j0.0469
S1 = 10 + j
10
sin(cos -1 (0.9)) = 10 + j4.843
0.9
S1 = V1 I 1*
But
I 1* =
S 1 11.111∠25.84°
=
V1 120.02 ∠0.02°
I 1 = 0.093∠ - 25.82° = 0.0837 − j0.0405
I = I 1 + I 2 = 0.2087 + j0.053
Vs = V1 + I (0.2 + j0.04)
Vs = (120.02 + j0.0469) + (0.2087 + j0.053)(0.2 + j0.04)
Vs = 120.06 + j0.0658
Vs = 120.06∠0.03° 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
S 2 = 16 + j
16
sin(cos -1 (0.85)) = 16 + j9.916
0.85
S3 =
(20)(0.6)
+ j20 = 15 + j20
sin(cos -1 (0.6)
S = 43 + j22.987 =
I *o =
1
V I *o
2
2 S 44 + j22.98
=
V
110
I o = 0.4513∠ - 27.58° A
Chapter 11, Solution 64.
I2
I1
8Ω
+
−
Is
120∠0º V
j12
Is + I2 = I1 or Is = I1 – I2
I1 =
But,
120
8 + j12
= 4.615 − j6.923
2500 − j400
S
S = VI ∗2 
→ I ∗2 =
=
= 20.83 − j3.333
V
120
or I 2 = 20.83 + j3.333
Is = I1 – I2 = –16.22 – j10.256 = 19.19∠–147.69˚ A.
Chapter 11, Solution 65.
C = 1 nF 
→
1
-j
= 4
= -j100 kΩ
jωC 10 × 10 -9
At the noninverting terminal,
4∠0° − Vo
Vo
=
100
- j100
Vo =
4
2
v o (t) =

→ Vo =
4
1+ j
∠ - 45°
4
2
cos(10 4 t − 45°)
2
2
 4 1   1 
Vrms
 
W
=
⋅
P=
R
 2 2   50 × 10 3 
P = 80 µW
Chapter 11, Solution 66.
As an inverter,
- Zf
- (2 + j4)
Vo =
Vs =
⋅ (4 ∠45°)
Zi
4 + j3
Io =
Vo
- (2 + j4)(4∠45°)
mA
mA =
(6 - j2)(4 + j3)
6 − j2
The power absorbed by the 6-kΩ resistor is
2
2
1
1  20 × 4 
-6
3
P = Io R = ⋅
 × 10 × 6 × 10
2
2  40 × 5 
P = 0.96 mW
Chapter 11, Solution 67.
ω = 2,
3H
10 //( − j 5) =

→
jωL = j 6,
0.1F

→
1
1
=
= − j5
jωC j 2 x0.1
− j 50
= 2 − j4
10 − j 5
The frequency-domain version of the circuit is shown below.
Z2=2-j4 Ω
Z1 =8+j6 Ω
I1
+
+
+
0.6∠20 o V
Io
Z 3 = 12Ω
Vo
-
(a) I 1 =
0.6∠20 o − 0 0.5638 + j 0.2052
=
= 0.06∠ − 16.87 o
8 + j6
8 + j6
1
S = Vs I *1 = (0.3∠20 o )(0.06∠ + 16.87 o ) = 14.4 + j10.8 mVA = 18∠36.86 o mVA
2
(b) Vo = −
Z2
Vs ,
Z1
P=
Io =
Vo
( 2 − j 4)
=−
(0.6∠20 o ) = 0.0224∠99.7 o
12(8 + j 6)
Z3
1
| I o | 2 R = 0.5(0.0224) 2 (12) = 2.904 mW
2
Chapter 11, Solution 68.
S = SR + SL + Sc
Let
S R = PR + jQ R =
where
1 2
I R + j0
2 o
1
S L = PL + jQ L = 0 + j I o2 ωL
2
1
1
S c = Pc + jQ c = 0 − j I o2 ⋅
ωC
2
S=
Hence,

1 2
1 

I o  R + jωL −

2 
ωC 
Chapter 11, Solution 69.
(a)
Given that Z = 10 + j12
tan θ =
12
10

→ θ = 50.19°
pf = cos θ = 0.6402
(b)
S=
V
2
2 Z*
=
(120) 2
= 295.12 + j354.09
(2)(10 − j12)
The average power absorbed = P = Re(S) = 295.1 W
(c)
For unity power factor, θ1 = 0° , which implies that the reactive power due
to the capacitor is Q c = 354.09
But
C=
Qc =
V2
1
= ωC V 2
2 Xc 2
2 Qc
(2)(354.09)
= 130.4 µF
2 =
(2π )(60)(120) 2
ωV
Chapter 11, Solution 70.
pf = cos θ = 0.8 
→
sin θ = 0.6
Q = S sin θ = (880)(0.6) = 528
If the power factor is to be unity, the reactive power due to the capacitor is
Q c = Q = 528 VAR
But
C=
2
Vrms
1
Q=
= ωC V 2
Xc
2

→ C =
2 Qc
ωV2
(2)(528)
= 69.45 µF
(2π)(50)(220) 2
Chapter 11, Solution 71.
P1 = Q1 = 150 x0.7071 = 106.065,
S 1 = 106.065 + j106.065,
Q2 = 50,
S2 =
Q2
,
0 .6
S 2 = 66.67 − j 50
S = S 1 + S 2 = 172.735 + j 56.06 = 181.6∠17.98 o ,
pf = cos17.98 o = 0.9512
Qc = P(tan θ 1 − tan θ 2 ) = 172.735(tan 17.98 o − 0) = 56.058
C=
Qc
56.058
=
= 10.33 µF
2
ωV rms 2πx60 x120 2
Chapter 11, Solution 72.
(a)
θ1 = cos -1 (0.76) = 40.54°
θ 2 = cos -1 (0.9) = 25.84°
Q c = P (tan θ1 − tan θ 2 )
Q c = (40)[ tan(40.54°) − tan(25.84°) ] kVAR
Q c = 14.84 kVAR
C=
P2 = 0.8S = 0.8
Qc
14840
=
= 2.734 mF
2
ω Vrms (2π )(60)(120) 2
50
= 66.67
0 .6
(b)
θ1 = 40.54° ,
θ 2 = 0°
Q c = (40)[ tan(40.54°) − 0 ] kVAR = 34.21 kVAR
C=
Qc
34210
= 6.3 mF
2
ω Vrms (2π)(60)(120) 2
Chapter 11, Solution 73.
(a)
S = 10 − j15 + j22 = 10 + j7 kVA
S = S = 10 2 + 7 2 = 12.21 kVA
(b)
S = V I*

→ I * =
S 10,000 + j7,000
=
240
V
I = 41.667 − j29.167 = 50.86∠ - 35° A
(c)
7
θ1 = tan -1   = 35° ,
10 
θ 2 = cos -1 (0.96) = 16.26°
Q c = P1 [ tan θ1 − tan θ 2 ] = 10 [ tan(35°) - tan(16.26°) ]
Q c = 4.083 kVAR
C=
(d)
Qc
4083
=
= 188.03 µF
2
ω Vrms (2π )(60)(240) 2
S 2 = P2 + jQ 2 ,
P2 = P1 = 10 kW
Q 2 = Q1 − Q c = 7 − 4.083 = 2.917 kVAR
S 2 = 10 + j2.917 kVA
But
I *2 =
S 2 = V I *2
S 2 10,000 + j2917
=
V
240
I 2 = 41.667 − j12.154 = 43.4∠ - 16.26° A
Chapter 11, Solution 74.
(a)
θ1 = cos -1 (0.8) = 36.87°
P1
24
=
= 30 kVA
cos θ1 0.8
S1 =
Q1 = S1 sin θ1 = (30)(0.6) = 18 kVAR
S1 = 24 + j18 kVA
θ 2 = cos -1 (0.95) = 18.19°
S2 =
P2
40
=
= 42.105 kVA
cos θ 2 0.95
Q 2 = S 2 sin θ 2 = 13.144 kVAR
S 2 = 40 + j13.144 kVA
S = S1 + S 2 = 64 + j31.144 kVA
 31.144 
 = 25.95°
θ = tan -1 
 64 
pf = cos θ = 0.8992
(b)
θ 2 = 25.95° ,
θ1 = 0°
Q c = P [ tan θ 2 − tan θ1 ] = 64 [ tan(25.95°) − 0 ] = 31.144 kVAR
C=
Qc
31,144
=
= 5.74 mF
2
ω Vrms (2π )(60)(120) 2
Chapter 11, Solution 75.
(a)
2
V
S1 =
Z1*
=
(240) 2
5760
=
= 517.75 − j323.59 VA
80 + j50 8 + j5
(240) 2
5760
S2 =
=
= 358.13 + j208.91 VA
120 − j70 12 − j7
S3 =
(240) 2
= 960 VA
60
S = S1 + S 2 + S 3 = 1835.88 − j114.68 VA
(b)
 114.68 
 = 3.574°
θ = tan -1 
1835.88 
pf = cos θ = 0.998
(c)
Q c = P [ tan θ 2 − tan θ1 ] = 18.35.88[ tan(3.574°) − 0 ]
Q c = 114.68 VAR
C=
Qc
114.68
=
= 6.336 µF
2
ω Vrms (2π )(50)(240) 2
Chapter 11, Solution 76.
The wattmeter reads the real power supplied by the current source. Consider the
circuit below.
4Ω
12∠0° V
-j3 Ω
+
−
3∠30° +
Vo
j2 Ω
12 − Vo Vo Vo
=
+
4 − j3
j2
8
8Ω
3∠30° A
Vo =
S=
36.14 + j23.52
= 0.7547 + j11.322 = 11.347 ∠86.19°
2.28 − j3.04
1
1
Vo I *o = ⋅ (11.347 ∠86.19°)(3∠ - 30°)
2
2
S = 17.021∠56.19°
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 Ω.
120 cos(2t ) 
→ 120∠0° ,
ω= 2
4H 
→
jωL = j8
1
0.1 F 
→
= -j5
jωC
Consider the following circuit.
6Ω
120∠0° V
Z = 15 || (-j5) =
j8 Ω
I
+
−
(15)(-j5)
= 1.5 − j4.5
15 − j5
I=
120
= 14.5∠ - 25.02°
(6 + j8) + (1.5 − j4.5)
S=
1
1 2
1
V I * = I Z = ⋅ (14.5) 2 (1.5 − j4.5)
2
2
2
S = 157.69 − j473.06 VA
The wattmeter reads
P = Re(S) = 157.69 W
Z
Chapter 11, Solution 78.
The wattmeter reads the power absorbed by the element to its right side.
ω= 4
2 cos(4t ) 
→ 2∠0° ,
1H 
→
jωL = j4
1
→
F 
12
1
= -j3
jωC
Consider the following circuit.
10 Ω
20∠0° V
I
+
−
Z = 5 + j4 + 4 || - j3 = 5 + j4 +
Z
(4)(- j3)
4 − j3
Z = 6.44 + j2.08
I=
20
= 1.207 ∠ - 7.21°
16.44 + j2.08
S=
1 2
1
I Z = ⋅ (1.207) 2 (6.44 + j2.08)
2
2
P = Re(S) = 4.691 W
Chapter 11, Solution 79.
The wattmeter reads the power supplied by the source and partly absorbed by the 40- Ω
resistor.
ω = 100,
j100x10x10 − 3 = j,

→
10 mH
500µF

→
1
1
=
= − j20
jωC 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 3(V1 − V2 ) V1 − V2
= 2I o + 1
+
=
+
40
j
20
20
j
10 = (7 − j40)V1 + (−6 + j40)V2
→
(1)
At node 2,
V1 − V 2 V1 − V 2
V
+
= 2
j
20
− j 20

→
0 = (20 + j )V1 − (19 + j )V 2
Solving (1) and (2) yields V1 = 1.5568 –j4.1405
I=
10 − V1
= 0.8443 + j 0.4141,
40
1
S = VI • = 4.2216 − j 2.0703
2
P = Re(S) = 4.222 W.
Chapter 11, Solution 80.
(a)
I=
V 110
=
= 17.19 A
Z 6.4
(2)
(b)
V 2 (110) 2
=
= 1890.625
S=
Z
6 .4
cos θ = pf = 0.825 
→ θ = 34.41°
P = S cos θ = 1559.76 ≅ 1.6 kW
Chapter 11, Solution 81.
kWh consumed = 4017 − 3246 = 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,
Q1 = 0
P2 = 30,000 x0.82 = 24,600,
Q2 = 30,000 sin(cos −1 0.82) = 17,171
S = S1 + S 2 = (P1 + P2 ) + j(Q1 + Q 2 ) = 29,600 + j17,171
S =| S |= 34.22 kVA
(b)
Q = 17.171 kVAR
(c )
pf =
P 29,600
=
= 0.865
S 34,220
Q c = P(tan θ1 − tan θ 2 )
[
]
= 29,600 tan(cos −1 0.865) − tan(cos −1 0.9) = 2833 VAR
(d)
C=
Qc
2833
=
= 130.46µ F
2
ωV rms 2πx60 x 240 2
Chapter 11, Solution 83.
1
1
(a) S = VI ∗ = (210∠60 o )(8∠ − 25 o ) = 840∠35 o
2
2
P = S cosθ = 840 cos 35 o = 688.1 W
(b) S = 840 VA
(c) Q = S sin θ = 840 sin 35 o = 481.8 VAR
(d) pf = P / S = cos 35 o = 0.8191 (lagging)
Chapter 11, Solution 84.
(a)
Maximum demand charge = 2,400 × 30 = $72,000
Energy cost = $0.04 × 1,200 × 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 × 10 3
Chapter 11, Solution 85.

→
j 2πx60 x15 x10 −3 = j 5.655
(a) 15 mH
We apply mesh analysis as shown below.
I1
+
Ix
120<0o V
-
10 Ω
In
30 Ω
Iz
120<0o V
10 Ω
+
Iy
-
j5.655 Ω
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)
S1 =
1
(120) I • x = 60 x 20 = 1200,
2
S2 =
1
(120) I • y = 1025.5 − j 308.5
2
S = S1 + S 2 = 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 
→ Z i = Z Th = Z *L
Z L = R + jωL = 75 + j (2π)(4.12 × 10 6 )(4 × 10 -6 )
Z L = 75 + j103.55 Ω
Z i = 75 − j103.55 Ω
Chapter 11, Solution 87.
Z = R ± jX
VR = I R
2

→ R =
Z = R 2 + X2
X = 2.5377 kΩ
VR
80
=
= 1 .6 k Ω
I
50 × 10 -3

→ X 2 = Z
2
− R 2 = (3) 2 − (1.6) 2
X
 2.5377 
 = 57.77°
θ = tan -1   = tan -1 
R 
 1.6 
pf = cos θ = 0.5333
Chapter 11, Solution 88.
(a)
S = (110)(2 ∠55°) = 220∠55°
P = S cos θ = 220 cos(55°) = 126.2 W
(b)
S = S = 220 VA
Chapter 11, Solution 89.
(a)
Apparent power = S = 12 kVA
P = S cos θ = (12)(0.78) = 9.36 kW
Q = S sin θ = 12 sin(cos -1 (0.78)) = 7.51 kVAR
S = P + jQ = 9.36 + j7.51 kVA
(b)
V
S= *
Z
2
V

→ Z =
S
2
*
=
(210) 2
(9.36 + j7.51) × 10 3
Z = 34.398 + j27.6 Ω
Chapter 11, Solution 90
Original load :
P1 = 2000 kW ,
S1 =
cos θ1 = 0.85 
→ θ1 = 31.79°
P1
= 2352.94 kVA
cos θ1
Q1 = S1 sin θ1 = 1239.5 kVAR
Additional load :
P2 = 300 kW ,
S2 =
cos θ 2 = 0.8 
→ θ 2 = 36.87°
P2
= 375 kVA
cos θ 2
Q 2 = S 2 sin θ 2 = 225 kVAR
Total load :
S = S1 + S 2 = (P1 + P2 ) + j (Q1 + Q 2 ) = P + jQ
P = 2000 + 300 = 2300 kW
Q = 1239.5 + 225 = 1464.5 kVAR
The minimum operating pf for a 2300 kW load and not exceeding the kVA rating of the
generator is
P
2300
cos θ =
=
= 0.9775
S1 2352.94
or
θ = 12.177°
The maximum load kVAR for this condition is
Q m = S1 sin θ = 2352.94 sin(12.177°)
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 θ
pf = cos θ =
P
2700
=
= 0.8182
S (220)(15)
Q = S sin θ = 220(15) sin(35.09°) = 1897.3
When the power is raised to unity pf, θ1 = 0° and Q c = Q = 1897.3
C=
Qc
1897.3
=
= 104 µF
2
ω Vrms (2π )(60)(220) 2
Chapter 11, Solution 92
(a)
Apparent power drawn by the motor is
P
60
Sm =
=
= 80 kVA
cos θ 0.75
Q m = S 2 − P 2 = (80) 2 − (60) 2 = 52.915 kVAR
Total real power
P = Pm + Pc + PL = 60 + 0 + 20 = 80 kW
Total reactive power
Q = Q m + Q c + Q L = 52.915 − 20 + 0 = 32.91 kVAR
Total apparent power
S = P 2 + Q 2 = 86.51 kVA
(b)
pf =
(c)
I=
P
80
=
= 0.9248
S 86.51
S 86510
=
= 157.3 A
V
550
Chapter 11, Solution 93
(a)
P1 = (5)(0.7457) = 3.7285 kW
S1 =
P1 3.7285
=
= 4.661 kVA
pf
0.8
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
Q 2 = 0 VAR
P3 = (10)(120) = 1.2 kW ,
S 3 = 1.2 + j0 kVA
Q 3 = 0 VAR
Q 4 = 1.6 kVAR ,
S4 =
cos θ 4 = 0.6 
→ sin θ 4 = 0.8
Q4
= 2 kVA
sin θ 4
P4 = S 4 cos θ 4 = (2)(0.6) = 1.2 kW
S 4 = 1.2 − j1.6 kVA
S = S1 + S 2 + S 3 + S 4
S = 7.3285 + j1.196 kVA
Total real power = 7.3285 kW
Total reactive power = 1.196 kVAR
(b)
 1.196 
 = 9.27°
θ = tan -1 
 7.3285 
pf = cos θ = 0.987
Chapter 11, Solution 94
cos θ1 = 0.7 
→ θ1 = 45.57°
S1 = 1 MVA = 1000 kVA
P1 = S1 cos θ1 = 700 kW
Q1 = S1 sin θ1 = 714.14 kVAR
For improved pf,
cos θ 2 = 0.95 
→ θ 2 = 18.19°
P2 = P1 = 700 kW
S2 =
P2
700
=
= 736.84 kVA
cos θ 2 0.95
Q 2 = S 2 sin θ 2 = 230.08 kVAR
P1 = P2 = 700 kW
θ1
θ2
Q2
S2
S1
Q1
(a)
Qc
Reactive power across the capacitor
Q c = Q1 − Q 2 = 714.14 − 230.08 = 484.06 kVAR
Cost of installing capacitors = $30 × 484.06 = $14,521.80
(b)
Substation capacity released = S1 − S 2
= 1000 − 736.84 = 263.16 kVA
Saving in cost of substation and distribution facilities
= $120 × 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 = R s − jXc
ZL = R L + jX 2
Source impedance
Load impedance
For maximum load transfer
Z L = Z *s 
→ R s = R L , X c = X L
Xc = XL
or
ω=

→
1
LC
1
= ωL
ωC
= 2π f
f=
(b)
1
2π LC
=
1
2π (80 × 10 -3 )(40 × 10 -9 )
Vs2
(4.6) 2
P=
=
= 529 mW
4 R L (4)(10)
= 2.814 kHz
(since Vs is in rms)
Chapter 11, Solution 96
ZTh
+
−
VTh
(a)
ZL
VTh = 146 V, 300 Hz
Z Th = 40 + j8 Ω
Z L = Z *Th = 40 − j8 Ω
(b)
P=
VTh
2
8 R Th
(146) 2
=
= 66.61 W
(8)(40)
Chapter 11, Solution 97
Z T = (2)(0.1 + j) + (100 + j20) = 100.2 + j22 Ω
I=
Vs
240
=
Z T 100.2 + j22
2
P = I R L = 100 I
2
(100)(240) 2
=
= 547.3 W
(100.2) 2 + (22) 2
Chapter 12, Solution 1.
(a)
If Vab = 400 , then
Van =
400
3
∠ - 30° = 231∠ - 30° V
Vbn = 231∠ - 150° V
Vcn = 231∠ - 270° V
(b)
For the acb sequence,
Vab = Van − Vbn = Vp ∠0° − Vp ∠120°
 1
3
Vab = Vp 1 + − j  = Vp 3∠ - 30°
2 
 2
i.e. in the acb sequence, Vab lags Van by 30°.
Hence, if Vab = 400 , then
Van =
400
3
∠30° = 231∠30° V
Vbn = 231∠150° V
Vcn = 231∠ - 90° V
Chapter 12, Solution 2.
Since phase c lags phase a by 120°, this is an acb sequence.
Vbn = 160∠(30° + 120°) = 160∠150° V
Chapter 12, Solution 3.
Since Vbn leads Vcn by 120°, this is an abc sequence.
Van = 208∠(130° + 120°) = 208∠ 250° V
Chapter 12, Solution 4.
Vbc = Vca ∠120° = 208∠140° V
Vab = Vbc ∠120° = 208∠260° V
Van =
Vab
3 ∠30°
=
208∠260°
3 ∠30°
= 120∠230° V
Vbn = Van ∠ - 120° = 120∠110° V
Chapter 12, Solution 5.
This is an abc phase sequence.
Vab = Van 3 ∠30°
or
Van =
Vab
3 ∠30°
=
420∠0°
3 ∠30°
= 242.5∠ - 30° V
Vbn = Van ∠ - 120° = 242.5∠ - 150° V
Vcn = Van ∠120° = 242.5∠90° V
Chapter 12, Solution 6.
Z Y = 10 + j5 = 11.18∠26.56°
The line currents are
Van
220 ∠0°
=
= 19.68∠ - 26.56° A
Ia =
Z Y 11.18∠26.56°
I b = I a ∠ - 120° = 19.68∠ - 146.56° A
I c = I a ∠120° = 19.68∠93.44° A
The line voltages are
Vab = 200 3 ∠30° = 381∠30° V
Vbc = 381∠ - 90° V
Vca = 381∠ - 210° V
The load voltages are
VAN = I a Z Y = Van = 220∠0° V
VBN = Vbn = 220∠ - 120° V
VCN = Vcn = 220∠120° V
Chapter 12, Solution 7.
This is a balanced Y-Y system.
440∠0° V
+
−
ZY = 6 − j8 Ω
Using the per-phase circuit shown above,
440∠0°
Ia =
= 44∠53.13° A
6 − j8
I b = I a ∠ - 120° = 44∠ - 66.87° A
I c = I a ∠120° = 44∠173.13° A
Chapter 12, Solution 8.
VL = 220 V ,
I an =
Vp
ZY
=
I L = 6.918 A
Z Y = 16 + j9 Ω
VL
3 ZY
=
220
3 (16 + j9)
= 6.918∠ - 29.36°
Chapter 12, Solution 9.
Ia =
Van
120 ∠0°
= 4.8∠ - 36.87° A
=
Z L + Z Y 20 + j15
I b = I a ∠ - 120° = 4.8∠ - 156.87° A
I c = I a ∠120° = 4.8∠83.13° A
As a balanced system, I n = 0 A
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
220 ∠0°
= 6.55∠36.53°
=
Z A + 2 27 − j20
Ib =
Vbn
220 ∠ - 120°
=
= 10 ∠ - 120°
ZB + 2
22
Ic =
Vcn
220 ∠120°
= 16.92 ∠97.38°
=
ZC + 2
12 + j5
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
- I n = (5.263 + j3.9) + (-5 − j8.66) + (-2.173 + j16.78)
I n = 1.91 − j12.02 = 12.17 ∠ - 81° A
Chapter 12, Solution 11.
Van =
Vbc
3 ∠ - 90°
VBC
=
3 ∠ - 90°
=
220∠10°
3 ∠ - 90°
Van = 127 ∠100° V
VAB = VBC ∠120° = 220∠130° V
VAC = VBC ∠ - 120° = 220∠ - 110° V
If I bB = 30 ∠60° , then
I aA = 30∠180° ,
I AB =
I aA
3 ∠ - 30°
I cC = 30 ∠ - 60°
=
30∠180°
3 ∠ - 30°
I BC = 17.32∠90° ,
= 17.32∠210°
I CA = 17.32 ∠ - 30°
I AC = -I CA = 17.32∠150° A
I BC Z = VBC
Z=
VBC
220 ∠0°
=
= 12.7 ∠ - 80° Ω
I BC 17.32 ∠90°
Chapter 12, Solution 12.
Convert the delta-load to a wye-load and apply per-phase analysis.
Ia
110∠0° V
ZY =
+
−
Z∆
= 20 ∠45° Ω
3
ZY
110∠0°
= 5.5∠ - 45° A
20∠45°
I b = I a ∠ - 120° = 5.5∠ - 165° A
Ia =
I c = I a ∠120° = 5.5∠75° A
Chapter 12, Solution 13.
First we calculate the wye equivalent of the balanced load.
ZY = (1/3)Z∆ = 6+j5
Now we only need to calculate the line currents using the wye-wye circuits.
110
= 6.471∠ − 61.93° A
2 + j10 + 6 + j5
110∠ − 120°
Ib =
= 6.471∠178.07° A
8 + j15
110∠120°
Ic =
= 6.471∠58.07° A
8 + j15
Ia =
Chapter 12, Solution 14.
We apply mesh analysis.
1 + j 2Ω
A
a
+
100∠0 o V
-
ZL
ZL
I3
n
100∠120 o V
+
c
I1
-
100∠120 o V
+
b
I2
B
C
Z L = 12 + j12Ω
1 + j 2Ω
1 + j 2Ω
For mesh 1,
− 100 + 100∠120 o + I 1 (14 + j16) − (1 + j 2) I 2 − (12 + j12) I 3 = 0
or
(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
I 1 = −3.161 − j19.3,
I 2 = −10.098 − j16.749,
I aA = I 1 = 19.58∠ − 99.3 A
o
I bB = I 2 − I 1 = 7.392∠159.8 o A
I cC = − I 2 = 19.56∠58.91o A
I 3 = −4.4197 − j12.016
Chapter 12, Solution 15.
Convert the delta load, Z ∆ , to its equivalent wye load.
Z Ye =
Z∆
= 8 − j10
3
Z p = Z Y || Z Ye =
(12 + j5)(8 − j10)
= 8.076∠ - 14.68°
20 − j5
Z p = 7.812 − j2.047
Z T = Z p + Z L = 8.812 − j1.047
Z T = 8.874 ∠ - 6.78°
We now use the per-phase equivalent circuit.
Vp
210
Ia =
,
where Vp =
Zp + ZL
3
Ia =
210
3 (8.874 ∠ - 6.78°)
= 13.66 ∠6.78°
I L = I a = 13.66 A
Chapter 12, Solution 16.
(a)
I CA = - I AC = 10∠(-30° + 180°) = 10∠150°
This implies that
I AB = 10 ∠30°
I BC = 10∠ - 90°
I a = I AB 3 ∠ - 30° = 17.32∠0° A
I b = 17.32∠ - 120° A
I c = 17.32∠120° A
(b)
Z∆ =
VAB 110 ∠0°
=
= 11∠ - 30° Ω
I AB 10 ∠30°
Chapter 12, Solution 17.
Convert the ∆-connected load to a Y-connected load and use per-phase analysis.
ZL
Van
ZY =
Ia =
Ia
+
−
ZY
Z∆
= 3 + j4
3
Van
120 ∠0°
=
= 19.931∠ - 48.37°
Z Y + Z L (3 + j4) + (1 + j0.5)
But
I AB =
I a = I AB 3 ∠ - 30°
19.931∠ - 48.37°
3 ∠ - 30°
= 11.51∠ - 18.37° A
I BC = 11.51∠ - 138.4° A
I CA = 11.51∠101.6° A
VAB = I AB Z ∆ = (11.51∠ - 18.37°)(15∠53.13°)
VAB = 172.6∠34.76° V
VBC = 172.6∠ - 85.24° V
VCA = 172.6∠154.8° V
Chapter 12, Solution 18.
VAB = Van 3 ∠30° = (440 ∠60°)( 3 ∠30°) = 762.1∠90°
Z ∆ = 12 + j9 = 15∠36.87°
I AB =
VAB 762.1∠90°
= 50.81∠53.13° A
=
Z ∆ 15∠36.87°
I BC = I AB ∠ - 120° = 50.81∠ - 66.87° A
I CA = I AB ∠120° = 50.81∠173.13° A
Chapter 12, Solution 19.
Z ∆ = 30 + j10 = 31.62 ∠18.43°
The phase currents are
Vab
173∠0°
I AB =
= 5.47 ∠ - 18.43° A
=
Z ∆ 31.62 ∠18.43°
I BC = I AB ∠ - 120° = 5.47 ∠ - 138.43° A
I CA = I AB ∠120° = 5.47 ∠101.57° A
The line currents are
I a = I AB − I CA = I AB 3 ∠ - 30°
I a = 5.47 3 ∠ - 48.43° = 9.474∠ - 48.43° A
I b = I a ∠ - 120° = 9.474∠ - 168.43° A
I c = I a ∠120° = 9.474∠71.57° A
Chapter 12, Solution 20.
Z ∆ = 12 + j9 = 15∠36.87°
The phase currents are
210∠0°
= 14∠ - 36.87° A
15∠36.87°
= I AB ∠ - 120° = 14∠ - 156.87° A
I AB =
I BC
I CA = I AB ∠120° = 14∠83.13° A
The line currents are
I a = I AB 3 ∠ - 30° = 24.25∠ - 66.87° A
I b = I a ∠ - 120° = 24.25∠ - 186.87° A
I c = I a ∠120° = 24.25∠53.13° A
Chapter 12, Solution 21.
(a)
I AC =
− 230∠120°
− 230∠120°
=
= 17.96∠ − 98.66° A(rms)
10 + j8
12.806∠38.66°
230∠ − 120 230∠0°
−
10 + j8
10 + j8
= 17.96∠ − 158.66° − 17.96∠ − 38.66°
= −16.729 − j6.536 − 14.024 + j11.220 = −30.75 + j4.684
= 31.10∠171.34° A
I bB = I BC + I BA = I BC − I AB =
(b)
Chapter 12, Solution 22.
Convert the ∆-connected source to a Y-connected source.
Vp
208
Van =
∠ - 30° =
∠ - 30° = 120 ∠ - 30°
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
+
−
Ia =
Z
Van
120 ∠30°
=
= 15.53∠ - 28.4° A
Z L + Z 7.723 − j0.2153
I b = I a ∠ - 120° = 15.53∠ - 148.4° A
I c = I a ∠120° = 15.53∠91.6° A
Chapter 12, Solution 23.
(a) I AB =
V AB
208
=
Z∆
25∠60 o
I a = I AB
208 3∠ − 30 o
3∠ − 30 =
= 14.411∠ − 90 o
o
25∠60
o
I L =| I a |= 14.41 A
 208 3 
 cos 60 o = 2.596 kW
(b) P = P1 + P2 = 3VL I L cosθ = 3 (208)

 25 
Chapter 12, Solution 24.
Convert both the source and the load to their wye equivalents.
Z∆
= 20 ∠30° = 17.32 + j10
ZY =
3
Van =
Vab
3
∠ - 30° = 240.2∠0°
We now use per-phase analysis.
1+jΩ
Van
Ia =
+
−
Ia
20∠30° Ω
Van
240.2
=
= 11.24∠ - 31° A
(1 + j) + (17.32 + j10) 21.37 ∠31°
I b = I a ∠ - 120° = 11.24∠ - 151° A
I c = I a ∠120° = 11.24∠89° A
But
I AB =
I a = I AB 3 ∠ - 30°
11.24 ∠ - 31°
3 ∠ - 30°
= 6.489∠ - 1° A
I BC = I AB ∠ - 120° = 6.489∠ - 121° A
I CA = I AB ∠120° = 6.489∠119° A
Chapter 12, Solution 25.
Convert the delta-connected source to an equivalent wye-connected source and
consider the single-phase equivalent.
Ia =
where
440 ∠(10° − 30°)
3 ZY
Z Y = 3 + j2 + 10 − j8 = 13 − j6 = 14.32 ∠ - 24°.78°
Ia =
440 ∠ - 20°
3 (14.32 ∠ - 24.78°)
= 17.74∠4.78° A
I b = I a ∠ - 120° = 17.74∠ - 115.22° A
I c = I a ∠120° = 17.74∠124.78° A
Chapter 12, Solution 26.
Transform the source to its wye equivalent.
Vp
Van =
∠ - 30° = 72.17 ∠ - 30°
3
Now, use the per-phase equivalent circuit.
Van
,
Z = 24 − j15 = 28.3∠ - 32°
I aA =
Z
I aA =
72.17 ∠ - 30°
= 2.55∠ 2° A
28.3∠ - 32°
I bB = I aA ∠ - 120° = 2.55∠ - 118° A
I cC = I aA ∠120° = 2.55∠122° A
Chapter 12, Solution 27.
Ia =
Vab ∠ - 30°
3 ZY
=
220∠ - 10°
3 (20 + j15)
I a = 5.081∠ - 46.87° A
I b = I a ∠ - 120° = 5.081∠ - 166.87° A
I c = I a ∠120° = 5.081∠73.13° A
Chapter 12, Solution 28.
Let
Vab = 400∠0°
Ia =
Van ∠ - 30°
3 ZY
=
400∠ - 30°
3 (30 ∠ - 60°)
= 7.7 ∠30°
I L = I a = 7.7 A
VAN = I a Z Y =
Van
3
∠ - 30° = 230.94∠ - 30°
Vp = VAN = 230.9 V
Chapter 12, Solution 29.
P = 3Vp I p cos θ ,
Vp =
VL
3
IL = Ip
,
P = 3 VL I L cos θ
IL =
P
3 VL cos θ
ZY =
Vp
Ip
=
=
VL
3 IL
5000
240 3 (0.6)
=
= 20.05 = I p
240
3 (20.05)
= 6.911
cos θ = 0.6 
→ θ = 53.13°
Z Y = 6.911∠ - 53.13° (leading)
Z Y = 4.15 − j5.53 Ω
S=
P 5000
=
= 8333
pf
0.6
Q = S sin θ = 6667
S = 5000 − j6667 VA
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 =
S=
3V 2 p
,
Z*p
Vp =
VL
3
V 2L
(208) 2
=
= 1.4421∠45 o kVA
Z * p 30∠ − 45 o
P = S cosθ = 1.02 kW
Chapter 12, Solution 31.
(a)
Pp = 6,000,
cosθ = 0.8,
Sp =
PP
= 6 / 0.8 = 7.5 kVA
cos θ
Q p = S P sin θ = 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
3(240) 2
=
,
S
(18 + j13.5) x10 3
6000
(b)
Pp = 3VL I L cosθ
(c )
We find C to bring the power factor to unity

→
Qc = Q p = 4.5 kVA
IL =

→
C=
Chapter 12, Solution 32.
S = 3 VL I L ∠θ
S = S = 3 VL I L = 50 × 10 3
IL =
5000
3 (440)
= 65.61 A
3 x 240 x0.8
Z P = 6.144 + j 4.608Ω
= 18.04 A
Qc
4500
=
= 207.2 µF
2
ωV rms 2πx60 x 240 2
For a Y-connected load,
I p = I L = 65.61 ,
Z =
Vp
Ip
=
Vp =
VL
3
=
440
= 254.03
254.03
= 3.872
65.61
Z = Z ∠θ ,
θ = cos -1 (0.6) = 53.13°
Z = (3.872)(cos θ + j sin θ)
Z = (3.872)(0.6 + j0.8)
Z = 2.323 + j3.098 Ω
Chapter 12, Solution 33.
S = 3 VL I L ∠θ
S = S = 3 VL I L
For a Y-connected load,
VL = 3 Vp
IL = Ip ,
S = 3 Vp I p
IL = Ip =
3
S
4800
=
= 7.69 A
3 Vp (3)(208)
VL = 3 Vp = 3 × 208 = 360.3 V
Chapter 12, Solution 34.
Vp =
Ia =
VL
3
Vp
ZY
=
220
3
200
=
3 (10 − j16)
= 6.73∠58°
I L = I p = 6.73 A
S = 3 VL I L ∠θ = 3 × 220 × 6.73∠ - 58°
S = 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
Z '' y =
1
Z ∆ = (60 + j 30) / 3 = 20 + j10
3
IL
+
230 V
-
Z’y
Z y = Z ' y // Z '' y = (40 + j10) //( 20 + j10) = 13.5 + j 5.5
IL =
230
= 14.61 − j 5.953 A
13.5 + j 5.5
(b) S = Vs I * L = 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 = 3V p I * p

→
I*p =
S
(0.75 + j 0.6614) x10 6
=
= 59.52 + j 52.49
3V p
3x 4200
PL =| I p | 2 Rl = (79.36) 2 (4) = 25.19 kW
(c) Vs = VL + I p (4 + j ) = 4.4381 − j 0.21 kV = 4.443∠ - 2.709 o kV
Chapter 12, Solution 37.
S=
P
12
=
= 20
pf 0.6
S = S∠θ = 20∠θ = 12 − j16 kVA
But
IL =
S = 3 VL I L ∠θ
20 × 10 3
3 × 208
S = 3 Ip
2
= 55.51 A
Zp
For a Y-connected load, I L = I p .
Zp =
S
3 IL
2
(12 − j16) × 10 3
=
(3)(55.51) 2
Z p = 1.298 − j1.731 Ω
Chapter 12, Solution 38.
As a balanced three-phase system, we can use the per-phase equivalent shown
below.
Ia =
110∠0°
110∠0°
=
(1 + j2) + (9 + j12) 10 + j14
Sp =
1
I
2 a
2
ZY =
1
(110) 2
⋅
⋅ (9 + j12)
2 (10 2 + 14 2 )
The complex power is
3 (110) 2
S = 3S p = ⋅
⋅ (9 + j12)
2 296
S = 551.86 + j735.81 VA
Chapter 12, Solution 39.
Consider the system shown below.
5Ω
a
100∠120°
c
+
−
+
− +
100∠-120°
−
100∠0°
A
5Ω
b
8Ω
B
I2
4Ω
-j6 Ω
I1
j3 Ω
I3
C
10 Ω
5Ω
For mesh 1,
100 = (18 − j6) I 1 − 5 I 2 − (8 − j6) I 3
(1)
100 ∠ - 120° = 20 I 2 − 5 I 1 − 10 I 3
20∠ - 120° = - I 1 + 4 I 2 − 2 I 3
(2)
For mesh 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∠ - 120° = -1.25 I 1 + 5 I 2 − 2.5 I 3
(6)
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
  44 − j18 - 38 + j15  I 1 

200
87.5 − j12.65 = 16.75 − j6 - 10.5 + j6  I 
 3 
 

∆ = 192.5 − j26.25 ,
∆ 1 = 900.25 − j935.2 ,
∆ 3 = 110.3 − j1327.6
I1 =
∆ 1 1298.1∠ - 46.09°
= 6.682 ∠ - 38.33° = 5.242 − j4.144
=
194.28∠ - 7.76°
∆
I3 =
∆ 3 1332.2∠ - 85.25°
=
= 6.857∠ - 77.49° = 1.485 − j6.694
∆
194.28∠ - 7.76°
We obtain I 2 from (6),
1
1
I 2 = 5∠ - 120° + I 1 + I 3
4
2
I 2 = (-2.5 − j4.33) + (1.3104 − j1.0359) + (0.7425 − j3.347)
I 2 = -0.4471 − j8.713
The average power absorbed by the 8-Ω resistor is
2
2
P1 = I 1 − I 3 (8) = 3.756 + j2.551 (8) = 164.89 W
The average power absorbed by the 4-Ω resistor is
2
P2 = I 3 (4) = (6.8571) 2 (4) = 188.1 W
The average power absorbed by the 10-Ω 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 ∠0°
Ia =
= 8.567 ∠ - 46.75°
(1 + j0.5) + (7 + j8)
For a wye-connected load,
I p = I a = I a = 8.567
S = 3 Ip
2
Z p = (3)(8.567) 2 (7 + j8)
P = Re(S) = (3)(8.567) 2 (7) = 1.541 kW
Chapter 12, Solution 41.
S=
P 5 kW
=
= 6.25 kVA
pf
0.8
But
IL =
S = 3 VL I L
S
3 VL
=
6.25 × 10 3
3 × 400
= 9.021 A
Chapter 12, Solution 42.
The load determines the power factor.
40
tan θ =
= 1.333 
→ θ = 53.13°
30
pf = cos θ = 0.6 (leading)
 7.2 
S = 7.2 − j (0.8) = 7.2 − j9.6 kVA
 0.6 
S = 3 Ip
But
Ip
2
=
2
Zp
S
(7.2 − j9.6) × 10 3
= 80
=
3Zp
(3)(30 − j40)
I p = 8.944 A
I L = I p = 8.944 A
VL =
S
3 IL
=
12 × 10 3
3 (8.944)
= 774.6 V
Chapter 12, Solution 43.
S = 3 Ip
2
Zp ,
I p = I L for Y-connected loads
S = (3)(13.66) 2 (7.812 − j2.047)
S = 4.373 − j1.145 kVA
Chapter 12, Solution 44.
For a ∆-connected load,
Vp = VL ,
IL = 3 Ip
S = 3 VL I L
IL =
S
3 VL
=
(12 2 + 5 2 ) × 10 3
3 (240)
= 31.273
At the source,
VL' = VL + I L Z L
VL' = 240∠0° + (31.273)(1 + j3)
VL' = 271.273 + j93.819
VL' = 287.04 V
Also, at the source,
S ' = 3VL' I *L
S ' = 3 (271.273 + j93.819)(31.273)
 93.819 
 = 19.078
θ = tan -1 
 271.273 
pf = cos θ = 0.9451
Chapter 12, Solution 45.
S = 3 VL I L ∠θ
IL =
IL =
S ∠-θ
3 VL
,
(635.6) ∠ - θ
3 × 440
S =
P 450 × 10 3
=
= 635.6 kVA
pf
0.708
= 834 ∠ - 45° A
At the source,
VL = 440 ∠0° + I L (0.5 + j2)
VL
VL
VL
VL
= 440 + (834 ∠ - 45°)(2.062 ∠76°)
= 440 + 1719.7 ∠31°
= 1914.1 + j885.7
= 2.109∠24.83° V
Chapter 12, Solution 46.
For the wye-connected load,
IL = Ip ,
VL = 3 Vp
S = 3 Vp I *p =
S=
VL
2
Z*
3 Vp
2
=
Z*
I p = Vp Z
3 VL
(110) 2
=
= 121 W
100
S = 3V I =
S=
2
Z*
For the delta-connected load,
Vp = VL ,
IL = 3 Ip ,
*
p p
3
3 Vp
Z*
2
=
3 VL
I p = Vp Z
2
Z*
(3)(110) 2
= 363 W
100
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 = 0.8 (lagging) 
→ θ = cos -1 (0.8) = 36.87°
S1 = 250 ∠36.87° = 200 + j150 kVA
pf = 0.95 (leading) 
→ θ = cos -1 (0.95) = -18.19°
S 2 = 300 ∠ - 18.19° = 285 − j93.65 kVA
pf = 1.0 
→ θ = cos -1 (1) = 0°
S 3 = 450 kVA
S T = S1 + S 2 + S 3 = 935 + j56.35 = 936.7 ∠3.45° kVA
S T = 3 VL I L
IL =
936.7 × 10 3
3 (13.8 × 10 3 )
= 39.19 A rms
pf = cos θ = cos(3.45°) = 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 Ω
ZA
40+j15 Ω
ZB
ZC
C
B
60 Ω
ZA =
(40 + j15)(18 − j12)
= 7.577 − j1.923
118 + j 3
ZB =
60(40 + j15).
= 20.52 − j 7.105
118 + j 3
ZC =
60(18 − j12)
= 8.992 − j 6.3303
118 + j 3
The system becomes that shown below.
C
B
a
2+j3
A
+
240<0o
240<120
+
c
o
-
I2
ZA
I1
o
240<-120
+
b
ZB
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 ) = 0
or
(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 ) = 0
or
(1)
− (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
(2)
I aA = I 1 = 24.34∠ − 12.64 o A,
I bB = I 2 − I 1 = 10.81∠ − 142.6 o A
I cC = − I 2 = 19.27∠141.9 o A
(b)
S a = (240∠0 o )(24.34∠12.64 o ) = 5841.6∠12.64 o
S b = (240∠ − 120 o )(10.81∠142.6 o ) = 2594.4∠22.6 o
S b = (240∠120 o )(19.27∠ − 141.9 o ) = 4624.8∠ − 21.9 o
S = S a + S b + S c = 12.386 + j 0.55 kVA = 12.4∠2.54 o kVA
Chapter 12, Solution 49.
(a) For the delta-connected load, Z p = 20 + j10Ω,
S=
V p = VL = 220 (rms) ,
3V 2 p
3 x 220 2
=
= 5808 + j 2904 = 6.943∠26.56 o kVA
*
(20 − j10)
Z p
(b) For the wye-connected load, Z p = 20 + j10Ω,
S=
V p = VL / 3 ,
3V 2 p
3 x 220 2
=
= 2.164∠26.56 o kVA
*
3(20 − j10)
Z p
Chapter 12, Solution 50.
S = S 1 + S 2 = 8(0.6 + j 0.8) = 4.8 + j 6.4 kVA,
Hence,
S 1 = 3 kVA
S 2 = S − S 1 = 1.8 + j 6.4 kVA
But S 2 =
Z*p =
3V 2 p
,
Z*p
Vp =
VL
3
V *L
240 2
=
(1.8 + j 6.4) x10 3
S2

→

→
S2 =
.V 2 L
Z*p
Z p = 2.346 + j8.34Ω
Chapter 12, Solution 51.
Apply mesh analysis to the circuit as shown below.
Za
+
150∠120°
−
150∠0°
i1
Zb
−
+
n
−
+
150∠-120°
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 ∠ - 120° + (Z b + Z c ) I 2 − Z b I 1 = 0
150 ∠ - 120° = (27 + j9) I 2 − (12 + j9) I 1
From (1) and (2),
  18 + j - 12 − j9 I 1 

150
150∠ - 120° =  - 12 − j9 27 + j9  I 
 2 
 

∆ = 414 − j27 ,
∆ 1 = 3780.9 + j3583.8 ,
(2)
∆ 2 = 579.9 − j1063.2
I1 =
∆ 1 5209.5∠43.47°
=
= 12.56 ∠47.2°
∆ 414.88∠ - 3.73°
I2 =
∆ 2 1211.1∠ - 61.39°
=
= 2.919 ∠ - 57.66°
∆
414.88∠ - 3.73°
I a = I 1 = 12.56∠47.2° A
I b = I 2 − I1 =
Ib =
∆ 2 − ∆ 1 - 3201 − j4647
=
∆
∆
5642.3∠235.44°
= 13.6∠239.17° A
414.88∠ - 3.73°
I c = - I 2 = 2.919∠122.34° A
Chapter 12, Solution 52.
Since the neutral line is present, we can solve this problem on a per-phase basis.
Van 120 ∠120°
=
= 6 ∠60°
Ia =
20 ∠60°
Z AN
Ib =
Vbn 120 ∠0°
=
= 4 ∠0°
30 ∠0°
Z BN
Vcn 120 ∠ - 120°
=
= 3∠ - 150°
40 ∠30°
Z CN
Ic =
Thus,
- In
- In
- In
- In
= Ia + Ib + Ic
= 6 ∠60° + 4 ∠0° + 3∠ - 150°
= (3 + j5.196) + (4) + (-2.598 − j1.5)
= 4.405 + j3.696 = 5.75∠40°
I n = 5.75∠ 220° 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∠0°
1
Ia =
⋅
= 3.608∠ - 60° A
40∠60°
3
Ib =
Ic =
250∠ - 120°
3
250∠120°
3
⋅
⋅
1
= 2.406∠ - 75° A
60∠ - 45°
1
= 7.217 ∠120° A
20∠0°
- In = Ia + Ib + Ic
- I n = (1.804 − j3.125) + (0.6227 − j2.324) + (-3.609 + j6.25)
I n = 1.1823 − j0.801 = 1.428 ∠ - 34.12° A
Chapter 12, Solution 54.
Consider the circuit shown below.
Ia
a
Vp∠0°
+
−
IAB
Vp∠120°
+
−
−
A
50 Ω
Vp∠-120°
j50 Ω
+
c
b
B
C
-j50 Ω
VAB = Vab = 100 × 3 ∠30°
I AB =
VAB 100 3 ∠30°
=
= 3.464∠ 30° A
50
Z AB
I BC =
VBC 100 3 ∠ - 90°
=
= 3.464∠0° A
50∠ - 90°
Z BC
I CA =
VCA 100 3 ∠150°
=
= 3.464∠60° A
50∠90°
Z CA
Chapter 12, Solution 55.
Consider the circuit shown below.
Ia
a
220∠0°
220∠120°
+
+
−
−
c
A
60 + j80
I1
−
220∠-120°
+
b
100 – j120
Ib
30 + j40
B
C
I2
Ic
For mesh 1,
220 ∠ - 120° − 220 ∠0° + (160 − j40) I 1 − (100 − j120) I 2 = 0
11 − 11∠ - 120° = (8 − j2) I 1 − (5 − j6) I 2
(1)
For mesh 2,
220 ∠120° − 220 ∠ - 120° + (130 − j80) I 2 − (100 − j120) I 1 = 0
11∠ - 120° − 11∠120° = - (5 − j6) I 1 + (6.5 − j4) I 2
(2)
From (1) and (2),
16.5 + j9.526  8 − j2 - 5 + j6  I 1 
 - j19.053  =  - 5 + j6 6.5 - j4  I 

 
 2 
∆ = 55 + j15 ,
∆ 1 = 31.04 − j99.35 ,
∆ 2 = 101.55 − j203.8
I1 =
∆ 1 104.08∠ - 72.65°
= 1.8257 ∠ - 87.91°
=
∆
57.01∠15.26°
I2 =
∆ 2 227.7 ∠ - 63.51°
=
= 3.994 ∠ - 78.77°
∆
57.01∠15.26°
I a = I 1 = 1.8257 ∠ - 87.91°
I b = I 2 − I1 =
∆ 2 − ∆ 1 70.51 − j104.45
=
= 2.211∠ - 71.23°
∆
55 + j15
I c = - I 2 = 3.994∠101.23°
SA = Ia
2
Z AN = (1.8257) 2 (60 + j80) = 199.99 + j266.7
SB = Ib
2
Z BN = (2.211) 2 (100 − j120) = 488.9 − j586.6
SC = Ic
2
Z CN = (3.994) 2 (30 + j40) = 478.6 + j638.1
S = S A + S B + S C = 1167.5 + j318.2 VA
Chapter 12, Solution 56.
(a)
Consider the circuit below.
a
A
440∠0° + −
440∠120°
+
j10 Ω
I1
b
−
B
− +
440∠-120°
I2
I3
-j5 Ω
20 Ω
C
c
For mesh 1,
440∠ - 120° − 440∠0° + j10 (I 1 − I 3 ) = 0
I1 − I 3 =
(440)(1.5 + j0.866)
= 76.21∠ - 60°
j10
(1)
For mesh 2,
440∠120° − 440∠ - 120° + 20 (I 2 − I 3 ) = 0
I3 − I2 =
(440)( j1.732)
= j38.1
20
For mesh 3,
j10 (I 3 − I 1 ) + 20 (I 3 − I 2 ) − j5 I 3 = 0
(2)
Substituting (1) and (2) into the equation for mesh 3 gives,
(440)(-1.5 + j0.866)
I3 =
= 152.42∠60°
j5
From (1),
I 1 = I 3 + 76.21∠ - 60° = 114.315 + j66 = 132∠30°
From (2),
I 2 = I 3 − j38.1 = 76.21 + j93.9 = 120.93∠50.94°
I a = I 1 = 132∠30° A
I b = I 2 − I 1 = -38.105 + j27.9 = 47.23∠143.8° A
I c = - I 2 = 120.9∠230.9° A
(b)
2
S AB = I 1 − I 3 ( j10) = j58.08 kVA
2
S BC = I 2 − I 3 (20) = 29.04 kVA
2
S CA = I 3 (-j5) = (152.42) 2 (-j5) = -j116.16 kVA
S = S AB + S BC + S CA = 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 50Ω
I1
-
20 + j 30Ω
-
Vc
+
Vb
+
I2
60 − j 40Ω
Ib
Ic
(1)
(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
(2)
Solving (1) and (2) gives I 1 = 1.8616 − j 0.6084,
I 2 = 0.9088 − j1.722 .
I a = I 1 = 1.9585∠ − 18.1o A,
I b = I 2 − I 1 = −0.528 − j1.1136 = 1.4656∠ − 130.55 o A
I c = − I 2 = 1.947∠117.8 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∠–89.97° 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∠–34.56°, VBN = 214.1∠–81.49°, VCN = 49.91∠–50.59° 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∠–135.5° 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∠37° A, VBN = 230.8∠–133.4° 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∠120° A, IbB = 5.96∠–91.41° A
Chapter 12, Solution 63.
Let ω = 1 so that L = X/ω = 20 H, and C =
1
= 0.0333 F
ωX
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∠71.38° A, IbB = 6.781∠–142.6° A, IcC = 3.898∠–5.08° A
IAB = 3.547∠61.57° A, IAC = 1.357∠97.81° A, IBC = 3.831∠–164.9° 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∠98.66° A, IbB = 11.4∠–111.3 A, IcC = 6.581∠38.66° A
Chapter 12, Solution 66.
VL
Vp =
(b)
Because the load is unbalanced, we have an unbalanced three-phase
system. Assuming an abc sequence,
3
=
208
(a)
3
= 120 V
I1 =
120 ∠0°
= 2.5∠0° A
48
I2 =
120∠ - 120°
= 3∠ - 120° A
40
I3 =
120∠120°
= 2∠120° A
60


3
3
- I N = I 1 + I 2 + I 3 = 2.5 + (3)  - 0.5 − j  + (2)  - 0.5 + j 
2 
2 


IN = j
3
= j0.866 = 0.866∠90° A
2
Hence,
I1 = 2.5 A ,
(c)
I2 = 3 A ,
I3 = 2 A ,
P1 = I12 R 1 = (2.5) 2 (48) = 300 W
P2 = I 22 R 2 = (3) 2 (40) = 360 W
P3 = I 32 R 3 = (2) 2 (60) = 240 W
(d)
PT = P1 + P2 + P3 = 900 W
Chapter 12, Solution 67.
(a)
The power to the motor is
PT = S cos θ = (260)(0.85) = 221 kW
The motor power per phase is
1
Pp = PT = 73.67 kW
3
Hence, the wattmeter readings are as follows:
Wa = 73.67 + 24 = 97.67 kW
Wb = 73.67 + 15 = 88.67 kW
Wc = 73.67 + 9 = 83.67 kW
(b)
The motor load is balanced so that I N = 0 .
For the lighting loads,
Ia =
24,000
= 200 A
120
Ib =
15,000
= 125 A
120
Ic =
9,000
= 75 A
120
If we let
I N = 0.866 A
I a = I a ∠0° = 200∠0° A
I b = 125∠ - 120° A
I c = 75∠120° A
Then,
- I N = Ia + Ib + Ic


3
3
- I N = 200 + (125) - 0.5 − j  + (75) - 0.5 + j 
2 
2 


- I N = 100 − 86.602 A
I N = 132.3 A
Chapter 12, Solution 68.
(a)
S = 3 VL I L = 3 (330)(8.4) = 4801 VA
(b)
P = S cos θ 
→ pf = cos θ =
pf =
P
S
4500
= 0.9372
4801.24
(c)
For a wye-connected load,
I p = I L = 8.4 A
(d)
Vp =
VL
3
=
330
3
= 190.53 V
Chapter 12, Solution 69.
S 1 = 1.2(0.8 + j 0.6) = 0.96 + j 0.72 MVA,
S 2 = 2(0.75 − j 0.661) = 1.5 − 1.323 MVA,
S = S 1 + S 2 + S 3 = 3.26 − j 0.603 MVA,
pf =
S3 = 0.8 MVA
3.26
P
=
= 0.9833
S 3.3153
Qc = P(tan old − tan new ) = 3.26[tan(cos −1 0.9833) − tan(cos −1 0.99) = 0.1379 MVA
1
x0.1379 x10 6
3
C=
= 28 mF
2πx60 x6.6 2 x10 6
Chapter 12, Solution 70.
PT = P1 + P2 = 1200 − 400 = 800
Q T = P2 − P1 = -400 − 1200 = -1600
tan θ =
Q T - 1600
=
= -2 
→ θ = -63.43°
PT
800
pf = cos θ = 0.4472 (leading)
Zp =
VL 240
=
= 40
IL
6
Z p = 40 ∠ - 63.43° Ω
Chapter 12, Solution 71.
(a)
If Vab = 208∠0° , Vbc = 208∠ - 120° , Vca = 208∠120° ,
I AB =
Vab 208∠0°
=
= 10.4 ∠0°
20
Z Ab
I BC =
Vbc
208∠ - 120°
= 14.708∠ - 75°
=
Z BC 10 2 ∠ - 45°
I CA =
Vca
208∠120°
= 16 ∠97.38°
=
Z CA 13∠22.62°
I aA = I AB − I CA = 10.4∠0° − 16∠97.38°
I aA = 10.4 + 2.055 − j15.867
I aA = 20.171∠ - 51.87°
I cC = I CA − I BC = 16∠97.83° − 14.708∠ - 75°
I cC = 30.64 ∠101.03°
P1 = Vab I aA cos(θ Vab − θIaA )
P1 = (208)(20.171) cos(0° + 51.87°) = 2590 W
P2 = Vcb I cC cos(θ Vcb − θ IcC )
But
Vcb = -Vbc = 208∠60°
P2 = (208)(30.64) cos(60° − 101.03°) = 4808 W
(b)
PT = P1 + P2 = 7398.17 W
Q T = 3 (P2 − P1 ) = 3840.25 VAR
S T = PT + jQ T = 7398.17 + j3840.25 VA
S T = S T = 8335 VA
Chapter 12, Solution 72.
From Problem 12.11,
VAB = 220 ∠130° V
I aA = 30∠180° A
and
P1 = (220)(30) cos(130° − 180°) = 4242 W
VCB = -VBC = 220∠190°
I cC = 30∠ - 60°
P2 = (220)(30) cos(190° + 60°) = - 2257 W
Chapter 12, Solution 73.
Consider the circuit as shown below.
I1
Ia
240∠-60° V
+
−
Z
Z
240∠-120° V
Z
−
+
I2
Ib
Ic
Z = 10 + j30 = 31.62∠71.57°
Ia =
240∠ - 60°
= 7.59∠ - 131.57°
31.62∠71.57°
Ib =
240 ∠ - 120°
= 7.59∠ - 191.57°
31.62∠71.57°
I c Z + 240∠ - 60° − 240 ∠ - 120° = 0
Ic =
- 240
= 7.59∠108.43°
31.62∠71.57°
I 1 = I a − I c = 13.146∠ - 101.57°
I 2 = I b + I c = 13.146∠138.43°
P1 = Re [ V1 I 1* ] = Re [ (240∠ - 60°)(13.146 ∠101.57°) ] = 2360 W
P2 = Re [ V2 I *2 ] = Re [ (240 ∠ - 120°)(13.146∠ - 138.43°) ] = - 632.8 W
Chapter 12, Solution 74.
Consider the circuit shown below.
Z = 60 − j30 Ω
208∠0° V
208∠-60° V
+
−
I1
−
+
I2
Z
For mesh 1,
208 = 2 Z I 1 − Z I 2
For mesh 2,
- 208∠ - 60° = - Z I 1 + 2 Z I 2
Z
In matrix form,

