Engineering circuit-analysis solutions 7ed hayt

Engineering Circuit Analysis, 7
th
Edition Chapter Two Solutions 10 March 2006
1. (a) 12 μs (d) 3.5 Gbits (g) 39 pA
(b) 750 mJ (e) 6.5 nm (h) 49 k Ω
(c) 1.13 kΩ (f) 13.56 MHz
(i) 11.73 pA
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Engineering Circuit Analysis, 7
th
Edition Chapter Two Solutions 10 March 2006

2. (a) 1 MW (e) 33 μJ
(i) 32 mm
(b) 12.35 mm (f) 5.33 nW
(c) 47. kW (g) 1 ns
(d) 5.46 mA (h) 5.555 MW
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Engineering Circuit Analysis, 7
th
Edition Chapter Two Solutions 10 March 2006
3. (a) ()
⎛⎞
⎜⎟
⎝⎠
745.7 W
400 Hp =
1 hp
298.3 kW

(b) 12 ft =
12 in 2.54 cm 1 m
(12 ft) 3.658 m
1 ft 1 in 100 cm
⎛⎞⎛ ⎞⎛ ⎞
=
⎜⎟⎜ ⎟⎜ ⎟
⎝⎠⎝ ⎠⎝ ⎠

(c) 2.54 cm = 25.4 mm

(d)
()
1055 J
67 Btu =
1 Btu
⎛⎞
⎜⎟
⎝⎠
70.69 kJ

(e) 285.4
´10
-15
s = 285.4 fs
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Engineering Circuit Analysis, 7
th
Edition Chapter Two Solutions 10 March 2006
4. (15 V)(0.1 A) = 1.5 W = 1.5 J/s.

3 hrs running at this power level equates to a transfer of energy equal to

(1.5 J/s)(3 hr)(60 min/ hr)(60 s/ min) = 16.2 kJ

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Engineering Circuit Analysis, 7
th
Edition Chapter Two Solutions 10 March 2006

5.
Motor power = 175 Hp
(a)
With 100% efficient mechanical to electrical power conversion,
(175 Hp)[1 W/ (1/745.7 Hp)] = 130.5 kW
(b)
Running for 3 hours,
Energy = (130.5×10
3
W)(3 hr)(60 min/hr)(60 s/min) = 1.409 GJ
(c)
A single battery has 430 kW-hr capacity. We require
(130.5 kW)(3 hr) = 391.5 kW-hr therefore one battery is sufficient.
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Engineering Circuit Analysis, 7
th
Edition Chapter Two Solutions 10 March 2006

6. The 400-mJ pulse lasts 20 ns.
(a)
To compute the peak power, we assume the pulse shape is square:

400
Energy (mJ)
t (ns)
20

Then
P = 400×10
-3
/20×10
-9
= 20 MW.

(b)
At 20 pulses per second, the average power is

Pavg = (20 pulses)(400 mJ/pulse)/(1 s) = 8 W.
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Engineering Circuit Analysis, 7
th
Edition Chapter Two Solutions 10 March 2006

7. The 1-mJ pulse lasts 75 fs.
(a) To compute the peak power, we assume the pulse shape is square:

1
Energy (mJ)
t (fs)
75

Then
P = 1×10
-3
/75×10
-15
= 13.33 GW.

(b) At 100 pulses per second, the average power is

Pavg = (100 pulses)(1 mJ/pulse)/(1 s) = 100 mW.
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Engineering Circuit Analysis, 7
th
Edition Chapter Two Solutions 10 March 2006

8.
The power drawn from the battery is (not quite drawn to scale):

5 7 17 24
P (W)
10
6
t (min)

(a)
Total energy (in J) expended is
[6(5) + 0(2) + 0.5(10)(10) + 0.5(10)(7)]60 = 6.9 kJ.

(b)
The average power in Btu/hr is
(6900 J/24 min)(60 min/1 hr)(1 Btu/1055 J) = 16.35 Btu/hr.

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Engineering Circuit Analysis, 7
th
Edition Chapter Two Solutions 10 March 2006
9. The total energy transferred during the first 8 hr is given by

(10 W)(8 hr)(60 min/ hr)(60 s/ min) = 288 kJ

The total energy transferred during the last five mi
nutes is given by

300 s
0 10
+ 10
300
td t
⎡⎤
−
⎢⎥
⎣⎦
∫
=
300
2
0
10
+ 10 =
600
tt−
1.5 kJ

(a) The total energy transferred is 288 + 1.5 = 289.5 kJ

(b) The energy transferred in the last five mi
nutes is 1.5 kJ

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Engineering Circuit Analysis, 7
th
Edition Chapter Two Solutions 10 March 2006

10. Total charge
q = 18t
2
– 2t
4
C.

(a)
q(2 s) = 40 C.

(b) To find the m aximum charge within 0 ≤ t ≤ 3 s, we need to take the first and
second derivitives:

dq/dt = 36t – 8t
3
= 0, leading to roots at 0, ± 2.121 s

d
2
q/dt
2
= 36 – 24t
2

substituting
t = 2.121 s into the expression for d
2
q/dt
2
, we obtain a value of
–14.9, so that this root represents a maxim u
m.
Thus, we find a maximum charge
q = 40.5 C at t = 2.121 s.

(c)
The rate of charge accumulation at t = 8 s is
dq/dt|t = 0.8 = 36(0.8) – 8(0.8)
3
= 24.7 C/s.

(d) See Fig. (a) and (b).
0 0.5 1 1.5 2 2.5 3
-20
-10
0
10
20
30
40
50
60
70
time (s)
q (C)
0 0.5 1 1.5 2 2.5 3
-150
-100
-50
0
50
i (A )
time (t)
(a) (b)
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Engineering Circuit Analysis, 7
th
Edition Chapter Two Solutions 10 March 2006
11. Referring to Fig. 2.6c,

0 A, 3 2-
0 A, 3 2-
)(
3
5
1
⎩
⎨
⎧
>+
<+
=
−
te
te
ti
t
t

Thus,

(a)
i1(-0.2) = 6.155 A

(b)
i1 (0.2) = 3.466 A

(c)
To determine the instants at which i1 = 0, we must consider t < 0 and t > 0 separately:

for
t < 0, - 2 + 3e
-5t
= 0 leads to t = -0.2 ln (2/3) = +0.0811 s (impossible)

for
t > 0, -2 + 3e
3t
= 0 leads to t = (1/3) ln (2/3) = –0.135 s (impossible)

Therefore, the current is
never negative.

(d) The total charge passed left to right in the interval –0. 8 <
t < 0.1 s is

q(t) =
0.1
1
0.8
()itdt
−∫

=
00 .1
53
0.8 0
2 3 2 3
tt
edt ed
−
−
⎡⎤ ⎡−+ + −+
⎣⎦ ⎣
∫∫
t⎤
⎦

=
()
0
0.1
53
0
-0.8
3
2 2
5
tt
te te
−⎛⎞
−− + −+
⎜⎟
⎝⎠

= 33.91 C

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Engineering Circuit Analysis, 7
th
Edition Chapter Two Solutions 10 March 2006

12. Referring to Fig. 2.28,

(a) The average current over one period (10 s) is

i
avg = [-4(2) + 2(2) + 6(2) + 0(4)]/10 = 800 mA

(b) The total charge transferred over the interval 1 < t < 12 s is

∫
=
12
1
total
)( dttiq = -4(1) + 2(2) + 6(2) + 0(4) – 4(2) = 4 C

(c) See Fig. below
2 46
8
10 12
16
q (C)
t(s)
-16
8
16
-8
14

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Engineering Circuit Analysis, 7
th
Edition Chapter Two Solutions 10 March 2006
13. (a) VBA = –
-19
2 pJ
-1.602 10 C×
= 12.48 MV

(b) V
ED =
-19
0
-1.602 10 C×
= 0

(c) V
DC = –
-19
3 pJ
1.602 10 C×
= –18.73 MV

(d) It takes – 3 pJ to move +1.602x10
–19
C from D to C.
It takes 2 pJ to move –1.602x10
–19
C from B to C, or –2 pJ to move
+1.602x10
–19
C from B to C, or +2 pJ to move +1.602x10
–19
C from C to B.

Thus, it requires –3 pJ + 2 pJ = –1 pJ to m ove +1.602x10
–19
C from D to C to
B.

Hence, V
DB =
−
×
-19
1 pJ
1.602 10 C
= –6.242 MV.

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Engineering Circuit Analysis, 7
th
Edition Chapter Two Solutions 10 March 2006
14.

+
Voltmeter
–
+
V
1
–
-
Voltmeter
+
–
V
2
+
From the diagram, we see that V
2 = –V1 = +2.86 V.
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Engineering Circuit Analysis, 7
th
Edition Chapter Two Solutions 10 March 2006

15. (a) P
abs = (+3.2 V)(-2 mA) = –6.4 mW (or +6.4 mW supplied)

(b) P
abs = (+6 V)(-20 A) = –120 W (or +120 W supplied)

(d)
Pabs = (+6 V)(2 i x) = (+6 V)[(2)(5 A)] = +60 W

(e) P
abs = (4 sin 1000t V)(-8 cos 1000t mA) | t = 2 ms = +12.11 W

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Engineering Circuit Analysis, 7
th
Edition Chapter Two Solutions 10 March 2006

16. i = 3te
-100t
mA and v = [6 – 600t ] e
-100t
mV

(a)
The power absorbed at t = 5 ms is

P
abs = ()
[ ] W 3 6006
5
100100
μ
mst
tt
teet
=
−−
⋅−

= 0.01655 μ W = 16.55 nW

(b)
The energy delivered over the interval 0 < t <
∞ is

()∫∫
∞∞
−
−=
00
200
J 60063 μdtettdtP
t
abs

Making use of the relationship

1
0
!

+
∞
−
=∫ n
axn
a
n
dxex where n is a positive integer and a > 0,

we find the energy delivered to be = 18/(200)
2
- 1800/(200)
3
= 0

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Engineering Circuit Analysis, 7
th
Edition Chapter Two Solutions 10 March 2006

17. (a) P
abs = (40i )(3e
-100t
)| t = 8 ms =
[]
2
8
100
360
mst
te
=
−
= 72.68 W

(b) P
abs =
[] W36.34- 180- 2.0
2
8
100
==⎟
⎠
⎞
⎜
⎝
⎛
=
−
mst
tei
dt
di

(c)
Pabs = ()
mst
t
t
eidt
8
100
0
3 20 30
=
−
⎟
⎠
⎞
⎜
⎝
⎛
+∫

=
mst
t
t
tt
etdee
8
100
0
100100
60 3 90
=
−′−−
⎟
⎠
⎞
⎜
⎝
⎛
+′∫
= 27.63 W

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Engineering Circuit Analysis, 7
th
Edition Chapter Two Solutions 10 March 2006

18. (a) The short-circuit current is the value of the current at V = 0.

Reading from the graph, this corresponds to approximately 3.0 A.

(b) The open-circuit voltage is the value of the v
oltage at I = 0.

Reading from the graph, this corresponds to roughly 0.4875 V, estimating the curve as
hitting the x-axis 1 mm
behind the 0.5 V mark.

(c) We see that th e maximum current corresponds to zero voltage, and likewise, the
m
aximum voltage occu rs at zero curren t. The m aximum power point, therefore,
occurs somewhere between these two points. By trial and error,

P
max is roughly (375 mV)(2.5 A) = 938 mW, or just under 1 W.

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Engineering Circuit Analysis, 7
th
Edition Chapter Two Solutions 10 March 2006
19. (a) ()( )=
first 2 hours
P = 5 V 0.001 A 5 mW
()( ) =
next 30 minutes
P = ? V 0 A 0 mW
()( )−=−
last 2 hours
P = 2 V0.001 A 2 m W

(b) Energy = (5 V)(0.001 A)(2 hr)(60 m i
n/ hr)(60 s/ min) = 36 J

(c) 36 – (2)(0.001)(60)(60) = 21.6 J

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Engineering Circuit Analysis, 7
th
Edition Chapter Two Solutions 10 March 2006

20. Note that in the table below, only the –4-A sour
ce and the –3-A source are actually
“absorbing” power; the remaining sources are supplying power to the circuit.

Source Absorbed Power Absorbed Power
2 V source (2 V)(-2 A) - 4 W
8 V source (8 V)(-2 A) - 16 W
-4 A source (10 V)[-(-4 A)] 40 W
10 V source (10 V)(-5 A) - 50 W
-3 A source (10 V)[-(-3 A)] 30 W

The 5 powe r quantities sum to –4 – 16 + 40 – 50 + 30 = 0, as dem
anded from
conservation of energy.

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Engineering Circuit Analysis, 7
th
Edition Chapter Two Solutions 10 March 2006
21.
40 V
40 V
32 V
8 V
20 A
–12 A
–16 A

P
8V supplied = (8)(8) = 64 W (source of energy)

P
32V supplied = (32)(8) = 256 W (source of energy)

P
–16A supplied = (40)(–16) = –640 W

P
40V supplied = (40)(20) = 800 W (source of energy)

P
–12A supplied = (40)( –12) = –480 W

Check: = 64 + 256 – 640 + 800 – 480 = 0 (ok)
supplied power∑

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Engineering Circuit Analysis, 7
th
Edition Chapter Two Solutions 10 March 2006

22. We are told that V
x = 1 V, and from Fig. 2.33 w e see that the current flowing through
the dependent source (and hence th rough each element of t he circuit) is 5V
x = 5 A.
We will co mpute
absorbed power by using the current flowing into the po sitive
reference terminal of the appropriate voltage (passive sign convention), and we will
compute
supplied power by using the current flowing out of the pos itive reference
terminal of the appropriate voltage.

(a) The power absorbed by element “A” = (9 V)(5 A) = 45 W

(b) The power supplied by the 1-V source = (1 V)(5 A) = 5 W, and

the power supplied by the dependent source = (8 V)(5 A) = 40 W

(c) The sum of the supplied power = 5 + 40 = 45 W
The sum of the absorbed power is 45 W, so

yes, the su m of the power supplied = the sum of the power absorbed, as we
expect from the princip
le of conservation of energy.
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Engineering Circuit Analysis, 7
th
Edition Chapter Two Solutions 10 March 2006

23. We are asked to determine the voltage
vs, which is identical to the voltage labeled v1.
The only remaining reference to
v1 is in the expression for the current flowing through
the dependent source, 5
v1.

This current is equal to –
i2.
Thus,
5
v1 = -i2 = - 5 mA
Therefore
v1 = -1 mV

and so
vs = v1 = -1 mV

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Engineering Circuit Analysis, 7
th
Edition Chapter Two Solutions 10 March 2006
24. The voltage across the dependent source = v2 = –2ix = –2(–0.001) = 2 mV.
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Engineering Circuit Analysis, 7
th
Edition Chapter Two Solutions 10 March 2006

25. The battery delivers an energy of 460.8 W-hr over a period of 8 hrs.

(a)
The power delivered to the headlight is therefore (460.8 W-hr)/(8 hr) = 57.6 W

(b)
The current through the headlight is equal to the power it absorbs from the battery
divided by the voltage at which the power is supplied, or

I = (57.6 W)/(12 V) = 4.8 A

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Engineering Circuit Analysis, 7
th
Edition Chapter Two Solutions 10 March 2006

26. The supply voltage is 110 V, and the m axim
um dissipated power is 500 W. The fuse s
are specified in terms of current, so we need to determine the maximum current that
can flow through the fuse.

P = V I therefore Imax = Pmax/V = (500 W)/(110 V) = 4.545 A

If we choose the 5-A fuse, it will allow up to (110 V)(5 A) = 550 W of power to be
delive
red to the application (we must assume here that the fuse absorbs zero power, a
reasonable assumption in practice). This exceeds the specified maximum power.

If we choose the 4.5-A fuse instead, we will hav e a maximum current of 4.5 A. This
leads to a maximu
m power of (110)(4.5) = 495 W delivered to the application.

Although 495 W is less than the m aximum power allowed, this fuse will provide

adequate protection for the application circuitry. If a fault occurs and the application
circuitry attempts to draw too m uch power, 1000 W for example, the fuse will blow,
no current will f low, and the application c ircuitry will be p rotected. However, if the
application circuitry tries to draw its maximum rated power (500 W ), the fuse will
also blow. In practice, m ost equipm ent will not draw its m aximum rated powe r
continuously—although to be safe, we typically assume that it will.

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Engineering Circuit Analysis, 7
th
Edition Chapter Two Solutions 10 March 2006
27. (a) imax = 5/ 900 = 5.556 mA

imin = 5/ 1100 = 4.545 mA

(b)
p = v
2
/ R so

pmin = 25/ 1100 = 22.73 mW

pmax = 25/ 900 = 27.78 mW

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Engineering Circuit Analysis, 7
th
Edition Chapter Two Solutions 10 March 2006
28. p = i
2
R, so
p
min = (0.002)
2
(446.5) = 1.786 mW and (more relevant to our discussion)
p
max = (0.002)
2
(493.5) = 1.974 mW

1.974 mW would be a correct answer, although power ratings are typically expressed
as integers, so 2 mW might be m
ore appropriate.

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Engineering Circuit Analysis, 7
th
Edition Chapter Two Solutions 10 March 2006

29. (a)
Pabs = i
2
R = [20e
-12t
]
2 (1200) μW

= [20
e
-1.2
]
2 (1200) μW

= 43.54 mW

(b)
Pabs = v
2
/R = [40 cos 20t]
2 / 1200 W
keep in mind we
are using radians
= [40 cos 2]

2 / 1200 W

= 230.9 mW

(c)
Pabs = v i = 8t
1.5 W

= 253.0 mW

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Engineering Circuit Analysis, 7
th
Edition Chapter Two Solutions 10 March 2006

30. It’s probably best to begin this problem by sketching the voltage waveform:

60 40
20
t (ms)
v (V)
+10
-10
(a)
vmax = +10 V

(b)
vavg = [(+10)(20×10
-3
) + (-10)(20×10
-3
)]/(40×10
-3
) = 0

(c)
iavg = vavg /R = 0

(d)
R
v
p
abs
2
max
max =
= (10)
2
/ 50 = 2 W

(e)
⎥
⎦
⎤
⎢
⎣
⎡
⋅
−
+⋅
+
= 20
)10(
20
)10(
40
1

22
RR
p
avg
abs
= 2 W

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Engineering Circuit Analysis, 7
th
Edition Chapter Two Solutions 10 March 2006
31. Since we are inform ed that the same current must flow through each com ponent, we
begin by defining a current
I flowing out of the positive reference terminal of the
voltage source.

The power supplied by the voltage source is
Vs I.
The power absorbed by resistor
R1 is I
2
R1.
The power absorbed by resistor
R2 is I
2
R2.

Since we know that the total power supplied is equal to the total power absorbed,
we may write:
Vs I = I
2
R1 + I
2
R2
or
Vs = I R1 + I R2
Vs = I (R1 + R2)
By Ohm’s law,
I = /
R
2RV 2
so that
Vs =
()
21
2
2
RR
R
V
R
+
Solving for we find
2RV
()
21
2
s

2
RR
R
VV
R
+
=
Q.E.D.

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Engineering Circuit Analysis, 7
th
Edition Chapter Two Solutions 10 March 2006
32. (a)
-1.5 -1 -0.5 0 0.5 1 1.5 2 2.5
-4
-3
-2
-1
0
1
2
3
4
5
6
voltage (V)
current (mA)

(b) W e see from our answer to part (a) that this device has a reasonably linear
characteristic (a not unreasonable degree of experimental er
ror is evident in the data).
Thus, we choose to estimate the resistance using the two extreme points:

Reff = [(2.5 – (-1.5)]/[5.23 – (-3.19)] kΩ = 475 Ω

Using the last two points instead, we find
Reff = 469 Ω, so that we can state with some
certainty at least that a reasonable estimate of the resistance is approximately 470
Ω.

(c)
-1.5 -1 -0.5 0 0.5 1 1.5 2 2.5
-1.5
-1
-0.5
0
0.5
1
1.5
2
voltage (V)
c urrent (m A )

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Engineering Circuit Analysis, 7
th
Edition Chapter Two Solutions 10 March 2006
33. Top Left Circuit: I = (5/10) mA = 0.5 mA, and P10k = V
2
/10 mW = 2.5 mW

Top Right Circuit:
I = -(5/10) mA = -0.5 mA, and P10k = V
2
/10 mW = 2.5 mW

Bottom Left Circuit:
I = (-5/10) mA = -0.5 mA, and P10k = V
2
/10 mW = 2.5 mW

Bottom Right Circuit:
I = -(-5/10) mA = 0.5 mA, and P10k = V
2
/10 mW = 2.5 mW

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Engineering Circuit Analysis, 7
th
Edition Chapter Two Solutions 10 March 2006

34. The voltage
vout is given by

vout = -10
-3
vπ (1000)

= -
vπ

Since
vπ = vs = 0.01 cos 1000t V, we find that

vout = - vπ = -0.01 cos 1000t V

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Engineering Circuit Analysis, 7
th
Edition Chapter Two Solutions 10 March 2006
35. vout = -vπ = -vS = -2sin 5t V

vout (t = 0) = 0 V

vout (t = 0.324 s) = -2sin (1.57) = -2 V

(use care to em ploy r adian m ode on your calculator or convert 1.57 radians to
degrees)

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Engineering Circuit Analysis, 7
th
Edition Chapter Two Solutions 10 March 2006

36. 18 AWG wire has a resistance of 6.39
Ω / 1000 ft.

Thus, we require 1000 (53) / 6.39 = 8294 ft of wire.
(Or 1.57 miles. Or, 2.53 km).

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Engineering Circuit Analysis, 7
th
Edition Chapter Two Solutions 10 March 2006

37. We need to create a 47 0-
Ω resistor from 28 AWG wire, knowing that the am bient
tem perature is 108
o
F, or 42.22
o
C.

Referring to Table 2.3, 28 AWG wire is 65.3 m
Ω/ft at 20
o
C, and using the equation
provided we compute

R
2/R1 = (234.5 + T2)/(234.5 + T1) = (234.5 + 42.22)/(234.5 + 20) = 1.087

We thus find that 28 AWG wire is (1.087)(65.3) = 71.0 m
Ω/ft.

Thus, to repair the transmitter we will need

(470
Ω)/(71.0 × 10-3 Ω/ft) = 6620 ft (1.25 miles, or 2.02 km).

Note: This seems like a lot of wire to be washing up on shore. We
may find we don’t
have enough. In that case, perhaps we should take our cue from Eq. [6], and try to
squash a piece of the wire flat so that it has a very small cross-sectional area…..

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Engineering Circuit Analysis, 7
th
Edition Chapter Two Solutions 10 March 2006
38. We are given that the conductivity σ of copper is 5.8×10
7
S/m.

(a)
50 ft of #18 (18 AWG) copper wire, which has a diameter of 1.024 mm, will have
a resistance of
l/(σ A) ohms, where A = the cross-sectional area and l = 50 ft.

Converting the dimensional quantities to meters,

l = (50 ft)(12 in/ft)(2.54 cm/in)(1 m/100 cm) = 15.24 m

and

r = 0.5(1.024 mm)(1 m/1000 mm) = 5.12×10
-4
m

so that

A = π r
2
= π (5.12×10
-4
m)
2
= 8.236×10
-7
m
2

Thus,
R = (15.24 m)/[( 5.8×10
7
)( 8.236×10
-7
)] = 319.0 mΩ

(b)
We assume that the conductivity value specified also holds true at 50
o
C.

The cross-sectional area of the foil is

A = (33 μm)(500 μm)(1 m/10
6
μm)( 1 m/10
6
μm) = 1.65×10
-8
m
2

So that

R = (15 cm)(1 m/100 cm)/[( 5.8×10
7
)( 1.65×10
-8
)] = 156.7 mΩ

A 3-A current flowing through this copper in the direction specified would
lead to the dissipation of

I
2
R = (3)
2
(156.7) mW = 1.410 W

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Engineering Circuit Analysis, 7
th
Edition Chapter Two Solutions 10 March 2006
39. Since R = ρ l/ A, it follows that ρ = R A/ l.
From Table 2.4, we see that 28 AWG soft copper wire (cro
ss-sectional area = 0.0804
mm
2
) is 65.3 Ω per 1000 ft. Thus,

R = 65.3
Ω.

l = (1000 ft)(12 in/ft)(2.54 cm/in)(10 mm/cm) = 304,800 mm.

A = 0.0804 mm
2
.

Thus,
ρ = (65.3)(0.0804)/304800 = 17.23 μΩ/ mm

or

ρ = 1.723 μΩ
.
cm

which is in fact consistent with the representative data for copper in Table 2.3.

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Engineering Circuit Analysis, 7
th
Edition Chapter Two Solutions 10 March 2006
40. (a) From the text,
(1)
Zener diodes,
(2)
Fuses, and
(3)
Incandescent (as opposed to fluorescent) light bulbs

This last one requires a few facts to be put together. We have stated that temperature
can affect resistance—in
other words, if the temperature changes during operation, the
resistance will not rem ain constant and hence nonlinear behavior will be observed.
Most discrete resistors are rated for up to a sp ecific power in o rder to ensure th at
temperature variation during operation will no t significantly change th e res istance
value. Light bulbs, however, become rather warm when operating and can experience
a significant change in resistance.

(b) The energy is dissipated by the resistor, conv
erted to heat which is transferred to
the air surrounding the resistor. The resistor is unable to store the energy itself.

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Engineering Circuit Analysis, 7
th
Edition Chapter Two Solutions 10 March 2006
41. The quoted resistivity ρ of B33 copper is 1.7654 μΩ
.
cm.
A =
πr
2
= π(10
–3
)
2
= 10
–6
π m
2
. = 100 m. l

Thus, R =
ρ/ A = l
()
( )( )
6
6
1.7654 10 cm 1 m/100 cm 100 m
0.5619
10
π
−
−
×Ω
= Ω

And P = I
2
R = (1.5)
2
(0.5619) = 1.264 W.

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Engineering Circuit Analysis, 7
th
Edition Chapter Two Solutions 10 March 2006
42. We know that for any wire of cross-sectional area A and length l, the resistan ce is
given by R =
ρ l/ A.

If we keep
ρ fixed by choosing a material, and A fixed by choosing a wire gauge (e.g.
28 AWG), changing will change the resistance of our “device.”
l

A simple variable resistor concept, then: Leads to
connect to
circuit
Copper wire
Rotating sh ort
wire de ter-
mines length
of long wire
used in circuit.

But this is somewhat impractical, as the leads ma
y turn out to have alm ost the same
resistance unless we have a very long wire, which can also be im practical. One
improvement would be to replace the copper wire shown with a coil of insulated
copper wire. A s mall amount of insulation would then need to be removed from
where the moveable wire touches the coil so that electrical connection could be made.

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Engineering Circuit Analysis, 7
th
Edition Chapter Two Solutions 10 March 2006

43. (a) We need to plot the negative and positive voltage ranges s
eparately, as the positive
voltage range is, after all, exponential!

-0. 7 -0.6 -0.5 -0.4 -0.3 -0.2 -0. 1 0 0.1
-2
0
2
4
6
8
10
12
14
16
x 10
-6
voltage (V)
c urrent (A )
0 0.01 0.02 0.03 0.04 0.05 0.06 0.07
10
-8
10
-7
10
-6
10
-5
10
-4
voltage (V)
c urrent (A )
(b) To dete rmine the resistance of the device at V = 550 mV, we com pute the
corresponding current:

I = 10
-9
[e
39(0.55)
– 1] = 2.068 A

Thus,
R(0.55 V) = 0.55/2.068 = 266 mΩ

(c)
R = 1 Ω corresponds to V = I. Thus, we need to solve the transcendental equation

I = 10
-9
[e
39I
– 1]

Using a scientific calculator or the tried-and-true trial and error approach, we find that

I = 514.3 mA

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Engineering Circuit Analysis, 7
th
Edition Chapter Two Solutions 10 March 2006

44. We require a 10-Ω resistor, and are told it is f or a portable application, implying that
size, weight or both would be im
portant to consider when selecting a wire gauge. W e
have 10,000 ft of each of the gaug es listed in Table 2.3 w ith which to work. Quick
inspection of the values listed e liminates 2, 4 a nd 6 AW G wire as the ir respective
resistances are too low for only 10,000 ft of wire.

Using 12-AWG wire would require (10 Ω) / (1.
59 mΩ/ft) = 6290 ft.
Using 28-AWG wire, the narrowest available, would require

(10 Ω) / (65.3
mΩ/ft) = 153 ft.

Would the 28-AWG wire weight less? Again referring to Ta
ble 2.3, we see that the
cross-sectional area of 28-AWG wire is 0.0804 mm
2
, and that of 12-AWG wire is
3.31 mm
2
. The volume of 12-AWG wire required is therefore 6345900 mm
3
, and that
of 28-AWG wire required is only 3750 mm
3
.

The best (but not the only) choice for a portabl
e application is clear: 28-AWG wire!

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Engineering Circuit Analysis, 7
th
Edition Chapter Two Solutions 10 March 2006

45. Our target is a 100-Ω resistor. W
e see from the plot that at N D = 10
15
cm
-3
, μn ~ 2x10
3

cm
2
/V-s, yielding a resistivity of 3.121 Ω-cm.

At N
D = 10
18
cm
-3
, μn ~ 230 cm
2
/ V-s, yielding a resistivity of 0.02714 Ω-cm.

Thus, we s ee that the lower doping level clearly provides m a
terial with higher
resistivity, requiring less of the available area on the silicon wafer.

Since
R = ρL/A, where we know R = 100 Ω and ρ = 3.121 Ω-cm for a phosphorus
concentration of 10
15
cm
-3
, we need only define the resistor geometry to complete the
design.

We choose a geometry as shown in the figure; our contact area is arbitrarily chosen as
100 μm by 250 μ m
, so that only the length L remains to be specified. Solving,

() ()
4
R (100 )(100 m )(250 m )
L = A 80.1 m
3.121 cm 10 m/cm
μ μ
μ
ρ μ
Ω
==
Ω

Design summary (one possibility): N
D = 10
15
cm
-3
L = 80.1 μ m
Contact width = 100 μm
(Note: this is somewhat atypical; in the semi
conductor industry contacts are typically
made to the top and/o r bottom surface of a wafer. So, there’s more than one solu tion
based on geometry as well as doping level.)
250 μm
80.1 μm
contact
100 μm
contact
Wafer surface
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Engineering Circuit Analysis, 7
th
Edition Chapter Three Solutions 10 March 2006

1.

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Engineering Circuit Analysis, 7
th
Edition Chapter Three Solutions 10 March 2006

2. (a) six nodes; (b) nine branches.

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Engineering Circuit Analysis, 7
th
Edition Chapter Three Solutions 10 March 2006

3. (a) Four nodes; (b) five branches; (c) path, yes – loop, no.

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Engineering Circuit Analysis, 7
th
Edition Chapter Three Solutions 10 March 2006

4. (a) Five nodes;

( b) seven b
ranches;

( c) path, yes – loop, no.
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Engineering Circuit Analysis, 7
th
Edition Chapter Three Solutions 10 March 2006

5. (a) The number of nodes remains the same – four (4).

(b) The number of nodes is increased by one – to five (5).

(c) i) YES
ii) NO – does not return to starting point
iii) YES
iv) NO – does not return to starting point
v) NO – point B is crossed twice

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Engineering Circuit Analysis, 7
th
Edition Chapter Three Solutions 10 March 2006

6. (a) By KCL at the bottom node: 2 – 3 + i Z – 5 – 3 = 0
So i
Z = 9 A.

(b) If the left-most resistor has a value of 1 Ω, then 3 V appears acros
s the parallel
network (the ‘+’ reference terminal being the bottom node) Thus, the value of the
other resistor is given by

R =
3
600 m
(5)
=Ω
−−
.
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Engineering Circuit Analysis, 7
th
Edition Chapter Three Solutions 10 March 2006

7. (a) 3 A;

( b) –3 A;

( c) 0.
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Engineering Circuit Analysis, 7
th
Edition Chapter Three Solutions 10 March 2006

8. By KCL, we may write:
5 + i
y + iz = 3 + i x
(a) ix = 2 + i y + iz = 2 + 2 + 0 = 4 A

( b) i
y = 3 + i x – 5 – i z
i y = –2 + 2 – 2 i y

Thus, we find that i
y = 0.
(c) 5 + i
y + iz = 3 + i x → 5 + i x + ix = 3 + i x so ix = 3 – 5 = -2A.

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Engineering Circuit Analysis, 7
th
Edition Chapter Three Solutions 10 March 2006

9. Focusing our attention on the bottom left node, we see that i x = 1 A.

Focusing our attention next on the top right node, we see that i
y = 5 A.
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Engineering Circuit Analysis, 7
th
Edition Chapter Three Solutions 10 March 2006

10. We obtain the current each bulb draws by dividing its power rating by the operating
voltage (115 V):

I
100W = 100/115 = 896.6 mA

I
60W = 60/115 = 521.7 mA

I
40W = 347.8 mA

Thus, the total current draw is 1.739 A.

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Engineering Circuit Analysis, 7
th
Edition Chapter Three Solutions 10 March 2006

11. The DMM is connected in parallel with the 3 load resistors, across which develops the
voltage we wish to measure. If the DMM appears as a short, then all 5 A flows
through the DMM, and none through the resistors, resulting in a (false) reading of 0 V
for the circuit undergoing testing. If, instead, the DMM has an infinite internal
resistance, then no current is shunted away from the load resistors of the circuit, and a
true voltage reading results.

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Engineering Circuit Analysis, 7
th
Edition Chapter Three Solutions 10 March 2006

12. In either case, a bulb failure will adversely affect the sign.

Still, in the parallel-connected case, at least 10 (up to 11) of the other characters will
be lit, so th
e sign could be read and customers will know the restaurant is open for
business.

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Engineering Circuit Analysis, 7
th
Edition Chapter Three Solutions 10 March 2006

13. (a) v y = 1(3v x + iz)

v
x = 5 V and given that i z = –3 A, we find that

v
y = 3(5) – 3 = 12 V

( b) v
y = 1(3v x + iz) = –6 = 3v x + 0.5

Solving, we find that v
x = (–6 – 0.5)/3 = –2.167 V.

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Engineering Circuit Analysis, 7
th
Edition Chapter Three Solutions 10 March 2006

14. (a) i x = v1/10 + v 1/10 = 5

2v
1 = 50

so v
1 = 25 V.

By Ohm’s law, we see that i
y = v 2/10

also, using Ohm’s law in combination with KCL, we ma
y write

i
x = v2/10 + v 2/10 = i y + iy = 5 A

Thus, i
y = 2.5 A.
( b) From
part (a), i x = 2 v 1/ 10. Substituting the new value for v 1, we find that

i
x = 6/10 = 600 mA.
Since we have found that i
y = 0.5 i x, iy = 300 mA.
( c) no value – this is impossible.

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Engineering Circuit Analysis, 7
th
Edition Chapter Three Solutions 10 March 2006

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15. We begin by making use of the information given regarding the power generated by

the 5-A and the 40-V sources. The 5-A source supplies 100 W, so it must therefore
have a terminal voltage of 20 V. The 40-V source supplies 500 W, so it must therefore
provide a current I
X of 12.5 A.

(1) By KVL, –40 + (–110) + R(5) – 20 = 0

Thus, R = 34 Ω .

(2) By KVL, -V
G – (-110) + 40 = 0

So V
G = 150 V

Now that we know the voltage across the unknown conductance G, we need only to
find the current flowing through it to find its value by making use of Ohm’s law.

KCL provides us with the m
eans to find this current: The current flowing
into the “+”
terminal of the –110-V source is 12.5 + 6 = 18.5 A.

Then, I
x = 18.5 – 5 = 13.5 A

By Ohm’s law, I
x = G · VG

So G = 13.5/ 150 or G = 90 mS

Engineering Circuit Analysis, 7
th
Edition Chapter Three Solutions 10 March 2006

16. (a) -1 + 2 + 10i – 3.5 + 10i = 0

Solving, i = 125 mA

( b) +10 + 1i - 2 + 2
i + 2 – 6 + i = 0

Solving, we find that 4i = -4 or i = - 1 A.

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Engineering Circuit Analysis, 7
th
Edition Chapter Three Solutions 10 March 2006

17. Circuit I.

Starting at the bottom node and proceeding clockwise, we can write the KVL equation

+7 – 5 – 2 – 1(i) = 0

Which results in i = 0.

Circuit II.

Again starting with the bottom node and proceeding in a clockwise direction, we write
the KVL equation

-9
+4i + 4i = 0
(no current flows through either the -3 V source or the 2 Ω resistor)

Solving, we find that i = 9/8 A = 1.125 A.
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Engineering Circuit Analysis, 7
th
Edition Chapter Three Solutions 10 March 2006

18. Begin by defining a clockwise current i.

- v
S + v1 + v2 = 0 so v S = v1 + v2 = i(R 1 + R2)

and hence i =
12
S
v
RR+
.

Thus,
1
11
12
S
R
vRi v
RR
==
+
and
2
22
12
S
R
vRi v
RR
==
+
.

Q.E.D.

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Engineering Circuit Analysis, 7
th
Edition Chapter Three Solutions 10 March 2006

19. Given: (1) V d = 0 and (2) no current flows into either terminal of V d.

Calculate V
out by writing two KVL equations.

Begin by defining current i
1 flowing right through the 100 Ω resistor, and i 2 flowing
right through the 470 Ω resistor.

-5 + 100i
1 + Vd = 0 [1]

-5 + 100i
1 + 470i 2 + Vout = 0 [2]

Making use of the fact that in this case V
d = 0, we find that i1 = 5/100 A.

Making use of the fact that no current flows into the input terminals of the op am
p,
i
1 = i2. Thus, Eq. [2] reduces to

-5 + 570(5/100) + V
out = 0 or

V
out = -23.5 V (hence, the circuit is acting as a voltage amplifier.)

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Engineering Circuit Analysis, 7
th
Edition Chapter Three Solutions 10 March 2006

20. (a) By KVL, -2 + v x + 8 = 0

so that v
x = -6 V.

(b) By KCL at the top left node,

i
in = 1 + IS + vx/4 – 6

or i
in = 23 A

( c) By KCL at the top right node,

I
S + 4 v x = 4 - v x/4

So I
S = 29.5 A.

(d) The power provided by the dependent source is 8(4v
x) = -192 W.
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Engineering Circuit Analysis, 7
th
Edition Chapter Three Solutions 10 March 2006

21. (a) Working from left to right,
v
1 = 60 V
v
2 = 60 V
i
2 = 60/20 = 3 A
i
4 = v1/4 = 60/4 = 15 A
v1 = 60 V i 1 = 27 A
v2 = 60 V i 2 = 3 A
v3 = 15 V i 3 = 24 A
v4 = 45 V i 4 = 15 A
v5 = 45 V i 5 = 9 A
v 3 = 5i 2 = 15 V

By KVL, -60 + v
3 + v5 = 0
v
5 = 60 – 15 = 45 V
v
4 = v5 = 45

i
5 = v5/5 = 45/5 = 9 A
i
3 = i4 + i5 = 15 + 9 = 24 A
i
1 = i2 + i3 = 3 + 24 = 27

(b) It is now
e the power absorbed by each element:

p
1= -v1i1= -(60)(27) = -1.62 kW
p2= v2i2= (60)(3) = 180 W
p3= v3i3= (15)(24) = 360 W
p4= v4i4= (45)(15) = 675 W
p5= v5i5= (45)(9) = 405 W

and it is a simple matter to check that these values indeed sum to zero as they should.
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Engineering Circuit Analysis, 7
th
Edition Chapter Three Solutions 10 March 2006

22. Refer to the labeled diagram below.

Beginning from the left, we find

p
20V = -(20)(4) = -80 W

v
1.5 = 4(1.5) = 6 V therefore p 1.5 = (v 1.5)
2
/ 1.5 = 24 W.

v
14 = 20 – v 1.5 = 20 – 6 = 14 V therefore p 14 = 14
2
/ 14 = 14 W.

i
2 = v 2/2 = v 1.5/1.5 – v 14/14 = 6/1.5 – 14/14 = 3 A

Therefore v
2 = 2(3) = 6 V and p 2 = 6
2
/2 = 18 W.

v
4 = v 14 – v2 = 14 – 6 = 8 V therefore p 4 = 8
2
/4 = 16 W

i
2.5 = v2.5/ 2.5 = v 2/2 – v 4/4 = 3 – 2 = 1 A

Therefore v
2.5 = (2.5)(1) = 2.5 V and so p 2.5 = (2.5)
2
/2.5 = 2.5 W.

I
2.5 = - IS, thefore IS = -1 A.

KVL allows us to write -v
4 + v2.5 + vIS = 0

so V
IS = v4 – v2.5 = 8 – 2.5 = 5.5 V and p IS = -VIS IS = 5.5 W.