  2 Z - Z  I 1 
208
 - 208∠ - 60° =  - Z 2 Z  I 

 
 2 
∆ = 3Z 2 ,
∆ 1 = (208)(1.5 + j0.866) Z ,
∆ 2 = (208)( j1.732) Z
I1 =
∆ 1 (208)(1.5 + j0.866)
=
= 1.789∠56.56°
∆
(3)(60 − j30)
I2 =
∆ 2 (208)( j1.732)
=
= 1.79∠116.56°
∆
(3)(60 − j30)
P1 = Re [ V1 I 1* ] = Re [ (208)(1.789∠ - 56.56°) ] = 208.98 W
P2 = Re [ V2 (- I 2 ) * ] = Re [ (208∠ - 60°))(1.79∠63.44°) ] = 371.65 W
Chapter 12, Solution 75.
(a)
I=
V 12
=
= 20 mA
R 600
(b)
I=
V 120
=
= 200 mA
R 600
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
 P2 R

2
Power loss = I 2 R =  120
2
P R
 240 2
Since
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 = PT − Pload − Pline ,
But
pf = 0.85
PT = 3600 cos θ = 3600 × pf = 3060
Pg = 3060 − 2500 − (3)(80) = 320 W
Chapter 12, Solution 78.
cos θ1 =
51
= 0.85 
→ θ1 = 31.79°
60
Q1 = S1 sin θ1 = (60)(0.5268) = 31.61 kVAR
P2 = P1 = 51 kW
cos θ 2 = 0.95 
→ θ 2 = 18.19°
S2 =
P2
= 53.68 kVA
cos θ 2
Q 2 = S 2 sin θ 2 = 16.759 kVAR
Q c = Q1 − Q 2 = 3.61 − 16.759 = 14.851 kVAR
For each load,
Q c1 =
C=
Qc
= 4.95 kVAR
3
Q c1
4950
= 67.82 µF
2 =
ωV
(2π )(60)(440) 2
Chapter 12, Solution 79.
Consider the per-phase equivalent circuit below.
Ia
2Ω
a
A
+
−
Van
ZY = 12 + j5 Ω
n
Ia =
N
Van
255∠0°
=
= 17.15∠ - 19.65° A
Z Y + 2 14 + j5
Thus,
I b = I a ∠ - 120° = 17.15∠ - 139.65° A
I c = I a ∠120° = 17.15∠100.35° A
VAN = I a Z Y = (17.15∠ - 19.65°)(13∠22.62°) = 223∠2.97° V
Thus,
VBN = VAN ∠ - 120° = 223∠ - 117.63° V
VCN = VAN ∠120° = 223∠122.97° V
Chapter 12, Solution 80.
S = S1 + S 2 + S 3 = 6[0.83 + j sin(cos −1 0.83)] + S 2 + 8(0.7071 − j 0.7071)
S = 10.6368 − j 2.31 + S 2 kVA
(1)
But
S = 3VL I L ∠θ = 3 (208)(84.6)(0.8 + j 0.6) VA = 24.383 + j18.287 kVA
(2)
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.
Chapter 12, Solution 81.
pf = 0.8 (leading) 
→ θ1 = -36.87°
S1 = 150 ∠ - 36.87° kVA
pf = 1.0 
→ θ 2 = 0°
S 2 = 100 ∠0° kVA
pf = 0.6 (lagging) 
→ θ3 = 53.13°
S 3 = 200∠53.13° kVA
S 4 = 80 + j95 kVA
S = S1 + S 2 + S 3 + S 4
S = 420 + j165 = 451.2∠21.45° kVA
S = 3 VL I L
IL =
S
3 VL
=
451.2 × 10 3
3 × 480
= 542.7 A
For the line,
S L = 3 I 2L Z L = (3)(542.7) 2 (0.02 + j0.05)
S L = 17.67 + j44.18 kVA
At the source,
S T = S + S L = 437.7 + j209.2
S T = 485.1∠25.55° kVA
VT =
ST
3 IL
=
485.1 × 10 3
3 × 542.7
= 516 V
Chapter 12, Solution 82.
S 1 = 400(0.8 + j 0.6) = 320 + j 240 kVA,
S2 = 3
V 2p
Z*p
For the delta-connected load, V L = V p
(2400) 2
S 2 = 3x
= 1053.7 + j842.93 kVA
10 − j8
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
3VL
=
Vp =
VL
3
(320 + j 240) x10 3
3 (2400)
,
I 1 = 76.98 − j 57.735
For I2, convert the load to wye.
I 2 = I p 3∠ − 30 o =
2400
3∠ − 30 o = 273.1 − j 289.76
10 + j8
I = I 1 + I 2 = 350 − j 347.5
Vs = VL + Vline = 2400 + I (3 + j 6) = 5.185 + j1.405 kV

→
| Vs |= 5.372 kV
Chapter 12, Solution 83.
S1 = 120 x746 x0.95(0.707 + j 0.707) = 60.135 + j 60.135 kVA,
S 2 = 80 kVA
S = S1 + S 2 = 140.135 + j 60.135 kVA
But | S |= 3VL I L

→
IL =
|S|
3VL
=
152.49 x10 3
3 x 480
= 183.42 A
Chapter 12, Solution 84.
We first find the magnitude of the various currents.
For the motor,
S
IL =
3 VL
=
4000
440 3
= 5.248 A
For the capacitor,
Q c 1800
=
= 4.091 A
IC =
VL
440
For the lighting,
440
Vp =
= 254 V
3
I Li =
PLi 800
=
= 3.15 A
Vp 254
Consider the figure below.
Ia
a
IC
+
Vab
b
I1
Ib
−
-jXC
I2
Ic
c
I3
ILi
In
R
n
If Van = Vp ∠0° ,
Vab = 3 Vp ∠30°
Vcn = Vp ∠120°
IC =
Vab
= 4.091∠120°
-j X C
I1 =
Vab
= 4.091∠(θ + 30°)
Z∆
where θ = cos -1 (0.72) = 43.95°
I 1 = 5.249 ∠73.95°
I 2 = 5.249 ∠ - 46.05°
I 3 = 5.249∠193.95°
I Li =
Vcn
= 3.15∠120°
R
Thus,
I a = I 1 + I C = 5.249∠73.95° + 4.091∠120°
I a = 8.608∠93.96° A
I b = I 2 − I C = 5.249∠ - 46.05° − 4.091∠120°
I b = 9.271∠ - 52.16° A
I c = I 3 + I Li = 5.249∠193.95° + 3.15∠120°
I c = 6.827 ∠167.6° A
I n = - I Li = 3.15∠ - 60° A
Chapter 12, Solution 85.
Let
ZY = R
Vp =
VL
3
=
240
3
= 138.56 V
Vp2
27
P = Vp I p =
= 9 kW =
2
R
R=
Vp2
=
P
(138.56) 2
= 2.133 Ω
9000
Z Y = 2.133 Ω
Thus,
Chapter 12, Solution 86.
Consider the circuit shown below.
1Ω
a
A
+
−
120∠0° V rms
I1
24 – j2 Ω
1Ω
n
N
I2
+
−
120∠0° V rms
15 + j4 Ω
1Ω
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,
120  26 − j2
- 1  I 1 
120 =  - 1
17 + j4 I 2 
  
∆ = 449 + j70 ,
∆ 1 = (120)(18 + j4) ,
∆ 2 = (120)(27 − j2)
I1 =
∆ 1 120 × 18.44 ∠12.53°
= 4.87 ∠3.67°
=
∆
454.42 ∠8.86°
I2 =
∆ 2 120 × 27.07 ∠ - 4.24°
=
= 7.15∠ - 13.1°
∆
454.42 ∠8.86°
I aA = I 1 = 4.87 ∠3.67° A
I bB = - I 2 = 7.15∠166.9° A
I nN = I 2 − I 1 =
I nN =
∆ 2 − ∆1
∆
(120)(9 − j6)
= 2.856∠ - 42.55° A
449 + j70
Chapter 12, Solution 87.
L = 50 mH 
→ jωL = j (2π)(60)(5010 -3 ) = j18.85
Consider the circuit below.
1Ω
115 V
+
−
I1
20 Ω
2Ω
15 + j18.85 Ω
115 V
+
−
I2
30 Ω
1Ω
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 = 0
In matrix form,
- 20
 23 - 2
  I 1  115
 - 2 33
 I  = 115
- 30

 2  
- 20 - 30 65 + j18.85 I 3   0 
∆ = 12,775 + j14,232 ,
∆ 2 = (115)(1825 + j471.3) ,
(1)
(2)
(3)
∆ 1 = (115)(1975 + j659.8)
∆ 3 = (115)(1450)
I1 =
∆ 1 115 × 2082∠18.47°
=
= 12.52∠ - 29.62°
∆
19214∠48.09°
I2 =
∆ 2 115 × 1884.9 ∠14.48°
= 11.33∠ - 33.61°
=
∆
19124 ∠48.09°
I n = I 2 − I1 =
∆ 2 − ∆ 1 (115)(-150 − j188.5)
=
= 1.448∠ - 176.6° A
∆
12,775 + j14,231.75
S 1 = V1 I *1 = (115)(12.52∠ 29.62°) = 1252 + j711.6 VA
S 2 = V2 I *2 = (115)(1.33∠33.61°) = 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).
Is = I 1 + I2
and
Zeq = Vs/Is
Applying KVL to each branch gives,
Vs = jωL1I1 + jωMI2
(1)
Vs = jωMI1 + jω L2I2
(2)
 Vs   jωL1
 V  =  jωM
 s 
or
jωM   I1 
jωL 2  I 2 
∆ = –ω2L1L2 + ω2M2, ∆1 = jωVs(L2 – M), ∆2 = jωVs(L1 – M)
I1 = ∆1/∆, and I2 = ∆2/∆
Is = I1 + I2 = (∆1 + ∆2)/∆ = jω(L1 + L2 – 2M)Vs/( –ω2(L1L2 – M))
Zeq = Vs/Is = jω(L1L2 – M)/[jω(L1 + L2 – 2M)] = jωLeq
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
25x 60 − 19.36 2
=
mH = 24.31 mH
L1 + L 2 − 2M 25 + 60 − 2x19.36
Chapter 13, Solution 6.
V1 = (R1 + jωL1)I1 – jωMI2
V2 = –jωMI1 + (R2 + jωL2)I2
Chapter 13, Solution 7.
Applying KVL to the loop,
20∠30° = I(–j6 + j8 + j12 + 10 – j4x2) = I(10 + j6)
where I is the loop current.
I = 20∠30°/(10 + j6)
Vo = I(j12 + 10 – j4) = I(10 + j8)
= 20∠30°(10 + j8)/(10 + j6) = 22∠37.66° V
Chapter 13, Solution 8.
Consider the current as shown below.
j2
1Ω
10
+
–
I1
4Ω
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,
j2
10 1 + j6
 0  =  j2
4+
  
  I1 
j  I 2 
∆ = 2 + j25, and ∆2 = –j20
I2 = ∆2/∆ = –j20/(2 + j25)
Vo = –j3I2 = –60/(2 + j25) = 2.392∠94.57°
Chapter 13, Solution 9.
Consider the circuit below.
2Ω
8∠30o
+
–
I1
2Ω
j4
j4
I2
-j1
-j2V
+
–
For loop 1,
8∠30° = (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∠30° + (2 + j4)2 = (14 + j7)I2
I2 = (10.928 + j12)/(14 + j7) = 1.037∠21.12°
Vx = 2I2 = 2.074∠21.12°
(2)
Chapter 13, Solution 10.
Consider the circuit below.
jωM
jωL
jωL
Io
I1
Iin∠0o
I2
1/jωC
M = k L1 L 2 = L2 = L, I1 = Iin∠0°, I2 = Io
Io(jωL + R + 1/(jωC)) – jωLIin – (1/(jωC))Iin = 0
Io = j Iin(ωL – 1/(ωC)) /(R + jωL + 1/(jωC))
Chapter 13, Solution 11.
Consider the circuit below.
V2
R2
+–
I3
jωL1
R1
V1
+
–
I1
1/jωC
jωM
jωL2
I2
For mesh 1, V1 = I1(R1 + 1/(jωC)) – I2(1/jωC)) –R1I3
For mesh 2,
0 = –I1(1/(jωC)) + (jωL1 + jωL2 + (1/(jωC)) – j2ωM)I2 – jωL1I3 + jωMI3
For mesh 3,
or
–V2 = –R1I1 – jω(L1 – M)I2 + (R1 + R2 + jωL1)I3
V2 = R1I1 + jω(L1 – M)I2 – (R1 + R2 + jωL1)I3
Chapter 13, Solution 12.
Let ω = 1.
j4
j2
+
1V
-
j6
•
j8
I1
j10
I2
•
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
= jLeq
I1

→
Leq =
1
= 7.111 H
jI 1
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 ω = 1,
La = L1 − M = 18 − 10 = 8,
Lb = L2 − M = 20 − 10 = 10,
The equivalent circuit is shown below:
Lc = M = 10
12 Ω
j8 Ω
j10 Ω
2Ω
j10 Ω
-j6 Ω
Z
j4 Ω
Z=12 +j8 + j14//(2 + j4) = 13.195 + j11.244Ω
Chapter 13, Solution 14.
To obtain VTh, convert the current source to a voltage source as shown below.
j2
5Ω
j6 Ω
j8 Ω
-j3 Ω
2Ω
a
j10 V
+
–
+
VTh
I
–
b
Note that the two coils are connected series aiding.
ωL = ωL1 + ωL2 – 2ωM
jωL = 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∠34.11°
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
5Ω
j6 Ω
I1
a
j8 Ω
-j3 Ω
+
Vo
1A
2Ω
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∠50°
ZTh = Vo/1 = 2.332∠50° ohms
Chapter 13, Solution 15.
To obtain IN, short-circuit a–b as shown in Figure (a).
20 Ω
j20 Ω
20 Ω
a
j5
+
–
I1
j20 Ω
j5
j10 Ω
j10 Ω
60∠30o
1
+
–
IN
I1
I2
I2
b
(a)
For mesh 1,
a
b
(b)
60∠30° = (20 + j10)I1 + j5I2 – j10I2
or
For mesh 2,
12∠30° = (4 + j2)I1 – jI2
(1)
0 = (j20 + j10)I2 – j5I1 – j10I1
or
I1 = 2I2
(2)
12∠30° = (8 + j3)I2
Substituting (2) into (1),
IN = I2 = 12∠30°/(8 + j3) = 1.404∠9.44° 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.
8Ω
jΩ
-j2 Ω
a
• •
j4 Ω
+
80∠0 V
o
j6 Ω
I2
IN
I1
b
− 80 + (8 − j 2 + j 4) I 1 − jI 2 = 0

→
j 6 I 2 − jI 1 = 0

→
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.
8Ω
jΩ
-j2 Ω
2Ω
a
• •
j4 Ω
+
j6 Ω
I1
2V
I2
b
0 = (8 + j 2) I 1 − jI 2

→
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
= 1.894∠19.53o Ω
1
Chapter 13, Solution 17.
Z = -j6 // Zo
where
Z o = j20 +
Z=
144
= 0.5213 + j15.7
j30 − j2 + j5 + 4
− j6 xZ o
= 0.1989 − j9.7Ω
− j6 + Z o
Chapter 13, Solution 18.
Let ω = 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 )
= 2.215 + j 29.12Ω
4 + j7
We find VTh by looking at the circuit below.
-j4
j10
j25
j2
+
-j5
+
VTh
o
120<0
4+j6
-
-
VTh =
4+ j
(120) = 61.37∠ − 46.22 o V
4 + j + j6
Chapter 13, Solution 19.
Let ω = 1.
La = L1 − (− M ) = 40 + 25 = 65 H
Lb = L2 + M = 30 + 25 = 55 H,
L C = − M = −25
Thus, the T-section is as shown below.
j65 Ω
j55 Ω
-j25 Ω
Chapter 13, Solution 20.
Transform the current source to a voltage source as shown below.
k=0.5
4Ω
j10
8Ω
j10
I3
+
–
j12
I1
-j5
I2
20∠0o
+
–
k = M/ L1 L 2 or M = k L1 L 2
ωM = k ωL1ωL 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)
 j12 4 + j5 + j10   I1 
 20  =  + j10 8 + j5 I 
 2 
  
∆ = 107 + j60, ∆1 = –60 –j296, ∆2 = 40 – j100
I1 = ∆1/∆ = 2.462∠72.18° A
I2 = ∆2/∆ = 0.878∠–97.48° A
I3 = I1 – I2 = 3.329∠74.89° A
i1 = 2.462 cos(1000t + 72.18°) A
i2 = 0.878 cos(1000t – 97.48°) A
(2)
At t = 2 ms, 1000t = 2 rad = 114.6°
i1 = 0.9736cos(114.6° + 143.09°) = –2.445
i2 = 2.53cos(114.6° + 153.61°) = –0.8391
The total energy stored in the coupled coils is
w = 0.5L1i12 + 0.5L2i22 – Mi1i2
Since ωL1 = 10 and ω = 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∠30° = (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,
(2)
36∠30°  7 + j6 − 2 − j  I1 
 0  = − 2 − j 6 − j  I 
 2 

 
∆ = 48 + j35 = 59.41∠36.1°, ∆1 = (6 – j)36∠30° = 219∠20.54°
∆2 = (2 + j)36∠30° = 80.5∠56.56°, I1 = ∆1/∆ = 3.69∠–15.56°, I2 = ∆2/∆ = 1.355∠20.46°
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∠0° V
+
−
j80
I1
I2
100 Ω
Ib
−
+
j10Ia
Note the following,
Ia = I 1 – I3
Ib = I2 – I1
Ic = I 3 – I2
and
Io = I 3
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.
ω = 10
0.5 H converts to jωL1 = j5 ohms
1 H converts to jωL2 = j10 ohms
0.2 H converts to jωM = j2 ohms
25 mF converts to 1/(jωC) = 1/(10x25x10-3) = –j4 ohms
The frequency-domain equivalent circuit is shown below.
j2
j5
12∠0°
+
−
I1
j10
–j4
I2
5Ω
(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∠–100.89°
I1 = –j12 – 6 I2 = –j12 – 163.17∠–100.89 = 5.068∠52.54°
Hence,
i1 = 5.068cos(10t + 52.54°) A, i2 = 2.719cos(10t – 100.89°) A.
10t = 10x15x10-3 0.15 rad = 8.59°
At t = 15 ms,
i1 = 5.068cos(61.13°) = 2.446
i2 = 2.719cos(–92.3°) = –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)
ω = 4
1/4 F leads to 1/(jωC) = –j/(4x0.25) = –j
1||(–j) = –j/(1 – j) = 0.5(1 – j)
1 H produces jωM = j4
4 H produces j16
2 H becomes j8
j4
2Ω
j8
12∠0°
+
−
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∠–77.41°
Vo = I2(0.5)(1 – j) = 0.3217∠57.59°
vo = 321.7cos(4t + 57.6°) mV
(c)
From (2),
At t = 2s,
I1 = (0.5 + j7.5)I2/(–j4) = 0.855∠–81.21°
i1 = 0.885cos(4t – 81.21°) A, i2 = –0.455cos(4t – 77.41°) A
4t = 8 rad = 98.37°
i1 = 0.885cos(98.37° – 81.21°) = 0.8169
i2 = –0.455cos(98.37° – 77.41°) = –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∠0°, ω = 2
0.5 F converts to 1/(jωC) = –j
2 H becomes jωL = j4
j1
Io 4 Ω
1Ω
a
2Ω
–j1
12∠0°
+
−
j2
j2
j4
10 Ω
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∠17.9° ohms
Io = 12∠0°/(Zab + 4) = 12/(5.4355 + j0.4636) = 2.2∠–4.88°
io = 2.2sin(2t – 4.88°) A
Chapter 13, Solution 26.
M = k L1L 2
ωM = k ωL1ωL 2 = 0.6 20x 40 = 17
The frequency-domain equivalent circuit is shown below.
j17
50 Ω
200∠60°
–j30
I1
+
−
Io
j20
j40
I2
For mesh 1,
200∠60° = (50 – j30 + j20)I1 + j17I2 = (50 – j10)I1 + j17I2
10 Ω
(1)
For mesh 2,
0 = (10 + j40)I2 + j17I1
In matrix form,
(2)
j17   I1 
200∠60° 50 − j10
=


0
10 + j40 I 2 

  j17
∆ = 900 + j100, ∆1 = 2000∠60°(1 + j4) = 8246.2∠136°, ∆2 = 3400∠–30°
I2 = ∆2/∆ = 3.755∠–36.34°
Io = I2 = 3.755∠–36.34° 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∠150°. Thus,
Io = 3.755∠143.66° A
Chapter 13, Solution 27.
Zin = –j4 + j5 + 9/(12 + j6) = 0.6 + j.07 = 0.922∠49.4°
I1 = 12∠0°/0.922∠49.4° = 13∠–49.4° A
Chapter 13, Solution 28.
We find ZTh by replacing the 20-ohm load with a unit source as shown below.
j10 Ω
8Ω
-jX
• •
j12 Ω
j15 Ω
I2
+
1V
-
I1
For mesh 1,
0 = (8 − jX + j12) I 1 − j10 I 2
For mesh 2,
1 + j15I 2 − j10 I 1 = 0
(1)

→
I 1 = 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 =
1
12 + j8 − j1.5 X
=
− I 2 1.2 − j 0.8 − 0.1X
12 2 + (8 − 1.5 X ) 2
(1.2 − 0.1X ) + 0.8
2
2

→
Solving the quadratic equation yields X = 6.425
0 = 1.75 X 2 + 72 X − 624
Chapter 13, Solution 29.
30 mH becomes jωL = j30x10-3x103 = j30
50 mH becomes j50
Let X = ωM
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 = ωM
M = 38.127 mH
k = M/ L1 L 2 = 38.127/ 30x50 = 0.984
j38.127
10 Ω
165∠0°
+
−
I1
j30
j50
I2
20 Ω
165 = (10 + j30)I1 – j38.127I2
(1)
0 = (20 + j50)I2 – j38.127I1
(2)
165  10 + j30 − j38.127  I1 
 0  =  − j38.127 20 + j50   I 
  
 2 
In matrix form,
∆ = 154 + j1100 = 1110.73∠82.03°, ∆1 = 888.5∠68.2°, ∆2 = j6291
I1 = ∆1/∆ = 8∠–13.81°, I2 = ∆2/∆ = 5.664∠7.97°
i1 = 8cos(1000t – 13.83°), i2 = 5.664cos(1000t + 7.97°)
At t = 1.5 ms, 1000t = 1.5 rad = 85.94°
i1 = 8cos(85.94° – 13.83°) = 2.457
i2 = 5.664cos(85.94° + 7.97°) = –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)
jωLa = j30 – j10 = j20, jωLb = j20 – j10 = j10, jωLc = j10
Thus the Thevenin Equivalent of the linear transformer is shown below.
j40
25 Ω
j20
j10
j10
8Ω
–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’ = jωLb + ω2Mb2/(R + jωLb) = (jωLbR - ω2Lb2 + ω2Mb2)/(R + jωLb)
(1)
For the first stage, we have the circuit below.
La
LA
Zin’
Zin
Zin = jωLa + ω2Ma2/(jωLa + Zin)
= (–ω2La2 + ω2Ma2 + jωLaZin)/( jωLa + Zin)
(2)
Substituting (1) into (2) gives,
( jωL b R − ω 2 L2b + ω 2 M 2b )
− ω L + ω M + jωL a
R + jω L b
=
2 2
jωL b R − ω L b + ω 2 M 2b
jωL a +
R + jω L b
2
=
2
a
2
2
a
–Rω2La2 + ω2Ma2R – jω3LbLa + jω3LbMa2 + jωLa(jωLbR – ω2Lb2 + ω2Mb2)
jωRLa –ω2LaLb + jωLbR – ω2La2 + ω2Mb2
ω2R(La2 + LaLb – Ma2) + jω3(La2Lb + LaLb2 – LaMb2 – LbMa2)
Zin =
ω2(LaLb +Lb2 – Mb2) – jωR(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 Ω
1Ω
•
+
j12 Ω
o
1<0 V
-
8Ω
•
j10 Ω
I1
j4 Ω
I2
-j2 Ω
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

→
0 = − jI 1 + (2 + j 3) I 2
(2)
Solving (1) and (2) leads to I1=0.019 –j0.1068
Z=
1
= 1.6154 + j 9.077 = 9.219∠79.91o Ω
I1
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.
(a) n =
V2 2400
=
=5
V1
480
(b) S1 = I 1V1 = S 2 = I 2V2 = 50,000
(c ) I 2 =

→
I1 =
50,000
= 104.17 A
480
50,000
= 20.83 A
2400
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∠–53.13° or I2 = 17.391∠53.13° A
ZL = v2/I2 = 230∠0°/17.391∠53.13° = 13.235∠–53.13°
Zin = 2∠10° + 1323.5∠–53.13°
= 1.97 + j0.3473 + 794.1 – j1058.8
Zin = 1.324∠–53.05° kohms
Chapter 13, Solution 39.
Referred to the high-voltage side,
ZL = (1200/240)2(0.8∠10°) = 20∠10°
Zin = 60∠–30° + 20∠10° = 76.4122∠–20.31°
I1 = 1200/Zin = 1200/76.4122∠–20.31° = 15.7∠20.31° A
Since S = I1v1 = I2v2, I2 = I1v1/v2
= (1200/240)( 15.7∠20.31°) = 78.5∠20.31° A
Chapter 13, Solution 40.
n=
N2
500
1
=
= ,
N1 2000 4
P=
V 2 60 2
=
= 300 W
R
12
n=
V2
V1
→
V2 = nV1 =
1
(240) = 60 V
4
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∠0°/28 = 0.5 A and I2 = I1/n = 0.5/(–1/3) = –1.5 A
Chapter 13, Solution 42.
10 Ω
+ •
V1
+
120<0o V
-
I1
-j50 Ω
1:4
-
• +
V2
+
20 Ω
-
Vo
-
I2
Applying mesh analysis,
120 = 10I1 + V1
(1)
0 = (20 − j50)I 2 + V2
(2)
At the terminals of the transformer,
V2
=n=4
V1
→
I2
1
1
=− =−
I1
n
4
V2 = 4V1

→
(3)
I1 = −4I 2
(4)
Substituting (3) and (4) into (1) gives 120 = −40I 2 + 0.25V2
(5)
Solving (2) and (5) yields I 2 = −2.4756 − j0.6877
Vo = −20I 2 = 51.39∠15.52 o V
Chapter 13, Solution 43.
Transform the two current sources to voltage sources, as shown below.
10 Ω
+
20 V
+
–
Using mesh analysis,
I1
12 Ω
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∠0°
I2”/I1” = –1/n
But v2 = vm, v1 = –vm/n or I1” = vm/(Rn)
I2” = –I1”/n = –vm/(Rn2)
Hence,
i1(t) = (vm/Rn)cosωt A, and i2(t) = (–vm/(n2R))cosωt A
Chapter 13, Solution 45.
48 Ω
4∠–90˚
ZL = 8 −
+
−
j
= 8 − j4 , n = 1/3
ωC
Z
Z=
ZL
= 9 Z L = 72 − j36
n2
4∠ − 90°
4∠ − 90°
I=
=
= 0.03193∠ − 73.3°
48 + 72 − j36 125.28∠ − 16.7°
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,
P8Ω =
I2
72 = 0.5098x10 − 3 72 = 36.71 mW
2
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∠60°
+
−
I1
+
−
10∠30°/(–n) = –5∠30°
Zin = 10 + j16 + (1/4)(12 – j8) = 13 + j14
–16∠60° + ZinI1 – 5∠30° = 0 or I1 = (16∠60° + 5∠30°)/(13 + j14)
Hence,
(b)
I1 = 1.072∠5.88° A, and I2 = –0.5I1 = 0.536∠185.88° A
Switching a dot will not effect Zin but will effect I1 and I2.
I1 = (16∠60° – 5∠30°)/(13 + j14) = 0.625 ∠25 A
and I2 = 0.5I1 = 0.3125∠25° A
Chapter 13, Solution 47.
0.02 F becomes 1/(jωC) = 1/(j5x0.02) = –j10
We apply mesh analysis to the circuit shown below.
–j10
I3
10 Ω
3:1
+
10∠0°
+
–
I1
v1
−
+
v2
−
+
I2
vo
2Ω
−
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   I1   0 
− 0.333
 1
 10
6
− 10  I 2  = 10

− 10
10 − j10  I 3   0 
−4
(8)
I2 =
∆2
100 − j100
= 1.482∠32.9°
=
∆
− 20 − j93.33
vo = 2I2 = 2.963∠32.9° 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
(12)
0.333
0   I1   0 
 1
 10
−6
− 10  I 2  = 10

− 10
4
10 − j10  I 3   0 
I2 =
∆2
100 − j100
= 1.482∠–147.1°
=
∆
− 20 + j93.33
vo = 2I2 = 2.963∠–147.1° V
Chapter 13, Solution 48.
We apply mesh analysis.
8Ω
10 Ω
2:1
+
+
•
+
o
100∠0 V
-
I1
V1
•
V2
-
Ix
j6 Ω
I2
-j4 Ω
100 = (8 − j 4) I 1 − j 4 I 2 + V1
(1)
0 = (10 + j 2) I 2 − j 4 I 1 + V 2
(2)
V2
1
=n=
2
V1
(3)
But
I2
1
= − = −2
I1
n

→
V1 = 2V2

→
I 1 = −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
I x = I 1 + I 2 = 0.5 I 2 = 1.923∠157.4 o A
Chapter 13, Solution 49.
ω = 2,
1
F
20

→
1
= − j10
jω C
-j10
Ix
2Ω
I1
1:3
I2
1
+
12<0o V
-
2
+ •
V1
-
+
V2
•
-
6Ω
At node 1,
12 − V1 V1 − V2
=
+ I1

→ 12 = 2 I 1 + V1 (1 + j 0.2) − j 0.2V2
2
− j10
At node 2,
V − V2 V2
I2 + 1
=

→ 0 = 6 I 2 + j 0.6V1 − (1 + j 0.6)V2
− j10
6
At the terminals of the transformer, V2 = −3V1 ,
(1)
(2)
1
I 2 = − 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
4V1
=
= 0.937∠51.34 o
− j10
− j10
i x = 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∠0°/ZTh = 24∠0°/(8 – j1.5) = 24∠0°/8.14∠–10.62° = 8.95∠10.62° 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.
8Ω
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∠0°/(8 + j2) = 14.552∠–14.04°
Zp
Vp
+
−
1:n
I1
I2
Zs
Since both currents enter the dotted terminals as shown above,
I2 = –(1/n)I1 = –1.4552∠–14.04° = 1.4552∠166°
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.
2Ω
1:2
+
+
v1
46V
+
−
I1
v2
−
−
I2
10 Ω
5Ω
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∠20.04°
I1 = vs/Zin = 60∠90°/2.3168∠20.04° = 25.9∠69.96° A(rms)
I2 = I1/n = 12.95∠69.96° A(rms)
(b)
60∠90° = 2I1 + v1 or v1 = j60 –2I1 = j60 – 51.8∠69.96°
v1 = 21.06∠147.44° V(rms)
v2 = nv1 = 42.12∠147.44° V(rms)
vo = v2 = 42.12∠147.44° V(rms)
(c)
S = vsI1* = (60∠90°)(25.9∠–69.96°) = 1554∠20.04° VA
Chapter 13, Solution 58.
Consider the circuit below.
20 Ω
I3
20 Ω
1:5
+
80∠0°
+
–
I1
+
v1
−
v2
−
+
I2
vo
100 Ω
−
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.
2Ω
I3
8Ω
v1 I1 2 : 1
+
20∠0°
+
–
v1
−
I2
+
v2
−
v2
4Ω
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.
P8Ω = 0.5(8)|(20 – V1)/8|2 = 2.778 W
P2Ω = 0.5(2)I32 = 11.11 W, P4Ω = 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
4Ω
1:4
I1
+
120∠0°
+
–
v1
−
I2
5Ω
I2 ’
+
v2
10 Ω
10 Ω
−
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
2Ω
1:5
I1
+
24∠0°
+
–
v1
−
Io
14 Ω
Io ’
+
vo
60 Ω
90 Ω
−
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∠11.89°, where n =
5/2 = 2.5
I1 = 40/Zin = 40/7.767∠11.89° = 5.15∠–11.89°
S = 0.5vsI1* = (20∠0°)(5.15∠11.89°) = 103∠11.89° VA
(b)
I2 = –I1/n,
n = 2.5
I3 = –I2/n’,
n = 3
I3 = I1/(nn’) = 5.15∠–11.89°/(2.5x3) = 0.6867∠–11.89°
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∠0°/(3 – j) = 12/3.162∠–18.43° = 3.795∠18.43A
I2 = I1/n = 1.8975∠18.43° A
I3 = –I2/n2 = 632.5∠161.57° mA
Chapter 13, Solution 64.
We find ZTh at the terminals of Z by considering the circuit below.
10 Ω
1:n
+•
V1
• +
V2
–
+
–
I1
I2
20 Ω
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 = 0,
20(1 − n) I 2 + nV1 = 1
Solving these gives
I2 =
1
30n − 40n + 20

→
2
Z Th =
1
= 30n 2 − 40n + 20 = 7.5
I2
Solving the quadratic equation yields n=0.5 or 0.8333
Chapter 13, Solution 65.
40 Ω
10 Ω
I1
-
I2
+•
+
200 V
(rms)
50 Ω
I2
•
1
+
1:2
+
V1
-
V2
-
V3
-
1:3
I3
•
2
+
V4
-
20 Ω
•
At node 1,
200 − V1 V1 − V4
=
+ I1
10
40

→
200 = 1.25V1 − 0.25V4 + 10 I 1
(1)
At node 2,
V1 − V4 V4
=
+ I3
40
20

→
V1 = 3V4 + 40 I 3
(2)
At the terminals of the first transformer,
V2
= −2
V1

→
I2
= −1 / 2
I1
V2 = −2V1

→
(3)
I 1 = −2 I 2
(4)
For the middle loop,
− V2 + 50 I 2 + V3 = 0

→
V3 = V2 − 50 I 2
(5)
At the terminals of the second transformer,
V4
=3
V3
I3
= −1 / 3
I2

→
V4 = 3V3

→
(6)
I 2 = −3 I 3
(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 = −2 I 2 = −2(−3I 3 ) = 6 I 3 . Hence
200 = 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

→
I3 =
19
V4
210
(9)
Substituting (9) into (8) yields
200 = 13.452V4

→
V4 = 14.87
V 24
P=
= 11.05 W
20
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 N 1 + N 2
1
=
=
V2
N2
0 .4
(b)
S 2 = I 2V2 = 5,000
(c )
S 2 = S1 = I 1V1 = 5,000

→

→
V2 = 0.4V1 = 0.4 x 400 = 160 V
I2 =

→
5000
= 31.25 A
160
I2 =
5000
= 12.5 A
400
Chapter 13, Solution 68.
This is a step-up transformer.
I2
+
N1
2 – j6
I1
10 + j40
+
20∠30°
v2
N2
−
v1
+
−
−
For the primary circuit,
20∠30° = (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∠30° = (7.102 + j14.408)I1
I1 = 20∠30°/16.063∠63.76° = 1.245∠–33.76° A
I2 = 5I1/7 = 0.8893∠–33.76° A
Io = I1 – I2 = [(5/7) – 1]I1 = –2I1/7 = 0.3557∠146.2° 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∠0°
v2
j125 Ω
75 Ω
+
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∠0° rms = VTh.
To find ZTh, connect a 1-V source at the secondary terminals. We now have a
step-down transformer.
+
v1
j125 Ω
75 Ω
I2
I1
1∠0° 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) kΩ
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∠0°
+
−
+
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∠37.79°
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 = 7200 /(12470 3 ) = 1/3
(b)
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∠0°/(8 – j6) = 8.66∠36.87°
Ic = Ia∠120° = 8.66∠156.87° A
ILp = n 3 ILs
I1 = (1/3) 3 (8.66∠36.87°) = 5∠36.87°
I2 = I1∠–120° = 5∠–83.13° 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/cosθ = 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 Ω
2640V
j0.1 Ω
A
240V
j0.1 Ω
0.05 Ω
B
0.05 Ω
(b)
j0.1 Ω
C
Balanced
Load
60kVA
0.85pf
leading
Let VAN = |VAN|∠0° = 138.56∠0°
cosθ = pf = 0.85 or θ = 31.79°
IAA’ = IL∠θ = 144.34∠31.79°
VA’N’ = ZIAA’ + VAN
= 138.56∠0° + (0.05 + j0.1)(144.34∠31.79°)
= 138.03∠6.69°
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∠–8.53° A, I2 = 1.564∠27.49° 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∠–68.01° A, I2 = 406.8∠–77.86° mA, I3 = 1.336∠–54.92° 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∠62.92° 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∠13.96° mA, I2 = 29.54∠–143.8° mA,
I3 = 208.8∠24.4° 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∠83.32° V, V2 = 68.47∠46.4° V,
Io = 443.4∠–92.6° 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∠33.91° A, Vx = 15.14∠–34.21° 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∠–52.38° A, I2 = 2.019∠–52.11° A,
I3 = 1.338∠–52.2° 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 kΩ
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 (ω) =
Vo
R
jωRC
=
=
Vi R + 1 jωC 1 + jωRC
H (ω) =
jω ω 0
,
1 + jω ω 0
H = H (ω) =
where ω 0 =
ω ω0
1 + (ω ω0 ) 2
1
RC
φ = ∠H (ω) =
ω
π
− tan -1  
2
 ω0 
This is a highpass filter. The frequency response is the same as that for P.P.14.1
except that ω0 = 1 RC . Thus, the sketches of H and φ are shown below.
H
1
0.7071
0
ω0 = 1/RC
ω
φ
90°
45°
0
ω0 = 1/RC
ω
Chapter 14, Solution 2.
H (ω) =
R
1
1
=
=
,
R + jωL 1 + jωL R 1 + jω ω 0
H = H (ω) =
where ω0 =
R
L
ω
φ = ∠H (ω) = - tan -1  
 ω0 
1
1 + (ω ω0 ) 2
The frequency response is identical to the response in Example 14.1 except that
ω0 = R L . Hence the response is shown below.
H
1
0.7071
ω0 = R/L
0
φ
ω
ω0 = R/L
0°
ω
-45°
-90°
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 =
Vi
1 sC
Vi =
R + 1 sC
1 + sRC
ZTh
VTh
+
−
1
sC
+
Vo
−
Vo =
(b)
Vi
1 sC
⋅ VTh =
Z Th + 1 sC
(1 + sRC)(1 + sCZ Th )
H (s) =
Vo
1
1
=
=
Vi (1 + sCZ Th )(1 + sRC) (1 + sRC)(1 + sRC + sRC (1 + sRC))
H (s) =
1
s R C + 3sRC + 1
2
2
2
RC = (40 × 10 3 )(2 × 10 -6 ) = 80 × 10 -3 = 0.08
There are no zeros and the poles are at
- 0.383
s1 =
= - 4.787
RC
s2 =
- 2.617
= - 32.712
RC
Chapter 14, Solution 4.
(a)
R ||
1
R
=
jωC 1 + jωRC
R
Vo
R
1 + jωRC
H (ω) =
=
=
R
Vi
R + jωL (1 + jωRC)
jωL +
1 + jωRC
(b)
H (ω) =
R
- ω RLC + R + jωL
H (ω) =
jωC (R + jωL)
R + jωL
=
R + jωL + 1 jωC 1 + jωC (R + jωL)
H (ω) =
- ω 2 LC + jωRC
1 − ω 2 LC + jωRC
2
Chapter 14, Solution 5.
(a)
(b)
H (ω) =
Vo
1 jωC
=
Vi R + jωL + 1 jωC
H (ω) =
1
1 + jωRC − ω 2 LC
R ||
1
R
=
jωC 1 + jωRC
H (ω) =
Vo
jωL (1 + jωRC)
jωL
=
=
Vi
jωL + R (1 + jωRC) R + jωL (1 + jωRC)
jωL − ω 2 RLC
H (ω) =
R + jωL − ω 2 RLC
Chapter 14, Solution 6.
(a)
(b)
Using current division,
Io
R
H (ω) =
=
I i R + jωL + 1 jωC
H (ω) =
jω (20)(0.25)
jωRC
=
2
1 + jωRC − ω LC 1 + jω(20)(0.25) − ω2 (10)(0.25)
H (ω) =
jω5
1 + jω5 − 2.5 ω 2
We apply nodal analysis to the circuit below.
Io
Vx
Is
R
1/jωC
0.5 Vx
+ −
jωL
Is =
Vx Vx − 0.5Vx
+
R jωL + 1 jωC
But
Io =
0.5 Vx
jωL + 1 jωC

→ Vx = 2 I o ( jωL + 1 jωC)
Is
1
0 .5
= +
Vx R jωL + 1 jωC
Is
1
1
= +
2 I o ( jωL + 1 jωC) R 2 ( jωL + 1 jωC)
I s 2 ( jωL + 1 jωC)
=
+1
Io
R
Io
1
jωRC
=
=
I s 1 + 2 ( jωL + 1 jωC) R jωRC + 2 (1 − ω 2 LC)
jω
H (ω) =
jω + 2 (1 − ω2 0.25)
H (ω) =
H (ω) =
jω
2 + jω − 0.5 ω 2
Chapter 14, Solution 7.
(a)
0.05 = 20 log10 H
2.5 × 10 -3 = log10 H
H = 10 2.5×10 = 1.005773
-3
(b)
- 6.2 = 20 log10 H
- 0.31 = log10 H
H = 10 -0.31 = 0.4898
(c)
104.7 = 20 log10 H
5.235 = log10 H
H = 10 5.235 = 1.718 × 10 5
Chapter 14, Solution 8.
(a)
(b)
(c)
H = 0.05
H dB = 20 log10 0.05 = - 26.02 ,
φ = 0°
H = 125
H dB = 20 log10 125 = 41.94 ,
φ = 0°
H(1) =
j10
= 4.472∠63.43°
2+ j
H dB = 20 log10 4.472 = 13.01 ,
(d)
H(1) =
φ = 63.43°
3
6
+
= 3.9 − j1.7 = 4.254∠ - 23.55°
1+ j 2 + j
H dB = 20 log10 4.254 = 12.577 ,
φ = - 23.55°
Chapter 14, Solution 9.
H (ω) =
1
(1 + jω)(1 + jω 10)
H dB = -20 log10 1 + jω − 20 log10 1 + jω / 10
φ = - tan -1 (ω) − tan -1 (ω / 10)
The magnitude and phase plots are shown below.
HdB
0.1
1
10
ω
100
20 log 10
-20
1
1 + jω / 10
20 log10
-40
1
1 + jω
φ
0.1
-45°
1
10
ω
100
arg
1
1 + jω / 10
-90°
-135°
-180°
arg
1
1 + jω
Chapter 14, Solution 10.
H( jω) =
50
=
jω(5 + jω)
10
jω 

1 jω1 + 
5 

HdB
40
20 log1
20
10
0.1
-20
1
100


1
20 log

jω
 1+
5

 1 

20 log

 jω 
-40
φ
0.1






ω
1
10
-45°
ω
100
arg
1
1 + jω / 5
-90°
arg
-135°
1
jω
-180°
Chapter 14, Solution 11.
H (ω) =
5 (1 + jω 10)
jω (1 + jω 2)
H dB = 20 log10 5 + 20 log10 1 + jω 10 − 20 log10 jω − 20 log10 1 + jω 2
φ = -90° + tan -1 ω 10 − tan -1 ω 2
The magnitude and phase plots are shown below.
HdB
40
34
20
14
0.1
-20
1
10
100
ω
1
10
100
ω
-40
φ
90°
45°
0.1
-45°
-90°
Chapter 14, Solution 12.
T ( w) =
0.1(1 + jω )
,
jω (1 + jω / 10)
20 log 0.1 = −20
The plots are shown below.
|T|
(db)
20
ω
0
0.1
1
10
100
-20
-40
arg T
90o
ω
0
0.1
-90o
1
10
100
Chapter 14, Solution 13.
G (ω) =
(1 10)(1 + jω)
1 + jω
=
2
( jω) (10 + jω) ( jω) 2 (1 + jω 10)
G dB = -20 + 20 log10 1 + jω − 40 log10 jω − 20 log10 1 + jω 10
φ = -180° + tan -1ω − tan -1 ω 10
The magnitude and phase plots are shown below.
GdB
40
20
0.1
-20
1
10
100
ω
1
10
100
ω
-40
φ
90°
0.1
-90°
-180°
Chapter 14, Solution 14.
50
25
H (ω) =
1 + jω
 jω10  jω  2 
+   
jω1 +
25  5  

H dB = 20 log10 2 + 20 log10 1 + jω − 20 log10 jω
− 20 log10 1 + jω2 5 + ( jω 5) 2
 ω10 25 

φ = -90° + tan -1 ω − tan -1 
1 − ω2 5 
The magnitude and phase plots are shown below.
HdB
40
26
20
6
0.1
-20
1
10
100
ω
1
10
100
ω
-40
φ
90°
0.1
-90°
-180°
Chapter 14, Solution 15.
40 (1 + jω)
2 (1 + jω)
=
(2 + jω)(10 + jω) (1 + jω 2)(1 + jω 10)
H (ω) =
H dB = 20 log10 2 + 20 log10 1 + jω − 20 log10 1 + jω 2 − 20 log10 1 + jω 10
φ = tan -1 ω − tan -1 ω 2 − tan -1 ω 10
The magnitude and phase plots are shown below.
HdB
40
20
6
0.1
-20
1
10
100
ω
1
10
100
ω
-40
φ
90°
45°
0.1
-45°
-90°
Chapter 14, Solution 16.
G (ω) =
jω
jω 

100(1 + jω)1 + 
 10 
2
GdB
20
0.1
20 log jω
1
10
100
− 40 log
-20
-40
ω
jω
10
20 log(1/100)
-60
φ
90°
arg(jω)
ω
0.1
-90°
-180°
1
arg
10
100
arg
1
jω 

1 + 
 10 
2
1
1 + jω
Chapter 14, Solution 17.
G (ω) =
(1 4) jω
(1 + jω)(1 + jω 2) 2
G dB = -20log10 4 + 20 log10 jω − 20 log10 1 + jω − 40 log10 1 + jω 2
φ = -90° - tan -1ω − 2 tan -1 ω 2
The magnitude and phase plots are shown below.
GdB
20
0.1
-12
-20
-40
1
10
100
ω
φ
90°
0.1
1
10
100
ω
-90°
-180°
Chapter 14, Solution 18.
4 (1 + jω 2) 2
G (ω) =
50 jω (1 + jω 5)(1 + jω 10)
G dB = 20 log10 4 50 + 40 log10 1 + jω 2 − 20 log10 jω
− 20 log10 1 + jω 5 − 20 log10 1 + jω 10
where 20 log10 4 50 = -21.94
φ = -90° + 2 tan -1 ω 2 − tan -1 ω 5 − tan -1 ω 10
The magnitude and phase plots are shown below.
GdB
20
0.1
-20
1
10
100
ω
1
10
100
ω
-40
-60
φ
180°
90°
0.1
-90°
Chapter 14, Solution 19.
H (ω) =
jω
100 (1 + jω 10 − ω2 100)
H dB = 20 log10 jω − 20 log10 100 − 20 log10 1 + jω 10 − ω2 100
 ω 10 

φ = 90° − tan -1 
1 − ω2 100 
The magnitude and phase plots are shown below.
HdB
40
20
0.1
-20
1
10
100
ω
1
10
100
ω
-40
-60
φ
90°
0.1
-90°
-180°
Chapter 14, Solution 20.
10 (1 + jω − ω2 )
(1 + jω)(1 + jω 10)
N(ω) =
N dB = 20 − 20 log10 1 + jω − 20 log10 1 + jω 10 + 20 log10 1 + jω − ω2
 ω 
 − tan -1 ω − tan -1 ω 10
φ = tan -1 
1 − ω2 
The magnitude and phase plots are shown below.
NdB
40
20
0.1
-20
1
10
100
ω
1
10
100
ω
-40
φ
180°
90°
0.1
-90°
Chapter 14, Solution 21.
T(ω) =
jω (1 + jω)
100 (1 + jω 10)(1 + jω 10 − ω2 100)
TdB = 20 log10 jω + 20 log10 1 + jω − 20 log10 100
− 20 log10 1 + jω 10 − 20 log10 1 + jω 10 − ω2 100
 ω 10 

φ = 90° + tan -1 ω − tan -1 ω 10 − tan -1 
1 − ω2 100 
The magnitude and phase plots are shown below.
TdB
20
0.1
-20
1
10
100
ω
1
10
100
ω
-40
-60
φ
180°
90°
0.1
-90°
-180°
Chapter 14, Solution 22.
20 = 20 log10 k

→ k = 10
A zero of slope + 20 dB / dec at ω = 2 
→ 1 + jω 2
A pole of slope - 20 dB / dec at ω = 20 
→
1
1 + jω 20
A pole of slope - 20 dB / dec at ω = 100 
→
Hence,
H (ω) =
1
1 + jω 100
10 (1 + jω 2)
(1 + jω 20)(1 + jω 100)
10 4 ( 2 + jω)
H (ω) =
( 20 + jω)(100 + jω)
Chapter 14, Solution 23.
A zero of slope + 20 dB / dec at the origin

→
A pole of slope - 20 dB / dec at ω = 1 
→
1
1 + jω 1
A pole of slope - 40 dB / dec at ω = 10 
→
Hence,
H (ω) =
jω
(1 + jω)(1 + jω 10) 2
H (ω) =
100 jω
(1 + jω)(10 + jω) 2
jω
1
(1 + jω 10) 2
Chapter 14, Solution 24.
The phase plot is decomposed as shown below.
φ
90°
arg (1 + jω / 10)
45°
0.1
-45°
-90°
1
arg ( jω)
10
100
1000
ω


1

arg 
1 + jω / 100 
G (ω) =
k ′ (1 + jω 10)
k ′ (10)(10 + jω)
=
jω (1 + jω 100)
jω (100 + jω)
where k ′ is a constant since arg k ′ = 0 .
G (ω) =
Hence,
k (10 + jω)
,
jω (100 + jω)
where k = 10k ′ is constant
Chapter 14, Solution 25.
ω0 =
1
LC
=
1
(40 × 10 -3 )(1 × 10 -6 )
= 5 krad / s
Z(ω0 ) = R = 2 kΩ
 ω0
4 

Z(ω0 4) = R + j  L −
ω0 C 
 4

 5 × 10 3
4

⋅ 40 × 10 -3 −
Z(ω0 4) = 2000 + j 
(5 × 10 3 )(1 × 10 -6 ) 
 4
Z(ω0 4) = 2000 + j (50 − 4000 5)
Z(ω0 4) = 2 − j0.75 kΩ
 ω0
2 

Z(ω0 2) = R + j  L −
ω0 C 
 2
 (5 × 10 3 )

2

(40 × 10 -3 ) −
Z(ω0 2) = 2000 + j 
2
(5 × 10 3 )(1 × 10 -6 ) 

Z(ω0 4) = 2000 + j (100 − 2000 5)
Z(ω0 2) = 2 − j0.3 kΩ

1 

Z(2ω0 ) = R + j  2ω0 L −
2ω0 C 



1

Z(2ω0 ) = 2000 + j  (2)(5 × 10 3 )(40 × 10 -3 ) −
(2)(5 × 10 3 )(1 × 10 -6 ) 

Z(2ω0 ) = 2 + j0.3 kΩ

1 

Z(4ω0 ) = R + j  4ω0 L −
4ω0 C 



1

Z(4ω0 ) = 2000 + j  (4)(5 × 10 3 )(40 × 10 -3 ) −
3
-6
(4)(5 × 10 )(1 × 10 ) 

Z(4ω0 ) = 2 + j0.75 kΩ
Chapter 14, Solution 26.
(a)
fo =
(b)
B=
(c )
Q=
1
2π LC
=
1
2π 5 x10 −9 x10 x10 −3
R
100
=
= 10 krad/s
L 10 x10 −3
ωo L
R
=
L 10 6 10 x10 −3
=
= 14.142
3
50 0.1x10
LC R
1
Chapter 14, Solution 27.
At resonance,
Z = R = 10 Ω ,
B=
= 22.51 kHz
R
L
ω0 =
and
Q=
1
LC
ω 0 ω0 L
=
B
R
Hence,
L=
RQ (10)(80)
=
= 16 H
ω0
50
C=
1
1
=
= 25 µF
2
ω0 L (50) 2 (16)
B=
R 10
=
= 0.625 rad / s
L 16
Therefore,
R = 10 Ω ,
L = 16 H ,
C = 25 µF ,
B = 0.625 rad / s
Chapter 14, Solution 28.
Let R = 10 Ω .
L=
R 10
=
= 0.5 H
B 20
C=
1
1
=
= 2 µF
2
ω0 L (1000) 2 (0.5)
Q=
ω0 1000
=
= 50
B
20
Therefore, if R = 10 Ω then
C = 2 µF ,
L = 0.5 H ,
Q = 50
Chapter 14, Solution 29.
jω
Z
Z = jω +
1/jω
1
1
jω
+
j ω 1 + jω
jω

1  ω 2 + jω
Z = j ω −  +

ω  1 + ω2
Since v( t ) and i( t ) are in phase,
1
ω
Im(Z) = 0 = ω − +
ω 1 + ω2
ω4 + ω2 − 1 = 0
ω2 =
-1 ± 1+ 4
= 0.618
2
ω = 0.7861 rad / s
Chapter 14, Solution 30.
Select R = 10 Ω .
L=
R
10
=
= 0.05 H = 5 mH
ω0 Q (10)(20)
C=
1
1
=
= 0.2 F
2
ω0 L (100)(0.05)
B=
1
1
=
= 0.5 rad / s
RC (10)(0.2)
Therefore, if R = 10 Ω then
L = 5 mH , C = 0.2 F ,
B = 0.5 rad / s
Chapter 14, Solution 31.
X L = ωL
B=
→
L=
XL
ω
R ωR 2πx10 x10 6 x 5.6 x10 3
=
=
= 8.796 x10 6 rad/s
3
L XL
40 x10
Chapter 14, Solution 32.
Since Q > 10 ,
ω1 = ω0 −
B=
B
,
2
ω 2 = ω0 +
B
2
ω0 6 × 10 6
=
= 50 krad / s
Q
120
ω1 = 6 − 0.025 = 5.975 × 10 6 rad / s
ω2 = 6 + 0.025 = 6.025 × 10 6 rad / s
Chapter 14, Solution 33.
Q = ωo RC
Q=
→
R
ωo L
→
C=
L=
Q
80
=
= 56.84 pF
2πf o R 2πx5.6x10 6 x 40x10 3
R
40 x10 3
=
= 14.21 µH
2πf o Q 2πx 5.6 x10 6 x80
Chapter 14, Solution 34.
(a)
ωo =
1
LC
1
=
8x10
−3
x 60x10
−6
= 1.443 krad/s
1
1
=
= 3.33 rad/s
RC 5x10 3 x 60x10 − 6
(b)
B=
(c)
Q = ωo RC = 1.443x10 3 x 5x10 3 x 60x10 −6 = 432.9
Chapter 14, Solution 35.
At resonance,
1
R
Y=

→ R =
1
1
=
= 40 Ω
Y 25 × 10 -3
→ C =
Q = ω0 RC 
ω0 =
B=
1
LC

→ L =
Q
80
=
= 10 µF
ω0 R (200 × 10 3 )(40)
1
1
=
= 2.5 µH
2
10
ω0 C (4 × 10 )(10 × 10 -6 )
ω0 200 × 10 3
=
= 2.5 krad / s
Q
80
ω1 = ω0 −
B
= 200 − 2.5 = 197.5 krad / s
2
ω1 = ω0 +
B
= 200 + 2.5 = 202.5 krad / s
2
Chapter 14, Solution 36.
ω0 =
1
LC
= 5000 rad / s
Y(ω0 ) =
1
R

→ Z(ω0 ) = R = 2 kΩ
Y(ω0 4) =
 ω0
4 
1
 = 0.5 − j18.75 kS
+ j C−
ω0 L 
R
 4
Z(ω0 4) =
1
= 1.4212 + j53.3 Ω
0.0005 − j0.01875
Y(ω0 2) =
 ω0
2 
1
 = 0.5 − j7.5 kS
+ j C−
ω0 L 
R
 2
Z(ω0 2) =
Y(2ω0 ) =
1
= 8.85 + j132.74 Ω
0.0005 − j0.0075

1 
1
 = 0.5 + j7.5 kS
+ j  2ω0 L −
2ω0 C 
R

Z(2ω0 ) = 8.85 − j132.74 Ω
Y(4ω0 ) =

1 
1
 = 0.5 + j18.75 kS
+ j  4ω0 L −
4ω0 C 
R

Z(4ω0 ) = 1.4212 − j53.3 Ω
Chapter 14, Solution 37.
1 
L

)
 + jωLR  R + j(ωL −
1
C
ωC 



=
Z = jωL //( R +
)=
1
1 2
jωC
)
R+
+ jω L
R 2 + ( ωL −
jω C
ωC
jωL(R +
1
)
jω C
1 
L
 ωL −

ωC 
C
=0
1 2
2
R + ( ωL −
)
ωC
ωLR 2 +
Im(Z) =

→
Thus,
ω=
1
LC + R 2 C 2
Chapter 14, Solution 38.
Y
1
R − jωL
+ jωC = jωC + 2
R + jωL
R + ω2 L2
At resonance, Im(Y) = 0 , i.e.
ω 2 ( R 2 C 2 + LC) = 1
ω0 C −
ω0 L
=0
R 2 + ω02 L2
R 2 + ω02 L2 =
ω0 =
L
C
1 R2
−
=
LC L2
 50 
1


-3
-6 −
(40 × 10 )(10 × 10 )  40 × 10 -3 
2
ω0 = 4841 rad / s
Chapter 14, Solution 39.
(a)
B = ω 2 − ω1 = 2π(f 2 − f1 ) = 2π(90 − 86) x10 3 = 8πkrad / s
ωo =
B=
1
(ω1 + ω 2 ) = 2π(88) x10 3 = 176π
2
1
RC

→
1
C=
(b)
ωo =
(c )
ωo = 176π = 552.9krad / s
(d)
B = 8π = 25.13krad / s
(e)
Q=
LC

→
1
1
=
= 19.89nF
BR 8πx10 3 x 2x10 3
L=
1
ω2 o C
=
1
(176π) 2 x19.89x10 − 9
= 164.4H
ωo 176π
= 22
=
B
8π
Chapter 14, Solution 40.
(a)
L = 5 + 10 = 15 mH
ω0 =
1
LC
1
=
15x10
−3
x 20x10
−6
= 1.8257 k rad/sec
Q = ω0 RC = 1.8257 x10 3 x 25x10 3 x 20x10 −6 = 912.8
1
1
=
= 2 rad
3
RC 25x10 20x10 −6
B=
(b)
To increase B by 100% means that B’ = 4.
C′ =
Since C′ =
1
1
=
= 10 µF
RB′ 25x10 3 x 4
C1C 2
= 10µF and C1 = 20 µF, we then obtain C2 = 20 µF.
C1 + C 2
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 = 2+ 6 = 8Ω,
ω0 =
(b)
1
LC
=
1
0.4
L =1H,
C = 0.4 F
= 1.5811 rad / s
Q=
ω 0 L 1.5811
=
= 0.1976
R
8
B=
R
= 8 rad / s
L
This is a parallel RLC circuit.
3 µF and 6 µF 
→
C = 2 µF ,
(3)(6)
= 2 µF
3+ 6
R = 2 kΩ ,
L = 20 mH
ω0 =
1
LC
=
1
(2 × 10 -6 )(20 × 10 -3 )
= 5 krad / s
Q=
R
2 × 10 3
= 20
=
ω0 L (5 × 10 3 )(20 × 10 -3 )
B=
1
1
=
= 250 krad / s
3
RC (2 × 10 )(2 × 10 -6 )
Chapter 14, Solution 42.
(a)
Z in = (1 jωC) || (R + jωL)
Z in =
R + jωL
jωC
R + jωL +
1
jωC
=
R + jωL
1 − ω2 LC + jωRC
(R + jωL)(1 − ω2 LC − jωRC)
Z in =
(1 − ω2 LC) 2 + ω2 R 2 C 2
At resonance, Im(Z in ) = 0 , i.e.
0 = ωL(1 − ω2 LC) − ωR 2 C
ω2 LC = L − R 2 C
ω0 =
(b)
L − R 2C
=
LC
1 R2
−
C L
Z in = jωL || (R + 1 jωC)
Z in =
jωL (R + 1 jωC)
jωL (1 + jωRC)
=
R + jωL + 1 jωC (1 − ω2 LC) + jωRC
Z in =
(-ω2 RLC + jωL) [(1 − ω2 LC) − jωRC]
(1 − ω2 LC) 2 + ω2 R 2 C 2
At resonance, Im(Z in ) = 0 , i.e.
0 = ωL (1 − ω2 LC) + ω3 R 2 C 2 L
ω2 (LC − R 2 C 2 ) = 1
ω0 =
1
LC − R 2 C 2
Z in = R || ( jωL + 1 jωC)
(c)
R ( jωL + 1 jωC)
R (1 − ω2 LC)
=
R + jωL + 1 jωC (1 − ω 2 LC) + jωRC
R (1 − ω2 LC)[(1 − ω2 LC) − jωRC]
Z in =
(1 − ω2 LC) 2 + ω2 R 2 C 2
Z in =
At resonance, Im(Z in ) = 0 , i.e.
0 = R (1 − ω2 LC) ωRC
1 − ω2 LC = 0
ω0 =
1
LC
Chapter 14, Solution 43.
Consider the circuit below.
1/jωC
Zin
(a)
R1
jωL
Z in = (R 1 || jωL) || (R 2 + 1 jωC)
 R 1 jωL  
1 
 ||  R 2 +

Z in = 
jωC 
 R 1 + jωL  
R2
jωR 1 L 
1 

⋅ R 2 +
R 1 + jωL 
jωC 
Z in =
jR 1ωL
1
R2 +
+
jωC R 1 + jωL
Z in =
jωR 1 L (1 + jωR 2 C)
(R 1 + jωL)(1 + jωR 2 C) − ω2 LCR 1
Z in =
- ω2 R 1 R 2 LC + jωR 1 L
R 1 − ω2 LCR 1 − ω2 LCR 2 + jω (L + R 1 R 2 C)
(-ω2 R 1 R 2 LC + jωR 1 L)[R 1 − ω2 LCR 1 − ω2 LCR 2 − jω (L + R 1 R 2 C)]
Z in =
(R 1 − ω2 LCR 1 − ω2 LCR 2 ) 2 + ω2 (L + R 1 R 2 C) 2
At resonance, Im(Z in ) = 0 , i.e.
0 = ω3 R 1 R 2 LC (L + R 1 R 2 C) + ωR 1 L (R 1 − ω2 LCR 1 − ω2 LCR 2 )
0 = ω3 R 12 R 22 LC 2 + R 12 ωL − ω3 R 12 L2 C
0 = ω2 R 22 C 2 + 1 − ω2 LC
ω2 (LC − R 22 C 2 ) = 1
ω0 =
1
LC − R 22 C 2
ω0 =
1
(0.02)(9 × 10 -6 ) − (0.1) 2 (9 × 10 -6 ) 2
ω0 = 2.357 krad / s
(b)
At ω = ω0 = 2.357 krad / s ,
jωL = j(2.357 × 10 3 )(20 × 10 -3 ) = j47.14
R 1 || jωL =
R2 +
j47.14
= 0.9996 + j0.0212
1 + j47.14
1
1
= 0.1 +
= 0.1 − j47.14
jωC
j (2.357 × 10 3 )(9 × 10 -6 )
Z in (ω0 ) = (R 1 || jωL) || (R 2 + 1 jωC)
(0.9996 + j0.0212)(0.1 − j47.14)
(0.9996 + j0.0212) + (0.1 − j47.14)
Z in (ω0 ) =
Z in (ω0 ) = 1 Ω
Chapter 14, Solution 44.
We find the input impedance of the circuit shown below.
1
Z
jω(2/3)
1/jω
1/jωC
1
3
1
=
+ jω +
,
Z jω2
1 + 1 jωC
ω=1
C 2 + jC
1
jC
= -j1.5 + j +
= -j0.5 +
Z
1+ C2
1 + jC
v( t ) and i( t ) are in phase when Z is purely real, i.e.
0 = -0.5 +
C
1 + C2
1
C2
1
=
2 =
Z 1+ C
2

→ (C − 1) 2 = 1

→ Z = 2 Ω
V = Z I = (2)(10) = 20
v( t ) = 20 sin( t ) V ,
i.e.
Vo = 20 V
or
C = 1F
Chapter 14, Solution 45.
(a)
jω
,
1 + jω
1 || jω =
1 ||
1 jω
1
1
=
=
jω 1 + 1 jω 1 + jω
Transform the current source gives the circuit below.
jω
I
1 + jω
jω
1 + jω
+
−
1
1
1 + jω
+
Vo
−
1
jω
1 + jω
Vo =
⋅
I
1
jω 1 + j ω
1+
+
1 + jω 1 + jω
(b)
H (ω) =
Vo
jω
=
I
2 (1 + jω) 2
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.
ωo =
1
LC
→
(b)
(c )
C=
1
ω2 o L
=
1
(2πx15x10 3 ) 2 x10 x10 −3
Z = R, I = V/Z = 120/20 = 6 A
Q=
ωo L 2πx15x10 3 x10 x10 −3
=
= 15π = 47.12
R
20
= 11.26nF
Chapter 14, Solution 47.
H (ω) =
Vo
R
1
=
=
Vi R + jωL 1 + jωL R
H(0) = 1 and H(∞) = 0 showing that this circuit is a lowpass filter.
1
, i.e.
At the corner frequency, H(ωc ) =
2
1
=
2
1
ωc L 

1+ 
 R 
2

→ 1 =
ωc L
R
or
ωc =
R
L
Hence,
ωc =
R
= 2πf c
L
fc =
1 R
1 10 × 10 3
= 796 kHz
⋅ =
⋅
2π L 2π 2 × 10 -3
Chapter 14, Solution 48.
R ||
H (ω) =
1
jωC
jωL + R ||
1
jωC
R jωC
R + 1 jωC
H (ω) =
R jωC
jωL +
R + 1 jωC
H (ω) =
R
R + jωL − ω 2 RLC
H(0) = 1 and H(∞) = 0 showing that this circuit is a lowpass filter.
Chapter 14, Solution 49.
At dc, H(0) =
H(ω) =
Hence,
2
2
1
2
4
= 2.
2
H(0) =
2
2
4
=
4 + 100ωc2
4 + 100ωc2 = 8 
→ ωc = 0.2
H(2) =
4
2
=
2 + j20 1 + j10
H(2) =
2
101
= 0.199
In dB, 20 log10 H(2) = - 14.023
arg H(2) = -tan -110 = - 84.3°
Chapter 14, Solution 50.
H (ω) =
Vo
jωL
=
Vi R + jωL
H(0) = 0 and H(∞) = 1 showing that this circuit is a highpass filter.
H (ωc ) =
or
fc =
1
2
ωc =
=
1
 R 

1+ 
 ωc L 
2

→ 1 =
R
= 2πf c
L
1 R
1 200
⋅ =
⋅
= 318.3 Hz
2π L 2π 0.1
R
ωc L
Chapter 14, Solution 51.
H ′(ω) =
jωRC
jω
=
1 + jωRC jω + 1 RC
(from Eq. 14.52)
This has a unity passband gain, i.e. H(∞) = 1 .
1
= ωc = 50
RC
H ^ (ω) = 10 H ′(ω) =
H (ω) =
j10ω
50 + jω
j10ω
50 + jω
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.
ωc =
R
= 2πf c
L
R = 2πf c L = (2π)(10 5 )(40 × 10 -3 ) = 25.13 kΩ
Chapter 14, Solution 54.
ω1 = 2πf 1 = 20π × 10 3
ω2 = 2πf 2 = 22π × 10 3
B = ω2 − ω1 = 2π × 10 3
ω0 =
ω2 + ω1
= 21π × 10 3
2
Q=
ω0 21π
= 11.5
=
B
2π
ω0 =
1

→ L =
LC
1
ω02 C
L=
1
= 2.872 H
(21π × 10 ) (80 × 10 -12 )
B=
R
L
3 2

→ R = BL
R = (2π × 10 3 )(2.872) = 18.045 kΩ
Chapter 14, Solution 55.
ωc = 2πf c =
1
RC
→
R=
1
1
=
= 265.3kΩ
3
2πf c C 2πx 2x10 x300x10 −12
Chapter 14, Solution 56.
ωo =
1
LC
=
1
(25 × 10 )(0.4 × 10 − 6 )
−3
B=
R
10
=
= 0.4 krad / s
L 25 × 10 -3
Q=
10
= 25
0.4
= 10 krad / s
ω1 = ωo − B 2 = 10 − 0.2 = 9.8 krad / s
or
f1 =
9.8
= 1.56 kHz
2π
ω2 = ωo + B 2 = 10 + 0.2 = 10.2 krad / s
or
f2 =
10.2
= 1.62 kHz
2π
Therefore,
1.56 kHz < f < 1.62 kHz
Chapter 14, Solution 57.
(a)
From Eq 14.54,
R
R
sRC
L
H (s) =
=
=
2
R
1
1 1 + sRC + s LC
s2 + s +
R + sL +
L LC
sC
s
Since B =
R
and ω0 =
L
H (s) =
(b)
1
LC
,
sB
s + sB + ω02
2
From Eq. 14.56,
H (s) =
H (s) =
sL +
1
sC
1
R + sL +
sC
=
s2 +
s2 + s
1
LC
R
1
+
L LC
s 2 + ω02
s 2 + sB + ω02
Chapter 14, Solution 58.
(a)
Consider the circuit below.
I
Vs
+
−
R
I1
1/sC
+
1/sC
R
Vo
−
1 
1
R + 
1 
1
sC 
sC 
Z(s) = R +
||  R +  = R +
2
sC 
sC 
R+
sC
Z(s) = R +
1 + sRC
sC (2 + sRC)
1 + 3sRC + s 2 R 2 C 2
Z(s) =
sC (2 + sRC)
I=
Vs
Z
I1 =
Vs
1 sC
I=
2 sC + R
Z (2 + sRC)
Vo = I 1 R =
H (s) =
R Vs
sC (2 + sRC)
⋅
2 + sRC 1 + 3sRC + s 2 R 2 C 2
Vo
sRC
=
Vs 1 + 3sRC + s 2 R 2 C 2

3
s

1
RC
H (s) = 
3
1
3 2
s +
s+ 2 2

RC
R C
Thus, ω02 =
B=
(b)
1
R C2
2
or





ω0 =
1
= 1 rad / s
RC
3
= 3 rad / s
RC
Similarly,
Z(s) = sL + R || (R + sL) = sL +
R (R + sL)
2R + sL
R 2 + 3sRL + s 2 L2
Z(s) =
2R + sL
I=
Vs
,
Z
Vo = I 1 ⋅ sL =
I1 =
R Vs
R
I=
2R + sL
Z (2R + sL)
sLR Vs
2R + sL
⋅ 2
2R + sL R + 3sRL + s 2 L2
1  3R 

s
Vo
sRL
3 L 
H (s) =
=
=
3R
R2
Vs R 2 + 3sRL + s 2 L2
2
s +
s+ 2
L
L
Thus, ω0 =
B=
R
= 1 rad / s
L
3R
= 3 rad / s
L
Chapter 14, Solution 59.
1
ω0 =
(b)
R 2 × 10 3
B= =
= 2 × 10 4
L
0 .1
Q=
LC
=
1
(a)
(0.1)(40 × 10 -12 )
= 0.5 × 10 6 rad / s
ω0 0.5 × 10 6
=
= 250
B
2 × 10 4
As a high Q circuit,
B
ω1 = ω0 − = 10 4 (50 − 1) = 490 krad / s
2
ω 2 = ω0 +
(c)
B
= 10 4 (50 + 1) = 510 krad / s
2
As seen in part (b),
Q = 250
Chapter 14, Solution 60.
Consider the circuit below.
Ro
+
1/sC
Vi
+
−
R
Vo
sL
−

1  R (sL + 1 sC)
Z(s) = R || sL +  =

sC  R + sL + 1 sC
Z(s) =
R (1 + s 2 LC)
1 + sRC + s 2 LC
Vo
R (1 + s 2 LC)
Z
=
=
H=
Vi Z + R o R o + sRR o C + s 2 LCR o + R + s 2 LCR
R (1 + s 2 LC)
Z in = R o + Z = R o +
1 + sRC + s 2 LC
R o + sRR o C + s 2 LCR o + R + s 2 LCR
Z in =
1 + sRC + s 2 LC
s = jω
Z in =
R o + jωRR o C − ω2 LCR o + R − ω2 LCR
1 − ω2 LC + jωRC
Z in =
(R o + R − ω2 LCR o − ω2 LCR + jωRR o C)(1 − ω 2 LC − jωRC)
(1 − ω2 LC) 2 + (ωRC) 2
Im(Z in ) = 0 implies that
- ωRC [R o + R − ω2 LCR o − ω2 LCR ] + ωRR o C (1 − ω2 LC) = 0
R o + R − ω2 LCR o − ω2 LCR − R o + ω2 LCR o = 0
ω2 LCR = R
1
ω0 =
H=
LC
1
=
(1 × 10 )(4 × 10 -6 )
-3
R (1 − ω2 LC)
R o + jωRR o C + R − ω2 LCR o − ω2 LCR
H max = H(0) =
or
H max
R
Ro + R
 1

R  2 − LC 
R
ω

= H(∞) = lim
=
ω→ ∞ R o + R
RR o C
+j
− LC (R + R o ) R + R o
2
ω
ω
At ω1 and ω2 , H =
R
2 (R o + R )
1
2
1
2
0=
= 15.811 krad / s
=
=
=
1
2
H mzx
R (1 − ω2 LC)
R o + R − ω 2 LC (R o + R ) + jωRR o C
(R o + R )(1 − ω2 LC)
(ωRR o C) 2 + (R o + R − ω2 LC(R o + R )) 2
10 (1 − ω2 ⋅ 4 × 10 -9 )
(96 × 10 -6 ω) 2 + (10 − ω2 ⋅ 4 × 10 -8 ) 2
10 (1 − ω2 ⋅ 4 × 10 -9 )
(96 × 10 -6 ω) 2 + (10 − ω2 ⋅ 4 × 10 -8 ) 2
−
1
2
(10 − ω2 ⋅ 4 × 10 -8 )( 2 ) − (96 × 10 -6 ω) 2 + (10 − ω2 ⋅ 4 × 10 -8 ) 2 = 0
(2)(10 − ω2 ⋅ 4 × 10 -8 ) 2 = (96 × 10 -6 ω) 2 + (10 − ω2 ⋅ 4 × 10 -8 ) 2
(96 × 10 -6 ω) 2 − (10 − ω2 ⋅ 4 × 10 -8 ) 2 = 0
1.6 × 10 -15 ω4 − 8.092 × 10 -7 ω2 + 100 = 0
ω4 − 5.058 × 10 8 + 6.25 × 1016 = 0
 2.9109 × 10 8
ω2 = 
 2.1471 × 10 8
Hence,
ω1 = 14.653 krad / s
ω2 = 17.061 krad / s
B = ω2 − ω1 = 17.061 − 14.653 = 2.408 krad / s
Chapter 14, Solution 61.
(a)
V+ =
1 jωC
V,
R + 1 jωC i
V− = Vo
Since V+ = V− ,
1
V = Vo
1 + jωRC i
H (ω) =
(b)
V+ =
Vo
1
=
Vi 1 + jωRC
R
V,
R + 1 jωC i
Since V+ = V− ,
jωRC
V = Vo
1 + jωRC i
H (ω) =
Vo
jωRC
=
Vi 1 + jωRC
V− = Vo
Chapter 14, Solution 62.
This is a highpass filter.
(a)
H (ω) =
jωRC
1
=
1 + jωRC 1 − j ωRC
H (ω) =
1
,
1 − j ωc ω
H (ω) =
1
1
=
1 − j f c f 1 − j1000 f
H (f = 200 Hz) =
Vo =
(b)
1 − j5
Vo
1
=
1 − j5 Vi
Vo
1
=
1 − j0.5 Vi
120 mV
= 107.3 mV
1 − j0.5
Vo
1
=
1 − j0.1 Vi
H (f = 10 kHz) =
Vo =
1
= 2π (1000)
RC
= 23.53 mV
H (f = 2 kHz) =
Vo =
(c)
120 mV
ωc =
120 mV
1 − j0.1
= 119.4 mV
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:
C i R f = 10

→
Rf =
10
= 10MΩ
Ci
C i R i = 0.1
→
Ri =
0.1
= 100kΩ
Ci
Chapter 14, Solution 64.
Z f = R f ||
Rf
1
=
jωC f 1 + jωR f C f
Zi = R i +
1 + jωR i C i
1
=
jωC i
jωC i
Hence,
H (ω) =
Vo - Z f
- jωR f C i
=
=
Vi
Zi
(1 + jωR f C f )(1 + jωR i C i )
This is a bandpass filter. H(ω) is similar to the product of the transfer function
of a lowpass filter and a highpass filter.
Chapter 14, Solution 65.
V+ =
R
jωRC
Vi =
V
R + 1 jωC
1 + jωRC i
V− =
Ri
V
Ri + Rf o
Since V+ = V− ,
Ri
jωRC
Vo =
V
Ri + Rf
1 + jωRC i
H (ω) =
Vo 
R f   jωRC 


= 1 +
Vi 
R i   1 + jωRC 
It is evident that as ω → ∞ , 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 = R 2 R 3 ,
H (s) =
(c)
R4
s
⋅
R 3 + R 4 s + 1 R 2C
When R 3 → ∞ ,
H (s) =
- 1 R 1C
s + 1 R 2C
Chapter 14, Solution 67.
DC gain =
Rf 1
=
Ri 4

→ R i = 4R f
Corner frequency = ωc =
1
= 2π (500) rad / s
R f Cf
If we select R f = 20 kΩ , then R i = 80 kΩ and
C=
1
= 15.915 nF
(2π)(500)(20 × 10 3 )
Therefore, if R f = 20 kΩ , then R i = 80 kΩ and C = 15.915 nF
Chapter 14, Solution 68.
High frequency gain = 5 =
Corner frequency = ωc =
Rf
Ri

→ R f = 5R i
1
= 2π (200) rad / s
R i Ci
If we select R i = 20 kΩ , then R f = 100 kΩ and
C=
1
= 39.8 nF
(2π)(200)(20 × 10 3 )
Therefore, if R i = 20 kΩ , then R f = 100 kΩ and C = 39.8 nF
Chapter 14, Solution 69.
This is a highpass filter with f c = 2 kHz.
1
ωc = 2πf c =
RC
RC =
1
1
=
2πf c 4π × 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 kΩ , then R f = 25 kΩ , and C =
1
= 7.96 nF .
4000π × 10 4
Chapter 14, Solution 70.
(a)
H (s) =
Vo (s)
Y1 Y2
=
Vi (s) Y1 Y2 + Y4 (Y1 + Y2 + Y3 )
where Y1 =
H (s) =
(b)
1
1
= G 1 , Y2 =
= G 2 , Y3 = sC1 , Y4 = sC 2 .
R1
R2
G 1G 2
G 1 G 2 + sC 2 (G 1 + G 2 + sC1 )
G 1G 2
H(∞) = 0
= 1,
G 1G 2
showing that this circuit is a lowpass filter.
H ( 0) =
Chapter 14, Solution 71.
R = 50 Ω , L = 40 mH , C = 1 µF
L′ =
Km
Km
L 
→ 1 =
⋅ (40 × 10 -3 )
Kf
Kf
25K f = K m
C′ =
C
KmKf
10 6 K f =
(1)

→ 1 =
1
Km
Substituting (1) into (2),
1
10 6 K f =
25K f
K f = 0.2 × 10 -3
K m = 25K f = 5 × 10 -3
10 -6
KmKf
(2)
Chapter 14, Solution 72.
L′C′ =
LC
K f2

→ K f2 =
LC
L ′C′
(4 × 10 -3 )(20 × 10 -6 )
K =
= 4 × 10 -8
(1)(2)
2
f
K f = 2 × 10 -4
L′ L 2
= K
C′ C m

→ K 2m =
L′ C
⋅
C′ L
(1)(20 × 10 -6 )
K =
= 2.5 × 10 -3
(2)(4 × 10 -3 )
2
m
K m = 5 × 10 -2
Chapter 14, Solution 73.
R ′ = K m R = (12)(800 × 10 3 ) = 9.6 MΩ
L′ =
Km
800
L=
(40 × 10 -6 ) = 32 µF
Kf
1000
C′ =
C
300 × 10 -9
=
= 0.375 pF
K m K f (800)(1000)
Chapter 14, Solution 74.
R '1 = K m R 1 = 3x100 = 300Ω
R ' 2 = K m R 2 = 10 x100 = 1 kΩ
L' =
Km
10 2
L=
(2) = 200 µH
Kf
10 6
1
C
= 10 = 1 nF
C' =
K m K f 108
Chapter 14, Solution 75.
R ' = K m R = 20 x10 = 200 Ω
L' =
Km
10
L=
(4) = 400 µH
Kf
10 5
C' =
C
1
=
= 1 µF
K m K f 10x10 5
Chapter 14, Solution 76.
R ' = K m R = 50 x10 3
L' =
Km
L = 10 µH
Kf
C' = 40 pF =
C
KmKf
→
R=
50 x10 3
10 3
L = 10 x10 −6 x
→
= 50 Ω
10 6
10 3
= 10 mH
C = 40 x10 −12 x10 3 x10 6 = 40 mF
→
Chapter 14, Solution 77.
L and C are needed before scaling.
B=
R
L
ω0 =
(a)

→ L =
1
LC
R 10
=
=2H
B 5

→ C =
1
1
=
= 312.5 µF
2
ω0 L (1600)(2)
L′ = K m L = (600)(2) = 1200 H
C
3.125 × 10 -4
C′ =
=
= 0.5208 µF
Km
600
(b)
L′ =
L
2
= 3 = 2 mH
K f 10
C
3.125 × 10 -4
=
= 312.5 nF
C′ =
Kf
10 3
(c)
L′ =
Km
(400)(2)
= 8 mH
L=
10 5
Kf
C′ =
C
3.125 × 10 -4
=
= 7.81 pF
KmKf
(400)(10 5 )
Chapter 14, Solution 78.
R ′ = K m R = (1000)(1) = 1 kΩ
Km
10 3
L′ =
L = 4 (1) = 0.1 H
10
Kf
C′ =
C
1
=
= 0.1 µF
3
K m K f (10 )(10 4 )
The new circuit is shown below.
1 kΩ
+
I
1 kΩ
0.1 H
0.1 µF
1 kΩ
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 =
V2 = V1 − 3Vo =
Vo =

→ V2 = V1 − 3Vo
sL
V
sL + 1 sC 2
Combining (2) and (3)
(1)

→
Vo
V2
=
sL sL + 1 sC
(3)
sL + 1 sC
Vo
sL
s 2 LC
V
1 + 4s 2 LC 1
Substituting (3) and (4) into (1) gives
1 − V1 Vo
sC
=
=
V
R
sL 1 + 4s 2 LC 1
sRC
1 + 4s 2 LC + sRC
1 = V1 +
V =
V1
1 + 4s 2 LC 1
1 + 4s 2 LC
V1 =
(2)
1 + 4s 2 LC
1 + 4s 2 LC + sRC
(4)
Io =
1 − V1
sRC
=
R
R (1 + 4s 2 LC + sRC)
Z in =
1 1 + sRC + 4s 2 LC
=
sC
Io
Z in = 4sL + R +
1
sC
(5)
When R = 5 , L = 2 , C = 0.1 ,
Z in (s) = 8s + 5 +
10
s
At resonance,
Im(Z in ) = 0 = 4ωL −
or
(b)
ω0 =
1
2 LC
=
1
ωC
1
2 (0.1)(2)
= 1.118 rad / s
After scaling,
R′ 
→ K m R
4Ω 
→ 40 Ω
5Ω 
→ 50 Ω
L′ =
Km
10
L=
( 2 ) = 0 .2 H
Kf
100
C′ =
C
0.1
=
= 10 -4
K m K f (10)(100)
From (5),
Z in (s) = 0.8s + 50 +
ω0 =
1
2 LC
=
10 4
s
1
2 (0.2)(10 -4 )
= 111.8 rad / s
Chapter 14, Solution 80.
(a)
R ′ = K m R = (200)(2) = 400 Ω
L′ =
K m L (200)(1)
=
= 20 mH
Kf
10 4
C′ =
C
0.5
=
= 0.25 µF
K m K f (200)(10 4 )
The new circuit is shown below.
20 mH
a
Ix
0.25 µF
400 Ω
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
V2
+ 0 .5 I x =
sL
R
But, I x = sC V1 .
(1)
V1 − V2
V2
+ 0.5sC V1 =
sL
R
(2)
Solving (1) and (2),
sL + R
V1 = 2
s LC + 0.5sCR + 1
Z Th =
V1
sL + R
= 2
1 s LC + 0.5sCR + 1
At ω = 10 4 ,
Z Th
( j10 4 )(20 × 10 -3 ) + 400
=
( j10 4 ) 2 (20 × 10 -3 )(0.25 × 10 -6 ) + 0.5( j10 4 )(0.25 × 10 -6 )(400) + 1
Z Th =
400 + j200
= 600 − j200
0.5 + j0.5
Z Th = 632.5∠ - 18.435° ohms
Chapter 14, Solution 81.
(a)
1
(G + jωC)(R + jωL) + 1
1
= G + jωC +
=
R + jω L
R + jω L
Z
which leads to
Z=
jωL + R
2
− ω LC + jω(RC + LG) + GR + 1
ω R
+
C
LC
Z(ω) =
R
G  GR + 1

− ω2 + jω +  +
LC
L C
j
(1)
We compare this with the given impedance:
Z(ω) =
1000( jω + 1)
2
− ω + 2 jω + 1 + 2500
(2)
Comparing (1) and (2) shows that
1
= 1000
C

→
R G
+ =2
L C

→
C = 1 mF,
R/L = 1

→
R=L
G = C = 1 mS
GR + 1 10 −3 R + 1
2501 =
=
LC
10 −3 R
→
R = 0 .4 = L
Thus,
R = 0.4Ω, L = 0.4 H, C = 1 mF, G = 1 mS
(b) By frequency-scaling, Kf =1000.
R’ = 0.4 Ω, G’ = 1 mS
L' =
L
0.4
=
= 0.4mH ,
K f 10 3
C' =
C 10 −3
=
= 1µF
K f 10 − 3
Chapter 14, Solution 82.
C′ =
C
KmKf
Kf =
ω′c 200
=
= 200
ω
1
Km =
C 1
1
1
⋅
= -6 ⋅
= 5000
C′ K f 10 200
R ′ = K m R = 5 kΩ,
thus,
R ′f = 2R i = 10 kΩ
Chapter 14, Solution 83.
1µF

→
1
10 −6
C' =
C=
= 0.1 pF
K mKf
100 x10 5
5µF

→
C' = 0.5 pF
10 kΩ
→
R ' = K m R = 100x10 kΩ = 1 MΩ
20 kΩ
→
R ' = 2 MΩ
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 = Vo . The schematic is shown below. The circuit
is simulated for 100 < f < 10 kHz.
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
1
1 R2
f0 =
−
2π LC L2
R
400
10 7
=
=
,
L 240 × 10 -6
6
Since
R
1
<<
L
LC
1
1
1016
=
=
LC (240 × 10 -6 )(120 × 10 -12 ) 288
f0 ≅
1
2π LC
=
10 8
24π 2
= 938 kHz
R
1
<<
.
L
LC
The result remains the same.
If R is reduced to 40 Ω,
Chapter 14, Solution 94.
ωc =
1
RC
We make R and C as small as possible. To achieve this, we connect 1.8 k Ω and 3.3 k Ω
in parallel so that
1.8x 3.3
R=
= 1.164 kΩ
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
ωc =
=
= 114.55x10 6 rad/s
RC 1.164x10 3 x 7.5x10 −12
Chapter 14, Solution 95.
(a)
f0 =
1
2π LC
When C = 360 pF ,
f0 =
1
2π (240 × 10 -6 )(360 × 10 -12 )
= 0.541 MHz
When C = 40 pF ,
f0 =
1
2π (240 × 10 -6 )(40 × 10 -12 )
Therefore, the frequency range is
0.541 MHz < f 0 < 1.624 MHz
= 1.624 MHz
Q=
(b)
2πfL
R
At f 0 = 0.541 MHz ,
Q=
(2π )(0.541 × 10 6 )(240 × 10 -6 )
= 67.98
12
At f 0 = 1.624 MHz ,
(2π )(1.624 × 10 6 )(240 × 10 -6 )
= 204.1
Q=
12
Chapter 14, Solution 96.
Ri
Vi
L
V1
Vo
+
+
−
C1
C2
RL
Vo
−
Z2
Z1 = R L ||
Z2 =
Z1
RL
1
=
sC 2 1 + sR 2 C 2
1
1  sL + R L + s 2 R L C 2 L 

|| (sL + Z1 ) =
|| 
sC1
sC1 
1 + sR L C 2

1 sL + R L + s 2 R L C 2 L
⋅
sC1
1 + sR L C 2
Z2 =
sL + R L + s 2 R L C 2 L
1
+
sC1
1 + sR L C 2
sL + R L + s 2 R L LC 2
Z2 =
1 + sR L C 2 + s 2 LC1 + sR L C1 + s 3 R L LC1C 2
V1 =
Z2
V
Z2 + R i i
Vo =
Z1
Z2
Z1
V1 =
⋅
V
Z1 + sL
Z 2 + R 2 Z1 + sL i
Vo
Z2
Z1
=
⋅
Vi Z 2 + R 2 Z1 + sL
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
and
Z1
RL
=
Z1 + sL 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 )
where s = jω .
Chapter 14, Solution 97.
Ri
Vi
L
V1
Vo
+
+
−
C1
C2
RL
Vo
−
Z2
Z1

1  sL (R L + 1 sC 2 )
=
Z = sL ||  R L +
,
sC 2  R L + sL + 1 sC 2

V1 =
Z
V
Z + R i + 1 sC1 i
Vo =
RL
RL
Z
V1 =
⋅
V
R L + 1 sC 2
R L + 1 sC 2 Z + R i + 1 sC1 i
H (ω) =
Vo
RL
sL (R L + 1 sC 2 )
=
⋅
Vi R L + 1 sC 2 sL (R L + 1 sC 2 ) + (R i + 1 sC1 )(R L + sL + 1 sC 2 )
H (ω) =
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)
where s = jω .
Chapter 14, Solution 98.
B = ω2 − ω1 = 2π (f 2 − f 1 ) = 2π (454 − 432) = 44π
ω0 = 2πf 0 = QB = (20)(44π )
f0 =
s = jω
(20)(44π)
= (20)(22) = 440 Hz
2π
Chapter 14, Solution 99.
Xc =
C=
1
1
=
ωC 2πf C
1
1
10 -9
=
=
2πf X c (2π )(2 × 10 6 )(5 × 10 3 ) 20π
X L = ωL = 2πf L
XL
300
3 × 10 -4
=
=
L=
2πf (2π )(2 × 10 6 )
4π
f0 =
B=
1
2π LC
=
1
3 × 10 -4 10 -9
⋅
2π
4π
20π
= 1.826 MHz
 4π 
R
 = 4.188 × 10 6 rad / s
= (100) 
 3 × 10 -4 
L
Chapter 14, Solution 100.
ωc = 2πf c =
R=
1
RC
1
1
=
= 15.91 Ω
2πf c C (2π )(20 × 10 3 )(0.5 × 10 -6 )
Chapter 14, Solution 101.
ωc = 2πf c =
R=
1
RC
1
1
=
= 1.061 kΩ
2πf c C (2π )(15)(10 × 10 -6 )
Chapter 14, Solution 102.
(a)
When R s = 0 and R L = ∞ , we have a low-pass filter.
ωc = 2πf c =
fc =
1
RC
1
1
=
= 994.7 Hz
2πRC (2π)(4 × 10 3 )(40 × 10 -9 )
(b)
We obtain R Th across the capacitor.
R Th = R L || (R + R s )
R Th = 5 || (4 + 1) = 2.5 kΩ
fc =
1
1
=
2πR Th C (2π )(2.5 × 10 3 )(40 × 10 -9 )
f c = 1.59 kHz
Chapter 14, Solution 103.
H (ω) =
H (ω) =
H (ω) =
Vo
R2
,
=
Vi R 2 + R 1 || 1 sC
s = jω
R (R + 1 sC)
R2
= 2 1
R (1 sC) R 2 + R 1 (1 sC)
R2 + 1
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.
e at + e - at
2
1 1
1 
s
L [ cosh(at ) ] = 
+
= 2

2  s − a s + a  s − a2
(a)
cosh(at ) =
(b)
sinh(at ) =
e at − e - at
2
1 1
1 
a
L [ sinh(at ) ] = 
−
= 2

2  s − a s + a  s − a2
Chapter 15, Solution 2.
(a)
f ( t ) = cos(ωt ) cos(θ) − sin(ωt ) sin(θ)
F(s) = cos(θ) L [ cos(ωt ) ] − sin(θ) L [ sin(ωt ) ]
F(s) =
(b)
s cos(θ) − ω sin(θ)
s 2 + ω2
f ( t ) = sin(ωt ) cos(θ) + cos(ωt ) sin(θ)
F(s) = sin(θ) L [ cos(ωt ) ] + cos(θ) L [ sin(ωt ) ]
F(s) =
s sin(θ) − ω cos(θ)
s 2 + ω2
Chapter 15, Solution 3.
(a)
L [ e -2t cos(3t ) u ( t ) ] =
s+2
(s + 2 ) 2 + 9
(b)
L [ e -2t sin(4 t ) u ( t ) ] =
4
(s + 2) 2 + 16
(c)
Since L [ cosh(at ) ] =
s
s − a2
2
L [ e -3t cosh(2 t ) u ( t ) ] =
(d)
Since L [ sinh(at ) ] =
L [ e -4t sinh( t ) u ( t ) ] =
(e)
L [ e - t sin( 2t ) ] =
If
s+3
(s + 3 ) 2 − 4
a
s − a2
2
1
(s + 4) 2 − 1
2
(s + 1) 2 + 4
f (t) ←
→ F(s)
t f (t) ←
→
-d
F(s)
ds
Thus, L [ t e - t sin(2 t ) ] =
=
L [ t e -t sin( 2t ) ] =
[
-d
-1
2 ( (s + 1) 2 + 4)
ds
2
⋅ 2 (s + 1)
((s + 1) 2 + 4) 2
4 (s + 1)
((s + 1) 2 + 4) 2
Chapter 15, Solution 4.
(a)
G (s) = 6
(b)
F(s) =
s
s2 + 42
2
s2
+5
]
e −s =
e −2s
s+3
6se −s
s 2 + 16
Chapter 15, Solution 5.
(a)
L [ cos(2t + 30°) ] =
s cos(30°) − 2 sin(30°)
s2 + 4
L [ t 2 cos(2t + 30°) ] =
d 2  s cos(30°) − 1 
ds 2  s 2 + 4 
=

-1 
d d  3
s − 1 (s 2 + 4) 

ds ds  2


=
 3

-1
-2 
d  3 2
s − 1 (s 2 + 4) 
 (s + 4) − 2s 
ds  2

 2

 3

 3
 3

3

 2s 
 (8s 2 ) 

2
s
1
s
1
−
−
(- 2s )  2

 2 
 2


− 
+


= 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

=
+
3
2
2
2
s +4
s +4
(
=
)
(-3 3 s + 2)(s 2 + 4)
(s
2
+ 4)
3
L [ t 2 cos(2t + 30°) ] =
(
+
)
4 3 s3 − 8 s 2
(s
2
+ 4)
3
8 − 12 3 s − 6s 2 + 3s 3
( s 2 + 4) 3
4!
720
5 =
(s + 2)
(s + 2 ) 5
(b)
L [ 30 t 4 e - t ] = 30 ⋅
(c)

 2
d
2
L  2t u ( t ) − 4 δ( t )  = 2 − 4(s ⋅ 1 − 0) = 2 − 4s

 s
s
dt
)
(d)
2 e -(t-1) u ( t ) = 2 e -t u ( t )
2e
L [ 2 e -(t-1) u ( t ) ] =
s+1
(e)
Using the scaling property,
1
1
1
5
L [ 5 u ( t 2) ] = 5 ⋅
⋅
= 5⋅ 2⋅ =
1 2 s (1 2)
2s s
6
18
=
s + 1 3 3s + 1
(f)
L [ 6 e -t 3 u ( t ) ] =
(g)
Let f ( t ) = δ( t ) . Then, F(s) = 1 .
 dn

L  n δ( t )  =
 dt

 dn

L  n f ( t )  = s n F(s) − s n −1 f (0) − s n − 2 f ′(0) − "
 dt

 dn

L  n δ( t )  =
 dt

 dn

L  n f ( t )  = s n ⋅ 1 − s n −1 ⋅ 0 − s n − 2 ⋅ 0 − "
 dt

 dn

L  n δ( t )  = s n
 dt

Chapter 15, Solution 6.
(a)
L [ 2 δ( t − 1) ] = 2 e -s
(b)
L [ 10 u ( t − 2) ] =
10 - 2s
e
s
(c)
L [ ( t + 4) u ( t ) ] =
1 4
+
s2 s
(d)
L[ 2e
-t
u ( t − 4) ] = L [ 2 e
-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) ] = 2
s + 16
(a)
Since L [ cos(4t ) ] =
(b)
Since L [ t 2 ] =
L [ t 2 e -2 t ] =
2
2
1
[
]
L
=
u
(
t
)
,
,
s3
s
e -3s
2
[
]
,
and
L
−
=
u
(
t
3
)
s
(s + 2) 3
L [ t 2 e -2 t u ( t ) + u ( t − 3) ] =
2
e -3s
+
(s + 2 ) 3
s
Chapter 15, Solution 8.
(a)
L [ 2 δ(3t ) + 6 u (2t ) + 4 e -2 t − 10 e -3 t
]
1
1 1
4
10
= 2⋅ + 6⋅ ⋅
+
−
3
2 s 2 s+2 s+3
=
(b)
2 6
4
10
+ +
−
3 s s+2 s+3
t e -t u ( t − 1) = ( t − 1) e -t u ( t − 1) + e -t u ( t − 1)
t 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
e -(s+1)
e -(s+1)
+
=
+
(s + 1) 2
s + 1 (s + 1) 2 s + 1
L [ cos(2 ( t − 1)) u ( t − 1) ] =
s e -s
s2 + 4
(d)
Since sin(4 ( t − π)) = sin(4t ) cos(4π ) − sin( 4π) cos(4t ) = sin(4t )
sin(4t ) u ( t − π ) = sin(4 ( t − π )) u ( t − π )
L[ sin( 4 t ) [ u ( t ) − u ( t − π )] ]
= L[ sin( 4 t ) u ( t ) ] − L[ sin( 4( t − π )) u ( t − π) ]
=
4 e - πs
4
4
−
= 2
⋅ (1 − e -πs )
2
2
s + 16 s + 16 s + 16
Chapter 15, Solution 9.
(a)
f ( t ) = ( t − 4) u ( t − 2) = ( t − 2) u ( t − 2) − 2 u ( t − 2)
e -2s 2 e -2s
F(s) = 2 − 2
s
s
(b)
g( t ) = 2 e -4t u ( t − 1) = 2 e -4 e -4(t -1) u ( t − 1)
2 e -s
G (s) = 4
e (s + 4)
(c)
h ( t ) = 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)
h ( t ) = 5 cos(1) cos(2 t ) u ( t ) + 5 sin(1) sin(2t ) u ( t )
H(s) = 5 cos(1) ⋅
H(s) =
(d)
s
2
+ 5 sin(1) ⋅ 2
s +4
s +4
2
2.702 s 8.415
+
s2 + 4 s2 + 4
p( t ) = 6u ( t − 2) − 6u ( t − 4)
P(s) =
6 - 2s 6 -4s
e − e
s
s
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 ) = e -t cos( t ) − t e -t cos( t ) − t e -t sin( t )
(b)
G (s) =

s +1
d  s +1  d 
1
+ 
+ 

2
2
2
(s + 1) + 1 ds  (s + 1) + 1 ds  (s + 1) + 1
G (s) =
s +1
s 2 + 2s
2s + 2
s 2 (s + 2)
−
−
=
s 2 + 2s + 2 (s 2 + 2s + 2) 2 (s 2 + 2s + 2) 2 (s 2 + 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  s +1 
(s)(s 2 + 2s)
s 2 (s + 2)
=
=
ds  (s + 1) 2 + 1  (s 2 + 2s + 2) 2 (s 2 + 2s + 2) 2
Chapter 15, Solution 11.
(a)
Since L [ cosh(at ) ] =
F(s) =
(b)
s
s − a2
2
6 (s + 1)
6 (s + 1)
= 2
2
(s + 1) − 4 s + 2s − 3
Since L [ sinh(at ) ] =
L [ 3 e -2t sinh(4t ) ] =
a
s − a2
2
(3)(4)
12
= 2
2
(s + 2) − 16 s + 4s − 12
F(s) = L [ t ⋅ 3 e -2t sinh(4t ) ] =
-d
[ 12 (s 2 + 4s − 12) -1 ]
ds
F(s) = (12)(2s + 4)(s 2 + 4s − 12) -2 =
24 (s + 2)
(s + 4s − 12) 2
2
(c)
cosh( t ) =
1
⋅ (e t + e - t )
2
1
f ( t ) = 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)
s+2
Similarly, L [ 4 e -8 e -4(t -2) u ( t − 2)] =
4 e -(2s+8)
s+4
Therefore,
4 e -(2s+ 4) 4 e -(2s+8) e -(2s+ 6) [ (4 e 2 + 4 e -2 ) s + (16 e 2 + 8 e -2 )]
+
=
F(s) =
s+2
s+4
s 2 + 6s + 8
Chapter 15, Solution 12.
f ( t ) = te −2( t −1) e −2 u ( t − 1) = ( t − 1)e −2 e −2( t −1) u ( t − 1) + e −2 e −2( t −1) u ( t − 1)
f (s) = e − s
e−2
(s + 2) 2
+ e−2
e −s
e − (s + 2) 
1 
s + 3 − (s + 2)
=
e
1 +
=
s+2
s + 2  s + 2  (s + 2) 2
Chapter 15, Solution 13.
(a) tf (t )
←
→
−
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
1
= 2
2
( s + 1) + 1 s + 2s + 2
dF ( s 2 + 2s + 2)(0) − (1)(2s + 2)
=
ds
( s 2 + 2s + 2) 2
L (e −t t sin t ) = −
(c )
f (t )
t
dF
2( s + 1)
= 2
ds ( s + 2s + 2) 2
∞
←
→
∫ F (s)ds
s
Let f (t ) = sin βt , then F ( s ) =
∞
β
s +β2
2
β
1
s
 sin βt 
L
=∫ 2
ds = β tan −1
2

β
β
 t  s s +β
∞
s
=
π
2
− tan −1
s
β
Chapter 15, Solution 14.
 5t
0 < t <1
f (t) = 
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
β
s
Chapter 15, Solution 15.
f ( t ) = 10 [ u ( t ) − u ( t − 1) − u ( t − 1) + u ( t − 2)]
1 2
e -2s  10
F(s) = 10  − e -s +
=
(1 − e -s ) 2
s  s
s 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.
0
t<0
 t 2 0 < t < 1
f (t) = 
1 1< t < 3
 0
t>3
f ( t ) = 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)
F(s) =
2
2 -s e -3s
-s
−
−
(
1
e
)
e −
s3
s2
s
Chapter 15, Solution 18.
(a)
g ( t ) = 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
G (s) = (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
2
-s
+ 2 e = 2 (1 − e -s − e - 3s + e -4s )
2 (1 − e ) − 2 e
s
s
s
s
Chapter 15, Solution 19.
Since L[ δ( t )] = 1 and T = 2 , F(s) =
1
1 − e - 2s
Chapter 15, Solution 20.
Let
g 1 ( t ) = sin(πt ), 0 < t < 1
= sin( πt ) [ u ( t ) − u ( t − 1)]
= sin(πt ) u ( t ) − sin(πt ) u ( t − 1)
Note that sin(π ( t − 1)) = sin(πt − π) = - sin(πt ) .
So,
g1 ( t ) = sin( πt) u(t) + sin( π( t - 1)) u(t - 1)
G 1 (s) =
π
(1 + e -s )
s + π2
2
G 1 (s)
π (1 + e -s )
G (s) =
=
1 − e -2s (s 2 + π 2 )(1 − e - 2s )
Chapter 15, Solution 21.
T = 2π
Let

t 
f 1 ( t ) = 1 −  [ u ( t ) − u ( t − 1)]
 2π 

t
1
1 
f1 ( t ) = u ( t ) −
u(t) +
( t − 1) u ( t − 1) − 1 −  u ( t − 1)
 2π 
2π
2π

1
1
e -s
1  -s 1 [ 2π + (-2π + 1) e -s ] s + [ - 1 + e -s ]

e ⋅ =
F1 (s) = −
1
+
+
+
s 2πs 2 2πs 2 
2π 
s
2πs 2
F(s) =
[ 2π + (-2π + 1) e -s ] s + [ - 1 + e -s ]
F1 (s)
=
1 − e -Ts
2πs 2 (1 − e - 2 πs )
Chapter 15, Solution 22.
(a)
Let
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
G 1 (s) = 2 − 2 + e
s
s
s
G (s) =
G 1 (s)
, T =1
1 − e -sT
2 (1 − e -s + s e -s )
G (s) =
s 2 (1 − e -s )
(b)
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.
 2t
0 < t <1
h 1 (t) = 
 4 − 2t 1 < t < 2
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)
h 1 ( t ) = 2 t u ( t ) − 4 ( t − 1) u ( t − 1) − 2 ( t − 2) u ( t − 2)
2 4 -s 2 e -2s
2
= 2 (1 − e -s ) 2
2
2 − 2 e −
s
s
s
s
H 1 (s) =
2 (1 − e -s ) 2
H 0 (s) = 2
s (1 − e -2s )
1 2 (1 − e -s ) 2
H(s) = + 2
s s (1 − e - 2s )
Chapter 15, Solution 23.
(a)
Let
1 0 < t <1
f1 ( t ) = 
- 1 1 < t < 2
f 1 ( t ) = [ u ( t ) − u ( t − 1)] − [ u ( t − 1) − u ( t − 2)]
f 1 ( t ) = u ( t ) − 2 u ( t − 1) + u ( t − 2)
1
1
F1 (s) = (1 − 2 e -s + e -2s ) = (1 − e -s ) 2
s
s
F(s) =
F1 (s)
, T=2
(1 − e -sT )
(1 − e -s ) 2
F(s) =
s (1 − e - 2s )
(b)
Let
h 1 ( t ) = t 2 [ u ( t ) − u ( t − 2)] = t 2 u ( t ) − t 2 u ( t − 2)
h 1 ( t ) = t 2 u ( t ) − ( t − 2) 2 u ( t − 2) − 4 ( t − 2) u ( t − 2) − 4 u ( t − 2)
H 1 (s) =
2
4
4
- 2s
) − 2 e -2s − e -2s
3 (1 − e
s
s
s
H(s) =
H 1 (s)
, T=2
(1 − e -Ts )
H(s) =
2 (1 − e -2s ) − 4s e -2s (s + s 2 )
s 3 (1 − e - 2s )
Chapter 15, Solution 24.
(a)
10s 4 + s
2
s →∞ s + 6s + 5
f (0) = lim sF(s) = lim
s →∞
= lim
s →∞
10 +
1
s3
1 6 5
+ +
s 2 s3 s 4
=
10
=∞
0
10s 4 + s
=0
s → 0 s 2 + 6s + 5
f (∞) = lim sF(s) = lim
s →0
(b)
s2 + s
=1
2
s →∞ s − 4s + 6
f (0) = lim sF(s) = lim
s →∞
The complex poles are not in the left-half plane.
f (∞) does not exist
(c)
2s 3 + 7s
2
s →∞ (s + 1)(s + 2)(s + 2s + 5)
f (0) = lim sF(s) = lim
s →∞
2 7
+
0
s s3
= lim
= =0
s →∞ 
1  2  2 5  1
1 +  1 +  1 + + 2 
s 
s s 
 s 
2s 3 + 7s
0
f (∞) = lim sF(s) = lim
=
=0
2
s →0
s → 0 (s + 1)(s + 2)(s + 2s + 5)
10
Chapter 15, Solution 25.
(a)
(8)(s + 1)(s + 3)
s →∞ (s + 2)(s + 4)
f (0) = lim sF(s) = lim
s →∞
 1 3 
(8) 1 +  1 + 
s  s 
= lim 
=8
s →∞ 
2  4 
1 +  1 + 
s 
s

(8)(1)(3)
=3
s → 0 ( 2)( 4)
f (∞) = lim sF(s) = lim
s →0
(b)
6s (s − 1)
4
s →∞ s − 1
f (0) = lim sF(s) = lim
s →∞
1 1
6 2 − 4 
s
s  0
f (0) = lim
= =0
1
s →∞
1
1− 4
s
All poles are not in the left-half plane.
f (∞) does not exist
Chapter 15, Solution 26.
(a)
s 3 + 3s
=1
s →∞ s 3 + 4s 2 + 6
f (0) = lim sF(s) = lim
s →∞
Two poles are not in the left-half plane.
f (∞) does not exist
(b)
s 3 − 2s 2 + s
f (0) = lim sF(s) = lim
2
s →∞
s →∞ (s − 2)(s + 2s + 4)
2 1
+ 2
s
s
= lim
=1
s →∞ 
2  2 4 
1 −  1 + + 2 
 s  s s 
1−
One pole is not in the left-half plane.
f (∞) does not exist
Chapter 15, Solution 27.
(a)
f ( t ) = u(t ) + 2 e -t
(b)
G (s) =
3 (s + 4) − 11
11
= 3−
s+4
s+4
g( t ) = 3 δ(t ) − 11 e -4t
(c)
H(s) =
A = 2,
H(s) =
4
A
B
=
+
(s + 1)(s + 3) s + 1 s + 3
B = -2
2
2
−
s +1 s + 3
h ( t ) = 2 e -t − 2 e -3t
(d)
J (s) =
B=
12
A
B
C
=
+
2
2 +
(s + 2) (s + 4) s + 2 (s + 2)
s+4
12
= 6,
2
C=
12
=3
(-2) 2
12 = A (s + 2) (s + 4) + B (s + 4) + C (s + 2) 2
Equating coefficients :
s2 :
0= A+C 
→ A = -C = -3
s1 :
s0 :
0 = 6A + B + 4C = 2A + B 
→ B = -2A = 6
12 = 8A + 4B + 4C = -24 + 24 + 12 = 12
J (s) =
-3
6
3
+
2 +
s + 2 (s + 2)
s+4
j( t ) = 3 e -4t − 3 e -2t + 6 t e -2t
Chapter 15, Solution 28.
(a)
2(−2) 2(−4)
−2
4
+
F(s) = 2 + − 2 =
s+5 s+3 s+5
s+3
f ( t ) = (−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) = (A + B)s 2 + (2A + B + C)s + 5A + C
5A + C = 11; A = −B; − B + C = 3, B = C − 3 → A = 2; B = −2; C = 1
H(s) =
(
)
2
− 2s + 1
+
→ h ( t ) = 2e − t − 2e − t cos 2t + 1.5e − t sin 2t u ( t )
s + 1 s 2 + 2s + 5
Chapter 15, Solution 29.
V(s) =
2
As + B
; 2s 2 + 8s + 26 + As 2 + Bs = 2s + 26 → A = −2 and B = −6
+
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
2
v(t) = 2u ( t ) − 2e − 2 t cos 3t − e − 2 t sin 3t , t ≥ 0
3
Chapter 15, Solution 30.
(a) H1 (s) =
2(s + 2) + 2
(s + 2) 2 + 32
=
2(s + 2)
(s + 2) 2 + 3 2
h1 ( t ) = 2e −2 t cos 3t +
(b) H 2 (s) =
s2 + 4
(s + 1) 2 (s 2 + 2s + 5)
+
2
3
3 (s + 2) 2 + 32
2 −2 t
e
sin 3t
3
=
A
B
Cs + D
+
+
(s + 1) (s + 1) 2
(s 2 + 2s + 5)
s 2 + 4 = A(s + 1)(s 2 + 2s + 5) + B(s 2 + 2s + 5) + Cs(s + 1) 2 + D(s + 1) 2
or
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= A+C
s2 :
1 = 3A + B + 2C + D = A + B + D
s:

→
C = −A
0 = 7 A + 2B + C + 2D = 6A + 2B + 2D = 4A + 2
constant :
4 = 5A + 5B + D = 4A + 4B + 1