A quick check assures us that these power quantities sum to zero.
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Engineering Circuit Analysis, 7
th
Edition Chapter Three Solutions 10 March 2006

23. Sketching the circuit as described,

( a)
v 14 = 0. v 13 = v43 = 8 V
v
23 = -v 12 – v34 = -12 + 8 = -4 V
v
24 = v23 + v34 = -4 – 8 = -12 V

( b) v
14 = 6 V. v 13 = v14 + v43 = 6 + 8 = 14 V
v
23 = v13 – v12 = 14 – 12 = 2 V
v
24 = v23 + v34 = 2 – 8 = -6 V

( c) v
14 = -6 V. v 13 = v14 + v43 = -6 + 8 = 2 V
v
23 = v13 – v12 = 2 – 12 = -10 V
v
24 = v23 + v34 = -10 – 8 = -18 V
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Engineering Circuit Analysis, 7
th
Edition Chapter Three Solutions 10 March 2006

24. (a) By KVL, -12 + 5000I D + VDS + 2000ID = 0

Therefore, V
DS = 12 – 7(1.5) = 1.5 V.

( b) By KVL, - V
G + VGS + 2000ID = 0

Therefore, V
GS = VG – 2(2) = -1 V.

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Engineering Circuit Analysis, 7
th
Edition Chapter Three Solutions 10 March 2006

25. Applying KVL around this series circuit,

-120 + 30i
x + 40i x + 20i x + vx + 20 + 10i x = 0

where v
x is defined across the unknown element X, with the “+” reference on top.
Simplifying, we find that 100i
x + vx = 100

To solve further we require specific information about the element X and its
properties.

(a) if X is a 100-Ω resistor,

v
x = 100i x so we find that 100 i x + 100 i x = 100.
Thus
i
x = 500 mA and p x = vx ix = 25 W.
(b) If X is a 40-V independent voltage source such that v
x = 40 V, we find that

i
x = (100 – 40) / 100 = 600 mA and p x = vx ix = 24 W

(c) If X is a dependent voltage source such that v
x = 25ix,

i
x = 100/125 = 800 mA and p x = vx ix = 16 W.
(d) If X is a dependent voltage source so that v
x = 0.8v 1,
where v
1 = 40i x, we have
100 i
x + 0.8(40i x) = 100

or i
x = 100/132 = 757.6 mA and p x = vx ix = 0.8(40)(0.7576)
2
= 18.37 W.

(e) If X is a 2-A independent current source, arrow up,

100(-2) + v
x = 100

so that v
x = 100 + 200 = 300 V and p x = vx ix = -600 W

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Engineering Circuit Analysis, 7
th
Edition Chapter Three Solutions 10 March 2006

26. (a) We first apply KVL:
-20 + 10i
1 + 90 + 40i 1 + 2v 2 = 0

where v
2 = 10i 1. Substituting,
70 + 70 i
1 = 0

or i
1= -1 A.

(b) Applying KVL,
-20 + 10
i
1 + 90 + 40i 1 + 1.5v 3 = 0 [1]
where
v
3 = -90 – 10i 1 + 20 = -70 – 10 i 1

alternatively, we could write
v
3 = 40i 1 + 1.5v 3 = -80i 1

Using either expression in Eq. [1], we find i
1 = 1 A.

(c) Applying KVL,
-20 + 10i
1 + 90 + 40i 1 - 15 i 1 = 0
Solving, i
1 = - 2A.

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Engineering Circuit Analysis, 7
th
Edition Chapter Three Solutions 10 March 2006

27. Applying KVL, we find that

-20 + 10i
1 + 90 + 40i 1 + 1.8v 3 = 0 [1]

Also, KVL allows us to write
v
3 = 40i 1 + 1.8v 3

v
3 = -50i 1
So that we may write Eq. [1] as

50i
1 – 1.8(50)i 1 = -70

or i
1 = -70/-40 = 1.75 A.

Since v
3 = -50i 1 = -87.5 V, no further information is required to determine its value.

The 90-V source is absorbing (90)(i
1) = 157.5 W of power and the dependent source
is absorbing (1.8v
3)(i1) = -275.6 W of power.

Therefore, none of the conditions specified in (a) to (d) can be met by this circuit.
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Engineering Circuit Analysis, 7
th
Edition Chapter Three Solutions 10 March 2006

28. (a) Define the charging current i as flowing clockwise in the circuit provided.
By application of KVL,

-13 + 0.02i + Ri + 0.035i + 10.5 = 0

We know that we need a current i = 4 A, so we ma
y calculate the necessary resistance

R = [13 – 10.5 – 0.055(4)]/ 4 = 570 mΩ

(b) The
y consists of the power absorbed by the
0.035-Ω resistance (0.035i
2
), and the power absorbed by the 10.5-V ideal battery
(10.5i). Thus, we need to solve the quadratic equation

0.035i
2
+ 10.5i = 25

which has the solutions i = -302.4 A and i = 2.362 A.

In order to determine which of these two values should be used, we must recall that
the idea is to charge th
e battery, implying that it is absorbing power, or that i as
defined is positive. Thus, we choose i = 2.362 A, and, making use of the expression
developed in part (a), we find that

R = [13 – 10.5 – 0.055(2.362)]/ 2.362 = 1.003 Ω

(c) To obtain a voltage of 11 V across the battery, we apply KVL:

0.035i + 10.5 = 11 so that i = 14.29 A

From part (a), this m
eans we need

R = [13 – 10.5 – 0.055(14.29)]/ 14.29 = 119.9 mΩ

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Engineering Circuit Analysis, 7
th
Edition Chapter Three Solutions 10 March 2006

29. Drawing the circuit described, we also define a clockwise current i.

By KVL, we find that

-13 + (0.02 + 0.5 – 0.05)i + 0.035i + 10.5 = 0

or that i = (13 – 10.5)/0.505 = 4.951 A

and V
battery = 13 – (0.02 + 0.5)i = 10.43 V.

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Engineering Circuit Analysis, 7
th
Edition Chapter Three Solutions 10 March 2006

30. Applying KVL about this simple loop circuit (the dependent sources are still linear
elements, by the way, as they depend only upon a sum of voltages)

-40 + (5 + 25 + 20)i – (2 v
3 + v2) + (4v 1 – v2) = 0 [1]

where we have defined i to be flowing in the clockwise direction, and
v
1 = 5i, v 2 = 25i, and v 3 = 20i.

Performing the necessary substition, Eq. [1] becomes

50i - (40i + 25i) + (20i – 25i) = 40

so that i = 40/-20 = -2 A

Com
puting the absorbed power is now a straigh
tforward matter:

p
40V = (40)(-i) = 80 W
p5Ω = 5i
2
= 20 W
p25Ω = 25i
2
= 100 W
p20Ω = 20i
2
= 80 W
pdepsrc1= (2v 3 + v2)(-i) = (40i + 25i) = -260 W
pdepsrc2= (4v 1 - v2)(-i) = (20i - 25i) = -20 W
and we can easily verify that these quantities indeed sum
to zero as expected.
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Engineering Circuit Analysis, 7
th
Edition Chapter Three Solutions 10 March 2006

31. We begin by defining a clockwise current i.

(a) i = 12/(40 + R) mA, with R expressed in kΩ .

We want i
2
· 25 = 2
or
2 25
40
12
2
=⋅⎟
⎠
⎞
⎜
⎝
⎛
+R

Rearranging, we find a quadratic expression involving R:

R
2
+ 80R – 200 = 0

which has the solutions R = -82.43 kΩ and R = 2.426 kΩ. Only the latter is a
physical solution, so
R = 2.426 kΩ .

( b) We require i · 12 = 3.6 or i = 0.3 mA
From the circuit, we also see that i = 12/(15 + R + 25) mA
.
Substituting the desired value for i , we
find that the required value of R is R = 0.

( c)

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Engineering Circuit Analysis, 7
th
Edition Chapter Three Solutions 10 March 2006

32. By KVL,
–12 + (1 + 2.3 + R
wire segment) i = 0

The wire segment is a 3000–ft section of 28–AWG solid copper wire. Using Table
2.3, we compute its resistance as

(16.2 mΩ/ft)(3000 ft) = 48.6 Ω

which is certainly not negligib
le compared to the other resistances in the circuit!

Thus,
i = 12/(1 + 2.3 + 48.6) = 231.2 mA

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Engineering Circuit Analysis, 7
th
Edition Chapter Three Solutions 10 March 2006

33. We can apply Ohm’s law to find an expression for v o:

v
o = 1000(–gm vπ)

We do not have a value for v
π, but KVL will allow us to express that in terms of v o,
which we do know:

–10×10
–3
cos 5t + (300 + 50×10
3
) i = 0

where i is defined as flowing clockwise.

Thus, v
π = 50×10
3
i = 50×10
3
(10×10
–3
cos 5t ) / (300 + 50×10
3
)

= 9.940×10
–3
cos 5t V

and we by substitution we find that

v
o = 1000(–25×10
–3
)( 9.940×10
–3
cos 5t )

= –248.5 cos 5t mV

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Engineering Circuit Analysis, 7
th
Edition Chapter Three Solutions 10 March 2006

34. By KVL, we find that

–3 + 100 I
D + VD = 0

Substituting I
D = 3×10
–6
(e
V
D
/ 27×10
–3
– 1), we find that

–3 + 300×10
–6
(e
V
D
/ 27×10
–3
– 1) + VD = 0

This is a transcendental equation. Using a scientif
ic calculator or a numerical
software package such as MATLAB
®
, we find

V
D = 246.4 mV

Let’s assume such software-based assistance is unavailable. In that case, we need to
“guess” a value for V
D, substitute it into the right hand side of our equation, and see
how close the result is to the left hand side (in this case, zero).

GUESS RESULT
0 –3
1 3.648×10
12
0.5 3.308×10
4
0.25 0.4001
0.245 –0.1375
0.248 0.1732
0.246 –0.0377
oops
better
At this point, the error is getting much smaller, and our confidence is increasing
as to the value of V
D.

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Engineering Circuit Analysis, 7
th
Edition Chapter Three Solutions 10 March 2006

35. Define a voltage v x, “+” reference on the right, across the dependent current source.
Note that in fact v
x appears across each of the four elements. We first convert the 10
mS conductance into a 100–Ω resistor, and the 40–mS conductance into a 25–Ω
resistor.
(a) Applying KCL, we sum the currents flowing into the right–hand node:

5 – v
x / 100 – v x / 25 + 0.8 i x = 0 [1]
This represents one equation in two unknowns. A second equation to introduce at this
point is i
x = vx /25 so that Eq. [1] becomes

5 – v
x / 100 – v x / 25 + 0.8 (v x / 25) = 0
Solving for v
x, we find v x = 277.8 V. It is a simple matter now to compute the power
absorbed by each element:
P
5A = –5 v x = –1.389 kW
P100Ω = (v x)
2
/ 100 = 771.7 W
P25Ω = (v x)
2
/ 25 = 3.087 kW
Pdep= –v x(0.8 i x) = –0.8 (v x)
2
/ 25 = –2.470 kW
A quick check assures us that the calculated values sum to zero, as they should. (b) Again summ
ing the currents into the right–hand node,

5 – v
x / 100 – v x / 25 + 0.8 i y = 0 [2]
where i
y = 5 – v x/100
Thus, Eq. [2] becomes
5 – v
x / 100 – v x / 25 + 0.8(5) – 0.8 (i y) / 100 = 0
Solving, we find that v
x x = 155.2 V and i y = 3.448 A
So that

P
5A = –5 v x = –776.0 W
P100Ω = (v x)
2
/ 100 = 240.9 W
P25Ω = (v x)
2
/ 25 = 963.5 W
Pdep= –v x(0.8 i y) = –428.1 W
A quick check shows us that the calculated values sum to 0.3, which is reasonably
close to zero com
pared to the size of the terms (small roundoff errors accumulated).
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Engineering Circuit Analysis, 7
th
Edition Chapter Three Solutions 10 March 2006

36. Define a voltage v with the “+” reference at the top node. Applying KCL and
summing the currents flowing out of the top node,

v/5,000 + 4×10
–3
+ 3i 1 + v/20,000 = 0 [1]

This, unfortunately, is one equation in two unknowns, necessitating the search for a
second suitable equation. Returning to the circuit diagram, we observe that

i
1 = 3 i 1 + v/2,000

or i
1 = –v/40,000 [2]

Upon substituting Eq. [2] into Eq. [1], Eq. [1] becomes,

v/5,000 + 4×10
–3
– 3v/40,000 + v/20,000 = 0

Solving, we find that

v = –22.86 V
and
i
1 = 571.4 μ A

Since i
x = i1, we find that i x = 571.4 μ A.

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Engineering Circuit Analysis, 7
th
Edition Chapter Three Solutions 10 March 2006

37. Define a voltage v x with its “+” reference at the center node. Applying KCL and
summing the currents into the center node,

8 – v
x /6 + 7 – v x /12 – v x /4 = 0

Solving, v
x = 30 V.

It is now a straightforward matter to compute the power absorbed by each element:

P
8A = –8 v x = –240 W
P6Ω = (v x)
2
/ 6 = 150 W
P8A = –7 v x = –210 W
P12Ω = (v x)
2
/ 12 = 75 W
P4Ω = (v x)
2
/ 4 = 225 W

and a quick check verifies that the computed quantities sum to zero, as expected.

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Engineering Circuit Analysis, 7
th
Edition Chapter Three Solutions 10 March 2006

38. (a) Define a voltage v across the 1–kΩ resistor with the “+” reference at the top node.
Applying KCL at this top node, we find that

80×10
–3
– 30×10
–3
= v/1000 + v /4000
Solving,
v = (50×10
–3
)(4×10
6
/ 5×10
3
) = 40 V

and P
4kΩ = v
2
/4000 = 400 mW

(b) Once again, we first define a voltage v across the 1–kΩ res
istor with the “+”
reference at the top node. Applying KCL at this top node, we find that

80×10
–3
– 30×10
–3
– 20×10
–3
= v/1000
Solving,
v = 30 V

and P
20mA = v

· 20×10
–3
= 600 mW

(c) Once again, we first define a voltage v across the 1–kΩ resis
tor with the “+”
reference at the top node. Applying KCL at this top node, we find that

80×10
–3
– 30×10
–3
– 2i x = v/1000
where
i
x = v/1000
so that
80×10
–3
– 30×10
–3
= 2v /1000 + v /1000
and
v = 50×10
–3
(1000)/3 = 16.67 V

Thus, P
dep = v

· 2ix = 555.8 mW

(d) W
x = 60/1000 = 60 mA. KCL stipulates that (viewing currents
into and out of the top node)

80 – 30 + i
s = i x = 60
Thus, i
s = 10 mA

and P
60V = 60(–10) mW = –600 mW
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Engineering Circuit Analysis, 7
th
Edition Chapter Three Solutions 10 March 2006

39. (a) To cancel out the effects of both the 80-mA and 30-mA sources, i S must be set to

i
S = –50 mA.

(b) Define a current is flowing out of the
“+” reference terminal of the independent
voltage source. Interpret “no power” to mean “zero power.”

Summing the currents flowing into the top node and invoking KCL, we find that

80×10
-3
- 30×10
-3
- vS/1×10
3
+ i S = 0

Simplifying slightly, this becomes

50 - v
S + 10
3
iS = 0 [1]

We are seeking a value for v
S such that v S · iS = 0. Clearly, setting v S = 0 will achieve
this. From Eq. [1], we also see that setting v
S = 50 V will work as well.
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Engineering Circuit Analysis, 7
th
Edition Chapter Three Solutions 10 March 2006

40. Define a voltage v 9 across the 9-Ω resistor, with the “+” reference at the top node.

( a) Summ
ing the currents into the right-hand node and applying KCL,

5 + 7 = v
9 / 3 + v 9 / 9

Solving, we find that v
9 = 27 V. Since i x = v9 / 9, i x = 3 A.

( b) Again, w
e apply KCL, this time to the top left node:

2 – v
8 / 8 + 2i x – 5 = 0

Since we know from part (a) that i
x = 3 A, we may calculate v 8 = 24 V.

( c) p
5A = (v 9 – v8) · 5 = 15 W.

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Engineering Circuit Analysis, 7
th
Edition Chapter Three Solutions 10 March 2006

41. Define a voltage v x across the 5-A source, with the “+” reference on top.

Applying KCL at the top node then yields

5 + 5v
1 - vx/ (1 + 2) – v x/ 5 = 0 [1]

where v
1 = 2[v x /(1 + 2)] = 2 v x / 3.

Thus, Eq. [1] becomes
5 + 5(2 v
x / 3) – v x / 3 – v x / 5 = 0

or 75 + 50 v
x – 5 vx – 3 vx = 0, which, upon solving, yields v x = -1.786 V.

The power absorbed by the 5-Ω resistor is then simply (v
x)
2
/5 = 638.0 mW.

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Engineering Circuit Analysis, 7
th
Edition Chapter Three Solutions 10 March 2006

42. Despite the way it may appear at first glance, this is actually a simple node-pair
circuit. Define a voltage v across the elements, with the “+” reference at the top node.

Summ
ing the currents leaving the top node and applying KCL, we find that

2 + 6 + 3 + v/5 + v/5 + v/5 = 0

or v = -55/3 = -18.33 V. The power supplied by each source is then co
mputed as:

p
2A = -v(2) = 36.67 W
p
6A = -v(6) = 110 W
p
3A = -v(3) = 55 W

The power absorbed by each resistor is simp
ly v
2
/5 = 67.22 W for a total of 201.67 W,
which is the total power supplied by all sources. If instead we want the “power
supplied” by the resistors, we multiply by -1 to obtain -201.67 W. Thus, the sum of
the supplied power of each circuit element is zero, as it should be.

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Engineering Circuit Analysis, 7
th
Edition Chapter Three Solutions 10 March 2006

43. Defining a voltage Vx across the 10-A source with the “+” reference at the top node,
KCL tells us that 10 = 5 + I
1Ω, where I1Ω is defined flowing downward through the
1-Ω resistor.

Solving, we find that I
1Ω = 5 A, so that Vx = (1)(5) = 5 V.

So, we need to solve
V
x = 5 = 5(0.5 + Rsegment)
with R
segment = 500 mΩ.

From Table 2.3, we see that 28-AWG solid copper wire has a resistance of 65.3
mΩ/ft. Thus, the to
tal number of miles needed of the wire is

miles 101.450
ft/mi) /ft)(5280m (65.3
m 500
3-
×=
Ω
Ω

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Engineering Circuit Analysis, 7
th
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44. Since v = 6 V, we know the current through the 1-Ω resistor is 6 A, the current
through the 2-Ω resistor is 3 A, and the current through the 5-Ω resistor is 6/5
= 1.2 A, as shown below:

By KCL, 6 + 3 + 1.2 + i
S = 0 or i S = -10.2 A.
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45. (a) Applying KCL, 1 – i – 3 + 3 = 0 so i = 1 A.

(b) Look
it, we see 1 + 3 = 4 A flowing into the
unknown current source, which, by virtue of KCL, must therefore be a 4-A current
source. Thus, KCL at the node labeled with the “+” reference of the voltage v gives

4 – 2 + 7 – i = 0 or i = 9 A

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th
Edition Chapter Three Solutions 10 March 2006

46. (a) We may redraw the circuit as

Then, we see that v = (1)(1) = 1 V.

(b) W
Ω resistor into a single 7-A
current source. On the left, the two 1-A sources in series reduce to a single 1-A
source.

The new 1-A source and the 3-A source combine to yield a 4-A source in series with
the unknown current source which, by KCL, must be a 4-A current source.

At this point we have reduced the circuit to

Further simplification is possible, resulting in

From which we see clearly that v = (9)(1) = 9 V.

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47. (a) Combine the 12-V and 2-V series connected sources to obtain a new 12 – 2 = 10 V
source, with the “+” reference terminal at the to
p. The result is two 10-V sources in
parallel, which is permitted by KVL. Therefore,

i = 10/1000 = 10 mA.

(b) No current flows through the 6-V source, so we m
ay neglect it for this calculation.
The 12-V, 10-V and 3-V sources are connected in series as a result, so we replace
them with a 12 + 10 –3 = 19 V source as shown

Thus, i = 19/5 = 3.8 A.
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th
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48. We first combine the 10-V and 5-V sources into a single 15-V source, with the “+”
reference on top. The 2-A and 7-A current sources combine into a 7 – 2 = 5 A current
source (arrow pointing down); although these two current sources m
ay not appear to
be in parallel at first glance, they actually are.

Redrawing our circuit,

we see that v = 15 V (
note that we can completely the ignore the 5-A source here,
since we have a voltage source directly across the resistor). Thus,

P
16Ω = v
2
/16 = 14.06 W.
Returning to the original circuit, we see that the 2 A source is in parallel with both 16
Ω resistors, so that it has a voltage of 15 V across it as well (the same goe
s for the 7 A
source). Thus,

2Aabs
P = -15(2) = -30 W
7Aabs
P = -15(-7) = 105 W

Each resistor draws 15/16 A, so the 5 V and 10 V sources each see a current of

30/16 + 5 = 6.875 A flowing through them.

Thus,

5Vabs
P= -5(6.875) = -34.38 W
10Vabs
P = -10(6.875) = -68.75 W

which sum to -0.01 W, close enough to zero compared to the size of the terms

(roundoff error accumulated).
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th
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49. We can combine the voltage sources such that

i = v
S/ 14

( a) v S = 10 + 10 – 6 – 6 = 20 –12 = 8
Therefore
i = 8/14 = 571.4 mA
.
( b) v
S = 3 + 2.5 – 3 – 2.5 = 0 Therefore i = 0.
( c) v
S = -3 + 1.5 – (-0.5) – 0 = -1 V
Therefore
i = -1/14 = -71.43 mA.

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Engineering Circuit Analysis, 7
th
Edition Chapter Three Solutions 10 March 2006

50. We first simplify as shown, making use of the fact that we are told i x = 2 A to find the
voltage across the middle and right-most 1-Ω resistor
s as labeled.

By KVL, then, we find that v
1 = 2 + 3 = 5 V.
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Edition Chapter Three Solutions 10 March 2006

51. We see that to determine the voltage v we will need v x due to the presence of the
dependent current soruce. So, let’s begin with the right-hand side, where we find that

v
x = 1000(1 – 3) × 10
-3
= -2 V.

Returning to the left-hand side of the circuit, and summing currents into the top node,
we find that

(12 – 3.5) ×10
-3
+ 0.03 v x = v/10×10
3

or v = -515 V.

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Engineering Circuit Analysis, 7
th
Edition Chapter Three Solutions 10 March 2006

52. (a) We first label the circuit with a focus on determining the current flowing through
each voltage source:

Then the power absorbed by each voltage source is

P
2V = -2(-5) = 10 W
P
4V = -(-4)(4) = 16 W
P
-3V = -(-9)(-3) = -27 W

For the current sources,

So that the absorbed power is

P
-5A = +(-5)(6) = -30 W
P
-4A = -(-4)(4) = 16 W
P
3A = -(3)(7) = -21 W
P
12A = -(12)(-3) = 36 W

A quick check assures us that these absorbed powers sum to zero as they should.

(b) W
that the voltage across the –5-A source
drops to zero. Define V
x across the –5-A source such that the “+” reference terminal is
on the left. Then,
-2 + V
x – Vneeded = 0
or V
needed = -2 V.

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Engineering Circuit Analysis, 7
th
Edition Chapter Three Solutions 10 March 2006

53. We begin by noting several things:
(1) The bottom resistor has been shorted out;
(2) the far-right resistor is only connected by one terminal and therefore does
not affect the equi
valent resistance as seen from the indicated terminals;
(3) All resistors to the right of the top left resistor have been shorted.

Thus, from the indicated terminals, we only see the single 1-kΩ resistor, so that
R
eq = 1 kΩ .
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Edition Chapter Three Solutions 10 March 2006

54. (a) We see 1Ω || (1 Ω + 1 Ω ) || (1 Ω + 1 Ω + 1 Ω )
= 1 Ω || 2 Ω || 3 Ω
= 545.5 m Ω

( b) 1/R
eq = 1 + 1/2 + 1/3 + … + 1/N

Thus, R
eq = [1 + 1/2 + 1/3 + … + 1/N]
-1
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55. (a) 5 kΩ = 10 kΩ || 10 kΩ

( b) 57 333 Ω = 47 kΩ + 10 kΩ + 1 kΩ || 1kΩ || 1kΩ

( c) 29.5 kΩ = 47 kΩ || 47 kΩ + 10 kΩ || 10 kΩ + 1 kΩ
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0 V 1 A
5 Ω
7 Ω
5 Ω
1 A 2.917 Ω
56. (a) no simplification is possible using only source and/or resistor combination
techniques.

( b) W
e first simplify the circuit to

and then notice that the 0-V source is shorting out one of the 5-Ω resistors, so a
further simplification is possible, noting that 5 Ω || 7 Ω = 2.917 Ω:

Engineering Circuit Analysis, 7
th
Edition Chapter Three Solutions 10 March 2006

57. Req = 1 kΩ + 2 kΩ || 2 kΩ + 3 kΩ || 3 kΩ + 4 kΩ || 4 kΩ
= 1 k Ω + 1 k Ω + 1.5 kΩ + 2 kΩ

=
5.5 kΩ.

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58. (a) W
orking from right to left, we first see that we may combine several resistors as
100 Ω + 100 Ω || 100 Ω + 100 Ω = 250 Ω, yielding the following circuit:

R
eq →
100 Ω
100
Ω
100 Ω
100 Ω
100 Ω
250 Ω
100 Ω
100 Ω
10 Ω
20 Ω
5 Ω
14.4 Ω Req →
2 Ω
15 Ω 10 Ω
50 Ω
8 Ω 20 Ω 30 Ω
2 Ω
16.67 Ω
8 Ω 50 Ω

Next, we see 100 Ω + 100 Ω || 250 Ω + 100 Ω = 271.4 Ω,

and subsequently 100 Ω + 100 Ω || 271.4 Ω + 100 Ω = 273.1 Ω,
and, finally,
R
eq = 100 Ω || 273.1 Ω = 73.20 Ω .

(b) First, we com
bine 24 Ω || (50 Ω + 40 Ω ) || 60 Ω = 14.4 Ω , which leaves us with

Thus, R
eq = 10 Ω + 20 Ω || (5 + 14.4 Ω) = 19.85 Ω.

(c) First combine the 10-Ω and 40-Ω resistors and redraw the circuit:

We now see we have (10 Ω + 15 Ω ) || 50 Ω = 16.67 Ω. Redrawing once again,

where the eq
uivalent resistance is seen to be 2 Ω + 50 Ω || 16.67 Ω + 8 Ω = 22.5 Ω.

Engineering Circuit Analysis, 7
th
Edition Chapter Three Solutions 10 March 2006

59. (a) R eq = [(40 Ω + 20 Ω ) || 30 Ω + 80 Ω] || 100 Ω + 10 Ω = 60 Ω.

( b) R
eq = 80 Ω = [(40 Ω + 20 Ω ) || 30 Ω + R] || 100 Ω + 10 Ω
70 Ω = [(60 Ω || 30 Ω) + R] || 100 Ω
1/70 = 1/(20 + R) + 0.01
20+ R = 233.3 Ω therefore R = 213.3 Ω.

(c) R = [(40 Ω + 20 Ω ) || 30 Ω + R] || 100 Ω + 10
Ω
R – 10 Ω = [20 + R] || 100
1/(R – 10) = 1/(R + 20) + 1/ 100
3000 = R
2
+ 10R – 200

Solving, we find R = -61.79 Ω or R = 51.79 Ω.
Clearly, the first is not a physical solution, so R = 51.79 Ω.

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Engineering Circuit Analysis, 7
th
Edition Chapter Three Solutions 10 March 2006

60. (a) 25
Ω = 100 Ω || 100 Ω || 100 Ω || 100 Ω

( b) 60
Ω = [(100 Ω || 100 Ω ) + 100 Ω] || 100 Ω

( c) 40 Ω = (100 Ω + 100 Ω) || 100 Ω || 100 Ω
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Engineering Circuit Analysis, 7
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61. Req = [(5 Ω || 20 Ω) + 6 Ω] || 30 Ω + 2.5 Ω = 10 Ω
The source therefore provides a total of 1000 W a
nd a current of 100/10 = 10 A.

P
2.5Ω = (10)
2
· 2.5 = 250 W

V
30Ω = 100 - 2.5(10) = 75 V
P
30Ω = 75
2
/ 30 = 187.5 W

I
6Ω = 10 – V30Ω /30 = 10 – 75/30 = 7.5 A
P
6Ω = (7.5)
2
· 6 = 337.5 W

V
5Ω = 75 – 6(7.5) = 30 V
P
5Ω = 30
2
/ 5 = 180 W

V
20Ω = V5Ω = 30 V
P
20Ω = 30
2
/20 = 45 W

We check our results by verifying that the absorbed powers in fact add to 1000 W.
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62. To begin with, the 10-Ω and 15-Ω resistor
s are in parallel ( = 6 Ω), and so are the
20-Ω and 5-Ω resistor
s (= 4 Ω ).

Also, the 4-A, 1-A and 6-A current
sources are in parallel, so they can be combined
into a single 4 + 6 – 1 = 9 A current source as shown:

- vx +
9 A
14

Ω 6 Ω

6 Ω 4 Ω

Next, we note that (14 Ω + 6 Ω ) || (4 Ω + 6
Ω) = 6.667 Ω
so that
v
x = 9(6.667) = 60 V
and
i
x = -60/10 = -6 A.

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th
Edition Chapter Three Solutions 10 March 2006

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63. (a) W
orking from right to left, and borrowing x || y notation from resistance
calculations to indicate the operation xy/(x +
y),

G
in = {[(6 || 2 || 3) + 0.5] || 1.5 || 2.5 + 0.8} || 4 || 5 mS
= {[(1) + 0.5] || 1.5 || 2.5 + 0.8} || 4 || 5 mS

13.64 mS
100 mS
22.22 mS
Gin →
= {1.377} || 4 || 5
= 0.8502 mS = 850.2 mS

(b) The 50-mS and 40-mS conductances are in series, equivalent to (50(40)/90 =
22.22 mS. The 30-mS and 25-mS conductances are also in series, equivalent to 13.64
mS. Redrawing for clarity,

we see that G
in = 10 + 22.22 + 13.64 = 135.9 mS.

Engineering Circuit Analysis, 7
th
Edition Chapter Three Solutions 10 March 2006

64. The bottom four resistors between the 2-Ω resistor and the 30-V source are shorted
out. The 10-Ω and 40-Ω resist
ors are in parallel (= 8 Ω ), as are the 15-Ω and 60-Ω
(=12 Ω ) resistors. These combinations are in series.

Define a clockwise current I through the 1-Ω resistor:

I = (150 – 30)/(2 + 8 + 12 + 3 + 1 + 2) = 4.286 A

P
1Ω = I
2
· 1 = 18.37 W

To compute P
10Ω, consider that since the 10-Ω and 40-Ω resistors are in parallel, the
same voltage V
x (“+” reference on the left) appears across both resistors. The current I
= 4.286 A flows into this combination. Thus, V
x = (8)(4.286) = 34.29 V and

P
10Ω = (Vx)
2
/ 10 = 117.6 W.

P
13Ω = 0 since no current flows through that resistor.

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Engineering Circuit Analysis, 7
th
Edition Chapter Three Solutions 10 March 2006

65. With the meter being a short circuit and no current flowing through it, we can write

R
1i1 = R2i2
R
3i3 = RiR

And since i
1 = i3, and i 2 = iR, Eq [1] becomes
12
3
RR
RR
=, or
3
2
1
R
RR
R
= . Q.E.D.

→
11 2 2
33 R
RiRi
RiRi
=
[1]
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th
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66. The total resistance in the series string sums to 75 Ω.

The voltage dropped across the 2.2
Ω resistor is

2.2
2.2
V 10 293.3 mV
75
Ω
==
and the voltage dropped across the 47
Ω resistor is

47
47
V 10 6.267 mV
75
Ω
==

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Engineering Circuit Analysis, 7
th
Edition Chapter Three Solutions 10 March 2006

67. We first note that the 4.7 kΩ and 2.2 kΩ resistors can be combined into a single 1.5
k
Ω resistor, which is then in series with the 10 kΩ resistor. Next we note that the 33
k
Ω / 47 kΩ parallel combination can be replaced by a 19.39 kΩ resistance, which is
in series with the remaining 33 k
Ω resistor.

By voltage division, then, and noting that V
47kΩ is the same voltage as that across the
19.39 k
Ω resistance,
V
47kΩ =
19.39
2 607.0 mV
10 1.5 33 19.39
=
+++

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th
Edition Chapter Three Solutions 10 March 2006

68. (a) The current downward through the 33 Ω resistor is calculated more easily if we
first note that 134
Ω || 134 Ω = 67 Ω, and 67 Ω + 33 Ω = 100 Ω. Then,

33
1
100
12 571.4 mA
111
10 10 100
I
Ω
==
++

(b) The resistor flowing downward through either 134
Ω resistor is simply

571.4/2 = 285.7 mA (by current division).

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Engineering Circuit Analysis, 7
th
Edition Chapter Three Solutions 10 March 2006

69. We first note that 20 Ω || 60 Ω = 15 Ω, and 50 Ω || 30 Ω = 18.75 Ω.
Then, 15
Ω + 22 Ω + 18.75 Ω = 55.75 Ω.

Finally, we are left with two current sources, the series combination of 10
Ω + 15 Ω,
and 10
Ω || 55.75 Ω = 8.479 Ω.

Using current division on the simplified circuit,

()
15
1
10 15
30 8 22.12 A
11
10 15 8.479
I
Ω
+
=− =
+
+

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Engineering Circuit Analysis, 7
th
Edition Chapter Three Solutions 10 March 2006

70. One possible solution of many:

vS = 2(5.5) = 11 V
R
1 = R2 = 1 kΩ.
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Engineering Circuit Analysis, 7
th
Edition Chapter Three Solutions 10 March 2006

71. One possible solution of many:

iS = 11 mA
R
1 = R2 = 1 kΩ.
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th
Edition Chapter Three Solutions 10 March 2006

72. p15Ω = (v15)
2
/ 15×10
3
A

v15 = 15×10
3
(-0.3 v1)

where
v1 = [4 (5)/ (5 + 2)] · 2 = 5.714 V

Therefore
v15 = -25714 V and p15 = 44.08 kW.
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73. Replace the top 10 kΩ, 4 kΩ and 47 kΩ resistors with 10 kΩ + 4 kΩ || 47 kΩ =
13.69 k
Ω.

Define
vx across the 10 kΩ resistor with its “+” reference at the top node: then

vx = 5 · (10 kΩ || 13.69 kΩ) / (15 kΩ + 10 || 13.69 kΩ) = 1.391 V

ix = vx/ 10 mA = 139.1 μA

v15 = 5 – 1.391 = 3.609 V and p15 = (v15)
2
/ 15×10
3
= 868.3 μW.
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th
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74. We may combine the 12-A and 5-A current sources into a single 7-A current source
with its arrow oriented upwards. The left three resisto
rs may be replaced by a 3 +
6 || 13 = 7.105
Ω resistor, and the right three resistors may be replaced by a 7 + 20 || 4
= 10.33
Ω resistor.

By current division,
iy = 7 (7.105)/(7.105 + 10.33) = 2.853 A

We must now return to the original circuit. The current into the 6
Ω, 13 Ω parallel
combination is 7 –
iy = 4.147 A. By current division,

ix = 4.147 . 13/ (13 + 6) = 2.837 A

and
px = (4.147)
2
· 3 = 51.59 W
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Engineering Circuit Analysis, 7
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75. The controlling voltage v1, needed to obtain the power into the 47-kΩ resistor, can be
found separately as that network does not depend on the left-hand network.
The right-most 2 k
Ω resistor can be neglected.

By current division, then, in combination with Ohm’s law,

v1 = 3000[5×10
-3
(2000)/ (2000 + 3000 + 7000)] = 2.5 V

Voltage division gives the voltage across the 47-k
Ω resistor:

V 0.9228
16.67 47
7)0.5(2.5)(4

20 ||100 47
47
5.0
1 =
+
=
+
v
So that
p47kΩ = (0.9928)
2
/ 47×10
3
= 18.12 μW
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Engineering Circuit Analysis, 7
th
Edition Chapter Three Solutions 10 March 2006

76. The temptation to write an equation such as

v1 =
2020
20
10
+

must be fought!
Voltage division only applies to resistors connected in series, meaning that the
same
current must flow through
each resistor. In this circuit, i1 ≠ 0 , so we do not have the
same current flowing through both 20 k
Ω resistors.

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Engineering Circuit Analysis, 7
th
Edition Chapter Three Solutions 10 March 2006

77. (a)
)]R (R || [R R
)R (R || R
V
4321
432
S2
++
+
=
v
=
( )
()
4324321
432432
S
R R R)R (R R R
R R R)R (R R
V
++++
+++

=
()
)R (RR R R RR
)R (R R
V
4324321
432
S++++ +

(
b)
)]R (R || [R R
R
V
4321
1
S1++
=v

=
()
4324321
1
S
R R R)R (R R R
R
V
++++

=
()
)R (RR R R RR
)R R (R R
V
4324321
4321
S++++ ++

(
c)
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
++
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
=
432
2
1
1
4R R R
R

R

v
i

=
( )
[] )R R )(RR (RR )R R (RRR
R R R R R
V
43243243211
24321
S
++++++ ++

=
() )R (RR R R RR
R
V
4324321
2
S
++++

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teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

Engineering Circuit Analysis, 7
th
Edition Chapter Three Solutions 10 March 2006

78. (a) With the current source open-circuited, we find that

v1 =
V 8-
6000 || 3000 500
500
40=
+
−

(
b) With the voltage source short-circuited, we find that

i2 = ()
3 1/3000
3 10 400 A
1/500 1/3000 1/6000
μ
−
×=
+ +

i3 = ()
3 500
3 10 600 A
500 3000 || 6000
μ
−
×=
+

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Engineering Circuit Analysis, 7
th
Edition Chapter Three Solutions 10 March 2006

79. (a) The current through the 5-Ω resistor is 10/5 = 2 A. Define R as 3 || (4 + 5)
= 2.25
Ω. The current through the 2-Ω resistor then is given by

25.5
I

R) (2 1
1
I
S
S
=
++

The current through the 5-Ω resistor is

A 2
9 3
3

25.5
I

S
=⎟
⎠
⎞
⎜
⎝
⎛
+
so that I
S = 42 A.

(
b) Given that IS is now 50 A, the current through the 5-Ω resistor becomes

A 2.381
9 3
3

25.5
I

S
=⎟
⎠
⎞
⎜
⎝
⎛
+

Thus, vx = 5(2.381) = 11.90 V

(
c)
0.2381
I
9 3
3

5.25
5I

I

S
S
S
=
⎥
⎦
⎤
⎢
⎣
⎡
⎟
⎠
⎞
⎜
⎝
⎛
+
=
xv

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Engineering Circuit Analysis, 7
th
Edition Chapter Three Solutions 10 March 2006

80. First combine the 1 kΩ and 3 kΩ resistors to obtain 750 Ω.
By current division, the current through resistor R
x is

I
Rx =
750 R 2000
2000
1010
x
3
++
×
−

and we know that R
x · IRx = 9

so
x
x
R 2750
R 20
9
+
=

9 R
x + 24750 = 20 Rx or Rx = 2250 W. Thus,

P
Rx = 9
2
/ Rx = 36 mW.

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Engineering Circuit Analysis, 7
th
Edition Chapter Three Solutions 10 March 2006

81. Define R = R3 || (R4 + R5)

Then
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
+
=
2
SRR R
R
V v

=
( )
()
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
++++
+++
2543543
543543
SR R R R)R (R R
R R R)R (R R
V

=
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
++++
+
)R (RR )R R(R R
)R (R R
V
54354 32
543
S

Thus,
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
+
=
54
5
R5R R
R
vv

=
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
++++)R (RR )R R(R R
R R
V
54354 32
53
S

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Engineering Circuit Analysis, 7
th
Edition Chapter Three Solutions 10 March 2006

82. Define R1 = 10 + 15 || 30 = 20 Ω and R2 = 5 + 25 = 30 Ω.

(
a) Ix = I1 . 15 / (15 + 30) = 4 mA

(
b) I1 = Ix . 45/15 = 36 mA

(
c) I2 = IS R1 / (R1 + R2) and I1 = IS R2 / (R1 + R2)
So I
1/I2 = R2/R1
Therefore
I
1 = R2I2/R1 = 30(15)/20 = 22.5 mA

Thus, I
x = I1
.
15/ 45 = 7.5 mA

(
d) I1 = IS R2/ (R1 + R2) = 60 (30)/ 50 = 36 mA

Thus, I
x = I1 15/ 45 = 12 mA.
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Engineering Circuit Analysis, 7
th
Edition Chapter Three Solutions 10 March 2006

83. vout = -gm vπ (100 kΩ || 100 kΩ) = -4.762×10
3
gm vπ

where
vπ = (3 sin 10t) · 15/(15 + 0.3) = 2.941 sin 10t

Thus,
vout = -56.02 sin 10t V
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Engineering Circuit Analysis, 7
th
Edition Chapter Three Solutions 10 March 2006

84. vout = -1000gm vπ

where vπ =
V 10sin 2.679
0.3 3) || (15
3 || 15
10sin 3 tt =
+
therefore
vout = -(2.679)(1000)(38×10
-3
) sin 10t = -101.8 sin 10t V.

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Engineering Circuit Analysis, 7
th
Edition Chapter Four Solutions 10 March 2006
1. (a) 0.1 -0.3 -0.4 v 1 0
-0.5 0.1 0 v
2 = 4
-0.2 -0.3 0.4 v
3 6

Solving this matrix equation using a scientific calculator, v
2 = -8.387 V

( b) Using a scien
tific calculator, the determinant is equal to 32.

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Engineering Circuit Analysis, 7
th
Edition Chapter Four Solutions 10 March 2006
2. (a) 1 1 1 v A 27
-1 2 3 v
B = -16
2 0 4 v
C -6

Solving this matrix equation using a scientific calculator,

v
A = 19.57
v
B = 18.71
v
C = -11.29

( b) Using a scientific calculator,

1
1 1
-1 2 3 = 16
2 0 4
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Engineering Circuit Analysis, 7
th
Edition Chapter Four Solutions 10 March 2006
3.

(a) We begin by simplifying the equations prior to solution:

123
12
12
4 0.08 0.05 0.02
8 0.02 0.025 0.045
2 0.05 0.115 0.025
vvv
vv
vv
=−−
=− − +
−=− + −
3
3
v
v
⎤⎡⎤
⎥⎢⎥
⎥⎢⎥
⎥⎢⎥

Then, we can solve the matrix equation:

1
2
3
0.08 -0.05 -0.02 4
-0.02 -0.025 0.045 8
-0.05 0.115 -0.025 2
v
v
v
⎡⎤ ⎡
⎢⎥ ⎢
=
⎢⎥ ⎢
⎢⎥ ⎢
−
⎣⎦ ⎣⎦⎣⎦

to obtain v
1 = 264.3 V, v 2 = 183.9 V and v 3 = 397.4 V.

(b) We may solve the matrix equation d irectly using MATLAB, but a be tter check is to
invoke the sym
bolic processor:

>> e1 = '4 = v1/100 + (v1 - v2)/20 + (v1 - vx)/50';
>> e2 = '10 - 4 - (-2) = (vx - v1)/50 + (vx - v2)/40';
>> e3 = '-2 = v2/25 + (v2 - vx)/40 + (v2 - v1)/20';
>> a = solve(e1,e2,e3,'v1','v2','vx');
>> a.v1
ans = 82200/311 >> a.v2

ans =

57200/311

>> a.vx

ans =

123600/311

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Engineering Circuit Analysis, 7
th
Edition Chapter Four Solutions 10 March 2006
4. We select the bottom node as our reference terminal and define two nodal voltages:

Ref.

Next, we write the two required nodal equations:

Node 1:
112
1
23
vvv−
=+

Node 2:
22
3
13
vvv−
−= +
1

Which may be simplified to: 5v
1 – 2v 2 = 6
and -v
1 + 4v 2 = -9

Solving, we find that v
1 = 333.3 mV.

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Engineering Circuit Analysis, 7
th
Edition Chapter Four Solutions 10 March 2006
5. We begin by selecting the bottom node as the reference term inal, and defining
two nodal voltages V
A and VB, as shown. (Note if we cho ose the upp er right
node, v
1 becomes a nodal voltage and falls directly out of the solution.)

V
A VB
Ref.

We note that after completing nodal analysis, we can find v
1 as v1 = VA – VB.

At node A:
AA
VVV
B
10 5
4
−
=+ [1]

At node B:
BB
VVV
(6)
85
A
−
−− = + [2]

Simplifying, 3V
A – 2VB = 40 [1]
–8V
A + 13VB = 240 [2]

Solving, V
A = 43.48 V and VB = 45.22 V, so v 1 = –1.740 V.
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Engineering Circuit Analysis, 7
th
Edition Chapter Four Solutions 10 March 2006
6. By inspection, no current flows through the 2 Ω resistor, so i 1 = 0.

We next designate the bottom node as the reference terminal, and define V
A and
V
B as shown:

V
A VB
Ref.
At node A:
AAB
VVV
2 [1]
31
−
=+
At node B:
BBBA
VVVV
2 [
66 1
2]
−
−= + +

Note this y ields V
A and VB, not v 1, due to our choice of reference node. So, we
obtain v
1 by KVL: v 1 = VA – VB.

Simplifying Eqs. [1] and [2],

4V
A – 3VB = 6 [1]
–3V
A + 4VB = –6 [2]

Solving, V
A = 0.8571 V and VB = -0.8571 V, so v 1 = 1.714 V.

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Engineering Circuit Analysis, 7
th
Edition Chapter Four Solutions 10 March 2006
7. The bottom node has the largest number of branch connections, so we choose that as
our reference node. This also m akes v
P easier to find, as it will be a nodal voltag e.
Working from left to right, we name our nodes 1, P, 2, and 3.

NODE 1: 10 = v 1/ 20 + (v 1 – vP)/ 40 [1]

NODE P: 0 = (v P – v1)/ 40 + v P/ 100 + (v P – v2)/ 50 [2]

NODE 2: -2.5 + 2 = (v 2 – vP)/ 50 + (v 2 – v3)/ 10 [3]

NODE 3: 5 – 2 = v 3/ 200 + (v 3 – v2)/ 10 [4]

Simplifying,

60 v 1 - 20v P = 8000 [1]
-50 v
1 + 110 v P - 40v 2 = 0 [2]
- v
P + 6v 2 - 5v 3 = -25 [3]
-200v
2 + 210v 3 = 6000 [4]
Solving,
v
P = 171.6 V
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Engineering Circuit Analysis, 7
th
Edition Chapter Four Solutions 10 March 2006
8. The logical choice for a reference node is the bottom node, as then v x will
automatically become a nodal voltage.

NODE 1: 4 = v 1/ 100 + (v 1 – v2)/ 20 + (v 1 – vx)/ 50 [1]

NODE x: 10 – 4 – (-2) = (v x – v1)/ 50 + (v x – v2)/ 40 [2]

NODE 2: -2 = v 2 / 25 + (v 2 – vx)/ 40 + (v 2 – v1)/ 20 [3]

Simplifying,

4 = 0.0800 v 1 – 0.0500v 2 – 0.0200v x [1]
8 = -0.0200 v
1 – 0.02500v 2 + 0.04500v x [2]
-2 = -0.0500v
1 + 0.1150v 2 – 0.02500v x [3]
Solving,
v
x = 397.4 V.

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Engineering Circuit Analysis, 7
th
Edition Chapter Four Solutions 10 March 2006
9. Designate the node between the 3-Ω and 6-Ω resistors as node X, and the right-hand
node of the 6-Ω resistor as node Y. The bottom node is chosen as the reference node.

( a) Wr
iting the two nodal equations, then
NODE X: –10 = (v
X – 240)/ 3 + (v X – vY)/ 6 [1]
NODE Y: 0 = (v
Y – vX)/ 6 + v Y/ 30 + (v Y – 60)/ 12 [2]

Simplifying, -180 + 1440 = 9 v
X – 3 v Y [1]
10800 = - 360 v
X + 612 v Y [2]

Solving, v
X = 181.5 V and v Y = 124.4 V

Thus, v
1 = 240 – v X = 58.50 V and v 2 = vY – 60 = 64.40 V

( b) The power absorbed by the 6-Ω resistor is

(v
X – vY)
2
/ 6 = 543.4 W
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Engineering Circuit Analysis, 7
th
Edition Chapter Four Solutions 10 March 2006
10. Only one nodal equation is required: At the node where three resistors join,

0.02v 1 = (v x – 5 i 2) / 45 + (v x – 100) / 30 + (v x – 0.2 v 3) / 50 [1]

This, however, is one equation in four unknowns, the other three resulting from the
presence of the dependent sources. Thus, we require three additional equations:

i
2 = (0.2 v 3 - vx) / 50 [2]

v1 = 0.2 v 3 - 100 [3]

v3 = 50i 2 [4]

Simplifying,
v
1 – 0.2v 3 = -100 [3]
– v
3 + 50 i 2 = 0 [4]
–v
x + 0.2v 3 – 50 i 2 = 0 [2]
0.07556v
x – 0.02v 1 – 0.004v 3 – 0.111i 2 = 33.33 [1]

Solving, we find that v 1 = -103..8 V and i 2 = -377.4 mA.
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Engineering Circuit Analysis, 7
th
Edition Chapter Four Solutions 10 March 2006
11. If v 1 = 0, the dependent source is a short circuit and we may redraw the circuit as:

At NODE 1: 4 - 6 = v
1/ 40 + (v 1 – 96)/ 20 + (v 1 – V2)/ 10

Since v
1 = 0, this simplifies to

-2 = -96 / 20 - V
2/ 10

so that V
2 = -28 V.
20 Ω 10 Ω
40 Ω
+
v
1 = 0
-
.
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Engineering Circuit Analysis, 7
th
Edition Chapter Four Solutions 10 March 2006
12. We choose the bottom node as ground to make calculation of i 5 easier. The left-most
node is named “1”, the top node is nam ed “2”, the central node is name
d “3” and the
node between the 4-Ω and 6-Ω resistors is named “4.”

NODE 1: - 3 = v
1/2 + (v 1 – v2)/ 1 [1]
NODE 2: 2 = ( v
2 – v1)/ 1 + (v 2 – v3)/ 3 + (v 2 – v4)/ 4 [2]
NODE 3: 3 = v
3/ 5 + (v 3 – v4)/ 7 + (v 3 – v2)/ 3 [3]
NODE 4: 0 = v
4/ 6 + (v 4 – v3)/ 7 + (v 4 – v2)/ 4 [4]

Rearranging and grouping terms,

3v
1 – 2v 2 = -6 [1]
-12 v
1 + 19v 2 – 4v 3 – 3v 4 = 24 [2]
–35v
2 + 71v 3 – 15v 4 = 315 [3]
-42v
2 – 24v 3 + 94v 4 = 0 [4]

Solving, we find that v
3 = 6.760 V and so i 5 = v3/ 5 = 1.352 A.
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Engineering Circuit Analysis, 7
th
Edition Chapter Four Solutions 10 March 2006
13. We can redraw this circuit and eliminate the 2.2-kΩ resistor as no current flows
through it:

At NODE 2: 7×10
-3
– 5×10
-3
= (v 2 + 9)/ 470 + (v 2 – vx)/ 10×10
-3
[1]

At NODE x: 5×10
-3
– 0.2v 1 = (v x – v2)/ 10×10
3
[2]

The additional equation required by the presence of the depend
ent source and the fact
that its controlling variable is not one of the nodal voltages:

v
1 = v 2 – vx [3]

Eliminating the variable v1 and grouping terms, we obtain:

10,470 v
2 – 470 v x = –89,518
and
1999 v
2 – 1999 v x = 50

Solving, we find v
x = –8.086 V.

↓
9 V 7 mA
5 mA
0.2 v
1
10 k
Ω
470 Ω
+ v1 -
vx
v2
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Engineering Circuit Analysis, 7
th
Edition Chapter Four Solutions 10 March 2006
14. We need concern ourselves with the bottom part of this circuit only. W riting a single
nodal equation,
-4 + 2 = v/ 50

We find that v = -100 V.

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Engineering Circuit Analysis, 7
th
Edition Chapter Four Solutions 10 March 2006
15. We choose the bottom node as the reference terminal. Then:

Node 1:
11
2
21
vvv−
2
+−= [1]

Node 2:
2321 2 4
4
124
vvvv vv−−−
=++ [2]
Node 3:
32 3 34
2 [3]
2510
vv v vv−−
=++

Node 4:
4344 2
610 4
vvvvv−
0
−
=+ + [4]

Node 5:
557
1
21
vvv−
−= [5] +

Node 6:
6676 8
52 10
vvvvv−−
=+ +1
[6]

Node 7:
75 76 78
2 [7]
124
vv vv vv−−−
=++

Node 8:
8868 7
610 4
vvvvv−−
=+ +0
[8]

Note that Eqs. [1-4] may be solved independently of Eqs. [5-8].

Sim plifying,

to yield
12
1234
234
23 4
32 4 [1]
472 16 [2]

58 20 [3]
15 6 31 0 [4]
vv
vvvv
vvv
vv v
− =−
−+− − =
−+ − =
−−+ =
1
2
3
4
3.370 V
7.055 V
7.518 V
4.869 V
v
v
v
v
=
=
=
=

and

57
678
56 7 8
678
32 2 [5]
85 10 [6]

42 7 8 [7]
61531 0 [8]
vv
vvv
vv v v
vvv
− =−
−− =
−− + − =
−− + =
to yield
5
6
7
8
1.685 V
3.759 V
3.527 V
2.434 V
v
v
v
v
=
=
=
=

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Engineering Circuit Analysis, 7
th
Edition Chapter Four Solutions 10 March 2006
16. We choose the center node for our common terminal, since it connects to the largest
number of branches. W e name the left node “A”, the top node “B”, the right node
“C”, and the bottom
node “D”. We next form a supernode between nodes A and B.

At the supernode: 5 = (V
A – VD)/ 10 + VA/ 20 + (VB – VC)/ 12.5 [1]

At node C: V
C = 150 [2]

At node D: -10 = V
D/ 25 + (VD – VA)/ 10 [3]

Our supernode-related equation is V
B – VA = 100 [4]

Simplifiying and grouping terms,

0.15 V
A + 0.08 VB - 0.08 VC – 0.1 VD = 5 [1]
V
C = 150 [2]
-25 V
A + 35 VD = -2500 [3]
- V
A + VB = 100 [4]

Solving, we find that V
D = -63.06 V. Since v 4 = - VD,

v
4 = 63.06 V.
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Engineering Circuit Analysis, 7
th
Edition Chapter Four Solutions 10 March 2006
17. Choosing the bottom node as the reference terminal and naming the left node “1”, the
center node “2” and the right node “3”, we next form
a supernode about nodes 1 and
2, encompassing the dependent voltage source.

At the supernode, 5 – 8 = (v 1 – v2)/ 2 + v 3/ 2.5 [1]
At node 2, 8 = v
2 / 5 + (v 2 – v1)/ 2 [2]

Our supernode equation is v 1 - v3 = 0.8 v A [3]
Since v
A = v2, we can rewrite [3] as v 1 – v3 = 0.8v 2

Simplifying and collecting terms,

0.5 v 1 - 0.5 v 2 + 0.4 v 3 = -3 [1]
-0.5 v
1 + 0.7 v 2 = 8 [2]
v
1 - 0.8 v 2 - v 3 = 0 [3]

(a) Solving for v 2 = vA, we find that v A = 25.91 V

(b) The power absorbed by the 2.5-Ω resistor is
(v
3)
2
/ 2.5 = (-0.4546)
2
/ 2.5 = 82.66 mW.
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Engineering Circuit Analysis, 7
th
Edition Chapter Four Solutions 10 March 2006
18. Selecting the bottom node as the reference terminal, we name the left node “1”, the
middle node “2” and the right node “3.”

NODE 1: 5 = ( v 1 – v2)/ 20 + (v 1 – v3)/ 50 [1]

NODE 2: v 2 = 0.4 v 1 [2]

NODE 3: 0.01 v 1 = (v 3 – v2)/ 30 + (v 3 – v1)/ 50 [3]

Simplifying and collecting terms, we obtain

0.07 v 1 – 0.05 v 2 – 0.02 v 3 = 5 [1]
0.4 v
1 – v 2 = 0 [2]
-0.03 v
1 – 0.03333 v 2 + 0.05333 v 3 = 0 [3]

Since our choice of r eference ter minal m akes the contr olling var iable of both
dependent sources a nodal voltage, we have no need for an additional equation as we
mi
ght have expected.

Solving, we find that v 1 = 148.2 V, v 2 = 59.26 V, and v 3 = 120.4 V.

The power supplied by the dependent
current source is therefore

(0.01 v
1) • v3 = 177.4 W.
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Engineering Circuit Analysis, 7
th
Edition Chapter Four Solutions 10 March 2006
19. At node x: v x/ 4 + (v x – vy)/ 2 + (v x – 6)/ 1 = 0 [1]
At node y: (v
y – kv x)/ 3 + (v y – vx)/ 2 = 2 [2]

Our additional constraint is that v
y = 0, so we may simplify Eqs. [1] and [2]:

14 v
x = 48 [1]
-2k v
x - 3 v x = 12 [2]

Since Eq. [1] yields v
x = 48/14 = 3.429 V, we find that

k = (12 + 3 v
x)/ (-2 v x) = -3.250
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Engineering Circuit Analysis, 7
th
Edition Chapter Four Solutions 10 March 2006
20. Choosing the bottom node joining the 4-Ω resistor, the 2-A current sourcee and the
4-V voltage source as our reference node, we next name the other node of the 4-Ω
resistor node “1”, and the node joining the 2-Ω r
esistor and t he 2-A current source
node “2.” Finally, we create a supernode with nodes “1” and “2.”

At the supernode: –2 = v
1/ 4 + (v 2 – 4)/ 2 [1]
Our remaining equations: v
1 – v2 = –3 – 0.5i 1 [2]
and i
1 = (v 2 – 4)/ 2 [3]

Equation [1] simplifies to v
1 + 2 v 2 = 0 [1]
Combining Eqs. [2] and [3, 4 v
1 – 3 v 2 = –8 [4]

Solving these last two equations, we find that v
2 = 727.3 mV. Making use of Eq. [3],
we therefore find that
i
1 = – 1.636 A.
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Engineering Circuit Analysis, 7
th
Edition Chapter Four Solutions 10 March 2006
21. We first number the nodes as 1, 2, 3, 4, and 5 moving left to right. We next select
node 5 as the reference terminal. To simplify the analysis, we form a supernode from
nodes 1, 2, and 3.

At the supernode,

-4 – 8 + 6 = v 1/ 40 + (v 1 – v3)/ 10 + (v 3 – v1)/ 10 + v 2/ 50 + (v 3 – v4)/ 20 [1]

Note that since both ends of the 10-Ω resistor are connected to the supernode, the
related terms cancel each other out, and so could have been ignored.

At node 4: v
4 = 200 [2]

Supernode KVL equation: v
1 – v3 = 400 + 4v 20 [3]

Where the controlling voltage v
20 = v3 – v4 = v3 – 200 [4]

Thus, Eq. [1] becomes -6 = v
1/ 40 + v 2/ 50 + (v 3 – 200)/ 20 or, more simply,

4 = v
1/ 40 + v 2/ 50 + v 3/ 20 [1’]

and Eq. [3] becomes v
1 – 5 v3 = -400 [3’]

Eqs. [1’], [3’], and [5] are not sufficient, however, as we have four unknowns. At this
point we need to seek an additional equation, possibly in terms of v
2. Referring to the
circuit,
v
1 - v2 = 400 [5]
Rewriting as a matrix equation,

⎥
⎥
⎥
⎦
⎤
⎢
⎢
⎢
⎣
⎡
=
⎥
⎥
⎥
⎦
⎤
⎢
⎢
⎢
⎣
⎡
⎥
⎥
⎥
⎥
⎦
⎤
⎢
⎢
⎢
⎢
⎣
⎡
400
400-
4

0 1- 1
5- 0 1
20
1
50
1
40
1
3
2
1v
v
v

Solving, we find that
v
1 = 145.5 V, v 2 = -254.5 V, and v 3 = 109.1 V. Since v 20 = v3 – 200, we find that

v20 = -90.9 V.
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Engineering Circuit Analysis, 7
th
Edition Chapter Four Solutions 10 March 2006
22. We begin by naming the top left node “1”, the top right node “2”, the bottom node of
the 6-V source “3” and the top node of the 2-Ω resistor “4.” The reference node has
already been selected, and designated using a ground sym
bol.

By inspection, v 2 = 5 V.

Forming a supernode with nodes 1 & 3, we find

At the supernode: -2 = v 3/ 1 + (v 1 – 5)/ 10 [1]

At node 4: 2 = v 4/ 2 + (v 4 – 5)/ 4 [2]

Our supernode KVL equation: v 1 – v3 = 6 [3]

Rearranging, simplifying and collecting terms,

v1 + 10 v 3 = -20 + 5 = -15 [1]
and
v
1 - v3 = 6 [2]

Eq. [3] may be directly solved to obtain v 4 = 4.333 V.

Solving Eqs. [1] and [2], we find that

v1 = 4.091 V and v 3 = -1.909 V.

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Engineering Circuit Analysis, 7
th
Edition Chapter Four Solutions 10 March 2006
23. We begin by selecting the bottom node as the reference, naming the nodes as shown
below, and forming a supernode with nodes 5 & 6.

By inspection, v
4 = 4 V.

By KVL, v 3 – v4 = 1 so v 3 = -1 + v 4 = -1 + 4 or v 3 = 3 V.

At the supernode, 2 = v 6/ 1 + (v 5 – 4)/ 2 [1]

At node 1, 4 = v 1/ 3 therefore, v 1 = 12 V.

At node 2, -4 – 2 = (v 2 – 3)/ 4

Solving, we find that v 2 = -21 V

Our supernode KVL equation is v 5 - v6 = 3 [2]

Solving Eqs. [1] and [2], we find that

v5 = 4.667 V and v 6 = 1.667 V.

The power supplied by the 2-A source therefore is (v 6 – v2)(2) = 45.33 W.

4 A
2 A
1 V
4 V
3 V 4 Ω
3 Ω
2 Ω
1 Ω
v2
v1
v3 v4 v5
v6
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Engineering Circuit Analysis, 7
th
Edition Chapter Four Solutions 10 March 2006
24. We begin by selecting the bottom node as the reference, naming each node as shown
below, and forming two different supernodes as indicated.

By inspection, v
7 = 4 V and v = (3)(4) = 12 V. 1

At node 2: -4 – 2 = (v 2 – v3)/ 4 or v 2 -v3 = -24 [1]

At the 3-4 supernode:
0 = (v – v
3 2)/ 4 + (v – v)/ 6 or -6v + 6v 4 5 2 3 + 4v 4 – 4v = 0 [2] 5

At node 5:
0 = (v – v
5 4)/ 6 + (v – 4)/ 7 + (v – v 5 5 6 )/ 2 or -14v + 68v 4 5 – 42v = 48 [3] 6

At the 6-8 supernode: 2 = (v – v 6 5)/ 2 + v 8/ 1 or -v + v 5 6 + 2v 8 = 4 [4]

3-4 supernode KVL equation: v 3 - v4 = -1 [5]
6-8 supernode KVL equation: v
6 – v = 3 [6] 8

Rewriting Eqs. [1] to [6] in matrix form,

⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎦
⎤
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎣
⎡
=
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎦
⎤
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎣
⎡
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎦
⎤
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎣
⎡
3
1-
4
48
0
24-

1- 1 0 0 0 0
0 0 0 1- 1 0
2 1 1- 0 0 0
0 42- 68 14- 0 0
0 0 4- 4 6 6-
0 0 0 0 1- 1
8
6
5
4
3
2v
v
v
v
v
v

Solving, we find that
v
2 = -68.9 V, v 3 = -44.9 V, v = -43.9 V, v = -7.9 V, v = 700 mV, v = -2.3 V. 4 5 6 8

The power generated by the 2-A source is therefore (v 8 – v)(2) = 133.2 W. 6

v1
v2
v3
v4
v5
v6
v7
v8
Voltages in
volts.
Currents in
amperes.
Resistances
in ohms.

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Engineering Circuit Analysis, 7
th
Edition Chapter Four Solutions 10 March 2006
25. With the reference terminal already specified, we name the bottom terminal of the
3-mA source node “1,” the left term inal of the bottom 2.2-kΩ resistor node “2,” the

top term inal of the 3-mA source node “3,” the “+” reference term inal of the 9-V
source node “4,” and the “-” terminal of the 9-V source node “5.”

Since we know that 1 mA flows through the top 2.2-kΩ resistor, v = -2.2 V. 5
Also, we see that v 4 – v = 9, so that v5 4 = 9 – 2.2 = 6.8 V.
Proceeding with nodal analysis,

At node 1: -3×10
-3
= v 1/ 10×10
3
+ (v 1 – v2)/ 2.2×10
3
[1]

At node 2: 0 = (v 2 – v1)/ 2.2×10
3
+ (v 2 – v3)/ 4.7×10
3
[2]

At node 3: 1×10
3
+ 3×10
3
= (v 3 – v2)/ 4.7×10
3
+ v 3/3.3×10
3
[3]

Solving, v 1 = -8.614 V, v = -3.909 V and v 2 3 = 6.143 V.

Note that we could also have made use of the supernode approach here.

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Engineering Circuit Analysis, 7
th
Edition Chapter Four Solutions 10 March 2006
26. Mesh 1: –4 + 400i + 300i 1 1 – 300i 2 – 1 = 0 or 700i 1 – 300i = 5 2
Mesh 2: 1 + 500i 2 – 300i +2 – 2 = 0 or –300i + 500i 1 1 2 = –3.2

Solving, i = 5.923 mA and i = -2.846 mA.
1 2

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Engineering Circuit Analysis, 7
th
Edition Chapter Four Solutions 10 March 2006
27. (a) Define a clockwise m esh current i 1 in th e lef t-most m esh; a clockw ise m esh
curren t i in the central m esh, and note that i
2 y can be used as a m esh current for the
rem aining mesh.

Mesh 1: -10 + 7i – 2i
1 2 = 0
Mesh 2: -2i + 5i = 0
1 2
Mesh y: -2i + 9i 2 y = 0

Solve the resulting matrix equation:

to fi
nd that i
1
2
720 10
250 0
029 0
y
i
i
i
⎡⎤−⎡⎤
⎢⎥⎢⎥
−
⎢⎥⎢⎥
⎢⎥⎢⎥−
⎣⎦ ⎣⎦
⎡⎤
⎢⎥
=
⎢
⎥
⎢⎥
⎣⎦
1 = 1.613 A, and i y = 143.4 mA.

(b) The power supplied by the 10 V source is (10)(i
1) = 10(1.613) = 16.13 W.

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Engineering Circuit Analysis, 7
th
Edition Chapter Four Solutions 10 March 2006
28. Define three mesh currents as shown:

(a) The current through the 2 Ω res
istor is i 1.

Mesh 1: 5i
1 – 3i2 = 0 or 5i – 3i = 0 1 2
Mesh 2: –212 +8i –3i 2 1 = 0 or -3i 1 +8i2 = 212
Mesh 3: 8i
3 – 5i2 + 122 = 0 or –5i + 8i = –122 2 3

Solving, i = 20.52 A, i = 34.19 A and i
1 2 3 = 6.121 A.

(b) The current through the 5 Ω res
istor is i 3, or 6.121 A.

*** Note: since the problem statement did not specify a direction, only the current
m
agnitude is relevant, and its sign is arbitrary.

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Engineering Circuit Analysis, 7
th
Edition Chapter Four Solutions 10 March 2006
29. We begin by defining three clockwise m esh currents i, i 12 and i 3 in the lef t-most,
central, and right-most meshes, respectively. Then,

(a) Note that i
x = i2 – i3.

Mesh 1: i
1 = 5 A (by inspection)
Mesh 3: i
3 = –2 A (by inspection)

Mesh 2: –25i
1 + 75i 2 – 20i 3 = 0, or, making use of the above,

–125 + 75i + 40= 0 so that i = 1.133 A.
2 2

Thus, i
x = i2 – i3 = 1.133 – (–2) = 3.133 A.

(b) The power absorbed by the 25 Ω resis
tor is

P
25Ω = 25 (i 1 – i2)
2 2
= 25 (5 – 1.133) = 373.8 W.

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Engineering Circuit Analysis, 7
th
Edition Chapter Four Solutions 10 March 2006
30. Define three mesh currents as shown. Then,

– 40i Mesh 1: –2 + 80i
1 2 – 30i 3 = 0
Mesh 2: –40i
1 + 70i 2 = 0
Mesh 3: –30i
1 +70i = 0 3

Solving,
⎢⎥
1
2
3
80 40 30 2
40 70 0 0
30 0 70 0
i
i
i
−−⎡⎤ ⎡⎤⎡⎤
⎢⎥⎢⎥
−=
⎢⎥ ⎢⎥⎢⎥
⎢⎥ ⎢⎥⎢⎥−
⎣⎦ ⎣⎦⎣⎦

we find that i
2 = 25.81 mA and i 3 = 19.35 mA. Thus, i = i 3 – i2 = –6.46 mA.

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Engineering Circuit Analysis, 7
th
Edition Chapter Four Solutions 10 March 2006
31. Moving from left to right, we name the bottom three meshes, mesh “1”, mesh “2,”
and m esh “3.” In each of these th ree me
shes we define a clockwise curren t. The
remaining mesh current is clearly 8 A. We may then write:

MESH 1: 12 i 1 - 4 i = 100 2

MESH 2: -4 i + 9 i - 3 i = 0 1 2 3

MESH 3: -3 i + 18 i = -80 2 3

Solving this system of three (independent) equations in three unknowns, we find that

i2 = i x = 2.791 A.
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Engineering Circuit Analysis, 7
th
Edition Chapter Four Solutions 10 March 2006
32. We define four clockwise mesh currents. The top mesh current is labeled i 4. The
bottom left mesh current is labeled i, the bottom right mesh current is labeled i
1 3 , and
the remaining mesh current is labeled i. Define a voltage “v
2 4A ” across the 4-A current
source with the “+” reference terminal on the left.

By inspection, i
3 = 5 A and i a = i. 4

MESH 1: -60 + 2i – 2i + 6i = 0 or 2i 1 4 4 1 + 4i 4 = 60 [1]

MESH 2: -6 i 4 + v4A + 4i 2 – 4(5) = 0 or 4i 2 - 6i + v4 4A = 20 [2]

MESH 4: 2 i – 2i + 5i + 3i – 3(5) – v 4 1 4 4 4A = 0 or -2i + 10i - v 1 44A = 15 [3]

At this point, we are short an equation. Returning to the circuit diagram, we note that

i– i = 4 [4] 2 4

Collecting these equations and writing in matrix form, we have

⎥
⎥
⎥
⎥
⎦
⎤
⎢
⎢
⎢
⎢
⎣
⎡
=
⎥
⎥
⎥
⎥
⎦
⎤
⎢
⎢
⎢
⎢
⎣
⎡
⎥
⎥
⎥
⎥
⎦
⎤
⎢
⎢
⎢
⎢
⎣
⎡
4
15
20
60

0 1- 1 0
1- 10 0 2-
1 6- 4 0
0 4 0 2
A4
4
2
1v
i
i
i

Solving, i = 16.83 A, i = 10.58 A, i = 6.583 A and v 1 2 4 4A = 17.17 V.
Thus, the power dissipated by the 2-Ω res
istor is

2
(i1 – i4) • (2) = 210.0 W

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Engineering Circuit Analysis, 7
th
Edition Chapter Four Solutions 10 March 2006
33. We begin our analysis by defining three clockwise mesh currents. We will call the top
, the bottom left mesh current i, and the bottom right mesh current imesh current i
3 1 2 .

By inspection, i
1 = 5 A [1] and i 2 = -0.01 v 1 [2]

MESH 3: 50 i
3 + 30 i 3 – 30 i 2 + 20 i 3 – 20 i 1 = 0
– 30 i + 100 i or -20 i = 0 [3]
1 2 3

These three equations are insufficient, however, to solve for the unknowns. It would
be nice to be able to express the dependent source controlling variable v
1 in terms of
the mesh currents. Returning to the diagram, it can be seen that KVL around m esh 1
will yield
+ 20 i – 20 i- v
1 1 3 + 0.4 v 1 = 0
or v
1 = 20 i 1/ 0.6 – 20 i 3/ 0.6 or v 1 = (20(5)/ 0.6 - 20 i 3/ 0.6 [4]

Substituting Eq. [4] into Eq. [2] and then the modified Eq. [2] into Eq. [3], we find

/ 0.6 + 100 i-20(5) – 30(-0.01)(20)(5)/0.6 + 30(-0.01)(20) i = 0
3 3

Solving, we find that i
3 = (100 – 50)/ 90 = 555.6 mA

= 148.1 V, iThus, v
1 2 = -1.481 A, and the power generated by the dependent voltage
source is
0.4 v
1 (i2 – i1) = -383.9 W.

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Engineering Circuit Analysis, 7
th
Edition Chapter Four Solutions 10 March 2006
34. We begin by defining four clockwise mesh currents i, i, i and i 123 4 , in the meshes of
our circuit, starting at the left-m ost mesh. W
e also define a voltage v dep across the
dependent current source, with the “+” on the top.

By inspection, i 1 = 2A and i 4 = -5 A.

At Mesh 2: 10 i 2 - 10(2) + 20 i + v 2 dep = 0 [1]

At Mesh 3: - v dep + 25 i 3 + 5 i –5(-5) = 0 [2] 3

Collecting terms, we rewrite Eqs. [1] and [2] as

30 i2 + v dep = 20 [1]

30 i 3 – v dep = -25 [2]

This is only two equations but three unknowns, however, so we require an additional
equation. Returning to the circuit diagram , we note that it is possible to express th e
current of the dependent source in terms of mesh currents. (We might also choose to
obtain an expression for v
dep in terms of mesh currents using KVL around m esh 2 or
3.)

Thus, 1.5i x = i - i where i3 2 x = i 1 – i2 so -0.5 i 2 - i = -3 [3] 3

In matrix form,
⎥
⎥
⎥
⎦
⎤
⎢
⎢
⎢
⎣
⎡
=
⎥
⎥
⎥
⎦
⎤
⎢
⎢
⎢
⎣
⎡
⎥
⎥
⎥
⎦
⎤
⎢
⎢
⎢
⎣
⎡
3-
25-
20

0 1- 0.5-
1- 30 0
1 0 30
3
2
dep
v
i
i

Solving, we find that i 2 = -6.333 A so that i = i – i x 12 = 8.333 A.

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Engineering Circuit Analysis, 7
th
Edition Chapter Four Solutions 10 March 2006
35. We define a clockwise mesh current i in the bottom left mesh, a clockwise mesh 1
in the top left mesh, a clockwise mesh current icurrent i2 3 in the top right mesh, and a
clockwise mesh current i in the bottom right mesh.
4

MESH 1: -0.1 v a + 4700 i 1 – 4700 i 2 + 4700 i 1 – 4700 i 4 = 0 [1]

MESH 2: 9400 i 2 – 4700 i 1 – 9 = 0 [2]

MESH 3: 9 + 9400 i – 4700 i 3 4 = 0 [3]

MESH 4: 9400 i 4 – 4700 i 1 – 4700 i 3 + 0.1 i x = 0 [4]

The presen ce of the two dependent sources has led to the in troduction of tw o
additional unknowns (i and v
x a ) besides our four m esh currents. In a perfect world, it
would simplify the solution if we could express these tw o quantities in term s of the
mesh currents.

Referring to the circuit diagram, we see that i = i x 2 (easy enough) and that
v
a = 4700 i 3 (also straightforward). Thus, substituting these express ions into our
four mesh equations and creating a matrix equation, we arrive at:

⎥
⎥
⎥
⎥
⎦
⎤
⎢
⎢
⎢
⎢
⎣
⎡
=
⎥
⎥
⎥
⎥
⎦
⎤
⎢
⎢
⎢
⎢
⎣
⎡
⎥
⎥
⎥
⎥
⎦
⎤
⎢
⎢
⎢
⎢
⎣
⎡
0
9-
9
0

9400 4700- 0.1 4700-
4700- 9400 0 0
0 0 9400 4700-
4700- 470- 4700- 9400
4
3
2
1i
i
i
i

Solving,

i1 = 239.3 μ A, i = 1.077 mA, i = -1.197 mA and i = -478.8 μA. 2 3 4

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Engineering Circuit Analysis, 7
th
Edition Chapter Four Solutions 10 March 2006
36. We define a clockwise mesh current i in the upper right mesh, a clockwise m esh 3
in the lower left m esh, and a clockwise m esh current icurrent i1 2 in th e lower right
mesh.

MESH 1: -6 + 6 i - 2 = 0 [1] 1

MESH 2: 2 + 15 i – 12 i 2 3 – 1.5 = 0 [2]

MESH 3: i 3 = 0.1 v x [3]

Eq. [1] may be solved directly to obtain i 1 = 1.333 A.

It would help in the solution of Eqs. [2] and [3] if we could express the dependent
source controlling variable v
x in term s of m esh curren ts. Referring to the circuit
diagram, we see that v = (1)( i) = i, so Eq. [3] reduces to
x 1 1

i3 = 0.1 v x = 0.1 i = 133.3 mA. 1

As a result, Eq. [1] reduces to i 2 = [-0.5 + 12(0.1333)]/ 15 = 73.31 mA.
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Engineering Circuit Analysis, 7
th
Edition Chapter Four Solutions 10 March 2006
37. (a) Define a mesh current i in the second mesh. Then KVL allows us to write: 2

MESH 1: -9 + R i
1 + 47000 i 1 – 47000 i 2 = 0 [1]

MESH 2: 67000 i
2 – 47000 i 1 – 5 = 0 [2]

Given that i = 1.5 mA, we may solve Eq. [2] to find that
1

mA 1.127 mA
67
47(1.5) 5

2 =
+
=i
and so
3-
101.5
47(1.127) 47(1.5) - 9

×
+
=R = -5687 Ω.
(b) This value of R is unique; no ot
her value will satisfy both Eqs. [1] and [2].

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Engineering Circuit Analysis, 7
th
Edition Chapter Four Solutions 10 March 2006
38. Define three clockwise mesh currents i, i and i . The bottom 1-kΩ resistor can be 12 3
ignored, as no current flows through it.

MESH 1: -4 + (2700 + 1000 + 5000) i – 1000 i 1 2 = 0 [1]

MESH 2: (1000 + 1000 + 4400 + 3000) i – 1000 i 2 1 – 4400 i 3 + 2.2 – 3 = 0 [2]

MESH 3: (4400 + 4000 + 3000) i - 4400 i – 1.5 = 0 [3] 3 2

Combining terms,

8700 i 1 – 1000 i 2 = 4 [1]
–1000 i
1 + 9400 i 2 – 4400 i 3 = 0.8 [2]
– 4400 i
2 + 11400 i = 1.5 [3] 3
Solving,

i 1 = 487.6 μ A, i = 242.4 μ A and i = 225.1 μ A. 2 3

The power absorbed by each resistor may now be calculated:

P5k = 5000 (i)
2
= 1.189 mW 1
P2.7k = 2700 (i)
2
= 641.9 μ W 1
2
P1ktop = 1000 (i – i) = 60.12 μW 12
P1kmiddle = 1000 (i)
2
= 58.76 μ W 2
P1kbottom = 0 = 0
2
P4.4k = 4400 (i – i) = 1.317 μW 23
P3ktop = 3000 (i)
2
= 152.0 μ W 3
P4k = 4000 (i)
2
= 202.7 μ W 3
P3kbottom = 3000 (i)
2
= 176.3 μ W 2

Check: The sources supply a total of
4(487.6) + (3 – 2.2)(242.4) + 1.5(225.1) = 2482 μW.
The absorbed powers add to 2482
μW.
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th
Edition Chapter Four Solutions 10 March 2006
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39. (a) We begin by naming four mesh currents as depicted below:

Proceeding with mesh analysis, then, keeping in m
ind that I x = -i, 4

MESH 1: (4700 + 300) i - 4700 i = 0 [1] 1 2

MESH 2: (4700 + 1700) i – 4700 i 2 1 – 1700 i 3 = 0 [2]

Since we have a current source on the perimeter of mesh 3, we do not require a KVL
equation for that mesh. Instead, we may simply write

i3 = -0.03 v π [3a] where v π = 4700(i – i 12) [3b]

MESH 4: 3000 i 4 – 3000 i 3 + 1 = 0 [4]

Simplifying and combining Eqs. 3a and 3b,

5000 i 1 – 4700 i 2 = 0
–4700 i
1 + 6400 i 2 – 1700 i 3 = 0
–141 i
1 + 141 i 2 – i3 = 0
– 3000 i
3 + 3000 i = –1 4

Solving, we find that i 4 = -333.3 mA, so I = 333.3 μA. x

(b) At node “π” : 0.03 v π = vπ / 300 + v π / 4700 + v π /1700

Solving, we find that v π = 0, therefore no current flows through the dependent source.