→

→
A = −1 / 2, C = 1 / 2
B = 5 / 4, D = 1 / 4
H 2 (s) =
1 −2
5
2s + 1  1  − 2
5
2(s + 1) − 1 
= 
+
+
+
+



4  (s + 1) (s + 1) 2
(s 2 + 2s + 5)  4  (s + 1) (s + 1) 2
(s + 1) 2 + 2 2 ) 
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
 A
B  1 −s  1
1 
+
= e−s 
+

= e 
(s + 1)(s + 3)
 (s + 1) (s + 3) 
 (s + 1) (s + 3)  2
(
)
1 −( t −1)
e
+ e −3( t −1) u ( t − 1)
2
Chapter 15, Solution 31.
(a)
F(s) =
10s
A
B
C
=
+
+
(s + 1)(s + 2)(s + 3) s + 1 s + 2 s + 3
A = F(s) (s + 1) s= -1 =
- 10
= -5
2
B = F(s) (s + 2) s= -2 =
- 20
= 20
-1
C = F(s) (s + 3) s= -3 =
- 30
= -15
2
F(s) =
-5
20
15
+
−
s +1 s + 2 s + 3
f ( t ) = - 5 e -t + 20 e -2t − 15 e -3t
(b)
F(s) =
2s 2 + 4s + 1
A
B
C
D
+
+
3 =
2 +
(s + 1)(s + 2)
s + 1 s + 2 (s + 2)
(s + 2) 3
A = F(s) (s + 1) s= -1 = -1
D = F(s) (s + 2) 3
s = -2
= -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= A+B 
→ B = -A = 1
s2 :
s1 :
s0 :
F(s) =
2 = 6A + 5B + C = A + C 
→ C = 2 − A = 3
4 = 12A + 8B + 3C + D = 4A + 3C + D
4 = 6+A+ D 
→ 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
f(t) = -e - t + e -2t + 3 t e -2t −
t 2 -2t
e
2

t2 
f ( t ) = - e -t + 1 + 3 t −  e - 2t
2

(c)
F(s) =
s +1
A
Bs + C
=
+ 2
2
(s + 2)(s + 2s + 5) s + 2 s + 2s + 5
A = F(s) (s + 2) s= -2 =
-1
5
s + 1 = A (s 2 + 2s + 5) + B (s 2 + 2s) + C (s + 2)
Equating coefficients :
1
5
s2 :
0= A+B 
→ B = -A =
s1 :
s0 :
1 = 2A + 2B + C = 0 + C 
→ C = 1
1 = 5A + 2C = -1 + 2 = 1
F(s) =
-1 5
1 5⋅ s +1
-1 5
1 5 (s + 1)
45
+
+
2
2 =
2
2 +
s + 2 (s + 1) + 2
s + 2 (s + 1) + 2
(s + 1) 2 + 2 2
f ( t ) = - 0.2 e -2t + 0.2 e -t cos( 2t ) + 0.4 e -t sin( 2t )
Chapter 15, Solution 32.
(a)
F(s) =
8 (s + 1)(s + 3) A
B
C
= +
+
s (s + 2)(s + 4) s s + 2 s + 4
A = F(s) s s= 0 =
(8)(3)
=3
(2)(4)
B = F(s) (s + 2) s= -2 =
(8)(-1)
=2
(-4)
C = F(s) (s + 4) s= -4 =
(8)(-1)(-3)
=3
(-4)(-2)
F(s) =
3
2
3
+
+
s s+2 s+4
f ( t ) = 3 u(t ) + 2 e -2t + 3 e -4t
(b)
F(s) =
s 2 − 2s + 4
A
B
C
+
+
2 =
(s + 1)(s + 2)
s + 1 s + 2 (s + 2) 2
s 2 − 2s + 4 = A (s 2 + 4s + 4) + B (s 2 + 3s + 2) + C (s + 1)
Equating coefficients :
s2 :
1= A+ B 
→ B = 1 − A
1
s :
- 2 = 4A + 3B + C = 3 + A + C
0
s :
4 = 4A + 2B + C = -B − 2 
→ B = -6
A = 1− B = 7
F(s) =
C = -5 - A = -12
7
6
12
−
−
s + 1 s + 2 (s + 2) 2
f ( t ) = 7 e -t − 6 (1 + 2 t ) e -2t
(c)
F(s) =
s2 +1
A
Bs + C
=
+ 2
2
(s + 3)(s + 4s + 5) s + 3 s + 4s + 5
s 2 + 1 = A (s 2 + 4s + 5) + B (s 2 + 3s) + C (s + 3)
Equating coefficients :
s2 :
1= A+ B 
→ B = 1 − A
s1 :
0
s :
0 = 4A + 3B + C = 3 + A + C 
→ A + C = -3
1 = 5A + 3C = -9 + 2A 
→ A = 5
B = 1 − A = -4
F(s) =
C = -A − 3 = -8
4 (s + 2)
5
4s + 8
5
−
=
−
2
s + 3 (s + 2) + 1 s + 3 (s + 2) 2 + 1
f ( t ) = 5 e -3t − 4 e -2t cos(t )
Chapter 15, Solution 33.
(a)
F(s) =
6 (s − 1)
6
As + B
C
= 2
= 2
+
4
s −1
(s + 1)(s + 1) s + 1 s + 1
6 = A (s 2 + s) + B (s + 1) + C (s 2 + 1)
Equating coefficients :
s2 :
0= A+C 
→ A = -C
s1 :
0= A+B 
→ B = -A = C
0
6 = B + C = 2B 
→ B = 3
s :
A = -3 ,
F(s) =
B = 3,
C=3
3
- 3s + 3
3
- 3s
3
=
+ 2
+ 2
+ 2
s +1 s +1 s +1 s +1 s +1
f ( t ) = 3 e -t + 3 sin( t ) − 3 cos(t )
(b)
F(s) =
s e - πs
s2 +1
f ( t ) = cos(t − π ) u(t − π )
(c)
F(s) =
8
A
B
C
D
+
+
3 =
2 +
s (s + 1)
s s + 1 (s + 1)
(s + 1) 3
A = 8,
D = -8
8 = A (s 3 + 3s 2 + 3s + 1) + B (s 3 + 2s 2 + s) + C (s 2 + s) + D s
Equating coefficients :
s3 :
0= A+B 
→ B = -A
s2 :
0 = 3A + 2B + C = A + C 
→ C = -A = B
s1 :
s0 :
0 = 3A + B + C + D = A + D 
→ D = -A
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 )
Chapter 15, Solution 34.
(a)
F(s) = 10 +
s2 + 4 − 3
3
= 11 − 2
2
s +4
s +4
f ( t ) = 11 δ(t ) − 1.5 sin( 2t )
(b)
G (s) =
e -s + 4 e -2s
(s + 2)(s + 4)
Let
1
A
B
=
+
(s + 2)(s + 4) s + 2 s + 4
A =1 2
G (s) =
B=1 2
 1
1 
e -s  1
1 

 + 2 e -2s 

+
+
s + 2 s + 4
2 s + 2 s + 4
g( t ) = 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
A
B
C
= +
+
s (s + 3)(s + 4) s s + 3 s + 4
Let
A = 1 12 ,
B = 2 3,
C = -3 4
1 1 23
3 4  -2s
e
H(s) =  ⋅ +
−
12 s s + 3 s + 4 
1 2

3
h ( t ) =  + e - 3(t - 2) − e -4(t - 2)  u(t − 2)
 12 3

4
Chapter 15, Solution 35.
(a)
G (s) =
Let
B = -1
A = 2,
G (s) =
s+3
A
B
=
+
(s + 1)(s + 2) s + 1 s + 2
2
1
−
s +1 s + 2

→ g( t ) = 2 e - t − e -2t
F(s) = e -6s G (s) 
→ f ( t ) = g( t − 6) u ( t − 6)
-(t - 6)
-2(t - 6)
] u( t − 6)
f (t) = [ 2 e
−e
(b)
Let
G (s) =
A = 1 3,
1
A
B
=
+
(s + 1)(s + 4) s + 1 s + 4
B = -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)
Let
G (s) =
s
A
Bs + C
=
+ 2
2
(s + 3)(s + 4) s + 3 s + 4
A = - 3 13
s = A (s 2 + 4) + B (s 2 + 3s) + C (s + 3)
Equating coefficients :
s2 :
0= A+B 
→ B = -A
1 = 3B + C
s1 :
0
0 = 4A + 3C
s :
A = - 3 13 ,
13 G (s) =
B = 3 13 ,
C = 4 13
- 3 3s + 4
+
s + 3 s2 + 4
13 g( t ) = -3 e -3t + 3 cos(2t ) + 2 sin(2t )
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
A B
C
D
= + 2+
+
s (s + 2)(s + 3) s s
s+2 s+3
B = 1 6,
2
C =1 4,
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 = A+C+D
s3 :
2
s :
0 = 5A + B + 3C + 2D = 3A + B + C
1
s :
0 = 6 A + 5B
s0 :
1 = 6B 
→ B = 1 6
A = - 5 6 B = - 5 36
(b)
X(s) =
- 5 36 1 6 1 4
19
+ 2 +
−
s
s
s+2 s+3
x(t) =
-5
1
1
1
u(t ) + t + e - 2t − e - 3t
36
6
4
9
Y(s) =
1
A
B
C
+
+
2 =
s (s + 1)
s s + 1 (s + 1) 2
A = 1,
C = -1
1 = A (s 2 + 2s + 1) + B (s 2 + s) + C s
Equating coefficients :
0= A+B 
→ B = -A
s2 :
s1 :
s0 :
0 = 2A + B + C = A + C 
→ C = -A
1 = A, B = -1, C = -1
1
1
1
Y(s) = −
−
s s + 1 (s + 1) 2
y( t ) = u(t ) − e -t − t e -t
(c)
Z(s) =
A
B
Cs + D
+
+ 2
s s + 1 s + 6s + 10
A = 1 10 ,
B = -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+ B+C
s3 :
2
s :
0 = 7 A + 6 B + C + D = 6 A + 5B + D
1
s :
0 = 16A + 10B + D = 10A + 5B 
→ B = -2A
s0 :
1 = 10A 
→ A = 1 10
A = 1 10 ,
B = -2A = - 1 5 ,
C = A = 1 10 ,
D = 4A =
4
10
1
2
s+4
10 Z(s) = −
+ 2
s s + 1 s + 6s + 10
1
2
s+3
1
10 Z(s) = −
+
+
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
P(s) =
12
A Bs + C
= + 2
2
s (s + 4) s s + 4
A = P(s) s s=0 = 12 4 = 3
12 = A (s 2 + 4) + B s 2 + C s
Equating coefficients :
s0 :
12 = 4A 
→ A = 3
1
s :
0=C
2
s :
0= A+B 
→ B = -A = -3
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)
(b)
Let
G (s) =
2s + 1
As + B Cs + D
= 2
+
2
(s + 1)(s + 9) s + 1 s 2 + 9
2
2s + 1 = A (s 3 + 9s) + B (s 2 + 9) + C (s 3 + s) + D (s 2 + 1)
Equating coefficients :
0= A+C 
→ C = -A
s3 :
s2 :
0 = B+ D 
→ D = -B
s1 :
2 = 9A + C = 8A 
→ A = 2 8, C = - 2 8
0
s :
1 = 9B + D = 8B 
→ B = 1 8 , D = - 1 8
1  2s + 1  1  2s + 1 
− 

G (s) =  2
8  s + 1 8  s 2 + 9 
(c)
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
Let
9s2
36s + 117
H(s) = 2
= 9− 2
s + 4s + 13
s + 4s + 13
H(s) = 9 − 36 ⋅
s+2
3
2
2 − 15 ⋅
(s + 2) + 3
(s + 2) 2 + 3 2
h ( t ) = 9 δ(t ) − 36e -2 t cos( 3t ) − 15e -2 t sin( 3t )
Chapter 15, Solution 38.
(a)
F(s) =
s 2 + 4s
s 2 + 10s + 26 − 6s − 26
=
s 2 + 10s + 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
2
f ( t ) = δ(t ) − 6 e -t cos(5t ) + 4 e -t sin( 5t )
(b)
5s 2 + 7s + 29
A
Bs + C
= + 2
F(s) =
2
s (s + 4s + 29) s s + 4s + 29
5s 2 + 7s + 29 = A (s 2 + 4s + 29) + B s 2 + C s
Equating coefficients :
s0 :
29 = 29A 
→ A = 1
s1 :
7 = 4A + C 
→ C = 7 − 4A = 3
s2 :
5= A+B 
→ B = 5 − A = 4
B = 4,
A = 1,
C=3
4 (s + 2)
1
4s + 3
1
5
F(s) = + 2
= +
2
2 −
s s + 4s + 29 s (s + 2) + 5
(s + 2) 2 + 5 2
f ( t ) = u(t ) + 4 e -2t cos(5t ) − e -2t sin( 5t )
Chapter 15, Solution 39.
(a)
2s 3 + 4s 2 + 1
As + B
Cs + D
F(s) = 2
= 2
+ 2
2
(s + 2s + 17)(s + 4s + 20) s + 2s + 17 s + 4s + 20
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= A+C
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)
F(s) =
s2 + 4
As + B
Cs + D
= 2
+ 2
2
2
(s + 9)(s + 6s + 3) s + 9 s + 6s + 3
s 2 + 4 = A (s 3 + 6s 2 + 3s) + B (s 2 + 6s + 3) + C (s 3 + 9s) + D (s 2 + 9)
Equating coefficients :
s3 :
0= A+C 
→ C = -A
2
s :
1 = 6A + B + D
1
s :
0 = 3A + 6B + 9C = 6B + 6C 
→ B = -C = A
4 = 3B + 9D
s0 :
Solving these equations,
A = 1 12 ,
B = 1 12 ,
12 F(s) =
D = 5 12
s +1
-s+5
+ 2
2
s + 9 s + 6s + 3
s 2 + 6s + 3 = 0 
→
Let
G (s) =
E=
-s+5
s + 5.449
F=
-s+5
s + 0.551
G (s) =
C = - 1 12 ,
- 6 ± 36 - 12
= -0.551, - 5.449
2
-s+5
E
F
=
+
s + 6s + 3 s + 0.551 s + 5.449
2
s = -0.551
= 1.133
s = -5.449
= - 2.133
1.133
2.133
−
s + 0.551 s + 5.449
12 F(s) =
s
1
3
1.133
2.133
⋅ 2
−
2 +
2 +
s +3
3 s +3
s + 0.551 s + 5.449
2
f ( t ) = 0.08333 cos( 3t ) + 0.02778 sin( 3t ) + 0.0944 e -0.551t − 0.1778 e -5.449t
Chapter 15, Solution 40.
 4s 2 + 7s + 13 
A
Bs + C
Let H(s) = 
+
=
2
2
 (s + 2)(s + 2s + 5)  s + 2 s + 2s + 5
4s 2 + 7s + 13 = A(s 2 + 2s + 5) + B(s 2 + 2s) + C(s + 2)
Equating coefficients gives:
s2 :
4=A+B
s:
7 = 2A + 2B + C

→
C = −1
13 = 5A + 2C

→
5A = 15 or A = 3, B = 1
constant :
H(s) =
3
s −1
3
(s + 1) − 2
+
=
+
2
s + 2 s + 2s + 5 s + 2 (s + 1) 2 + 2 2
Hence,
h ( t ) = 3e −2 t + e − t cos 2t − e − t sin 2t = 3e −2 t + e − t (A cos α cos 2t − A sin α sin 2t )
where A cos α = 1,
A sin α = 1
Thus,
h(t) =
[ 2e
−t

→
A = 2,
α = 45 o
]
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 − λ)
h(λ )
0
t
t
y( t ) = ∫ (1)4λdλ = 2λ2 0 = 2t 2
0
t 1
2
λ
For 1 <t<3,
f ( t − λ)
h (λ )
0
1
t
0
1
1 t
2
t
t
y( t ) = ∫ (1)4λdλ + ∫ (1)(8 − 4λ )dλ = 2λ2 0 + (8λ − 2λ2 ) 1 = 8t − 2t 2 − 4
For 3 < t < 4
h (λ)
0
2
2
1
t-2
y( t ) = ∫ (8 − 4λ )λdλ = 8λ − 2λ2 t − 2 = 32 − 16t + 2t 2
t −2
Thus,

2t 2 , 0 < t < 1

 8t - 2t 2 − 4, 1 < t < 3
y( t ) = 
2
32 - 16t + 2t , 3 < t < 4

0, otherwise

2
f ( t − λ)
3
t
4
λ
Chapter 15, Solution 42.
(a)
For 0 < t < 1 , f1 ( t − λ) and f 2 (λ) overlap from 0 to t, as shown in Fig. (a).
y( t ) = f1 ( t ) ∗ f 2 ( t ) = ∫0
t
f1(t - λ)
λ2
(1)(λ) dλ =
2
f2(λ)
1
t-1
0 t
1
t2
=
2
t
0
1
0
λ
(a)
t-1 1
t
λ
(b)
For 1 < t < 2 , f1 ( t − λ) and f 2 (λ) overlap as shown in Fig. (b).
y( t ) = ∫t −1 (1)(λ) dλ =
1
λ2
2
1
t −1
=t−
t2
2
For t > 2 , there is no overlap.
Therefore,
 t 2 2,
0<t <1

2
y( t ) = t − t 2, 1 < t < 2
 0,
otherwise

(b)
For 0 < t < 1 , the two functions overlap as shown in Fig. (c).
y( t ) = f1 ( t ) ∗ f 2 ( t ) = ∫0 (1)(1) dλ = t
t
f1(t - λ)
f2(λ)
1
t-1
0 t
(c)
1
λ
1
0
t-1 1
(d)
t
λ
For 1 < t < 2 , the functions overlap as shown in Fig. (d).
y( t ) = ∫t −1 (1)(1) dλ = λ 1t −1 = 2 − t
1
For t > 2 , there is no overlap.
Therefore,
0<t<1
 t,

y( t ) =  2 − t , 1 < t < 2
 0,
otherwise

(c)
For t < -1 , there is no overlap. For - 1 < t < 0 , f1 ( t − λ) and f 2 (λ) overlap
as shown in Fig. (e).

 λ2
y( t ) = f 1 ( t ) ∗ f 2 ( t ) = ∫-1 (1)(λ + 1) dλ =  + λ -t1

2
t
y( t ) =
1 2
1
( t + 2t + 1) = ( t + 1) 2
2
2
f2(t - λ)
f1(λ)
1
-1
t
0
1
λ
1
-1
(e)
0 t
(f)
For 0 < t < 1 , the functions overlap as shown in Fig. (f).
y( t ) = ∫-1 (1)(λ + 1) dλ + ∫0 (1)(1 − λ) dλ
0
t

 λ2

λ2 
y( t ) =  + λ  0-1 +  λ −  0t
2

2

y( t ) =
1
(1 + 2t − t 2 )
2
For t > 1 , the two functions overlap.
1
λ
y( t ) = ∫-1 (1)(λ + 1) dλ + ∫0(1)(1 − λ) dλ
0
1
1 
1
λ2  1 1

y( t ) = + λ −  0 = + 1 − = 1
2 
2
2
2
Therefore,
0,
t < -1

 0.5(t 2 + 2t + 1), - 1 < t < 0

y( t ) = 
2
0.5(-t + 2t + 1), 0 < t < 1

1,
t >1
Chapter 15, Solution 43.
(a)
For 0 < t < 1 , x ( t − λ) and h (λ) overlap as shown in Fig. (a).
y( t ) = x ( t ) ∗ h ( t ) = ∫0 (1)(λ) dλ =
t
x(t - λ)
λ2
2
1
t
0
=
t2
2
1
h(λ)
t-1
0 t
1
λ
0
t-1 1
(a)
t
λ
(b)
For 1 < t < 2 , x ( t − λ) and h (λ) overlap as shown in Fig. (b).
y( t ) = ∫t −1 (1)(λ) dλ + ∫1 (1)(1) dλ =
1
t
λ2
2
1
t −1
For t > 2 , there is a complete overlap so that
y( t ) = ∫t −1 (1)(1) dλ = λ tt −1 = t − ( t − 1) = 1
t
+ λ 1t =
-1 2
t + 2t − 1
2
Therefore,

t 2 2,
0<t<1
 2
- (t 2) + 2t − 1, 1 < t < 2
y( t ) = 
1,
t>2


0,
otherwise
(b)
For t > 0 , the two functions overlap as shown in Fig. (c).
y( t ) = x ( t ) ∗ h ( t ) = ∫0 (1) 2 e -λ dλ = -2 e -λ
t
x(t-λ)
2
t
0
h(λ) = 2e-λ
1
0
t
λ
(c)
Therefore,
y( t ) = 2 (1 − e -t ), t > 0
(c)
For - 1 < t < 0 , x ( t − λ) and h (λ) overlap as shown in Fig. (d).
y( t ) = x ( t ) ∗ h ( t ) = ∫0 (1)(λ) dλ =
t +1
x(t - λ)
λ2
2
1
t-1 -1
t 0
t +1
0
=
1
( t + 1) 2
2
h(λ)
t+1 1
2
λ
(d)
For 0 < t < 1 , x ( t − λ) and h (λ) overlap as shown in Fig. (e).
y( t ) = ∫0 (1)(λ) dλ + ∫1 (1)(2 − λ) dλ
1
t +1
y( t ) =
λ2
2
1
0

-1
1
λ2 
+  2λ −  1t +1 = t 2 + t +
2
2
2

1
-1 t-1
0 t
1 t+1 2
λ
(e)
For 1 < t < 2 , x ( t − λ) and h (λ) overlap as shown in Fig. (f).
y( t ) = ∫t −1 (1)(λ) dλ + ∫1 (1)(2 − λ) dλ
1
y( t ) =
λ2
2
2
1
t −1

-1
1
λ2 
+  2λ −  12 = t 2 + t +
2
2
2

1
0
t-1 1
t
2 t+1
λ
(f)
For 2 < t < 3 , x ( t − λ) and h (λ) overlap as shown in Fig. (g).

λ2 

y( t ) = ∫t −1 (1)(2 − λ) dλ = 2λ − 
2

2
2
t −1
=
9
1
− 3t + t 2
2
2
1
0
1 t-1 2
(g)
t
t+1
λ
Therefore,
 ( t 2 2 ) + t + 1 2, - 1 < t < 0
 2
- ( t 2 ) + t + 1 2 , 0 < t < 2
y( t ) =  2
 ( t 2 ) − 3t + 9 2, 2 < t < 3

0,
otherwise
Chapter 15, Solution 44.
(a)
For 0 < t < 1 , x ( t − λ) and h (λ) overlap as shown in Fig. (a).
y( t ) = x ( t ) ∗ h ( t ) = ∫0 (1)(1) dλ = t
t
x(t - λ)
h(λ)
1
t-1
0 t
1
2
λ
-1
(a)
For 1 < t < 2 , x ( t − λ) and h (λ) overlap as shown in Fig. (b).
y( t ) = ∫t −1 (1)(1) dλ + ∫1 (-1)(1) dλ = λ 1t −1 − λ 1t = 3 − 2 t
1
t
For 2 < t < 3 , x ( t − λ) and h (λ) overlap as shown in Fig. (c).
y( t ) = ∫t −1 (1)(-1) dλ = -λ
2
2
t −1
= t−3
1
1
0
t-1 1
t
2
0
λ
-1
1 t-1
2 t
λ
-1
(b)
(c)
Therefore,
0<t <1
 t,
 3 − 2t , 1 < t < 2

y( t ) = 
2<t<3
 t − 3,
 0,
otherwise
(b)
For t < 2 , there is no overlap. For 2 < t < 3 , f1 ( t − λ) and f 2 (λ) overlap,
as shown in Fig. (d).
y( t ) = f 1 ( t ) ∗ f 2 ( t ) =

λ2

=  λt −
2

∫
t
2
(1)( t − λ) dλ
 t t2
 2 = − 2 t + 2
2

f1(t - λ)
f2(λ)
1
0
1 t-1 2
t
3
4
5
λ
4
5
λ
(d)
1
0
1
2 t-1 3
(e)
t
For 3 < t < 5 , f1 ( t − λ) and f 2 (λ) overlap as shown in Fig. (e).

1
λ2  t

y( t ) = ∫t −1 (1)( t − λ) dλ = λt −  t −1 =
2
2

t
For 5 < t < 6 , the functions overlap as shown in Fig. (f).

5
-1
λ2 
y( t ) = ∫t −1 (1)( t − λ) dλ =  λt −  5t −1 = t 2 + 5t − 12
2
2

1
0
1
2
3
4 t-1 5
(f)
Therefore,
 ( t 2 2 ) − 2t + 2,
2<t<3

1 2,
3<t<5

y( t ) =  2
- (t 2) + 5t − 12, 5 < t < 6

0,
otherwise
Chapter 15, Solution 45.
(a)
f ( t ) ∗ δ( t ) = ∫0 f (λ) δ( t − λ) dλ = f (λ) λ= t
t
f (t ) ∗ δ(t ) = f (t )
(b)
f ( t ) ∗ u ( t ) = ∫0 f (λ) u ( t − λ) dλ
t
1 λ< t
Since u ( t − λ) = 
0 λ> t
f ( t ) ∗ u( t ) =
∫
t
o
f ( λ ) dλ
t
λ
Alternatively,
L{ f ( t ) ∗ u ( t )} =
F(s)
s
t
 F(s) 
 = f ( t ) ∗ u ( t ) = ∫ f (λ) dλ
L−1 
o
 s 
Chapter 15, Solution 46.
(a)
Let y( t ) = x 1 ( t ) ∗ x 2 ( t ) = ∫0 x 2 ( t ) x 1 ( t − λ) dλ
t
For 0 < t < 3 , x 1 ( t − λ) and x 2 (λ) overlap as shown in Fig. (a).
y( t ) = ∫0 4 e -2λ e -(t -λ ) dλ = 4 e - t ∫0 e -λ dλ = 4 (e - t − e -2 t )
t
t
4
x2(λ)
x1(t-λ)
0
λ
3
t
(a)
For t > 3 , the two functions overlap as shown in Fig. (b).
y( t ) = ∫0 4 e -2 λ e -(t -λ ) dλ = 4 e - t ( - e -λ )
3
3
0
= 4 e - t (1 − e -3 )
4
3
0
(b)
t
λ
Therefore,
 4 (e - t − e -2t ), 0 < t < 3
y( t ) =  -t
-3
t>3
4e (1 − e ),
(b)
For 1 < t < 2 , x 1 (λ) and x 2 ( t − λ) overlap as shown in Fig. (c).
y( t ) = x 1 ( t ) ∗ x 2 ( t ) = ∫1 (1)(1) dλ = λ 1t = t − 1
t
x2(t - λ)
x1(λ)
1
0
t-1 1
t
2
λ
3
(c)
For 2 < t < 3 , the two functions overlap completely.
y( t ) = ∫t −1 (1)(1) dλ = λ tt −1 = t − ( t − 1) = 1
t
For 3 < t < 4 , the two functions overlap as shown in Fig. (d).
y( t ) = ∫t −1 (1)(1) dλ = λ 3t −1 = 4 − t
3
1
0
1
2 t-1 3
t
λ
(d)
Therefore,
 t − 1, 1 < t < 2
 1,
2<t<3

y( t ) = 
4 − t , 3 < t < 4
 0,
otherwise
(c)
For - 1 < t < 0 , x 1 ( t − λ) and x 2 (λ) overlap as shown in Fig. (e).
y( t ) = x1 ( t ) ∗ x 2 ( t ) = ∫-1 (1) 4 e -(t-λ) dλ
t
y( t ) = 4 e -t ∫-1 e λ dλ = 4 [ 1 − e -(t+1) ]
t
x1(t - λ)
1
-1
t
-1
x2(λ)
1
λ
(e)
For 0 < t < 1 ,
y( t ) = ∫-1 (1) 4 e -(t -λ) dλ + ∫0 (-1) 4 e -(t -λ) dλ
0
y( t ) = 4 e -t e λ
t
− 4 e -t e λ
0
-1
t
0
= 8 e -t − 4 e -(t+1) − 4
For t > 1 , the two functions overlap completely.
y( t ) = ∫-1 (1) 4 e -(t-λ ) dλ + ∫0 (-1) 4 e -(t-λ) dλ
0
y( t ) = 4 e -t e λ
Therefore,
1
0
-1
− 4 e -t e λ 10 = 8 e -t − 4 e -(t+1) − 4 e -(t−1)
[
]

4 1 − e -(t +1) ,
-1 < t < 0

-t
-(t +1)
− 4,
y( t ) =  8 e − 4 e
0<t<1
t
(t
+
1)
(t
−
1)
8 e − 4 e
− 4e
,
t >1

Chapter 15, Solution 47.
f1 ( t ) = f 2 ( t ) = cos( t )
L -1 [ F1 (s) F2 (s)] = ∫0 cos(λ) cos( t − λ) dλ
t
cos(A) cos(B) =
1
[ cos(A + B) + cos(A − B)]
2
L -1 [ F1 (s) F2 (s)] =
1 t
[cos( t ) + cos( t − 2λ )] dλ
2 ∫0
L -1 [ F1 (s) F2 (s)] =
1
1 sin( t − 2λ)
cos(t ) ⋅ λ 0t + ⋅
2
2
-2
t
0
L-1 [ F1 (s) F2 (s)] = 0.5 t cos(t ) + 0.5 sin( t )
Chapter 15, Solution 48.
(a)
Let G (s) =
2
2
=
s + 2s + 5 (s + 1) 2 + 2 2
2
g( t ) = e -t sin(2 t )
F(s) = G (s) G (s)
f ( t ) = L -1 [ G (s) G (s)] = ∫0 g (λ) g ( t − λ) dλ
t
f ( t ) = ∫0 e -λ sin( 2λ) e -( t −λ ) sin( 2( t − λ)) dλ
t
sin(A) sin(B) =
f (t) =
1
[ cos(A − B) − cos(A + B)]
2
1 - t t -λ
e ∫ e [ cos(2t ) − cos(2( t − 2λ))] dλ
2 0
t
e -t
e -t
-2 λ
f (t) =
cos(2 t ) ∫0 e dλ −
2
2
f (t) =
e -t
e -2λ
cos(2 t ) ⋅
2
-2
t
0
−
e -t
2
∫
∫
t
0
t
0
e -2λ cos(2 t − 4λ) dλ
e -2λ [ cos(2 t ) cos(4λ) + sin( 2 t ) sin( 4λ)] dλ
t
1 -t
e -t
-2 t
f ( t ) = e cos(2 t ) (-e + 1) − cos(2 t ) ∫0 e -2 λ cos(4λ) dλ
4
2
−
t
e -t
sin( 2 t ) ∫0 e -2λ sin( 4λ) dλ
2
f (t) =
1 -t
e cos(2t ) (1 − e -2 t )
4
 e -2λ

e -t
(- 2cos(4λ) − 4 sin(4λ)) 0t
− cos(2t ) 
2
 4 + 16

−
f (t) =
(b)
Let
 e -2λ

e -t
(- 2sin(4λ) + 4 cos(4λ)) 0t
sin(2t ) 
2
 4 + 16

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) =
2
,
s +1
Y(s) =
x ( t ) = 2 e -t u ( t ) ,
s
s+4
y( t ) = cos(2t ) u ( t )
F(s) = X(s) Y(s)
f ( t ) = L -1 [ X(s) Y(s)] = ∫0 y(λ) x ( t − λ) dλ
∞
f ( t ) = ∫0 cos(2λ) ⋅ 2 e -(t−λ ) dλ
t
f ( t ) = 2 e -t ⋅
eλ
(cos(2λ) + 2 sin(2λ)) 0t
1+ 4
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
Chapter 15, Solution 49.
Let x(t) = u(t) – u(t-1) and y(t) = h(t)*x(t).
 4(1 − e − s ) 
 4 1 e −s 
y( t ) = L−1 [H(s)X(s)] = L−1 
( −
) = L−1 

s 
 s(s + 2) 
 s + 2 s
But
1
A
B
1 1
1 
= +
=  −
s(s + 2) s s + 2 2  s s + 2 
1
1
e−s e−s 
Y(s) = 2 −
−
+

s
s + 2 
 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 ′(0)] + 2 [ s V(s) − v(0)] + 10 V(s) =
s 2 V(s) − s + 2 + 2s V(s) − 2 + 10 V(s) =
3s
s +4
2
3s
s +4
2
3s
s 3 + 7s
(s + 2s + 10) V(s) = s + 2
=
s + 4 s2 + 4
2
V(s) =
s 3 + 7s
As + B
Cs + D
= 2
+ 2
2
2
(s + 4)(s + 2s + 10) s + 4 s + 2s + 10
s 3 + 7s = A (s 3 + 2s 2 + 10s) + B (s 2 + 2s + 10) + C (s 3 + 4s) + D (s 2 + 4)
Equating coefficients :
1= A+C 
→ C = 1 − A
s3 :
2
s :
0 = 2A + B + D
1
7 = 10A + 2B + 4C = 6A + 2B + 4
s :
0 = 10B + 4D 
→ D = -2.5 B
s0 :
Solving these equations yields
9
12
A=
,
B=
,
26
26
C=
17
,
26
D=
- 30
26
V(s) =
1  9s + 12
17s − 30 
+
2
2
26  s + 4 s + 2s + 10 
V(s) =

s +1
47
2
1  9s
+ 6⋅ 2
+ 17 ⋅

2
2
2 −
2
2 
(s + 1) + 3
(s + 1) + 3 
s +4
26  s + 4
v( t ) =
47
17
6
9
sin( 2t ) + e -t cos( 3t ) − e -t sin( 3t )
cos( 2t ) +
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) = s10+ 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 = 5,

→
V(s) =
B = 0,
Hence,
v( t ) = (5e − t − 3e −3t )u ( t )
2s 2 + 16s + 24
(s + 1)(s + 2)(s + 3)
C = −3
Chapter 15, Solution 52.
Take the Laplace transform of each term.
[s
2
I(s) − s i(0) − i ′(0)] + 3 [ s I(s) − i(0)] + 2 I(s) + 1 = 0
(s 2 + 3s + 2) I(s) − s − 3 − 3 + 1 = 0
I(s) =
s+5
A
B
=
+
(s + 1)(s + 2) s + 1 s + 2
A = 4,
I(s) =
B = -3
4
3
−
s +1 s + 2
i( t ) = (4 e -t − 3 e -2t ) u(t )
Chapter 15, Solution 53.
Take the Laplace transform of each term.
[s
2
Y(s) − s y(0) − y ′(0)] + 5 [ s Y(s) − y(0)] + 6 V(s) =
(s 2 + 5s + 6) Y(s) − s − 4 − 5 =
(s 2 + 5s + 6) Y(s) = s + 9 +
Y(s) =
s
s +4
2
s
s +4
2
s
s + (s + 9)(s 2 + 4)
=
s2 + 4
s2 + 4
s 3 + 9s 2 + 5s + 36
A
B
Cs + D
=
+
+ 2
2
(s + 2)(s + 3)(s + 4) s + 2 s + 3 s + 4
A = (s + 2) Y(s) s= -2 =
27
,
4
B = (s + 3) Y(s) s= -3 =
- 75
13
When s = 0 ,
36
A B D
= + +
(2)(3)(4) 2 3 4

→ D =
5
26
When s = 1 ,
46 + 5 A B C D
= + + +
(12)(5) 3 4 5 5
Thus, Y(s) =

→ C =
1
52
27 4 75 13 1 52 ⋅ s + 5 26
−
+
s+2 s+3
s2 + 4
y( t ) =
27 - 2t 75 - 3t 1
5
e − e + cos( 2t ) + sin( 2t )
4
13
52
52
Chapter 15, Solution 54.
Taking the Laplace transform of the differential equation gives
[ s 2 V(s) − s v(0) − v′(0)] + 3[ s V(s) − v(0)] + 2 V(s) =
(s 2 + 3s + 2) V(s) =
5
2−s
−1 =
s+3
s+3
V(s) =
2−s
2−s
=
2
(s + 3)(s + 3s + 2) (s + 1)(s + 2)(s + 3)
V(s) =
A
B
C
+
+
s +1 s + 2 s + 3
A = 3 2,
V(s) =
B = -4 ,
C=5 2
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
s+3
Chapter 15, Solution 55.
Take the Laplace transform of each term.
[s
3
Y(s) − s 2 y(0) − s y′(0) − y′′(0)] + 6 [ s 2 Y(s) − s y(0) − y′(0)]
+ 8 [ s Y(s) − y(0)] =
s +1
(s + 1) 2 + 2 2
Setting the initial conditions to zero gives
(s 3 + 6 s 2 + 8s) Y(s) =
Y(s) =
A=
s +1
s + 2s + 5
2
(s + 1)
A
B
C
Ds + E
= +
+
+ 2
2
s (s + 2)(s + 4)(s + 2s + 5) s s + 2 s + 4 s + 2s + 5
1
,
40
B=
1
,
20
C=
-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

12 
 4 s + s  V(s) = 8
V(s) =
8s
2s
= 2
4s + 12 s + 3
2
v( t ) = 2 cos
(
3t
)
Chapter 15, Solution 57.
Take the Laplace transform of each term.
[ s Y(s) − y(0)] + 9 Y(s) =
s
s
s +4
2
s2 + 9 
s
s2 + s + 4

 Y(s) = 1 + 2
= 2
s +4
s +4
 s 
s 3 + s 2 + 4s
As + B Cs + D
Y(s) = 2
=
+
(s + 4)(s 2 + 9) s 2 + 4 s 2 + 9
s 3 + s 2 + 4s = 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= A+C
Solving these equations gives
A = 0,
Y(s) =
B = - 4 5,
C = 1,
D=9 5
-4 5 s+9 5 -4 5
95
s
= 2
+ 2
+ 2
+ 2
2
s +4 s +9 s +4 s +9 s +9
y( t ) = - 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) = e − 2s
s
V(s) =
(s + 3)e −2s − 3e −2s
(s + 3) 2 + 1

→
V(s) =
se −2s
s 2 + 6s + 10
Hence,
v(t ) = e −3( t − 2) cos(t − 2) − 3e −3( t − 2) sin(t − 2)  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
6
s+2
 6

(s 2 + 4s + 3) Y(s) = s 
− 1
s + 2 
Y(s) =
s ( 4 − s)
( 4 − s) s
=
2
(s + 2)(s + 4s + 3) (s + 1)(s + 2)(s + 3)
Y(s) =
A
B
C
+
+
s +1 s + 2 s + 3
A = 2 .5 ,
Y(s) =
B = 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
4
2 [ s X(s) − x (0)] + 5 X(s) + X(s) + = 2
s
s s + 16
(2s 2 + 5s + 3) X(s) = 2s − 4 +
4s
2s 3 − 4s 2 + 36s − 64
=
s 2 + 16
s 2 + 16
2s 3 − 4s 2 + 36s − 64
s 3 − 2s 2 + 18s − 32
X(s) =
=
(2s 2 + 5s + 3)(s 2 + 16) (s + 1)(s + 1.5)(s 2 + 16)
X(s) =
A
B
Cs + D
+
+ 2
s + 1 s + 1.5 s + 16
A = (s + 1) X(s) s= -1 = -6.235
B = (s + 1.5) X(s) s = -1.5 = 7.329
When s = 0 ,
B D
- 32
= A+
+
1.5 16
(1.5)(16)

→ 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+ B+C 
→ C = -0.0935
X(s) =
- 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 ) =
1s
1
1
= 2
=
1 + s + 1 s s + s + 1 (s + 1 2) 2 + ( 3 2) 2
 3 
e - t 2 sin 
t 
2
3


2
i( t ) = 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 = 0
8
s
2
4+
s
Vx −
Vx (4s + 8) −
(16s + 32)
+ (2s 2 + 4s)Vx + s 2 Vx = 0
s
Vx (3s 2 + 8s + 8) =
16s + 32
s




s+2
0.25
− 0.125 
− 0.125

Vx = −16
+
= −16
+
 s
8
4
8
4
s(3s 2 + 8s + 8)
s+ − j
s+ + j


3 
3
3
3

v x = (−4 + 2e − (1.3333 + j0.9428) t + 2e − (1.3333 − j0.9428) t )u ( t ) V
2 2
vx = 4u ( t ) − e − 4 t / 3 cos
 3
 6 − 4t / 3  2 2
t  −
e
sin 
2

 3
Chapter 16, Solution 3.
s
+
1/2
5/s
1/8
Vo
−
Current division leads to:
1




5
5
1 5
2
=
=
Vo =
8s1 1
 10 + 16s 16(s + 0.625)
 + +s
2 8

(
)
vo(t) = 0.3125 1 − e −0.625t u ( t ) V

t  V

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 
 1 
10



= 2
s  s + 1  s + 6s + 10  s + 1 
Vo (s) =
10 s
s + 6 + 10
Vo (s) =
10
A
Bs + C
=
+ 2
(s + 1)(s + 6s + 10) s + 1 s + 6s + 10
2
10 = A (s 2 + 6s + 10) + B (s 2 + s) + C (s + 1)
Equating coefficients :
s2 :
0= A+B 
→ B = -A
s1 :
0
s :
0 = 6A + B + C = 5A + C 
→ C = -5A
10 = 10A + C = 5A 
→ A = 2, B = -2, C = -10
Vo (s) =
2 (s + 3)
2
2s + 10
2
4
=
−
− 2
2
2 −
s + 1 s + 6s + 10 s + 1 (s + 3) + 1
(s + 3) 2 + 12
v o ( t ) = 2 e -t − 2 e -3t cos(t ) − 4 e -3t sin( t ) V
Chapter 16, Solution 5.
Io
1
s+2
s
2
2
s





1 
1
2s
2s
= 1 
 =
V=

s + 2  1 1 s  s + 2  s 2 + s + 2  (s + 2)(s + 0.5 + j1.3229)(s + 0.5 − j1.3229)
 + + 
s 2 2
Io =
Vs
s2
=
2
(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)
+
=
+
s+2
s + 0.5 + j1.3229
s + 0.5 − j1.3229
(
)
i o ( t ) = e − 2 t + 0.3779e − 90° e − t / 2 e − j1.3229 t + 0.3779e 90° 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
5
s+2
10/s
s
Use current division.
Io =
5
5s
5(s + 1)
5
s+2
=
=
−
2
2
2
10 s + 2 s + 2s + 10 (s + 1) + 3
(s + 1) 2 + 3 2
s+2+
s
5
i o ( t ) = 5e − t cos 3t − e − t sin 3t
3
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
2s
2s 2 + 2s + 1
Z = 1 + // 2s = 1 + s
= 1+
=
1
s
1 + 2s 2
1 + 2s 2
+ 2s
s
V
2
1 + 2s 2
2s 2 + 1
A
Bs + C
=
=
+
Ix = =
x
2
2
2
Z s + 1 2s + 2s + 1 (s + 1)(s + s + 0.5) (s + 1) (s + s + 0.5)
2s 2 + 1 = A(s 2 + s + 0.5) + B(s 2 + s) + C(s + 1)
s2 :
2=A+B
0 = A+B+C = 2+C
s:
constant :
Ix =

→
1 = 0.5A + C or 0.5A = 3
C = −2

→
A = 6, B = -4
6
4s + 2
6
4(s + 0.5)
−
=
−
2
s + 1 (s + 0.5) + 0.75 s + 1 (s + 0.5) 2 + 0.866 2
[
]
i x ( t ) = 6 − 4e − 0.5t cos 0.866 t u ( t ) A
Chapter 16, Solution 8.
1
1 (1 + 2s) s 2 + 1.5s + 1
=
(a) Z = + 1 //(1 + 2s) = +
s
s 2 + 2s
s(s + 1)
1 1 1
1
3s 2 + 3s + 2
= + +
=
(b)
1
Z 2 s
2s(s + 1)
1+
s
Z=
2s(s + 1)
3s 2 + 3s + 2
Chapter 16, Solution 9.
The s-domain form of the circuit is shown in Fig. (a).
(a)
2 (s + 1 s)
2 (s 2 + 1)
Z in = 2 || (s + 1 s) =
=
2 + s + 1 s 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) 2 (s + 2)
=
3+ 2 s
3s + 2
1 + 2 || (1 + 2 s) =
5s + 6
3s + 2
 5s + 6 

s ⋅
 3s + 2 
 5s + 6 
s (5s + 6)
=
Z in = s || 
= 2
 3s + 2 
 5s + 6  3s + 7s + 6

s +
 3s + 2 
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
Vx
=
2 + 1/ s 2 + 1/ s
ZTh =

→
Vx = −
(2s + 1)
3s
Vx
(2s + 1)
=−
1
3s
To find VTh, consider the circuit below.
1/s
Vy
+
2
s +1
2
Vo
-
Applying KCL gives
2
V
+ 2Vo = o
s +1
2

→
Vo = −
4
3(s + 1)
2Vo
1
But − Vy + 2Vo • + Vo = 0
s
2
4  s + 2  − 4(s + 2)
VTh = Vy = Vo (1 + ) = −

=
s
3(s + 1)  s  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 
4
=  2 +  I1 − 2 I 2
s 
s
(1)
-4
= -2 I1 + (s + 2) I 2
s+2
(2)
Put equations (1) and (2) into matrix form.
 1 s   2 + 4 s - 2   I1 
 - 4 (s + 2)  =  - 2
s + 2  I 2 

 
s 2 − 4s + 4
,
s (s + 2)
∆=
2 2
(s + 2s + 4) ,
s
I1 =
∆1
1 2 ⋅ (s 2 − 4s + 4)
A
Bs + C
=
=
+ 2
2
∆ (s + 2)(s + 2s + 4) s + 2 s + 2s + 4
∆1 =
∆2 =
-6
s
1 2 ⋅ (s 2 − 4s + 4) = A (s 2 + 2s + 4) + B (s 2 + 2s) + C (s + 2)
Equating coefficients :
s2 :
1 2= A+B
1
s :
- 2 = 2A + 2B + C
2 = 4 A + 2C
s0 :
Solving these equations leads to
A = 2,
B = -3 2,
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
C = -3
i1 ( t ) = [ 2 e -2t − 1.5 e -t cos(1.732t ) − 0.866 sin(1.732t )] u(t ) A
I2 =
s
-3
∆2 - 6
=
⋅
=
2
2
s 2 (s + 2s + 4) (s + 1) + ( 3 ) 2
∆
i 2 (t) =
-3
3
e - t sin( 3t ) = - 1.732 e -t sin(1.732t ) u(t ) A
Chapter 16, Solution 12.
We apply nodal analysis to the s-domain form of the circuit below.
s
10/(s + 1)
+
−
Vo
1/(2s)
4
10
− Vo 3 V
s +1
o
+ =
+ 2sVo
s
s
4
(1 + 0.25s + s 2 ) Vo =
Vo =
10
10 + 15s + 15
+ 15 =
s +1
s +1
15s + 25
A
Bs + C
=
+ 2
2
(s + 1)(s + 0.25s + 1) s + 1 s + 0.25s + 1
3/s
A = (s + 1) Vo
s = -1
=
40
7
15s + 25 = A (s 2 + 0.25s + 1) + B (s 2 + s) + C (s + 1)
Equating coefficients :
s2 :
0= A+B 
→ B = -A
1
s :
15 = 0.25A + B + C = -0.75A + C
0
25 = A + C
s :
A = 40 7 ,
B = - 40 7 ,
C = 135 7
- 40 135
40
3
1
s+
+
s
 155 2 
40 1
40
7
7
7
2
2

+
=
−
+
⋅
Vo =
2
2
2
7
+
7
s
1
7
s +1  1


3  1
 1
3
3
3
s +  +
s +  +
s +  +
 2
 2
 2
4
4
4
v o (t) =
 3  (155)(2)
 3 
40 - t 40 - t 2
e − e cos 
t  +
e - t 2 sin 
t 
7
7
 2  (7)( 3 )
 2 
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,
Vo
Vo
1
s +1
=
+
=
V
s + 2 2s + 1 2 + 1 s 2s + 1 o
1
Vo =
2s + 1
(s + 1)(s + 2)
Io =
Vo
1
A
B
=
=
+
2s + 1 (s + 1)(s + 2) s + 1 s + 2
A = 1,
Io =
B = -1
1
1
−
s +1 s + 2
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).
1Ω
4Ω
+
5V
+
−
vc(0)
io
−
(a)
i o (0 − ) = 5 A , v c (0 − ) = 0 V
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
+
+ +
=0
1
2s s 4 + 4 s
5/s
4/s
1
s 
15 5 
V
− = 1 + +
s s  2s 4 (s + 1)  o
5s 2 + 6s + 2
10 4s 2 + 4s + 2s + 2 + s 2
Vo
Vo =
=
4s (s + 1)
s
4s (s + 1)
Vo =
40 (s + 1)
5s 2 + 6s + 2
Vo 5
4 (s + 1)
5
+
+ =
2
2s s s (s + 1.2s + 0.4) s
5 A
Bs + C
Io = + + 2
s s s + 1.2s + 0.4
Io =
4 (s + 1) = A (s 2 + 1.2s + 0.4) + B s s + C s
Equating coefficients :
s0 :
4 = 0.4A 
→ A = 10
s1 :
2
s :
4 = 1.2A + C 
→ C = -1.2A + 4 = -8
0= A+B 
→ B = -A = -10
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 ( 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
s+2
5
Vo −
Vo − 3Vx Vo − 0
s+2 =0
+
+
s/4
5/s
10
5s
5s
= 0 = (2s 2 + s + 40)Vo − 120Vx −
40Vo − 120Vx + 2s 2 Vo + sVo −
s+2
s+2
5
5
→ Vo = Vx +
s+2
s+2
But, Vx = Vo −
We can now solve for Vx.
5 
5s

(2s 2 + s + 40) Vx +
=0
 − 120Vx −
s + 2
s+2

2(s 2 + 0.5s − 40)Vx = −10
Vx = − 5
(s 2 + 20)
s+2
(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.
2Ω
+
−
1Ω
1F
Vo/2
Vo
+
−
(a)
+
−
1H
io
3V
Hence,
i L (0) = i o =
-3
= -1 A ,
3
v o = -1 V
 - 1
v c (0) = -(2)(-1) −   = 2.5 V
2
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 +
+
=0
s+2 
s
s
s
2
But,
Vo = I o = I 2
 1
 1 1
5
2.5
 2 +  I1 +  −  I 2 =
−

2 s
s
s+2 s
(1)
For mesh 2,
V

1
1
2.5
1 + s +  I 2 − I1 + 1 − o −
=0

s
s
2
s
1
1
1
2.5
- I1 +  + s +  I 2 =
−1
2
s
s
s
(2)
Put (1) and (2) in matrix form.
 1
 5
1 1 
2.5 
−   I1  
−
2 + s
2 s   s+2 s 



=


1
1
1
2
.
5




+ s +  I 2  
−1 
 s
 s

2
s
3
∆ = 2s + 2 + ,
s
Io = I2 =
∆ 2 = -2 +
4
5
+
s s (s + 2)
∆2
- 2s 2 + 13
A
Bs + C
=
=
+ 2
2
∆
(s + 2)(2s + 2s + 3) s + 2 2s + 2s + 3
- 2s 2 + 13 = 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 = 5.429
0.7143 3.429 s − 5.429 0.7143 1.7145 s − 2.714
−
=
−
s+2
2s 2 + 2s + 3
s+2
s 2 + s + 1.5
0.7143 1.7145 (s + 0.5) (3.194)( 1.25 )
Io =
−
+
s+2
(s + 0.5) 2 + 1.25 (s + 0.5) 2 + 1.25
Io =
[
]
i o ( t ) = 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
1
s
I1
I2
1
4
For mesh 3,
 1
2
1
+  s +  I 3 − I1 − s I 2 = 0
s +1  s
s
For the supermesh,
 1
1 
1 +  I1 + (1 + s) I 2 −  + s  I 3 = 0
 s
s 
But
I1 = I 2 − 4
Substituting (3) into (1) and (2) leads to

 1
 1
1
 2 + s +  I 2 − s +  I 3 = 4 1 + 

 s
 s
s
 1
 1
-4
2
- s +  I 2 + s +  I 3 =
−
 s
 s
s s +1
Adding (4) and (5) gives
2
2 I2 = 4 −
s +1
I2 = 2 −
1
s +1
(1)
(2)
(3)
(4)
(5)
i o ( t ) = i 2 ( t ) = ( 2 − e -t ) u(t ) A
Chapter 16, Solution 18.
3 e −s 3
= (1 − e − s )
vs(t) = 3u(t) – 3u(t–1) or Vs = −
s
s
s
1Ω
Vs
+
+
−
1/s
2Ω
Vo
−
V
Vo − Vs
+ sVo + o = 0 → (s + 1.5)Vo = Vs
2
1
Vo =
3
2 
2
−s
(1 − e − s ) =  −
(1 − e )
s(s + 1.5)
s
s
+
1
.
5


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,
V1 1
4 (s + 2) − V1
+2=
+ + sVo
s s
2
 1 1
1
2
+ 2 =  +  V1 + + s Vo
2 s
s
s+2
But
Vo = V1 + 2 I and
Vo = V1 +
2 (V1 + 1)
s
I=
(1)
V1 + 1
s

→ V1 =
Vo − 2 s s Vo − 2
=
(s + 2) s
s+2
Substituting (2) into (1)
2
1  2s + 1  s 
2 

 Vo −
+ 2− =
+ s Vo
s+2
s  s  s + 2 
s + 2 
2
1 2 (2s + 1)  2s + 1  
+s V
+2− +
= 
s+2
s s (s + 2)  s + 2   o
2s 2 + 9s 2s + 9 s 2 + 4s + 1
=
=
Vo
s (s + 2)
s+2
s+2
Vo =
2s + 9
A
B
=
+
s + 4s + 1 s + 0.2679 s + 3.732
2
A = 2.443 ,
Vo =
B = -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.
1
2/s
1/s
Vo
+ −
1/(s + 1)
+
−
1
At the main non-reference node, KCL gives
1 (s + 1) − 2 s − Vo Vo Vo 1
=
+
+
1+1 s
1
s s
s
s +1
− 2 − s Vo = (s + 1)(s + 1 s) Vo +
s +1
s
s
s +1
−
− 2 = (2s + 2 + 1 s) Vo
s +1
s
Vo =
- 2s 2 − 4s − 1
(s + 1)(2s 2 + 2s + 1)
Vo =
- s − 2s − 0.5
A
Bs + C
=
+ 2
2
(s + 1)(s + s + 0.5) s + 1 s + s + 0.5
A = (s + 1) Vo
s = -1
=1
- s 2 − 2s − 0.5 = A (s 2 + s + 0.5) + B (s 2 + s) + C (s + 1)
Equating coefficients :
s2 :
-1 = A + B 
→ B = -2
s1 :
s0 :
Vo =
-2 = A+ B+C 
→ C = -1
- 0.5 = 0.5A + C = 0.5 − 1 = -0.5
2 (s + 0.5)
1
2s + 1
1
=
−
− 2
s + 1 s + s + 0.5 s + 1 (s + 0.5) 2 + (0.5) 2
v o ( t ) = [ 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
V − Vo s
s
= 1
+ Vo
1
s
2
At node 2,
V1 − Vo Vo
= + sVo
s
2

→

→
10 = ( s + 1)V1 + (
s2
− 1)Vo
2
s
V1 = Vo ( + s 2 + 1)
2
(1)
(2)
Substituting (2) into (1) gives
s2
10 = ( s + 1)( s + s / 2 + 1)Vo + ( − 1)Vo = s ( s 2 + 2s + 1.5)Vo
2
2
Vo =
A
Bs + C
10
= + 2
s ( s + 2s + 1.5) s s + 2s + 1.5
2
10 = A( s 2 + 2 s + 1.5) + Bs 2 + Cs
s2 :
0 = A+ B
s:
0 = 2A + C
constant :
10 = 1.5 A

→
Vo =
A = 20 / 3, B = -20/3, C = -40/3

0.7071
20  1
s+2
s +1
 20  1
− 1.414
=
− 2
 −
2
2
2
2 


3  s s + 2 s + 1.5  3  s ( s + 1) + 0.7071
( s + 1) + 0.7071 
Taking the inverse Laplace tranform finally yields
v o (t) =
[
]
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
1
At node 1,
V
V − V2
12
= 1+ 1
s +1 1
4s
At node 2,
V1 − V2 V2 s
=
+ V2
4s
2 3
V2
2
→
3/s
12
1 V

= V1 1 +  − 2
s +1
 4s  4s
(1)
4

V1 = V2  s 2 + 2s + 1
3

(2)
→
Substituting (2) into (1),
 4
12
1  1  4
7
3

= V2  s 2 + 2s + 11 +  −  =  s 2 + s + V2
s +1
3
2
 4s  4s   3
 3
V2 =
9
7
9
(s + 1)(s 2 + s + )
4
8
9 = A(s 2 +
=
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=A+B
s:
0=
7
3
A+B+C = A+C
4
4
constant :
9=
9
3
A + C= A 
→
8
8

→
3
C=− A
4
A = 24, B = -24, C = -18
V2 =
3
24s + 18
24
24(s + 7 / 8)
+
=
−
7
23
7
9
7
23
(s + 1)
(s + ) 2 +
(s 2 + s + )
(s + ) 2 +
8
64
4
8
8
64
24
−
(s + 1)
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
Es + F
3
 = D +
V1 =
7
9
7
9
(s + 1)
(s + 1)(s 2 + s + )
(s 2 + s + )
4
8
4
8
7
9
4

9 s 2 + 2s + 1 = D(s 2 + s + ) + E(s 2 + s) + F(s + 1)
4
8
3

Equating coefficients:
s2 :
s:
constant :
V1 =
Thus,
12 = D + E
18 =
9=
7
3
D + E + F or 6 = D + F
4
4
9
3
D + F or 3 = D 
→
8
8
8
+
(s + 1)