Hence, I = 333.3 μA as found in part (a). x

(c) V/ Isx has units of resistance. It can be thought of as the resistance “seen” by the
voltage source V…. more on this in Chap. 5….
s

Engineering Circuit Analysis, 7
th
Edition Chapter Four Solutions 10 March 2006
40. We begin by naming each mesh and the three undefined voltage sources as shown
below:

MESH 1: –V
z + 9i – 2i 1 2 – 7i 4 = 0

MESH 2: –2i 1 + 7i 2 – 5i3 = 0

– 5iMESH 3: Vx 2 + 8i 3 – 3i 4 = 0

MESH 4: Vy – 7i – 3i1 3 + 10i 4 = 0

Rearranging and setting i 1 – i2 = 0, i 2 – i3 = 0, i 1 – i4 = 0 and i – i 43 = 0,

9i1 - 2i -7i = V2 4 z
-2i1 + 7i - 5i 2 3 = 0
+ 8i -5i
2 3 – 3i 4 = - Vx
-7i1 -3i + 10i 3 4 = - Vy

Since i 1 = i2 = i = i, these equations produce: 3 4

V z = 0
0 = 0
-V = 0
x
-V y = 0
This is a unique solution. Therefore, the request that nonzero values be
found cannot be satisfied.
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Engineering Circuit Analysis, 7
th
Edition Chapter Four Solutions 10 March 2006
41. The “supermesh” concept is not required (or helpful) in solving this problem, as there
are no current sources s hared between m eshes. Starting with the left-most m esh and
moving right, we define four clockwise me
sh currents i
1, i, i and i23 4 . By inspection,
we see that i = 2 mA.
1

MESH 2: -10 + 5000i + 4 + 1000i = 0 [1] 2 3

MESH 3: -1000i + 6 + 10,000 – 10,000i 3 4 = 0 [2]

MESH 4: i 4 = -0.5i [3] 2

Reorganising, we find

5000 i 2 + 1000 i 3 = 6 [1]
9000 i
3 – 10,000 i 4 = -6 [2]
0.5 i
2 + i 4 = 0 [3]

We could either subtitute Eq. [3] into Eq. [2] to reduce the number of equations, or
simply go ahead and solve the system of Eqs. [1-3]. Ei
ther way, we find that

i1 = 2 mA, i = 1.5 mA, i = -1.5 mA and i = -0.75 mA. 2 3 4

The power generated by each source is:

P2mA = 5000(i – i 12)(i) = 5 mW 1
P4V = 4 (-i 2) = -6 mW
P
6V = 6 (-i 3) = 9 mW
P
depV = 1000 i (i – i 332 ) = 4.5 mW
P
depI = 10,000(i – i 34)(0.5 i ) = -5.625 mW 2

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Engineering Circuit Analysis, 7
th
Edition Chapter Four Solutions 10 March 2006
42. This circuit does not require the supermesh technique, as it does not contain any
current sources. Redrawing the circuit so its planar nature and me
sh structure are
clear,

MESH 1: -20 + i
1 – i2 + 2.5 i A = 0 [1]

MESH 2: 2 i
2 + 3 i 2 + i2 – 3 i 3 – i1 = 0 [2]

MESH 3: -2.5 i
A + 7 i 3 – 3 i 2 = 0 [3]

Combining terms and making use of the fact that i
A = - i3,

i
1 – i2 – 2.5 i 3 = 20 [1]
- i
1 + 6i 2 – 3 i 3 = 0 [2]
–3 i
2 + 9.5 i 3 = 0 [3]

Solving, i
1 = 30.97 A, i 2 = 6.129 A, and i 3 = 1.936 A. Since i A = - i 3,

i
A = –1.936 A.
20 V
2.5 i
A
1
Ω
4 Ω
2 Ω
iA →
i2
i3
i1
•
•
3 Ω

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Engineering Circuit Analysis, 7
th
Edition Chapter Four Solutions 10 March 2006
43. Define four mesh currents

By inspection, i
1 = -4.5 A.

We form a supermesh with meshes 3 and 4 as defined above.

MESH 2: 2.2 + 3 i + 4 i 2 2 + 5 – 4 i = 0 [1] 3

SUPERME SH: 3 i 4 + 9 i 4 – 9 i 1 + 4 i 3 – 4 i 2 + 6 i 3 + i3 – 3 = 0 [2]

Superm esh KCL equation: i 4 - i = 2 [3] 3

Simplifying and combining terms, we may rewrite these three equations as:

7 i 2 – 4 i 3 = -7.2 [1]
-4 i
2 + 11 i + 12 i = -37.5 [2] 3 4
- i + i = 2 [3] 3 4

Solving, we find that i 2 = -2.839 A, i 3 = -3.168 A, and i = -1.168 A. 4

The power supplied by the 2.2-V source is then 2.2 (i – i) = -3.654 W. 12
i1
i2
i4i3

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Engineering Circuit Analysis, 7
th
Edition Chapter Four Solutions 10 March 2006
44. We begin by defining six mesh currents as depicted below:

i5
i6
i3i2i1 i4

We form a supermesh with meshes 1 and 2 since they share a current source. •
• We form a second supermesh with meshes 3 and 4 since th ey also share a curren t
source.

1, 2 Supermesh:
(4700 + 1000 + 10,000) i
1 – 2200 i 5 + (2200 + 1000 + 4700) i 2 – 1000 i 3 = 0 [1]

3, 4 Supermesh:
– 1000 i (4700 + 1000 + 2200) i
3 2 – 2200 i 6 + (4700 + 10,000 + 1000) i = 0 [2] 4

MESH 5: (2200 + 4700) i – 2200 i 5 2 + 3.2 – 1.5 = 0 [3]

MESH 6: 1.5 + (4700 + 4700 + 2200) c – 2200 i 3 = 0 [4]

1, 2 Supermesh KCL equation: i 1 – i2 = 3×10
-3
[5]

3, 4 Supermesh KCL equation: i 4 – i3 = 2×10
-3
[6]

We can simplify these equations prior to solution in several ways. Choosing to retain
six equations,
15,700 i
1 + 7900 i 2 - 1000 i 3 -2200 i5 = 0 [1]
- 1000 i
2 + 7900 i 3 + 15,700 i 4 -2200 i = 0 [2] 6
- 2200 i 2 + 6900 i 5 = -1.7 [3]
+ 11,600 i - 2200 i
3 6 = -1.5 [4]
i
1 – i 2 = 3×10
-3
[5]
- i
3 + i = 2×10
-3
[6] 4

Solving, we find that i
4 = 540.8 mA. Thus, the voltage across the 2-mA source is

(4700 + 10,000 + 1000) (540.8×10
-6
) = 8.491 V

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Engineering Circuit Analysis, 7
th
Edition Chapter Four Solutions 10 March 2006
45. We define a mesh current i a in the left-hand mesh, a mesh current i in the top right 1
mesh, and a mesh current i in the bottom right mesh (all flowing clockwise). 2

The left-most mesh can be analysed separately to determine the controlling voltage v a,
as KCL assures us that no current flows through either the 1-Ω or 6-Ω resistor.

Thus, -1.8 + 3 i a – 1.5 + 2 i a = 0, which may be solved to find i a = 0.66 A. Hence,
v
a = 3 i a = 1.98 V.

Forming one supermesh from the remaining two meshes, we may write:

-3 + 2.5 i + 3 i 1 2 + 4 i 2 = 0

and the supermesh KCL equation: i 2 – i1 = 0.5 v a = 0.5(1.98) = 0.99

Thus, we have two equations to solve:

2.5 i1 + 7 i 2 = 3
+ i-i = 0.99
1 2

Solving, we find that i 1 = -413.7 mA and the voltage across the 2.5-Ω res istor
(arbitrarily assuming the left terminal is the “+” reference) is 2.5 i
1 = -1.034 V.

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Engineering Circuit Analysis, 7
th
Edition Chapter Four Solutions 10 March 2006
46. There are only three meshes in this circuit, as the botton 22-mΩ resistor is not
connected connected at its left term inal. Thus, we define three m esh currents, i, i 12,
and i, beginning with the left-most mesh.
3

We next create a superm esh from meshes 1 and 2 (note that m esh 3 is independent,
and can be analysed separately).

Thus, -11.8 + 10×10
-3
i1 + 22×10
-3
i2 + 10×10
-3
i2 + 17×10
-3
i1 = 0

and applying KCL to obtain an equation containing the current source,

i1 – i2 = 100

Combining terms and simplifying, we obtain

27×10-3 i 1 + 32×10
-3
i = 11.8 2
i 1 – i 2 = 100

Solving, we find that i 1 = 254.2 A and i = 154.2 A. 2

The final mesh current is easily found: i = 13×10
3
/ (14 + 11.6 + 15) = 320.2 A. 3

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Engineering Circuit Analysis, 7
th
Edition Chapter Four Solutions 10 March 2006
47. MESH 1: -7 + i – i = 0 [1] 12
MESH 2: i 2 – i1 + 2i 2 + 3i – 3i 2 3 = 0 [2]
MESH 3: 3i
3 – 3i2 + xi +2i – 7 = 0 [3] 3 3

Grouping terms, we find that

i
1 – i2 = 7 [1]
+ 6i – 3i-i = 0 [2]
1 2 3
-3i + (5 + x)i 2 3 = 7 [3]

This, unfortunately, is four unknowns but only three equations. However, we have not
yet made use of the fact that we are trying to obtain i
2 = 2.273 A. Solving these “four”
equations, we find that

– 5 i)/ ix = (7 + 3 i
2 33 = 4.498 Ω.
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Engineering Circuit Analysis, 7
th
Edition Chapter Four Solutions 10 March 2006
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7 V
7 A
1 Ω
2 Ω
3 Ω
300 mΩ
2 Ω
i2 i3
i1
48. We begin by redrawing the circuit as instructed, and define three mesh currents:

B y inspection, i
3 = 7 A.

MESH 1: -7 + i – i = 0 or i 12 1 – i2 = 7 [1]

MESH 2: (1 + 2 + 3) i – i –3(7) = 0 or -i + 6i = 21 [2] 21 1 2

There is no need for supermesh techniques for this situation, as the only current
source lies on the outside perimeter of a mesh- it is not shared between m
eshes.

Solving, we find that i = 12.6 A, i = 5.6 A and i 1 2 3 = 7 A.

Engineering Circuit Analysis, 7
th
Edition Chapter Four Solutions 10 March 2006
49. (a) We are asked for a voltage, and have one current source and one voltage source.
Nodal analysis is probably best then- the nodes can be name
d so that the desired
voltage is a nodal voltage, or, at worst, we have one supernode equation to solve.

Name the top left node “1” and the top right node “x”; designate the bottom node as
the reference terminal. Next, form a supernode with nodes “1” and “x.”

At the supernode: 11 = v 1/ 2 + v x/ 9 [1]

and the KVL Eqn: v 1 – v = 22 [2] x

Rearranging, 11(18) = 9 v 1 + 2 v [1] x
22 = v 1 – v [2] x

Solving, v = 0 x

( b) We are asked for a voltage, and so may suspect that nodal analysis is preferrable;
with two current sources and only one voltage source (easily dealt w
ith using the
supernode technique), nodal analysis does seem to have an edge over m esh analysis
here.

Name the top left node “x,” the top right node “y” and designate the bottom node as
the reference node. Forming a supernode from nodes “x” and “y,”

At the supernode: 6 + 9 = v / 10 + v x y / 20 [1]
and the KVL Eqn: v
y – v = 12 [2] x

Rearranging, 15(20) = 2 v x + v [1] y
and 12 = - v x + v [2] y

Solving, we find that v x = 96 V.

(c) We are asked for a voltage, but would have to subtract two nodal voltages (not
much harder than invoking Ohm ’s law). On the other hand, the dependent current
source depends on the desired unknown, which would lead to the need for another
equation if invoking mesh analysis. Trying nodal analysis,

0.1 v x = (v 1 – 50) / 2 + v x / 4 [1]

referring to the circuit w e see tha t v = v x 1 – 100. Rearranging so that we m ay
eliminate v in Eq. [1], we obtain v = v
1 1 x + 100. Thus, Eq. [1] becomes

0.1 v x = (v x + 100 – 50)/ 2 + v x / 4

and a little algebra yields v x = -38.46 V.
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Engineering Circuit Analysis, 7
th
Edition Chapter Four Solutions 10 March 2006
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Both ends of the
resistor are
connected to the
supernode, so we
could actually just
ignore it…
(a) (b)
Ref.
50.
v1

(a) We begin by noting that it is a voltage that is required; no curren
t values are
requested. This is a three-mesh circuit, or a four-node circuit, depending on your
perspective.
Either app roach requires three equations…. Except that applying the
supernode technique reduces the number of needed equations by one.
At the 1, 3 supernode:
– 80)/ 10 + (v – v0 = (v
1 1 3 )/ 20 + (v 3 – v)/ 20 + v1 3 / 40 + v/ 30 3
and v 3 - v1 = 30

We simplify these two equations and collect terms, yielding
0.1 v
1 + 0.05833 v 3 = 8
+ v - v
1 3 = 30

Solving, we find that v 3 = 69.48 V

( b) Mesh analysis would be straightfo
rward, requiring 3 equations and a (trivial)
application of O hm’s law to obta in the f inal answer. Nodal analysis, on the other
hand, would require on
ly two equations, and the desired voltage will be a nodal
voltage.

At the b, c, d supernode: 0 = (v – 80)/ 10 + v b d / 40 + v c/ 30

and: v
d – v = 30 vb c – vd = 9

Simplify and collect terms: 0.1 v
b + 0.03333 v c + 0.025 v d = 80
- v + v
b d = 30
v
c - v = 9 d
Solving, v (= v) = 67.58 V d 3

(c) We are now faced with a dependent current source whose value depends on a
mesh current. Mesh analysis in th is situation requires 1 supermesh, 1 KCL equation
and Ohm
’s law. Nodal analysis requires 1 supernode, 1 KVL equation, 1 other nodal
equation, and one equation to express i
1 in term s of nodal voltages. Thus, m esh
analysis has an edg e here. Define th e left m esh as “1,” the top m esh as “2”, and the
bottom mesh as “3.”

Mesh 1: -80 + 10 i + 20 i 1 1 – 20 i 2 + 30 i – 30 i 1 3 = 0
2, 3 supermesh: 20 i
2 – 20 i 1 – 30 + 40 i + 30 i – 30 i 3 3 1 = 0
and: i
2 - i = 5 i3 1

Rewriting, 60 i 1 – 20 i 2 – 30 i
3 = 80
+ 20 i + 70 i -50 i
1 2 3 = 30
5 i
1 – i 2 + i = 0 3
= 4.727 A so vSolving, i3 3 = 40 i 3 = 189 V.

Engineering Circuit Analysis, 7
th
Edition Chapter Four Solutions 10 March 2006
51. This circuit consists of 3 meshes, and no dependent sources. Therefore 3 simultaneous
equations and 1 subtraction operation would be required to solve for the two desired
currents. On the other hand, if we use nodal analysis, forming a supernode about the

30-V source would lead to 5 – 1 – 1 = 3 sim ulataneous equations as well, plus several
subtraction and division operations to find the currents. Thus, m esh analysis has a
slight edge here.

Define three clockwise mesh currents: i a in the left-most mesh, i b in the top right
mesh, and i in the bottom right mesh. Then our mesh equations will be:
c

Mesh a: -80 + (10 + 20 + 30) i a – 20 i b – 30 i c = 0 [1]
– 12 i – 20 iMesh b: -30 + (12 + 20) i
b c a = 0 [2]
Mesh c:
(12 + 40 + 30) i c – 12 i – 30 i b a = 0 [3]

Simplifying and collecting terms,

60 ia – 20 i – 30 i b c = 80 [1]
-20 i
a + 32 i b – 12 i c = 30 [2]
-30 i
a – 12 i + 82 i b c = 0 [3]

Solving, we find that i a = 3.549 A, i = 3.854 A, and i b c = 1.863 A. Thus,

i1 = ia = 3.549 A and i 2 = i a – ic = 1.686 A.
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Engineering Circuit Analysis, 7
th
Edition Chapter Four Solutions 10 March 2006
52. Approaching this problem using nodal analysis would require 3 separate nodal
equations, plus one equation to deal with the dependent source, plus subtraction and
division steps to actually find the current i
10. Mesh analy sis, on the o ther hand, will
require 2 mesh/supermesh equations, 1 KCL equation, and one subtraction step to find
i
10. Thus, mesh analys is has a clear edge. Define three clockwise m esh currents: i 1 in
the bottom left mesh, i in the top mesh, and i in the bottom right mesh.
2 3

MESH 1: i 1 = 5 mA by inspection [1]

SUPERMESH: i 1 – i2 = 0.4 i 10
i 1 – i2 = 0.4(i 3 – i2)
i
1 – 0.6 i 2 – 0.4 i 3 = 0 [2]

MESH 3: -5000 i – 10000 i 1 2 + 35000 i 3 = 0 [3]

Sim plify: 0.6 i 2 + 0.4 i = 5×10
-3
[2] 3
+ 35000 i-10000 i2 3 = 25 [3]

Solving, we find i 2 = 6.6 mA and i = 2.6 mA. Since i 3 10 = i 3 – i2, we find that

i10 = -4 mA.
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Engineering Circuit Analysis, 7
th
Edition Chapter Four Solutions 10 March 2006
53. For this circuit problem, nodal analysis will require 3 simultaneous nodal equations,
then subtraction/ division steps to obtain the desired currents. Mesh analysis requ
ires
1 mesh equation, 1 supermesh equation, 2 simple KCL equations and one subtraction
step to dete rmine the currents. If either technique has an edge in this situa tion, it’s
probably mesh analysis. Thus, define four clockwise mesh equations: i
a in the bottom
left mesh, i in the top lef t mesh, i
b c in the top right m esh, and i d in the bottom right
mesh.

At the a, b, c supermesh: -100 + 6 i a + 20 i b + 4 i c + 10 i c – 10 i d = 0 [1]

Mesh d: 100 + 10 id – 10 i
c + 24 i d = 0 [2]

KCL: - i a + ib = 2 [3]
+ iand - i
b c = 3 i 3 = 3 i a [4]

Collecting terms & simplifying,

6 i a + 20 i + 14 i b c – 10 i = 100 [1] d
-10 i c + 34 i = -100 [2] d
- i a + i b = 2 [3]
-3 i
a – i + ib c = 0 [4]
Solving,

ia = 0.1206 A, i = 2.121 A, i b c = 2.482 A, and i d = -2.211 A. Thus,

i3 = ia = 120.6 mA and i 10 = ic – i = 4.693 A. d

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Engineering Circuit Analysis, 7
th
Edition Chapter Four Solutions 10 March 2006
54. With 7 nodes in this circuit, nodal analysis will require the solution of three
simultaneous nodal equations (assum ing we make use of the supernode technique)
and one KV
L equation. Mesh analysis will require the solution of three simultaneous
mesh equations (one m esh current can be found by inspection), plus several
subtraction and m ultiplication oper ations to f inally de termine the voltage a t th e
central nod e. Either will probab ly require a com parable am ount of algebraic
manoeuvres, so we go with nodal analysis, as the desired unknown is a direct result of
solving the simultaneous equations. Define the nodes as:

v1
v2
v3
v4
v5
v6

NODE 1: -2 ×10
-3
= (v 1 – 1.3)/ 1.8×10
3
→ v = -2.84 V. 1

2, 4 Supernode:
-3
2.3×10 = (v 2 – v5)/ 1x10
3
+ (v 4 – 1.3)/ 7.3×10
3
+ (v 4 – v5)/ 1.3×10
3 3
+ v/ 1.5×104

KVL equation: -v + v 2 4 = 5.2

Node 5: 0 = (v 5 – v2)/ 1x10
3
+ (v 5 – v4)/ 1.3x10
3
+ (v 5 – 2.6)/ 6.3x10
3

Simplifying and collecting terms,

14.235 v + 22.39 v – 25.185 v = 35.275 [1] 2 4 5
-v2 + v = 5.2 [2] 4
-8.19 v2 – 6.3 v + 15.79 v = 3.38 [3] 4 5

Solving, we find the voltage at the central node is v 4 = 3.460 V.
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Engineering Circuit Analysis, 7
th
Edition Chapter Four Solutions 10 March 2006
55. Mesh analysis yields current values directly, so use that approach. We therefore
in the left-most mesh, then i, idefine four clockwise mesh currents, starting
with i 1 23
and i moving towards the right.
4

Mesh 1: -0.8 i + (2 + 5) i – 5 i = 0 [1] x 1 2

Mesh 2: i 2 = 1 A by inspection [2]

Mesh 3: (3 + 4) i 3 – 3(1) – 4(i ) = 0 [3] 4

Mesh 4: (4 + 3) i 4 – 4 i 3 – 5 = 0 [4]

Simplify and collect terms, noting that i = i – i = i – 1 x 12 1

-0.8(i 1 – 1) + 7 i – 5(1) = 0 yields i = 677.4 mA 1 1

Thus, [3] and [4] become: 7 i 3 – 4 i 4 = 3 [3]
-4 i + 7 i = 5 [4]
3 4

Solving, we find that i = 1.242 A and i 3 4 = 1.424 A. A map of individual branch
currents can now be drawn:

↑
677.4 mA
677.4 mA
→
-322.6 mA
↓
-242.0 mA
↓
↑
182.0 mA
↑
-1.424 A

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Engineering Circuit Analysis, 7
th
Edition Chapter Four Solutions 10 March 2006
56. If we choose to perform mesh analysis, we require 2 simultaneous equations (there are
four m eshes, but one m esh current is known, and we can employ the superm
esh
technique around the left two m eshes). In order to find the voltage across the 2-m A
source we will need to write a KVL equation, however. Using nodal analysis is less
desirable in this case, as there will be a large num ber of nodal equations needed.
Thus, we de fine four clockwise m esh currents i
1, i, i and i23 4 starting with the lef t-
most mesh and moving towards the right of the circuit.

At the 1,2 supermesh: 2000 i 1 + 6000 i 2 – 3 + 5000 i 2 = 0 [1]
and i
1 – i2 = 2×10
-3
[2]

by inspection, i 4 = -1 m A. However, this as well as any equation for mesh
four are unnecessary: w e already have two equations in two unknowns and i and i
1 2
are sufficient to enable us to find the voltage across the current source.

Simplifying, we obtain 2000 i 1 + 11000 i 2 = 3 [1]
1000 i
1 - 1000 i 2 = 2 [2]

Solving, i = 1.923 mA and i = -76.92 μA. 1 2

Thus, the voltage across the 2-mA source (“+” reference at the top of the source) is

v = -2000 i 1 – 6000 (i 1 – i2) = -15.85 V.

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Engineering Circuit Analysis, 7
th
Edition Chapter Four Solutions 10 March 2006
57. Nodal analysis will require 2 nodal equations (one being a “supernode” equation), 1
KVL equation, and subtraction/division operations to obtain the desired current. Mesh
analysis s
imply requ ires 2 “supermesh” equations and 2 KCL equations, with the
desired current being a m esh current. Thus, we define four clockwise mesh currents
i
a, i, ibc, i starting with the left-most mesh and proceeding to the right of the circuit. d

At the a, b supermesh: -5 + 2 i a + 2 i b + 3 i b – 3 i c = 0 [1]

At the c, d supermesh: 3 i c – 3 i + 1 + 4 i = 0 [2] b d

and i a - i = 3 [3] b
i c - i = 2 [4] d

Simplifying and collecting terms, we obtain

2 ia + 5 i – 3 i b c = 5 [1]
-3 i + 3 i
b c + 4 i = -1 [2] d
ia - i = 3 [3] b
ic - i = 2 [4] d

Solving, we find i a = 3.35 A, i = 350 mA, i b c = 1.15 A, and i d = -850 mA. As i 1 = ib,

i1 = 350 mA.
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Engineering Circuit Analysis, 7
th
Edition Chapter Four Solutions 10 March 2006
58. Define a voltage v at the top node of the current source Ix 2 , and a clockwise mesh
current i in the right-most mesh.
b

We want 6 W dissipated in the 6-Ω resistor, which leads to the requirement i = 1 A. b
Applying nodal analysis to the circuit,

I 1 + I2 = (v – v x 1)/ 6 = 1

so our requirement is I1 + I2 = 1. There is no constraint on the value of v other than 1
we are told to select a nonzero value.

= 500 mA and vThus, we choose I = I1 2 1 = 3.1415 V.
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Engineering Circuit Analysis, 7
th
Edition Chapter Four Solutions 10 March 2006
59. Inserting the new 2-V source with “+” reference at the bottom, and the new 7-mA
source with the arrow pointing down, we define four clockwise me
sh currents i 1, i2, i3,
i
4 starting with the left-most mesh and proceeding towards the right of the circuit.

Mesh 1: (2000 + 1000 + 5000) i – 6000 i 1 2 – 2 = 0 [1]

2, 3 Supermesh:
– 6000 i2 + (5000 + 5000 + 1000 + 6000) i
2 1 + (3000 + 4000 + 5000) i – 5000 i 3 4
= 0 [2]
and i
2 - i = 7×10
-3
[3] 3

Mesh 4: i 4 = -1 mA by inspection [4]

Simplifying and combining terms,

8000 i 1 – 6000 i 2 = 2 [1]
1000 i
2 – 1000 i 3 = 7 [4]
-6000 i + 17000 i
1 2 + 12000 i 3 = -7 [2]

Solving, we find that

i1 = 2.653 A, i 2 = 3.204 A, i = -3.796 A, i = -1 mA 3 4

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Engineering Circuit Analysis, 7
th
Edition Chapter Four Solutions 10 March 2006
60. This circuit is e asily analyzed by mesh analysis; it’s planar, and af ter combining the
2A and 3 A sources into a single 1 A source, supermesh analysis is simple.

, i, i and i First, define clockwise m esh currents i
x12 3 starting from the left-most mesh
and moving to the right. Next, com bine the 2 A and 3 A sources tem porarily into a 1
A source, arrow pointing upwards. Then, define four nodal voltages, V, V, V
123 and
V moving from left to right along the top of the circuit.
4

= –5 i At the left-most mesh, i [1]
x 1
– 2i For the supermesh, we can write 4i1 x + 2 + 2i 3 = 0 [2]
– i and the corresponding KCL equation: i = 1 [3]
31

Substituting Eq. [1] into Eq. [2] and simplifying,

14 i
1 + 2i 3 = –2
– i
1 + i3 = 1

Solving, i = –250 mA and i
1 3 = 750 mA.
Then, i = -5 i
x 1 = 1.35 A and i = i – 2 = –2.25 A 2 1

Nodal voltages are straightfoward to find, then:

V = 2i = 1.5 V
4 3
V = 2 + V = 3.5 V 3 4
V = 2 i = 2 i – V 2 3 1 or V + V = 3 V 2 1 3
V = 2 i = 2 i – V 1 2 x or V + V = 5.5 V 1 x 2

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Engineering Circuit Analysis, 7
th
Edition Chapter Four Solutions 10 March 2006
61. Hand analysis:

Define three clockwise m esh currents: i in the bottom left mesh, i 1 2 in the top m esh,
and i in the bottom right mesh.
3

MESH 1: i 1 = 5 mA by inspection [1]

SUPERMESH: i 1 – i2 = 0.4 i 10
i 1 – i2 = 0.4(i 3 – i2)
i
1 – 0.6 i 2 – 0.4 i 3 = 0 [2]

MESH 3: -5000 i – 10000 i 1 2 + 35000 i 3 = 0 [3]

Sim plify: 0.6 i 2 + 0.4 i = 5×10
-3
[2] 3
+ 35000 i-10000 i2 3 = 25 [3]

Solving, we find i 2 = 6.6 mA and i = 2.6 mA. Since i 3 10 = i 3 – i2, we find that

i10 = -4 mA.

PSpice simulation results:

i
10
→

Summary: The current entering the right-hand node of the 10-kΩ resistor R2 is
equal to 4.000 mA. Since this current is –i
10, i10 = -4.000 mA as found by hand.
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Engineering Circuit Analysis, 7
th
Edition Chapter Four Solutions 10 March 2006
62. Hand analysis:
Define the nodes as:

v1
v2
v3
v4
v5
v6

NODE 1: -2 ×10
-3
= (v 1 – 1.3)/ 1.8×10
3
→ v = -2.84 V. 1

2, 4 Supernode:
-3
2.3×10 = (v 2 – v5)/ 1x10
3
+ (v 4 – 1.3)/ 7.3×10
3
+ (v 4 – v5)/ 1.3×10
3 3
+ v/ 1.5×104

KVL equation: -v + v 2 4 = 5.2

Node 5: 0 = (v 5 – v2)/ 1x10
3
+ (v 5 – v4)/ 1.3x10
3
+ (v 5 – 2.6)/ 6.3x10
3

Simplifying and collecting terms,

14.235 v + 22.39 v – 25.185 v = 35.275 [1] 2 4 5
-v2 + v = 5.2 [2] 4
-8.19 v2 – 6.3 v + 15.79 v = 3.38 [3] 4 5

Solving, we find the voltage at the central node is v 4 = 3.460 V.

PSpice simulation results:

Summary: The voltage at the center node is found to be 3.460 V, which is in
agreement with our hand calculation.
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Engineering Circuit Analysis, 7
th
Edition Chapter Four Solutions 10 March 2006
63. Hand analysis:
At the 1,2 supermesh: 2000 i
1 + 6000 i 2 – 3 + 5000 i 2 = 0 [1]
and i
1 – i2 = 2×10
-3
[2]

by inspection, i 4 = -1 m A. However, this as well as any equation for mesh
four are unnecessary: w e already have two equations in two unknowns and i and i
1 2
are sufficient to enable us to find the voltage across the current source.

Simplifying, we obtain 2000 i 1 + 11000 i 2 = 3 [1]
1000 i
1 - 1000 i 2 = 2 [2]

Solving, i = 1.923 mA and i = -76.92 μA. 1 2

Thus, the voltage across the 2-mA source (“+” reference at the top of the source) is

v = -2000 i 1 – 6000 (i 1 – i2) = -15.85 V.

PSpice simulation results:

Summary: Again arbitrarily selecting the “+” reference as the top node of the
2-mA current source, we find the voltage across it is –5.846 – 10 = -15.846 V, in
agreement with our han
d calculation.
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Engineering Circuit Analysis, 7
th
Edition Chapter Four Solutions 10 March 2006
64. Hand analysis:
Define a voltage v at the top node of the current source I
x 2 , and a clockwise mesh
current i in the right-most mesh.
b

We want 6 W dissipated in the 6-Ω resistor, which leads to the requirement i = 1 A. b
Applying nodal analysis to the circuit,

I 1 + I2 = (v – v x 1)/ 6 = 1

so our requirement is I1 + I2 = 1. There is no constraint on the value of v other than 1
we are told to select a nonzero value.

= 500 mA and vThus, we choose I = I1 2 1 = 3.1415 V.

PSpice simulation results:

Summary: We see from the labeled schematic above that our choice for I 1, I and 2
lead to 1 A through the 6-Ω resistor, or 6 W dissipated in that resistor, as desired. V1
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Engineering Circuit Analysis, 7
th
Edition Chapter Four Solutions 10 March 2006
65. Hand analysis:
Define node 1 as the top left node, and node 2 as the node joining the three 2-Ω
resistors. Place the “+” reference term inal of the 2-V source at the right.
The right-
most 2-Ω resistor has therefore been shorted out. Applying nodal analysis then,

= (v – vNode 1: -5 i
1 1 2 )/ 2 [1]

– vNode 2: 0 = (v
2 1)/ 2 + v 2/ 2 + (v – 2)/ 2 [2] 2

and, i
1 = (v – 2)/ 2 [3] 2

Simplifying and collecting terms,

v
1 + v = 10 [1] 2
+ 3 v-v1 2 = 2 [2]

Solving, we find that v
1 = 3.143 V and v 2 = 1.714 V.

Defining clockwise m esh currents i
a, i, ibc, id starting with the left-most m esh and
proceeding right, we may easily determine that

i
a = -5 i 1 = 714.3 mA
i
b = -142.9 mA
i
c = i – 2 = -2.143 A 1
id = 3 + i c = 857.1 mA

PSpice simulation results:

Summary: The simulation results agree with the hand calculations.
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Engineering Circuit Analysis, 7
th
Edition Chapter Four Solutions 10 March 2006
66. (a) One possible circuit configuration of many that would satisfy the requirements:

↑ 5 V
3 A 2 v x
100 Ω
50 Ω
20 Ω
10 Ω
+ v
x -

At node 1: -3 = (v
1 – 5)/ 100 + (v – v2)/ 50 [1] 1

At node 2: 2 v x = (v2 – v 1)/ 50 + v / 30 [2] 2

and, v x = 5 – v [3] 1

Simplifying and collecting terms,

150 v 1 – 100 v 2 = -14750 [1]

2970 v 1 + 80 v 2 = 15000 [2]

Solving, we find that v 1 = 1.036 V and v 2 = 149.1 V.
The current through the 100-Ω resistor is sim
ply (5 – v )/100 = 39.64 mA 1
– vThe current through the 50-Ω resistor is (v 1 2)/ 50 = -2.961 A,
and the current through the 20-Ω and 10-Ω
series combination is v 2/ 30 = 4.97 A.
Finally, the dependent source generates a current of 2 v = 7.928 A.
x
(b) PSpice simulation results

Summary: The simulated results agree with the hand calculations.

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Engineering Circuit Analysis, 7
th
Edition Chapter Four Solutions 10 March 2006
67. One possible solution of many:

+ 5 V -
Choose R so that 3R = 5; then the voltage across the current source will be 5 V, and
so will the voltage across the resistor R.

R = 5/3 Ω. To construct this from 1-Ω resistors, note that

5/3 Ω = 1 Ω + 2/3 Ω = 1 Ω + 1 Ω || 1 Ω || 1Ω + 1Ω || 1Ω || 1Ω

* Solution to Problem 4.57

.OP

V1 1 0 DC 10
I1 0 4 DC 3
R1 1 2 1
R2 2 3 1
R3 2 3 1
R4 2 3 1
R5 3 4 1
R6 3 4 1
R7 3 4 1

.END
**** SMALL SIGNAL BIAS SOLUTION TEMPERATURE = 27.000 DEG C
******************************************************************************
NODE VOLTAGE NODE VOLTAGE NODE VOLTAGE NODE VOLTAGE
( 1) 10.0000 ( 2) 7.0000 ( 3) 6.0000 ( 4) 5.0000

VOLTAGE SOURCE CURRENTS
NAME CURRENT

V1 -3.000E+00
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Engineering Circuit Analysis, 7
th
Edition Chapter Four Solutions 10 March 2006
68. We first name each node, resistor and voltage source:

We next write an appropriate input deck for SPICE:

And obtain the following output:

We see from this simulation result that the voltage v
5 = 2.847 V.

1 2 3
4
5
0
R1
R2
R3
R4
R5
R6
R7
R8
R9
R10
V1
* Solution to Problem 4.58

.OP

V1 1 0 DC 20
R1 1 2 2
R2 2 0 3
R3 2 3 4
R4 2 4 10
R5 3 0 5
R6 3 4 6
R7 3 5 11
R8 4 0 7
R9 4 5 8
R10 5 0 9

.END
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Engineering Circuit Analysis, 7
th
Edition Chapter Four Solutions 10 March 2006
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69. One possible solution of many:

v1
v2
v3
All resistors are
1 Ω, except R1,
which represents
5 1-Ω resistors
in series.

R1
R2
R3
R4
R5

Verify: v
1 = 9(4/9) = 4 V
v
2 = 9(3/9) = 3 V
v
3 = 9(2/9) = 2 V

SPICE INPUT DECK:
* Solution to Problem 4.59

.OP

V1 1 0 DC 9
R1 1 2 5
R2 2 3 1
R3 3 4 1
R4 4 5 1
R5 5 0 1

.END

**** 07/29/01 21:36:26 *********** Evaluation PSpice (Nov 1999) **************

* Solution to Problem 4.59
**** SMALL SIGNAL BIAS SOLUTION TEMPERATURE = 27.000 DEG C
***********************************************************************
*******

NODE VOLTAGE NODE VOLTAGE NODE VOLTAGE NODE
VOLTAGE

( 1) 9.0000 ( 2) 4.0000 ( 3) 3.0000 ( 4) 2.0000

(
5) 1.0000
1 2
3
4
5
0
R1
R4
R5
R2
R3

Engineering Circuit Analysis, 7
th
Edition Chapter Four Solutions 10 March 2006
70. (a) If only two bulbs are not lit (and thinking of each bulb as a resistor), the bulbs
must be in parallel- otherwise, the burned out bulbs, acting as short circuits, would
prevent current from flowing to the “good” bulbs.

(b) In a parallel connected circuit, each bulb “sees” 115 VAC. Therefore, the
individual bulb current is 1 W / 115 V = 8.696 mA. The resistance of each “good
”
bulb is V/I = 13.22 kΩ. A simplified, electrically-equivalent model for this cir cuit
would be a 115 VAC source connected in parallel to a resistor R such that
eq

1/R
eq = 1/13.22×10
3
+ 1/13.22×10
3
+ …. + 1/13.22×10
3
(400 – 2 = 398 terms)
= 33.22 Ω . W
e expect the source to provide 398 W. or Req
* Solution to Problem 4.60

.OP

V1 1 0 AC 115 60
R1 1 0 33.22

.AC LIN 1 60 60
.PRINT AC VM(1)IM(V1)

.END

**** 07/29/01 21:09:32 *********** Evaluation PSpice (Nov 1999) **************

* Solution to Problem 4.60

**** SMALL SIGNAL BIAS SOLUTION TEMPERATURE = 27.000 DEG C
******************************************************************************

NODE VOLTAGE NODE VOLTAGE NODE VOLTAGE NODE VOLTAGE
( 1) 0.0000

VOLTAGE SOURCE CURRENTS
NAME CURRENT

V1 0.000E+00

TOTAL POWER DISSIPATION 0.00E+00 WATTS

**** 07/29/01 21:09:32 *********** Evaluation PSpice (Nov 1999) **************
* Solution to Problem 4.60

**** AC ANALYSIS TEMPERATURE = 27.000 DEG C
******************************************************************************

FREQ VM(1) IM(V1)
6.000E+01 1.150E+02 3.462E+00

This calculated power is not the value
sought. It is an artifact of the use of
an ac s ource, which requires that we
perform an ac analysis. T he supplied
power is th en separately computed as
(1.15×10)(3.462) = 398.1 W.
2

(c) The in herent serie s res istance of the wire connections
leads to a voltage dro p
which increases the f urther one is from the voltage source. Thus, th e furthest bulbs
actually have less than 115 VAC across them , so they draw slightly less current and
glow more dimly.
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Engineering Circuit Analysis, 7
th
Edition Chapter Five Solutions 10 March 2006

1. Define percent error as 100 [e
x
– (1 + x)]/ e
x

x 1 + x e
x
% error 0.001 1.001 1.001 5×10
-5
0.005 1.005 1.005 1×10
-3
0.01 1.01 1.010 5×10
-3
0.05 1.05 1.051 0.1
0.10 1.10 1.105 0.5
0.50 1.50 1.649 9
1.00 2.00 2.718 26
5.00 6.00 148.4 96

Of course, “reasonable” is a very subjective term. However, if we choose x < 0.1, we
ensure th
at the error is less than 1%.

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Engineering Circuit Analysis, 7
th
Edition Chapter Five Solutions 10 March 2006

2. (a) Short-circuit the 10 V source.
Note that 6 || 4 = 2.4 Ω. By voltage division, the voltage across the 6 Ω
resist
or is then

2.4
4 1.778 V
32.4
=
+

So that
1
1.778
0.2963 A
6
i′==
.

(b) Short-circuit the 4 V source.
Note that 3 || 6 = 2 Ω . By voltage division, the voltage across the 6 Ω
resist
or is then

2
10 3.333 V
6
−=−

So that
1
3.333
0.5556 A
6
i
−
′′==−
.

(c)
112
-259.3 mAiii′′′=+=
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Engineering Circuit Analysis, 7
th
Edition Chapter Five Solutions 10 March 2006

3. Open circuit the 4 A source. Then, since

(7 + 2) || (5 + 5) = 4.737 Ω, we can calculate
1
(1)(4.737)v
′= = 4.737 V.

To find the total current flowing through the 7 Ω resi
stor, we first determine the total
voltage v
1 by continuing our superposition procedure. The contribution to v 1 from the
4 A source is found by first open-circuiting the 1 A source, then noting that current
division yields:
520
4 1.053 A
5 (5 7 2) 19
==
+++

Then, 9.477 V. Hence,
1
(1.053)(9)v′′==
111
vvv
′′′=+= 14.21 V.
We may now find the total current flowing downward through the 7 Ω resistor
as

14.21/7 = 2.03 A.

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Engineering Circuit Analysis, 7
th
Edition Chapter Five Solutions 10 March 2006

4. One approach to this problem is to write a set of mesh equations, leaving the voltage
source and current source as variables which can be set to zero.

We first rename the voltage source as V
x. We next define three clockwise mesh
currents in the bottom three meshes: i
1, iy and i 4. Finally, we define a clockwise mesh
current i
3 in the top mesh, noting that it is equal to –4 A.

Our general mesh equations are then:

-V
x + 18i1 – 10iy = 0
–10i
1 + 15iy – 3i4 = 0
–3i
y + 16i4 – 5i3 = 0

** Set V
x = 10 V, i3 = 0. Our mesh equations then become:

18i
1 – 10 = 10
y
i′
–10i 1 + 15 – 3i
y
i′ 4 = 0
– 3 + 16i
y
i′ 4 = 0

Solving, = 0.6255 A.
y
i′

** Set V
x = 0 V, i3 = – 4 A. Our mesh equations then become:

18i
1 – 10 = 0
y
i′′
–10i 1 + 15 – 3i
y
i′′ 4 = 0
– 3 + 16i
y
i′′ 4 = -20

Solving, = –0.4222 A.
y
i′′

Thus, i
y = + = 203.3 mA.
y
i′
y
i′′

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Engineering Circuit Analysis, 7
th
Edition Chapter Five Solutions 10 March 2006

5. We may solve this problem without writing circuit equations if we first realise that the
current i
1 is composed of two terms: one that depends solely on the 4 V source, and
one that depends solely on the 10 V source.

This may be written as i
1 = 4K1 + 10K2, where K1 and K2 are constants that depend
on the circuit topology and resistor values.

We may not change K
1 or K2, as only the source voltages may be changed. If we
increase both sources by a factor of 10, then i
1 increases by the same amount.

Thus, 4 V → 40 V and 10 V →
100 V.

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Engineering Circuit Analysis, 7
th
Edition Chapter Five Solutions 10 March 2006

6. i A, vB “on”, v C = 0: i x = 20 A
i
A, vC “on”, v B = 0: i x = -5 A
i
A, vB, vC “on” : i x = 12 A

so, we can write i x’ + ix” + ix”’ = 12
i
x’ + ix” = 20
i
x’ + i x”’ = - 5

In matrix form,
⎥
⎥
⎥
⎦
⎤
⎢
⎢
⎢
⎣
⎡
=
⎥
⎥
⎥
⎦
⎤
⎢
⎢
⎢
⎣
⎡
′′′
′′
′
⎥
⎥
⎥
⎦
⎤
⎢
⎢
⎢
⎣
⎡
5-
20
12

1 0 1
0 1 1
1 1 1
x
x
x
i
i
i

(a) with i A on only, the response i x = ix’ = 3 A.
(b) with v
B on only, the response i x = ix” = 17 A.
(c) with v
C on only, the response i x = ix”’ = -8 A.
(d) i
A and v C doubled, v B reversed: 2(3) + 2(-8) + (-1)(17) = -27 A.
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Engineering Circuit Analysis, 7
th
Edition Chapter Five Solutions 10 March 2006

7. One source at a time:
The contribution from the 24-V source may be found by shorting the 45-V source and
open-circuiting the 2-A source. Applying voltage division,

vx’ =
V 10
182010
20
24
30||452010
20
24 =
++
=
++

We find the contribution of the 2-A source by shorting both voltage sources and
applying current division:

vx” =
V 8.333
182010
10
220 =
⎥
⎦
⎤
⎢
⎣
⎡
++

Finally, the contribution from the 45-V source is found by open-circuiting the 2-A
source and shorting the 24-V source. Defining v
30 across the 30-Ω resistor with the
“+” reference on top:

0 = v 30/ 20 + v 30/ (10 + 20) + (v 30 – 45)/ 45

solving, v 30 = 11.25 V and hence v x”’ = -11.25(20)/(10 + 20) = -7.5 V

Adding the individual contributions, we find that v x = vx’ + v x” + v x”’ = 10.83 V.
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Engineering Circuit Analysis, 7
th
Edition Chapter Five Solutions 10 March 2006

8. The contribution of the 8-A source is found by shorting out the two voltage sources
and employing simple current division:

i3' =
A 5-
3050
50
8 =
+
−

The contribution of the voltage sources may be found collectively or individually. The
contribution of the 100-V source is found by open-circuiting the 8-A source and
shorting the 60-V source. Then,

i3" =
A 6.25
30||60||)3050(
100
=
+

The contribution of the 60-V source is found in a similar way as i 3"' = -60/30 = -2 A.
The total response is i
3 = i3' + i3" + i3"' = -750 mA.
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Engineering Circuit Analysis, 7
th
Edition Chapter Five Solutions 10 March 2006

9. (a) By current division, the contribution of the 1-A source i 2’ is
i
2’ = 1 (200)/ 250 = 800 mA.

The contribution of the 100-V source is i
2” = 100/ 250 = 400 mA.

The contribution of the 0.5-A source is found by current division once the 1-A source
is open-circuited and the voltage source is shorted. Thus,

i
2”’ = 0.5 (50)/ 250 = 100 mA

Thus, i
2 = i2’ + i2” + i2”’ = 1.3 A

(b) P
1A = (1) [(200)(1 – 1.3)] = 60 W
P
200 = (1 – 1.3)
2
(200) = 18 W
P
100V = -(1.3)(100) = -130 W
P
50 = (1.3 – 0.5)
2
(50) = 32 W
P
0.5A = (0.5) [(50)(1.3 – 0.5)] = 20 W

Check: 60 + 18 + 32 + 20 = +130.
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Engineering Circuit Analysis, 7
th
Edition Chapter Five Solutions 10 March 2006

10. We find the contribution of the 4-A source by shorting out the 100-V source and
analysing the resulting circuit:

4 = V 1' / 20 + (V1' – V')/ 10 [1]

0.4 i 1' = V1'/ 30 + (V' – V1')/ 10 [2]

where i 1' = V1'/ 20

Simplifying & collecting terms, we obtain 30 V1' – 20 V' = 800 [1]
-7.2 V
1' + 8 V' = 0 [2]

Solving, we find that V' = 60 V. Proceeding to the contribution of the 60-V source, we
analyse the following circuit after defining a clockwise mesh current i
a flowing in the
left mesh and a clockwise mesh current i
b flowing in the right mesh.

30 i
a – 60 + 30 i a – 30 i b = 0 [1]
i
b = -0.4 i 1" = +0.4 i a [2]

Solving, we find that i
a = 1.25 A and so V" = 30(i a – ib) = 22.5 V.

Thus, V = V' + V" = 82.5 V.

'
'
"
"
'
"
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Engineering Circuit Analysis, 7
th
Edition Chapter Five Solutions 10 March 2006

11. (a) Linearity allows us to consider this by viewing each source as being scaled by
25/ 10. This means that the response (v
3) will be scaled by the same factor:

25 iA'/ 10 + 25 i B'/ 10 = 25 vB 3'/ 10

∴ v3 = 25v 3'/ 10 = 25(80)/ 10 = 200 V

(b) i
A' = 10 A, i B' = 25 A → v 4' = 100 V
i
A" = 10 A, i B" = 25 A → v 4" = -50 V
i
A = 20 A, i B = -10 A → v 4 = ?

We can view this in a somewhat abstract form: the currents i
A and i B multiply
the same circuit parameters regardless of their value; the result is v
4.

Writing in matrix form, , we can solve to find
⎥
⎦
⎤
⎢
⎣
⎡
=
⎥
⎦
⎤
⎢
⎣
⎡
⎥
⎦
⎤
⎢
⎣
⎡
50-
100

b
a

10 25
25 10

a = -4.286 and b = 5.714, so that 20a – 10b leads to v
4 = -142.9 V

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Engineering Circuit Analysis, 7
th
Edition Chapter Five Solutions 10 March 2006

12. With the current source open-circuited and the 7-V source shorted, we are left with
100k || (22k + 4.7k) = 21.07 kΩ.

Thus, V 3V = 3 (21.07)/ (21.07 + 47) = 0.9286 V.

In a similar fashion, we find that the contribution of the 7-V source is:

V 7V = 7 (31.97) / (31.97 + 26.7) = 3.814 V

Finally, the contribution of the current source to the voltage V across it is:

V 5mA = (5×10
-3
) ( 47k || 100k || 26.7k) = 72.75 V.

Adding, we find that V = 0.9286 + 3.814 + 72.75 = 77.49 V.
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Engineering Circuit Analysis, 7
th
Edition Chapter Five Solutions 10 March 2006

13. We must find the current through the 500-kΩ resistor using superposition, and then
calculate the dissipated power.

The contribution from the current source may be calculated by first noting that
1M || 2.7M || 5M = 636.8 kΩ. Then,

i60μA =
A 43.51
0.636830.5
3
1060
6
μ=⎟
⎠
⎞
⎜
⎝
⎛
++
×
−

The contribution from the voltage source is found by first noting that 2.7M || 5M =
1.753 MΩ. The total cu
rrent flowing from the voltage source (with the current source
open-circuited) is –1.5/ (3.5 || 1.753 + 1) μA = -0.6919 μA. The current flowing
through the 500-kΩ resistor due to the voltage source acting alone is then

i1.5V = 0.6919 (1.753)/ (1.753 + 3.5) mA = 230.9 nA.

The total current through the 500-kΩ resistor is then i 60μA + i1.5V = 43.74 μA and the

dissipated power is (43.74×10
-9
)
2
(500×10
3
) = 956.6 μW.

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Engineering Circuit Analysis, 7
th
Edition Chapter Five Solutions 10 March 2006

14. We first determine the contribution of the voltage source:

'
↑
I
1'
'

Via mesh analysis, we write: 5 = 18000 I
1' – 17000 Ix'
-6 I
x' = -17000 Ix' + 39000 Ix'
Solving, we find I
1' = 472.1 mA and Ix' = 205.8 mA, so V' = 17×10
3
(I1' - Ix')
= 4.527 V. We proceed to find the contribution of the curren
t source:

Vx" V" Via supernode: -20×10
-3
= Vx"/ 22×10
3
+ V"/ 0.9444×10
3
[1]
and V" – V
x" = 6Ix" or V" – Vx" = 6 Vx"/ 22×10
3
[2]
Solving, we find that V" = -18.11 V. Thus, V = V' + V" = -13.58 V.

The m
aximum power is V
2
/ 17×10
3
= V
2
/ 17 mW = 250 mW, so
V =
( )(17)(250) 65.19 V - -18.11′== . Solving, we find = 83.3 V.
max
V′
The 5-V source may then be increased by a factor of 83.3/ 4.527, so that its

maximum positive value is 92 V; past this value, and the resistor will overheat.
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Engineering Circuit Analysis, 7
th
Edition Chapter Five Solutions 10 March 2006

15. It is impossible to identify the individual contribution of each source to the power
dissipated in the resistor; superposition cannot be used for such a purpose.

Simplifying the circuit, we may at least determine the total power dissipated in the
resistor:

i →

Via superposition in one step, we may write

i =
mA 195.1
2.12
2.1
2 -
1.22
5
=
++

Thus,
P
2Ω = i
2

.
2 = 76.15 mW
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Engineering Circuit Analysis, 7
th
Edition Chapter Five Solutions 10 March 2006

16. We will analyse this circuit by first considering the combined effect of both dc
sources (left), and then finding the effect of the single ac sour
ce acting alone (right).

I
B
I
B
V
1
V
2
V
3
v
x'

1, 3 supernode: V
1/ 100 + V1/ 17×10
3
+ (V1 – 15)/ 33× 10
3
+ V3/ 10
3
= 20 IB [1]

and: V 1 – V3 = 0.7 [2]

Node 2: -20 I B = (V2 – 15)/ 1000 [3]

We require one additional equation if we wish to have I B as an unknown:

20 I B + IB = V3/ 1000 [4]

Simplifying and collecting terms,

10.08912 V 1 + V3 – 20× 10
3
IB = 0.4545 [1]

V 1 - V3 = 0.7 [2]

V 2 + 20× 10
3
IB = 15 [3]

-V 3 + 21× 103 I B = 0 [4]

Solving, we find that IB = -31.04 μA.

To analyse the right-hand circuit, we first find the Thévenin equivalent to the left of
the wire marked i
B', noting that the 33-kΩ and 17-kΩ resistors are now in parallel. We
find that V
TH = 16.85 cos 6t V by voltage division, and R TH = 100 || 17k || 33k =
99.12 Ω . We may now proceed:

20 i
B' = vB x' / 1000 + (v x' – 16.85 cos 6t )/ 99.12 [1]
20 i
B' + iB B'' = v x'/ 1000 [2]

Solving, we find that i
B' = 798.6 cos 6t mA. Thus, adding our two results, we find the
complete current is
i
B = iB B' + IB = -31.04 + 798.6 cos 6t μA.

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Engineering Circuit Analysis, 7
th
Edition Chapter Five Solutions 10 March 2006

17.

We first consider the effect of the 2-A source separately, using the left circuit:

Vx' =
V 1.765
143
3
2 5=
⎥
⎦
⎤
⎢
⎣
⎡
+

Next we consider the effect of the 6-A source on its own using the right circuit:

Vx" =
V 15.88
89
9
6 5=
⎥
⎦
⎤
⎢
⎣
⎡
+

Thus, V x = Vx' + Vx" = 17.65 V.

(b) PSpice verification (DC Sweep)
The DC sweep results
below
confirm
that V
x' =
1.765 V

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Engineering Circuit Analysis, 7
th
Edition Chapter Five Solutions 10 March 2006

18.

'
(a) Beginning with the circuit on the left, we find the contribution of the 2-V source to
V
x:
50
2V
100
V
V4
xx
x −′
+
′
=′−
which leads to V
x' = 9.926 mV.

The circuit on the right yields the contribution of the 6-A source to Vx:

50
V
100
V
V4
xx
x ′′
+
′′
=′′−
which leads to V
x" = 0.

Thus, Vx = Vx' + Vx" = 9.926 mV.

(b) PSpice verification.

As can be seen from the two separate PSpice simulations, our hand calculations are
correct; the pV-scale voltage in
the second simulation is a result
of numerical inaccuracy.
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Engineering Circuit Analysis, 7
th
Edition Chapter Five Solutions 10 March 2006

19.
V 2.455 V so
0
2
15V
3
V
1
12V
x
xxx=
=
+
++
−

V 0.5455 V so
0
2
10V
3
V
1
6V
x
xxx=′
=
+′
+
′
+
−′

V 1.909 V so
0
2
5V
3
V
1
6V
x
xxx=′′
=
+′′
+
′′
+
−′′

Adding, we find that V
x' + Vx" = 2.455 V = Vx as promised.

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Engineering Circuit Analysis, 7
th
Edition Chapter Five Solutions 10 March 2006

20. (a) We first recognise that the two current sources are in parallel, and hence may be
replaced by a single –7 A source (arrow directed downward). This source is in parallel
with a 10 kΩ resistor. A simple source transformation th erefore yields a 10 kΩ
resistor in series with a (–7)(10,000) = –70,000 V source (+ reference on top):

(b) This circuit requires several source transforma
tions. First, we convert the 8 V
source and 3 Ω resistor to an 8/3 A current source in parallel with 3 Ω. This yields a
circuit with a 3 Ω and 10 Ω parallel combination, which may be replaced with a
2.308 Ω resistor. We may now convert the 8/3 A current source and 2.308 Ω resistor
to a (8/3)(2.308) = 6.155 V voltage source in series with a 2.308 Ω resistor. This
modified circuit contains a series combination of 2.308 Ω and 5 Ω ; performing a
source transformation yet again, we obtain a current source with value (6.155)/(2.308
+ 5) = 0.8422 A in parallel with 7.308 Ω and in parallel with the remaining 5 Ω
resistor. Since 7.308 Ω || 5 Ω = 2.969 Ω, our solution is:

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Engineering Circuit Analysis, 7
th
Edition Chapter Five Solutions 10 March 2006

21. (a) First we note the three current sources are in parallel, and may be replaced by a
single current source having value 5 – 1 + 3 = 7 A, arrow pointing upwards. This
source is in parallel with the 10 Ω resistor and the 6 Ω resistor. Performing a source
transformation on the current source and 6 Ω resistor, we obtain a voltage source
(7)(6) = 42 V in series with a 6 Ω resistor and in series with the 10 Ω resistor:

+
v
–

(b) By voltage division,
v = 42(10)/16 = 26.25 V.

(c) Once the 10 Ω resistor is involved in a source transforma
tion, it disappears, only
to be replaced by a resistor having the same value – but whose current and voltage can
be different. Since the quantity
v appearing across this resistor is of interest, we
cannot involve the resistor in a transformation.

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Engineering Circuit Analysis, 7
th
Edition Chapter Five Solutions 10 March 2006

22. (a) [120 cos 400t] / 60 = 2 cos 400t A. 60 || 120 = 40 Ω.

[2 cos 400t] (40) = 80 cos 400t V. 40 + 10 = 50 Ω.

[80 cos 400t]/ 50 = 1.6 cos 400t A. 50 || 50 = 25 Ω .

25 Ω 1.6 cos 400t A

(b) 2k || 3k + 6k = 7.2 kΩ. 7.2k || 12k = 4.5 kΩ

(20)(4.5) = 90 V.
4.5 kΩ
3.5 kΩ
8 kΩ

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Engineering Circuit Analysis, 7
th
Edition Chapter Five Solutions 10 March 2006

23. We can ignore the 1-kΩ resistor, at least when performing a source transformation on
this circuit, as the 1-mA source will pump 1 mA through
whatever value resistor we
place there. So, we need only combine the 1 and 2 mA sources (which are in parallel
once we replace the 1-kΩ resistor with a 0-Ω resistor). The current through the 5.8-
kΩ resistor is then simply given by voltage division:

mA 1.343
5.84.7
4.7
103
3-
=
+
×=i

The power dissipated by the 5.8-kΩ resistor is then i
2

.
5.8×10
3
= 10.46 mW.

(Note that we did not “transform” either sour
ce, but rather drew on the relevant
discussion to understand why the 1-kΩ resistor could be omitted.)
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Engineering Circuit Analysis, 7
th
Edition Chapter Five Solutions 10 March 2006

24. We may ignore the 10-kΩ and 9.7-kΩ resistors, as 3-V will appear across them
regardless of their value. Performing a quick source transformation on the 10-kΩ
resistor/ 4-mA
current source combination, we replace them with a 40-V source in
series with a 10-kΩ resistor:

I = 43/ 15.8 mA = 2.722 mA. Therefore, P
5.8Ω = I
2.
5.8×10
3
= 42.97 mW.

Ω
Ω
I →

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Engineering Circuit Analysis, 7
th
Edition Chapter Five Solutions 10 March 2006

25. (100 kΩ )(6 mA) = 0.6 V

0.6 V

470 k || 300 k = 183.1 kΩ
(-3 – 0.6)/ 300×10
3
= -12 μ A
(183.1 kΩ )(-12 μA) = -2.197 V

Solving, 9 + 1183.1×10
3
I – 2.197 = 0, so I = -5.750 μA. Thus,

P
1MΩ = I
2

.
10
6
= 33.06 μW.
183.1
-2.197 V
← I

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Engineering Circuit Analysis, 7
th
Edition Chapter Five Solutions 10 March 2006

26. (1)(47) = 47 V. (20)(10) = 200 V. Each voltage source “+” corresponds to its
corresponding current source’s arrow head.

Using KVL on the simplified circuit above,

47 + 47×10
3
I1 – 4 I1 + 13.3×10
3
I1 + 200 = 0

Solving, we find that I1 = -247/ (60.3×10
3
– 4) = -4.096 mA.

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Engineering Circuit Analysis, 7
th
Edition Chapter Five Solutions 10 March 2006

27. (a) (2 V1)(17) = 34 V1

34
← I

Analysing the simplified circuit above,

34 V1 – 0.6 + 7 I + 2 I + 17 I = 0 [1] and V1 = 2 I [2]

Substituting, we find that I = 0.6/ (68 + 7 + 2 + 17) = 6.383 mA. Thus,

V1 = 2 I = 12.77 mV
(b)

V
1 is the top-labelled voltage (12.77 mV).
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Engineering Circuit Analysis, 7
th
Edition Chapter Five Solutions 10 March 2006

28. (a) 12/ 9000 = 1.333 mA. 9k || 7k = 3.938 kΩ. → (1.333 mA)(3.938 kΩ ) = 5.249
V.

5.249 V
3.938

5.249/ 473.938×10
3
= 11.08 μA

473.9 kΩ
10 kΩ
11.08 μA

473.9 k || 10 k = 9.793 kΩ. (11.08 mA)(9.793 kΩ ) = 0.1085 V

I
x = 0.1085/ 28.793×10
3
= 3.768 μA.
0.1085 V 17 kΩ
11.793 kΩ

(b)

Ix
↓

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Engineering Circuit Analysis, 7
th
Edition Chapter Five Solutions 10 March 2006

29. First, (-7 μA)(2 MΩ) = -14 V, “+” reference down. 2 MΩ + 4 MΩ = 6 MΩ.
+14 V/ 6 ΜΩ = 2.333 μA, arrow pointing up; 6 M || 10 M = 3.75 MΩ.

2.333
3.75 MΩ

(2.333)(3.75) = 8.749 V. R
eq = 6.75 MΩ
∴ I
x = 8.749/ (6.75 + 4.7) μA = 764.1 nA.
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Engineering Circuit Analysis, 7
th
Edition Chapter Five Solutions 10 March 2006

30. To begin, note that (1 mA)(9 Ω) = 9 mV, and 5 || 4 = 2.222 Ω.

The above circuit may not be further simp
lified using only source transformation
techniques.
9
15
2.222 Ω

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Engineering Circuit Analysis, 7
th
Edition Chapter Five Solutions 10 March 2006

31. Label the “–” terminal of the 9-V source node x and the other terminal node x'. The
9-V source will force the voltage across these two terminals to be –9 V regardless of
the value of the current source and resistor to its left. These two components ma
y
therefore be neglected from the perspective of terminals
a & b. Thus, we may draw:

7.25 A
2 Ω

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Engineering Circuit Analysis, 7
th
Edition Chapter Five Solutions 10 March 2006

32. Beware of the temptation to employ superposition to compute the dissipated power- it
won’t work
!

Instead, define a current I flowing into the bottom terminal of the 1-MΩ resistor
.
Using superposition to co
mpute this current,

I = 1.8/ 1.840 + 0 + 0 μA = 978.3 nA.
Thus,
P
1MΩ = (978.3×10
-9
)
2
(10
6
) = 957.1 nW.

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Engineering Circuit Analysis, 7
th
Edition Chapter Five Solutions 10 March 2006

33. Let’s begin by plotting the experimental results, along with a least-squares fit to part
of the data:

Least-squares fit results:

Voltage (V) Current (mA)
1.567 1.6681
1.563 6.599
1.558 12.763

We see from the figure that we cannot draw a very good line through all data points
representing currents from 1 mA to 20 mA. We
have therefore chosen to perform a
linear fit for the three lower voltages only, as shown. Our model will not be as
accurate at 1 mA; there is no way to know if our model will be accurate at 20 mA,
since that is beyond the range of the experimental data.

Modeling this system as an ideal voltage source in series with a resistance
(representing the internal resistance of the battery) and a varying load resistance, we
may write the following two equations based on the linear fit to the data:

1.567 = Vsrc – Rs (1.6681×10
-3
)
1.558 = V
src – Rs (12.763×10
-3
)

Solving, Vsrc = 1.568 V and Rs = 811.2 mΩ. It should be noted that depending on the
line fit to the experimental data, these values can change somewhat, particularly the
series resistance value.

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Engineering Circuit Analysis, 7
th
Edition Chapter Five Solutions 10 March 2006

34. Let’s begin by plotting the experimental results, along with a least-squares fit to part
of the data:

Least-squares fit results:

Voltage (V) Current (mA)
1.567 1.6681
1.563 6.599
1.558 12.763

We see from the figure that we cannot draw a very good line through all data points
representing currents from 1 mA to 20 mA. We
have therefore chosen to perform a
linear fit for the three lower voltages only, as shown. Our model will not be as
accurate at 1 mA; there is no way to know if our model will be accurate at 20 mA,
since that is beyond the range of the experimental data.

Modeling this system as an ideal current source in parallel with a resistance R p
(representing the internal resistance of the battery) and a varying load resistance, we
may write the following two equations based on the linear fit to the data:

1.6681×10
-3
= Isrc – 1.567/ Rp

12.763×10
-3
= Isrc – 1.558/ Rp

Solving, Isrc = 1.933 A and Rs = 811.2 mΩ. It should be noted that depending on the
line fit to the experimental data, these values can change somewhat, particularly the
series resistance value.
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Engineering Circuit Analysis, 7
th
Edition Chapter Five Solutions 10 March 2006

35. Working from left to right,

2 μA – 1.8 μA = 200 nA, arrow up.
1.4 MΩ + 2.7 MΩ = 4.1 MΩ

A transformation to a voltage source yields (200 nA)(4.1 MΩ) = 0.82 V in series with

4.1 MΩ + 2 MΩ = 6.1 MΩ, as shown below:

Then, 0.82 V/ 6.1 MΩ = 134.4 nA, arrow up.
6.1 MΩ || 3 MΩ = 2.011 MΩ
4.1 μA + 134.4 nA = 4.234 mA
, arrow up.
(4.234 μA) (2.011 MΩ) = 8.515 V.
The final circuit is an 8.515 V voltage source
in series with a 2.011 MΩ resistor, as shown:

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Engineering Circuit Analysis, 7
th
Edition Chapter Five Solutions 10 March 2006

36. To begin, we note that the 5-V and 2-V sources are in series:

3

Next, noting that 3 V/ 1 Ω = 3 A, and 4 A – 3 A = +1 A (arrow down), we obtain:

The left-hand resistor and the current
source are easily transformed into a
1-V source in series with a 1-Ω
resistor:

By voltage division, the voltage across the
5- Ω resis
tor in the circuit to the right is:

(-1)
2 5||2
5||2
+
= -0.4167 V.

Thus, the power dissipated by the 5-Ω resistor is (-0.4167)
2
/ 5 = 34.73 mW.

± -1 V

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Engineering Circuit Analysis, 7
th
Edition Chapter Five Solutions 10 March 2006

37. (a) We may omit the 10 Ω resistor from the circuit, as it does not affect the voltage or
current associated with R
L since it is in parallel with the voltage source. We are thus
left with an 8 V source in series with a 5 Ω resistor. These may be transformed to an
8/5 A current source in parallel with 5 Ω, in parallel with R
L.

(b)
We see from simulating both circuits simultaneously that the
voltage across R
L is the same (4 V).

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Engineering Circuit Analysis, 7
th
Edition Chapter Five Solutions 10 March 2006

38. (a) We may begin by omitting the 7 Ω and 1 Ω resistors. Performing the indicated
source transformations, we find a 6/4 A source in parallel with 4 Ω, and a 5/10 A
source in parallel with 10 Ω . These are both in parallel with the series combination of
the two 5 Ω resistors. Since 4 Ω || 10 Ω = 2.857 Ω, and 6/4 + 5/10 = 2 A, we may
further simplify the circuit to a single current source (2 A) in parallel with 2.857 Ω
and the series combination of two 5 Ω resistors. Simple current division yields the
current flowing through the 5 Ω resistors:

I =
2(2.857)
0.4444 A
2.857 10
=
+

The power dissipated in either
of the 5 Ω resistors is then I
2
R = 987.6 mW.

(b) We note that PSpice will NOT allow the 7 Ω resistor to be left floating! For both
circu
its simulated, we observe 987.6 mW of power dissipated for the 5 Ω resistor,
confirming our analytic solution.

(c) Neither does. No current flows through the 7 Ω re
sistor; the 1 Ω resistor is in
parallel with a voltage source and hence cannot affect any other part of the circuit.

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Engineering Circuit Analysis, 7
th
Edition Chapter Five Solutions 10 March 2006

39. We obtain a 5v3/4 A current source in parallel with 4 Ω, and a 3 A current source in
parallel with 2 Ω . We now have two dependent current sources in parallel, which may
be combined to yield a single –0.75v
3 current source (arrow pointing upwards) in
parallel with 4 Ω . Selecting the bottom node as a reference terminal, and naming the
top left node V
x and the top right node Vy, we write the following equations:

–0.75v
3 = Vx/4 + (Vx – Vy)/3
3 = V
y/2 + (Vy – Vx)/3
v
3 = Vy – Vx

Solving, we find that v
3 = –2 V.

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Engineering Circuit Analysis, 7
th
Edition Chapter Five Solutions 10 March 2006

40. (a) RTH = 25 || (10 + 15) = 25 || 25 = 12.5 Ω .

V
TH = Vab =
⎟
⎠
⎞
⎜
⎝
⎛
++
+
+⎟
⎠
⎞
⎜
⎝
⎛
++ 251015
1015
100
251510
25
50 = 75 V.

(b) If R
ab = 50 Ω,

P
50Ω =
W72
50
1

12.550
50
75
2
=⎟
⎠
⎞
⎜
⎝
⎛
⎥
⎦
⎤
⎢
⎣
⎡
⎟
⎠
⎞
⎜
⎝
⎛
+

(c) If R
ab = 12.5 Ω ,

P
12.5Ω =
W112.5
12.5
1

12.55.21
12.5
75
2
=⎟
⎠
⎞
⎜
⎝
⎛
⎥
⎦
⎤
⎢
⎣
⎡
⎟
⎠
⎞
⎜
⎝
⎛
+

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Engineering Circuit Analysis, 7
th
Edition Chapter Five Solutions 10 March 2006

41. (a) Shorting the 14 V source, we find that RTH = 10 || 20 + 10 = 16.67 Ω .

Next, we find V
TH by determining VOC (recognising that the right-most 10 Ω resistor
carries no current, hence we have a simple voltage divider):
V
TH = VOC =
10 10
14 9.333 V
10 10 10
+
=
++

Thus, our Thevenin equivalent is a 9.333 V source in series with a 16.67 Ω resistor
,
which is in series with the 5 Ω resistor of interest.

(b) I
5Ω = 9.333/ (5 + 16.67) = 0.4307 A. Thus,

P
5Ω = (0.4307)
2 .
5 = 927.5 mW

(c) We see from the PSpice simulation that keeping four significant digits in
calcu
lating the Thévenin equivalent yields at least 3 digits’ agreement in the results.

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Engineering Circuit Analysis, 7
th
Edition Chapter Five Solutions 10 March 2006

42. (a) Replacing the 7 Ω resistor with a short circuit, we find

I
SC = 15 (8)/ 10 = 12 A.

Removing the short circuit, and open-circuiting the 15 A source, we see that

R
TH = 2 + 8 = 10 Ω.

Thus, V
TH = ISCRTH = (12)(10) = 120 V.

Our Thévenin equivalent is therefore a 120 V source in series with 10 Ω.

(b) As found above, I
N = ISC = 12 A, and RTH = 10 Ω.

(c) Using the Thévenin equivalent circuit, we may find
v1 using voltage division:

v1 = 120 (7)/17 = 49.41 V.

Using the Norton equivalent circuit and a comb
ination of current division and
Ohm’s law, we find

v1 =
10
7 12 49.41 V
17
⎛⎞
=
⎜⎟
⎝⎠

As expected, the results are equal.

(d) Employing the more convenient Thévenin equivalent mo
del,

v1 = 120(1)/17 = 7.059 V.

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Engineering Circuit Analysis, 7
th
Edition Chapter Five Solutions 10 March 2006

43. (a) RTH = 10 mV/ 400 μA = 25 Ω

(b) R
TH = 110 V/ 363.6×10
–3
A = 302.5 Ω

(c) Increased current leads to increased filam
ent temperature, which results in a
higher resistance (as measured). This means the Thévenin equivalent must apply to
the specific current of a particular circuit – one model is not suitable for all operating
conditions (the light bulb is nonlinear).

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Engineering Circuit Analysis, 7
th
Edition Chapter Five Solutions 10 March 2006

44. (a) We begin by shorting both voltage sources, and removing the 1 Ω resistor of
interest. Looking into the terminals where the 1 Ω resistor had been connected, we see
that the 9 Ω resistor is shorted out, so that

R
TH = (5 + 10) || 10 + 10 = 16 Ω .

To continue, we return to the original circuit and replace the 1 Ω resi
stor with a short
circuit. We define three clockwise mesh currents:
i1 in the left-most mesh, i2 in the
top-right mesh, and isc in the bottom right mesh. Writing our three mesh equations,

– 4 + 9
i1 – 9i2 + 3 = 0
– 9
i1 + 34i2 – 10isc = 0
– 3 – 10
i2 + 20isc = 0

Solving using MATLAB:

>> e1 = '-4 + 9*i1 - 9*i2 + 3 = 0';
>> e2 = '-9*i1 + 34*i2 - 10*isc = 0';
>> e3 = '-3 + 20*isc - 10*i2 = 0';
>> a = solv
e(e1,e2,e3,'i1','i2','isc');

we find
isc = 0.2125 A, so IN = 212.5 mA and VTH = INRTH = (0.2125)(16) = 3.4 V.

(b) Working with the Thévenin equivalent circuit, I
1Ω = VTH/(RTH + 1) = 200 mA.
Thus, P
1Ω = (0.2)2.1 = 40 mW.

Switching to the Nort on equivalent, we find I1Ω by current division:
I
1Ω = (0.2125)(16)/(16+1) = 200 mA. Once again, P 1Ω = 40 mW (as
expected).

(c) As we can see from simulating the original circuit simultaneously with its
Thevenin an
d Norton equivalents, the 1 Ω resistor does in fact dissipate 40 mW, and
either equivalent is equally applicable. Note all three SOURCES provide a different
amount of power in total.

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Engineering Circuit Analysis, 7
th
Edition Chapter Five Solutions 10 March 2006

45. (a) Removing terminal c, we need write only one nodal equation:

0.1 =
15
5V

12
2V
bb
−
+
−
, which may be solved to
yield Vb = 4 V. Therefore, Vab = VTH = 2 – 4
= -2 V.
R
TH = 12 || 15 = 6.667 Ω. We may then
calculate I
N as IN = VTH/ RTH

= -300 mA (arrow pointing upwards).

(b) Removing terminal
a, we again find RTH = 6.667 Ω, and only need write a single
nodal equation; in fact, it is identical to that written for the circuit above, and we once
again find that V
b = 4 V. In this case, VTH = Vbc = 4 – 5 = -1 V, so IN = -1/ 6.667
= –150 mA (arrow pointing upwards).
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Engineering Circuit Analysis, 7
th
Edition Chapter Five Solutions 10 March 2006

46. (a) Shorting out the 88-V source and open-circuiting the 1-A source, we see looking
into the terminals x and x' a 50-Ω resistor in para
llel with 10 Ω in parallel with
(20 Ω + 40 Ω), so
R
TH = 50 || 10 || (20 + 40) = 7.317 Ω

Using superposition to determine the voltage Vxx' across the 50-Ω resistor, we find

Vxx' = VTH =
⎥
⎦
⎤
⎢
⎣
⎡
++
+
⎥
⎦
⎤
⎢
⎣
⎡
++
+
)10||50(2040
40
10)||(1)(50
)]4020(||50[10
)4020(||50
88

=
⎥
⎦
⎤
⎢
⎣
⎡
++
+
⎥
⎦
⎤
⎢
⎣
⎡
333.82040
40
(1)(8.333)
27.37
27.27
88
= 69.27 V

(b) Shorting out the 88-V source and open-circuiting the 1-A source, we see looking

into the terminals y and y' a 40-Ω resistor in para
llel with [20 Ω + (10 Ω || 50 Ω)]:

RTH = 40 || [20 + (10 || 50)] = 16.59 Ω

Using superposition to determine the voltage Vyy' across the 1-A source, we find

Vyy' = VTH = (1)(RTH) +
⎟
⎠
⎞
⎜
⎝
⎛
+
⎥
⎦
⎤
⎢
⎣
⎡
+ 4020
40

27.2710
27.27
88

= 59.52 V
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Engineering Circuit Analysis, 7
th
Edition Chapter Five Solutions 10 March 2006

47. (a) Select terminal b as the reference terminal, and define a nodal voltage V1 at the
top of the 200-Ω resistor. Then,

0 =
200
V
100
VV
40
20V
1TH11
+
−
+
−
[1]

1.5 i1 = (VTH – V1)/ 100 [2]

where i1 = V1/ 200, so Eq. [2] becomes 150 V1/ 200 + V1 - VTH = 0 [2]

Simplifying and collecting terms, these equations may be re-written as:

(0.25 + 0.1 + 0.05) V1 – 0.1 VTH = 5 [1]
(1 + 15/ 20) V
1 – VTH = 0 [2]

Solving, we find that VTH = 38.89 V. To find RTH, we short the voltage source and
inject 1 A into the port:
0 =
200
V
40
V
100
VV
11in1
++
−
[1]
1.5 i1 + 1 =
100
VV
1in−
[2]

i1 = V1/ 200 [3] 1 A
+

V
in

-
V
1
Ref.

Combining Eqs. [2] and [3] yields
1.75 V 1 – Vin = -100 [4]

Solving Eqs. [1] & [4] then results in V
in = 177.8 V, so that RTH = Vin/ 1 A = 177.8 Ω .

(b) Adding a 100-Ω loa
d to the original circuit or our Thévenin equivalent, the
voltage across the load is
V
100Ω =
V 14.00
177.8100
100
V
TH
=⎟
⎠
⎞
⎜
⎝
⎛
+ , and so P100Ω = (V100Ω)
2
/ 100 = 1.96 W.
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Engineering Circuit Analysis, 7
th
Edition Chapter Five Solutions 10 March 2006

48. We inject a current of 1 A into the port (arrow pointing up), select the bottom terminal
as our reference terminal, and define the nodal voltage V
x across the 200-Ω resistor.

Then, 1 = V 1/ 100 + (V1 – Vx)/ 50 [1]
-0.1 V
1 = Vx/ 200 + (Vx – V1)/ 50 [2]

which may be simplified to

3 V1 – 2 Vx = 100 [1]
16 V
1 + 5 Vx = 0 [2]

Solving, we find that V1 = 10.64 V, so RTH = V1/ (1 A) = 10.64 Ω.

Since there are no independent sources present in the original network, I N = 0.
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Engineering Circuit Analysis, 7
th
Edition Chapter Five Solutions 10 March 2006

49. With no independent sources present, VTH = 0.

We decide to inject a 1-A current into the port:

Ref.
vx
vf
•
A
Node ‘
x’: 0.01 vab = vx/ 200 + (vx – vf)/ 50 [1]
Supernode: 1 =
vab/ 100 + (vf – vx) [2]
and:
vab – vf = 0.2 vab [3]

Rearranging and collecting terms,

-2 vab + 5 vx – 4 vf = 0 [1]

vab – 2 vx + 2 vf = 100 [2]
0.8
vab - vf = 0 [3]

Solving, we find that vab = 192.3 V, so RTH = vab/ (1 A) = 192.3 Ω.
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Engineering Circuit Analysis, 7
th
Edition Chapter Five Solutions 10 March 2006

50. We first find RTH by shorting out the voltage source and open-circuiting the current
source.

Looking into the terminals
a & b, we see
R
TH = 10 || [47 + (100 || 12)]
= 8.523 Ω.

Returning to the original circuit, we decide to perform nodal analysis to obtain V TH:

-12×10
3
= (V1 – 12)/ 100×10
3
+ V1/ 12×10
3
+ (V1 – VTH)/ 47×10
3
[1]

12×10
3
= VTH/ 10×10
3
+ (VTH – V1)/ 47×10
3
[2]

Rearranging and collecting terms,

0.1146 V1 - 0.02128 VTH = -11.88 [1]
-0.02128 V
1 + 0.02128 VTH = 12 [2]

Solving, we find that VTH = 83.48 V.
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Engineering Circuit Analysis, 7
th
Edition Chapter Five Solutions 10 March 2006

51. (a) RTH = 4 + 2 || 2 + 10 = 15 Ω.
(b) same as above: 15 Ω.
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Engineering Circuit Analysis, 7
th
Edition Chapter Five Solutions 10 March 2006

52. For Fig. 5.78a, IN = 12/ ~0 → ∞ A in parallel with ~ 0 Ω.

For Fig. 5.78b, V
TH = (2)(~∞) → ∞ V in series with ~∞ Ω.
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Engineering Circuit Analysis, 7
th
Edition Chapter Five Solutions 10 March 2006

53. With no independent sources present, VTH = 0.

Connecting a 1-V source to the port and measuring the current that flows as a result,

I = 0.5 V
x + 0.25 Vx = 0.5 + 0.25 = 0.75 A.

R TH = 1/ I = 1.333 Ω.

The Norton equivalent is 0 A in parallel with 1.333 Ω .

+
-
← I

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Engineering Circuit Analysis, 7
th
Edition Chapter Five Solutions 10 March 2006

54. Performing nodal analysis to determine VTH,

100 ×10
-3
= Vx/ 250 + Voc/ 7.5×10
3
[1]

and V x – Voc = 5 ix

where ix = Vx/ 250. Thus, we may write the second equation as

0.98 V x – Voc = 0 [2]

Solving, we find that Voc = VTH = 23.72 V.

In order to determine RTH, we inject 1 A into the port:

V
ab/ 7.5×10
3
+ Vx/ 250 = 1 [1]

and V x – Vab = 5 ix = 5Vx/ 250 or

-V ab + (1 – 5/ 250) Vx = 0 [2]

Solving, we find that Vab = 237.2 V. Since RTH = Vab/ (1 A), RTH = 237.2 Ω.

Finally, IN = VTH/ RTH = 100 mA.
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Engineering Circuit Analysis, 7
th
Edition Chapter Five Solutions 10 March 2006

55. We first note that VTH = Vx, so performing nodal analysis,

-5 V x = Vx/ 19 which has the solution Vx = 0 V.