→
3
F = 6− D
4
D = 8, E = 4, F = 0
4s
8
4(s + 7 / 8)
7/2
=
+
−
7
9
7
23
7
23
(s + 1)
(s 2 + s + )
(s + ) 2 +
(s + ) 2 +
4
8
8
64
8
64
[
]
v1 ( t ) = 8e − t + 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
V V
+ + 5C = +
+ sCV
s s
R sL
s
1 
6 + 5 sC CV  2
s +

=
+
RC LC 
s
s 
V=
But
5s + 6 C
s + s RC + 1 LC
2
1
1
=
= 8,
RC 10 80
V=
1
1
=
= 20
LC 4 80
5s + 480
5 (s + 4)
(230)(2)
=
2
2 +
s + 8s + 20 (s + 4) + 2
(s + 4) 2 + 22
2
v( t ) = 5 e -4t cos( 2t ) + 230 e -4t sin( 2t ) V
I=
V
5s + 480
=
sL 4s (s 2 + 8s + 20)
I=
1.25s + 120
A
Bs + C
= + 2
2
s (s + 8s + 20) s s + 8s + 20
A = 6,
I=
B = -6 ,
C = -46.75
6 6s + 46.75 6
6 (s + 4)
(11.375)(2)
− 2
= −
2
2 −
s s + 8s + 20 s (s + 4) + 2
(s + 4) 2 + 22
i( t ) = 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.
+
4Ω
9A
5Ω
vo
-
v o (0) = 5x
4
(9) = 20
4+5
For t > 0, we have the Laplace transform of the circuit as shown below after
transforming the current source to a voltage source.
4Ω
16 Ω
Vo
+
36V
10A
2/s
5Ω
-
Applying KCL gives
36 − Vo
sV
V
+ 10 = o + o
20
2
5
Thus,

→
Vo =
[
3.6 + 20s A
B
= +
,
s(s + 0.5) s s + 0.5
]
v o ( t )= 7.2 − 12.8e −0.5t u ( 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 + = 0
s s
s + 16 
2
I=
4s 2 + 32
(s 2 + 6s + 9)(s 2 + 16)
V=
9
2 2
36s 2 + 288
I+ = +
s
s 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 = -0.2016 ,
(1)
(2)
(3)
(4)
(5)
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 0 − Vo
=
+ (0 − Vo ) sC
R1
R2
 1

+ sC  ( - Vo )
Vs = R 1 
R2

Vo
-1
=
Vs sR 1C + R 1 R 2
But
R 1 20
=
= 2,
R 2 10
So,
Vo
-1
=
Vs s + 2
Vs = 3 e -5t
R 1C = (20 × 103 )(50 × 10-6 ) = 1

→ Vs = 3 (s + 5)
+
Vo
−
Vo =
-3
(s + 2)(s + 5)
- Vo =
3
A
B
=
+
(s + 2)(s + 5) s + 2 s + 5
A = 1,
Vo =
B = -1
1
1
−
s+5 s+2
v o ( t ) = ( e -5t − e -2t ) u(t )
Chapter 16, Solution 27.
Consider the following circuit.
2s
10/(s + 3)
+
−
I1
s
2s
1
I2
1
For mesh 1,
10
= (1 + 2s) I1 − I 2 − s I 2
s+3
10
= (1 + 2s) I1 − (1 + s) I 2
s+3
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,
10 (s + 3)   2s + 1 - (s + 1)  I1 
 =  - (s + 1) 2 (s + 1)  I 

0
 2 
 

∆ = 3s 2 + 4s + 1
(1)
(2)
∆1 =
20 (s + 1)
s+3
∆2 =
10 (s + 1)
s+3
Thus
I1 =
20 (s + 1)
∆1
=
∆ (s + 3)( 3s 2 + 4s + 1)
I2 =
10 (s + 1)
∆2
I
=
= 1
2
∆
(s + 3)( 3s + 4s + 1) 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
 2
I1 = - 1 +  I 2
 s
Substituting (2) into (1) gives
 2
6
- (s 2 + 5s + 2)
I2
= -(1 + 2s)1 +  I 2 + s I 2 =
 s
s
s
or
I2 =
-6
s + 5s + 2
2
(2)
Vo = 2 I 2 =
- 12
- 12
=
s + 5s + 2 (s + 0.438)(s + 4.561)
2
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
= -2.91 ,
4.123
B=
- 12
= 2.91
- 4.123
- 2.91
2.91
+
s + 0.438 s + 4.561
Vo (s) =
v o ( t ) = 2.91 [ e -4.561t − e 0.438t ] u(t ) V
Chapter 16, Solution 29.
Consider the following circuit.
1
10/(s + 1)
Let
Io
1:2
+
−
4/s
Z L = 8 ||
4 (8)(4 s)
8
=
=
s 8 + 4 s 2s + 1
When this is reflected to the primary side,
Zin = 1 +
ZL
, n=2
n2
Zin = 1 +
2
2s + 3
=
2s + 1 2s + 1
Io =
10 1
10 2s + 1
⋅
=
⋅
s + 1 Zin s + 1 2s + 3
8
Io =
10s + 5
A
B
=
+
(s + 1)(s + 1.5) s + 1 s + 1.5
A = -10 ,
I o (s) =
B = 20
- 10
20
+
s + 1 s + 1.5
[
]
i o ( t ) = 10 2 e -1.5t − e − t u(t ) A
Chapter 16, Solution 30.
Y(s) = H(s) X(s) ,
X(s) =
4
12
=
s + 1 3 3s + 1
Y(s) =
12 s 2
4 8s + 4 3
−
2 =
(3s + 1)
3 (3s + 1) 2
Y(s) =
4 8
s
4
1
− ⋅
⋅
2 −
3 9 (s + 1 3)
27 (s + 1 3) 2
Let G (s) =
-8
s
⋅
9 (s + 1 3) 2
Using the time differentiation property,

-8 d
- 8  -1
g( t ) =
⋅ ( t e -t 3 ) =  t e -t 3 + e -t 3 

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) 
→ X(s) =
1
s
y( t ) = 10 cos(2t ) 
→ Y(s) =
H(s) =
10s
s2 + 4
Y(s) 10s 2
=
X(s) s 2 + 4
Chapter 16, Solution 32.
(a)
Y(s) = H(s) X(s)
=
s+3
1
⋅
s + 4s + 5 s
=
s+3
A
Bs + C
= + 2
s (s + 4s + 5) s s + 4s + 5
2
2
s + 3 = A (s 2 + 4s + 5) + Bs 2 + Cs
Equating coefficients :
3 = 5A 
→ A = 3 5
s0 :
s1 :
1 = 4A + C 
→ C = 1 − 4A = - 7 5
s2 :
0= A+B 
→ 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

→ X(s) =
Y(s) = H(s) X(s) =
Y(s) =
6
(s + 2) 2
s+3
6
⋅
s + 4s + 5 (s + 2) 2
2
6 (s + 3)
A
B
Cs + D
=
+
2
2
2 + 2
(s + 2) (s + 4s + 5) s + 2 (s + 2) s + 4s + 5
Equating coefficients :
s3 :
0= A+C 
→ 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 ,
(1)
(2)
(3)
(4)
D = -18
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 ) = [ 6 e -2t + 6 t e -2t − 6 e -2t cos(t ) − 6 e -2t sin( t )] u(t )
Chapter 16, Solution 33.
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
H(s) = s Y(s) = 4 +
X(s) =
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 Vo Vo
=
+
s+2
4 10 s
 1

 1
1 s
1
Vs = (s + 2) 
+ +  Vo = 1 + (s + 2) + (s 2 + 2s)  Vo
 4

 s + 2 4 10 
10
Vs =
1
( 2s 2 + 9s + 30) Vo
20
20
Vo
= 2
Vs 2s + 9s + 30
Chapter 16, Solution 35.
Consider the following circuit.
I
Vs
2/s
s
V1
+
+
−
2I
Vo
−
At node 1,
2I + I =
V1
,
s+3
where I =
Vs − V1
2s
3
3⋅
Vs − V1
V
= 1
2s
s+3
3s
V1
3s
= Vs − V1
2
s+3 2
 1
3s
3s 

+  V1 = Vs
s + 3 2 
2
V1 =
3s (s + 3)
V
3s 2 + 9s + 2 s
Vo =
9s
3
V
V1 = 2
3s + 9s + 2 s
s+3
H(s) =
9s
Vo
= 2
Vs 3s + 9s + 2
Chapter 16, Solution 36.
From the previous problem,
3I =
I=
But
V1
3s
V
= 2
s + 3 3s + 9s + 2 s
s
V
3s + 9s + 2 s
Vs =
2
3s 2 + 9s + 2
Vo
9s
V
s
3s 2 + 9 s + 2
I= 2
⋅
Vo = o
3s + 9 s + 2
9s
9
H(s) =
Vo
=9
I
Chapter 16, Solution 37.
(a)
Consider the circuit shown below.
3
Vs
+
−
2s
+
I1
Vx
2/s
I2
+
−
4Vx
−
For loop 1,
 2
2
Vs = 3 +  I1 − I 2

s
s
(1)
For loop 2,

2
2
4Vx +  2s +  I 2 − I1 = 0

s
s
But,
2
Vx = (I1 − I 2 )  
s
So,

2
2
8
(I1 − I 2 ) +  2s +  I 2 − I1 = 0

s
s
s
0=

6
-6
I1 +  − 2s  I 2

s
s
(2)
In matrix form, (1) and (2) become
 Vs  3 + 2 s
- 2 s  I1 
 0  =  - 6 s 6 s − 2s  I 
 2 
  

  6  2 
2  6
∆ = 3 +  − 2s  −   

  s  s 
s  s
∆=
18
− 6s − 4
s
6

∆ 1 =  − 2s  Vs ,
s

∆2 =
6
V
s s
I1 =
∆1
(6 s − 2s)
=
V
∆ 18 s − 4 − 6s s
I1
3 s−s
s2 − 3
=
= 2
Vs 9 s − 2 − 3 3s + 2s − 9
(b)
I2 =
∆2
∆
Vx =
2
2  ∆1 − ∆ 2 

( I1 − I 2 ) = 
s
s ∆ 
Vx =
2 s Vs (6 s − 2s − 6 s) - 4Vs
=
∆
∆
6 s Vs - 3
I2
=
=
Vx
- 4Vs
2s
Chapter 16, Solution 38.
(a)
Consider the following circuit.
Is
1
V1
s
Vo
Io
+
Vs
+
−
1/s
1/s
1
Vo
−
At node 1,
Vs − V1
V1 − Vo
= s V1 +
1
s

1
1
Vs = 1 + s +  V1 − Vo

s
s
(1)
At node o,
V1 − Vo
= s Vo + Vo = (s + 1) Vo
s
V1 = (s 2 + s + 1) Vo
(2)
Substituting (2) into (1)
Vs = (s + 1 + 1 s)(s 2 + s + 1)Vo − 1 s Vo
Vs = (s 3 + 2s 2 + 3s + 2)Vo
H 1 (s) =
(b)
Vo
1
= 3
2
Vs s + 2s + 3s + 2
I s = Vs − V1 = (s 3 + 2s 2 + 3s + 2)Vo − (s 2 + s + 1)Vo
I s = (s 3 + s 2 + 2s + 1)Vo
H 2 (s) =
(c)
(d).
Io =
Vo
1
= 3
2
Is
s + s + 2s + 1
Vo
1
H 3 (s) =
I o Vo
1
=
= H 2 (s) = 3
2
Is
Is
s + s + 2s + 1
H 4 (s) =
I o Vo
1
=
= H 1 (s) = 3
2
Vs Vs
s + 2s + 3s + 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.

1
Vo = -I o  R + 

sC 
Va = Vb = Vs =
H(s) =
- Io
sC
Vo R + 1 sC
=
= sRC + 1
Vs
1 sC
Chapter 16, Solution 40.
(a)
(b)
H(s) =
Vo
R
R L
=
=
Vs R + sL s + R L
h(t) =
R - Rt L
e
u( t )
L
v s (t) = u(t) 
→ Vs (s) = 1 s
Vo =
R L
R L
A
B
Vs =
= +
s+R L
s (s + R L) s s + R L
A = 1,
B = -1
1
1
Vo = −
s s+R L
v o ( t ) = 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 ) 
→
H(s) =
2
s +1
v i (t) = 5 u(t) 
→ Vi (s) = X(s) = 5 s
Y(s) =
10
A
B
= +
s (s + 1) s s + 1
A = 10 ,
Y(s) =
B = -10
10 10
−
s s +1
y( t ) = 10 (1 − e -t ) u(t )
Chapter 16, Solution 42.
2s Y(s) + Y(s) = X(s)
(2s + 1) Y(s) = X(s)
H(s) =
Y(s)
1
1
=
=
X(s) 2s + 1 2 (s + 1 2)
h ( t ) = 0.5 e -t 2 u(t )
Chapter 16, Solution 43.
1Ω
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,
 v ′   0 1   v C  0 
v 
+   u ( t ) ; i( t ) = [0 1]  C  + [0]u ( t )
 C =



 i 
 i ′  − 1 − 1  i  1
Chapter 16, Solution 44.
1/8 F
1H
4u ( t )
+
−
+
vx
2Ω
4Ω
−
First select the inductor current iL and the capacitor voltage vC to be the state
variables.
Applying KCL we get:
v
− iL + x +
2
v 'C
= 0; or v 'C = 8i L − 4v x
8
i 'L = 4u ( t ) − v x
v 'C
v 'C
v x = vC + 4
= vC +
= v C + 4i L − 2v x ; or v x = 0.3333v C + 1.3333i L
8
2
v 'C = 8i L − 1.3333v C − 5.333i L = −1.3333v C + 2.666i L
i 'L = 4u ( t ) − 0.3333v C − 1.3333i L
Now we can write the state equations.
 v 'C   − 1.3333
2.666   v C  0
0.3333  v C 
+   u ( t ); v x = 
 ' =



 
1.3333  i L 
 i L  − 0.3333 − 1.3333  i L  4
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
−
+
+ i L = 0 or v 'C = 4i L + 2 v o
4
2
i 'L = v o − v 2
v o = − v C + v1
v 'C = 4i L − 2 v C + 2 v1
i 'L = − v C + v1 − v 2
 i ′  0 − 1  i L  1 − 1  v1 ( t ) 
i 
 v (t) 
+
; v o ( t ) = [0 − 1]  L  + [1 0]  1 
 L ′=






 v 2 ( t )
v C 
 v C  4 − 2  v C  2 0   v 2 ( t )
Chapter 16, Solution 46.
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 = 0 or v 'C = −0.25v C + i L + i s
4
i 'L = − v C + v s
Thus,
 v '  − 0.25 1  v '  0 1  v s 
1  v C  0 0  v s 
C +
=
;
v
(
t
)

 'C  = 

o
0   i  + 0 0   i 
0  i 'L  1 0  i s 
  L  
 s 
 i L   − 1
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  = 0 or v 'C = 4i L − 2 v C + 2 v1
 4



Loop 2:

v 'C  '

2 iL −
+ i + v 2 = 0 or
 L

4


4i − 2v C + 2v1
− v 2 = − v C + v1 − v 2
i 'L = −2i L + L
2
i1 =
i ′ 
 L ′=
 v C 
4i L − 2 v C + 2 v1
= i L − 0.5v C + 0.5v1
4
0 − 1  i L  1 − 1  v1 ( t ) 
 4 − 2  v  +  2 0   v ( t )  ;

 C 
 2 
 i1 ( t )  1 − 0.5  i L  0.5 0  v1 ( t ) 
+
i ( t ) = 1
0   v C   0 0  v 2 ( t )
2  
Chapter 16, Solution 48.
Let x1 = y(t). Thus, x1' = y ' = x 2 and x '2 = y′′ = −3x1 − 4 x 2 + z( t )
This gives our state equations.
 x1'   0
1   x 1  0 
x 
+   z( t ); y( t ) = [1 0] 1  + [0]z( t )
 ' =



x 2 
 x 2  − 3 − 4  x 2  1
Chapter 16, Solution 49.
Let x1 = y( t ) and x 2 = x1' − z = y ' − z or y ' = x 2 + z
Thus,
x '2 = y ′′ − z ' = −6x1 − 5( x 2 + z) + z ' + 2z − z ' = −6x1 − 5x 2 − 3z
This now leads to our state equations,
 x1'   0
1   x1   1 
x 
+   z( t ); y( t ) = [1 0]  1  + [0]z( t )
 ' =



 x 2  − 6 − 5  x 2  − 3
x 2 
Chapter 16, Solution 50.
Let x1 = y(t), x2 = x1' , and x 3 = x '2 .
Thus,
x "3 = −6x1 − 11x 2 − 6x 3 + z( t )
We can now write our state equations.
 x1'   0
1
0   x 1  0
 x1 
 '  





0
1   x 2  + 0 z( t ); y( t ) = [1 0 0] x 2  + [0]z( t )
x 2  =  0
 x '  − 6 − 11 − 6  x  1
 x 3 
 3   
 3  
Chapter 16, Solution 51.
We transform the state equations into the s-domain and solve using Laplace
transforms.
1
sX(s) − x (0) = AX(s) + B 
s
Assume the initial conditions are zero.
1
(sI − A)X(s) = B 
s
s + 4 − 4
X(s) = 
s 
 2
−1
4  0 
0  1 
s
1
 2  s  = 2
 2 s + 4  ( 2 / s ) 
   s + 4s + 8 


1
−s−4
= +
s(s 2 + 4s + 8) s s 2 + 4s + 8
1
1
−2
− (s + 2)
−s−4
+
= +
= +
2
2
2
2
s (s + 2) + 2
s (s + 2) + 2
(s + 2) 2 + 2 2
8
Y(s) = X1 (s) =
(
)
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,
1 
s + 2
X(s) = 

 − 2 s + 4
X1 =
=
3s + 8
2
2
s((s + 3) + 1 )
−1
=
s + 4 − 1  3 / s 
1 1  1 / s 
1

 4 0  2 / s  = 2
s + 2 4 / s 


 s + 6s + 10  2
0.8 − 0.8s − 1.8
+
s
(s + 3) 2 + 12
0.8
s+3
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 )
X2 =
=
4s + 14
s((s + 3) 2 + 12
=
1.4 − 1.4s − 4.4
+
s
(s + 3) 2 + 12
1.4
s+3
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 ) = −2x1 ( t ) − 2x 2 ( t ) + 2u ( t )
= (−2.4 + 4.4e − 3t cos t − 0.8e − 3t sin t )u ( t )
y 2 ( t ) = x1 ( t ) − 2u ( 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 1
1
+ sC Vo + o =  + sC +  Vo
R
sL  R
sL 
Is
Vo =
Io =
1
1
+ sC +
R
sL
=
sRL Is
sL + R + s 2 RLC
Vo
sL Is
= 2
R s RLC + sL + R
H(s) =
Io
sL
s RC
= 2
= 2
Is s RLC + sL + R s + s RC + 1 LC
The roots
s1, 2 =
-1
1
1
±
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(s) = H1 (s) H 2 (s) =
H 2 (s) =
1
s+4
3
(s + 1)(s + 4)
 A
B 
+
h ( t ) = L-1 [ H(s)] = L-1 
 s + 1 s + 4 
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 − 1 sC − 1
=
=
,
Vs
R
sRC
H(s) =
Vo
1
= 2 2 2
Vs s R C
h(t) =
t
R C2
Vo
−1
=
Vo1 sRC
2
lim h ( t ) = ∞ , i.e. the output is unbounded.
t →∞
Hence, the circuit is unstable.
Chapter 16, Solution 56.
1
sL ⋅
1
sC = sL
sL ||
=
1 1 + s 2 LC
sC
sL +
sC
sL
2
V2
sL
= 1 + s LC = 2
sL
V1
s RLC + sL + R
R+
2
1 + s LC
V2
=
V1
1
RC
1
1
s2 + s ⋅
+
RC LC
s⋅
Comparing this with the given transfer function,
1
1
2=
,
6=
RC
LC
If R = 1 kΩ ,
C=
1
= 500 µF
2R
L=
1
= 333.3 H
6C
Chapter 16, Solution 57.
The circuit in the s-domain is shown below.
R1
Vi
L
V1
+
+
−
C
R2
Vx
−
Z
Z=
(1 sC) ⋅ (R 2 + sL)
R 2 + sL
1
|| (R 2 + sL) =
=
sC
R 2 + sL + 1 sC
1 + s 2 LC + sR 2 C
V1 =
Z
V
R1 + Z i
Vo =
R2
R2
Z
⋅
V1 =
V
R 2 + sL
R 2 + sL R 1 + Z i
R 2 + sL
Vo
R2
R2
1 + s 2 LC + sR 2 C
Z
=
⋅
=
⋅
R 2 + sL
Vi R 2 + sL R 1 + Z R 2 + sL
R1 +
1 + s 2 LC + sR 2 C
Vo
R2
= 2
Vi s R 1 LC + sR 1 R 2 C + R 1 + R 2 + sL
R2
Vo
R 1 LC
=
R2
Vi
1  R1 + R 2
+
s 2 + s
+
 L R 1C  R 1 LC
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 = 4 Ω and R 2 = 1 Ω ,
1
1
5=

→ LC =
4 LC
20
6=
1
1
+
L 4C
25 =
5
4 LC
(2)

→ LC =
1
20
Substituting (1) into (2),
1
6 = 20 C +

→ 80 C 2 − 24 C + 1 = 0
4C
Thus, C =
1
,
4
1
20
(1)
When C =
1
,
4
L=
1
1
= .
20 C 5
When C =
1
,
20
L=
1
= 1.
20 C
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
- Y3
=
Vs Y1 + Y2
Let
Y1 = sC1 ,
Y2 = 1 R 1 ,
Y3 = sC 2
Vo
- sC 2
- sC 2 C1
=
=
Vs sC1 + 1 R 1 s + 1 R 1C1
Comparing this with the given transfer function,
C2
1
= 1,
= 10
R 1 C1
C1
If R 1 = 1 kΩ ,
C1 = C 2 =
1
= 100 µF
10 4
Chapter 16, Solution 59.
Consider the circuit shown below. We notice that V3 = Vo and V2 = V3 = Vo .
Y4
Y1
Vin
Y2
V2
V1
+
−
−
+
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 (Y1 Y2 + Y22 + Y2 Y4 + Y1 Y3 + Y2 Y3 + Y3 Y4 − Y22 − Y2 Y4 )
Vo
Y1 Y2
=
Vin 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
1
Vo
R 1R 2
=
sC1 sC1
1
Vin
+
+
+ s 2 C1 C 2
R 1R 2 R 1 R 2
1
Vo
R 1 R 2 C1C 2
=
 R1 + R 2 
Vin
1
+
s2 + s ⋅
 R 1 R 2 C 2  R 1 R 2 C1 C 2
Choose R 1 = 1 kΩ , then
1
= 10 6
R 1 R 2 C1 C 2
and
R1 + R 2
= 100
R 1R 2 C 2
We have three equations and four unknowns. Thus, there is a family of solutions. One
such solution is
R 2 = 1 kΩ , C1 = 50 nF , C 2 = 20 µF
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 µF
1
,
R1
R 1 = 10 kΩ
Y2 =
Y3 = Y2 ,
Y4 = sC 2 ,
C 2 = 1 µF
Vo
- sC1 R 1
- sC1 R 1
=
=
2
Vi 1 R 1 + sC 2 (sC1 + 2 R 1 ) 1 + sC 2 R 1 (2 + sC1 R 1 )
Vo
- sC1 R 1
= 2
Vi s C1C 2 R 12 + s ⋅ 2C 2 R 1 + 1
Vo
- s (0.5 × 10 -6 )(10 × 10 3 )
=
Vi s 2 (0.5 × 10 -6 )(1 × 10 -6 )(10 × 10 3 ) 2 + s (2)(1 × 10 -6 )(10 × 10 3 ) + 1
Vo
- 100 s
= 2
Vi s + 400 s + 2 × 10 4
Therefore,
a = - 100 ,
b = 400 ,
c = 2 × 10 4
Chapter 16, Solution 68.
(a)
Let
Y(s) =
K (s + 1)
s+3
K (s + 1)
K (1 + 1 s)
= lim
=K
s →∞
s →∞ 1 + 3 s
s+3
Y(∞) = lim
i.e.
0.25 = K .
Hence, Y(s) =
(b)
s+1
4 (s + 3)
Consider the circuit shown below.
t=0
Vs = 8 V
+
−
I
YS
Vs = 8 u ( t ) 
→ Vs = 8 s
I=
Vs
8 s + 1 2 (s + 1)
= Y(s) Vs (s) = ⋅
=
Z
4s s + 3 s (s + 3)
I=
A
B
+
s s+3
A = 2 3,
i( t ) =
B= -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 = -Vi
R i
Vo =
- 1 sC
1
V1 =
V
R
sCR i
Io =
Vo
Vo
=
R sR 2 C
Vo
= sR 2 C
Io
Vo
= sL, when L = R 2 C
Io
Chapter 17, Solution 1.
(a)
This is periodic with ω = π which leads to T = 2π/ω = 2.
(b)
y(t) is not periodic although sin t and 4 cos 2πt 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
ω = 1 or T = 2π/ω = 2π.
(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. ω = 2 or T = 2π/ω = π.
(e)
The frequency ratio 0.6|0.4 = 1.5 makes z(t) periodic.
ω = 0.2π or T = 2π/ω = 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.
ω = 1 = 2π/T or T = 2π.
(b)
ω = 2 or T = 2π/ω = π.
(c)
f3(t) = 4 sin2 600π t = (4/2)(1 – cos 1200π t)
ω = 1200π or T = 2π/ω = 2π/(1200π) = 1/600.
(d)
f4(t) = ej10t = cos 10t + jsin 10t. ω = 10 or T = 2π/ω = 0.2π.
Chapter 17, Solution 3.
T = 4, ωo = 2π/T = π/2
g(t) = 5,
10,
0,
0<t<1
1<t<2
2<t<4
T
1
0
0
2
ao = (1/T) ∫ g( t )dt = 0.25[ ∫ 5dt + ∫ 10dt ] = 3.75
an = (2/T)
1
T
∫ g( t ) cos(nωo t )dt = (2/4)[
0
1
∫ 5 cos(
0
2
nπ
nπ
t )dt + ∫ 10 cos( t )dt ]
1
2
2
2
1
nπ
2
2
nπ
t + 10
sin
t ] = (–1/(nπ))5 sin(nπ/2)
= 0.5[ 5
sin
2 0
nπ
nπ
2 1
an =
bn = (2/T)
(5/(nπ))(–1)(n+1)/2,
0,
T
∫ g( t ) sin(nω t )dt =
o
0
(2/4)[
0
1
= 0.5[
nπ
1
∫ 5 sin( 2
n = odd
n = even
2
t )dt + ∫ 10 sin(
1
nπ
t )dt ]
2
2
nπ
nπ
− 2x5
2 x10
t –
t ] = (5/(nπ))[3 – 2 cos nπ + cos(nπ/2)]
cos
cos
2 0
2 1
nπ
nπ
Chapter 17, Solution 4.
f(t) = 10 – 5t, 0 < t < 2, T = 2, ωo = 2π/T = π
ao = (1/T)
an = (2/T)
=
T
2
2
0
0
0
2
∫ f ( t )dt = (1/2) ∫ (10 − 5t )dt = 0.5[10t − (5t / 2)] = 5
T
∫ f ( t ) cos(nωo t )dt = (2/2)
0
2
∫ (10) cos(nπt )dt –
0
2
2
∫ (10 − 5t ) cos(nπt )dt
0
2
∫ (5t ) cos(nπt )dt
0
2
−5
5t
sin nπt = [–5/(n2π2)](cos 2nπ – 1) = 0
= 2 2 cos nπt +
n π
nπ
0
0
bn = (2/2)
2
∫ (10 − 5t ) sin(nπt )dt
0
2
2
0
0
∫ (10) sin(nπt )dt – ∫ (5t ) sin(nπt )dt
=
2
2
−5
5t
cos nπt = 0 + [10/(nπ)](cos 2nπ) = 10/(nπ)
= 2 2 sin nπt +
n π
nπ
0
0
f(t) = 5 +
Hence
10 ∞ 1
∑ sin(nπt )
π n =1 n
Chapter 17, Solution 5.
T = 2π,
ω = 2π / T = 1
T
1
1
a o = ∫ z( t )dt = [1xπ − 2 xπ] = −0.5
T
2π
0
an =
T
π
2π
0
0
π
T
π
2π
0
0
2
1
1
z( t ) cos nωo dt = ∫ 1 cos ntdt −
∫
T
π
π
2
1
1
b n = ∫ z( t ) cos nωo dt = ∫ 1sin ntdt −
π
π
T
Thus,
z( t ) = − 0.5 +
∞
6
sin nt
n =1 nπ
∑
n =odd
1
∫ 2 cos ntdt = nπ sin ..nt
2
π
2π
−
sin nt π = 0
0 nπ
 6
1
2
2π 
π
, n = odd
∫ 2 sin ntdt = − nπ cos nt 0 + nπ cos nt π =  nπ
0, n = even
π
Chapter 17, Solution 6.
T = 2, ωo =
ao =
2π
=π
2
1 2
1
6
y( t )dt = (4 x1 + 2 x1) = = 3
∫
2 0
2
2
Since this is an odd function, a n = 0.
bn =
1
2
2 2
y( t ) sin( nωo t )dt = ∫ 4 sin( nπt )dt + ∫ 2 sin( nπt )dt
∫
0
1
2 0
=
−4
−4
2
2
1
2
(cos(2nπ) − cos(nπ))
(cos(nπ) − 1) −
cos(nπt ) 1 =
cos(nπt ) 0 −
nπ
nπ
nπ
nπ
=
2
2
4
0,
n = even
(1 − cos(nπ)) = 4
(1 − cos(nπ)) =
(1 − cos(nπ)) −
nπ
nπ
nπ
, n = odd
nπ
4 ∞ 1
y( t ) = 3 +
∑ sin(nπt )
π n =1 n
n = odd
Chapter 17, Solution 7.
T = 12,
ω = 2π / T =
π
,
6
a0 = 0
T
4
10
0
−2
4
1
2
a n = ∫ f ( t ) cos nωo dt = [ ∫ 10 cos nπt / 6dt + ∫ (−10) cos nπt / 6dt ]
6
T
=
10
10
4
10 10
[2 sin 2nπ / 3 + sin nπ / 3 − sin 5nπ / 3]
sin nπt / 6 − 2 −
sin nπt / 6 4 =
nπ
nπ
nπ
T
4
10
0
−2
4
1
2
b n = ∫ f ( t ) sin nωo dt = [ ∫ 10 sin nπt / 6dt + ∫ (−10) sin nπt / 6dt ]
6
T
=−
10
10
4
10 10
[cos 5nπ / 3 + cos nπ / 3 − 2 sin 2nπ / 3]
cos nπt / 6 − 2 +
cos nπnt / 6 4 =
nπ
nπ
nπ
f (t) =
∞
∑ (a n cos nπt / 6 + b n sin nπt / 6)
n =1
where an and bn are defined above.
Chapter 17, Solution 8.
f ( t ) = 2(1 + t ), - 1 < t < 1,
T
T = 2,
1
ωo = 2π / T = π
1
1
a o = ∫ f ( t )dt = ∫ 2( t + 1)dt = t 2 + t
T
2
−1
0
T
1
=2
−1
1
1
an =
 1

2
2
t
1
f ( t ) cos nωo dt = ∫ 2( t + 1) cos nπtdt = 2
cos nπt +
sin nπt +
sin nπt  = 0
∫
T
2
nπ
nπ
 n 2π2
 −1
−1
0
bn =


t
1
4
2
2
1
sin nπt − cos nπt − cos nπt  = − cos nπ
f ( t ) sin nωo dt = ∫ 2( t + 1) sin nπtdt = 2 −
∫
2
2
nπ
nπ
nπ
T
2
 n π
 −1
T
1
0
−1
f (t) = 2 −
1
4 ∞ (−1) n
cos nπt
∑
π n =1 n
Chapter 17, Solution 9.
f(t) is an even function, bn=0.
T = 8,
ao =
ω = 2π / T = π / 4
2
T
 10 4
1
2
(
)
f
t
dt
=
 ∫ 10 cos πt / 4dt + 0 = ( ) sin πt / 4
∫
8 0
T 0
 4 π
2
0
=
10
π
= 3.183
an =
4
T
T /2
∫
f (t ) cos nω o dt =
0
2
2
40
[ 10 cos πt / 4 cos nπt / 4dt +0] = 5∫ [cos πt (n + 1) / 4 + cos πt (n − 1) / 4]dt
8 ∫0
0
For n = 1,
2
2
2

a1 = 5∫ [cos πt / 2 + 1]dt = 5 sin πt / 2dt + t  = 10
π
0
0
For n>1,
2
20
20
20
20
π (n + 1)t
π (n − 1)
π (n + 1)
π (n − 1)
an =
sin
sin
sin
sin
+
=
+
π (n + 1)
π (n − 1)
π (n + 1)
π (n − 1)
4
4
2
2
0
a2 =
10
π
sin π +
20
π
sin π / 2 = 6.3662,
a3 =
20
10
sin 2π + sin π = 0
4π
π
Thus,
a 0 = 3.183,
a1 = 10,
a 2 = 6.362,
a3 = 0,
b1 = 0 = b2 = b3
Chapter 17, Solution 10.
T = 2,
ωo = 2π / T = π
cn =
T
 − jnπt 1 2e − jnπt 2 
2
1
1  1 − jnπt
− jnωo t
− jnπt  1 4e
h
(
t
)
e
dt
4
e
dt
(
2
)
e
dt
=
=
+
−
∫1
 2  − jnπ 0 − − jnπ 1 
T∫
2  ∫0


0
cn =
 6j
j
j
−
, n = odd
4e − jπn − 4 − 2e − j2nπ + 2e − jnπ =
[6 cos nπ − 6] =  nπ
,
2nπ
2nπ
 0, n = even
[
]
Thus,
f (t ) =
∞
 − j6  jnπt

e
n =−∞  nπ 
∑
n =odd
Chapter 17, Solution 11.
T = 4,
ω o = 2π / T = π / 2
T
1
1
1 0
c n = ∫ y( t )e − jnωo t dt =  ∫ ( t + 1)e − jnπt / 2 dt + ∫ (1)e − jnπt / 2 dt 
0
T
4  −1

0
cn =
=
1  e − jnπt / 2
2 − jnπt / 2 0
2 − jnπt / 2 1 
−
e
e
 2 2 (− jnπt / 2 − 1) −
−1 jnπ
0
4  − n π / 4
jnπ

1 4
2
4
2 jnπ / 2
2 − jnπ / 2
2 
e jnπ / 2 ( jnπ / 2 − 1) +
e
e
−
+
−
+

4  n 2 π 2 jnπ n 2 π 2
jnπ
jnπ
jnπ 
But
e jnπ / 2 = cos nπ / 2 + j sin nπ / 2 = j sin nπ / 2,
cn =
1
n 2π2
e − jnπ / 2 = cos nπ / 2 − j sin nπ / 2 = − j sin nπ / 2
[1 + j( jnπ / 2 − 1) sin nπ / 2 + nπ sin nπ / 2]
y( t ) =
∞
∑
n = −∞
1
2 2
n π
[1 + j( jnπ / 2 − 1) sin nπ / 2 + nπ sin nπ / 2]e jnπt / 2
Chapter 17, Solution 12.
A voltage source has a periodic waveform defined over its period as
v(t) = t(2π - t) V,
for all 0 < t < 2π
Find the Fourier series for this voltage.
v(t) = 2π t – t2, 0 < t < 2π, T = 2π, ωo = 2π/T = 1
ao =
T
(1/T) ∫ f ( t )dt =
0
1 2π
1
(πt 2 − t 3 / 3)
(2πt − t 2 )dt =
∫
0
2π
2π
2π
0
=
4π 3
2π 2
(1 − 2 / 3) =
2π
3
2π
2 T
1  2π
2πt

an = ∫ (2πt − t 2 ) cos(nt )dt =  2 cos(nt ) +
sin(nt )
0
T
n
π n
0
bn =
=
[
−
1
2nt cos(nt ) − 2 sin(nt ) + n 2 t 2 sin( nt )
3
πn
=
−4
2
1
(1 − 1) − 3 4nπ cos(2πn ) = 2
2
πn
n
n
]
2π
0
2 T
1
(2nt − t 2 ) sin( nt )dt = ∫ (2nt − t 2 ) sin(nt )dt
∫
T 0
π
2π
2n 1
1
π
(sin(nt ) − nt cos(nt )) 0 − 3 (2nt sin(nt ) + 2 cos(nt ) − n 2 t 2 cos(nt ))
2
0
π n
πn
=
Hence,
f(t) =
− 4 π 4π
+
=0
n
n
2π 2 ∞ 4
− ∑ 2 cos(nt )
3
n =1 n
Chapter 17, Solution 13.
T = 2π, ωo = 1
T
ao = (1/T) ∫ h( t )dt =
0
=
an = (2/T)
2π
1 π
[ ∫ 10 sin t dt + ∫ 20 sin( t − π) dt ]
π
2π 0
[
]
1
30
π
2π
− 10 cos t 0 − 20 cos( t − π) π =
2π
π
T
∫ h( t ) cos(nω t )dt
0
o
π
= [2/(2π)]  ∫ 10 sin t cos( nt )dt +
 0
∫
2π
π
20 sin( t − π) cos( nt )dt 

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 – π) = sin t cos π – cost sin π = –sin t
sin(t – π)cos(nt) = –sin(t)cos(nt)
an =
2π
1  π
10∫ [sin([1 + n ]t ) + sin([1 − n ]t )]dt − 20∫ [sin([1 + n ]t ) + sin([1 − n ]t )]dt 

π
2π  0
5
=
π
 cos([1 + n ]t ) cos([1 − n ]t )  π  2 cos([1 + n ]t ) 2 cos([1 − n ]t )  2 π 
−
+
 −
 +
 
1+ n
1− n
1+ n
1− n
0 
 π 

But,
3
3 cos([1 + n ]π) 3 cos([1 − n ]π) 
 3
−
1 + n + 1 − n −

1+ n
1− n
5
π
an =
[1/(1+n)] + [1/(1-n)] = 1/(1–n2)
cos([n–1]π) = cos([n+1]π) = cos π cos nπ – sin π sin nπ = –cos nπ
an = (5/π)[(6/(1–n2)) + (6 cos(nπ)/(1–n2))]
= [30/(π(1–n2))](1 + cos nπ) = [–60/(π(n–1))], n = even
= 0,
n = odd
T
bn = (2/T) ∫ h ( t ) sin nωo t dt
0
π
2π
0
π
= [2/(2π)][ ∫ 10 sin t sin nt dt + ∫ 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)]
π
bn = (5/π){[(sin([1–n]t)/(1–n)) – (sin([1+n]t)/ (1 + n )] 0
2π
+ [(2sin([1-n]t)/(1-n)) – (2sin([1+n]t)/ (1 + n )] π }
=
Thus,
5
π
 sin([1 − n ]π) sin([1 + n ]π) 
+
 −
 = 0
1− n
1+ n
h(t) =
30 60 ∞ cos( 2kt )
−
∑
π
π k = 1 ( 4k 2 − 1)
Chapter 17, Solution 14.
Since cos(A + B) = cos A cos B – sin A sin B.
∞
10

 10
cos(nπ / 4) cos( 2nt ) − 3
sin(nπ / 4) sin( 2nt ) 
f(t) = 2 + ∑  3
n +1

n =1  n + 1
Chapter 17, Solution 15.
(a)
Dcos ωt + Esin ωt = A cos(ωt - θ)
where
f(t) = 10 +
A =
D 2 + E 2 , θ = tan-1(E/D)
A =
16
1
+ 6 , θ = tan-1((n2+1)/(4n3))
2
( n + 1)
n
∞
∑
n =1
(b)
2
2

16
1
−1 n + 1 

cos 10nt − tan
+
( n 2 + 1) 2 n 6
4n 3 

Dcos ωt + Esin ωt = A sin(ωt + θ)
where
D 2 + E 2 , θ = tan-1(D/E)
A =
f(t) = 10 +
∞
∑
n =1

16
1
4n 3 
−1

sin 10nt + tan
+
( n 2 + 1) 2 n 6
n 2 + 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
π2
8
π2


 t + 1
 t + 1 1
 t + 1 1
cos π 2  + 9 cos 3π 2  + 25 cos 5π 2  + "








  πt π  1

 5πt 5π 
 3πt 3π  1
cos 2 + 2  + 9 cos 2 + 2  + 25 cos 2 + 2  + "





 

v2(t) = −
8
π2
  π t  1  3 πt 

1
 5 πt 
sin 2  + 9 sin 2  + 25 sin 2  + "




  

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 nπ(-t) sin nπ(-t) = - cos nπt sin nπt,
odd
(d)
sin2 n(-t) = (-sin πt)2 = sin2 πt,
even
(e)
e t,
neither odd nor even.
Chapter 17, Solution 18.
(a)
T = 2 leads to ωo = 2π/T = π
f1(-t) = -f1(t), showing that f1(t) is odd and half-wave symmetric.
(b)
T = 3 leads to ωo = 2π/3
f2(t) = f2(-t), showing that f2(t) is even.
(c)
T = 4 leads to ωo = π/2
f3(t) is even and half-wave symmetric.
Chapter 17, Solution 19.
This is a half-wave even symmetric function.
ao = 0 = bn, ωo = 2π/T π/2
an =
4
T
∫
T/2
0
4t 

1 − T  cos(nωo t )dt
= [4/(nπ)2](1 − cos nπ)
f (t) =
8
π2
∞
∑
n = odd
= 8/(n2π2),
=
0,
1
 nπt 
cos

2
n
 2 
Chapter 17, Solution 20.
This is an even function.
bn = 0, T = 6, ω = 2π/6 = π/3
ao =
2
T
∫
T/2
0
f ( t )dt =
3
2 2
−
(
4
t
4
)
dt
4 dt 
∫
∫


1
2
6
n = odd
n = even
2
1 2
( 2 t − 4 t ) + 4(3 − 2) = 2
1

3 
=
4
T
an =
∫
T/4
0
f ( t ) cos( nπt / 3)dt
2
= (4/6)[ ∫ ( 4 t − 4) cos( nπt / 3)dt +
1
∫
3
2
4 cos( nπt / 3)dt ]
2
3
16  3
3
16  9
 nπt 
 nπt 
 nπt 
 nπt  3t
=
sin
sin
sin
cos

 +
−
+
2 2


6  nπ  3  2
6 n π
 3  nπ  3  nπ  3 1
= [24/(n2π2)][cos(2nπ/3) − cos(nπ/3)]
f(t) = 2 +
Thus
24 ∞ 1
∑
π 2 n =1 n2
  2πn 
 nπt 
 πn  
cos 3  − cos 3   cos 3 





 
At t = 2,
f(2) = 2 + (24/π2)[(cos(2π/3) − cos(π/3))cos(2π/3)
+ (1/4)(cos(4π/3) − cos(2π/3))cos(4π/3)
+ (1/9)(cos(2π) − cos(π))cos(2π) + -----]
= 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, ωo = 2π/T = π/2.
f(t) = 2 − 2t,
= 0,
0<t<1
1<t<2
1

t2 
2 1
ao =
2(1 − t )dt =  t −  = 0.5
2 0
4 ∫0

an =
4
T
∫
T/2
0
f ( t ) cos( nωo t )dt =
4 1
 nπt 
2(1 − t ) cos
dt
∫
4 0
 2 
= [8/(π2n2)][1 − cos(nπ/2)]
∞
1
+
2
f(t) =
8
∑n π
n=1
2
2

 nπt 
 nπ  
1 − cos 2   cos 2 


 

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 ωo = π/2.
ao =
an =
2
T
∫
4
T
T2
0
∫
f ( t )dt =
T2
0
1
2 1
4 tdt = t 2 = 1
∫
0
0
4
f ( t ) cos(ωo nt )dt =
4 1
4 t cos( nπt / 2)dt
4 ∫0
1
2t
 4

sin( nπt / 2)
= 4  2 2 cos( nπt / 2) +
nπ
n π
0
an =
16
8
sin( nπ / 2)
(cos( nπ / 2) − 1) +
2 2
nπ
n π
Chapter 17, Solution 23.
f(t) is an odd function.
f(t) = t, −1< t < 1
ao = 0 = an, T = 2, ωo = 2π/T = π
bn =
=
4
T
∫
T/2
0
f ( t ) sin( nωo t )dt =
4 1
t sin( nπt )dt
2 ∫0
2
[sin(nπt ) − nπt cos(nπt )] 10
2
n π
2
= −[2/(nπ)]cos(nπ) = 2(−1)n+1/(nπ)
f(t) =
2
π
( −1) n + 1
sin( nπt )
n
n =1
∞
∑
Chapter 17, Solution 24.
(a)
This is an odd function.
ao = 0 = an, T = 2π, ωo = 2π/T = 1
bn =
4
T
∫
T/2
0
f ( t ) sin(ωo nt )dt
f(t) = 1 + t/π,
bn =
4
2π
∫
π
0
0<t<π
(1 + t / π) sin( nt )dt
π
=
2 1
1
t

− cos( nt ) + 2 sin( nt ) −
cos( nt )

π n
n π
nπ
0
= [2/(nπ)][1 − 2cos(nπ)] = [2/(nπ)][1 + 2(−1)n+1]
a2 = 0, b2 = [2/(2π)][1 + 2(−1)] = −1/π = −0.3183
(b)
ωn = nωo = 10 or n = 10
a10 = 0, b10 = [2/(10π)][1 − cos(10π)] = −1/(5π)
Thus the magnitude is A10 =
and the phase is
2
a 210 + b10
= 1/(5π) = 0.06366
φ10 = tan−1(bn/an) = −90°
∞
(c)
f(t) =
2
∑ nπ [1 − 2 cos(nπ)] sin(nt ) π
n =1
f(π/2) =
∞
2
∑ nπ [1 − 2 cos(nπ)] sin(nπ / 2) π
n =1
For n = 1,
f1 = (2/π)(1 + 2) = 6/π
For n = 2,
f2 = 0
For n = 3,
f3 = [2/(3π)][1 − 2cos(3π)]sin(3π/2) = −6/(3π)
For n = 4,
f4 = 0
For n = 5,
f5 = 6/(5π), ----
Thus, f(π/2) = 6/π − 6/(3π) + 6/(5π) − 6/(7π) --------= (6/π)[1 − 1/3 + 1/5 − 1/7 + --------]
f(π/2) ≅ 1.3824
which is within 8% of the exact value of 1.5.
(d)
From part (c)
f(π/2) = 1.5 = (6/π)[1 − 1/3 + 1/5 − 1/7 + - - -]
(3/2)(π/6) = [1 − 1/3 + 1/5 − 1/7 + - - -]
or π/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, ωo = 2π/3.
bn =
4
T
∫
T/2
0
f ( t ) sin( nωo t )dt =
4 1
t sin(2πnt / 3)dt
3 ∫0
1
3t
4 9
 2πnt 
 2πnt 
=
cos
sin
−

2 2

3  4π n
 3  2nπ
 3  0
=
f(t) =
3
4 9
 2 πn  
 2 πn 
cos
sin

−
2 2

3  4π n
 3 
 3  2nπ
∞
 3
∑ π n
n =1
2

2
2
 2 πn 
 2π n    2π t 
cos
sin
−

  sin
 3  nπ
 3   3 
Chapter 17, Solution 26.
T = 4, ωo = 2π/T = π/2
ao =
1 1
1 T
1 dt +
=
f
(
t
)
dt
4  ∫0
T ∫0
an =
2 T
f ( t ) cos( nωo t )dt
T ∫0
an =
2 2
1 cos( nπt / 2)dt +
4  ∫1
∫
3
1
∫
3
2
2 dt + ∫ 1 dt  = 1

3
4
2 cos( nπt / 2)dt + ∫ 1 cos( nπt / 2)dt 

3
4
2
3
4
2
nπt 
nπt
2
nπt
4
sin
+
= 2  sin
sin
+

2 3 
2 2 nπ
2 1 nπ
 nπ
=
4
nπ
nπ 
 3nπ
sin 2 − sin 2 
bn =
2 T
f ( t ) sin( nωo t )dt
T ∫0
=
nπt
2 2
dt +
1 sin
∫

1
2
4
∫
3
2
2 sin
nπt
dt +
2
∫
4
3
1 sin
nπt 
dt 
2

2
3
4
 2
nπt 
nπt
2
nπt
4
cos
−
cos
−
cos
= 2−

2 3 
2 2 nπ
2 1 nπ
 nπ
=
4
[cos(nπ) − 1]
nπ
Hence
f(t) =
1+
∞
4
∑ nπ [(sin( 3nπ / 2) − sin(nπ / 2)) cos( nπt / 2) + (cos( nπ) − 1) sin(nπt / 2)]
n =1
Chapter 17, Solution 27.
(a)
(b)
odd symmetry.
ao = 0 = an, T = 4, ωo = 2π/T = π/2
f(t)
= t, 0 < t < 1
= 0,
1<t<2
1
nπt 
nπt 2 t
nπt
4 1
 4
bn =
cos
dt =  2 2 sin
t sin
−
∫
2  0
2
nπ
2
4 0
n π
=
nπ
nπ
2
4
−
−0
cos
sin
2
2
2
nπ
n π
2
= 4(−1)(n−1)/2/(n2π2),
n = odd
−2(−1)n/2/(nπ),
n = even
a3 = 0, b3 = 4(−1)/(9π2) = 0.045
(c)
b1 = 4/π2, b2 = 1/π, b3 = −4/(9π2), b4 = −1/(2π), b5 = (25π2)
Frms =
a 2o +
1
∑ (a 2n + b 2n )
2
Frms2 = 0.5Σbn2 = [1/(2π2)][(16/π2) + 1 + (16/(8π2)) + (1/4) + (16/(625π2))]
= (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
∫ 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, ωo = 2π/2 = π
an =
4
T
∫
T/2
0
f ( t ) cos( nωo t )dt =
4 1
( 2 − 2 t ) cos( nπt )dt
2 ∫0
1
t
1
1

= 4  sin( nπt ) − 2 2 cos( nπt ) −
sin( nπt )
nπ
n π
 nπ
0
= [4/(n2π2)][1 − cos(nπ)] =
8/(n2π2),
0,
n = odd
n = even
1
bn = 4 ∫ (1 − t ) sin( nπt )dt
0
1
t
1
 1

= 4 −
cos( nπt )
cos( nπt ) − 2 2 sin( nπt ) +
nπ
n π
 nπ
0
= 4/(nπ), n = odd
f(t) =
∞

∑  n
k =1
8
4

cos( nπt ) +
sin(nπt )  , n = 2k − 1
2
nπ
π

2
Chapter 17, Solution 29.
This function is half-wave symmetric.
T = 2π, ωo = 2π/T = 1, f(t) = −t, 0 < t < π
For odd n,
an =
2
T
bn =
2
π
∫
π
∫
π
0
0
( − t ) cos( nt )dt = −
( − t ) sin( nt )dt = −
2
[cos(nt ) + nt sin(nt )] 0π = 4/(n2π)
2
n π
2
[sin(nt ) − nt cos(nt )] 0π = −2/n
2
n π
Thus,
∞
1
 2

f(t) = 2∑  2 cos( nt ) − sin(nt ) ,
n

k =1  n π
n = 2k − 1
Chapter 17, Solution 30.
1
cn =
T
(a)
T/2
∫
f ( t )e − jnωo t dt =
−T / 2
T/2
1  T/2
f ( t ) cos nω o tdt − j∫
f ( t ) sin nω o tdt 
∫

−T / 2
T  −T / 2

The second term on the right hand side vanishes if f(t) is even. Hence
cn =
(b)
(1)
2
T
T/2
∫ f (t ) cos nωo tdt
0
The first term on the right hand side of (1) vanishes if f(t) is odd. Hence,
j2
cn = −
T
T/2
∫ f (t ) sin nωo tdt
0
Chapter 17, Solution 31.
If h ( t ) = f (αt ),
T' = T / α
an '=
Let αt = λ, ,
2π
2π
= αωo
=
T' T / α
T'
T'
0
0
T
αT ' = T
2α
f (λ) cos nωo λdλ / α = a n
T ∫
0
Similarly,
ωo ' =
2
2
h ( t ) cos nωo ' tdt = ∫ f (αt ) cos nωo ' tdt
∫
T'
T'
d t = dλ / α ,
an '=

→
bn ' = bn
Chapter 17, Solution 32.
When is = 1 (DC component)
i = 1/(1 + 2) = 1/3
ωn = 3n, Is = 1/n2∠0°
For n ≥ 1,
I = [1/(1 + 2 + jωn2)]Is = Is/(3 + j6n)
1
∠0°
2
1
n
=
∠ − tan(2n )
=
3 1 + 4n 2 ∠ tan −1 (6n / 3) 3n 2 1 + 4n 2
Thus,
i(t) =
1
+
3
∞
∑
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
jnπVo
=0
+ o +
10
j2nπ
4

5 

1 + j 2.5nπ −  Vo = Vs
nπ  


Vo =
Vs
5 

1 + j 2.5nπ − 
nπ 

A n ∠Θ n =
An =
4
nπ
1
5 

1 + j 2.5nπ − 
nπ 

=
4
nπ + j(2.5n 2 π 2 − 5)
 2.5n 2 π 2 − 5 

; Θ n = − tan −1 


2 2
2 2
2
n
π
n π + (2.5n π − 5)


4
v o (t) =
∞
∑ A n sin(nπt + Θ n ) V
n =1
Chapter 17, Solution 34.
For any n, V = [10/n2]∠(nπ/4), ω = n.
1 H becomes jωnL = jn and 0.5 F becomes 1/(jωnC) = −j2/n
2Ω
jn
+
+
−
V
−j2/n
Vo
−
Vo = {−j(2/n)/[2 + jn − j(2/n)]}V = {−j2/[2n + j(n2 − 2)]}[(10/n2)∠(nπ/4)]
=
=
20∠((nπ / 4) − π / 2)
n
2
4n + (n 2 − 2) 2 ∠ tan −1 ((n 2 − 2) / 2n )
2
20
n
2
n +4
2
vo(t) =
∠[(nπ / 4) − (π / 2) − tan −1 ((n 2 − 2) / 2n )]
∞
∑
n =1

nπ π
n2 −
cos nt +
− − tan −1
4
2
2n
n2 + 4

20
n2
2


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).
1Ω
1H
+
vS
+
−
1F
1Ω
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, ωo = 2π/3.
vs(t)
ao =
an =
=
= 1 for all 0 < t < 1
= 2 for all 1 < t < 1.5
2 1
1dt +
3  ∫0
4
2dt  =
 3
1.5
∫
1
4 1
cos(2nπt / 3)dt +
3  ∫0
1.5
∫
1
2 cos(2nπt / 3)dt 

4 3
6
2
1
1.5 
sin(
2
n
t
/
3
)
π
+
π
sin(
2
n
t
/
3
)
=
−
sin(2nπ / 3)
0
1

3  2nπ
2nπ
nπ
4 2 ∞ 1
vs(t) =
− ∑ sin(2nπ / 3) cos(2nπt / 3)
3 π n =1 n
Now consider this circuit,
1Ω
j2nπ/3
+
vS
+
−
-j3/(2nπ)
1Ω
vo
−
Let Z = [-j3/(2nπ)](1)/(1 – j3/(2nπ)) = -j3/(2nπ - j3)
Therefore, vo = Zvs/(Z + 1 + j2nπ/3). Simplifying, we get
vo =
− j9 v s
12nπ + j( 4n 2 π 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)
3 ∞
 2nπt

vo(t) =  + ∑ A n cos
+ Θ n   volts
 3

 8 n =1
where A n =
6
sin( 2nπ / 3)
nπ
3 
 nπ
−
and Θ n = 90 o − tan −1 

2
3
2
n
π
2 2



 4n π
− 6 
16n 2 π 2 + 

 3
where we can further simplify An to this, A n =
9 sin( 2nπ / 3)
nπ 4n 4 π 4 + 81
Chapter 17, Solution 36.
∞
vs(t) =
∑A
n =1
n = odd
n
cos( nt − θ n )
where θn = tan−1[(3/(nπ))/(−1/(nπ))] = tan−1(−3) = 100.5°
An =
πn
nπ
1
9
1
+ 2 2 sin 2
=
9 + sin 2
2
2
2
nπ
n π
n π
2
ωn = n and 2 H becomes jωnL = 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∠−θn
Vo = j2n An∠−θn/(1 + j4n)
Io = Vo/j = [2n An∠−θn]/ 1 + 16n 2 ∠tan−14n
1 
nπ 
 9 + sin 2
2n
nπ 
2 
∠−100.5° − tan−14n
=
2
1 + 16n
Since sin(nπ/2) = (−1)(n−1)/2 for n = odd, sin2(nπ/2) = 1
2 10
∠ − 100.5° − tan −1 4n
π
Io =
1 + 16n 2
io(t) =
2 10 ∞
∑
π n =1
n = odd
1
1 + 16n
2
cos(nt − 100.5° − tan −1 4n )
Chapter 17, Solution 37.
From Example 15.1,
vs(t) = 5 +
20 ∞ 1
∑ sin(nπt ),
π k =1 n
n = 2k − 1
For the DC component, the capacitor acts like an open circuit.
Vo = 5
For the nth harmonic,
Vs = [20/(nπ)]∠0°
10 mF becomes 1/(jωnC) = −j/(nπx10x10−3) = −j100/(nπ)
100
5
Vs
100∠ − 90° + tan −1
−
j
100
20
nπ
nπ
=
=
vo =
2
2
100
nπ 25 + n π
− j
+ 20 20nπ − j100 nπ
nπ
− j
vo(t) =
100
π
∑
1
n 25 + n 2 π 2
sin(nπt − 90° + tan −1
5
)
nπ
Chapter 17, Solution 38.
1 2 ∞ 1
v s ( t ) = + ∑ sin nπt ,
2 π k =1n
Vo =
jω n
Vs ,
1 + jω n
For dc, ω n = 0,
Vo =
ω n = nπ
Vs = 0.5,
For nth harmonic, Vs =
n = 2k + 1
Vo = 0
2
∠ − 90 o
nπ
2
2∠ − tan −1 nπ
∠90 o =
1 + n 2 π 2 ∠ tan −1 nπ nπ
1 + n 2π2
nπ∠90 o
v o (t) =
∞
∑
k =1
2
2 2
1+ n π
•
cos(nπt − tan −1 nπ),
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 ∞ 1
vs(t) = 5 +
n = 2k − 1
∑ sin(nπt ),
π k =1 n
T = 2, ωo = 2π//T = π, ωn = nωo = nπ
For the DC component,
For the kth harmonic,
io = 5/(20 + 40) = 1/12
Vs = (10/(nπ))∠0°
100 mH becomes jωnL = jnπx0.1 = j0.1nπ
50 mF becomes 1/(jωnC) = −j20/(nπ)
I 20 Ω
VS
40 Ω
Io
−j20/(nπ)
+
−
j0.1nπ
Z
j20
( 40 + j0.1nπ)
n
π
Let Z = −j20/(nπ)||(40 + j0.1nπ) =
j20
−
+ 40 + j0.1nπ
nπ
−
− j20( 40 + j0.1nπ
2nπ − j800
=
2 2
− j20 + 40nπ + j0.1n π
40nπ + j(0.1n 2 π 2 − 20)
=
Zin = 20 + Z =
I =
802nπ + j( 2n 2 π 2 − 1200)
40nπ + j(0.1n 2 π 2 − 20)
Vs
400nπ + j( n 2 π 2 − 200)
=
Z in
nπ[802nπ + j( 2n 2 π 2 − 1200)]
j20
I
nπ
−
Io =
=
=
j20
−
+ ( 40 + j0.1nπ)
nπ
− j20I
40nπ + j(0.1n 2 π 2 − 20)
=
− j200
nπ[802nπ + j( 2n 2 π 2 − 1200)]
200∠ − 90° − tan −1{(2n 2 π 2 − 1200) /(802nπ)}
nπ (802) 2 + ( 2n 2 π 2 − 1200) 2
Thus
io(t) =
where
1
200
+
π
20
∞
∑I
k =1
θ n = 90° + tan −1
In =
n
sin(nπt − θ n ) ,
2n 2 π 2 − 1200
802nπ
1
n (804nπ) + (2n 2 π 2 − 1200)
2
n = 2k − 1
Chapter 17, Solution 40.
T = 2, ωo = 2π/T = π
1
ao =
T
an =
2
T
1