Thus, V TH (and hence IN) = 0. (Assuming RTH ≠ 0)

To find R TH, we inject 1 A into the port, noting that RTH = Vx/ 1 A:

-5 Vx + 1 = Vx/ 19

Solving, we find that Vx = 197.9 mV, so that RTH = RN = 197.9 mΩ.
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Engineering Circuit Analysis, 7
th
Edition Chapter Five Solutions 10 March 2006

56. Shorting out the voltage source, we redraw the circuit with a 1-A source in place of
the 2-kΩ resistor:

Noting that 300 Ω || 2 MΩ ≈ 300 Ω,

0 = ( vgs – V)/ 300 [1]

1 – 0.02 vgs = V/ 1000 + (V – vgs)/ 300 [2]

Simplifying & collecting terms,

vgs – V = 0 [1]

0.01667 vgs + 0.00433 V = 1 [2]

Solving, we find that vgs = V = 47.62 V. Hence, RTH = V/ 1 A = 47.62 Ω.

1 A
Ref.
V

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Engineering Circuit Analysis, 7
th
Edition Chapter Five Solutions 10 March 2006

57. We replace the source vs and the 300-Ω resistor with a 1-A source and seek its
voltage:

By nodal analysis, 1 = V1/ 2×10
6
so V1 = 2×10
6
V.

Since V = V 1, we have Rin = V/ 1 A = 2 MΩ.
1 A
+

V

-
Ref.
V
1 V2
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Engineering Circuit Analysis, 7
th
Edition Chapter Five Solutions 10 March 2006

58. Removing the voltage source and the 300-Ω resistor, we replace them with a 1-A
source and seek the voltage that develops across its term
inals:

1 A
+

V

-
Ref.

We select the bottom node as our reference terminal, and define nodal voltages V
1
and V2. Then,

1 = V 1 / 2×10
6
+ (V1 – V2)/ rπ [1]

0.02 vπ = (V2 – V1)/ rπ + V2/ 1000 + V2/ 2000 [2]

where vπ = V1 – V2

Simplifying & collecting terms,

(2 ×10
6
+ rπ) V1 – 2×10
6
V2 = 2×10
6
rπ [1]

-(2000 + 40 rπ) V1 + (2000 + 43 rπ) V2 = 0 [2]

Solving, we find that V1 = V =
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
++×
+
×
π
πr 14.33 666.7 102
r 14.33 666.7
102
6
6
.

Thus, R TH = 2×10
6
|| (666.7 + 14.33 rπ) Ω.

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Engineering Circuit Analysis, 7
th
Edition Chapter Five Solutions 10 March 2006

59. (a) We first determine vout in terms of vin and the resistor values only; in this case,
V
TH = vout. Performing nodal analysis, we write two equations:

() ()
1
0
din dod
if
vv vvv
RR R
−− −−−
=+ +
[1] and
( )()
0
od o d
fo
vv vAv
RR
+−
=+
[2]

Solving using MATLAB, we obtain:

>> e1 = 'vd/Ri + (vd + vin)/R1 + (vd + vo)/Rf = 0';
>> e2 = '(vo + vd)/Rf + (vo – A*vd)/Ro = 0';
>> a = solve(e1,e2,'vo','vd');
>> pretty(a.vo)

Ri vin (–Ro + Rf A)
– -------------------------------------
----------
R1 Ro + Ri Ro + R1 Rf + Ri Rf + R1 Ri
+ A R1 Ri

Thus, V
TH =
11 1 1
()
in i o f
oio f if i i
vRR ARRRRRRRRRRRARR
−
+++++
, which in the limit of A → ∞,
approaches –R
f/R1.

To find R
TH, we short out the independent source vin, and squirt 1 A into the terminal
marked
vout, renamed VT. Analyzing the resulting circuit, we write two nodal
equations:

( )
1
0
dTdd
if
vvvv
RR R
−−−
=−+
[1] and
( )( )
1 [2]
Td T d
fo
vv vAv
RR
+−
=+

Solving using MATLAB:

>> e1 = 'vd/R1 + vd/Ri + (vd + VT)/Rf = 0';
>> e2 = '1 = (VT + vd)/Rf + (VT - A*vd)/Ro';
>> a = solve(e1,e2,'vd','
VT');
>> pretty(a.VT)

Ro (Ri Rf + R1 Rf + R1 Ri)
------------------------------------------------
--------------
Ri Ro + R1 Ro + Ri
Rf + R1 Rf + R1 Ri + A R1 Ri

Since V
T/1 = VT, this is our Thévenin equivalent resistance (R TH).
+
vout
–
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Engineering Circuit Analysis, 7
th
Edition Chapter Five Solutions 10 March 2006

60. Such a scheme probably would lead to maximum or at least near-maximum power
transfer to our home. Since we pay the utility comp
any based on the power we use,
however, this might not be such a hot idea…
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Engineering Circuit Analysis, 7
th
Edition Chapter Five Solutions 10 March 2006

61. We need to find the Thévenin equivalent resistance of the circuit connected to R L, so
we short the 20-V source and open-circuit the 2-A source; by inspection, then

RTH = 12 || 8 + 5 + 6 = 15.8 Ω

Analyzing the original circuit to obtain V1 and V2 with RL removed:

V
1 = 20 8/ 20 = 8 V; V2 = -2 (6) = -12 V.

We define V
TH = V1 – V2 = 8 + 12 = 20 V. Then,

P
R
L|max =
W6.329
4(15.8)
400

R 4
V
L
2
TH
==
V1
V2
+ VTH -
Ref.
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Engineering Circuit Analysis, 7
th
Edition Chapter Five Solutions 10 March 2006

62. (a) RTH = 25 || (10 + 15) = 12.5 Ω

Using superposition, Vab = VTH =
50
1015
100
251015
25
50
+
+
++
= 75 V.

(b) Connecting a 50-Ω resistor,

P load = W90
50 12.5
75

R R
V
2
loadTH
2
TH
=
+
=
+

(c) Connecting a 12.5-Ω resistor,

P load =
()
W112.5
12.5 4
75

R 4
V
2
TH
2
TH
==
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Engineering Circuit Analysis, 7
th
Edition Chapter Five Solutions 10 March 2006

63. (a) By inspection, we see that i10 = 5 A, so
V
TH = Vab = 2(0) + 3 i10 + 10 i10 = 13 i10 = 13(5) = 65 V.

To find R
TH, we open-circuit the 5-A source, and connect a 1-A source between
terminals
a & b:

A simple KVL equation yields V
x = 2(1) + 3i10 + 10 i10.
Since
i10 = 1 A in this circuit, Vx = 15 V.

We thus find the Thevenin equivalent resistance is 15/1 = 15 Ω.

(b) P
max =
22
TH
TH
V65
70.42 W
4 R 4(15)
==
+

V
x

–
1 A

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Engineering Circuit Analysis, 7
th
Edition Chapter Five Solutions 10 March 2006

64.
(a) Replacing the resistor R
L with a 1-A source, we seek the voltage that develops across
its terminals with the independent voltage source shorted:

11 1
11
1
1
10 20 40 0 [1] 30 20 0 [1]
and 1 [2] 1 [2]
Solving, 400mA
So 40 16V and 16
1A
−+ + = ⇒ + =
−= ⇒ −=
=
== ==Ω
xx
xx
TH
iii ii
ii ii
i
V
Vi R

(b) Removing the resistor RL from the original circuit, we seek the resulting open-circuit
voltage:
+

VTH

-

1
110 50
0[ 1]
20 40
50
where
40
150 5
so [1] becomes 0
20 2 40 40
50
0
20 80
04 50
550
or 10V
−−
=+
−
=
−−⎛⎞⎛⎞
=− +
⎜⎟⎜⎟
⎝⎠⎝⎠
−
=+
=+−
=
=
TH TH
TH
TH TH TH
TH TH
TH TH
TH
TH
ViV
V
i
VV V
VV
VV
V
V
0

Thus, if
16 ,
5V
2
= =Ω
==
+
L
LTH
LT H
RTH
LTH
RR
RV
VV
RR

=
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Engineering Circuit Analysis, 7
th
Edition Chapter Five Solutions 10 March 2006

65.

(a) 2.5A=
N
I

2
20 80
2A
=
=
i
i

R 20N Ω
↓
2 A

By current division,

22.5
20
=
+
N
N
R
R

Solving, 80==
NTH
RR Ω
Thus, 2.5 80 200V== ×=
TH OC
VV

(b)
2 2
max
200
125W
80
=
44
==
×
TH
THV
P
R
Ω

(c)
80==
LTH
RR

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Engineering Circuit Analysis, 7
th
Edition Chapter Five Solutions 10 March 2006

66.

By Voltage ÷,

=
+
N
RN
N
R
II
RR

So

Solving, and 1.7A=
N
I 33.33
= Ω
N
R

(a) If

(b) If

(c) If

is a maximum,
33.33
33.33
1.7 850mA
33.33 33.33
33.33 28.33V
== Ω
=× =
+
==
LL
LN
L
LLvi
RR
i
vi
iis a maximum
; ma
x when 0
So 1.7A
0V
== Ω
+
=
=
L
N
LN L
NL
L
L
R
ii R
RR
i
v
0.2= [1]
250
0.5 [2]
80
+
=
+
N
N
N
N
N
N
R
I
R
R
I
R
10W to 250 corresp to 200mA.
20W to 80 corresp to 500mA.
RN R

IN
Ω
Ω
is av maximum
()
So is a maximum when is a maximum, which occurs at .
Then 0 and 1.7 56.66V
=
=∞
==×=
L
LNNL
LN L L
LLN
VIRR
vR R R
ivR
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Engineering Circuit Analysis, 7
th
Edition Chapter Five Solutions 10 March 2006

67. There is no conflict with our derivation concerning maximum power. While a dead
short across the battery terminals will indeed res
ult in maximum current draw from
the battery, and power is indeed proportional to i
2
, the power delivered to the load is
i
2
RLOAD = i
2
(0) = 0 watts. This is the minimum, not the maximum, power that the
battery can deliver to a load.
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Engineering Circuit Analysis, 7
th
Edition Chapter Five Solutions 10 March 2006

68. Remove R E: =
TH E inRRR
bottom node:
3
3
1310 [1]
300 70 10
− ππ
π −−
−× = +
×Vv Vv
v

at other node:
33
0
10 10 300 70 10
ππ π
− −
=++
××vvVv
V
1 A
V
[2]

Simplifying and collecting terms,

210×10
5
= 70×10
3
V + 300V + 63000 v π – 70x10
3
vπ - 300 v π

35
33
33
or 70.3 10 7300 210 10 [1]
0 2100 70 10 70 10 300 300
or 69.7 10 72.4 10 0 [2]
solving, 331.9 So 331.9
π
ππ π
π×− =×
=+×−×+−
−× +× =
==Ω
TH E
Vv
vvVv
Vv
VVRR
V
2
i
2
0

Next, we determine v TH using mesh analysis:

33
12
70.3 10 70 10 0 [1]−+ × − × =
s
vii

33
213
80 10 70 10 0 [2]×−×+=
E
iiRi
and:
3
32
310 [3]
−
π
−=×ii v
or ii
33
32
3 10 (10 10 )
−
−=× ×
or
32 30−=ii i
or

2331 0 [3]−+=ii

Solving :
33
1
33
2
3
70.3 10 70 10 0
70 10 80 10 0
0311
⎡⎤×−×
⎡⎤⎡⎤
⎢⎥ ⎢⎥⎢⎥
−× × =
⎢⎥ ⎢⎥⎢⎥
⎢⎥ ⎢⎥⎢⎥−
⎣⎦⎣⎦⎣⎦
s
E
iv
Ri
i

We seek i
3:

3
3 63
21.7 10
7.24 10 21.79 10
−×
=
×+ ×
s
Ev
i
R

So
3
3 63
21.7 10
7.24 10 21.79 10
−×
== =
×+ ×
E
OC TH E s
ER
VV Ri v
R
22
32
8 263
62
2
63 2
21.7 10 8
8
8 7.24 10 21.79 10
331.9
331.9
11.35 10 (331.9 )
(7.24 10 21.79 10 )
Ω
⎡⎤⎡ ⎤−×
==
⎢⎥⎢ ⎥
+×+×
⎡ ⎤⎣⎦⎣ ⎦
⎢ ⎥
+⎣ ⎦
×+
=
×+ ×
TH E
TH E
E
E
E
EVR
P
RR
R
R
R
vs
R
vs

This is maximized by setting
.=∞
E
R
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Engineering Circuit Analysis, 7
th
Edition Chapter Five Solutions 10 March 2006

69. Thévenize the left-hand network, assigning the nodal voltage V x at the free end of
right-most 1-k Ω re
sistor.
A single nodal equation:
3
3
40 10
710
−
×=
×
x
ocV

So
280V==
TH x
ocVV
R
TH = 1 k + 7 k = 8 kΩ

Select R 1 = RTH = 8 kΩ.

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Engineering Circuit Analysis, 7
th
Edition Chapter Five Solutions 10 March 2006

1 MΩ
100 kΩ
850 MΩ
70.

6
65
1
56
2 6
66
3 6
1 850 0.1 851.1 10
10 10
117.5
10 850 10
99.87
851.1 10
850 10 10
998.7
851.1 10
=++=+ += ×
×
== =Ω
××
== = Ω
×
××
== = Ω
×
ABC
AB
BC
CADR R R
RR
R
DD
RR
Rk
D
RR
Rk
D

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Engineering Circuit Analysis, 7
th
Edition Chapter Five Solutions 10 March 2006

71.

R1 R2
R3

12 23 31
2
3
1
0.1 0.4 0.4 0.9 0.9 0.1
0.49
1.225
544.4m
4.9
=++
=×+×+×
=Ω
== Ω
== Ω
==Ω
A
B
C
NRRRRRR
N
R
R
N
R
R
N
R
R

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Engineering Circuit Analysis, 7
th
Edition Chapter Five Solutions 10 March 2006

72.

1
2:1 6 3 10
61 63 31
0.6, 1.8 , 0.3
10 10 10
:5 1 4 10
51 14 54
0.5 , 0.4 , 2
10 10 10
1.8 2 0.5 4.3
0.3 0.6 0.4 1.3
1.3 4.3 0.9982
0.9982 0.6 2 3.598
3.598 6 2.249
Δ++=Ω
×× ×
===
Δ++=Ω
×× ×
== =
=Ω
=
Ω
=Ω
++= Ω
=Ω
++
++

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Engineering Circuit Analysis, 7
th
Edition Chapter Five Solutions 10 March 2006

73.

2
2
622336 36
36 36 36
6, 18, 12
62 3
12 4 3 , 6 12 4
431825
18
32.16
25
18
42.88
25
3
40.48
25
9.48 2.16 9.48 2.88 2.88 2.16 54
54 54
18.75 5.696
2.88 9.48
54
25
2.16
75 18.75 15
100 25 20
(15 20) 5.696
×+×+×= Ω
=Ω = Ω = Ω
=Ω Ω=Ω
++ = Ω
×= Ω
×= Ω
×= Ω
×+×+×=Ω
=Ω =Ω
=Ω
=Ω
=Ω
+= 4.899
5 4.899 9.899
Ω
∴=+ = Ω
in
R

96

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Engineering Circuit Analysis, 7
th
Edition Chapter Five Solutions 10 March 2006

74. We begin by converting the Δ-connected network consisti ng of the 4-, 6-, and 3-Ω
resistors to an equivalent Y-connected network:

RA
RB
RC

1
2
3
64313
64
1.846
13
43
923.1m
13
36
1.385
13
=++= Ω
×
===Ω
×
===
×
===Ω
AB
BC
CA
D
RR
R
D
RR
R
D
RR
R
D
Ω

Then network becomes:

1.846 Ω 0.9231 Ω
1.385 Ω

Then we may write

12 [13.846 (19.385 6.9231)]
7.347
=+
=Ω
inR

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Engineering Circuit Analysis, 7
th
Edition Chapter Five Solutions 10 March 2006

75.
1
2
3
112 4
12 1
42
21 1
42
11
0.25
4
++=Ω
×
= =Ω
×
= =Ω
×
= =Ω
R
R
R

0.5

0.5
0.25

Next, we convert the Y-connected network on the left to a Δ-connected network:

2
1 0.5 0.5 2 2 1 3.5×+×+×=Ω

3.5
7
0.5
3.5
1.75
2
3.5
3.5
1
==Ω
==
==Ω
A
B
CR
R
R
Ω

After this procedure, we have a 3.5-Ω resistor in parallel with the 2.5-Ω resistor. Replacing
them with a 1.458-
Ω resistor, we may redraw the circuit:

7
1.75 0.25
1.458

This circuit may be easily analysed to find:
12 1.458
5.454V
1.75 1.458
0.25 1.458 1.75
1.045
×
==
+
=+
=Ω
oc
THV
R

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Engineering Circuit Analysis, 7
th
Edition Chapter Five Solutions 10 March 2006

76. We begin by converting the Y-network to a Δ-connected network:

2
1.1 1.1 1.1 3
3
3
1
3
3
1
3
3
1
=++=Ω
==Ω
==Ω
==Ω
A
B
C
N
R
R
R

3 3
3 Ω

Next, we note that 13 0.75=Ω , and hence have a simple Δ-network. This is easily
converted to a Y-connected network:

1
2
3
0.75 3 3 6.75
0.75 3
0.3333
6.75
33
1.333
6.75
30.75
0.3333
6.75
++= Ω
×
== Ω
×
== Ω
×
==
Ω
R
R
R

1 A

1.333 0.3333
1.667
1/3
1
1/3 1 1/3
11
131 5
0.2A
200mA
= +
=Ω
==×
++
==
++
=
=
N
NSC
R
II

Analysing this final
circuit,

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Engineering Circuit Analysis, 7
th
Edition Chapter Five Solutions 10 March 2006

77. Since 1 V appears across the resistor associated with I1, we know that I1 = 1 V/ 10 Ω
= 100 mA. From the perspective of the open term inals, the 10-Ω resisto
r in parallel
with the voltage source has no influence if we replace the “dependent” source with a
fixed 0.5-A source:

Then, we may write:
-1 + (10 + 10 + 10) i
a – 10 (0.5) = 0
so that i
a = 200 mA.

We next find that VTH = Vab = 10(-0.5) + 10(i a – 0.5) + 10(-0.5) = -13 V.

To determine R TH, we first recognise that with the 1-V source shorted, I 1 = 0 and
hence the dependent current source is dead. Thus, we ma
y write R TH from inspection:

RTH = 10 + 10 + 10 || 20 = 26.67 Ω.
0.5 A

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Engineering Circuit Analysis, 7
th
Edition Chapter Five Solutions 10 March 2006

78. (a) We begin by splitting the 1-kΩ resistor into two 500-Ω resistors in series. We then
have two related Y-connected networks, each with a 500-Ω resis
tor as a leg.
Converting those networks into Δ -connected networks,

Σ = (17)(10) + (1)(4) + (4)(17) = 89×106 Ω
2

89/0.5 = 178 kΩ; 89/ 17 = 5.236 kΩ; 89/4 = 22.25 kΩ

Following this conversion, we find that we have two 5.235 kΩ resistors in parallel,
and a 178-kΩ resis
tor in parallel with the 4-kΩ resistor. Noting that 5.235 k || 5.235 k
= 2.618 kΩ and 178 k || 4 k = 3.912 kΩ, we may draw the circuit as:

We next attack the Y-connected network in the center:

178 kΩ
233.6 kΩ
27.49 kΩ
27.49 kΩ
3.912 kΩ

Σ = (22.25)(22.25) + (22.25)(2.618) + (2.618)(22.25) = 611.6×106 Ω
2

611.6/ 22.25 = 27.49 kΩ; 611.6/ 2.618 = 233.6 kΩ

Noting that 178 k || 27.49 k = 23.81 kΩ and 27.49 || 3.912 = 3.425 kΩ, we are left
with a simple Δ -connected network. To convert this to the requested Y-network,

312.6 Ω
21.33 kΩ 3.068 kΩ
Σ = 23.81 + 233.6 + 3.425 = 260.8 kΩ

(23.81)(233.6)/ 260.8 = 21.33 kΩ
(233.6)(3.425)/ 260.8 = 3.068 kΩ
(3.425)(23.81)/ 260.8 = 312.6 Ω

(b)

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Engineering Circuit Analysis, 7
th
Edition Chapter Five Solutions 10 March 2006

79. (a) Although this network may be simplified, it is not possible to replace it with a
three-resistor equivalent.

(b) See (a).

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Engineering Circuit Analysis, 7
th
Edition Chapter Five Solutions 10 March 2006

80. First, replace network to left of the 0.7-V source with its Thévenin equivalent:

15
20 2.609V
100 15
100 15 13.04
=× =
+
== Ω
kk
TH
THV
Rk

Redraw:

Analysing the new circuit to find I
B, we note that I C = 250 I B:

3
3
4
2.609 13.04 10 0.7 5000( 250 ) 0
2.609 0.7
1.505 A
13.04 10 251 5000
250 3.764 10 A
376.4 A
−
−+×++ + =
−
== μ
×+×
==×
=μ
BB
B
CB
II
I
II
B
I

2.609 V
13.04 kΩ

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Engineering Circuit Analysis, 7
th
Edition Chapter Five Solutions 10 March 2006

81. (a) Define a nodal voltage V1 at the top of the current source IS, and a nodal voltage
V
2 at the top of the load resistor RL. Since the load resistor can safely dissipate 1 W,
and we know that
P
R
L =
1000
V
2
2

then
V 31.62 V
max
2
= . This corresponds to a load resistor (and hence lamp) current
of 32.62 mA, so we may treat the lamp as a 10.6-Ω resistor.

Proceeding with nodal analysis, we may write:

IS = V1/ 200 + (V1 – 5 Vx)/ 200 [1]

0 = V2/ 1000 + (V2 – 5 Vx)/ 10.6 [2]

Vx = V1 – 5 Vx or Vx = V1/ 6 [3]

Substituting Eq. [3] into Eqs. [1] and [2], we find that

7 V1 = 1200 IS [1]
-5000 V
1 + 6063.6 V2 = 0 [2]

Substituting
V 31.62 V
max
2
= into Eq. [2] then yields V1 = 38.35 V, so that

IS|
max
= (7)(38.35)/ 1200 = 223.7 mA.

(b) PSpice verification.

The lamp current does not exceed 36 mA in the range of operation allowed
(i.e. a load power of < 1 W.) The simulation result shows that the load will dissipate
slightly more than 1 W for a source current magnitude of 224 mA, as predicted by
hand analysis.
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Engineering Circuit Analysis, 7
th
Edition Chapter Five Solutions 10 March 2006

82. Short out all but the source operating at 10
4
rad/s, and define three clockwise mesh
currents i
1, i2, and i 3 starting with the left-most mesh. Then

608 i 1 – 300 i 2 = 3.5 cos 10
4
t [1]
-300 i
1 + 316 i 2 – 8 i 3 = 0 [2]
-8 i
2 + 322 i 3 = 0 [3]

Solving, we find that i 1(t) = 10.84 cos 10
4
t mA
i
2(t) = 10.29 cos 10
4
t mA
i
3(t) = 255.7 cos 10
4
t μA

Next, short out all but the 7 sin 200t V source, and and define three clockwise mesh
currents i
a, ib, and i c starting with the left-most mesh. Then

608 i a – 300 i b = -7 sin 200t [1]
-300 i
a + 316 i b – 8 i c = 7 sin 200t [2]
-8 i
b + 322 i c = 0 [3]

Solving, we find that i a(t) = -1.084 sin 200t mA
i
b(t) = 21.14 sin 200t mA
i
c(t) = 525.1 sin 200t μA

Next, short out all but the source operating at 10
3
rad/s, and define three clockwise
mesh currents i
A, iB, and i C starting with the left-most mesh. Then

608 i A – 300 i B = 0 [1] B
-300 i A + 316 i B – 8 iB C = 0 [2]
-8 i
B + 322 i C = -8 cos 10
4
t [3]

Solving, we find that i A(t) = -584.5 cos 10
3
t μA
i
B(t) = -1.185 cos 10t mA B
3
i C(t) = -24.87 cos 10
3
t mA

We may now compute the power delivered to each of the three 8-Ω speakers:

p
1
= 8[i 1 + ia + iA]
2
= 8[10.84× 10
-3
cos 10
4
t -1.084× 10
-3
sin 200t -584.5×10
-6
cos 10
3
t]
2
p2 = 8[i 2 + ib + iB]
2
= 8[10.29× 10
-3
cos 10
4
t +21.14× 10
-3
sin 200t –1.185× 10
-3
cos 10
3
t]
2
p3 = 8[i 3 + ic + iC]
2
= 8[255.7× 10
-6
cos 10
4
t +525.1×10
-6
sin 200t –24.87× 10
-3
cos 10
3
t]
2

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Engineering Circuit Analysis, 7
th
Edition Chapter Five Solutions 10 March 2006

83. Replacing the DMM with a possible Norton equivalent (a 1-MΩ resistor in parallel
with a 1-A source):

1 A
V
in

We begin by noting that 33 Ω || 1 MΩ ≈ 33
Ω. Then,

0 = (V 1 – Vin)/ 33 + V1/ 275×10
3
[1]
and
1 - 0.7 V
1 = Vin/ 10
6
+ Vin/ 33×10
3
+ (Vin – V1)/ 33 [2]

Simplifying and collecting terms,

(275× 10
3
+ 33) V1 - 275× 10
3
Vin = 0 [1]

22.1 V
1 + 1.001 Vin = 33 [2]

Solving, we find that V
in = 1.429 V; in other words, the DMM sees 1.429 V across its
terminals in response to the known current of 1 A it’s supplying. It therefore thinks
that it is connected to a resistance of 1.429 Ω.

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Engineering Circuit Analysis, 7
th
Edition Chapter Five Solutions 10 March 2006

84. We know that the resistor R is absorbing maximum power. We might be tempted to
say that the resistance of the cylinder is therefore 10 Ω, but this is wrong: The larger
we make the cylinder res
istance, the small the power delivery to R:

P
R = 10 i
2
= 10
2
10
120
⎥
⎥
⎦
⎤
⎢
⎢
⎣
⎡
+
cylinder
R

Thus, if we are in fact delivering the maximum possible power to the resistor from
the
120-V source, the resistance of the cylinder must be zero.

This corresponds to a temperature of absolute zero using the equation given.
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Engineering Circuit Analysis, 7
th
Edition Chapter Five Solutions 10 March 2006

85. We note that the buzzer draws 15 mA at 6 V, so that it may be modeled as a 400-Ω
resistor. One possible solution of many, then, is:

Note: construct the 18-V source from 12 1.5-V batteries in series, and the two 400-Ω
resistors can be fabricated by soldering 400 1-Ω resistors in series, although there’s
probably a much better alternative…
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Engineering Circuit Analysis, 7
th
Edition Chapter Five Solutions 10 March 2006

86. To solve this problem, we need to assume that “45 W” is a designation that applies
when 120 Vac is applied directly to a particular lamp.
This corresponds to a current
draw of 375 mA, or a light bulb resistance of 120/ 0.375 = 320 Ω.

Original wiring scheme New
wiring scheme

3

In the original wiring scheme, Lamps 1 & 2 draw (40)2 / 320 = 5 W of power each,
and Lamp 3 draws (80)2 / 320 = 20 W of power. Therefore, none of the lamps is
running at its maximum rating of 45 W. We require a circuit which will deliver the
same intensity after the lamps are reconnected in a Δ configuration. Thus, we need a
total of 30 W from the new network of lamps.

There are several ways to accomplish this, but the simplest may be to just use one
120-Vac source connected to the left port in series with a resistor whose value is
chosen to obtain 30 W delivered to the three lamps.

In other words,

30
320
213.3Rs
213.3
06
2
320
213.3Rs
213.3
120
22
=
⎥
⎦
⎤
⎢
⎣
⎡
+
+
⎥
⎦
⎤
⎢
⎣
⎡
+

Solving, we find that we require Rs = 106.65 Ω , as confirmed by the PSpice
simulation below, which shows that both wiring configurations lead to one lamp with
80-V across it, and two lamps with 40 V across each.

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Engineering Circuit Analysis, 7
th
Edition Chapter Five Solutions 10 March 2006

87.
Maximum current rating for the LED is 35 mA. •

•
•
•
Its resistance can vary between 47 and 117 Ω.
A 9-V battery must be used as a power source.
Only standard resistance values may be used.

One possible current-limiting scheme is to connect a 9-V battery in series with a
resistor R
limiting and in series with the LED.
From KVL,

LEDlimitingR R
9
+
I
LED =

The maximum value of this current will occur at the minimum LED resistance, 47 Ω.
Thus, we solve
35×10
-3
=
47 R
9
limiting+

to obtain Rlimiting ≥ 210.1 Ω to ensure an LED current of less than 35 mA. This is not a
standard resistor value, however, so we select

Rlimiting = 220 Ω .

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Engineering Circuit Analysis, 7
th
Edition Chapter Six Solutions 10 March 2006

1. This is an inverting amplifier, therefore,
in
f
out
V
R
R
V
1
−=
So:

a) VV
out 303
10
100
−=×−=
b) V
M
M
V
out 5.25.2
1
1
−=×−= c) VV
out 42.11
3.3
7.4
=−×−=

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Engineering Circuit Analysis, 7
th
Edition Chapter Six Solutions 10 March 2006

2. This is also an inverting amplifier. The loading resistance R s only affects the output
current drawn from the op-amp. Therefore,

a) VV
out 05.75.1
10
47
−=×−=
b) VV
out9=
c) mVV
out680−=
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Engineering Circuit Analysis, 7
th
Edition Chapter Six Solutions 10 March 2006

3. For this inverting amplifier,
ininoutvv
k
k
v 10
1
10
−=×−=. Therefore,
a) tvv
inout 5sin2010−
=−=

-25
-20
-15
-10
-5
0
5
10
15
20
25

2π / 5 6π / 5 4π /5 2π 8π / 5

b) tvv
inout 5sin51010
−−=−=

-16
-14
-12
-10
-8
-6
-4
-2
0

-5 V
-10 V
-15 V
2π / 5 6π / 5 4π /5 2π 8π / 5

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Engineering Circuit Analysis, 7
th
Edition Chapter Six Solutions 10 March 2006

4. For this inverting amplifier,
inin
f
outvv
R
R
v 1.0
1
−=−=, hence,
a) tvv
inout 4cos1.0−=−=

-1.5
-1
-0.5
0
0.5
1
1.5

b) tvv
inout 4cos4.05.11.0−−=−=

-2
-1.8
-1.6
-1.4
-1.2
-1
-0.8
-0.6
-0.4
-0.2
0

π / 2 3π / 2 π 2π
π / 2 3π / 2 π 2π
-1.5 V
-1.9 V
-1.1 V
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Engineering Circuit Analysis, 7
th
Edition Chapter Six Solutions 10 March 2006

5. One possible solution is by using an inverting amplifier design, we have

in
in
f
outV
R
R
V −=
⇔
5
9
=−=
in
out
in
f
V
V
R
R

Using standard resistor values, we can have R
f=9.1kΩ and R in=5.1kΩ

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Engineering Circuit Analysis, 7
th
Edition Chapter Six Solutions 10 March 2006

6. One possible solution is by using an inverting amplifier design, and a -5V input to
give a positive output voltage:

The resistance values are given by:
5
20
=
in
f
R
R

Giving possible resistor values R
f = 20 kΩ and R in = 5.1 kΩ

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Engineering Circuit Analysis, 7
th
Edition Chapter Six Solutions 10 March 2006

7. To get a positive output that is smaller than the input, the easiest way is to use
inverting amplifier with an inverted voltage supply to give a negative voltage:

The resistances are given by:
5
5.1
=
in
f
R
R

Giving possible resistor values R
f = 1.5 kΩ and R in = 5.1 kΩ

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Engineering Circuit Analysis, 7
th
Edition Chapter Six Solutions 10 March 2006

8. Similar to question 7, the following is proposed:

The resistances are given by:
9
3
=
in
f
R
R

Giving possible resistor values R
f = 3.0kΩ and R in = 9.1 kΩ

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Engineering Circuit Analysis, 7
th
Edition Chapter Six Solutions 10 March 2006

9. This circuit is a non-inverting amplifier, therefore,
in
f
out
V
R
R
V )1(
1
+=
So:

a) V 71.1300)
10
47
1( =×+=mV
out
b) V 35.1)
1
1
1( =×+=
M
M
V
out
c)
V 42.21)
3.3
7.4
1( −=−×+=
outV

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Engineering Circuit Analysis, 7
th
Edition Chapter Six Solutions 10 March 2006

10. This is again a non-inverting amplifier. Similar to question 9, we have:

a) V 14.1)7.41(200=
+
×=mV
out
b) V 189)11(−

×+=
outV
c) V 78.01008.7=
×=mV
out

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Engineering Circuit Analysis, 7
th
Edition Chapter Six Solutions 10 March 2006

11.
ininoutvvv 2)
1
1
1( =+= for this non inverting amplifier circuit, therefore:
a) tvv
inout 10sin82==

-10
-8
-6
-4
-2
0
2
4
6
8
10

b) tvv
inout 10sin5.022+==

1
1.2
1.4
1.6
1.8
2
2.2
2.4
2.6

2π / 5 6π / 5 4π /5 2π 8π / 5
2 V
1.5 V
2.5 V
2π / 5 6π / 5 4π /5 2π 8π / 5
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Engineering Circuit Analysis, 7
th
Edition Chapter Six Solutions 10 March 2006

12.
inin
in
f
out
vv
R
R
v 5.1)1(=+= for this non inverting op-amp circuit. Hence,

a) tvv
inout 2cos35.1
==
-4
-3
-2
-1
0
1
2
3
4

π / 2 2π

b) tvv
inout 2cos5.165.1+==
0
1
2
3
4
5
6
7
8

6 V
5.5 V
7.5 V
π / 2 2π
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Engineering Circuit Analysis, 7
th
Edition Chapter Six Solutions 10 March 2006

13. The first step is to perform a simple source transf
ormation, so that a 0.15-V source in
series with a 150-Ω resistor is connected to the in
verting pin of the ideal op amp.

Then, v
out =
() V 2.2- 15.0
150
2200
=−
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Engineering Circuit Analysis, 7
th
Edition Chapter Six Solutions 10 March 2006

14. In order to deliver 150 mW to the 10-kΩ resistor, we need v out =
V. 38.73 )1010)(15.0(
3
=× Writing a nodal equation at the inverting input,
we find
1000
5
R
5
0
outv
−
+=

Using vout = 38.73, we find that R = 148.2 Ω .
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Engineering Circuit Analysis, 7
th
Edition Chapter Six Solutions 10 March 2006

15. Since the 670-Ω switch requires 100 mA to activate, the voltage delivered to it by our
op amp circuit must be (670)(0.1) = 67 V. The microphone acts as the input to the
circuit, and
provides 0.5 V. Thus, an amplifier circuit having a gain = 67/0.5 = 134 is
required.

One possible solution of many: a non-inverting op am
p circuit with the microphone
connected to the non-inverting input terminal, the switch connected between the op
amp output pin and ground, a feedback resistor R
f = 133 Ω , and a resistor R1 = 1 Ω .
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Engineering Circuit Analysis, 7
th
Edition Chapter Six Solutions 10 March 2006

16. We begin by labeling the nodal voltages v- and v+ at the inverting and non-inverting
input terminals, respectively. Since no current can fl
ow into the non-inverting input,
no current flows through the 40-kΩ resistor; hence,
v+ = 0. Therefore, we know that
v- = 0 as well.

Writing a single nodal equation at the non-inverting input then leads to
22000
) - (

100
) - (
0
out-S-vvvv
+=
or
22000
-

100
-
0
outSvv
+=
Solving,

vout = -220 vS
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Engineering Circuit Analysis, 7
th
Edition Chapter Six Solutions 10 March 2006

17. We first label the nodal voltage at the output pin Vo. Then, writing a single nodal
equation at the inverting input terminal of the op amp,

17000
V-4

1000
3-4
0
o
+=

Solving, we find that Vo = 21 V. Since no current can flow through the 300-kΩ
resistor, V
1 = 21 as well.
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Engineering Circuit Analysis, 7
th
Edition Chapter Six Solutions 10 March 2006

18. A source transformation and some series combinations are well worthwhile prior to
launching into the analysis. With 5 kΩ || 3 kΩ = 1.875 kΩ and (1 m
A)(1.875 kΩ) =
1.875 V, we may redraw the circuit as

Ω
Ω
V2

This is now a simple inverting amplifier with gain – R
f/ R1 = -75.33/ 1.975 = -38.14.

Thus, V 2 = -38.14(3.975) = -151.6 V.
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Engineering Circuit Analysis, 7
th
Edition Chapter Six Solutions 10 March 2006

19. This is a simple inverting amplifier, so we may write

()( ) V 3sin 14- 3sin22
1000
2000-

out ttv+=+=
vout(t = 3 s) = -5.648 V.

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Engineering Circuit Analysis, 7
th
Edition Chapter Six Solutions 10 March 2006

20. We first combine the 2 MΩ and 700 kΩ resistors into a 518.5 kΩ resistor.

We are left with a simple non-inverting amplifier having a gain of
1 + 518.5/ 250 = 3.074. Thus,

vout = (3.074) vin = 18 so vin = 5.856 V.
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Engineering Circuit Analysis, 7
th
Edition Chapter Six Solutions 10 March 2006

21. This is a simple non-inverting amplifier circuit, and so it has a gain of 1 + R f/ R1.
We want
vout = 23.7 cos 500t V when the input is 0.1 cos 500t V, so a gain of 23.7/0.1
= 237 is required.
One possible solution of many: R
f = 236 kΩ and R 1 = 1 kΩ .
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Engineering Circuit Analysis, 7
th
Edition Chapter Six Solutions 10 March 2006

22. Define a nodal voltage V- at the inverting input, and a nodal voltage V+ at the non-
inverting input. Then,
At the non-inverting input: -3×10
-6
=
6
101.5
V
×
+
[1]

Thus, V + = -4.5 V, and we therefore also know that V- = -4.5 V.

At the inverting input:
7
out-
6
-R
V - V

R
V
0+= [2]

Solving and making use of the fact that V- = -4.5 V,

vout =
() V 1
R
R
4.5- 4.5 - 5.4
R
R
6
7
6
7
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
+=−

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Engineering Circuit Analysis, 7
th
Edition Chapter Six Solutions 10 March 2006

23. (a) B must be the non-inverting input: that yields a gain of 1 + 70/10 = 8 and an
output of 8 V for a 1-V input.

(b) R
1 = ∞, R A = 0. We need a gain of 20/10 = 2, so choose R2 = RB = 1 Ω .

(c)
A is the inverting input since it has the feedback connection to the output pin.
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Engineering Circuit Analysis, 7
th
Edition Chapter Six Solutions 10 March 2006

24. It is probably best to first perform a simple source transformation:

(1 mA)(2 kΩ ) = 2 V.

Since no current can flow into the non-inverting input pin, we
know that V + = 2 V,
and therefore also that V
- = 2 V. A single nodal equation at the inverting input yields:

13000
-2

1000
3-2
0
outv
+=
which yields
vout = -11 V.
2 V
3 V
vout
13 k
Ω
1 kΩ
2 kΩ
V
+
V-
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Engineering Circuit Analysis, 7
th
Edition Chapter Six Solutions 10 March 2006

25. We begin by find the Thévenin equivalent to the left of the op amp:

Vth = -3.3(3) vπ = -9.9 vπ =
1100
1000
9.9
Sv
− = -9 vS

R th = 3.3 kΩ, so we can redraw the circuit as:

which is simply a classic inverting op amp circuit with gain of -100/3.3 = -30.3.

Thus, vout = (-30.3)( -9 vS) = 272.7 vS

For vS = 5 sin 3t mV, vout = 1.364 sin 3t V, and vout(0.25 s) = 0.9298 V.
vout
-9 vS
3.3 kΩ
100 kΩ
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Engineering Circuit Analysis, 7
th
Edition Chapter Six Solutions 10 March 2006

26. We first combine the 4.7 MΩ and 1.3 kΩ resistors: 4.7 MΩ || 1.3 kΩ = 1.30 kΩ.
Next, a source transformation yields (3×10
-6
)(1300) = 3.899 mV which appears in
series with the 20 mV source and the 500-Ω resistor. Thus, we may redraw the circuit
as

23.899 mV
-6 V
370 Ω
37.7 kΩ
1.8 kΩ
vout
500 Ω

Since no current flows through the 1.8 kΩ resistor
, V + = 23.899 mV and hence
V
- = 23.899 mV as well. A single nodal equation at the inverting input terminal yields

3
out
-3-3
1037.7
- 1023.899

500
1023.899
0
×
×
+
×
= v
Solving,
vout = 1.826 V
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Engineering Circuit Analysis, 7
th
Edition Chapter Six Solutions 10 March 2006

27. We first combine the 4.7 MΩ and 1.3 kΩ resistors: 4.7 MΩ || 1.3 kΩ = 1.30 kΩ.
Next, a source transformation yields (27×10
-6
)(1300) = 35.1 mV which appears in
series with the 20 mV source and the 500-Ω resistor. Thus, we may redraw the circuit
as

Since no current flows through the 1.8 kΩ resistor
, V + = 55.1 mV and hence
V
- = 55.1 mV as well. A single nodal equation at the inverting input terminal yields

3
out
-3-3
1037.7
- 101.55

500
101.55
0
×
×
+
×
= v
Solving,
vout = 4.21 V
55.1 mV
-6 V
370 Ω
37.7 kΩ
1.8 kΩ
vout
500 Ω

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Engineering Circuit Analysis, 7
th
Edition Chapter Six Solutions 10 March 2006

28. The 3 mA source, 1 kΩ resistor and 20 kΩ resistor may be replaced with a –3 V
source (“+” reference up) in series with a 21 kΩ resistor. No current flows through
either
1 MΩ resistor, so that the voltage at each of the four input terminals is
identically zero. Considering each op amp circuit separately,

V 14.29
21
100
(-3)-
AMLEFTOPout
==
P
v
V 50-
10
100
(5)-
AMPOP RIGHout
==v

v
x =
PP
vv
AMOP RIGHoutAMLEFTOPout
- = 14.29 + 50 = 64.29 V.
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Engineering Circuit Analysis, 7
th
Edition Chapter Six Solutions 10 March 2006

29. A general summing amplifier with N input sources:

1. v
a = vb = 0
2. A single nodal equation at the inverting input leads to:

N
Na
2
2a
1
1a
f
aR
...
R

R

R
0
vvvvvvvv
out
−
++
−
+
−
+
−
=
Simplifying and making use of the fact that
va = 0, we may write this as
∏∏∏∏
====
+++=
⎥
⎦
⎤
⎢
⎣
⎡
−
N
1
i
N
N
N
1
i
2
2
N
1
i
1
1
N
1
i
f
R
R
... R
R
R
R
R
R
1
iii
out
i
vvv
v

or simply
N
N
2
2
1
1
fR
...
R

R

R
vvvv
out
+++=−
Thus,
v
out =
∑
=
N
1
fR
R-
i i
i
v

v
1
v2
vN
vout
RN
R1
R2
Rf
va
vb

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Engineering Circuit Analysis, 7
th
Edition Chapter Six Solutions 10 March 2006

30. A general difference amplifier:

Writing a nodal equation at the inverting input,
v1
v2
R1
R2
R3
R4
vout
f
a
1
1aR
-

R
-
0
outvvvv
+=
Writing a nodal equation at the non-inverting input,
2
2b
3
b
R
-

R
0
vvv
+=
Simplifying and collecting terms, we m
ay write
(R
f + R1) va – R1 vout = Rf v1 [1]
(R
2 + R3) vb = R3 v2 [2]

From Eqn. [2], we have v
b =
2
32
3
RR
R
v
+

Since va = vb, we can now rewrite Eqn. [1] as

( )
2
32
31f
1f1
RR
RRR
R Rvvv
out
+
+
−=−
and hence
2
32
1f
1
3
1
1
f

RR
RR

R
R

R
R
- vvv
out ⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
+
+
+=

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Engineering Circuit Analysis, 7
th
Edition Chapter Six Solutions 10 March 2006

31. In total darkness, the CdS cell has a resistance of 100 kΩ, and at a light intensity L of
6 candela it has a resistance of 6 kΩ. Thus, we m
ay compute the light-dependent
resistance (assuming a linear response in the range between 0 and 6 candela) as R
CdS =
-15L + 100 Ω.

Our design requirement (using the standard inverting op amp circuit shown) is that the
voltage across the load is 1.5 V at 2 candela, and less th
an 1.5 V for intensities greater
than 2 candela.

Thus, v out(2 candela) = -RCdS vS/ R1 = -70 VS/ R1 = 1.5 (R 1 in kΩ ).

Pick R 1 = 10 kΩ. Then v S = -0.2143 V.

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Engineering Circuit Analysis, 7
th
Edition Chapter Six Solutions 10 March 2006

32. We want Rf/ Rinstrument = 2K, and Rf/ Rvocal = 1K, where K is a constant not specified.
Assuming K = 1, one possible solution of ma
ny is:

v
out
vocals
microphone

instruments
microphone
R
vocal = 1
Ω
Rinstruments = 2 Ω
R
f = 2
Ω

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Engineering Circuit Analysis, 7
th
Edition Chapter Six Solutions 10 March 2006

33. One possible solution of many:

vS
vout
2 V
1 kΩ
99 kΩ
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Engineering Circuit Analysis, 7
th
Edition Chapter Six Solutions 10 March 2006

34. To get the average voltage value, we want
3
321vvv
v
out
++
= . This voltage stays
positive and therefore a one stage summing circuit (which inverts the voltage) is not
sufficient. Using the cascade setup as shown figure 6.15 and modified for three inputs
we have:

The nodal equation at the inverting input of the first op-amp gives
13
3
2
2
1
1 f
o
R
v
R
v
R
v
R
v
−
=++

If we assume R
1=R2=R3=R, then
R
vvv
Rv
f
321
0
1 ++
−=
Using the nodal equation at the inverting input of the second op-amp, we have:
R
vvv
R
R
R
v
R
v
fo
f
out 321
44
1
2
++
−
==
−

Or,
R
vvv
R
RR
v
ff
out
321
4
2
1 ++
=

For simplicity, we can take R
f2 = Rf1 = R4= Rx, then, to give a voltage average,

3
321321vvv
R
vvv
Rv
xout
++
=
++
=

I.e. R
x/R = 3. Therefore, the circuit can be completed with R 1 = R2= R3 = 30 kΩ and
R
f2 = Rf1 = R4 = 10 kΩ
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Engineering Circuit Analysis, 7
th
Edition Chapter Six Solutions 10 March 2006

35. The first stage is to subtract each voltage signal from the scale by the voltage
corresponding to the weight of the pallet (V
tare). This can be done by using a
differential amplifier:

The resistance of R can be arbitary as long as they resistances of each resistor is th
e
same and the current rating is not exceeded. A good choice would be R = 10 kΩ.
The output voltage of the differential amps from each of the scale, V
1 – V4 (now gives
the weight of the items only), is then added by using a two stage summing amplifier:

The output is given by:
R
vvvv
R
RR
v
in
ff
out
43212
1 +++
=

Therefore, to get the sum of the voltages v
1 to v4, we only need to set all resistances to
be equal, so setting R
f2 = Rf1 = Rin = R =10 kΩ would give an output that is
proportional to the total weight of the items
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Engineering Circuit Analysis, 7
th
Edition Chapter Six Solutions 10 March 2006

36. a) Using a difference amplifier, we can provide a voltage that is the difference
between the radar gun output and police speedometer output, which is proportional to
the speed difference between the targeted car and the police car. Note that since a
positive voltage is required which the police car is slower, the police speedometer
voltage would be feed into the inverting input:

Again, R can be arbitary as long as they are equal and doesn’t give an excessive current. 10 kΩ is a good choice here.

b) To convert to kph (km per hour) from mph (miles per hou r), it is noted that 1 mph
= 1.609 kph. Therefore, the voltage output from each device must be multiplied by
1.609. This can be done by using a non-inverting amplifier, which has an output given by:

inin
in
f
outvv
R
R
v 609.1)1(=+=

This gives Rf/Rin =0.609 ≈ 61/100, i.e. R
f = 6.2 kΩ and R in = 10 kΩ

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Engineering Circuit Analysis, 7
th
Edition Chapter Six Solutions 10 March 2006

37. v out of stage 1 is (1)(-20/ 2) = -10 V.

v
out of stage 2 is (-10)(-1000/ 10) = 1000 V

Note: in reality, the output voltage will be limited
to a value less than that used to
power the op amps.
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Engineering Circuit Analysis, 7
th
Edition Chapter Six Solutions 10 March 2006

38. We have a difference amplifier as the first amplifier stage, and a simple voltage
follower as the second stage. We therefore need only to find the output voltage of the
first stage: v
out will track this voltage. Using voltage division, then, we find that the
voltage at the non-inverting input pin of the first op amp is:

⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
+
32
3
2
RR
R
V

and this is the voltage at the inverting input terminal also. Thus, we may write a
single nodal equation at the inverting input of the first op amp:

⎥
⎥
⎦
⎤
⎢
⎢
⎣
⎡
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
+
+
⎥
⎥
⎦
⎤
⎢
⎢
⎣
⎡
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
+
=
1 out
32
3
2
f
1
32
3
2
1
V -
RR
R
V
R
1
V -
RR
R
V
R
1
0
Stage

which may be solved to obtain:

1
1
f
2
32
3
1
f
1
outoutV
R
R
- V
RR
R
1
R
R
V V
+
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
+==
Stage

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Engineering Circuit Analysis, 7
th
Edition Chapter Six Solutions 10 March 2006

39. The output of the first op amp stage may be found by realising that the voltage at the
non-inverting input (and hence the voltage at the
inverting input) is 0, and writing a
ingle nodal equation at the inverting input:

7
3 - 0

1
2 - 0

47
V - 0
0
1 stageout
++= which leads to
1 steageout
V = -114.1 V

This voltage appears at the input of the second op amp stage, which has a gain of
–3/ 0.3 = 10. Thus, the output of the second op amp stage is –10(-114.1) = 1141 V.
This voltage appears at the input of the final op amp stage, which has a gain of
–47/ 0.3 = -156.7.

Thus, the output of the circuit is –156.7(1141) = -178.8 kV, which is completely and
utterly ridiculous.
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Engineering Circuit Analysis, 7
th
Edition Chapter Six Solutions 10 March 2006

40. The output of the top left stage is –1(10/ 2) = -5 V.
The output of the middle left stage is –2(10/ 2) = -10 V.
The output of the bottom right stage is –3(10/ 2) = -15 V.

These three voltages are the input to a summing amplifier such that

Vout =
() 10 15105
100
R
=−−−−

Solving, we find that R = 33.33 Ω .
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Engineering Circuit Analysis, 7
th
Edition Chapter Six Solutions 10 March 2006

41. Stage 1 is configured as a voltage follower: the output voltage will be equal to the
input voltage. Using voltage division, the voltage at the non-inverting input (and
hence at the inverting input, as well), is

V 1.667
50100
50
5 =
+

The second stage is wired as a voltage follower also, so vout = 1.667 V.
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Engineering Circuit Analysis, 7
th
Edition Chapter Six Solutions 10 March 2006

42. a) Since the voltage supply is higher than the Zener voltage of the diode, the diode is
operating in the breakdown region. This means V
2 = 4.7 V, and assuming ideal op-
amp, V
1 = V2= 4.7 V. This gives a nodal equation at the inverting input:

k
V
k1.1
7.4
1
7.4
3
−
=

Solving this gives V
3 = 9.87 V

b) PSpice simulation gives:

It can be seen that all voltage values are very close to what was calculated. The
voltage output V
3 is 9.88Vinstead of 9.87 V. This can be explained by the fact that the
operating voltage is slightly higher than the breakdown voltage, and also the non-ideal
characteristics of the op-amp.

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Engineering Circuit Analysis, 7
th
Edition Chapter Six Solutions 10 March 2006

43. The following circuit can be used:

The circuit is governed by the equations:
13)1(V
R
R
V
in
f
+=
And
diodeVVV
==
21

Since the diode voltage is 5.1 V, and the desired output voltage is 5.1 V, we have
Rf/Rin = 0. In other words, a voltage follower is needed with R
f = 0Ω, and R in can be
arbitary – R
in =100 kΩ would be sufficient.

The resistor value of R2 is determined by:
ref
diodes
I
VV
R
−
=
2
At a voltage of 5.1 V, the current is 76 mA, as described in the problem. This gives
R
2≈ 51 Ω using standard resistor values.

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Engineering Circuit Analysis, 7
th
Edition Chapter Six Solutions 10 March 2006

44. For the Zener diode to operate in the breakdown region, a voltage supply greater than
the breakdown voltage, in this case 10 V is needed. With only 9 V batteries, the
easiest way is the stack two battery to give a 18 V power supply. Also, as the input is
inverted, an inverting amplifier would be needed. Hence we have the following
circuit:

in
in
f
out
V
R
R
V−=

Here, the input voltage is the diode voltage = 10 V, and the desired output voltage is -
2.5 V. This gives R
f/Rin = 25 / 100 = 50 / 200, or Rf = 51 kΩ and R in = 200 kΩ using
standard values. Note that large values are chosen so that most current flow through
the Zener diode to provide sufficient current for breakdown condition.

The resistance R is given by R = (18-10) V /25 mA = 320 Ω = 330 Ω using standard
values.

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Engineering Circuit Analysis, 7
th
Edition Chapter Six Solutions 10 March 2006

45. For a 20 V Zener diode, three 9 V batteries giving a voltage of 27 V would be needed.
However, because the required voltage is smaller than the Zener voltage, a non-
inverting amplifier can not be used. To use a inverting amplifier to give a positive
voltage, we first need to invert the input to give a negative input:

In this circuit, the diode is flipped but so is the power supply, therefore keeping the
diode in the breakdown region, giving V
in = -20 V. Then, using the inverting amp
equation, we have R
f / Rin = 12/20 giving Rf = 120 kΩ and R in = 200 kΩ using
standard resistor values.

The resistance R is then given by R = (-20 - -27) V / 12.5 mA = 560 Ω using standard
resistor values.

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Engineering Circuit Analysis, 7
th
Edition Chapter Six Solutions 10 March 2006

46. a) using inverting amplifier:

Rf/Rin = 5/3.3 giving Rf = 51 kΩ and R in = 33 kΩ. R = (9-3.3) V / 76 mA = 75 Ω using
standard resistor values.

b) To give a voltage output of +2.2 V instead, the same setup can be used, with supply
and diode inverted:

Correspondingly, the resistor values needs to be changed:
Rf / Rin = 3.3/2.2 giving R
f = 33 kΩ and R in = 22 kΩ . R would be the same as before
as the voltage difference between supply and diode stays the same i.e. R = 75 Ω.

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Engineering Circuit Analysis, 7
th
Edition Chapter Six Solutions 10 March 2006

47. The following setup can be used:

I Is

I
s = I = 10 V / R = 25 mA assuming ideal op-amp. This gives R = 400 Ω . Again,
taking half of max current rating as the operating current, we get R
1 = (12 – 10) V / 25
mA = 80 Ω = 82 Ω using standard values.

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Engineering Circuit Analysis, 7
th
Edition Chapter Six Solutions 10 March 2006

48. Using the following current source circuit, we have:

I Is

I
s = I = 12.5 mA = 20 V / R, assuming ideal op-amp. This gives R = 1.6 kΩ and R 1 =
(27 – 20) / 12.5 mA = 560 Ω.

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Engineering Circuit Analysis, 7
th
Edition Chapter Six Solutions 10 March 2006

49. In this situation, we know that there is a supply limit at ±15 V, which is lower than the
zener diode voltage. Therefore, previous designs need to be modified to suit this
application. One possible solution is shown here:

I Is

Here, we have Is = I = (27-20)/ R = 75 mA, assuming infinite op-amp input
resistance. This gives R = 93.3 ≈ 91 Ω using standard values. We also have R1 = (27-
20)/ 12.5 mA = 560 Ω.

Now look at the range of possible loads. The maximum output voltage is
approximately equal to the supply voltage, i.e. 15 V. Therefore, the minimum load is
given by R
L = (20 – 15) V / 75 mA = 66.67 Ω. Similarly, the maximum load is given
by RL = (20 - -15) V/ 75 mA = 466.67 Ω. i.e. this design is suitable for
466.67 Ω > R
L > 66.67 Ω.

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Engineering Circuit Analysis, 7
th
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50.

(a)
va = vb = 1 nV ∴ vd = 0 and vout = 0. Thus, P8Ω = 0 W.

(b)
va = 0, vb = 1 nV ∴ vd = -1 nV

vout = (2× 10
5
)(-1×10
-9
)
875
8
+
= -19.28 μ V. Thus, P
8Ω =
8
2
out
v
= 46.46 pW.
(c)
va = 2 pV, vb = 1 fV ∴ vd = 1.999 pV

vout = (2× 10
5
)(1.999× 10
-12
)
875
8
+
= 38.53 nV. Thus, P8Ω =
8
2
out
v
= 185.6 aW.

(c)
va = 50 μ V, vb = -4 μ V ∴ vd = 54 μ V
vout = (2×10
5
)(54×10
-6
)
875
8
+
= 1.041 V. Thus, P8Ω =
8
2
out
v
= 135.5 mW.
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Engineering Circuit Analysis, 7
th
Edition Chapter Six Solutions 10 March 2006

51.

Writing a nodal equation at the “-
vd” node,

f
outd
1
Sd
in
dR
- -

R
V - -

R
-
0
vvvv
++= [1]

or (R 1Rf + RinRf + RinR1) vd + RinR1vout = -RinRfVS [1]
Writing a nodal equation at the “
vout” node,

f
dout
o
doutR
)(- -

R
A - -
0
vvvv
+= [2]
Eqn. [2] can be rewritten as:
()
out
fo
of
d
AR - R
R R-
vv
+
= [2]
so that Eqn. [1] becomes:
vout =
( )
ino1o1infin1f1in
SofinRR RR RR RR RR RAR
V R - ARR
-
+++++

where for this circuit, A = 10
6
, Rin = 10 TΩ, R o = 15 Ω , R f = 1000 kΩ, R 1 = 270 kΩ.

(a) –3.704 mV; (b) 27.78 mV; (c) –3.704 V.
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th
Edition Chapter Six Solutions 10 March 2006

52. vout = Avd = () V 2 sin 1080
R16
R
A
15
i
i
t×
+

(a) A = 10
5
, Ri = 100 MΩ , R o value irrelevant. vout = 8 sin 2t nV

(b) A = 10
6
, Ri = 1 TΩ, R o value irrelevant. vout = 80 sin 2t nV
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Engineering Circuit Analysis, 7
th
Edition Chapter Six Solutions 10 March 2006

53.

(a) Find
vout/ vin if Ri = ∞, R o = 0, and A is finite.

The nodal equation at the inverting input is

100
- -

1
- -
0
outdindvvvv
+= [1]

At the output, with Ro = 0 we may write vout = Avd so vd = vout/ A. Thus, Eqn. [1]
becomes
100

100A

A
0
outout
in
outvv
v
v
+++=
from which we find
A 101
100A-

in
out +
=v
v
[2]

(b) We want the value of A such that vout/ vin = -99 (the “ideal” value would be –100
if A were infinite). Substituting into Eqn. [2], we find

A = 9999

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Engineering Circuit Analysis, 7
th
Edition Chapter Six Solutions 10 March 2006

54. (a) δ = 0 V ∴ vd = 0, and P8Ω = 0 W.

(b) δ = 1 nV, so vd = 5 – (5 + 10
-9
) = -10
-9
V
Thus,

vout = (2×10
5
)vd
758
8
+
= -19.28 μV and P
8Ω = (vout)
2
/ 8 = 46.46 pW.
(c)
δ = 2.5 μ V, so vd = 5 – (5 + 2.5×10
-6
) = -2.5×10
-6
V
Thus,
vout = (2×10
5
)vd
758
8
+
= -48.19 mV and P
8Ω = (vout)
2
/ 8 = 290.3 μW.

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th
Edition Chapter Six Solutions 10 March 2006

55.
AD549

Writing a single nodal equation at the output, we find that

o
dout
i
inoutR
A -

R
-
0
vvvv
+= [1]

Also, vin – vout = vd, so Eqn. [1] becomes
0 = (
vout – vin) Ro + (vout – Avin + Avout) Ri
and

vout =
()
inv
A
R )1(R
AR R
io
io++ +

To within 4 significant figures (and more, actually), when
vin = -16 mV, vout = -16 mV (this is, after all, a voltage follower circuit).
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Engineering Circuit Analysis, 7
th
Edition Chapter Six Solutions 10 March 2006

56. The Voltage follower with a finite op-amp model is shown below:

Nodal equation at the op-amp output gives:

o
outd
i
inoutR
VAV
R
VV
−
=
−

But in this circuit, V
d=Vin – Vout. Substitution gives:

o
outin
o
outoutin
i
inoutR
VAAV
R
VVVA
R
VV
)1()(
+−
=
−−
=
−

Further rearranging gives:

iniinooutioutoAVRVRVARVR
+=++ )1(
⇔
in
io
io
out
V
ARR
ARR
V
)1(++
+
=

This is the expression for the voltage follower in non-ideal situation. In the case of
ideal op-amp, A → ∞, and so A+1 → A. This means the denominator and the
numerator would cancel out to give V
out = Vin, which is exactly what we expected.

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Engineering Circuit Analysis, 7
th
Edition Chapter Six Solutions 10 March 2006

57. a) By definition, when the op-amp is at common mode, v
out = A
CMv
in. Therefore, a model that can represent this is:

This model relies on that fact that A
CM is much smaller than the differential gain A, and therefore when the inputs
are different, the contribution of A
CM is negligible. When the inputs are the same, however, the differential term
Av
d vanishes, and so v
out = A
CMv
2, which is correct.

b) The voltage source in the circuit now becomes 10
5
v
d+10v
2. Assuming R
o = 0, the circuit in figure 6.25 becomes:

The circuit is governed by the following equations:
abdvvv−= so
R
vv
R
vvv
aabd 1
51010 −
=
−+

2/
2vv
b= (from voltage divider)

Rearranging gives:
122
5)2/(10)2/(10vvvvvv
aaa−=−+−
12
5
251050000 vvvvvv
aaa−=−+−
)210(
50005
5
12
+
+
=
vv
v
a
Then, the output is given by:
22
555)2/(101010vvvvvv
abdout+−×=+=

in this case, v
1 = 5 + 2 sin t, v 2 = 5. This gives va=0.50004v2+9.99980×10
-6
v1. Thus

v out = 10
5
(-0.00004v2 - 9.99980×10
-6
v1)+5v2 = 1.00008v2 - 0.99998v1 = 0.0005 –
1.99996 sin t

c) If the common mode gain is 0, than the equation for v
a becomes
)210(
50000
5
12
+
+
=
vv
v
a
Giving v
a=0.49999v2+9.99980×10
-6
v1. The output equation becomes
)2/(1010
2
55
adoutvvvv−×===0.99998v2-0.99998v1 = 1.99996 sin t
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Engineering Circuit Analysis, 7
th
Edition Chapter Six Solutions 10 March 2006

58. Slew rate is the rate at which output voltage can respond to changes in the input. The
higher the slew rate, the faster the op-amp responds to changes. Limitation in slew
rate – i.e. when the change in input is faster than the slew rate, causes degradation in
performance of the op-amp as the change is delayed and output distorted.

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Engineering Circuit Analysis, 7
th
Edition Chapter Six Solutions 10 March 2006

59. a) V2 = 4.7 V from the Zener diode, V1 = V2 = 4.7 V assuming ideal op-amp, and V3
is given by the nodal equation at the inverting input:
k
V
k
VV
17.4
113
=
−

Solving gives V
3 = 26.79 V

b) The simulation result is shown below

There are considerable discrepancies between calculated and simulated volta ges. In particular, V
1 = 3.090 V is
considerably lower than the expected 4.7 V. This is due to the non-ideal characteristics of uA741 which has a finite
input resistance, inducing a voltage drop between the two input pins. A more severe limitation, however, is the
supply voltage. Since the supply voltage is 18V, the output cannot exceed 18 V. This is consistent with the
simulation result which gives V
3 = 17.61 V but is quite different to the calculated value as the mathematical model
does not account for supply limitations.

c) By using a DC sweep, the voltage from the diode (i.e. V
2) was monitored as the
battery voltage changes from 12 V to 4V.

It can be seen that the diode
voltage started dropping when
batteries drop below 10 V.
However, the diode can still be
considered as operating in the
breakdown region, until it hit
the knee of the curve. This
occurs at V
supply = 5.28 V

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Engineering Circuit Analysis, 7
th
Edition Chapter Six Solutions 10 March 2006

60. The ideal op amp model predicts a gain v out/ vin = -1000/ 10 = -100, regardless of the
value of v
in. In other words, it predicts an input-output characteristic such as:

vout (V)
v
in (V)
-100

1

From the PSpice simulation result shown below, we see that the ideal op amp model
is reasonab
ly accurate for |v
in| × 100 < 15 V (the supply voltage, assuming both have
the same magnitude), but the onset of saturation is at ±14.5 V, or |v
in| ~ 145 mV.
Increasing |v
in| past this value does not lead to an increase in | v out|.

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Engineering Circuit Analysis, 7
th
Edition Chapter Six Solutions 10 March 2006

61. Positive voltage supply, negative voltage supply, inverting in
put, ground, output pin.
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Engineering Circuit Analysis, 7
th
Edition Chapter Six Solutions 10 March 2006

62. This op amp circuit is an open-loop circuit; there is no external feedback path from
the output terminal to either input. Thus, the output s hould be the open-loop gain
times the dif
ferential input voltage, minus any resistive losses.

From the simulation results below, we see that all three op amps saturate at a voltage
ma
gnitude of approximately 14 V, corresponding to a differential input voltage of 50
to 100 μ V, except in the interest case of the LM 324, which may be showing some
unexpected input offset behavior.

op amp onset of
negat
ive
satur
ati
on
negative
satur
atio
n
volta
ge
onset of
positi
ve
satur
atio
n
positive
satur
atio
n
volta
ge
μA 741 -92 μV -14.32 V 54.4 mV 14.34 V
LM 324 41.3 μV -14.71 V 337.2 mV 13.87 V
LF 411 -31.77 μV -13.81 V 39.78 mV 13.86 V

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Engineering Circuit Analysis, 7
th
Edition Chapter Six Solutions 10 March 2006

63. This is a non-inverting op amp circuit, so we expect a gain of 1 + 1000/4.7 = 213.8.
With ±15 V DC supplies, we need to sweep the input just above and just below this
value divided by the gain to ensure that we see the saturation regions. Performi
ng the
indicated simulation and a DC sweep from –0.1 V to +0.1 V with 0.001 V steps, we
obtain the following input-output characteristic:

Using the cursor tool, we see that the linear region is in the range of
–68.2 mV < V
in < 68.5 mV.
The simulation predicts a gain of 7.103 V/ 32.87 mV = 216.1, which is reasonably
close to the value predicted using the ideal op amp m odel.
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Engineering Circuit Analysis, 7
th
Edition Chapter Six Solutions 10 March 2006

64. To give a proper simulation, the inputs are grounded to give an input of 0. This gives:

As can be seen, a current of 18.57 mA is drawn from the uA741. Assum
ing the output
voltage from the op-amp before R
o is 0, we have Ro = (1-18.57m)/18.57m = 52.9 Ω.
This is close to the value given in table 6.3. There is difference between the two as here we are still using the assumption that the voltage output is independent to the
loading circuit. This is illustrated by the fact that as the supplied voltage to the 1 ohm
resistor changes, the voltage at the output pin actually increases, and is always higher
than the voltage provided by the battery, as long as the supplied to the op-amp is greater than the battery voltage. When the supply voltage drops to 1V, the output
current increased greatly and gave an output resistance of only 8 Ω. This suggests that
the inner workings of the op-amp depend on both the supply and the loading.

For LF411, a current of 25.34 mA is drawn from the op-amp. This gives a output
resistance of 38.4 Ω. This value is qu
ite different to the 1 Ω figure given in the table.

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Engineering Circuit Analysis, 7
th
Edition Chapter Six Solutions 10 March 2006

65. Based on the detailed model of the LF 411 op amp, we can write the following nodal
equation at the inverting input:

o
6
dd
4
dx
in
dR10
- A

10
-

R
-
0
+
++=
vvvvv

Substituting values for the LF 411 and simplifying, we make appropriate
approximations and then solve for v
d in terms of v x, finding that

199.9
-
10199.9
10-

x
x6
6
dv
vv =
×
=

With a gain of –1000/10 = -100 and supply voltage magnitudes of 15 V, we are
effectively limited to values of |v
x| < 150 mV.

For v x = -10 mV, PSpice predicts v d = 6 μV, where the hand calculations based on the
detailed model predict 50 μV, which is about one order of magnitude larger. For the
same input voltage, PSpice predicts an input current of -1 μ A, whereas the hand
calculations predict 99.5v
x mA = -995 nA (which is reasonably close).
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Engineering Circuit Analysis, 7
th
Edition Chapter Six Solutions 10 March 2006

66. (a) The gain of the inverting amplifier is –1000. At a sensor voltage of –30 mV, the
predicted output voltage (assuming an ideal op amp) is +30 V. At a sensor voltage of
+75 mV, the predicted output voltage (again assuming an ideal op amp) is –75 V.
Since the op amp is being powered by dc sources with voltage magnitude equal to 15
V, the output voltage range will realistically be limited to the range
–15 < V
out < 15 V.

(b) The peak input voltage is 75 mV. Therefore, 15/ 75×10
-3
= 200, and we should set
the resistance ratio R
f/ R1 < 199 to ensure the op amp does not saturate.
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Engineering Circuit Analysis, 7
th
Edition Chapter Six Solutions 10 March 2006

67. (a)

We see from the simulation result that nega
tive saturation begins at V in = –4.72 V,
and positive saturation begins at V
in = +4.67 V.

(b) Using a 1 pΩ resistor between the output pin and ground, we obtain an output

current of 40.61 mA, slightly larger than the expected 35 mA, but not too far off.
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Engineering Circuit Analysis, 7
th
Edition Chapter Six Solutions 10 March 2006

68. We assume that the strength of the separately-broadcast chaotic “noise” signal is
received at the appropriate intensity such that
it may precisely cancel out the chaotic
component of the total received signal; otherwise, a variable-gain stage would need to
be added so that this could be adjusted by the user. We also assume that the signal
frequency is separate from the “carrier” or broadcast frequency, and has already been
separated out by an appropriate circuit (in a similar fashion, a radio station
transmitting at 92 MHz is sending an audio signal of between 20 and 20 kHz, which
must be separated from the 92 MHz frequency.)

One possible solution of many (all resistan
ces in ohms):

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Engineering Circuit Analysis, 7
th
Edition Chapter Six Solutions 10 March 2006

69. One possible solution of many:

This circuit produces an output equal to the average of V1, V2, and V3, as shown in the
simulation result: V
average = (1.45 + 3.95 + 7.82)/ 3 = 4.407 V.

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Engineering Circuit Analysis, 7
th
Edition Chapter Six Solutions 10 March 2006

70. Assuming ideal situations (ie slew rate = infinite)

a)

-5
0
5
10
15
20
-5 -3 -1 1 3 5
V
active (V)
V
out
(V)

b)
-5
0
5
10
15
20
-5 -3 -1 1 3 5
V
active (V)
V
out
(V)

18 V
18 V

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Engineering Circuit Analysis, 7
th
Edition Chapter Six Solutions 10 March 2006

71. a)

-15
-10
-5
0
5
10
15
-2-1012
V
active (V)
V
out
(V)

b) The simulation is performed using the fo
llowing circuit:

12 V
-12 V

Where RL = load resistor which is needed for the voltage probe to perform properly.
The battery is swept from -2V to +2 V and the voltage sweep is displayed on the next
page.

It can be seen that the sweep is very much identical to what was expected, with a
discontinuity at 0V. The only difference is the voltage levels which are +11.61V and –
11.61 V instead of ±12 V. This is because the output of an op-amp or comparator can
never quite reach the supplied voltage.

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Engineering Circuit Analysis, 7
th
Edition Chapter Six Solutions 10 March 2006

72. a)

-15
-10
-5
0
5
10
15
-5 -3 -1 1 3 5
V
active (V)
V
out
(V)

b)
-15
-10
-5
0
5
10
15
-5 -3 -1 1 3 5
V
2 (V)
V
out
(V)

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Engineering Circuit Analysis, 7
th
Edition Chapter Six Solutions 10 March 2006

73. The following comparator setup would give a logic 0 for voltages below 1.5 V and
logic 1 for voltages above 1.5 V

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Engineering Circuit Analysis, 7
th
Edition Chapter Six Solutions 10 March 2006

74. The voltage output of the circuit is given by

−+−
+
+
= v
R
R
v
RR
RR
R
R
v
out
1
2
34
12
3
4 )
/1
/1
(

a) When R
1 = R3 and R2 = R4, the equation reduces to
)(
3 4
−+
−=vv
R
R
v
out

When v
+ = v-, vout = 0, thus ACM = 0. Hence CMRR = ∞

b) If R
1, R2, R3 and R4 are all different, then when v+ = v- = v,
v
R
R
RR
RR
R
R
v
out ))
/1
/1
((
1
2
34
12
3 4
−
+
+
=
Simplifying the algebra gives
v
RRRR
RRRR
v
out
4131
4341
+
−
=

If v
+ and v- are different, it turns out that it is impossible to separate vout and vd
completely. Therefore, it is not possible to obtain A or CMRR in symbolic form.

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Engineering Circuit Analysis, 7
th
Edition Chapter Six Solutions 10 March 2006

75. a) The voltage at node between R1 and R2 is
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
+
=
21
2
1
RR
R
VV
ref
by treating it as a voltage divider. Similarly, the voltage at node between R
Gauge and
R
3 is:
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
+
=
Gauge
refRR
R
VV
3
3
2

Therefore, the output voltage is
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
+
−
+
=−=
Gauge
refoutRR
R
RR
R
VVVV
3
3
21
2
21

b) If R
1 = R2 = R3 = Rgauge then the two terms in the bracket cancels out, giving V out =
0.
c) The amplifier has a maximum gain of 1000 and minimum gain of 2. Therefore to
get a voltage of 1V at maximum loading, the voltage input into the amplifier must fall
between 0.001 and 0.5, i.e. 0.5 > V
out > 0.001.

To simplify the situation, let R
1 = R2 = R3 = R, then at maximum loading,
⎟
⎠
⎞
⎜
⎝
⎛
++
−=
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
Δ++
−
+
=
mkR
R
RRR
R
RR
R
VV
Gauge
refout 5052
1
12

Using this we can set up two inequalities according to the two limits. The first one is:
001.0
505
12
6≥⎟
⎠
⎞
⎜
⎝
⎛
++
−
mkR
R

Solving gives
05.5000
12
999.5+
≥R
R

R≥38.4998
Similarly, the lower gain limit gives:
05.5000
12
5.5+
≤R
R

⇒ R ≤4230

This gives 4998.38 > R > 4230. Using standard resistor values, the only possible
resistor values are R = 4.3 kΩ and R = 4.7 kΩ.

If we take R = 4.7 kΩ , then

1855.0
5057.4
7.4
2
1
12=⎟
⎠
⎞
⎜
⎝
⎛
++
−=
mkk
k
V
out

Giving a gain of 5.388. This means a resistor value of R = 50.5/(5.388 – 1) = 11.5 kΩ
or 11kΩ using standard value is needed between pin 1 and pin 8 of the amplifier.
PROPRIETARY MATERIAL . © 2007 The McGraw-Hill Companies, Inc. Limited distribution permitted only to
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Engineering Circuit Analysis, 7
th
Edition Chapter Seven Solutions 10 March 2006

1.
dv
iC
dt
=

(a) i = 0
(DC)

(b)
() () ()
6
10 10 115 2 120 sin120 613sin120 mA
dv
iC t t
dt ππ π
−
==−× =−
(c)
()()
63
10 10 4 10 40 nA
ttdv
iC e e
dt
−−−−
==−× × =−

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Engineering Circuit Analysis, 7
th
Edition Chapter Seven Solutions 10 March 2006

2.
dv
iC
dt
=

60
66
06
vt
−
=+=
−
t− , therefore
6
4.7 10
dv
iC
−
dt
==−× μA

t (s)
i (μA)
–4.7

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Engineering Circuit Analysis, 7
th
Edition Chapter Seven Solutions 10 March 2006

3.
dv
iC
dt
=

(a)
()
3
30 therefore 10 30 1 mA
tt tdv dv
ete i te
dt dt
−− − −
⎡⎤=− = =−
⎣⎦

(b)

()
55
35
4 5 sin100 100 cos100
therefore 10 4 100cos100 5sin100 mA
tt
tdv
etet
dt
dv
iett
dt
−−
−−
⎡⎤=− +
⎣⎦
== −
PROPRIETARY MATERIAL . © 2007 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers
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Engineering Circuit Analysis, 7
th
Edition Chapter Seven Solutions 10 March 2006

4.
21
2
WCV=

(a) ()
61
2000 10 1600 1.6 J
2
−⎛⎞
×=
⎜⎟
⎝⎠

(b)
() ()
2
31
25 10 35 15.3 J
2
−⎛⎞
×=
⎜⎟
⎝⎠

(c)
()()
2
41
10 63 198 mJ
2
−⎛⎞
=
⎜⎟
⎝⎠

(d)
() ()
31
2.2 10 2500 2.75 J
2
−⎛⎞
×=
⎜⎟
⎝⎠

(e)
()( )
21
55 2.5 171.9 J
2
⎛⎞
=
⎜⎟
⎝⎠

(f) () ()
2
31
4.8 10 50 6 J
2
−⎛⎞
×=
⎜⎟
⎝⎠

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and educators for course preparation. If you are a student using this Manual, you are using it without permission.

Engineering Circuit Analysis, 7
th
Edition Chapter Seven Solutions 10 March 2006

5. (a)
()
pF
d
A
C954.6
10100
1054.7810854.8
6
612
=
×
××
==
−
−−
ε

(b)
( )
3
2
12
21 1012
, 16.96 kV
2 6.954 10
E
Energy E CV V
C
−
−
×
=∴== =
×

(c)
()
()
pF
V
E
CCVE 500
100
105.222
2
1
2
6
2
2
=
×
==∴=
−

() ( )
()
1
6
612
.62.636
1054.78
1010010500
−
−
−−
=
×
××
==∴= mpF
A
Cd
d
A
Cε
ε

12
12
0
636.62 10
\Relative permittivity : 71.9
8.854 10ε
ε
−
−
×
==
×

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Engineering Circuit Analysis, 7
th
Edition Chapter Seven Solutions 10 March 2006

6. (a) For V A = –1V, ()
()( )
()()
()157.0
101106.1
10854.88.1122
2419
12
0
+
××
×
=−=
−
−
Abi
s
VV
qN
K
W
ε

m
9
10281.45
−
×=

( )( )
fFC
j 307.2
10281.45
10110854.88.11
9
1212
=
×
××
=
−
−−

(b) For V
A = –5V,
()
()( )
()()
() 557.0
101106.1
10854.88.1122
2419
12
0
+
××
×
=−=
−
−
Abi
s
VV
qN
K
W
ε

m
9
10289.85
−
×=

( )( )
fFC
j 225.1
10289.85
10110854.88.11
9
1212
=
×
××
=
−
−−

(c) For V
A = –10V,

()
()( )
()()
() 1057.0
101106.1
10854.88.1122
2419
12
0
+
××
×
=−=
−
−
Abi
s
VV
qN
K
W
ε

m
9
10491.117
−
×=

( )( )
aFC
j 239.889
10491.117
10110854.88.11
9
1212
=
×
××
=
−
−−

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Engineering Circuit Analysis, 7
th
Edition Chapter Seven Solutions 10 March 2006

7. We require a capacitor that may be manually varied between 100 and 1000 pF by rotation
of a knob. Let’s choose an air dielectric for simplicity of construction, and a series of 11
half-pla
tes:

Constructed as shown, the half-plates
are in parallel, so that each of the 10 pairs must
have a capacitance of 1000/ 10 = 100 pF when rotated such that they overlap completely.
If
we arbitrarily select an area of 1 cm
2
for each half-plate, then the g ap spacing between
each plate is d = εA/C = (8.854×10
-14
F/cm)(1 cm
2
)/ (100×10
-12
F) = 0.8854 mm. This is
tight, but not impossible to achieve. The final step is to determine the amount of overlap
which corresponds to 100 pF for the total capacitor structure. A capacitance of 100 pF is
equal to 10% of the capacitance when all of the plate areas are aligned, so we need a pie-
shaped wedge having an area of 0.1 cm
2
. If the m iddle figure above corresponds to an
angle of 0
o
and the case of perfect alignm ent (maximum capacitance) corresponds to an
angle of 180
o
, we need to set out minimum angle to be 18
o
.
Side view with no
overlap between plates
Top view
Side view with a small overlap between plates.
fixed

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Engineering Circuit Analysis, 7
th
Edition Chapter Seven Solutions 10 March 2006

8. (a) Energy stored
3
0
210
55
0 3
3 1.080
5
tt
t
t
dv
vC C e e dt J
dt
μ
−
−−×⎛⎞
=⋅ = ⋅− =− ⎜⎟
⎝⎠
∫∫

(b) V
max = 3 V
Max. energy at t=0,
JEmJCVμ5.499%3735.1
2
1
max
2=∴==
V at 37% E
max = 1.825 V

()
sstetv
t
2486.23825.1
5
⇒≈=∴==
−

(c) Ae
dt
dv
Ciμ593.141
5
3
10300
5
2.1
6
−=
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
−×==
−
−

(d) ( ) WviPμ6.24210658.120011.2
6
−=×−==
−

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and educators for course preparation. If you are a student using this Manual, you are using it without permission.