1 1
t2 
v
(
t
)
dt
(
2
2
t
)
dt
t
= 1/ 2
=
−
=
−

∫0
2 ∫0
2  0

T
T
∫ v( t ) cos(nπt )dt =
0
1
∫ 2(1 − t ) cos(nπt )dt
0
1
1
t
1

sin( nπt )
= 2  sin( nπt ) − 2 2 cos( nπt ) −
n π
nπ
 nπ
0
2
= 2 2 (1 − cos nπ) =
n π
bn =
n = even
0,
4
4
, n = odd = 2
2
n π
π ( 2n − 1) 2
2
1
2 T
v ( t ) sin( nπt )dt = 2 ∫ (1 − t ) sin( nπt )dt
∫
0
T 0
1
1
t
2
 1

cos( nπt ) − 2 2 sin( nπt ) +
cos( nπt ) =
= 2−
n π
nπ
 nπ
 0 nπ
vs(t) =
1
+
2
∑A
where φn = tan −1
n
cos( nπt − ϕ n )
π( 2n − 1) 2
, An =
2n
4
16
+ 4
2
n π
π ( 2n − 1) 4
2
For the DC component, vs = 1/2. As shown in Figure (a), the capacitor acts
like an open circuit.
1Ω
0.5V
+
−
Vx
i
− +
2Vx
+
Vx
−
(a)
Vo
+
3Ω
Vo
−
1Ω
Vx
− +
2Vx
Vo
+
VS
+
−
Vo
3Ω
(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).
ωn = nπ, Vs = An∠–φ, 1/(jωnC) = –j4/(nπ)
At the supernode,
(Vs – Vx)/1 = –[nπ/(j4)]Vx + Vo/3
Vs = [1 + jnπ/4]Vx + Vo/3
But
(3)
–Vx – 2Vx + Vo = 0 or Vo = 3Vx
Substituting this into (3),
Vs = [1 + jnπ/4]Vx + Vx = [2 + jnπ/4]Vx
= (1/3)[2 + jnπ/4]Vo = (1/12)[8 + jnπ]Vo
Vo = 12Vs/(8 + jnπ) =
Vo =
12
64 + n π
2
2
12A n ∠ − φ
64 + n 2 π 2 ∠ tan −1 (nπ / 8)
4
16
+ 4
∠[tan −1 (nπ / 8) − tan −1 (π(2n − 1) /(2n ))]
2
4
n π
π (2n − 1)
2
Thus
vo(t) =
where
Vn =
3
+
4
∞
∑V
n =1
n
cos( nπt + θ n )
12
64 + n 2 π 2
4
16
+ 4
2
n π
π ( 2n − 1) 4
2
θn = tan–1(nπ/8) – tan–1(π(2n – 1)/(2n))
Chapter 17, Solution 41.
For the full wave rectifier,
T = π, ωo = 2π/T = 2, ωn = nωo = 2n
Hence
vin(t) =
2 4 ∞
1
− ∑ 2
cos (2nt )
π π n =1 4n − 1
For the DC component,
Vin = 2/π
The inductor acts like a short-circuit, while the capacitor acts like an open circuit.
Vo = Vin = 2/π
For the nth harmonic,
Vin = [–4/(π(4n2 – 1))]∠0°
2 H becomes jωnL = j4n
0.1 F becomes 1/(jωnC) = –j5/n
Z = 10||(–j5/n) = –j10/(2n – j)
Vo = [Z/(Z + j4n)]Vin = –j10Vin/(4 + j(8n – 10))
= −

j10
4∠0° 
 −

4 + j(8n − 10)  π(4n 2 − 1) 
=
40∠{90° − tan −1 (2n − 2.5)}
π(4n 2 − 1) 16 + (8n − 10) 2
Hence
vo(t) =
2
+
π
∞
∑A
n =1
n
cos( 2nt + θ n )
where
An =
20
π( 4n 2 − 1) 16n 2 − 40n + 29
θn = 90° – tan–1(2n – 2.5)
Chapter 17, Solution 42.
20 ∞ 1
vs = 5 +
∑ sin nπt, n = 2k - 1
π k =1n
Vs − 0
= jω n C(0 − Vo )
R
→
Vo =
j
Vs , ω n = nω o = nπ
ω n RC
For n = 0 (dc component), Vo=0.
For the nth harmonic,
1∠90 o 20
20
10 5
o
Vo =
∠ − 90 =
=
nπRC nπ
n 2 π 2 x10 4 x 40 x10 −9 2n 2 π 2
Hence,
v o (t) =
10 5 ∞ 1
∑
2π 2 k =1 n 2
cos nπt , n = 2k - 1
Alternatively, we notice that this is an integrator so that
v o (t) = −
1
10 5 ∞ 1
v
dt
=
∑ cos nπt, n = 2k - 1
s
RC ∫
2π 2 k =1n 2
Chapter 17, Solution 43.
a 02 +
(a)
Vrms =
(b)
Irms =
(c)
P = VdcIdc +
1 ∞ 2
1
(a n + b 2n ) = 30 2 + (20 2 + 10 2 ) = 33.91 V
∑
2 n =1
2
1
6 2 + (4 2 + 2 2 ) = 6.782 A
2
1
∑ Vn I n cos(Θ n − Φ 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 = 27.19 + 3.535 + 0 = 30.73 W
2
(a)
p = vi =
(b)
The power spectrum is shown below.
p
27.19
3.535
0
1
2
3
ω
Chapter 17, Solution 45.
ωn = 1000n
jωnL = j1000nx2x10–3 = j2n
1/(jωnC) = –j/(1000nx40x10–6) = –j25/n
Z = R + jωnL + 1/(jωnC) = 10 + j2n – j25/n
I = V/Z
For n = 1, V1 = 100, Z = 10 + j2 – j25 = 10 – j23
I1 = 100/(10 – j23) = 3.987∠73.89°
For n = 2, V2 = 50, Z = 10 + j4 – j12.5 = 10 – j8.5
I2 = 50/(10 – j8.5) = 3.81∠40.36°
For n = 3, V3 = 25, Z = 10 + j6 – j25/3 = 10 – j2.333
I3 = 25/(10 – j2.333) = 2.435∠13.13°
Irms = 0.5 3.987 2 + 3.812 + 2.435 2 = 3.014 A
1 3
p = VDCIDC + ∑ Vn I n cos(θ n − φ n )
2 n =1
= 0 + 0.5[100x3.987cos(73.89°) + 50x3.81cos(40.36°)
+ 25x2.435cos(13.13°)]
= 0.5[110.632 + 145.16 + 59.28] = 157.54 watts
Chapter 17, Solution 46.
(a)
This is an even function
Irms =
f(t) =
1 T 2
f ( t )dt =
T ∫0
2 − 2t,
0,
2 T/2 2
f ( t )dt
T ∫0
0 < t <1
1< t < 2
T = 4, ωo = 2π/T = π/2
Irms2 =
1
2 1
4(1 − t ) 2 dt = 2( t − t 2 + t 3 / 3)
∫
0
4 0
= 2(1 – 1 + 1/3) = 2/3 or
Irms
(b)
= 0.8165 A
From Problem 16.14,
an = [8/(n2π2)][1 – cos(nπ/2)], ao = 0.5
a1 = 8/π2, a2 = 4/π2, a3 = 8/(9π2), a4 = 0, a5 = 9/(25π2), a6 = 4/(9π2)
Irms =
ao +
1 ∞ 2
∑ An ≅
2 n =1
64 64 16 
1
1 
+ 4  64 + 16 +
+
+  = 0.66623
81 625 81 
4 2π 
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
5Ω
10 Ω
Is
For AC, ω = 100
For Is = 0.5∠–60°
jωL = j100x80x10–3 = j8
In = 5Is/(5 + 10 + j8)
I1 = 10∠–60°/(15 + j8) or |I1| = 10/ 15 2 + 8 2
For Is = 0.5∠–120°
I2 = 2.5∠–120°/(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,
ωn = 10n, n = 1,2
1/(jωnC) = –j/(10nx100x10–6) = (–j/n) kΩ
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∠45° = 14.311∠–18.43° mV
For n = 2,
V2 = [–j2/(4 – j)]12∠–60° = 5.821∠–135.96° mV
v(t) = 40 + 0.014311cos(10t – 18.43°) + 0.005821cos(20t – 135.96°) V
(b)
p = VDCIDC +
1 ∞
∑ Vn I n cos(θ n − φ n )
2 n =1
= 20x40 + 0.5x10x0.014311cos(45° + 18.43°)
+0.5x12x0.005821cos(–60° + 135.96°)
= 800.1 mW
Chapter 17, Solution 49.
(a)
T
2π 
π
1 2
1 
1
 ∫ 1dt + ∫ 4dt  =
Z rms = ∫ z ( t )dt =
(5π) = 2.5
T
2π 
2π

0
π

0
2
Z rms = 1.581
(b)
Z 2 rms = a 2 o +
1 ∞ 2
1 1 ∞ 36
1
1
2
(
a
b
)
+
=
+ ∑
= +
n
n
∑
2 n =1
4 2 n =1n 2 π 2 4 18π 2
Z rms = 1.7086
(c )
 1.7086 
%error = 
− 1 x100 = 8.071

 1.581
1

 1 1 1
+ ... = 2.9193
1 + + + +

 4 9 16 25
Chapter 17, Solution 50.
cn =
=
1
T
∫
T
0
f ( t )e − jωo nt dt,
ωo =
2n
=π
1
1 1 − jnπt
te
dt
2 ∫−
Using integration by parts,
u = t and du = dt
dv = e–jnπtdt which leads to v = –[1/(2jnπ)]e–jnπt
t
e − jnπt
cn = −
2 jnπ
1
+
−1
[
1 1 − jnπt
e
dt
2 jnπ ∫−1
]
j − jnπ
1
e
e − jnπt
=
+ e jnπt +
2 2
2
nπ
2n π ( − j)
1
−1
= [j/(nπ)]cos(nπ) + [1/(2n2π2)](e–jnπ – ejnπ)
j( −1) n
2j
j( −1) n
cn =
sin( nπ) =
+
nπ
2n 2 π 2
nπ
Thus
∞
f(t) =
∑ c n e jnωot =
n = −∞
∞
∑ ( −1)
n = −∞
n
j jnπt
e
nπ
Chapter 17, Solution 51.
T = 2,
ωo = 2π / T = π
T
cn =
cn =
(
2
)
1
1 2 − jnπt
1 e − jnπt
2
− jnωo t
f
(
t
)
e
dt
=
t
e
dt
=
− n 2 π 2 t 2 + 2 jnπt + 2 0
T∫
2∫
2 (− jnπ) 3
0
0
1
j2n 3 π 3
f (t) =
(−4n 2 π 2 + j4nπ) =
∞
∑
n = −∞ n
2
2 2
π
2
n 2π2
(1 + jnπ)e jnπt
(1 + jnπ)
Chapter 17, Solution 52.
cn =
=
1
T
∫
T
0
f ( t )e − jωo nt dt,
ωo =
2n
=π
1
1 1 − jnπt
te
dt
2 ∫−
Using integration by parts,
u = t and du = dt
dv = e–jnπtdt which leads to v = –[1/(2jnπ)]e–jnπt
t
cn = −
e − jnπt
2 jnπ
1
1 1 − jnπt
e
dt
2 jnπ ∫−1
+
−1
[
]
j − jnπ
1
=
+ e jnπt +
e
e − jnπt
2 2
2
nπ
2n π ( − j)
1
−1
= [j/(nπ)]cos(nπ) + [1/(2n2π2)](e–jnπ – ejnπ)
cn =
j( −1) n
2j
j( −1) n
+
sin(
n
π
)
=
nπ
2n 2 π 2
nπ
Thus
f(t) =
∞
∑c e
n
jnωo t
=
n = −∞
∞
∑ ( −1)
n = −∞
n
j jnπt
e
nπ
Chapter 17, Solution 53.
ωo = 2π/T = 2π
cn =
∫
T
0
1
e − t e − jnωo t dt = ∫ e −(1+ jnωo ) t dt
0
−1
e − (1 + j2 nπ ) t
=
1 + j2πn
1
=
0
[
= [1/(j2nπ)][1 – e–1(cos(2πn) – jsin(2nπ))]
= (1 – e–1)/(1 + j2nπ) = 0.6321/(1 + j2nπ
0.6321e j2 nπt
f(t) = ∑
n = −∞ 1 + j2nπ
∞
]
−1
e − (1 + j 2 n π ) − 1
1 + j2nπ
Chapter 17, Solution 54.
T = 4, ωo = 2π/T = π/2
cn =
1
T
∫
T
0
f ( t )e − jωo nt dt
=
1  1 − jnπt / 2
2e
dt +
4  ∫0
=
j
2e − jnπ / 2 − 2 + e − jnπ − e − jnπ / 2 − e − j2 nπ + e − jnπ
2nπ
=
j
3e − jnπ / 2 − 3 + 2e − jnπ
2nπ
f(t) =
∫
2
1
1e − jnπt / 2 dt −
∫
4
2
1e − jnπt / 2 dt 

[
[
∞
∑c e
n
]
]
jnωo t
n = −∞
Chapter 17, Solution 55.
T = 2π, ωo = 2π/T = 1
1
T
cn =
But
i(t) =
cn =
1
2π
∫
π
0
∫
T
0
i( t )e − jnωo t dt
sin( t ),
0,
0<t<π
π < t < 2π
sin( t )e − jnπt dt =
1
2π
∫
π
0
1 jt
(e − e − jt )e − jnt dt
2j
π
1  e jt (1 − n )
e − jt (1 + n ) 
=
+
4πj  j(1 − n ) j(1 + n )  0
=−
1  e jπ (1− n ) − 1 e − jπ ( n + 1) − 1
+
4  1 − n
1 + n 
=
[
1
e jπ (1 − n ) − 1 + ne jπ (1 − n ) − n + e − jπ (1 + n ) − 1 − ne − jπ (1+ n ) + n
2
4π( n − 1)
]
But ejπ = cos(π) + jsin(π) = –1 = e–jπ
1 + e − jnπ
1
− jnπ
− jnπ
− jnπ
− jnπ
cn =
−e
−e
− ne
+ ne
−2 =
2 π(1 − n 2 )
4 π( n 2 − 1)
[
]
Thus
i(t) =
∞
1 + e − jnπ
∑ 2π(1 − n
n = −∞
2
)
e jnπt
Chapter 17, Solution 56.
co = ao = 10, ωo = π
co = (an – jbn)/2 = (1 – jn)/[2(n2 + 1)]
∞
f(t) = 10 +
(1 − jn) jnπt
e
2
+ 1)
∑ 2(n
n = −∞
n≠0
Chapter 17, Solution 57.
ao = (6/–2) = –3 = co
cn = 0.5(an –jbn) = an/2 = 3/(n3 – 2)
∞
f(t) = − 3 +
∑n
n = −∞
n≠0
3
3
e j50nt
−2
Chapter 17, Solution 58.
cn = (an – jbn)/2, (–1) = cos(nπ), ωo = 2π/T = 1
cn = [(cos(nπ) – 1)/(2πn2)] – j cos(nπ)/(2n)
Thus
f(t) =
π
cos(nπ )  jnt
 cos(nπ ) − 1
−j
+ ∑
e
2
4
2n 
 2πn
Chapter 17, Solution 59.
For f(t), T = 2π, ωo = 2π/T = 1.
ao = DC component = (1xπ + 0)/2π = 0.5
For
h(t), T = 2, ωo = 2π/T = π.
ao = (3x1 – 2x1)/2 = 0.5
Thus by replacing ωo = 1 with ωo = π and multiplying the magnitude by five,
we obtain
∞
1
j5e j( 2n +1) πt
h(t) =
− ∑
2 n = −∞ ( 2n + 1)π
n≠0
Chapter 17, Solution 60.
From Problem 16.17,
ao = 0 = an, bn = [2/(nπ)][1 – 2 cos(nπ)], co = 0
cn = (an – jbn)/2 = [j/(nπ)][2 cos(nπ) – 1], n ≠ 0.
Chapter 17, Solution 61.
ωo = 1.
(a)
f(t) = ao +
∑A
n
cos(nω o t − φ n )
= 6 + 4cos(t + 50°) + 2cos(2t + 35°)
+ cos(3t + 25°) + 0.5cos(4t + 20°)
= 6 + 4cos(t)cos(50°) – 4sin(t)sin(50°) + 2cos(2t)cos(35°)
– 2sin(2t)sin(35°) + cos(3t)cos(25°) – sin(3t)sin(25°)
+ 0.5cos(4t)cos(20°) – 0.5sin(4t)sin(20°)
= 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 ∞ 2
∑ An
2 n =1
a o2 +
frms2 = 62 + 0.5[42 + 22 + 12 + (0.5)2] = 46.625
frms = 6.828
Chapter 17, Solution 62.
(a) ωo = 20 = 2π / T
(b) f ( t ) = a o +

→
T=
2π
= 0.3141s
20
∞
∑ A n cos(nωo t + φ n ) = 3 + 4 cos(20t + 90 o ) +5.1cos(40t + 90 o ) + ...
n =1
f ( t ) = 3 − 4 sin 20 t − 5.1sin 40t − 2.7 sin 60 t − 1.8 sin 80t − ....
Chapter 17, Solution 63.
This is an even function.
T = 3, ωo = 2π/3, bn = 0.
f(t) =
ao =
an =
1, 0 < t < 1
2, 1 < t < 1.5
1.5
2 T/2
2 1
f ( t )dt =  ∫ 1dt + ∫ 2 dt  = (2/3)[1 + 1] = 4/3
∫

1
T 0
3  0
1.5
4 T/2
4 1
f ( t ) cos(nωo t )dt =  ∫ 1cos(2nπt / 3)dt + ∫ 2 cos(2nπt / 3)dt 
∫

1
T 0
3  0
4 3
6
 2nπt 
 2nπt 
= 
sin 
sin 
 +

3  2nπ  3  0 2nπ  3  1
1
= [–2/(nπ)]sin(2nπ/3)
1.5



4 2 ∞ 1  3nπ   2nπt 
f2(t) = − ∑ sin 
 cos

3 π n =1 n  3   3 
ao = 4/3 = 1.3333, ωo = 2π/3, an = –[2/(nπ)]sin(2nπt/3)
An =
a 2n + b 2n =
2
 2nπ 
sin 

nπ  3 
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
2π
4π
2π
4π
6π
ω
φn
0
6π
ω
-180o
Chapter 17, Solution 65.
an = 20/(n2π2), bn = –3/(nπ), ωn = 2n
An = a 2n + b 2n =
=
400
9
+ 2 2
4 4
n π
n π
3
44.44
1 + 2 2 , n = 1, 3, 5, 7, 9, etc.
nπ
n π
n
1
3
5
7
9
An
2.24
0.39
0.208
0.143
0.109
φn = tan–1(bn/an) = tan–1{[–3/(nπ)][n2π2/20]} = tan–1(–nx0.4712)
n
1
3
5
7
9
∞
φn
–25.23°
–54.73°
–67°
–73.14°
–76.74°
–90°
2.24
0
2
6
10
14
18
ωn
–30°
–25.23°
An
0.39
0.208
0
2
6
10
–60°
–54.73°
0.0.143 0.109
14
18
ωn
–90°
φn
–67°
–73.14°
–76.74°
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, ωo = 2π/T = 2π/5
f(t) is an odd function. ao = 0 = an
4 T/2
4 10
f ( t ) sin(nωo t )dt = ∫ 10 sin(0.4nπt )dt
∫
T 0
5 0
bn =
1
20
8x 5
[1 − cos(0.4nπ)]
= −
cos(0.4πnt ) =
2 nπ
nπ
0
f(t) =
20 ∞ 1
∑ [1 − cos(0.4nπ)]sin(0.4nπt )
π n =1 n
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 ,
φ = tan–1(bn/an)
A1 =
6 2 + 8 2 = 10,
φ1 = tan–1(6/8) = 36.87°
A2 =
3 2 + 4 2 = 5,
φ2 = tan–1(3/4) = 36.87°
i(t) = {4 + 10cos(100πt – 36.87°) – 5cos(200πt – 36.87°)} 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
-
vs =
vo
-
Aτ 2A ∞ 1
nπτ
+
sin
cos nωo t
∑
T
T n =1n
T
1
jω n C
1
Vs =
Vs ,
Vo =
1
1 + jω n RC
R+
jω n C
For n=0, (dc component), Vo = Vs =
Aτ
T
ω n = nωo = 2nπ / T
(1)
For the nth harmonic,
2A
nπτ
sin
∠ − 90 o
2
2 2
−1
nT
T
1 + ω n R C ∠ tan ω n RC
1
Vo =
When n=1, | Vo |=
2A
nπτ
sin
•
T
T
•
1
1+
(2)
2
4π
R 2C 2
T
From (1) and (2),
Aτ
2A
π
sin
= 50 x
T
T
10
1+
1
1+
4π 2 2 2
R C
T
4π 2 2 2
R C = 1010
T
→
C=

→
1+
4π 2 2 2 30.9
R C =
= 3.09 x10 4
T
τ
T
10 −2 x 3.09 x10 4
10 5 =
= 24.59 mF
2πR
4πx10 3
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 ∞ 1
vo(t) = 5 + ∑ sin( nπt ) , n = 2k – 1
π k =1 n
T = 2, ωo = 2π/T = 2π, ωn = nωo = 2nπ
jωnL = j2nπ; Z = R||10 = 10R/(10 + R)
Vo = ZVs/(Z + j2nπ) = [10R/(10R + j2nπ(10 + R))]Vs
Vo =
10R∠ − tan −1{(nπ / 5R )(10 + R )}
100R 2 + 4n 2 π 2 (10 + R ) 2
Vs
Vs = [20/(nπ)]∠0°
The source current Is is
20
(10 + R )
Vs
Vs
nπ
Is =
=
=
10R
Z + j2nπ
10R + j2nπ(10 + R )
+ j2nπ
10 + R
20
∠ − tan −1{( nπ / 3)(10 + R )}
nπ
100R 2 + 4n 2 π 2 (10 + R ) 2
(10 + R )
=
1
∑ Vsn I sn cos(θ n − φ n )
2
ps = VDCIDC +
For the DC case, L acts like a short-circuit.
Is =
5(10 + R )
5
, Vs = 5 = Vo
=
10R
10R
10 + R

 −1  π

tan  (10 + R )  

2 (10 + R ) cos

25(10 + R ) 1  20 
5


+  
ps =
10R
2  π 
100R 2 + 4π 2 (10 + R ) 2




 2π

(10 + R ) 2 cos tan −1  (10 + R )  

 5

 10 

+ 
+ "
2
2
2

π
 
100R + 16π (10 + R )


2
ps =
=
VDC 1 ∞ Von
+ ∑
R
2 n =1 R

25 1 
100R
100R
+ 
+
+ "
2
2
2
2
2
2
R 2 100R + 4π (10 + R )
100R + 10π (10 + R )

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 7 25(10 + R )
= x
R 10
10R
100 = 70 + 7R which leads to R = 30/7 = 4.286 Ω
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 ωo = 2.
T = 2π/ωo = 2π/2 = π
(b) The average value is the DC component = –2
Vrms =
(c)
ao +
1 ∞ 2
(a n + b 2n )
∑
2 n =1
1
2
Vrms
= (−2) 2 + (10 2 + 8 2 + 6 2 + 3 2 + 12 ) = 121.5
2
Vrms = 11.02 V
Chapter 17, Solution 78.
2
(a)
2
2
Vn ,rms
VDC
Vn2 VDC
1
p =
+ ∑
=
+∑
R
2
R
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
= 21R = 105
R
VDC = 10.25 V
Chapter 17, Solution 79.
From Table 17.3, it is evident that an = 0,
bn = 4A/[π(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–jnπ]/[2π(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 ∞
1
−
cos(nωo t )
∑
2
π
π 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
∫
0
T
From the Fourier series above
|co| = 2A/2, |cn| = 4A/[π(4n2 – 1)]
3T
n
0
1
2
3
4
ωo
0
2ωo
4ωo
6ωo
8ωo
|cn|
2A/π
2A/(3π)
2A/(15π)
2A/(35π)
2A/(63π)
(c)
81.1%
(d)
0.72%
2|cn|2
4A2/(π2)
8A2/(9π2)
2A2/(225π2)
8A2/(1225π2)
8A2/(3969π2)
Chapter 17, Solution 82.
2
VDC
1 ∞ Vn2
P =
+ ∑
R
2 n =1 R
Assuming V is an amplitude-phase form of Fourier series. But
|An| = 2|Cn|, co = ao
|An|2 = 4|Cn|2
Hence,
∞
c o2
c 2n
+ 2∑
P =
R
n =1 R
Alternatively,
2
Vrms
P =
R
where
2
Vrms
= a o2 +
∞
∞
1 ∞ 2
2
2
A
=
c
+
2
c
=
c 2n
∑
∑
∑
n
o
n
2 n =1
n =1
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 ) = δ( t + 2) − δ( t + 1) − δ( t − 1) + δ( t − 2)
jωF(ω) = e j2 ω − e jω − e − jω + e − jω2
= 2 cos 2ω − 2 cos ω
F(ω) =
2[cos 2ω − cos ω]
jω
Chapter 18, Solution 2.
t,
f (t) = 
0,
0 < t <1
otherwise
f ”(t)
f ‘(t)
1
δ(t)
0
t
t
1
–δ’(t-1)
-δ(t-1)
-δ(t-1)
f"(t) = δ(t) - δ(t - 1) - δ'(t - 1)
Taking the Fourier transform gives
-ω2F(ω) = 1 - e-jω - jωe-jω
F(ω) =
(1 + jω)e jω − 1
ω2
1
or F(ω) = ∫ t e − jωt dt
0
eax
But ∫ x e dx = 2 (ax − 1) + c
a
ax
F(ω) =
e − jω
(− jω)
2
(− jωt − 1) 10 =
[
]
1
(1 + jω)e − jω − 1
2
ω
Chapter 18, Solution 3.
f (t) =
1
t , − 2 < t < 2,
2
1
f ' (t) = , − 2 < t < 2
2
1 jωt
e − jωt
t e dt =
(− jωt − 1) 2− 2
2
−2 2
2(− jω)
F(ω) = ∫
F(ω) =
2
[
]
[
]
=−
1
e − jω2 (− jω2 − 1) − e jω2 ( jω2 − 1)
2
2ω
=−
1
− jω2(e jω2 + e jω2 ) + e jω2 − e − jω2
2
2ω
=−
1
(− jω4 cos 2ω + j2 sin 2ω)
2ω 2
j
(sin 2ω − 2ω cos 2ω)
ω2
Chapter 18, Solution 4.
2δ(t+1)
g’
2
–1
0
1
t
–2
–2δ(t–1)
4δ(t)
2δ’(t+1)
g”
–1
0
–2δ(t+1)
1
t
–2
–2δ(t–1)
–2δ’(t–1)
g ′′ = −2δ( t + 1) + 2δ′( t + 1) + 4δ( t ) − 2δ( t − 1) − 2δ′( t − 1)
( jω) 2 G (ω) = −2e jω + 2 jωe jω + 4 − 2e − jω − 2 jωe − jω
= −4 cos ω − 4ω sin ω + 4
G (ω) =
4
ω2
(cos ω + ω sin ω − 1)
Chapter 18, Solution 5.
h’(t)
1
0
–1
t
1
–2δ(t)
h”(t)
1
δ(t+1)
1
–1
–2δ’(t)
t
0
–δ(t–1)
h ′′( t ) = δ( t + 1) − δ( t − 1) − 2δ′( t )
( jω) 2 H(ω) = e jω − e − jω − 2 jω = 2 j sin ω − 2 jω
H(ω) =
2j 2j
−
sin ω
ω ω2
Chapter 18, Solution 6.
0
F(ω) = ∫ (−1)e
− jωt
−1
1
dt + ∫ te − jωt dt
0
0
1
−1
0
Re F(ω) = − ∫ cos ωtdt + ∫ t cos ωtdt
=−
 1
 1
1
t
1
0
sin ωt −1 + 
cos ωt + sin ωt  0 =
(cos ω − 1)
2
ω
ω
ω2

ω
Chapter 18, Solution 7.
(a)
f1 is similar to the function f(t) in Fig. 17.6.
f 1 ( t ) = f ( t − 1)
Since F(ω) =
2(cos ω − 1)
jω
2e − jω (cos ω − 1)
F1 (ω) = e F(ω) =
jω
jω
Alternatively,
f 1' ( t ) = δ( t ) − 2δ( t − 1) + δ( t − 2)
jωF1 (ω) = 1 − 2e − jω + e − j2 ω = e − jω (e jω − 2 + e jω )
= e − jω (2 cos ω − 2)
F1(ω) =
(b)
2e − jω (cos ω − 1)
jω
f2 is similar to f(t) in Fig. 17.14.
f2(t) = 2f(t)
F2(ω) =
4(1 − cos ω)
ω2
Chapter 18, Solution 8.
1
(a)
F(ω) = ∫ 2e
dt + ∫ (4 − 2 t )e − jωt dt
0
=
1
2 − jωt 1
4 − jωt 2
2 − jωt
2
e
+
e
−
e
(− jωt − 1) 1
1
0
2
− jω
− jω
−ω
F(ω) =
(b)
2
− jωt
2
ω
2
+
2 − jω 2
4 − j2ω 2
e
+
−
e
−
(1 + j2ω)e − j2ω
2
jω
jω jω
ω
g(t) = 2[ u(t+2) – u(t-2) ] - [ u(t+1) – u(t-1) ]
G (ω) =
4 sin 2ω 2 sin ω
−
ω
ω
Chapter 18, Solution 9.
(a)
y(t) = u(t+2) – u(t-2) + 2[ u(t+1) – u(t-1) ]
Y(ω) =
2
4
sin 2ω + sin ω
ω
ω
1
(b) Z(ω) = ∫ (−2 t )e − jωt dt =
0
− 2e − jωt
− ω2
2 2e − j ω
1
(− jωt − 1) 0 =
−
(1 + jω)
2
2
ω
Chapter 18, Solution 10.
(a)
x(t) = e2tu(t)
X(ω) = 1/(2 + jω)
(b)
e − t , t > 0
e −( t ) =  t
e , t < 0
1
0
1
−1
−1
0
Y(ω) = ∫ y( t )e jωt dt = ∫ e t e jωt dt + ∫ e − t e − jωt dt
ω
=
e (1− jω) t
1 − jω
=
 cos ω + jsin ω cos ω − jsin ω 
2
+
− e −1 

2
1 − jω
1 + jω
1+ ω


Y(ω) =
0
−1
+
e − (1+ jω) t
− (1 + jω)
1
0
[
2
1 − e −1 (cos ω − ω sin ω)
2
1+ ω
Chapter 18, Solution 11.
f(t) = sin π t [u(t) - u(t - 2)]
2
F(ω) = ∫ sin πt e − jωt dt =
0
(
)
1 2 j πt
e − e − j πt e − jωt dt
∫
0
2j
=
1  2 + j( − ω + π ) t
+ e − j( ω + π ) t )dt 
(e

2 j  ∫0
=
1 
1
e − j( ω+ π ) t 2 
e − j( ω− π ) t 02 +

0
− j(ω + π) 
2 j  − j(ω − π)
=
1  1 − e − j2 ω 1 − e − j2 ω 


+
2  π − ω
π + ω 
=
1
2π + 2πe − j2 ω
2
2(π − ω )
(
2
F(ω) =
(
)
)
π
e − jω 2 − 1
2
ω −π
2
Chapter 18, Solution 12.
(a)
∞
2
0
0
F(ω) = ∫ e t e − jωt dt = ∫ e (1− jω) t dt
=
1
e (1− jω) t
1 − jω
2
0
=
e 2− jω 2 − 1
1 − jω
]
(b)
0
1
−1
0
H(ω) = ∫ e − jωt dt + ∫ (−1)e − jωt dt
=−
=
(
)
(
)
1
1 − jω
1
1 − e jω +
e −1 =
(−2 + 2 cos ω)
jω
jω
jω
− 4 sin 2 ω / 2
 sin ω / 2 
= jω

jω
 ω/ 2 
2
Chapter 18, Solution 13.
(a) We know that F[cos at ] = π[δ(ω − a ) + δ(ω + a )] .
Using the time shifting property,
F[cos a ( t − π / 3a )] = πe − jωπ / 3a [δ(ω − a ) + δ(ω + a )] = πe − jπ / 3δ(ω − a ) + πe jπ / 3δ(ω + a )
(b) sin π( t + 1) = sin πt cos π + cos πt sin π = − sin πt
g(t) = -u(t+1) sin (t+1)
Let x(t) = u(t)sin t, then X(ω) =
1
( jω) 2 + 1
=
1
1 − ω2
Using the time shifting property,
G (ω) = −
1
1 − ω2
e jω =
e jω
ω2 − 1
(c ) Let y(t) = 1 + Asin at, then Y(ω) = 2πδ(ω) + jπA[δ(ω + a ) − δ(ω − a )]
h(t) = y(t) cos bt
Using the modulation property,
1
H(ω) = [Y(ω + b) + Y(ω − b)]
2
H(ω) = π[δ(ω + b) + δ(ω − b)] +
jπA
[δ(ω + a + b) − δ(ω − a + b) + δ(ω + a − b) − δ(ω − a − b)]
2
4
(d) I(ω) = ∫ (1 − t )e − jωt dt =
0
e − j ωt e − j ωt
1
e − j4ω e − j4ω
4
−
(− jωt − 1) 0 =
−
−
( j4ω + 1)
− jω − ω 2
jω
ω2
ω2
Chapter 18, Solution 14.
(a)
cos(3t + π) = cos 3t cos π − sin 3t sin π = cos 3t (−1) − sin 3t (0) = − cos(3t )
f ( t ) = −e − t cos 3t u ( t )
F(ω) =
− (1 + jω )
(1 + jω)2 + 9
(b)
g(t)
1
-1
1
t
-1
g’(t)
π
-1
1
t
-π
g ' ( t ) = π cos πt[u ( t − 1) − u ( t − 1)]
g" ( t ) = −π 2 g( t ) − πδ( t + 1) + πδ( t − 1)
− ω 2 G (ω) = − π 2 G (ω) − πe jω + πe − jω
(π 2 − ω2 )G(ω) = −π(e jω − e − jω ) = −2 jπ sin ω
G(ω) =
2 jπ sin ω
ω2 − π 2
Alternatively, we compare this with Prob. 17.7
f(t) = g(t - 1)
F(ω) = G(ω)e-jω
G (ω) = F(ω)e jω =
− j2π sin ω
ω2 − π 2
=
G(ω) =
(c)
π
(e − jω − e jω )
2
ω −π
2
2 jπ sin ω
π 2 − ω2
cos π( t − 1) = cos πt cos π + sin πt sin π = cos πt (−1) + sin πt (0) = − cos πt
Let x ( t ) = e −2( t −1) cos π( t − 1)u ( t − 1) = −e 2 h ( t )
and
y( t ) = e −2 t cos(πt )u ( t )
Y(ω) =
2 + jω
(2 + jω) 2 + π 2
y( t ) = x ( t − 1)
Y(ω) = X(ω)e − jω
X(ω) =
(2 + jω)e jω
(2 + jω)2 + π 2
X(ω) = −e 2 H(ω)
H(ω) = −e −2 X(ω)
=
(d)
− (2 + jω)e jω− 2
(2 + jω)2 + π 2
Let x ( t ) = e −2 t sin( −4t )u (− t ) = y(− t )
p( t ) = − x ( t )
where y( t ) = e 2 t sin 4t u ( t )
Y (ω) =
2 + jω
(2 + jω)2 + 4 2
X(ω) = Y(−ω) =
2 − jω
(2 − jω)2 + 16
p(ω) = −X(ω) =
(e)
Q(ω) =
jω − 2
(jω − 2 )2 + 16

8 − jω 2
1 
e
+ 3 − 2 πδ(ω) + e − jω2
jω
jω 

6 jω 2
e + 3 − 2πδ(ω)e − jω 2
jω
Q(ω) =
Chapter 18, Solution 15.
(a)
F(ω) = e j3ω − e − jω3 = 2 j sin 3ω
(b)
Let g( t ) = 2δ( t − 1), G (ω) = 2e − jω
t
F(ω) = F  ∫ g ( t ) dt 
 −∞

=
G (ω)
+ πF(0)δ(ω)
jω
2e − j ω
=
+ 2πδ(−1)δ(ω)
jω
=
(c)
F [δ(2t )] =
F(ω) =
2e − jω
jω
1
⋅1
2
1
1
1 jω
⋅ 1 − jω = −
3
2
3 2
Chapter 18, Solution 16.
(a) Using duality properly
t →
−2
ω2
−2
→ 2π ω
t2
4
→ − 4π ω
t2
or
4
F(ω) = F  2  = − 4π ω
t 
(b)
e
−at
2a
a + ω2
2
2a
a + t2
2π e
2
8
a + t2
2
4π e
−a ω
−2 ω
 8 
−2 ω
= 4π e
G(ω) = F 
2 
4+t 
Chapter 18, Solution 17.
(a)
Since H(ω) = F (cos ω0 t f ( t ) ) =
where F(ω) = F [u (t )] = πδ(ω) +
H(ω) =
1
[F(ω + ω0 ) + F(ω − ω0 )]
2
1
, ω0 = 2
jω
1
1
1 
+ πδ(ω − 2) +
πδ (ω + 2) + (

2
j ω + 2)
j (ω − 2) 
=
π
[δ(ω + 2) + δ(ω − 2)] − j  ω + 2 + ω − 2 
2
2  (ω + 2)(ω − 2) 
H(ω) =
(b)
π
[δ(ω + 2) + δ(ω − 2 )] − 2jω
2
ω −4
G(ω) = F [sin ω0 t f ( t )] =
j
[F(ω + ω0 ) − F(ω − ω0 )]
2
where F(ω) = F [u (t )] = πδ (ω) +
G (ω) =
=
=
1
jω

j
1
1
πδ(ω + 10) +
− πδ(ω − 10) −

2
j(ω + 10)
j(ω − 10 ) 
jπ
[δ(ω + 10) − δ (ω − 10)] + j  j − j 
2
2  ω − 10 ω + 10 
jπ
[δ(ω + 10) − δ(ω − 10 )] − 2 10
2
ω − 100
Chapter 18, Solution 18.
Let f (t ) = e − t u (t )
f (t ) cos t
Hence Y(ω) =
F(ω) =
1
j + jω
1
[F(ω − 1) + F(ω + 1)]
2

1
1
1
+


2 1 + j (ω − 1) 1 + j (ω + 1) 
=
1  1 + jω + j + 1 + jω − j 


2  [1 + j(ω − 1)][1 + j (ω + 1)]
=
1 + jω
1 + jω + j + jω − j − ω 2 + 1
=
1 + jω
2 jω − ω 2 + 2
Chapter 18, Solution 19.
∞
F(ω) = ∫ f ( t )e jωt dt =
−∞
F(ω) =
1 1 j2 πt
(
e + e − j2 πt )e − jωt dt
∫
0
2
[
]
1 1 − j( ω + 2 π ) t
e
+ e − j(ω− 2 π )t dt
∫
0
2
1

1
1
1
e − j( ω − 2 π ) t 
e − j( ω + 2 π ) t +
= 
2  − j (ω + 2π )
− j(ω − 2π )
0
1  e − j( ω+ 2 π ) − 1 e − j( ω− 2 π ) − 1 
=− 
+

2  j (ω + 2π)
j(ω − 2π ) 
But
e j2 π = cos 2π + j sin 2π = 1 = e − j2 π
1  e − jω − 1  1
1 

+
F(ω) = − 

2
j  ω + 2π ω − 2π 
=
(
)
jω
e − jω − 1
2
ω − 4π
2
Chapter 18, Solution 20.
(a)
F (cn) = cnδ(ω)
(
)
F c n e jnωo t = c n δ(ω − nωo )
(b)
 ∞

F  ∑ c n e jnωo t  =
 n = −∞

n = −∞
T = 2π
ωo =
cn =
∞
∑ c δ(ω − nω )
n
o
2π
=1
T
1  π
1 T
− jnt
f (t ) e − jnωo t dt =
 ∫0 1⋅ e dt + 0 
∫
0


2π
T
=
1  1 jnt
− e
2π  jn
π
0

j
 =
(
e − jnπ − 1)
 2πn
But e − jnπ = cos nπ + j sin nπ = cos nπ = (−1) n
cn =
[
]

j
(− 1)n − 1 =  0−,j ,
2nπ
 nπ
n = even
n = odd , n ≠ 0
for n = 0
cn =
1 π
1
1 dt =
∫
0
2π
2
Hence
f (t) =
∞
1
j jnt
− ∑
e
2 n = −∞ nπ
n ≠0
n = odd
F(ω) =
∞
1
j
δω − ∑
δ(ω − n )
2
n = −∞ nπ
n≠0
n = odd
Chapter 18, Solution 21.
Using Parseval’s theorem,
∞
∫− ∞ f
2
( t )dt =
1 ∞
| F(ω) | 2 dω
∫
−
∞
2π
If f(t) = u(t+a) – u(t+a), then
∞
∫−∞
a
f 2 ( t )dt = ∫ (1) 2 dt = 2a =
−a
or
2
4πa π
 sin aω 
∫− ∞  aω  dω = 4a 2 = a as required.
∞
2
1 ∞
 sin aω 
4a 2 
 dω
∫
2 π −∞
 aω 
Chapter 18, Solution 22.
F [f ( t ) sin ωo t ] = ∫ f ( t )
∞
(e
−∞
=
jω o t
)
− e − j ω o t − j ωt
e dt
2j
∞
1 ∞
f ( t )e − j(ω− ωo )t dt − ∫ e − j(ω+ ωo )t dt 
∫


−
∞
−
∞
2j 
1
[F(ω − ω o ) − F(ω + ωo )]
2j
=
Chapter 18, Solution 23.
(a) f(3t) leads to
F [f (− 3t )] =
1
10
30
⋅
=
3 (2 + jω / 3)(5 + jω / 3) (6 + jω)(15 + jω)
30
(6 − jω)(15 − jω)
(b) f(2t)
1
10
20
⋅
=
2 (2 + jω / 2)(15 + jω / 2) (4 + jω)(10 + jω)
20e − jω / 2
(4 + jω)(10 + jω)
f(2t-1) = f [2(t-1/2)]
1
1
F(ω + 2) + F(ω + 2 )
2
2
(c) f(t) cos 2t
5
=
[2 + j(ω + 2)][5 + j(ω + 2)]
(d) F [f ' (t )] = jω F(ω) =
(e)
∫ f (t ) dt
t
−∞
+
5
[2 + j(ω − 2 )[5 + j(ω − 2)]]
jω10
(2 + jω)(5 + jω)
F(ω)
+ πF(0 )δ(ω)
j(ω)
=
10
x10
+ πδ(ω)
jω(2 + jω)(5 + jω)
2x5
=
10
+ πδ(ω)
jω(2 + jω)(5 + jω)
Chapter 18, Solution 24.
(a) X (ω) = F(ω) + F [3]
= 6πδ(ω) +
(
)
j − jω
e −1
ω
(b) y(t ) = f (t − 2 )
Y(ω) = e − jω2 F(ω) =
je − j2ω − jω
e −1
ω
(c) If h(t) = f '(t)
H(ω) = jωF(ω) = jω
(
(
)
)
j − jω
e − 1 = 1 − e − jω
ω
3 3 
3 3 
2 
5 
(d) g(t ) = 4f  t  + 10f  t , G (ω) = 4 x F ω  + 10x F ω 
2 2 
5 5 
3 
3 
= 6⋅
=
j
3
ω
2
(
(e
− j3ω / 2
)
−1 +
)
(
)
6 j − j3ω / 5
e
−1
3
ω
5
(
)
j4 − j3ω / 2
j10 − j3ω / 5
e
−1 +
e
−1
ω
ω
Chapter 18, Solution 25.
(a) F(s ) =
A=
10
A
B
= +
, s = jω
s(s + 2) s s + 2
10
10
= 5, B =
= −5
2
−2
F(ω) =
5
5
−
jω jω + 2
f(t) =
5
sgn(t ) − 5e −2 t u(t )
2
jω − 4
A
B
=
+
( jω + 1)( jω + 2) jω + 1 jω + 2
(b) F(ω) =
F(s ) =
s−4
A
B
=
+
, s = jω
(s + 1)(s + 2) s + 1 s + 2
A = 5, B = 6
F(ω) =
6
−5
+
1 + jω 2 + jω
(
)
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,

→
X(ω) = 2
sin ω
ω
G (ω) = 2u (ω + 1) − 2u (ω − 1)

→
g( t ) =
2 sin t
πt
Chapter 18, Solution 27.
(a) Let F(s ) =
100
A
B
= +
, s = jω
s (s + 10) s s + 10
A=
100
100
= 10, B =
= −10
− 10
10
F(ω) =
10
10
−
jω jω + 10
f(t) = 5 sgn(t ) − 10e −10 t u(t )
(b) G (s ) =
A=
10s
A
B
=
+
, s = jω
(2 − s )(3 + s ) 2 − s s + 3
20
− 30
= 4, B =
= −6
5
5
G (ω) =
4
6
−
= − jω + 2 jω + 3
g(t) = 4e 2 t u(− t ) − 6e −3 t u(t )
(c) H (ω) =
( j ω)
60
2
+ j40ω + 1300
=
60
( jω + 20)2 + 900
h(t) = 2e −20 t sin( 30t ) u(t )
1 ∞ δ(ω)e jωt dω
1 1 1
y (t ) =
= π⋅ = π
∫
−
∞
(2 + jω)( jω + 1) 2 2 4
2π
Chapter 18, Solution 28.
(a)
f (t) =
=
(b)
(c)
1 ∞ 10δ(ω + 2) jωt
10
e − j2 t
e
d
ω
=
2π ∫−∞ jω( jω + 1)
2π (− j2)(− j2 + 1)
j5 e − j2 t
( −2 + j)e − j2 t
=
2π
2π 1 − j2
1 ∞ 20δ(ω − 1)e jωt
20
e jt
f (t) =
dω =
2π ∫−∞ (2 + jω)(3 + 5ω)
2π (2 + j)(3 + j)
=
(d)
1 1
1
=
= 0.05
2 (5)(2) 20
f (t) =
=
πδ(ω) e jωt
1 ∞
1 ∞
j ωt
F
(
ω
)
e
d
ω
=
dω
2π ∫−∞
2π ∫−∞ (5 + jω)(2 + jω)
20e jt
(1 − j)e jt
=
2π(5 + 5 j)
π
Let
F(ω) =
5πδ(ω)
5
+
= F1 (ω) + F2 (ω)
(5 + jω) jω(5 + jω)
f1 ( t ) =
1 ∞ 5πδ(ω) jωt
5π 1
e dω =
⋅ = 0.5
∫
−
∞
2π
5 + jω
2π 5
F2 (s) =
5
A
B
= +
, A = 1, B = −1
s(5 + s) s s + 5
F2 (ω) =
f 2 (t) =
1
1
−
jω jω + 5
1
1
sgn( t ) − e −5 t = − + u ( t ) − e 5 t
2
2
f ( t ) = f 1 ( t ) + f 2 ( t ) = u( t ) − e − 5 t
Chapter 18, Solution 29.
(a)
f(t) = F -1 [δ(ω)] + F -1 [4δ(ω + 3) + 4δ(ω − 3)]
=
(b)
1 4 cos 3t
1
(1 + 8 cos 3t )
+
=
π
2π
2π
If h ( t ) = u ( t + 2) − u ( t − 2)
H(ω) =
2 sin 2ω
ω
g( t ) =
G (ω) = 4H(ω)
g(t) =
(c)
1 8 sin 2 t
⋅
2π
t
4 sin 2t
πt
Since
cos(at) ↔πδ(ω + a ) + πδ(ω − a )
Using the reversal property,
2π cos 2ω ↔ πδ( t + 2) + πδ( t − 2)
or F -1 [6 cos 2ω] = 3δ(t + 2) + 3δ(t − 2)
Chapter 18, Solution 30.
(a)
y( t ) = sgn( t )
H(ω) =
(b) X(ω) =
Y(ω) =
2
,
jω
Y(ω) 2(a + jω)
2a
=
= 2+
X(ω)
jω
jω
1
,
1 + jω
H(ω) =

→
Y(ω) =
X(ω) =

→
1
a + jω
h ( t ) = 2δ( t ) + a[u ( t ) − u (− t )]
1
2 + jω
1 + jω
1
= 1−
2 + jω
2 + jω
(c ) In this case, by definition, h ( t ) =

→
h ( t ) = δ( t ) − e − 2 t u ( t )
y( t ) = e −at sin bt u ( t )
Chapter 18, Solution 31.
Y(ω) =
(a)
X(ω) =
1
(a + jω)
2
H(ω) =
,
Y(ω)
1
=
H(ω) a + jω
1
a + jω

→
x ( t ) = e − at u ( t )
(b)
By definition, x ( t ) = y( t ) = u ( t + 1) − u ( t − 1)
(c )
Y(ω) =
X(ω) =
Y(ω)
jω
1
a
= −
=
H(ω) 2(a + jω) 2 2(a + jω)
1
H(ω) =
,
(a + jω)
2
jω

→
x(t) =
Chapter 18, Solution 32.
(a)
e − jω
jω + 1
and F(− ω)
e − ( t −1) u ( t − 1)
Since
F1 (ω) =
f(-t)
e jω
− jω + 1
f 1 (t ) = e − (− t −1) u (− t − 1)
f1(t) = e (t +1 )u(− t − 1)
(b)
From Section 17.3,
2
t +1
2πe
2
If F2 (ω) = 2e
f2(t) =
−ω
, then
2
π t +1
(
2
−ω
)
a
1
δ( t ) − e − at u ( t )
2
2
(b)
By partial fractions
F3 (ω) =
1
( jω + 1)2 ( jω − 1)2
Hence f 3 (t ) =
=
(d)
1
1
1
1
4
4
=
+ 4 +
− 4
2
2
( jω + 1) ( jω + 1) ( jω − 1) jω − 1
(
)
1 −t
te + e − t + te t − e t u (t )
4
1
(t + 1)e −t u(t ) + 1 (t − 1)e t u(t )
4
4
f 4 (t ) =
1
1 ∞
1 ∞ δ(ω)e jωt
jωt
(
)
F
ω
e
d
ω
=
=
1
∫
∫
2π − ∞
2π − ∞ 1 + j2ω
2π
Chapter 18, Solution 33.
(a)
Let x (t ) = 2 sin πt[u (t + 1) − u (t − 1)]
From Problem 17.9(b),
4 jπ sin ω
π 2 − ω2
Applying duality property,
X(ω) =
f (t ) =
(b)
1
2 j sin (− t )
X(− t ) = 2 2
2π
π −t
f(t) =
2 j sin t
t 2 − π2
F(ω) =
j
(cos 2ω − j sin 2ω) − j (cos ω − j sin ω)
ω
ω
j j2 ω
e − jω e j 2 ω
− jω
= (e − e ) =
−
ω
jω
jω
f (t ) =
1
1
sgn (t − 1) − sgn (t − 2)
2
2
But sgn( t ) = 2u ( t ) − 1
f (t ) = u (t − 1) −
1
1
− u (t − 2 ) +
2
2
= u(t − 1) − u(t − 2 )
Chapter 18, Solution 34.
First, we find G(ω) for g(t) shown below.
g (t ) = 10[u (t + 2 ) − u (t − 2 )] + 10[u (t + 1) − u (t − 1)]
g ' (t ) = 10[δ(t + 2 ) − δ(t − 2 )] + 10[δ(t + 1) − δ(t − 1)]
The Fourier transform of each term gives
g(t)
20
10
–2
–1
0
1
t
2
g ‘(t)
10δ(t+2)
10δ(t+1)
–2
–1
0
1
–10δ(t-1)
2
–10δ(t-2)
jωG (ω) = 10(e jω2 − e − jω2 ) + 10(e jω − e − jω )
= 20 j sin 2ω + 20 j sin ω
G (ω) =
20 sin 2ω 20 sin ω
+
= 40 sinc(2ω) + 20 sinc(ω)
ω
ω
Note that G(ω) = G(-ω).
t
F(ω) = 2πG (− ω)
f (t ) =
1
G (t )
2π
= (20/π)sinc(2t) + (10/π)sinc(t)
Chapter 18, Solution 35.
(a)
x(t) = f[3(t-1/3)]. Using the scaling and time shifting properties,’’
X(ω) =
(b)
1
1
e − jω / 3
e − jω / 3 =
3 2 + jω / 3
(6 + jω)
Using the modulation property,
 1 1
1
1
1
1
1 
= 
+
Y(ω) = [F(ω + 5) + F(ω − 5)] = 
+

2
2  2 + j(ω + 5) 2 + j(ω − 5)  2  jω + 7 jω − 3 
jω
2 + jω
(c )
Z(ω) = jωF(ω) =
(d)
H(ω) = F(ω)F(ω) =
(e)
I(ω) = j
1
(2 + jω) 2
d
(0 − j)
1
F(ω) = j
=
dω
(2 + jω) 2 (2 + jω) 2
Chapter 18, Solution 36.
H(ω) =
Vo (ω)
Vi (ω)
Vo (ω) = H(ω)Vi (ω) =
10Vi (ω)
2 + jω
(a)
vi = 4δ(t)
Vo (ω) =
Vi(ω) = 4
40
2 + jω
v o ( t ) = 40e −2 t u (t )
vo(2) = 40e–4 = 0.7326 V
(b)
v i = 6e − t u (t )
Vi (ω) =
6
1 + jω
Vo (ω) =
60
(2 + jω)(1 + jω)
Vo (s ) =
60
A
B
=
+
, s = jω
(s + 2)(s + 1) s + 1 s + 2
A=
60
60
= 60, B =
= −60
−1
1
Vo (ω) =
60
60
−
1 + jω 2 + jω
[
(2) = 60[e
]
] = 60 (0.13533 − 0.01831)
v o ( t ) = 60 e − t − e −2 t u ( t )
vo
= 7.021 V
(c)
−2
− e −4
vi(t) = 3 cos 2t
Vi(ω) = π[δ(ω + 2) + δ(ω- 2)]
Vo =
10π[δ(ω + 2) + δ(ω − 2 )]
2 + jω
v o (t) =
1 ∞
Vo (ω)e jωt dω
2π ∫−∞
jωt
∞ δ(ω − 2 )e
δ(ω + 2 ) jωt
e dω + 5 ∫
= 5∫
dω
−∞
− ∞ 2 + jω
2 + jω
∞
=
[
5e − j2 t 5e + j2 t
5
+
=
e − j(2 t − 45° ) + e j(2 t − 45° )
2 − j2 2 + j2 2 2
5
cos(2 t − 45°)
2
=
v o (2 ) =
5
5
cos(229.18° − 45°)
cos(4 − 45°) =
2
2
vo(2) = –3.526 V
Chapter 18, Solution 37.
2 jω =
j2ω
2 + jω
By current division,
j2ω
I (ω)
j2ω
2 + jω
H(ω) = o
=
=
j2ω
I s (ω)
j2ω + 8 + j4ω
4+
2 + jω
H(ω) =
jω
4 + j3ω
Chapter 18, Solution 38.
Vi (ω) = πδ(ω) +
Vo (ω) =
]
1
jω
5 
1 
10
 πδ (ω) + 
Vi (ω) =
5 + jω 
jω 
10 + jω2
Let Vo (ω) = V1 (ω) + V2 (ω) =
5πδ(ω)
5
+
5 + jω jω(5 + jω)
V2 (ω) =
5
A
B
= +
s(s + 5) s s + 5
V2 (ω) =
1
1
−
jω 5 + j ω
V1 =
5πδ(ω)
5 + jω
A = 1, B = -1, s = jω
1
v 2 (t ) = sgn( t ) − e − 5 t
2
v 1 (t ) =
1 ∞ 5πδ (ω) jωt
e dω
2π ∫−∞ 5 + jω
5π 1
⋅ = 0.5
2π 5
v1(t) =
v 0 (t ) = v1 (t ) + v 2 (t ) = 0.5 + 0.5 sgn (t ) − e −5 t
sgn (t ) = −1 + 2u (t )
But
v o (t ) = +0.5 − 0.5 + u (t ) − e −5 t u (t ) = u(t ) − e −5t u(t )
Chapter 18, Solution 39.
∞
Vs (ω) =
∫ (1 − t )e
− jωt
dt =
−∞
I(ω) =
Vs (ω)
3
10 + jωx10
−3
=
1
1
1 − jω
+
−
e
2
jω ω
ω2
10 3
 1
1
1 − jω 
 +

e
−
2
10 + jω  jω ω
ω2

6
Chapter 18, Solution 40.
v( t ) = δ( t ) − 2δ( t − 1) + δ( t − 2)
− ω 2 V(ω) = 1 − 2e − jω + e jω2
V(ω) =
Now
1 − 2e − jω + e − jω2
− ω2
Z(ω) = 2 +
1 1 + j2ω
=
jω
jω
I=
=
V(ω) 2e jω − e jω2 − 1
jω
=
⋅
2
Z(ω)
1 + j2ω
ω
1
(
0.5 + 0.5e − jω2 −e − jω )
jω(0.5 + jω)
1
A
B
= +
s(s + 0.5) s s + 0.5
But
I(ω) =
i(t) =
A = 2, B = -2
2
2
(
(0.5 + 0.5e− jω2 − e− jω )
0.5 + 0.5e jω2 − e − jω ) −
jω
0.5 + jω
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 + jω
V
1/s
0.5s
−
V−
1
2V
2 + jω
+ jω V +
−2=0
2
jω
(
)
jω
− 4ω 2 + j9ω
jω − 2ω + 4 V = j4ω +
=
2 + jω
2 + jω
V(ω) =
2
2 jω(4.5 + j2ω)
(2 + jω)(4 − 2ω 2 + jω)
2
Chapter 18, Solution 42.
By current division, I o =
(a)
2
⋅ I(ω)
2 + jω
For i(t) = 5 sgn (t),
10
jω
2
10
20
Io =
⋅
=
2 + jω jω jω(2 + jω)
I(ω) =
Let I o =
20
A
B
= +
, A = 10, B = −10
s(s + 2) s s + 2
I o (ω) =
10
10
−
j ω 2 + jω
io(t) = 5 sgn( t ) − 10e −2 t u(t )A
i(t)
(b)
i’(t)
4
4δ(t)
1
1
t
i' ( t ) = 4δ( t ) − 4δ( t − 1)
jω I(ω) = 4 − 4e − jω
(
4 1 − e − jω
I(ω) =
jω
Io =
)
 1
8(1 − e − jω )
1 
(1 − e − jω )
= 4 −
ω
+
ω
jω(2 + jω)
j
2
j


=
4
4
4e − j ω 4e − j ω
−
−
+
jω 2 + jω
jω
2 + jω
t
–4δ(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.
20 mF
1
1
50
=
=
,
−
3
jωC j20x10 ω jω

→
Vo =
Vo =
50
50
40
=
Is •
,
50
jω (s + 1.25)(s + 5)
40 +
jω
i s = 5e − t

→
Is =
1
5 + jω
s = jω
A
B
40 
1
1 
+
=
−

s + 1.25 s + 5 3  jω + 1.25 jω + 5 
v o (t) =
40 −1.25t
(e
− e −5 t ) u ( t )
3
Chapter 18, Solution 44.
1H
jω
We transform the voltage source to a current source as shown in Fig. (a) and then
combine the two parallel 2Ω resistors, as shown in Fig. (b).
Io
+
Vs/2
2
2 Vo
Io
+
Vs/2
1 Vo
jω
−
(a)
2 2 = 1Ω, I o =
Vo = jω I o =
V
1
⋅ s
1 + jω 2
jω Vs
2(1 + jω)
−
(b)
jω
v s ( t ) = 10δ(t ) − 10δ( t − 2)
jω Vs (ω) = 10 − 10e − j2 ω
Vs (ω) =
(
10 1 − e − j2ω
jω
)
(
)
5 1 − e − j2 ω
5
5
Hence Vo =
=
−
e − j2 ω
1 + jω
1 + jω 1 + jω
v o ( t ) = 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
jω
1
2 + jω +
jω
(2) =
2
( jω + 1)
2

→
v o ( t ) = 2te − t u ( t )
Chapter 18, Solution 46.
1
F
4
1
jω
2H
3δ( t )
1
4
=
− j4
ω
jω2
3
1
1 + jω
e − t u(t)
The circuit in the frequency domain is shown below:
2Ω
Vo
Io(ω)
–j4/ω
1/(1+jω)
+
−
3
+
−
j2ω
At node Vo, KCL gives
1
− Vo
3 − Vo
V
1 + jω
+
= o
− j4
2
j2ω
ω
2
j2Vo
− 2Vo + jω3 − jωVo = −
ω
1 + jω
2
+ jω3
1 + jω
Vo =
j2
2 + jω −
ω
2 + jω3 − 3ω 2
V
1 + jω
I o (ω) = o =
j2 
j2ω

j2ω 2 + jω − 
ω

Io(ω) =
2 + jω 2 − 3ω 2
4 − 6ω 2 + j(8ω − 2ω 3 )
Chapter 18, Solution 47.
Transferring the current source to a voltage source gives the circuit below:
1/(jω)
2Ω
+
8V
+
−
Vo
−
1Ω
jω/2
jω
4 + j3ω
jω
Let Z in = 2 + 1
= 2+ 2 =
jω 2 + jω
2
1+
2
By voltage division,
1
8
8
jω
Vo (ω) =
⋅8 =
=
1
jω(4 + j3ω)
1 + jωZ in
+ Z in
1+
jω
2 + jω
=
8(2 + jω)
2 + jω + jω4 − 3ω 2
=
8(2 + jω)
2 + jω5 − 3ω 2
Chapter 18, Solution 48.
0.2F
1
j5
=−
jωC
ω
As an integrator,
RC = 20 x 10 3 x 20 x 10 −6 = 0.4
vo = −
1 t
v i dt
RC ∫o
Vo = −

1  Vi
+ πVi (0)δ(ω)

RC  jω

=−
Io =

1 
2
+ πδ (ω)
 (
0 .4  j ω 2 + j ω )



Vo
2
mA = −0.125 
+ π δ (ω)
20
 jω (2 + jω)

=−
0.125 0.125
+
− 0.125πδ (ω)
jω
2 + jω
i o ( t ) = −0.125 sgn( t ) + 0.125e − 2 t u (t ) −
0.125
πδ (ω)e jωt dt
∫
2π
= 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:
jω
j2ω
jω
+
VS
+
−
i1
i2
2Ω
1 Ω vo
−
Vs = π[δ (ω + 1) + δ (ω − 2)]
For mesh 1, − Vs + (2 + j2ω)I1 − 2I 2 − jωI 2 = 0
Vs = 2 (1 + jω) I1 − (2 + jω)I 2
0 = (3 + jω)I 2 − 2I1 − jωI1
For mesh 2,
I1 =
(3 + ω)I 2
(2 + ω)
(2)
Substituting (2) into (1) gives
Vs = 2
(1)
2 (1 + jω)(3 + jω)I 2
− (2 + jω)I 2
2 + jω
[(
) (
)]
Vs (2 + ω) = 2 3 + j4ω − ω 2 − 4 + j4ω − ω 2 I 2
= I 2 (2 + j4ω − ω
2
)
I2 =
(s + 2)Vs
s 2 + 4s + 2
Vo = I 2 =
( jω + 2) π [δ (ω + 1) + δ (ω − 1)]
( jω)2 + jω4 + 2
1 ∞
v o (ω)e jωt dω
∫
−
∞
2π
v o (t) =
=∫
, s = jω
∞
−∞
1
( jω + 2) e jωt δ (ω + 1)dω 1 ( jω + 2)e jωt δ(ω − 1)dω
2
+2
2
( jω) + jω4 + 2
( jω)2 + jω4 + 2
1
(− j + 2)e jt 1 ( j + 2)e jt
= 2
+ 2
− 1 − j4 + 2 − 1 + j4 + 2
1
1
(2 − j)(1 + j4)
(2 − j)(1 − j4)e jt
v o (t) = 2
e jt + 2
17
17
=
1
(6 + j7 )e jt + 1 (6 − j7 )e jt
34
34
= 0.271e − j ( t −13.64° ) + 0.271e j ( t −13.64° )
vo(t) = 0.542 cos(t − 13.64°)V
Chapter 18, Solution 50.
Consider the circuit shown below:
j0.5ω
1Ω
VS
+
−
i1
jω
i2
jω
+
1Ω
vo
−
For loop 1,
For loop 2,
− 2 + (1 + jω)I1 + j0.5ωI 2 = 0
(1)
(1 + jω)I 2 + j0.5ωI1 = 0
(2)
From (2),
I1 =
(1 + jω)I 2
− j0.5ω
= −2
(1 + jω)I 2
jω
Substituting this into (1),
− 2(1 + jω)I 2 jω
2=
+
I2
jω
2
3 

2 jω = − 4 + j4ω − ω 2 I 2
2 

I2 =
2 jω
4 + j4ω − 1.5ω 2
Vo = I 2 =
− 2 jω
4 + j4ω + 1.5( jω)
2
4
jω
3
Vo =
8
8ω
2
+j
+ ( jω )
3
3
=

4
− 4 + jω 

3
  8 
4
 + jω  + 
  3 
3
2
2
+
16
3
2
  8 
4
 + jω  + 
  3 
3
2
 8 
 8 
t  u(t ) + 5.657e − 4t / 3 sin
t u(t ) V
Vo ( t ) = − 4e − 4t / 3 cos
3
3




Chapter 18, Solution 51.
1
1
1
jω
Z = 1 //
=
=
1
jω 1 +
1 + jω
jω
1
Z
2
1
2
1 + jω
∗
=
Vo =
Vo =
1
Z+2
1 + jω 3 + 2 jω 1 + jω
2+
1 + jω
1
=
, s = jω
(s + 1)(s + 1.5)
Vo =
A
B
2
2
+
=
−
s + 1 s + 1.5 s + 1 s + 1.5
∞
W=
→
v o ( t ) = 2(e − t − e −1.5t )u ( t )
∞
∫
−∞
f 2 ( t )dt = 4 ∫ (e − t − e −1.5t ) 2 dt
∞
= 4 ∫ (e
0
∞
− 2t
− 2e
0
− 2.5 t
+e
− 3t
 e − 2t
e − 2.5t e − 3t 
+2
−
)dt = 4
 −2
2
.
5
3 

0
1 2 1
W = 4( −
+ ) = 0.1332 J
2 2.5 3
Chapter 18, Solution 52.
∞
J = 2 ∫ f 2 ( t ) dt =
0
1 ∞
2
F(ω) dω
∫
0
π
∞
1 ∞
1
1
1 π
dω =
tan −1 (ω / 3) =
= (1/6)
=
2
2
∫
0
π 9 +ω
3π
3π 2
0
Chapter 18, Solution 53.
J =
∫
∞
0
∞
F(ω) dω = 2π∫ f 2 ( t ) dt
2
−∞
f(t) =
e 2t ,
t<0
e −2 t ,
t>0
 e 4t
0
∞
J = 2π  ∫ e 4 t dt + ∫ e − 4 t dt  = 2π
 −∞

0
 4
0
−∞
e −4 t
+
−4

 = 2π[(1/4) + (1/4)] = π
0 

∞
Chapter 18, Solution 54.
W1Ω =
∫
∞
−∞
∞
∞
0
0
f 2 ( t ) dt = 16 ∫ e − 2 t dt = − 8e − 2 t
= 8J
Chapter 18, Solution 55.
f(t) = 5e2e–tu(t)
F(ω) = 5e2/(1 + jω), |F(ω)|2 = 25e4/(1 + ω2)
W1Ω
1 ∞
25e 4
2
=
F
(
ω
)
d
ω
=
π ∫0
π
∫
∞
0
= 12.5e4 = 682.5 J
or
W1Ω =
∫
∞
−∞
∞
f 2 ( t ) dt = 25e 4 ∫ e − 2 t dt = 12.5e4 = 682.5 J
0
Chapter 18, Solution 56.
W1Ω =
But,
∫
∞
−∞
∞
f 2 ( t ) dt = ∫ e − 2 t sin 2 (2 t ) dt
0
sin2(A) = 0.5(1 – cos(2A))
∞
1
25e 4
d
ω
=
tan −1 (ω)
2
π
1+ ω
0
W1Ω =
∫
∞
0
e
−2 t
1 e −2 t
0.5[1 − cos(4 t )]dt =
2 −2
∞
0
∞
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.
W1Ω =
or
∫
∞
−∞
0
0
−∞
−∞
i 2 ( t ) dt = ∫ 4e 2 t dt = 2e 2 t
= 2J
I(ω) = 2/(1 – jω), |I(ω)|2 = 4/(1 + ω2)
∞
W1Ω =
4
4π
1 ∞
4 ∞
1
2
= 2J
dω = tan −1 (ω) =
ω
ω
=
I
(
)
d
2
∫
∫
−
∞
−
∞
2
π
π
2π
2π
(1 + ω )
0
In the frequency range, –5 < ω < 5,
5
4
4
4
W =
tan −1 ω = tan −1 (5) = (1.373) = 1.7487
π
π
π
0
W/ W1Ω = 1.7487/2 = 0.8743 or 87.43%
Chapter 18, Solution 58.
ωm = 200π = 2πfm which leads to fm = 100 Hz
(a)
ωc = πx104 = 2πfc 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.
10
6
−
V (ω) 2 + jω 4 + jω
5
3
H(ω) = o
=
=
−
Vi (ω)
2
2 + jω 4 + jω
 5
3  4

Vo (ω) = H(ω)Vi (ω) = 
−
 2 + jω 4 + jω  1 + jω
20
12
=
−
, s = jω
(s + 1)(s + 2) (s + 1)(s + 4)
Using partial fraction,
A
B
C
D
16
20
4
+
+
+
=
−
+
s + 1 s + 2 s + 1 s + 4 1 + jω 2 + jω 4 + jω
Vo (ω) =
Thus,
(
)
v o ( t ) = 16e − t − 20e −2 t + 4e −4 t u ( t ) V
Chapter 18, Solution 60.
2
+
Is
1/jω
jω
V
−
V = jωI s
1
jω
1
+ 2 + jω
jω
=
jωI s
1 − ω 2 + j2ω
Since the voltage appears across the inductor, there is no DC component.
V1 =
2π∠90°8
1 − 4π 2 + j4π
V2 =
=
50.27∠90°
= 1.2418∠ − 71.92°
− 38.48 + j12.566
4π∠90°5
2
1 − 16π + j8π
=
62.83∠90°
= 0.3954∠ − 80.9°
− 156.91 + j25.13
v( t ) = 1.2418 cos( 2πt − 41.92°) + 0.3954 cos( 4πt + 129.1°) 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.
ω = 3.4 kHz
fs = 2ω = 6.8 kHz
Chapter 18, Solution 66.
ω = 4.5 MHz
fc = 2ω = 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(ω2/2)
f(t)
F(ω)
A
ω
t
–T/2
T/2
G(ω)
ω
–ωm/2
According to the duality property,
Aτsinc(τt/2)
becomes 2πArect(τ)
g(t) = sinc(200πt) becomes 2πArect(τ)
where Aτ = 1 and τ/2 = 200π or T = 400π
i.e. the upper frequency ωu = 400π = 2πfu or fu = 200 Hz
The Nyquist rate = fs = 200 Hz
The Nyquist interval = 1/fs = 1/200 = 5 ms
ωm/2
Chapter 18, Solution 68.
The total energy is
WT =
∫
∞
−∞
v 2 ( t ) dt
Since v(t) is an even function,
WT =
∫
∞
0
2500e
−4 t
e −4 t
dt = 5000
−4
∞
= 1250 J
0
V(ω) = 50x4/(4 + ω2)
1 5
1 5 (200) 2
2
|
V
(
ω
)
|
d
ω
=
dω
2π ∫1
2π ∫1 (4 + ω 2 ) 2
W =
But
∫ (a
2
1  x
1
1

+ tan −1 ( x / a ) + C
dx = 2  2
2 2
2
a
+x )
2a  x + a

5
2x10 4 1  ω
1

W =
+
tan −1 (ω / 2)
2

π 8 4 + ω
2
1
= (2500/π)[(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 ∞
1 ∞ 400
2
F(ω) dω =
dω
∫
2π − ∞
2π ∫−∞ 4 2 + ω 2
[
400
(1 / 4) tan −1 (ω / 4)
π
]
∞
0
=
100 π
= 50
π 2
[
1 2
400
2
F(ω) dω =
(1 / 4) tan −1 (ω / 4)
∫
0
2π
2π
]
2
0
= [100/(2π)]tan–1(2) = (50/π)(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).
1Ω
4Ω
I2 = 0
+
I1
6Ω
V1
Io
+
2Ω
V2
−
−
(a)
z 11 =
V1
= 1 + 6 || (4 + 2) = 4 Ω
I1
Io =
1
I ,
2 1
z 21 =
V2 = 2 I o = I 1
V2
= 1Ω
I1
To get z 22 and z 12 , consider the circuit in Fig. (b).
I1 = 0
1Ω
4Ω
Io '
+
+
6Ω
V1
2Ω
V2
−
−
(b)
z 22 =
V2
= 2 || (4 + 6) = 1.667 Ω
I2
Io' =
2
1
I2 = I2 ,
2 + 10
6
z 12 =
V1
= 1Ω
I2
V1 = 6 I o ' = I 2
I2
Hence,
4
1 
[z ] = 
Ω
 1 1.667 
Chapter 19, Solution 2.
Consider the circuit in Fig. (a) to get z 11 and z 21 .
1Ω
Io '
1Ω
1Ω
+
I1
1Ω
V1
1Ω
1Ω
Io
+
1Ω
V2
−
−
1Ω
1Ω
1Ω
(a)
z 11 =
V1
= 2 + 1 || [ 2 + 1 || (2 + 1) ]
I1
(1)(11 4)

11
3
= 2 + = 2.733
z 11 = 2 + 1 ||  2 +  = 2 +

15
4
1 + 11 4
Io =
1
1
Io' = Io'
1+ 3
4
Io' =
1
4
I1 = I1
1 + 11 4
15
Io =
1 4
1
⋅ I1 = I1
4 15
15
V2 = I o =
z 21 =
I2 = 0
1
I
15 1
V2
1
=
= z 12 = 0.06667
I 1 15
1Ω
To get z 22 , consider the circuit in Fig. (b).
I1 = 0
1Ω
1Ω
1Ω
1Ω
+
+
1Ω
V1
1Ω
1Ω
V2
−
−
1Ω
1Ω
1Ω
1Ω
(b)
z 22 =
V2
= 2 + 1 || (2 + 1 || 3) = z 11 = 2.733
I2
Thus,
 2.733 0.06667 
[z ] = 
Ω
 0.06667 2.733 
Chapter 19, Solution 3.
(a)
To find z 11 and z 21 , consider the circuit in Fig. (a).
-j Ω
Io
+
I1
V1
I2 = 0
jΩ
1Ω
−
V1
j (1 − j)
= j || (1 − j) =
= 1+ j
I1
j +1− j
By current division,
j
Io =
I = j I1
j +1− j 1
V2
−
(a)
z 11 =
+
I2
V2 = I o = jI 1
z 21 =
V2
=j
I1
To get z 22 and z 12 , consider the circuit in Fig. (b).
-j Ω
I1 = 0
+
+
jΩ
V1
1Ω
−
I2
V2
−
(b)
z 22 =
V2
= 1 || ( j − j) = 0
I2
V1 = j I 2
z 12 =
V1
=j
I2
Thus,
 1+ j j 
[z ] = 
Ω
0 
 j
(b)
To find z 11 and z 21 , consider the circuit in Fig. (c).
jΩ
-j Ω
+
I1
+
1Ω
V1
−
I2 = 0
V2
-j Ω
1Ω
(c)
−
z 11 =
V1
-j
= j + 1 + 1 || (-j) = 1 + j +
= 1.5 + j0.5
1− j
I1
V2 = (1.5 − j0.5) I 1
z 21 =
V2
= 1.5 − j0.5
I1
To get z 22 and z 12 , consider the circuit in Fig. (d).
I1 = 0
jΩ
-j Ω
+
+
1Ω
V1
−
V2
-j Ω
1Ω
(d)
z 22 =
V2
= -j + 1 + 1 || (-j) = 1.5 - j1.5
I2
V1 = (1.5 − j0.5) I 2
z 12 =
V1
= 1.5 − j0.5
I2
Thus,
 1.5 + j0.5 1.5 − j0.5 
[z ] = 
Ω
 1.5 − j0.5 1.5 − j1.5 
Chapter 19, Solution 4.
Transform the Π network to a T network.
Z1
Z3
Z2
−
I2
Z1 =
(12)( j10)
j120
=
12 + j10 − j5 12 + j5
Z2 =
- j60
12 + j5
Z3 =
50
12 + j5
The z parameters are
z 12 = z 21 = Z 2 =
(-j60)(12 - j5)
= -1.775 - j4.26
144 + 25
z 11 = Z1 + z 12 =
( j120)(12 − j5)
+ z 12 = 1.775 + j4.26
169
z 22 = Z 3 + z 21 =
(50)(12 − j5)
+ z 21 = 1.7758 − j5.739
169
Thus,
 1.775 + j4.26 - 1.775 − j4.26 
[z ] = 
Ω
 - 1.775 − j4.26 1.775 − j5.739 
Chapter 19, Solution 5.
Consider the circuit in Fig. (a).
1
s
I2 = 0
Io
+
I1
V1
1
1/s
1/s
+
V2
−
−
(a)
 1 
1

1 + s + 

1 
1
1   s + 1 
s
z 11 = 1 || || 1 + s +  =
|| 1 + s +  =
1 
1
s 
s
s  1 
1+

 +1+ s +
s
 s + 1
s
1
s
s2 + s +1
s 3 + 2s 2 + 3s + 1
z 11 =
Io =
Io =
1 ||
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
I1
1
V2 = I o = 3
s
s + 2s 2 + 3s + 1
z 21 =
V2
1
= 3
2
I 1 s + 2s + 3s + 1
Consider the circuit in Fig. (b).
s
1
I1 = 0
+
+
1
V1
1/s
1/s
−
−
(b)
z 22 =
z 22
V2 1 
1 1 
1 

= || 1 + s + 1 ||  = || 1 + s +
s s 
s + 1
I2 s 
 1 
1 
1
 1 + s +

1
s
+
+
 s 
s + 1
s +1
=
=
s
1
1
1+ s + s2 +
+1+ s +
s +1
s
s +1
z 22 =
s 2 + 2s + 2
s 3 + 2s 2 + 3s + 1
z 12 = z 21
Hence,
V2
I2


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 
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 Ω
Vo
20 Ω
+
V1
+
−
30 Ω
0.5 V2
V2
−
(a)
V1 − Vo
Vo
,
= 0.5 V2 +
10
50
where V2 =
30
3
Vo = Vo
20 + 30
5
3  V
V1 = Vo + 5  Vo  + o = 4.2 Vo
5  5
I1 =
V1 − Vo 3.2
=
V = 0.32 Vo
10
10 o
z 11 =
4.2 Vo
V1
=
= 13.125 Ω
I 1 0.32 Vo
z 21 =
0.6 Vo
V2
=
= 1.875 Ω
I 1 0.32 Vo
To obtain z 22 and z 12 , use the circuit in Fig. (b).
10 Ω
20 Ω
I2
+
V1
0.5 V2
−
(b)
30 Ω
+
−
V2
V2
= 0.5333 V2
30
V2
1
=
=
= 1.875 Ω
I 2 0.5333
I 2 = 0.5 V2 +
z 22
V1 = V2 − (20)(0.5 V2 ) = -9 V2
z 12 =
- 9 V2
V1
=
= -16.875 Ω
I 2 0.5333 V2
Thus,
 13.125 - 16.875 
[z ] = 
Ω
1.875 
 1.875
Chapter 19, Solution 7.
To get z11 and z21, we consider the circuit below.
I1
20 Ω
I2=0
100 Ω
+
+
V1
vx
50 Ω
60 Ω
-
-
12vx
+
V1 − Vx Vx Vx + 12Vx
=
+
20
50
160
V − Vx
81 V1
I1 = 1
( )
=
20
121 20

→
→
Vx =
z11 =
40
V1
121
V1
= 29.88
I1
+
V2
-
13Vx
57
57 40
57 40 20x121
) − 12Vx = − Vx = − (
)V1 = − (
)
I1
160
8
8 121
8 121
81
V
= −70.37 I1 
→ z 21 = 2 = −70.37
I1
V2 = 60(
To get z12 and z22, we consider the circuit below.
I1=0
20 Ω
I2
100 Ω
+
+
V1
50 Ω
vx
60 Ω
-
-
12vx
+
V2
-
+
Vx =
50
1
V2 = V2 ,
100 + 50
3
z 22 =
V2
= 1 / 0.09 = 11.11
I2
I2 =
V2 V2 + 12Vx
+
= 0.09V2
150
60
11.11
1
I 2 = 3.704I 2
V1 = Vx = V2 =
3
3

→
Thus,
 29.88 3.704
[z] = 
Ω
− 70.37 11.11
V
z12 = 1 = 3.704
I2
Chapter 19, Solution 8.
To get z11 and z21, consider the circuit below.
j4 Ω
I1 -j2 Ω
•
5Ω
I2 =0
•
j6 Ω
+
j8 Ω
+
V2
V1
10 Ω
-
-
V1 = (10 − j2 + j6)I1
V2 = −10I1 − j4I1
V
z11 = 1 = 10 + j4
I1

→

→
z 21 =
V2
= −(10 + j4)
I1
To get z22 and z12, consider the circuit below.
j4 Ω
I1=0 -j2 Ω
•
j6 Ω
5Ω
•
j8 Ω
I2
+
+
V2
V1
10 Ω
-
-
V2 = (5 + 10 + j8)I 2
V1 = −(10 + j4)I 2

→

→
z 22 =
V2
= 15 + j8
I2
V
z12 = 1 = −(10 + j4)
I2
Thus,
 (10 + j4) − (10 + j4)
[z] = 
Ω
− (10 + j4) (15 + j8) 
Chapter 19, Solution 9.
It is evident from Fig. 19.5 that a T network is appropriate for realizing the z
parameters.
6Ω
R2
R1
2Ω
4Ω
R3
R 1 = z 11 − z 12 = 10 − 4 = 6 Ω
R 2 = z 22 − z 12 = 6 − 4 = 2 Ω
R 3 = z 12 = z 21 = 4 Ω
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 Ω
I1
10 Ω
I2
+
+
+
−
20 I2
V1
+
−
5 I1
V2
−
−
(b)
z 11 − z 12 = 1 +
2
1
= 1+
s
0.5 s
z 22 − z 12 = 2s
z 12 =
I1
1
s
1Ω
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
1Ω
j5 Ω
3Ω
j1 Ω
5Ω
5-j2
-j2 Ω
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
−
2Ω
−
(a)
I1
8Ω
2Ω
+
V1
+
4Ω
(b)
V1 = (8 + 4 || 4) I 1 = 10 I 1
V2
−
−
From Fig. (b),
I2
Io
2Ω
By current division,
Io =
1
I ,
2 1
V2 = 2 I o = I 1
V2
I1
=
= 0 .1
V1 10 I 1
Chapter 19, Solution 13.
This is a reciprocal two-port so that the circuit can be represented by the circuit below.
40 Ω
120∠0° V
rms
+
−
I1
50 Ω
10 I2
20 Ω
+
−
+
−
30 I1
We apply mesh analysis.
For mesh 1,
- 120 + 90 I 1 + 10 I 2 = 0 
→ 12 = 9 I 1 + I 2
For mesh 2,
30 I 1 + 120 I 2 = 0 
→ I 1 = -4 I 2
Substituting (2) into (1),
- 12
12 = -36 I 2 + I 2 = -35 I 2 
→ I 2 =
35
2
1
1  12 
2
P = I 2 R =   (100) = 5.877 W
2
2  35 
I2
100 Ω
(1)
(2)
Chapter 19, Solution 14.
To find Z Th , consider the circuit in Fig. (a).
I2
I1
+
ZS
+
−
V1
Vo = 1
−
(a)
V1 = z 11 I 1 + z 12 I 2
V2 = z 21 I 1 + z 22 I 2
(1)
(2)
But
V1 = - Z s I 1
V2 = 1 ,
0 = (z 11 + Z s ) I 1 + z 12 I 2
Hence,

→ I 1 =
- z 12
I
z 11 + Z s 2
 - z 21 z 12

1=
+ z 22  I 2
 z 11 + Z s

Z Th =
V2
z z
1
=
= z 22 − 21 12
z 11 + Z s
I2 I2
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 = z 11 I 1
V2 = z 21 I 1 =
VTh = V2 =

→ I 1 =
Vs
z 11 + Z s
z 21 Vs
z 11 + Z s
z 21 Vs
z 11 + Z s
Chapter 19, Solution 15.
(a) From Prob. 18.12,
ZTh = z 22 −
z12z 21
80x 60
= 120 −
= 24
z11 + Zs
40 + 10
ZL = ZTh = 24Ω
(b) VTh =
z 21
80
Vs =
(120) = 192
z11 + Zs
40 + 10
Pmax =
V 2Th 192 2
=
= 192 W
8R Th 8x 24
Chapter 19, Solution 16.
As a reciprocal two-port, the given circuit can be represented as shown in Fig. (a).
5Ω
15∠0° V
10 – j6 Ω
+
−
4 – j6 Ω
a
j6 Ω
(a)
j4 Ω
b
At terminals a-b,
Z Th = (4 − j6) + j6 || (5 + 10 − j6)
Z Th = 4 − j6 +
j6 (15 − j6)
= 4 − j6 + 2.4 + j6
15
Z Th = 6.4 Ω
VTh =
j6
(15∠0°) = j6 = 6∠90° V
j6 + 5 + 10 − j6
The Thevenin equivalent circuit is shown in Fig. (b).
6.4 Ω
+
6∠90° V
+
−
Vo
j4 Ω
−
(b)
From this,
Vo =
j4
( j6) = 3.18∠148°
6.4 + j4
v o ( t ) = 3.18 cos( 2t + 148°) V
Chapter 19, Solution 17.
To obtain z 11 and z 21 , consider the circuit in Fig. (a).
4Ω
+
I1
V1
Io '
2Ω
Io
I2 = 0
+
V2
8Ω
−
−
6Ω
(a)
In this case, the 4-Ω and 8-Ω resistors are in series, since the same current, I o , passes
through them. Similarly, the 2-Ω and 6-Ω resistors are in series, since the same current,
I o ' , passes through them.
V1
(12)(8)
= (4 + 8) || (2 + 6) = 12 || 8 =
= 4 .8 Ω
I1
20
z 11 =
Io =
But
8
2
I1 = I1
8 + 12
5
Io' =
3
I
5 1
- V2 − 4 I o + 2 I o ' = 0
V2 = -4 I o + 2 I o ' =
z 21 =
-8
6
-2
I1 + I1 =
I
5
5
5 1
V2 - 2
=
= -0.4 Ω
I1
5
To get z 22 and z 12 , consider the circuit in Fig. (b).
4Ω
I1 = 0
+
V1
2Ω
+
V2
8Ω
−
−
6Ω
(b)
z 22 =
V2
(6)(14)
= (4 + 2) || (8 + 6) = 6 || 14 =
= 4 .2 Ω
I2
20
z12 = z 21 = -0.4 Ω
Thus,
 4.8 - 0.4 
[z ] = 
Ω
 - 0.4 4.2 
We may take advantage of Table 18.1 to get [y] from [z].
∆ z = (4.8)(4.2) − (0.4) 2 = 20
I2
y 11 =
z 22 4.2
=
= 0.21
20
∆z
y 12 =
- z 12 0.4
=
= 0.02
20
∆z
y 21 =
- z 21 0.4
=
= 0.02
20
∆z
y 22 =
z 11 4.8
=
= 0.24
20
∆z
Thus,
 0.21 0.02 
[y ] = 
S
 0.02 0.24 
Chapter 19, Solution 18.
To get y 11 and y 21 , consider the circuit in Fig.(a).
6Ω
I1
3Ω
I2
+
+
−
V1
6Ω
3Ω
V2 = 0
−
(a)
V1 = (6 + 6 || 3) I 1 = 8 I 1
I1 1
=
V1 8
y 11 =
I2 =
-6
- 2 V1 - V1
I1 =
=
6+3
3 8
12
y 21 =
I 2 -1
=
V1 12
To get y 22 and y 12 , consider the circuit in Fig.(b).
I1
6Ω
Io
3Ω
I2
+
V1 = 0
6Ω
3Ω
−
(b)
+
−
V2
I2
1
1
1
=
=
=
V2 3 || (3 + 6 || 6) 3 || 6 2
y 22 =
I1 =
- Io
,
2
I1 =
- I 2  - 1  1  - V2
=   V2  =
 6  2  12
6
y 12 =
Io =
3
1
I2 = I2
3+ 6
3
I1
-1
=
= y 21
V2 12
Thus,
 1 -1 
 8 12 
[y ] = 
S
 -1 1 
 12 2 
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)
2s
1 
2
V1 =  || 2  I 1 =
I1 =
I
s 
2 + (1 s)
2s + 1 1
y 11 =
I2 =
I 1 2s + 1
=
= s + 0 .5
V1
2
- I1
- V1
(- 1 s )
I1 =
=
(1 s) + 2
2s + 1
2
V2 = 0
−
y 21 =
I2
= -0.5
V1
To get y 22 and y 12 , refer to the circuit in Fig.(b).
1
I1
I2
+
V1 = 0
1/s
s
1
−
(b)
V2 = (s || 2) I 2 =
y 22 =
I1 =
2s
I
s+2 2
I2 s + 2
1
=
= 0 .5 +
V2
s
2s
- V2
-s
-s s+ 2
I2 =
⋅
V2 =
s+2
s + 2 2s
2
y 12 =
I1
= -0.5
V2
Thus,
 s + 0.5
- 0.5 
[y ] = 
S
 - 0.5 0.5 + 1 s 
Chapter 19, Solution 20.
To get y11 and y21, consider the circuit below.
+
−
V2
3ix
2Ω
I1
I2
+
V1
I1
+
ix
4Ω
6Ω
V2 =0
-
-
Since 6-ohm resistor is short-circuited, ix = 0
V1 = I1(4 // 2) =
I2 = −
8
I1
6
→
I
y11 = 1 = 0.75
V1
4
2 6
1
I1 = − ( V1) = − V1
4+2
3 8
2
→
I
y 21 = 2 = −0.5
V1
To get y22 and y12, consider the circuit below.
3ix
2Ω
I1
+
V1=0
ix
4Ω
-
+
6 Ω V2
-
I2
ix =
V2
,
6
V
V
I 2 = i x − 3i x + 2 = 2
2
6
V
I1 = 3i x − 2 = 0
2
I
1
y 22 = 2 = = 0.1667
V2 6
→
I
y12 = 1 = 0
V2

→
Thus,
0 
 0.75
[ y] = 
S
− 0.5 0.1667 
Chapter 19, Solution 21.
To get y 11 and y 21 , refer to Fig. (a).
I1
0.2 V1
V1
I2
+
+
−
V1
5Ω
10 Ω
V2 = 0
−
(a)
At node 1,
I1 =
V1
+ 0.2 V1 = 0.4 V1
5
I 2 = -0.2 V1

→ y 11 =
I1
= 0 .4
V1
I2
= -0.2
V1

→ y 21 =
To get y 22 and y 12 , refer to the circuit in Fig. (b).
I1
0.2 V1
V1
I2
+
V1 = 0
5Ω
10 Ω
+
−
V2
−
(b)
Since V1 = 0 , the dependent current source can be replaced with an open circuit.
V2 = 10 I 2
y 12 =

→ y 22 =
I2
1
=
= 0 .1
V2 10
I1
=0
V2
Thus,
 0.4
0
[y ] = 
S
 - 0.2 0.1
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
2Ω
V1
Vo
3Ω
+
+
V1
+
−
Vx
I2
1Ω
Vx/2
−
V2 = 0
−
(a)
At node 1,
I1 =
V1 V1 − Vo
+
1
2
At node 2,
V1 − Vo V1 Vo
+
=
2
2
3

→ I 1 = 1.5 V1 − 0.5 Vo

→ 1.2 V1 = Vo
Substituting (2) into (1) gives,
(1)
(2)
I 1 = 1.5 V1 − 0.6 V1 = 0.9 V1
I2 =
- Vo
= -0.4 V1
3

→ y 11 =

→ y 21 =
I1
= 0 .9
V1
I2
= -0.4
V1
To get y 22 and y 12 refer to the circuit in Fig. (b).
I1
2Ω
V1
+
+
V1 = 0
Vx
−
−
3Ω
1Ω
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 
→ y 22 =
= = 0 .2
V2 5
I1
I 1 = - I 2 = -0.2 V2 
→ y 12 =
= -0.2
V2
Thus,
 0.9 - 0.2 
[y ] = 
S
 - 0.4 0.2 
(b)
To get y 11 and y 21 refer to Fig. (c).
jΩ
Io '
I1
Io
1Ω
1Ω
Io''
I2
+
V1
+
−
-j Ω
V2 = 0
−
Zin
(c)

-j 
 = j || (1.5 − j0.5)
Z in = j || (1 + 1 || -j) = j || 1 +
 1− j
=
j (1.5 − j0.5)
= 0.6 + j0.8
1.5 + j0.5
V1 = Z in I 1
Io =

→ y 11 =
I1
1
1
=
=
= 0.6 − j0.8
V1 Z in 0.6 + j0.8
j
I ,
1.5 + j0.5 1
I o '' =
Io' =
1.5 − j0.5
I
1.5 + j0.5 1
I1
I
-j
Io =
= 1
1− j
(1 − j)(1.5 + j0.5) 2 − j
- I2 = Io + Io
'
''
(1.5 − j0.5) 2
2+ j
=
I1 +
I = (1.2 − j0.4) I 1
2 .5
5 1
- I 2 = (1.2 − j0.4)(0.6 − j0.8) V1 = (0.4 − j1.2) V1
y 21 =
I2
= -0.4 + j1.2 = y 12
V1
To get y 22 refer to the circuit in Fig.(d).
jΩ
I1
1Ω
1Ω
I2
+
+
−
-j Ω
V1 = 0
−
(d)
Z out = j || (1 + 1 || - j) = 0.6 + j0.8
y 22 =
1
= 0.6 − j0.8
Z out
Thus,
 0.6 − j0.8 - 0.4 + j1.2 
[y ] = 
S
 - 0.4 + j1.2 0.6 − j0.8 
Zout
V2
Chapter 19, Solution 23.
(a)
1
1
− y12 = 1 // =
s s +1
y11 + y12 = 1

→

→
y 22 + y12 = s
y12 = −
1
s +1
y11 = 1 − y12 = 1 +

→
1
s+2
=
s +1 s +1
y 22 = s − y12 = s +
s + 2
 s +1
[ y] = 
 −1
 s + 1
1
s2 + s +1
=
s +1
s +1
−1 
s +1 

s 2 + s + 1
s + 1 
(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)
I 2 = y 21V1 + y 22 V2
(4)
From (1) and (3)
Vs − V1 = y11V1 + y12 V2

→
Vs = (1 + y11 )V1 + y12 V2
(5)
From (2) and (4),
− 0.5V2 = y 21V1 + y 22 V2
→
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

→
V2 =
2/s


1
(1 + y11)(0.5 + y 22 )
 y12 −
y 21


2/s
V2 =
−
2
1
 2s + 3  1 s + s + 1 
+ (s + 1)
 +
s +1
s + 1 
 s + 1  2
=
2(s + 1)
s(2s3 + 6s 2 + 7.5s + 3.5)
Chapter 19, Solution 24.
Since this is a reciprocal network, a Π network is appropriate, as shown below.
Y2
Y1
Y3
(a)
4Ω
1/4 S
1/4 S
1/8 S
(b)
4Ω
8Ω
(c)
Y1 = y 11 + y 12 =
Y2 = - y 12 =
1 1 1
− = S,
2 4 4
Z1 = 4 Ω
1
S,
4
Y3 = y 22 + y 21 =
Z2 = 4 Ω
3 1 1
− = S,
8 4 8
Z3 = 8 Ω
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).
4Ω
2Ω
V1
+
−
1
2
+
Vx
1Ω
2 Vx
I2
+
V2 = 0
−
−
(a)
At node 1,
V1 − Vx
V
V
+ 2 Vx = x + x
2
1
4

→ 2 V1 = -Vx
But
I1 =
V1 − Vx V1 + 2 V1
=
= 1.5 V1
2
2
Also,
I2 +
Vx
= 2 Vx
4
y 21 =
(1)

→ y 11 =
I1
= 1 .5
V1

→ I 2 = 1.75 Vx = -3.5 V1
I2
= -3.5
V1
To get y 22 and y 12 , consider the circuit in Fig.(b).
4Ω
2Ω
1
2
+
I1
1Ω
Vx
2 Vx
I2
+
−
V2
−
(b)
At node 2,
I 2 = 2 Vx +
V2 − Vx
4
(2)
At node 1,
2 Vx +
V2 − Vx Vx Vx 3
=
+
= Vx
4
2
1
2

→ V2 = -Vx
Substituting (3) into (2) gives
1
I 2 = 2 Vx − Vx = 1.5 Vx = -1.5 V2
2
y 22 =
I1 =
I2
= -1.5
V2
- Vx V2
=
2
2

→ y 12 =
I1
= 0 .5
V2
(3)
Thus,
 1.5 0.5 
[y ] = 
S
 - 3.5 - 1.5 
Chapter 19, Solution 27.
Consider the circuit in Fig. (a).
4Ω
I1
I2
+
+
−
V1
0.1 V2
−
+
10 Ω
20 I1
V2 = 0
−
(a)
V1 = 4 I 1

→ y 11 =
I 2 = 20 I 1 = 5 V1
I1
I1
=
= 0.25
V1 4 I 1

→ y 21 =
I2
=5
V1
Consider the circuit in Fig. (b).
I1
4Ω
I2
+
V1 = 0
0.1 V2
−
+
10 Ω
20 I1
+
−
V2
−
(b)
4 I 1 = 0.1 V2
I 2 = 20 I 1 +

→ y 12 =
I 1 0 .1
=
= 0.025
4
V2
V2
= 0.5 V2 + 0.1 V2 = 0.6 V2
10

→ y 22 =
I2
= 0 .6
V2
Thus,
 0.25 0.025 
[y ] = 
S
0.6 
 5
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
1
= = 0.25 ,
h11 4
y 12 =
- h 12 0.1
=
= 0.025
4
h 11
y 21 =
h 21 20
=
= 5,
4
h 11
y 22 =
∆ h 0 .4 + 2
=
= 0 .6
4
h 11
as above.
Chapter 19, Solution 28.
We obtain y 11 and y 21 by considering the circuit in Fig.(a).
1Ω
4Ω
I2
+
I1
V1
+
6Ω
(a)
Z in = 1 + 6 || 4 = 3.4
I2 =
V2 = 0
−
−
y 11 =
2Ω
I1
1
=
= 0.2941
V1 Z in
 - 6  V1  - 6
-6
I 1 =    =
V
 10  3.4  34 1
10
y 21 =
I2 - 6
=
= -0.1765
V1 34
To get y 22 and y 12 , consider the circuit in Fig. (b).
1Ω
I1
4Ω
Io
+
+
6Ω
V1 = 0
2Ω
−
V2
I2
−
(b)
1
6  (2)(34 7) 34 V2

= 2 || (4 + 6 || 1) = 2 ||  4 +  =
=
=
y 22
7  2 + (34 7) 24 I 2

y 22 =
24
= 0.7059
34
I1 =
-6
I
7 o
Io =
I1 =
-6
V
34 2

→ y 12 =
2
14
7
I2 =
I2 =
V
2 + (34 7)
48
34 2
I1 - 6
=
= -0.1765
V2 34
Thus,
 0.2941 - 0.1765
[y ] = 
S
 - 0.1765 0.7059 
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
2Ω
8.5 Ω
5.667 Ω
+
8.5 V
+
−
1.889 Ω
V
2Ω
−
(d)
V=
(2 || 1.889)(8.5)
(0.9714)(8.5)
=
= 0.5454
2 || 1.889 + 8.5 + 5.667 0.9714 + 14.167
P=
V 2 (0.5454) 2
=
= 0.1487 W
R
2
Chapter 19, Solution 29.
(a)
Transforming the ∆ subnetwork to Y gives the circuit in Fig. (a).
1Ω
1Ω
Vo
+
10 A
+
2Ω
V1
−
-4 A
V2
−
(a)
It is easy to get the z parameters
z 12 = z 21 = 2 , z 11 = 1 + 2 = 3 ,
z 22 = 3
∆ z = z 11 z 22 − z 12 z 21 = 9 − 4 = 5
y 11 =
z 22 3
= = y 22 ,
∆z 5
y 12 = y 21 =
- z 12 - 2
=
∆z
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
I 1 = 10 =
3
2
V1 − V2
5
5
I 2 = -4 =
-2
3
V1 + V2
5
5
10 = V1 − 1.5 V2

→ 50 = 3 V1 − 2 V2

→ - 20 = -2 V1 + 3 V2

→ V1 = 10 + 1.5 V2
Substituting (2) into (1),
50 = 30 + 4.5 V2 − 2 V2
(1)
(2)

→ V2 = 8 V
V1 = 10 + 1.5 V2 = 22 V
(b)
For direct circuit analysis, consider the circuit in Fig. (a).
For the main non-reference node,
Vo
10 − 4 =

→ Vo = 12
2
10 =
V1 − Vo
1

→ V1 = 10 + Vo = 22 V
-4=
V2 − Vo
1

→ V2 = Vo − 4 = 8 V
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 Ω ,
z 22 = 100 Ω
∆ z = z 11 z 22 − z 12 z 21 = 6000 − 3600 = 2400
h 11 =
∆ z 2400
=
= 24 ,
100
z 22
h 12 =
z 12
60
=
= 0 .6
z 22 100
h 21 =
- z 21
= -0.6 ,
z 22
h 22 =
1
= 0.01
z 22
Thus,
 24 Ω
0.6 
[h] = 

 - 0.6 0.01 S 
(b)
Similarly,
z 11 = 30 Ω
z 12 = z 21 = z 22 = 20 Ω
∆ z = 600 − 400 = 200
h11 =
200
= 10
20
h 21 = -1
h12 =
20
=1
20
h 22 =
1
= 0.05
20
Thus,
 10 Ω
1 
[h] = 

 - 1 0.05 S 
Chapter 19, Solution 31.
We get h11 and h 21 by considering the circuit in Fig. (a).
1Ω
2Ω
V3
V4
1Ω
I2
+
I1
2Ω
V1
4 I1
−
(a)
At node 1,
I1 =
V3 V3 − V4
+
2
2

→ 2 I 1 = 2 V3 − V4
(1)

→ 16 I 1 = -2 V3 + 6 V4
(2)
At node 2,
V3 − V4
V
+ 4 I1 = 4
2
1
8 I 1 = -V3 + 3 V4
Adding (1) and (2),
18 I 1 = 5 V4 
→ V4 = 3.6 I 1
V3 = 3 V4 − 8 I 1 = 2.8 I 1
V1 = V3 + I 1 = 3.8 I 1
h11 =
V1
= 3 .8 Ω
I1
I2 =
- V4
= -3.6 I 1
1

→ h 21 =
I2
= -3.6
I1
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 .
1Ω
I1
1Ω
2Ω
I2
+
2Ω
V1
+
−
4 I1 = 0
V2
−
(b)
V1 =
2
2
V2 = V2
2 + 2 +1
5
I2 =
V2
V2
=
2 + 2 +1 5

→ h12 =

→ h 22 =
V1
= 0 .4
V2
I2 1
= = 0 .2 S
V2 5
Thus,
 38 Ω 0.4 
[h] = 

 - 3.6 0.2 S 
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)


s 
1
I
V1 = 1 + s + s ||  I 1 = 1 + s + 2


s + 1 1
s
h11 =
V1
s
= s +1+ 2
I1
s +1
By current division,
- I1
I2
-1 s
-1
I2 =
I1 =

→ h 21 =
= 2
s +1 s
s +1
I1 s + 1
To get h 22 and h12 , refer to Fig. (b).
I1 = 0
1
s
s
I2
+
+
−
1/s
V1
V2
−
(b)
V1 =
V2
V1
1s
1
V2 = 2

→ h12 =
= 2
s +1 s
s +1
V2 s + 1
 1
V2 = s +  I 2
 s

→ h 22 =
I2
1
s
=
= 2
V2 s + 1 s s + 1
Thus,

s
 s + 1 + s2 + 1
[h] = 
-1

2
s +1
(b)
1 
s +1 
s 

s2 + 1 
2
To get g11 and g 21 , refer to Fig. (c).
I1
1
s
s
I2 = 0
+
V1
+
−
1/s
V2
−
(c)

1
V1 = 1 + s +  I 1

s
V2 =

→ g 11 =
I1
1
s
=
= 2
V1 1 + s + 1 s s + s + 1
V1
V2
1s
1
V1 = 2

→ g 21 =
= 2
1+ s +1 s
s + s +1
V1 s + s + 1
To get g 22 and g 12 , refer to Fig. (d).
1
I1
s
s
+
I2
+
1/s
V1 = 0
−
V2
I2
−
(d)

(s + 1) s 
 1

I2
V2 = s + || (s + 1)  I 2 = s +
 s

 1+ s +1 s 
g 22 =
I1 =
V2
s +1
=s+ 2
s + s +1
I2
- I2
I1
-1 s
-1
I2 = 2

→ g 12 =
= 2
1+ s +1 s
s + s +1
I2 s + s +1
Thus,

 2
[g ] =  s
 2
s

s
-1
2
+s+1
s +s+1 
s+1 
1
s+ 2

+s+1
s +s+1 
Chapter 19, Solution 33.
To get h11 and h21, consider the circuit below.
4Ω
j6 Ω
+
I1
V1 = 5 //( 4 + j6)I1 =
Also, I 2 = −
-j3 Ω
5Ω
V1
-
5
I1
9 + j6
I2
5(4 + j6)I1
9 + j6
+
V2=0
-
V
h11 = 1 = 3.0769 + j1.2821
I1
I
h 21 = 2 = −0.3846 + j0.2564
I1

→
To get h22 and h12, consider the circuit below.
4Ω
j6 Ω
I2
I1
+
V1 =
-j3 Ω
5Ω
V1
-
5
V2
9 + j6
+
-

→
V2 = − j3 //(9 + j6)I 2
h12 =

→
+
V2
V1
5
=
= 0.3846 − j0.2564
V2 9 + j6
I
1
9 + j3
h 22 = 2 =
=
V2 − j3 //(9 + j6) − j3(9 + j6)
= 0.0769 + j0.2821
Thus,
 3.0769 + j1.2821 0.3846 − j0.2564
[h ] = 

− 0.3846 + j0.2564 0.0769 + j0.2821
Chapter 19, Solution 34.
Refer to Fig. (a) to get h11 and h 21 .
300 Ω
10 Ω
50 Ω
2
1
+
I1
V1
−
−
+
100 Ω
Vx
I2
+
+
10 Vx
V2 = 0
−
−
(a)
At node 1,
I1 =
Vx Vx − 0
+
100
300

→ 300 I 1 = 4 Vx
300
I = 75 I 1
4 1
Vx =
V1 = 10 I 1 + Vx = 85 I 1
But
(1)

→ h11 =
V1
= 85 Ω
I1
At node 2,
I2 =
0 + 10 Vx Vx
Vx Vx
75
75
−
=
−
=
I1 −
I = 14.75 I 1
50
300
5 300 5
300 1
h 21 =
I2
= 14.75
I1
To get h 22 and h 12 , refer to Fig. (b).
300 Ω
I1 = 0 10 Ω
50 Ω
1
+
V1
−
+
Vx
−
+
100 Ω
−
(b)
10 Vx
2
I2
+
−
V2
At node 2,
I2 =
V2 V2 + 10 Vx
+
400
50
But
Vx =
V2
100
V2 =
400
4
Hence,
400 I 2 = 9 V2 + 20 V2 = 29 V2
h 22 =

→ 400 I 2 = 9 V2 + 80 Vx
I2
29
=
= 0.0725 S
V2 400
V1 = Vx =
V2
4

→ h 12 =
V1 1
= = 0.25
V2 4
 85 Ω
0.25 
[h] = 

 14.75 0.0725 S 
To get g 11 and g 21 , refer to Fig. (c).
300 Ω
I1
10 Ω
50 Ω
1
V1
Vx
I2 = 0
+
+
+
−
2
100 Ω
−
+
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

→ 350 I 1 = 14.5 Vx
(2)

→ 10 I 1 = V1 − Vx
(3)
Substituting (3) into (2) gives
350 I 1 = 14.5 V1 − 145 I 1
g 11 =

→ 495 I 1 = 14.5 V1
I 1 14.5
=
= 0.02929 S
V1 495
At node 2,
 11

V2 = (50) 
Vx  − 10 Vx = -8.4286 Vx
 350 
 14.5 
= -8.4286 V1 + 84.286 I 1 = -8.4286 V1 + (84.286) 
 V1
 495 
V2 = -5.96 V1

→ g 21 =
V2
= -5.96
V1
To get g 22 and g 12 , refer to Fig. (d).
300 Ω
I1
Io
10 Ω
+
V1 = 0
+
+
Vx
−
Io
50 Ω
−
+
100 Ω
10 Vx
I2
V2
−
−
(d)
10 || 100 = 9.091
I2 =
But
V2 + 10 Vx
V2
+
50
300 + 9.091
309.091 I 2 = 7.1818 V2 + 61.818 Vx
(4)
9.091
V = 0.02941 V2
309.091 2
(5)
Vx =
Substituting (5) into (4) gives
309.091 I 2 = 9 V2
g 22 =
V2
= 34.34 Ω
I2
Io =
34.34 I 2
V2
=
309.091 309.091
I1 =
- 34.34 I 2
- 100
Io =
110
(1.1)(309.091)
g 12 =
I1
= -0.101
I2
Thus,
 0.02929 S - 0.101 
[g ] = 
34.34 Ω 
 - 5.96
Chapter 19, Solution 35.
To get h11 and h 21 consider the circuit in Fig. (a).
1Ω
I1
1:2
4Ω
+
+
V1
V2 = 0
−
−
(a)
ZR =
I2
4
4
=1
2 =
n
4
V1 = (1 + 1) I 1 = 2 I 1

→ h 11 =
V1
= 2Ω
I1
I1 - N 2
I 2 -1
=
= -2 
→ h 21 =
=
= -0.5
I2
N1
I1
2
To get h 22 and h 12 , refer to Fig. (b).
1Ω
I1 = 0
4Ω
1:2
I2
+
−
+
V1
V2
−
(b)
Since I 1 = 0 , I 2 = 0 .
Hence,
h 22 = 0 .
At the terminals of the transformer, we have V1 and V2 which are related as
V2 N 2
V1 1
=
=n=2 
→ h12 =
= = 0 .5
V1 N 1
V2 2
Thus,
 2 Ω 0.5 
[h] = 

 - 0.5 0 
Chapter 19, Solution 36.
We replace the two-port by its equivalent circuit as shown below.
4Ω
I1
16 Ω
2 I1
+
10 V
+
−
V1
I2
+
3 V2
+
−
−
-2 I1
100 Ω V2
−
100 || 25 = 20 Ω
V2 = (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 Ω
I1 =
1
,
14
V2 =
V1 = 16 I 1 + 3 V2 =
40
14
136
14
-8
 100 
I2 = 
 (2 I 1 ) =
70
 125 
(a)
V2
40
=
= 0.2941
V1 136
(b)
I2
= - 1.6
I1
(c)
I1
1
=
= 7.353 × 10 -3 S
V1 136
(d)
V2 40
=
= 40 Ω
1
I1
Chapter 19, Solution 37.
(a)
We first obtain the h parameters. To get h11 and h 21 refer to Fig. (a).
6Ω
3Ω
I2
+
I1
V1
+
6Ω
3Ω
−
−
(a)
3 || 6 = 2
V1 = (6 + 2) I 1 = 8 I 1
V2 = 0

→ h11 =
V1
=8Ω
I1
I2 =
-6
-2
I1 =
I
3+ 6
3 1

→ h 21 =
I2 - 2
=
I1
3
To get h 22 and h12 , refer to the circuit in Fig. (b).
6Ω
I1 = 0
3Ω
I2
+
6Ω
V1
+
−
3Ω
V2
−
(b)
3 || 9 =
9
4
V2 =
9
I
4 2
V1 =
6
2
V2 = V2
6+3
3

→ h 22 =
I2 4
=
V2 9

→ h12 =
V1 2
=
V2 3

2 
8 Ω 3 
[h] =  - 2 4 
S

3 9 
The equivalent circuit of the given circuit is shown in Fig. (c).
I1
8Ω
I2
+
10 V
+
−
2/3 V2
+
−
-2/3 I1
9/4 Ω V2
−
(c)
2
8 I 1 + V2 = 10
3
(1)
5Ω
V2 =
I1 =
2  9  2  45  30
I 5 ||  = I   =
I
3 1  4  3 1  29  29 1
29
V
30 2
(2)
Substituting (2) into (1),
 29 
2
(8)   V2 + V2 = 10
 30 
3
V2 =
(b)
300
= 1.19 V
252
By direct analysis, refer to Fig.(d).
6Ω
3Ω
+
10 V
+
−
6Ω
3Ω
V2
5Ω
−
(d)
10
-A current source. Since
6
6 || 6 = 3 Ω , we combine the two 6-Ω resistors in parallel and transform
10
× 3 = 5 V voltage source shown in Fig. (e).
the current source back to
6
Transform the 10-V voltage source to a
3Ω
3Ω
+
5V
+
−
V2
−
(e)
3 || 5 =
(3)(5) 15
=
8
8
3 || 5 Ω
V2 =
15 8
75
(5) =
= 1.19 V
6 + 15 8
63
Chapter 19, Solution 38.
We replace the two-port by its equivalent circuit as shown below.
200 Ω
I1 800 Ω
I2
+
10 V
+
−
V1
+
10-4 V2
+
−
50 I1
200 kΩ V2
−
Z in =
−
Vs
,
I1
200 || 50 = 40 kΩ
V2 = -50 I 1 (40 × 10 3 ) = (-2 × 10 6 ) I 1
For the left loop,
Vs − 10 -4 V2
= I1
1000
Vs − 10 -4 (-2 × 10 6 I 1 ) = 1000 I 1
Vs = 1000 I 1 − 200 I 1 = 800I 1
Z in =
Vs
= 800 Ω
I1
Alternatively,
Z in = Z s + h11 −
h12 h 21 Z L
1 + h 22 Z L
(10 -4 )(50)(50 × 10 3 )
Z in = 200 + 800 −
= 800 Ω
1 + (0.5 × 10 -5 )(50 × 10 3 )
50 kΩ
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 Ω
-
8x8
= 3.2
20
R1 =
4 x8
= 1.6 = R 2 ,
8+8+ 4
V2 =
13.2
V1 = 0.8919V1
13.2 + 1.6
→
V1 = I1(1.6 + 3.2 + 10) = 14.8I1

→
R3 =
g 21 =
V2
= 0.8919
V1
I
1
= 0.06757
g11 = 1 =
V1 14.8
To get g22 and g12, consider the circuit below.
1.6 Ω
1.6 Ω
I1
+
V1=0
V2
13.2 Ω
-
I2
I1 = −
13.2
I2 = −0.8919I 2
13.2 + 1.6
V2 = I 2 (1.6 + 13.2 // 1.6) = 3.027I 2
I
g12 = 1 = −0.8919
I2

→

→
g 22 =
V2
= 3.027
I2
0.06757 − 0.8919
[g ] = 
3.027 
 0.8919
Chapter 19, Solution 40.
To get g 11 and g 21 , consider the circuit in Fig. (a).
-j6 Ω
I1
j10 Ω
I2 = 0
+
+
−
V1
12 Ω
V2
−
(a)
V1 = (12 − j6) I 1
g 21 =

→ g 11 =
I1
1
=
= 0.0667 + j0.0333 S
V1 12 − j6
12 I 1
V2
2
=
=
= 0.8 + j0.4
V1 (12 − j6) I 1 2 − j
To get g 12 and g 22 , consider the circuit in Fig. (b).
I1
-j6 Ω
j10 Ω
I2
+
V1 = 0
12 Ω
−
(b)
I2
I1 =
- 12
I
12 - j6 2

→ g 12 =
I1
- 12
=
= - g 21 = -0.8 − j0.4
I 2 12 - j6
V2 = ( j10 + 12 || -j6) I 2
g 22 =
V2
(12)(-j6)
= j10 +
= 2.4 + j5.2 Ω
12 - j6
I2
 0.0667 + j0.0333 S - 0.8 − j0.4 
[g ] = 
0.8 + j0.4
2.4 + j5.2 Ω 

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
- g 21
=
I 1 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)
 V2 
V2 = g 21 Vs − [ g 11 Z s Z L + ∆ g Z s + g 22 ]

 ZL 
V2 [ Z L + g 11 Z s Z L + ∆ g Z s + g 22 ]
ZL
= g 21 Vs
V2
g 21 Z L
=
Vs Z L + g 11 Z s Z L + ∆ g Z s + g 22
V2
g 21 Z L
=
Vs Z L + g 11 Z s Z L + g 11 g 22 Z s − g 21 g 12 Z s + g 22
V2
g 21 Z L
=
Vs (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 Ω
I1
I2
+
+
100 Ω
V1
+
−
-0.5 I2
0.5 I1
V2
−
−
(a)
(b)
The network is shown in Fig. (b).
2Ω
I1
+
V1
s
I2
+
10 Ω
+
−
−
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

→ A =
I1 = 0 
→ C =
V1
=1
V2
I1
=0
V2
To get B and D , consider the circuit in Fig. (b).
Z
I1
I2
+
V1
+
−
V2 = 0
−
(b)
V1 = Z I 1 ,
B=
- V1 - Z I 1
=
=Z
I2
- I1
D=
- I1
=1
I2
Hence,
1 Z
[T] = 

0 1 
I 2 = - I1
(b)
To find A and C , consider the circuit in Fig. (c).
I1
I2
+
V1
+
−
Z
V2
−
(c)
V1 = V2

→ A =
V1 = Z I 1 = V2
V1
=1
V2

→ C =
I1
1
= =Y
V2 Z
To get B and D , refer to the circuit in Fig.(d).
I2
+
I1
+
Y
V1
−
−
(d)
V1 = V2 = 0
B=
- V1
= 0,
I2
V2 = 0
I 2 = - I1
D=
- I1
=1
I2
Thus,
 1 0
[T] = 

Y 1
Chapter 19, Solution 44.
To determine A and C , consider the circuit in Fig.(a).
j15 Ω
Io
-j10 Ω
I1
-j20 Ω
Io '
V1
+
−
I2 = 0
Io
20 Ω
+
V2
−
(a)
V1 = [ 20 + (- j10) || ( j15 − j20) ] I 1


(-j10)(-j5) 
10 
V1 =  20 +
I 1 =  20 − j  I 1


- j15 
3

'
I o = I1
 - j10 
2
 I 1 =   I 1
I o = 
3
 - j10 − j5 
V2 = (-j20) I o + 20 I o ' = − j
A=
C=
40 
40

I1 + 20I1 =  20 − j I1
3 
3

V1 (20 − j10 3) I 1
=
= 0.7692 + j0.3461
40 
V2

 20 − j I1
3 

I1
=
V2
1
40
20 − j
3
= 0.03461 + j0.023
To find B and D , consider the circuit in Fig. (b).
j15 Ω
I1
-j10 Ω
-j20 Ω
I2
+
+
−
V1
20 Ω
V2 = 0
−
(b)
We may transform the ∆ subnetwork to a T as shown in Fig. (c).
Z1 =
( j15)(-j10)
= j10
j15 − j10 − j20
Z2 =
40
(-j10)(-j20)
= -j
3
- j15
Z3 =
( j15)(-j20)
= j20
- j15
I1
j10 Ω
j20 Ω
I2
+
V1
+
−
20 – j40/3 Ω
V2 = 0
−
(c)
- I2 =
D=
20 − j40 3
3 − j2
I1 =
I
20 − j40 3 + j20
3+ j 1
- I1
3+ j
= 0.5385 + j0.6923
=
3 − j2
I2

( j20)(20 − j40 3) 
V1 =  j10 +
I
20 − j40 3 + j20  1

V1 = [ j10 + 2 (9 + j7) ] I 1 = j I 1 (24 − j18)
B=
- V1 - j I 1 (24 − j18) 6
=
= (-15 + j55)
- (3 - j2)
I2
13
I1
3+ j
B = -6.923 + j25.385 Ω
 0.7692 + j0.3461 - 6.923 + j25.385 Ω 
[T] = 

 0.03461 + j0.023 S 0.5385 + j0.6923 
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
→
→
I
1
C= 1 =
V2 R1
To obtain B and D, consider the circuit below.
A=
V1 R1 + R 2 + sL
=
V2
R1
I1
sL
1/sC
I2
+
+
V1
R1
V2=0
-
R2
I2 = −
R1
R1 +
1
sC
I1 = −
sR1C
I1
1 + sR1C

→
I
1 + sR1C
D=− 1 =
I2
sR1C
R1 



sC

I1 = − [(1 + sR1C)(R 2 + sL) + R1 ] (1 + sR1C) I 2
V1 = R 2 + sL +
1 
1 + sR1C
sR1C

R1 +


sC 

V
1
[R1 + (1 + sR1C)(R 2 + sL)]
B=− 1 =
I 2 sR1C
Chapter 19, Solution 46.
To get A and C , refer to the circuit in Fig.(a).
I1
V1
1Ω
+
−
1Ω
1
+
Vo
I2 = 0
2
Ix
2Ω
+
4 Ix
V2
−
−
(a)
At node 1,
I1 =
Vo Vo − V2
+
2
1

→ 2 I 1 = 3 Vo − 2 V2
(1)
At node 2,
Vo − V2
4 Vo
= 4Ix =
= 2 Vo
1
2

→ Vo = -V2
(2)
From (1) and (2),
2 I 1 = -5 V2
But
I1 =

→ C =
V1 − Vo
= V1 + V2
1
- 2.5 V2 = V1 + V2
A=
I1 - 5
=
= -2.5 S
V2
2

→ V1 = -3.5 V2
V1
= -3.5
V2
To get B and D , consider the circuit in Fig. (b).
I1
V1
1Ω
+
+
−
1Ω
1
Vo
I2
2
Ix
2Ω
+
4 Ix
V2 = 0
−
−
(b)
At node 1,
I1 =
Vo Vo
+
2
1
I2 +
Vo
+ 4Ix = 0
1

→ 2 I 1 = 3 Vo
(3)
At node 2,
→ I 2 = -3 Vo
– I 2 = Vo + 2 Vo = 0 
Adding (3) and (4),
2 I1 + I 2 = 0 
→ I 1 = -0.5 I 2
D=
- I1
= 0 .5
I2
(4)
(5)
But
I1 =
V1 − Vo
1

→ V1 = I 1 + Vo
(6)
Substituting (5) and (4) into (6),
-1
-1
-5
V1 = I 2 + I 2 =
I
2
3
6 2
B=
- V1 5
= = 0.8333 Ω
I2
6
Thus,
 - 3.5 0.8333 Ω 
[T] = 
- 0.5 
 - 2.5 S
Chapter 19, Solution 47.
To get A and C, consider the circuit below.
6Ω
I1
1Ω
+
V1
-
V1 − Vx Vx Vx − 5Vx
=
+
1
2
10
4Ω
+
Vx
2Ω
-

→
V2 = 4(−0.4Vx ) + 5Vx = 3.4Vx
V − Vx
I1 = 1
= 1.1Vx − Vx = 0.1Vx
1
I2=0
+
5Vx
+
V2
-
-
V1 = 1.1Vx

→
→
A=
V1
= 1.1 / 3.4 = 0.3235
V2
I
C = 1 = 0.1 / 3.4 = 0.02941
V2
Chapter 19, Solution 48.
(a)
Refer to the circuit below.
I2
I1
+
V1
+
−
[T]
V2
ZL
−
V1 = 4 V2 − 30 I 2
I 1 = 0.1 V2 − I 2
(1)
(2)
When the output terminals are shorted, V2 = 0 .
So, (1) and (2) become
V1 = -30 I 2
and
I1 = - I 2
Hence,
V1
= 30 Ω
Z in =
I1
(b)
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)
V1
= 40 Ω
I1
When the output port is terminated by a 10-Ω load, V2 = -10 I 2 .
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
= 35 Ω
I1
Alternatively, we may use Z in =
A ZL + B
CZL + D
Chapter 19, Solution 49.
To get A and C , refer to the circuit in Fig.(a).
1/s
I1
I2 = 0
+
V1
+
−
1Ω
1/s
1/s
1Ω
V2
−
(a)
1s
1
1
1 || =
=
s 1+1 s s +1
V2 =
1 || 1 s
V
1 s + 1 || 1 s 1
1
V2
s
s +1
A=
=
=
1
V1 1
2s + 1
+
s s +1
 1  1
 1   2s + 1 
1 

 ||  +
 = I1
 || 
V1 = I 1 
 s + 1  s s + 1
 s + 1   s (s + 1) 
 1   2s + 1 

⋅

V1  s + 1   s (s + 1) 
2s + 1
=
=
1
2s + 1
(s + 1)(3s + 1)
I1
+
s + 1 s (s + 1)
But
V1 = V2 ⋅
2s + 1
s
V2 2s + 1
2s + 1
⋅
=
s
(s + 1)(3s + 1)
I1
Hence,
C=
V2 (s + 1)(3s + 1)
=
I1
s
To get B and D , consider the circuit in Fig. (b).
1/s
I1
I2
+
V1
+
−
1Ω
1/s
1/s
1Ω
V2 = 0
−
(b)
I
 1 1
 1
V1 = I 1 1 || ||  = I 1 1 ||  = 1
 s s
 2s  2s + 1
-1
I
-s
s +1 1
I2 =
=
I
1
1 2s + 1 1
+
s +1 s
D=
- I 1 2s + 1
1
=
= 2+
I2
s
s
I
 1  2s + 1 

I2 = 2
V1 = 
 2s + 1  - s 
-s
Thus,

2

2s + 1
[T] = 
(s + 1)( 3s + 1)

s


1
2+ 
s
1
s
Chapter 19, Solution 50.
To get a and c, consider the circuit below.

→ B =
- V1 1
=
I2
s
I1=0
2
s
I2
+
+
4/s
V1
V2
-
V1 =
-
4/s
4
V2 =
V2
2
s + 4/s
s +4

→
a = V2
V1
= 1 + 0.25s 2
V2 = (s + 4 / s)I 2 or
V2
(1 + 0.25s 2 )V1
I2 =
=
s + 4/s
s + 4/s

→
I 2 s + 0.25s3
c=
=
V1
s2 + 4
To get b and d, consider the circuit below.
I1
2
s
I2
+
+
V1=0
4/s
V2
-
I1 =
− 4/s
2I
I2 = − 2
2 + 4/s
s+2
-

→
I
d = − 2 = 1 + 0.5s
I1
4
(s 2 + 2s + 4)
V2 = (s + 2 // )I2 =
I2
s
s+2
=−
(s 2 + 2s + 4)( s + 2)
I1
s+2
2

→
V
b = − 2 = 0.5s 2 + s + 2
I1
 0.25s 2 + 1 0.5s 2 + s + 2


[ t ] =  0.25s 2 + s
0.5s + 1 

 s2 + 4


Chapter 19, Solution 51.
To get a and c , consider the circuit in Fig. (a).
jΩ
I1 = 0
1Ω
-j3 Ω
I2
+
j2 Ω
V1
+
−
jΩ
V2
−
(a)
V2 = I 2 ( j − j3) = -j2 I 2
V1 = -jI 2
a=
V2 - j2 I 2
=
=2
V1
- jI 2
c=
I2
1
=
=j
V1 - j
To get b and d , consider the circuit in Fig. (b).
jΩ
I1
1Ω
-j3 Ω
I2
+
V1 = 0
j2 Ω
jΩ
−
(b)
For mesh 1,
0 = (1 + j2) I1 − j I 2
or
I 2 1 + j2
=
= 2− j
I1
j
+
−
V2
d=
- I2
= -2 + j
I1
For mesh 2,
V2 = I 2 ( j − j3) − j I 1
V2 = I 1 (2 − j)(- j2) − j I 1 = (-2 − j5) I 1
b=
- V2
= 2 + j5
I1
Thus,
 2 2 + j5 
[t ] = 

 j -2+ j
Chapter 19, Solution 52.
It is easy to find the z parameters and then transform these to h parameters and T
parameters.
 R1 + R 2
[z ] = 
 R2

R 2 + R 3 
R2
∆ z = (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   R 1 R 2 + R 2 R 3 + R 3 R 1
z 22  
R2 + R3
=
- R2
1
 
z 22  
R2 + R3
R2 
R2 + R3 

1

R2 + R3 
Thus,
h 11 = R 1 +
R 2R 3
,
R2 + R3
h 12 =
R2
= - h 21 ,
R2 + R3
as required.
(b)
 z 11
z
[T] =  21
1

 z 21
∆ z   R1 + R 2
z 21   R 2
 
z 22 =
1
 
z 21   R 2
R 1R 2 + R 2 R 3 + R 3 R 1 

R2

R2 + R3

R2

h 22 =
1
R2 + R3
Hence,
A = 1+
R3
R1
R1
1
, B = R3 +
(R 2 + R 3 ) , C =
, D = 1+
R2
R2
R2
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),
I
D
V2 = 1 + I 2
C C
Comparing (2) and (5),
1
z 21 = ,
C
(3)
(4)
(5)
z 22 =
D
C
Substituting (5) into (3),
 AD

A
V1 = I1 + 
− B I 2
 C

C
=
A
AD − BC
I1 +
I2
C
C
Comparing (6) and (1),
A
z11 =
C
Thus,
A

[Z] =  C
1

C
(6)
z 12 =
∆T 
C
D

C
AD − BC ∆ T
=
C
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 
z11

V2 =
∆
z 21
V1 + z I 2
z11
z11
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
z
1
, g 12 = 12 , g 21 = 21 ,
g 11 =
z 11
z 11
z 11
(4)
g 22 =
∆z
z 11
as required.
Chapter 19, Solution 56.
(a) ∆ y = (2 + j)(3 − j) + j4 = 7 + j5
 y 22 / ∆ y
[z] = 
− y 21 / ∆ y
− y12 / ∆ y  0.2162 − j0.2973 − 0.2703 − j0.3784
=
Ω
y11 / ∆ y  0.0946 − j0.0676 0.2568 − j0.0405 
− y12 / y11   0.4 − j0.2 − 0.8 − j1.6
 1 / y11
(b) [h ] = 
=

 y 21 / y11 ∆ y / y11  − 0.4 + j0.2 3.8 + j0.6 
− y11 / y12
(c ) [ t ] = 
 − ∆ y / y12
− 1 / y12   − 0.25 + j0.5
j0.25 
=

− y 22 / y12  − 1.25 + j1.75 0.25 + j0.75
Chapter 19, Solution 57.
∆ T = (3)(7) − (20)(1) = 1
A

[z ] =  C
1

C
∆T 
C =  3 1Ω
D   1 7 

C
D

[y ] =  B
-1

B
- ∆T  
B = 
A  
 
B  
B

[h] =  -D1

D
1 
∆ T   20
Ω

D = 7
7 
C   -1
1 
S
 
D
7
7 
C

[g ] =  A
1

A
- ∆T 
A =
B 

A 
D
∆
[t ] =  CT

 ∆T
7
20
-1
20
-1
20
3
20
1
3S
 1

3


S

-1 
3 
20 
Ω

3
B 
∆ T   7 20 Ω 
A  =  1 S
3 

∆T 
Chapter 19, Solution 58.
The given set of equations is for the h parameters.
1 Ω
2 
∆ h = (1)(0.4) − (2)(-2) = 4.4
[h] = 

 - 2 0.4 S
(a)
 1
h
[y ] =  11
h
 21
 h11
(b)


[T] = 


- ∆h
h 21
- h 22
h 21
- h12
h11
∆h
h11

  1 -2 
=
S
  - 2 4.4 

- h11
h 21
-1
h 21

  2.2 0.5 Ω 
= 

  0.2 S 0.5 

Chapter 19, Solution 59.
∆ g = (0.06)(2) − (-0.4)(2) = 0.12 + 0.08 = 0.2
(a)


[z] = 


1
g 11
g 21
g 11
∆g
(b)


[y ] = 


(c)
 g 22
 ∆
g
[h] =  - g
 21
 ∆ g
- g 12 
2 
∆ g   10 Ω
=
g 11   - 1 0.3 S 
∆ g 
(d)


[T] = 


g 22 
10 Ω 
g 21   5
= 
∆g
1 
  0.3 S
g 21 
g 22
- g 21
g 22
1
g 21
g 11
g 21
- g 12 
g 11   16.667 6.667 

∆ g =  3.333 3.333  Ω

g 11 
g 12
g 22
1
g 22

  0.1 - 0.2 
=
S
  - 0.1 0.5 

Chapter 19, Solution 60.
∆ y = y 11 y 22 − y 12 y 21 = 0.3 − 0.02 = 0.28
(a)


[z ] = 


y 22
∆y
- y 21
∆y
- y 12
∆y
y 11
∆y

  1.786 0.7143 
= 
Ω
  0.3571 2.143 

- y 12 
y 11   1.667 Ω 0.3333 

∆ y =  - 0.1667 0.4667 S 

y 11 
(b)


[h] = 


1
y 11
y 21
y 11
(c)


[t ] = 


- y 11
y 12
- ∆y
y 12
-1 
5Ω
y 12   3
=
- y 22   1.4 S 2.5 
y 12 
Chapter 19, Solution 61.
(a)
To obtain z 11 and z 21 , consider the circuit in Fig. (a).
1Ω
Io
1Ω
1Ω
+
I1
+
1Ω
V1
−
 2 5
V1 = I 1 [1 + 1 || (1 + 1) ] = I 1 1 +  = I 1
 3 3
Io =
V1 5
=
I1 3
1
1
I1 = I1
1+ 2
3
- V2 + I o + I 1 = 0
1
4
V2 = I 1 + I 1 = I 1
3
3
V2
−
(a)
z 11 =
I2 = 0
z 21 =
V2 4
=
I1
3
To obtain z 22 and z 12 , consider the circuit in Fig. (b).
1Ω
1Ω
I1
1Ω
+
+
1Ω
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


[z ] = 


5
3
4
3
4
3
Ω
5
3
(b)


[h] = 


∆z
z 22
- z 21
z 22
z 12
z 22
1
z 22
 
 
= 
 
 
(c)


[T] = 


z 11
z 21
1
z 21
∆z
z 21
z 22
z 21
 
 
=
 
 
Chapter 19, Solution 62.
Consider the circuit shown below.
3
4 
Ω
5
5 
-4 3 
S
5
5 
5
4
3
S
4
3 
Ω
4 
5 

4 
I1
10 kΩ
a
40 kΩ
+
−
+
I2
+
50 kΩ
b
30 kΩ
V1
Ib
20 kΩ
−
Since no current enters the input terminals of the op amp,
V1 = (10 + 30) × 10 3 I 1
But
Va = Vb =
V2
−
(1)
30
3
V1 = V1
40
4
Vb
3
V
3 =
20 × 10
80 × 10 3 1
which is the same current that flows through the 50-kΩ resistor.
Ib =
Thus,
V2 = 40 × 10 3 I 2 + (50 + 20) × 10 3 I b
V2 = 40 × 10 3 I 2 + 70 × 10 3 ⋅
V2 =
3
V
80 × 10 3 1
21
V + 40 × 10 3 I 2
8 1
V2 = 105 × 10 3 I 1 + 40 × 10 3 I 2
From (1) and (2),
 40 0 
[z ] = 
 kΩ
 105 40 
∆ z = z 11 z 22 − z 12 z 21 = 16 × 10 8
(2)

A B 
[T] = 
=
C D 

∆z
z 21
z 22
z 21
z 11
z 21
1
z 21

  0.381 15.24 kΩ 
=
0.381 
  9.52 µS

Chapter 19, Solution 63.
To get z11 and z21, consider the circuit below.
1:3
I1
•
+
4Ω
V1
I2=0
• +
+
V’1
V’2
-
-
+
9Ω
-
V2
-
ZR =
9
n2
= 1,
V1 = (4 // ZR )I1 =
n = 3
4
I1
5

→
V2 = V2 ' = nV1' = nV1 = 3(4 / 5)I1
V
z11 = 1 = 0.8
I1

→
z 21 =
V2
= 2.4
I1
To get z21 and z22, consider the circuit below.
I1=0
1:3
•
+
V1
4Ω
• +
+
V’1
V’2
-
-
-
Z R ' = n 2 ( 4 ) = 36 ,
I2
+
9Ω
V2
-
n =3
V2 = (9 // ZR ' )I 2 =
V1 =
9x36
I2
45
V2 V2
=
= 2.4I 2
n
3
→
→
z 22 =
V2
= 7.2
I2
V
z 21 = 1 = 2.4
I2
Thus,
0.8 2.4
[z] = 
Ω
2.4 7.2
Chapter 19, Solution 64.
1
-j
=
= - j kΩ
3
jωC (10 )(10 -6 )
1 µF 
→
Consider the op amp circuit below.
40 kΩ
I1
20 kΩ
Vx
10 kΩ
1
+
−
+
2
-j kΩ
V1
I2
+
V2
−
−
At node 1,
V1 − Vx Vx Vx − 0
=
+
20
-j
10
V1 = (3 + j20) Vx
(1)
At node 2,
Vx − 0 0 − V2
=
10
40
But
I1 =
V1 − Vx
20 × 10 3
Substituting (2) into (3) gives

→ Vx =
-1
V
4 2
(2)
(3)
I1 =
V1 + 0.25 V2
= 50 × 10 -6 V1 + 12.5 × 10 -6 V2
20 × 10 3
(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
 50 × 10 -6
[y ] = 
1

12.5 × 10 -6 
S
0.75 + j5 
∆ y = (77.5 + j25. − 12.5) × 10 -6 = (65 + j250) × 10 -6


[h] = 


1
y 11
y 21
y 11
- y 12 
- 0.25 
y 11   2 × 10 4 Ω

∆ y =  2 × 10 4 1.3 + j5 S 

y 11 
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.
4 2
For N a ,
[z a ] = 

2 2
For N b ,
 2 1
[z b ] = 

 1 1
6 3
[z ] = [z a ] + [z b ] = 

3 3
∆ z = 18 − 9 = 9
 z 22
 ∆
[y ] =  z
-z
 21
 ∆z
- z 12  
∆z  
=
z 11
 
∆z  
1
3
-1
3
-1 
3 S
2 

3 
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 × 10 -3 )(10 × 10 -3 ) − 0 = 20 × 10 -6


[z a ] = 


y 22
∆y
- y 21
∆y
- y 12
∆y
y 11
∆y

  500 Ω
0 
=
100 Ω 
  0

 500 0   100 100   600 100 
[z ] = 
+
=

 0 100   100 100   100 200 
i.e.
V1 = z 11 I 1 + z 12 I 2
V2 = z 21 I 1 + z 22 I 2
or
V1 = 600 I 1 + 100 I 2
V2 = 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 = -300 I 2
(4)
From (2) and (4),
100 I 1 + 200 I 2 = -300 I 2
I 1 = -5I 2
(5)
Substituting (1) and (5) into (3),
Vs = 600 I 1 + 100 I 2 + 60 I 1
= (660)(-5) I 2 + 100 I 2
= -3200I 2
(6)
From (4) and (6),
Vo
- 300 I 2
=
= 0.09375
V2 - 3200 I 2
Chapter 19, Solution 67.
The y parameters for the upper network is
 2 -1 
[y ] = 
,
 -1 2 


[z a ] = 


- y 12
∆y
y 11
∆y
y 22
∆y
- y 21
∆y
 2
 
= 3
 1
  3
∆ y = 4 −1 = 3
1
3

2
3
1 1
[z b ] = 

1 1
5 3 4 3
[z ] = [z a ] + [z b ] = 

4 3 5 3
∆z =
25 16
−
=1
9
9


[T] = 


z 11
z 21
1
z 21
∆z
z 21
z 22
z 21

  1.25 0.75 Ω 
=

  0.75 S 1.25 

Chapter 19, Solution 68.
 4 -2
For the upper network N a , [y a ] = 

-2 4 
 2 -1 
and for the lower network N b , [y b ] = 

1 2 
For the overall network,
 6 -3
[y ] = [y a ] + [y b ] = 

 -3 6 
∆ y = 36 − 9 = 27


[h] = 


1
y 11
y 21
y 11
- y 12   1
-1 
y 11   6 Ω 2 

∆y =  -1 9 
 
S
2 
y 11   2
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 =
1s 4
=
n2 s
 2s + 4 

4
I
V1 = (2 + Z R ) I 1 =  2 +  I 1 = 
 s  1

s
I1
s
=
V1 2 (s + 2)
y 11 =
I2 =
- s V1
- I1
= -2 I 1 =
s+2
n
y 21 =
I2
-s
=
V1 s + 2
To get y 22 and y 12 , consider the circuit in Fig. (b).
I1
2Ω
1/s
2:1
I2
+
+
V1 =0
V2
−
−
I2
(b)
1
1
Z R ' = (n 2 )(2) =   (2) =
4
2
s + 2
1

 1 1
I
V2 =  + Z R '  I 2 =  +  I 2 = 
 2s  2
s

 s 2
y 22 =
I2
2s
=
V2 s + 2
 - 1  2s 
 -s 
 V2 = 
V
I 1 = - n I 2 =  
 2  s + 2 
s + 2 2
y 12 =
I1
-s
=
V2 s + 2

s
 2 (s + 2)
[y a ] = 
-s

 s+2
-s 
s+2 
2s 

s+2 
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 I 1

→ y 11 =
I1 1
=
V1 2
I 2 = - I1 =
- V1
2
I 2 -1
=
V1 2

→ y 21 =
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 = - I 2 ⋅
y 12 =
2s
I
s+2 2

→ y 22 =
I2 s + 2
=
2s
V2
- V2
 - s  s + 2 
-s

 V2 =
=
s + 2  s + 2  2s 
2
I1 - 1
=
2
V2
12
-1 2 
[y b ] = 

 - 1 2 (s + 2) 2s 
 s+1
 s+2
[y ] = [y a ] + [y b ] = 
 - (3s + 2)
 2 (s + 2)
- (3s + 2)
2 (s + 2)
5s 2 + 4s + 4
2s (s + 2)





Chapter 19, Solution 70.
We may obtain the g parameters from the given z parameters.
 25 20 
[z a ] = 
∆ z a = 250 − 100 = 150
,
 5 10 
 50 25 
[z b ] = 
,
 25 30 
∆ z b = 1500 − 625 = 875


[g ] = 


1
z 11
z 21
z 11
- z 12
z 11
∆z
z 11





 0.04 - 0.8 
[g a ] = 
,
6 
 0.2
 0.02 - 0.5 
[g b ] = 

 0.5 17.5 
 0.06 S - 1.3 
[g ] = [g a ] + [ g b ] = 
23.5 Ω 
 0.7
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 = −2I 2
2
Comparing this with
V1 = AV2 − BI2 ,
I1 = CV2 − DI 2
shows that
0.5 0
[Tb1] = 

 0 2
To get A and C for Tb2 , consider the circuit below.
I1
+
V1
-
4Ω
I2 =0
5Ω
2Ω
+
V2
-
V1 = 9I1,
A=
V2 = 5I1
V1
= 9 / 5 = 1.8,
V2
I
C = 1 = 1 / 5 = 0.2
V2
Chapter 19, Solution 72.
Consider the network shown below.
I1
Ia1
+
Va1
+
V1
Ia2
Na
Ib1
+
Vb1
−
+
Va2
+
V2
Ib2
Nb
Va1 = 25 I a1 + 4 Va 2
I a 2 = - 4 I a1 + Va 2
Vb1 = 16 I b1 + Vb 2
I b 2 = - I b1 + 0.5 Vb 2
V1 = Va1 + Vb1
V2 = Va 2 = Vb 2
I 2 = I a 2 + I b2
I 1 = 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
(10)
I b1 = I a1 = I 1
(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
 57.67 3.333 
[z ] = 
Ω
 3.333 0.6667 
Alternatively,
 25 4 
[h a ] = 
,
-4 1
 16 1 
[h b ] = 

 - 1 0.5 
 41 5 
[h] = [h a ] + [h b ] = 

 - 5 1.5 
 ∆h
 h
[z ] =  22
-h
 21
 h 22
as obtained previously.
h12
h 22
1
h 22
∆ h = 61.5 + 25 = 86.5

  57.67 3.333 
=
Ω
3
.
333
0
.
6667




Chapter 19, Solution 73.
From Example 18.14 and the cascade two-ports,
2 3
[Ta ] = [Tb ] = 

1 2
 2 3  2 3   7 12 Ω 
[T] = [Ta ][Tb ] = 
= 

7 
 1 2  1 2   4 S
When the output is short-circuited, V2 = 0 and by definition
V1 = - B I 2 ,
I1 = - D I 2
Hence,
V1 B 12
Z in =
= =
Ω
I1 D 7
Chapter 19, Solution 74.
From Prob. 18.35, the transmission parameters for the circuit in Figs. (a) and (b) are
1 Z
[Ta ] = 
,
0 1 
 1 0
[Tb ] = 

1 Z 1 
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
1 0 
[T1 ] = 
,
1 1 
1 s
[T2 ] = 
,
 0 1
1 0 
[T3 ] = 

s 1
[T4 ] = [T2 ] ,
[T5 ] = [T1 ] ,
[T6 ] = [T3 ]
 1 0  1 0 
[T] = [T1 ][T2 ][T3 ][T4 ][T5 ][T6 ] = [T1 ][T2 ][T3 ][T4 ]


 1 1  s 1 
0
0
1 s 1
 1
= [T1 ][T2 ][T3 ] 
= [T1 ][T2 ][T3 ][T4 ] 




 0 1   s +1 1 
 s +1 1 
 1 0   s2 + s +1 s 
= [T1 ][T2 ] 


1
 s 1   s +1
s 
 1 s   s2 + s +1
= [T1 ] 
 3


2
2
 0 1   s + s + 2s + 1 s + 1 
 1 0   s 4 + s 3 + 3s 2 + 2s + 1 s 3 + 2s 
=


3
2
s2 +1 
 1 1   s + s + 2s + 1
 s 4 + s 3 + 3s 2 + 2s + 1

s 3 + 2s
[T] =  4

3
2
3
2
 s + 2s + 4s + 4s + 2 s + s + 2s + 1 
Note that AB − CD = 1 as expected.
Chapter 19, Solution 75.
(a) We convert [za] and [zb] to T-parameters. For Na, ∆ z = 40 − 24 = 16 .
∆ z / z 21   2
4 
z / z
=
[Ta ] =  11 21


 1 / z 21 z 22 / z 21  0.25 1.25
For Nb, ∆ y = 80 + 8 = 88 .
− y 22 / y 21 − 1 / y 21   − 5 − 0.5
[Tb ] = 
=

 − ∆ y / y 21 − y11 / y 21  − 44 − 4 
− 17 
 − 186
[T] = [Ta ][Tb ] = 

− 56.25 − 5.125
We convert this to y-parameters. ∆ T = AD − BC = −3.
 D / B − ∆ T / B 0.3015 − 0.1765
=
[ y] = 

A / B  0.0588 10.94 
− 1 / B
(b)
The equivalent z-parameters are
A / C ∆ T / C  3.3067 0.0533
[z] = 
=

 1 / C D / C  − 0.0178 0.0911
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
→
I 2 = −Vo / ZL
(3)
From (2) and (3) ,
V
Vo = z 21I1 − z 22 o
ZL
→

 1
z
I1 = Vo 
+ 22 
 z 21 ZL z 21 
(4)
Substituting (3) and (4) into (1) gives
Vi  z11 z11z 22  z12
−
=
+
= −194.3
Vo  z 21 z 21ZL  ZL
→
Vo.
= −0.0051
Vi
Chapter 19, Solution 76.
To get z11 and z21, we open circuit the output port and let I1 = 1A so that
V
V
z11 = 1 = V1, z 21 = 2 = 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
V
z12 = 1 = V1,
I2
z 22 =
V2
= V2
I2
The schematic is shown below. After it is saved and run, we obtain
z12 = V1 = 1.122,
z 22 = V2 = 3.849
Thus,
3.949 1.122 
[z] = 
Ω
1.122 3.849
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($N_0001)
1.592 E–01
9.488 E–01
–1.616 E+02
From this we obtain
h11 = V1/1 = 0.9488∠–161.6°
h21 = I2/1 = 0.3163∠–161.6°.
(b)
In this case, we set I1 = 0 and V2 = 1V. The schematic is shown below. In the
AC Sweep box, we set Total Pts = 1, Start Freq = 0.1592, and End Freq = 0.1592.
After simulation, we obtain an output file which includes
FREQ
VM($N_0001)
VP($N_0001)
1.592 E–01
3.163 E–.01
1.842 E+01
FREQ
IM(V_PRINT2)
IP(V_PRINT2)
1.592 E–01
9.488 E–01
–1.616 E+02
From this,
h12 = V1/1 = 0.3163∠18.42°
h21 = I2/1 = 0.9488∠–161.6°.
Thus,
0.9488∠ − 161.6° 0.3163∠18.42° 
[h] = 

0.3163∠ − 161.6° 0.9488∠ − 161.6°
Chapter 19, Solution 78
For h11 and h21, short-circuit the output port and let I1 = 1A. f = ω / 2π = 0.6366 . The
schematic is shown below. When it is saved and run, the output file contains the
following:
FREQ
IM(V_PRINT1)IP(V_PRINT1)
6.366E-01
FREQ
1.202E+00
1.463E+02
VM($N_0003) VP($N_0003)
6.366E-01
3.771E+00
-1.350E+02
From the output file, we obtain
I 2 = 1.202∠146.3o ,
V1 = 3.771∠ − 135o
so that
V
h11 = 1 = 3.771∠ − 135o ,
1
I
h 21 = 2 = 1.202∠146.3o
1
For h12 and h22, open-circuit the input port and let V2 = 1V. The schematic is shown
below. When it is saved and run, the output file includes:
FREQ
VM($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
I 2 = 0.3727∠ − 153.4o ,
V1 = 1.202∠ − 33.69o
so that
V
h12 = 1 = 1.202∠ − 33.69o ,
1
I
h 22 = 2 = 0.3727∠ − 153.4o
1
Thus,
3.771∠ − 135o
[h ] = 
 1.202∠146.3
1.202∠ − 33.69o 

0.3727∠ − 153.4o 
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∠–136.7°/1 = 4.669∠–136.7°
z21 = V2/I1 = 2.53∠–108.4°/1 = 2.53∠–108.4°.
(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∠–108.4°/1 = 2.53∠–108..4°
Thus,
z22 = V2/I2 = 1.789∠–153.4°/1 = 1.789∠–153.4°.
4.669∠ − 136.7° 2.53∠ − 108.4° 
[z] = 

 2.53∠ − 108.4° 1.789∠ − 153.4°
Chapter 19, Solution 80
To get z11 and z21, we open circuit the output port and let I1 = 1A so that
V
z11 = 1 = V1,
I1
z 21 =
V2
= V2
I1
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
V
z12 = 1 = V1,
I2
z 22 =
V2
= V2
I2
The schematic is shown below. After it is saved and run, we obtain
z12 = V1 = 3.704,
z 22 = V2 = 11.11
Thus,
 29.88 3.704
[z] = 
Ω
− 70.37 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
1.5 − 0.5
[Y] = 

 3.5 1.5 
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,
 3.8 0.4 
[h] = 

 3.6 0.25
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
= 0.3235,
V2
I
1
C= 1 =
= 0.02941
V2 34
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.
V
2.5
= 1.1765,
B=− 1 =
I 2 2.125
I
1
= 0.4706
D=− 1 =
I 2 2.125
Thus,
 0.3235 1.1765 
[T ] = 

0.02941 0.4706
Chapter 19, Solution 84
(a)
Since A =
V1
V2
and C =
I 2 =0
I1
V2
, we open-circuit the output port and let V1
I 2 =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
A B 
1.4 − 0.8
 C D = 1.4 − 1.8 




Thus
Chapter 19, Solution 85
(a)
Since A =
V1
V2
and C =
I 2 =0
I1
V2
, we let V1 = 1 V and openI 2 =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($N_0002)
–7.159 E+01
From this, we obtain
A =
1
1
=
= 1.581∠71.59°
V2 0.6325∠ − 71.59°
C =
(b)
I1
0.6325∠18.43°
= 1∠90° = j
=
V2 0.6325∠ − 71.59°
Similarly, since B =
V1
I2
and D = −
V2 = 0
I1
I2
, we let V1 = 1 V and shortV2 = 0
circuit the output port. The schematic is shown below. Again, we set Total Pts = 1, Start
Freq = 0.1592, and End Freq = 0.1592 in the AC Sweep box. After simulation, we get
an output file which includes the following results:
FREQ
1.592 E–01
IM(V_PRINT1)
5.661 E–04
IP(V_PRINT1)
8.997 E+01
FREQ
1.592 E–01
IM(V_PRINT3)
9.997 E–01
IP(V_PRINT3)
–9.003 E+01
From this,
B = −
1
1
=−
= −1∠90° = − j
I2
0.9997∠ − 90°
D = −
I1
5.661x10 −4 ∠89.97°
=−
= 5.561x10–4
I2
0.9997∠ − 90°
−j
A B 
1.581∠71.59°

 C D = 
−4 
j
5.661x10 



Chapter 19, Solution 86
(a)
By definition, g11 =
I1
V1
, g21 =
I 2 =0
V1
V2
.
I 2 =0
We let V1 = 1 V and open-circuit the output port. The schematic is shown below. After
simulation, we obtain
g11 = I1 = 2.7
g21 = V2 = 0.0
(b)
Similarly,
g12 =
I1
I2
, g22 =
V1 = 0
V2
I2
V1 = 0
We let I2 = 1 A and short-circuit the input port. The schematic is shown below. After
simulation,
g12 = I1 = 0
g22 = V2 = 0
 2.727S 0
[g] = 
0
 0
Thus
Chapter 19, Solution 87
(a)
Since
a =
V2
V1
and c =
I1 = 0
I2
V1
,
I1 = 0
we open-circuit the input port and let V2 = 1 V. The schematic is shown below. In the
AC Sweep box, set Total Pts = 1, Start Freq = 0.1592, and End Freq = 0.1592. After
simulation, we obtain an output file which includes
FREQ
1.592 E–01
IM(V_PRINT2)
5.000 E–01
IP(V_PRINT2)
1.800 E+02
FREQ
1.592 E–01
VM($N_0001)
5.664 E–04
VP($N_0001)
8.997 E+01
From this,
a =
1
= 1765∠ − 89.97°
5.664x10 −4 ∠89.97°
c =
0.5∠180°
= −882.28∠ − 89.97°
5.664x10 − 4 ∠89.97°
(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 ∠ − 90.1°
= –j1765
0.5∠180°
= j888.28
5.664x10 − 4 ∠ − 90.1°
 − j1765 − j1765
[t] = 

 j888.2 j888.2 
Chapter 19, Solution 88
To get Z in , consider the network in Fig. (a).
Rs
I1
I2
+
Vs
+
−
+
Two-Port
V1
RL
V2
−
−
Zin
(a)
I 1 = y 11 V1 + y 12 V2
I 2 = y 21 V1 + y 22 V2
But
I2 =
- V2
= y 21 V1 + y 22 V2
RL
V2 =
- y 21 V1
y 22 + 1 R L
(1)
(2)
(3)
Substituting (3) into (1) yields
 - y 21 V1 
,
I 1 = y 11 V1 + y 12 ⋅ 
 y 22 + 1 R L 
 ∆ y + y 11 YL 
 V1 ,
I1 = 
 y 22 + YL 
or
YL =
∆ y = y 11 y 22 − y 12 y 21
Z in =
y 22 + YL
V1
=
I 1 ∆ y + y 11 YL
Ai =
y
I2
y V + y 22 V2
= 21 1
= y 21 Z in +  22
I1
I1
 I1
= y 21 Z in −
y 22 y 21 Z in  y 22 + YL
=
∆ +y Y
y 22 + YL
11
L
 y
Ai =
y 21 YL
∆ y + y 11 YL
Av =
V2
- y 21
=
V1 y 22 + YL
From (3),
1
RL
  - y 21 V1
 
  y 22 + YL




 y 21 − y 22 y 21

y 22 + YL




To get Z out , consider the circuit in Fig. (b).
I1
I2
+
Rs
V1
+
−
Two-Port
V2
−
(b)
Z out =
But
V2
V2
=
I 2 y 21 V1 + y 22 V2
V1 = - R s I 1
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
y 12 V2
=
1 + y 11 R s
Rs
- y 12 R s
V1
=
V2 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
=
Z out =
y 22
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 × 16 × 10 -6 − 2.6 × 10 -4 × 72) ⋅ 10 5
Av =
- 72 ⋅ 10 5
= - 1613
2640 + 1824
dc gain = 20 log A v = 20 log (1613) = 64.15
Chapter 19, Solution 90
(a)
Z in = h ie −
h re h fe R L
1 + h oe R L
10 -4 × 120 R L
1500 = 2000 −
1 + 20 × 10 -6 R L
500 =
12 × 10 -3
1 + 2 × 10 -5 R L
500 + 10 -2 R L = 12 × 10 -3 R L
500 × 10 2 = 0.2 R L
R L = 250 kΩ
(b)
Av =
- h fe R L
h ie + (h ie h oe − h re h fe ) R L
- 120 × 250 × 10 3
Av =
2000 + (2000 × 20 × 10 -6 − 120 × 10 -4 ) × 250 × 10 3
- 30 × 10 6
= - 3333
Av =
2 × 10 3 + 7 × 10 3
Ai =
(c)
h fe
120
=
= 20
1 + h oe R L 1 + 20 × 10 -6 × 250 × 10 3
Z out =
R s + h ie
600 + 2000
=
(R s + h ie ) h oe − h re h fe (600 + 2000) × 20 × 10 -6 − 10 -4 × 120
Z out =
2600
kΩ = 65 kΩ
40
Vc Vc
=
Vb Vs
Av =

→ Vc = A v Vs = -3333 × 4 × 10 -3 = - 13.33 V
Chapter 19, Solution 91
R s = 1.2 kΩ ,
R L = 4 kΩ
Av =
- h fe R L
h ie + (h ie h oe − h re h fe ) R L
Av =
- 80 × 4 × 10 3
1200 + (1200 × 20 × 10 -6 − 1.5 × 10 -4 × 80) × 4 × 10 3
Av =
- 32000
= - 25.64
1248
(b)
Ai =
h fe
80
=
= 74.074
1 + h oe R L 1 + 20 × 10 -6 × 4 × 10 3
(c)
Z in = h ie − h re A i
(a)
Z in = 1200 − 1.5 × 10 -4 × 74.074 ≅ 1.2 kΩ
(d)
Z out =
R s + h ie
(R s + h ie ) h oe − h re h fe
Z out =
1200 + 1200
2400
=
= 51.282 kΩ
-6
-4
2400 × 20 × 10 − 1.5 × 10 × 80 0.0468
Chapter 19, Solution 92
Due to the resistor R E = 240 Ω , 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
IE = Ib + Ic
(1)
Vb = h ie I b + h re Vc + (I b + I c ) R E
(2)
Vc
RE + 1
(3)
I c = h fe I b +
But
hoe
h oe
Vc = - I c R L
(4)
Substituting (4) into (3),
I c = h fe I b −
or
Ai =
RL
RE + 1
Ic
h oe
I c h fe (1 + R E h oe )
=
Ib
1 + h oe (R L
100(1 + 240x30 x10 −6 )
Ai =
1 + 30 × 10 -6 (4,000 + 240)
A i = 79.18
From (3) and (5),
(5)
RL
Ic =
h fe (1 + R E )h oe
Vc
I b = h fe I b +
1 + h oe (R L + R E )
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 =
Vc (h ie + R E )
V
+ h re Vc − c R E
RL


1   h fe (1 + R E h oe )
 
 R E +
− h fe 
h oe  1 + h oe (R L + R E )


V
(h ie + R E )
R
1
+ h re − E
= b =
A v Vc 
RL

1   h fe (1 + R E h oe )
 
 R E +
− h fe 
h oe  1 + h oe (R L + R E )


1
=
Av 
1
 240 +
30 x10 −6

(7)
(4000 + 240)
240
+ 10 -4 −
−6
4000

 100(1 + 240 x 30 x10 )
− 100

-6
  1 + 30 × 10 × 4240

1
= −6.06x10 −3 + 10 -4 − 0.06 = -0.066
Av
A v = –15.15
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
 h (1 + R E h oe )

Vb = (h ie + R E ) I b + (R E − h re R L )  fe
I b 
 1 + h oe (R L + R E ) 
Z in =
Vb
h (R − h re R L )(1 + R E h oe )
= h ie + R E + fe E
Ib
1 + h oe (R L + R E )
(100)(240 × 10 -4 × 4 × 10 3 )(1 + 240x30x10 −6 )
Z in = 4000 + 240 +
1 + 30 × 10 -6 × 4240
Z in = 12.818 kΩ
(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
−
+
−
Vc
1V
−
(b)
Zout
From the input loop,
I b (R s + h ie ) + h re Vc + R E (I b + I c ) = 0
But
So,
Vc = 1
I b (R s + h ie + R E ) + h re + R E I c = 0
(9)
From the output loop,
Ic =
or
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
Ib =
−
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 −

h

(R s + R E + h ie ) oe
h fe 

=0
1 + R E h oe
R s + R E + h ie
R + R E + h ie
Ic + R E Ic = s
h fe
1 + R E h oe
 h oe

 h fe

 − h re

 R + R E + h ie 
(h oe h fe )  s
 − h re
1 + R E h oe 

Ic =
R E + (R s + R E + h ie ) h fe
Z out =
Z out =
Z out =
R E h fe + R s + R E + h ie
1
=
I c  R s + R E + h ie 

 h oe − h re h fe
 1 + R E h oe 
240 × 100 + (1200 + 240 + 4000)
1200 + 240 + 4000 
-6
-4
 1 + 240 x 30 x10 −6  × 30 × 10 − 10 × 100


24000 + 5440
= 193.7 kΩ
0.152
Chapter 19, Solution 93
We apply the same formulas derived in the previous problem.
(h ie + R E )
R
1
=
+ h re − E
Av 
RL

1   h fe (1 + R E h oe )
 R E +
 
− h fe 
h oe  1 + h oe (R L + R E )


1
=
Av
200
(2000 + 200)
+ 2.5 × 10 -4 −
3800
150(1 + 0.002)

− 150
(200 + 10 5 ) 
 1 + 0.04

1
= −0.004 + 2.5 × 10- 4 − 0.05263 = -0.05638
Av
A v = –17.74
h fe (1 + R E h oe )
150(1 + 200x10 −5 )
Ai =
=
= 144.5
1 + h oe (R L + R E ) 1 + 10 -5 × (200 + 3800)
Z in = h ie + R E +
h fe (R E − h re R L )(1 + R E h oe )
1 + h oe (R L + R E )
Z in = 2000 + 200 +
(150)(200 − 2.5 × 10 -4 × 3.8 × 10 3 )(1.002)
1.04
Z in = 2200 + 28966
Z in = 31.17 kΩ
Z out =
Z out =
R E h fe + R s + R E + h ie
 R s + R E + h ie 

 h oe − h re h fe
 1 + R E h oe 
33200
200 × 150 + 1000 + 200 + 2000
=
-5
- 0.0055
 3200 × 10 
-4
 − 2.5 × 10 × 150

 1.002 
Z out = –6.148 MΩ
Chapter 19, Solution 94
We first obtain the ABCD parameters.
 200 0 
[h] = 
Given
,
 100 10 -6 


[T] = 


∆h
h 21
- h 22
h 21
- h11
h 21
-1
h 21
∆ h = h11 h 22 − h12 h 21 = 2 × 10 -4

  - 2 × 10 -6
=
-8
  - 10

-2 

- 10 -2 
The overall ABCD parameters for the amplifier are
 - 2 × 10 -6
- 2  - 2 × 10 -6
-2
[T] = 
-8
-2 
-8
- 10  - 10
- 10 -2
 - 10
∆ T = 2 × 10 -12 − 2 × 10 -12 = 0
B

[h] =  D
-1

D
∆T
D
C
D

0 
  200
 =  - 10 -4 10 -6 

  2 × 10 -8
≅
  10 -10
2 × 10 -2 

10 -4 
Thus,
h ie = 200 ,
Av =
h re = 0 ,
h fe = -10 -4 ,
h oe = 10 -6
(10 4 )(4 × 10 3 )
= 2 × 10 5
200 + (2 × 10 -4 − 0) × 4 × 10 3
Z in = h ie −
h re h fe R L
= 200 − 0 = 200 Ω
1 + h oe R L
Chapter 19, Solution 95
Let Z A =
1
s 4 + 10s 2 + 8
=
s 3 + 5s
y 22
Using long division,
5s 2 + 8
ZA = s + 3
= s L1 + Z B
s + 5s
i.e.
L1 = 1 H
and
ZB =
5s 2 + 8
s 3 + 5s
as shown in Fig (a).
L1
ZB
y22 = 1/ZA
(a)
1
s 3 + 5s
YB =
=
Z B 5s 2 + 8
Using long division,
YB = 0.2s +
where
C 2 = 0 .2 F
3.4s
= sC 2 + YC
5s 2 + 8
and
YC =
3.4s
5s 2 + 8
as shown in Fig. (b).
L1
C2
Yc = 1/ZC
(b)
ZC =
1
5s 2 + 8 5s
8
1
=
=
+
= s L3 +
YC
3.4s
3.4 3.4s
s C4
i.e. an inductor in series with a capacitor
5
L3 =
= 1.471 H and
3.4
C4 =
3.4
= 0.425 F
8
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)
1Ω
1
2.613s + 2.613s
H(s) =
s 4 + 3.414s 2 + 1
1+
2.613s 3 + 2.613s
3
which indicates that
-1
2.613s + 2.613s
s 4 + 3.414s + 1
=
2.613s 3 + 2.613s
y 21 =
y 22
3
We seek to realize y 22 . By long division,
2.414s 2 + 1
y 22 = 0.383s +
= s C 4 + YA
2.613s 3 + 2.613s
i.e.
C 4 = 0.383 F
YA =
and
2.414s 2 + 1
2.613s 3 + 2.613s
as shown in Fig. (b).
L1
YA
L3
C2
C4
y22
(b)
1
2.613s 3 + 2.613s
=
ZA =
YA
2.414s 2 + 1
By long division,
Z A = 1.082s +
i.e.
L 3 = 1.082 H
1.531s
= s L3 + Z B
2.414s 2 + 1
and
ZB =
1.531s
2.414s 2 + 1
as shown in Fig.(c).
L1
ZB
L3
C2
C4
(c)
YB =
i.e.
1
1
1
= 1.577s +
= s C2 +
1.531s
ZB
s L1
C 2 = 1.577 F
and
L1 = 1.531 H
Thus, the network is shown in Fig. (d).
1.531 H
1.577 F
1.082 H
0.383 F
1Ω
(d)
Chapter 19, Solution 97
Hence,
s3
s3
s 3 + 12s
H(s) = 3
=
6s 2 + 24
(s + 12s) + (6s 2 + 24)
1+ 3
s + 12s
y 22 =
6s 2 + 24
1
=
+ ZA
3
s + 12s s C 3
where Z A is shown in the figure below.
(1)
C1
C3
L2
ZA
y22
We now obtain C 3 and Z A using partial fraction expansion.
Let
6s 2 + 24
A Bs + C
=
+
s (s 2 + 12) s s 2 + 12
6s 2 + 24 = A (s 2 + 12) + Bs 2 + Cs
Equating coefficients :
s0 :
24 = 12A 
→ A = 2
1
s :
0=C
2
s :
6= A+B 
→ B = 4
Thus,
6s 2 + 24
2
4s
= + 2
2
s (s + 12) s s + 12
(2)
Comparing (1) and (2),
1 1
C3 = = F
A 2
But
1
s 2 + 12 1
3
=
= s+
4s
4
ZA
s
(3)
1
1
= sC1 +
ZA
s L2
(4)
Comparing (3) and (4),
1
1
C1 = F
and
L2 = H
4
3
Therefore,
C1 = 0.25 F ,
L 2 = 0.3333 H ,
C 3 = 0.5 F
Chapter 19, Solution 98
∆ h = 1 − 0 .8 = 0 .2
 − ∆ h / h 21 − h11 / h 21   − 0.001
[Ta ] = [Tb ] = 
=
−6
− h 22 / h 21 − 1 / h 21  − 2.5x10
− 10 
− 0.005
2.6x10−5
0.06 
[T] = [Ta ][Tb ] = 

−8
5x10−5 
1.5x10
We now convert this to z-parameters
A / C ∆ T / C 1.733x103
[z] = 
=
7
 1 / C D / C  6.667 x10
1000
I1
0.0267 

3.33x103 
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
→
I 2 = −Vo / ZL
(3)
Substituting (3) into (2) gives

 1
z
I1 = Vo 
+ 22 
 z 21 z 21ZL 
We substitute (3) and (4) into (1)
(4)

 1
z
z
Vs = (1000 + z11)
+ 22  Vo − 12 Vo
ZL
 z11 z 21ZL 
= 7.653x10− 4 − 2.136 x10−5 = 744µV
Chapter 19, Solution 99
Z ab = Z1 + Z 3 = Z c || (Z b + Z a )
Z1 + Z 3 =
Z c (Z a + Z b )
Za + Zb + Zc
(1)
Z cd = Z 2 + Z 3 = Z a || (Z b + Z c )
Z2 + Z3 =
Z a (Z b + Z c )
Za + Zb + Zc
(2)
Z ac = Z1 + Z 2 = Z b || (Z a + Z c )
Z1 + Z 2 =
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.