Engineering Circuit Analysis, 7
th
Edition Chapter Seven Solutions 10 March 2006

9. (a) ()
( )
( ) mV
C
v 421.33101
2
14159.3
.
1047
1
101
2
.
1
2
3
6
2
3
=×
×
=×=
−
−
− π

(b) ()
( )
( ) mV
C
v421.33101
2
14159.3
.
1047
1
0101
2
.
1
2
3
6
2
3
=×
×
=⎟
⎠
⎞
⎜
⎝
⎛
+×=
−
−
−π

(c) () ( ) ( ) mV
C
v132.50101
4
3
.
1047
1
101
4
101
2
.
1
2
3
6
2
3
2
3
=⎟
⎠
⎞
⎜
⎝
⎛
×
×
=⎟
⎠
⎞
⎜
⎝
⎛
×+×=
−
−
−−πππ

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Engineering Circuit Analysis, 7
th
Edition Chapter Seven Solutions 10 March 2006

10.
C
t
C
idt
C
V
ms
ms
426.0
cos
10711
200
0
3
200
0
=
⎥
⎥
⎦
⎤
⎢
⎢
⎣
⎡
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛×
−==
−
∫
π
π

( )
FC
C
CVEμ30181
1032
10086.181
2
10086.181
103
2
1
6
99
62
=
×
×
=∴
×
=×==
−
−
−
−

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Engineering Circuit Analysis, 7
th
Edition Chapter Seven Solutions 10 March 2006

11.

(a)

(b)
2722 7
max11
2 10 (5 3cos 200 ) 10 64 6.4 J
22
cc cwcv tw
μ
−−
==×× + ∴=×=

(c)
6 100 3 3 100 100
01
10 8 10 10 40( 0.01)( 1) 400(1 )V
0.2
−− −
=× × =×− −= −∫
t
tt
c
ved t e
t
e

(d)

100
500 400 V
t
c
ve
−
=−

0.12sin 400 mA=−
cit
t
26
0.2 F, 5 3cos 200 V; 0.2 10 (3)( 2)200sin 200 cos200
−
==+ ∴=×−
cccv ti tμ
∴
PROPRIETARY MATERIAL . © 2007 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers
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Engineering Circuit Analysis, 7
th
Edition Chapter Seven Solutions 10 March 2006

0.1
0
0.2
0.1
0.9
22
(0) 250V, 2mF ( ) (0.1) 250 500 5
(0.1) 500V; (0.2) 500 10 1000V
(0.6) 1750V, (0.9) 2000V
0.9 1: 2000 500 10 2000 5000( 0.9)
2100 2000 5000( 0.9) 0.92 0.9 0
== =+
∴= = =
∴= =
∴<< = + = + −
∴= = + − ∴= ∴ << ∫
∫
∫cc
cc
cc
t
c
cvcav d t
vv d t
vv
tv dt t
vt t .92s
12.

t

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Engineering Circuit Analysis, 7
th
Edition Chapter Seven Solutions 10 March 2006

13.

(a)

(b)
2 6 2 2 1000 500
6 500 50011
C 10 2 10 200 V
22
C 10 ( 200)( 500) 0.1
200
R2
0.1
−− − −
−−−
==×=× ∴=±
′== ± − =
−
∴== =Ω
m
tt
c
tt
wv v ev e
iv e e
v
k
i
2 1000 1000
1000 1000
R0
0
P R 0.01 2000 20 W
W 20 0.02

0.02J
∞
−−
∞
−−
== × =

∴ == −∫
tt
R
tt
iee
edt e
=

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Engineering Circuit Analysis, 7
th
Edition Chapter Seven Solutions 10 March 2006

14. (a) Left circuit:
By Voltage division, () V
kk
k
V
C 877.05
17.4
1
=+
=
Right circuit:

()
VV
3
2
2//11
1 ==
By Voltage Division, VVVV
C
3
1
3
1
2 −=∴=

(b)
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Engineering Circuit Analysis, 7
th
Edition Chapter Seven Solutions 10 March 2006

15.
di
vL
dt
=

(a) v = 0 since i = cons
tant (DC)

(b)
() ()
8
10 115 2 120 sin120 613sin120 Vvt tπ ππ=− μ
−
=−
(c)
() ()
83 6
240 pV
tt 6
10 115 2 24 10 e
−− −−
=− × = −ve

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Engineering Circuit Analysis, 7
th
Edition Chapter Seven Solutions 10 March 2006

16.
di
vL
dt
=

()
()
9
99
3
60 10
6 10 6 10 10
06 10
it
−
−−
−
⎡⎤−×
=+ ×= ×⎢⎥
−×⎢⎥⎣⎦
6
t
−
− , therefore

()()
12 6 18
10 10 10 V = 1 aV
di
vL
dt
−− −
==− =− −

t (s)
v (aV)
–1

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Engineering Circuit Analysis, 7
th
Edition Chapter Seven Solutions 10 March 2006

17.
di
vL
dt
=

(a) () ()
69
5 10 30 10 150 1 fV
tt tdi
Le te
dt
−−−
te⎡⎤=× × −
⎣⎦
− −
= −

(b)

()()
()
5
e 100 20cos100 sin100 pV
t
ve t t
−
=−
635 5
5 10 4 10 5 sin100 100 cos100
therefor
ttdi
Le t e t
dt
−−− −
⎡⎤=× × − +
⎣⎦

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Engineering Circuit Analysis, 7
th
Edition Chapter Seven Solutions 10 March 2006

18.
21
2
WL=I. Maximum energy corresponds to m a
ximum current flow, so

() ()
2
3
max1
5 10 1.5
2
W
−
=× 5.625 mJ=

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Engineering Circuit Analysis, 7
th
Edition Chapter Seven Solutions 10 March 2006

19.

(a)

(b)
max
P P ( 100)( 5) 500W at 40 ms
−
=∴ =− −= =
L
LLi L
vt

(c)
minP 100( 5) 500W at 20 and 40 ms
++
=−=− =
L t

(d)
2211
W L W (40ms) 0.2( 5) 2.5J
22
=∴ =×−=
L
LL i

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Engineering Circuit Analysis, 7
th
Edition Chapter Seven Solutions 10 March 2006

20.

3 100 100
-100t 100 1
max
100
max
100 100
L5010, 0: 0; 0 80 mA0.08 A
i =0.08e 8 0.08 8 , , 0.01 , 0.08 0.01
0.2943mA; 0.05 (0.004 0.4 )
( 0.4) 100 (0.004 0.4 ) 0.4 0.4 40 ,
−− −
−−
−
−−
=× < = > = =
′∴−∴===×
′∴= = = −
′∴= − − − ∴− = −
tt
t
m
t
tt
titite te
te t t s i e
ivi et
ve e t t
2
max
0.8
0.02
40
(0.004 0.008) 0.5413mV this is minimum 0.004Vat t=0
−
==
=−=− ∴=
ts
ve v
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Engineering Circuit Analysis, 7
th
Edition Chapter Seven Solutions 10 March 2006

21.

(a)
22
0: 0.4 A 10 5 4 4 V
′
>= ∴=+=+
si n s s
ti tv iitt

(b)
2
01
0.1 40 5 4 4 5A
5
′=+ +=++∫
t
in s
iv tdttt

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Engineering Circuit Analysis, 7
th
Edition Chapter Seven Solutions 10 March 2006

22. 20cos1000 V, L 25mH, (0) 0==
LL
vti =
==∫
t
L
it dt t

(a)
0
40 20 cos 1000 0.8sin 1000 A 8sin 2000 W∴=tp

(b)
32 2
8sin 1000 mJ=
1
25100.64sin 1000
2
wt t
−
=× × ×

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Engineering Circuit Analysis, 7
th
Edition Chapter Seven Solutions 10 March 2006

23.

(a)
0
0 10 ms: 2 5 100 2 500 (10ms) 3A, (8ms) 2A<< =−+ =−+ ∴ = =∫
t
LL
ti dt ti i
L

(b)

(c) If the circuit has been connected for a long time
, L appears like short circuit.

() VVV 80100
82
8
8 =
+
=
Ω

A
V
I 10
2
20
2 =
Ω
=
Ω

A
V
i
x 1
80
80
=
Ω
=∴
0.02
4
0.01
40 .024
0.01
2
(0) 0 (10ms) 500 0.01 5A (20ms) 5 5 10 (0.02 )
(20ms) 5 5 10 (0.02 0.5 ) 5 5 10 (0.0002 0.00015) 7.5A
1
0.2 7.5 5.625J
2
LL L
L
Lii i td
it t
w
=∴ = × = ∴ =+ −t
∴ =+× − =+× − =
∴=× × =
∫
PROPRIETARY MATERIAL . © 2007 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers
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Engineering Circuit Analysis, 7
th
Edition Chapter Seven Solutions 10 March 2006

24. After a very long time connected only to DC sources, the inductors act as short circuits.
The circuit may thus be redrawn as

And we find that
80
1009
1 A
8028
80
9
x
i
⎛⎞
⎜⎟ ⎛⎞
==⎜⎟ ⎜⎟
+⎝⎠⎜⎟+
⎝⎠

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Engineering Circuit Analysis, 7
th
Edition Chapter Seven Solutions 10 March 2006

25.
2
L 5H,V 10( )V, (0) 0.08A
−−
==− =
tt
LL
ee i

(a)
12
(1) 10( ) 2.325 V
Lvee
−− +
=−=

(b)

(c) ( ) 1.08A
L
i∞=
22
0
0
22
0.08 0.2 10( ) 0.08 2( 0.5 )
0.08 2( 0.5 1 0.5) 1.08 2 (1)

0.4796A=
−− − −
−− −−
=+ − =+−+
=+−+ +−=+−∴∫
t
tt t tt
L
tt tt
L L
ie edtee
e eeiie
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Engineering Circuit Analysis, 7
th
Edition Chapter Seven Solutions 10 March 2006

26.

(a)

(b)
40 12
120 40 5
12 20 40 12 20 40
200 100
100V
33
=× +××
++ ++
=+=
xv

15 12120 15
40 40 5
12 15 60 15 60 15 12 60
120 1 6.667
40 200
12 12 5 66.667
40 2060V=
=××+×
++ +
=×
×+
+
=+
xv
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Engineering Circuit Analysis, 7
th
Edition Chapter Seven Solutions 10 March 2006

27.
(a)
21
51.6 6.4J
2
Lw=×× =

(b)
621
20 10 100 0.1J
2
cw
−
=× × × =

(c) Left to right (magnitudes): 100, 0, 100, 116, 16, 16, 0 (V)

(d) Left to right (magnitudes): 0, 0, 2, 2, 0.4, 1.6, 0 (A)

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Engineering Circuit Analysis, 7
th
Edition Chapter Seven Solutions 10 March 2006

28.
(a) 2
52
400 V, 0; (0) 0.5A; 0.4
1
400 0.16 64V, 10 64 20.48mJ
2
−
=>==
=× = ××=
sL
ccvt ti ts
vw

(b)

(c)
0.4
24 45
0
4 , P 100 16 1600 320 0.4 3.277J==×∴= =×= ∫RR R
it tw tdt
0.4
23
0
2 1
0.5 0.1 400 0.5 40 0.4 1.3533A
3
1
10 1.3533 9.1581J
2
=+ =+×× =
∴=×× =
∫L
L
it dt
w

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Engineering Circuit Analysis, 7
th
Edition Chapter Seven Solutions 10 March 2006

29. (a)
()
W
R
V
PWP 4.0
10
2
;0
22
107
====
ΩΩ

(b) PSpice verification
We see from the PSpice
simulation that the
voltage across the 10-Ω
resistor is –2 V, so that
it is dissipating 4/10
= 400 mW.

The 7-Ω resistor has
zero volts across its
terminals, and hence
dissipates zero power.
Both results agree with the hand calculations.
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Engineering Circuit Analysis, 7
th
Edition Chapter Seven Solutions 10 March 2006

30. (a) We find RTH by first short-circuiting the voltage source, removing the inductor, and
looking into the open terminals.

Sim plifying the network from the right, 3 || 6 + 4 = 6 Ω, which is in parallel with 7 Ω.
6 || 7 + 5 = 8.23 Ω. Thus, R
TH = 8.23 || 8 = 4.06 Ω. To find V TH, we remove the inductor:

+ V
TH –
REF
V
1 V2 V3

Writing the nodal equations required:

(V
1 – 9)/3 + V1/6 + (V1 – V2)/4 = 0
(V
2 – V1)/4 + V2/7 + (V2 – V3)/5 = 0
V
3/8 + (V3 – V2)/5 = 0

Solving, V
3 = 1.592 V, therefore VTH = 9 – V3 = 7.408 V.

(b) i
L = 7.408/4.06 = 1.825 A (inductor acts like a short circuit to DC).

(c)
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Engineering Circuit Analysis, 7
th
Edition Chapter Seven Solutions 10 March 2006

31.
⎟
⎟
⎟
⎟
⎠
⎞
⎜
⎜
⎜
⎜
⎝
⎛
+
+≡
μμ
μ10
1
10
1
1
10
equivC in series with
μ10in series with
⎟
⎟
⎟
⎟
⎠
⎞
⎜
⎜
⎜
⎜
⎝
⎛
+
+
μμ
μ10
1
10
1
1
10

F
μ286.4≡
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Engineering Circuit Analysis, 7
th
Edition Chapter Seven Solutions 10 March 2006

32. ()() ( )( ) pHpppppppL
equiv
6.1797777//77777777//77
&
=++++≡
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Engineering Circuit Analysis, 7
th
Edition Chapter Seven Solutions 10 March 2006

33. (a) Assuming all resistors have value R, all inductors have value L, and all capacitors
have value C,

(b) At dc, 20 μF is open ci
rcuit; 500μ H is short circuit.
Using voltage division,
() V
kk
k
V
x 6.39
1510
10
=+
=

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Engineering Circuit Analysis, 7
th
Edition Chapter Seven Solutions 10 March 2006

34. (a) As all resistors have value R, all inductors value L, and all capacitors value C,

(b) as L is short circ
uit at dc. 0 V
x
V=

+

V
x

-

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Engineering Circuit Analysis, 7
th
Edition Chapter Seven Solutions 10 March 2006

35. Cequiv = { [(100 n + 40 n) || 12 n] + 75 n} || {7 μ + (2 μ || 12 μ)}

nFC
equiv 211.85≡

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Engineering Circuit Analysis, 7
th
Edition Chapter Seven Solutions 10 March 2006

36. Lequiv = {[ (17 p || 4 n) + 77 p] || 12 n} + {1 n || (72 p + 14 p)}

pHL
equiv 388.172≡

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Engineering Circuit Analysis, 7
th
Edition Chapter Seven Solutions 10 March 2006

37. () FCC
xT μ171100161477=++++=−

() ()() JVCCE
xTCC
xT
μμ375.5345.2171
2
1
2
1
22
==−=
−

() nJJEEE
xTTxCCCC
425375.5348.534
=−=−=
−
μ

()
()
nF
n
CVCnE
xxC
x
136
5.2
2425
2
1
425
2
2
==⇒==∴

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Engineering Circuit Analysis, 7
th
Edition Chapter Seven Solutions 10 March 2006

38.
(a) For all L = 1.5H, HL
equiv 75.2
5.1
1
5.1
1
5.1
1
1
5.1
1
5.1
1
1
5.1 =
⎟
⎟
⎟
⎟
⎠
⎞
⎜
⎜
⎜
⎜
⎝
⎛
++
+
⎟
⎟
⎟
⎟
⎠
⎞
⎜
⎜
⎜
⎜
⎝
⎛
+
+=

(b) For a general network of this type, having N stages (and all L values equiv),

∑
=
−
=
n
N
N
N
equiv
NL
L
L
1
1

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Engineering Circuit Analysis, 7
th
Edition Chapter Seven Solutions 10 March 2006

39.
(a) HL
equiv 3
3
1
3
1
3
1
1
2
1
2
1
1
1 =
⎟
⎟
⎟
⎟
⎠
⎞
⎜
⎜
⎜
⎜
⎝
⎛
++
+
⎟
⎟
⎟
⎟
⎠
⎞
⎜
⎜
⎜
⎜
⎝
⎛
+
+=

(b) For a network of this type having 3 stages,

()
()
()
()
()
()
2
32
22
33
3
22
2
1
3
1
3
33
1
2
22
1
1 ++=
+
+
+
+
+=
equivL

Extending for the general case of N stages,

N
1
N
1
1
3
1
3
1
3
1
1

2
1
2
1
1
1
K
K
+
++
++
+
+
+=
equiv
L
N
N(1/N)
1
)3/1(3
1

)2/1(2
1
1 =++++=K

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Engineering Circuit Analysis, 7
th
Edition Chapter Seven Solutions 10 March 2006

40.
()( )
pF
pp
pp
C
equiv 231.0
25.03
25.03
=
+
=

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Engineering Circuit Analysis, 7
th
Edition Chapter Seven Solutions 10 March 2006

41.
()()
nH
n
nn
L
equiv 6291.0
6.2
3.03.2
&
&
&&
==

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Engineering Circuit Analysis, 7
th
Edition Chapter Seven Solutions 10 March 2006

42. (a) Use 2 x 1μH in series with 4 x 1μH in parallel.

(b) Use 2 x 1 μH in paralle
l, in series with 4 x 1μH in parallel.

(c) Use 5 x 1 μH in parallel, in series with 4 x 1μH in parallel.

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Engineering Circuit Analysis, 7
th
Edition Chapter Seven Solutions 10 March 2006

43.

(a)

(b) L 10H L 11.379H=∴ =
eq

(c)
10 10 55
R 10 :10 10 10 , 10 10 10
33 3
55
R 30 11.379
3
=Ω = ++ =
∴ ==Ω
eq
1
C 10F: 5.4545
1/30 1/10 1/ 20
10
C 5.4545

8.788F
3
==
++
∴ =+
eq
=

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Engineering Circuit Analysis, 7
th
Edition Chapter Seven Solutions 10 March 2006

44.

(a)
17 1
:1 , 1.3125F
1/4 1/2 3 3/7 1/2
15 5
: ,C 4 4.833F
1/5 1 6 6
+== =
++
==+=
+
eq
eqoc c
sc
:L 6 1 3 3.857H
: L (3 2 1) 4 2.2 4
=+=
=+
=
eq
eq
oc
sc

1.4194H=
(b)
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Engineering Circuit Analysis, 7
th
Edition Chapter Seven Solutions 10 March 2006

45.

(a)

(b)

(c)

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Engineering Circuit Analysis, 7
th
Edition Chapter Seven Solutions 10 March 2006

46.
200
1
60 mA, (0) 20mA
t
s
ie i
−
==

(a)
200
200
6 4 2.4H L 2.4 0.06( 200)
or 28.8 V
−
−
′=∴= =×−
=−
t
eq s
t
vi e
ve

(b)

(c)

200 200 200
21
60 24 4 36 4mA( 0)
tt t
s
iii e e e t
−− −
=−= − += + >
200 200
1
20014 .8
28.8 0.02 ( 1) 0.02
6 200
24 4mA( 0)
−−
−
=− + = −+
=−>∫
t
tt
o
t
ied t e
et

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Engineering Circuit Analysis, 7
th
Edition Chapter Seven Solutions 10 March 2006

47.
80
1
100 , (0) 20V
t
s
veVv
−
==

(a)
68 0 3
C 0.8 10 ( 80)100
80−− −
′==×−
eq siv e

6.4 10 A
−
=−×
tt
e

(b)
6 3 80 80
1
80
1 6400
10 ( 6.4 10 ) 20 ( 1) 20
80
80 60V
−− −
−
=−× += −+
∴=−
∫
t
tt
o
t
ve dte
ve

(c) 6
380 80
2
80
10 1600
( 6.4 10 ) 80 ( 1) 80
48 0
20 60V
−− −
−
−× += −+
=+ ∫
t
tt
o
t
ve dte
e

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Engineering Circuit Analysis, 7
th
Edition Chapter Seven Solutions 10 March 2006

48.

(a)
6
3
510 0
20 10
1
20
10 8 10
−
−−−
′+× + =
−
++
×
∫
cs cL
c
t
Lc
L
ovv vv
v
vv
vdt

=
(b)
20 206
3
206
1
20 ( ) 12
510
1
( ) 12 10 8 10 0
510
−
−
−
+−+=
×
′−−++× =
× ∫
∫
t
Ls
o
t
LL
o
iii dtv
iidt i i
L

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Engineering Circuit Analysis, 7
th
Edition Chapter Seven Solutions 10 March 2006

49.

3
33
( ): 30mA: 0.03 20 0.6V, 0.6V
9V: 9V, 20mA: 0.02 20 0.4V
0.04cos10 : 0
() 9.2V
():30mA,20mA,
9V: 0; 0.04cos10 : 0.06 0.04( 1000)sin10 2.4sin10 V
×= =
== −×=
=
∴=
== −× −=
cc
cc
c
c
L
LL
vt v
vv
tv
vt
vt
vt v t
3
t

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Engineering Circuit Analysis, 7
th
Edition Chapter Seven Solutions 10 March 2006

50. We begin by selecting the bottom node as the reference and assigning four nodal
voltages:

1, 4 Supernode: 20× 10
-3
e
-20t
=
( )∫
′−×+
−
t
t
tde
0
20
4
321
40V100.02
50
V-V
[1]

and: V
1 – V4 = 0.2 Vx or 0.8V1 + 0.2 V2 – V4 = 0 [2]

Node 2:
dt
de
t
26-
20
212
V
10
100
40 - V

50
V - V
0 ++=
−
[3]
Ref.
V1
V2
V3
V4

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Engineering Circuit Analysis, 7
th
Edition Chapter Seven Solutions 10 March 2006

51. (a) R,R0,A 0 C
also 0 R 0 RC
1
RA 0,
11A
A
AR
1111A
AARCA
1A 1A
Ao r

A
RC
0
RC
′=∞ = =∞∴ = ∴ =
′++=∴=−
−+ − =
−+
=− ∴ = ∴ =
+
∴= − =− + −
++
′′ ′ ′∴=−− + +=
∫∫
io i s
s
ii i
oiio i
o
soo
so oo os viv
iv v v
viv
vvvvi v
v
v idt v v dt
c
vv vv vv
(b)
= +
∫
oo
s
v idtv
c

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Engineering Circuit Analysis, 7
th
Edition Chapter Seven Solutions 10 March 2006

52. Place a current source in parallel with a 1-MΩ resistor on the positive input of a buffer
with output voltage, v. This feeds into an integrator stage with input resistor, R
2, of
1-MΩ and feedback capacitor, C
f, of 1 μF.

sec
10602.1
19ions
dt
dv
Ci
fc
f
××==
−

sec
10602.1
101101
0
19
66ionsVV
dt
dv
C
VV
a
c
f
af −
×+
×
−
=+
×
−
=

sec
10602.1
101
0
19
6
2ionsV
dt
dv
C
R
V
f
c
f
−
×+
×
−
=+
−
=

Integrating current with respect to t,
()( )0'
1
0
2
ffcc
t
f
VVCvdt
R
−=∫

fcf
VC
R
ions
=
××
−
2
19
10602.1

ions
C
Vions
CR
R
VVVV
f
out
f
outoutac
f
×××
−
=⇒×××
−
=⇒−=
−− 1919
2
1
10602.1
1
10602.1

R
1 = 1 MΩ, C f = 1μF
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Engineering Circuit Analysis, 7
th
Edition Chapter Seven Solutions 10 March 2006

53. R 0.5M , C 2 F, R , R 0, cos10 1V=Ω= =∞== −
ioo vtμ

(a)
11
Eq. (16) is: 1 (0)
ARCA
11 1 1
11 (10sin10)1
ARCAA A
111
1 10sin10 cos10 Let A 2000
AAA
10.005sin10 0.0005 0.0005c
⎛⎞⎛⎞
+=− +−
⎜⎟ ⎜ ⎟
⎝⎠ ⎝⎠
⎛⎞⎛⎞ ⎛⎞ ⎛
′

1
cos10
A
⎞
∴+=− +∴+− =−+ −
⎜⎟ ⎜⎟ ⎜⎜⎟
⎝⎠ ⎝⎠ ⎝⎝⎠
⎛⎞

(b) Let A 10sin10 V
svt=∞∴ =
⎟
⎠
∴=+ +− =
⎜⎟ ⎝⎠
∴=+−
∫
t
o
osc
o
o
os s
s
s v
vvd tv
v
vv tv t
vtt
vt os10t

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Engineering Circuit Analysis, 7
th
Edition Chapter Seven Solutions 10 March 2006

54. Create a op-amp based differentiator using an ideal op amp with input capacitor C 1 and
feedback resistor R
f followed by inverter stage with unity gain.

min/
1
60
1
rpm
mV
dt
dvs
CR
R
R
V
fout ×=+=
R
fC1=60 so choose Rf = 6 MΩ and C 1 = 10 μF.
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Engineering Circuit Analysis, 7
th
Edition Chapter Seven Solutions 10 March 2006

55. (a) ∫
−
+=
f
outaR
VV
vdt
L
1
0

'
1
,0
0
dtv
L
R
V
R
V
dtv
L
VV
t
s
f
out
f
out
La
∫∫
−
=⇒=∴==

(b) In practice, capacitors are usually used as capacitor values are m
ore readily
available than inductor values.
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Engineering Circuit Analysis, 7
th
Edition Chapter Seven Solutions 10 March 2006

56. One possible solution:

1
3
1
3
0
11
1
we want 1 V for 1 mV over 1 s.
11
In other words, 1 10
out in
f
out in
0
f f
vvd t
RC
vv
dt
RCR C
−
−
=−
==
=− =−∫
∫
vin
Neglecting the sign (we can reverse terminals of output connection if needed),
we therefore need R
1Cf = 10
–3
.

Arbitrarily selecting C
f = 1 μF, we find R 1 = 1 kΩ.

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Engineering Circuit Analysis, 7
th
Edition Chapter Seven Solutions 10 March 2006

57. One possible solution of many:

vin
100 mV
maximum .
60s
100 mV
In other words, 1 V =
60s
600
in
out
in
out
dv
vRC
dt
dv
dt
vRC
or RC
=−
⎛⎞
=
⎜⎟
⎝⎠
⎛⎞
=
⎜⎟
⎝⎠
=

Arbitrarily selecting C = 1000 μF, we find that R = 600 kΩ.

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Engineering Circuit Analysis, 7
th
Edition Chapter Seven Solutions 10 March 2006

58. One possible solution of many:

vin
100 mV
At 1 litre/s, .
s
100 mV
In other words, 1 V =
1s
10
in
out
in
out
dv
vRC
dt
dv
dt
vRC
or RC
=−
⎛⎞
=
⎜⎟
⎝⎠
⎛⎞
=
⎜⎟
⎝⎠
=

Arbitrarily selecting C = 10 μF, we find that R = 1 MΩ.

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Engineering Circuit Analysis, 7
th
Edition Chapter Seven Solutions 10 March 2006

59. One possible solution:

vin

The power into a 1 Ω load is I
2
, therefore energy = W = I
2
Δt.

()
2
1
2
331
we want 1 mV for 1 mV (corresponding to 1 A ).
, 10 10 , so 1
out
f
out in
vIdt
RC
vv
Thus RC RC
−−
=
==
==∫

Arbitrarily selecting C = 1 μF, we find that we need R = 1 MΩ.

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Engineering Circuit Analysis, 7
th
Edition Chapter Seven Solutions 10 March 2006

60. One possible solution of many:

vin
in
out
dv
vRC
dt
=−
Input: 1 mV = 1 mph, 1 mile = 1609 metres. Thus, on the input side, we see 1 mV corresponding to 1609/3600 m/
s.
Output: 1 mV per m/s
2
. Therefore,

2.237 1
0.447
out
vR
so RC
==
=
C

Arbitrarily selecting C = 1 μF, we find that R = 447 kΩ .

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Engineering Circuit Analysis, 7
th
Edition Chapter Seven Solutions 10 March 2006

61.

(a)

(b)
20 206
3
206
1
20 ( ) 12
510
1
( ) 12 10 8 10 0
510
−
−
−
+−+=
×
′−−++× =
× ∫
∫
t
cs
o
t
cc
o
vvv dti
vvdt v v

c
(c)
6
3
510 0
20 10
1
20
10 8 10
−
−−−
′+× + =
−
++ =
×
∫
Ls Lc
L
t
cL
c
oii ii
i
ii
idt

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Engineering Circuit Analysis, 7
th
Edition Chapter Seven Solutions 10 March 2006

62.

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Engineering Circuit Analysis, 7
th
Edition Chapter Seven Solutions 10 March 2006

63.

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Engineering Circuit Analysis, 7
th
Edition Chapter Seven Solutions 10 March 2006

64.

(a)

(b) “Let i
s = 100e
-80t
A and i 1 (0) = 20 A in the circuit of (new) Fig. 7.62.

(a) Determine v(t) fo
r all t.
(b) Find i
1(t) for t ≥ 0.
(c) Find v
2(t) for t ≥ 0.”

(c) (a)
68
-80t
L 1 4 0.8 H ( ) L 0.8 10 100( 80) V
v(t) 6.43 mV
−−
′== ∴= =× × −
0
∴ =−
t
eq eq s
vt i rμ

(b)
6 3 80 80 80
11 6400
( ) 10 6.4 10 20 ( ) ( 1) 80 60A
80
−− − −
=−× +∴= −= −∫
t
tt
o
t
e dt it e
i it it e
−
=− ∴ = +
it e

(c) it
80
212
() () () 20 60A
t
s

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Engineering Circuit Analysis, 7
th
Edition Chapter Seven Solutions 10 March 2006

65.

In creating the dual of the original circuit, we have lost both v
s and v out. However, we
may write the dual of the original transfer function: i
out/ is. Performing nodal analysis,

)V - (V G V
L
1

21in
0
1
1
S
+′=∫
t
tdi [1]

i out = Ai d = GfV2 + Gin (V2 – V1) [2]

Dividing, we find that

)V - (V G V
L
1
V G )V - (V G

21in
0
1
1
2f12in
S
out+′
+
=
∫
t
td
i
i

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Engineering Circuit Analysis, 7
th
Edition Chapter Seven Solutions 10 March 2006

66. IL = 4/10 = 400 mA.
21
2
L
WLI= = 160 mJ

PSpice verification:

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Engineering Circuit Analysis, 7
th
Edition Chapter Seven Solutions 10 March 2006

67. IL = 4/(4/3) = 3 A.
21
2
L
WLI= = 31.5 J

PSpice verification:

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Engineering Circuit Analysis, 7
th
Edition Chapter Seven Solutions 10 March 2006

68. We choose the bottom node as the reference node, and label the nodal voltage at the top
of the dependent source V
A.

Then, by KCL,

4
0.8
100 20 25 25
A AA
VVV−
++=
A
V

Solving, we find that V
A = 588 mV.

Therefore, V
C, the voltage on the capacitor, is 588 mV (no DC current can flow through
the 75 Ω resistor due to the presence of the capacitor.)

Hence, the energy stored in the capacitor is
()(
2
2311
10 0.588
22
CV
−
= ) = 173 μJ

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Engineering Circuit Analysis, 7
th
Edition Chapter Seven Solutions 10 March 2006

69. By inspection, noting that the capacitor is acting as an open circuit,

the current through the 4 kΩ resistor
is 8 mA. Thus, Vc = (8)(4) = 32 V.

Hence, the energy stored in the capacitor =
() ()
2
2611
5 10 32 2.56 mJ
22
CV
−
=× =

PSpice verification:

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Engineering Circuit Analysis, 7
th
Edition Chapter Seven Solutions 10 March 2006

70. C1 = 5 nF, Rf = 100 MΩ.

()() ()
98
1
5 10 10 30cos100 15cos10 V
s
out f
dv
vRC t t
dt
−
=− =− × =−

Verifying with PSpice, choosing the LF411 and ±18 V supplies:

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Engineering Circuit Analysis, 7
th
Edition Chapter Seven Solutions 10 March 2006

71. PSpice verification

w = ½ Cv
2
= 0.5 (33×10
-6
)[5 cos (75×10
-2
)]
2
= 220.8 μJ. This is in agreement with the
PSpice simulation results shown below.

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Engineering Circuit Analysis, 7
th
Edition Chapter Seven Solutions 10 March 2006

72. PSpice verification

w = ½ Li
2
= 0.5 (100×10
-12
)[5 cos (75×10
-2
)]
2
= 669.2 pJ. This is in agreement with
the PSpice simulation results shown below.

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Engineering Circuit Analysis, 7
th
Edition Chapter Seven Solutions 10 March 2006

73. ∫
+
−
= dtv
LR
VV
f
L
sa
1
0
1

,0==
baVV
∫
+
−
= dtv
LR
V
f
L
s
1
0
1

dt
dVs
R
L
VVVV
outoutaL
f
1
0 =−=−=

() LetRLtA
dt
d
R
L
dt
dVs
R
L
V
f
ff
out
π _;2102cos
1
3
11
=⇒−=−=
R = 1 Ω and L = 1 H.

PSpice Verification: clearly, something rather odd is occuring in the simulation of this
particular circuit, since the output is not a pure sinusoid, but a comb
ination of several
sinusoids.

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Engineering Circuit Analysis, 7
th
Edition Chapter Seven Solutions 10 March 2006

74. PSpice verification

w = ½ Cv
2
= 0.5 (33×10
-6
)[5 cos (75×10
-2
) - 7]
2
= 184.2 μJ. This is in reasonable
agreement with the PSpice simulation results shown below.

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Engineering Circuit Analysis, 7
th
Edition Chapter Seven Solutions 10 March 2006

75. PSpice verification

w = ½ Li
2
= 0.5 (100×10
-12
)[5 cos (75×10
-2
) - 7]
2
= 558.3 pJ. This is in agreement with
the PSpice simulation results shown below.

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Engineering Circuit Analysis, 7
th
Edition Chapter Eight Solutions 10 March 2006

1.
9
4.7 10
() (0) 2 mA
Rt
tL
it i e e
−
−×
==

(a)
i(100 ps) =
(
)
91 2
4.7 10 100 10
2 = 1.25 mAe
−
−× ×

(b)
i(212.8 ps) =
(
)
91 2
4.7 10 212.8 10
2 = 736 Ae μ
−
−× ×

(c)
vR = –iR

vR(75 ps) = ()
( )
912
4.7 10 75 10
2 4700 = 6.608 Ve
−
−× ×
−−
(d) v
L(75 ps) = v R(75 ps) = –6.608 V

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Engineering Circuit Analysis, 7
th
Edition Chapter Eight Solutions 10 March 2006

2.

2
/21
100 mJ at 0.
2
, ( 0) 0.1 316 mA
and ( ) (0) 316 mA
R
t
tL
WLi t
Thus i
it i e e
−
−
== =
==
==

(a) At t = 1 s,

1/ 2
( ) 316 mA = 192 mAit e
−
=

(b) At t = 5 s,
5/2
( ) 316 mA = 25.96 mAit e
−
=

(c) At t = 10 s,
10 / 2
( ) 316 mA = 2.13 mAit e
−
=

(d) At t = 2 s, . Thus, the energy rema
ining is
1
( ) 316 mA = 116.3 mAit e
−
=

21
(2) (2) 13.53 mJ
2
WLi==

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Engineering Circuit Analysis, 7
th
Edition Chapter Eight Solutions 10 March 2006

3. We know that
100
3
() (0) 2 10
R
t
L
it i e e
−
−
==×
t
L
−
, and that i(500 μs) = 735.8 μA.

Thus,
()
( )
66
66
33
100 500 10 100 500 10
(500 10 ) 735.8 10
ln ln
210 210
L
i
−−
−−
−−
−×−×
==
⎛⎞⎛ ⎞××
⎜⎟ ⎜
⎟
××
⎝⎠⎝⎠
= 50 mH

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Engineering Circuit Analysis, 7
th
Edition Chapter Eight Solutions 10 March 2006

4. We know that
3
310
() (0) 1.5
RR
tt
L
it i e e
−
−−
×
== , and that i(2) = 551.8 mA.

Thus,
33
3 10 (2) 3 10 0.5518
ln ln
21.5 2 1.5
i
R
−−
×× ⎛⎞ ⎛ ⎞
=−
⎜⎟ ⎜
⎟
⎝⎠ ⎝ ⎠
=−
= 1.50 mΩ

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Engineering Circuit Analysis, 7
th
Edition Chapter Eight Solutions 10 March 2006

5. We know that
3
310
() (0) 1.5
RR
tt
L
it i e e
−
−−
×
== , and that W(0) = 1 J; W(10
–3
) = 100 mJ.
At t = 0,
() ()
2
31
3 10 0 1 therefore (0) 25.82 A.
2
ii
−
⎡⎤×= =
⎣⎦

At t = 1 ms
,
()()
2
33 31
3 10 10 0.1 therefore (10 ) 8.165 A.
2
ii
−− −
⎡⎤×= =
⎣⎦

Thus,
33
3 10 ( ) 3 10 8.165
ln ln 3.454
(0) 0.001 25.82
it
R
ti
−−
⎛⎞×× ⎛⎞
=− =− = Ω
⎜⎟ ⎜⎟
⎝⎠⎝⎠

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Engineering Circuit Analysis, 7
th
Edition Chapter Eight Solutions 10 March 2006

6.

(a)
Since the inductor current can’t change instantaneously, we simply need to find i L while
the switch is closed. The inductor is shorting out both of the resistors, so i
L(0
+
) = 2 A.

(b)
The instant after the switch is thrown, we know that 2 A flows through the inductor. By
KCL, the simple circuit must have 2 A flowing through the 20-Ω resistor as well. Thus,

v = 4(20) = 80 V.
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Engineering Circuit Analysis, 7
th
Edition Chapter Eight Solutions 10 March 2006

7. (a) Prior to the switch being thrown, the 12-Ω resistor is isolated and we have a simple
two-resistor current divider (the inductor is actin
g like a short circuit in the DC circuit,
since it has been connected in this fashion long enough for any transients to have
decayed). Thus, the current i
L through the inductor is simply 5(8)/ (8 + 2) = 4 A.
The voltage v must be 0 V.

(b) The instant just after the switch is thrown, the inductor current must rem
ain the same,
so i
L = 4 A. KCL requires that the same current must now be flowing through the 12-Ω
resistor, so v = 12(-4) = -48 V.
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Engineering Circuit Analysis, 7
th
Edition Chapter Eight Solutions 10 March 2006

8.

6
36
3
10 / 4 1.25
10 10
(a) (0) 4.5mA, R/L
410
4
4.5 mA (5 ) 4.5
−
−−
===
×

∴=∴=
L
t
LL
ie ise μ
i
= 1.289 mA.
(b) i
SW( 5 μs) = 9 – 1.289 = 7.711 mA.
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Engineering Circuit Analysis, 7
th
Edition Chapter Eight Solutions 10 March 2006

9.

(a)
80 / 0.2
400100
(0) 2A ( ) 2
50
2A,0
−
−
==∴=
=>
t
LL
t
ii t
et

(b)
4
(0.01) 2 36.63mA
Lie
−
==

(c)
11400 400
1
2 1, 2, 1.7329ms
−
===
tt
eet
e

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Engineering Circuit Analysis, 7
th
Edition Chapter Eight Solutions 10 March 2006

10. (a)

5 0 [1]
2 so Eq. [1] can be written as
2
5 0 or
2
2.5 2.5 0
R
R
R
R
R
di
Li
dt
vi
v
d
v
L
dt
dv
v
dt
+=
=−
−⎛⎞
⎜⎟
−⎛⎞⎝⎠
−=
⎜⎟
⎝⎠
+=

(b) Characteristic equation is 2.5
s + 2.5 = 0, or s + 1 = 0
Solving,
s = –1, ()
t
R
vt Ae
−
=

(c) At t = 0
–
, i(0
–
) = 5 A = i(0
+
). Thus, ( ) 10 , 0
t
R
vt e t
−
=−>
()
2
(0 ) 10 6.667 V
3
R
v
−
==
v (0 ) 10 V
R
+
=−

1
(1) 10 3.679 V
R
ve
−
=− = −
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Engineering Circuit Analysis, 7
th
Edition Chapter Eight Solutions 10 March 2006

11.

(a)
/ II
, , 10 10 2.303;
I
II
100, 4.605; 1000, 6.908
−
== =∴==
== = =ll
t oo
o
ooit t
en n
ii
tt
ii
τ
ττ

τ τ
(b)
// 1
11
11(/I) ()
,; at /1,
I(/) ()
Now, ( 1) ( 1) ,
I
At 0, ( 1) 2 / 2
−−
−−
−−
== −== −
⎛⎞
=−+=− −+ = = ⎜⎟
⎝⎠
=−=∴=∴=
tt o
o
odiid
eete
dt d
ti
ymx
b ex e x y
yex ex t
ττ
τ
τ
τ
τ

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Engineering Circuit Analysis, 7
th
Edition Chapter Eight Solutions 10 March 2006

12. Reading from the graph current is at 0.37 at 2 ms

A 10I
ms 2
0
=
=
∴τ

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Engineering Circuit Analysis, 7
th
Edition Chapter Eight Solutions 10 March 2006

13. w = ½ Li
2
, so an initial energy of 15 mJ in a 10-mH inductor corresponds to an initial
inductor current of 1.732 A. For R = 1 kΩ,
τ = L/R = 10 μs, so i L(t) = 1.732 e
–0.1t
A.
For R = 10 kΩ,
τ = 1 μs so i L(t) = 1.732 e
-t
. For R = 100 kΩ, τ = 100 ns or 0.1 μs so
i
L(t) = 1.732 e
-10t
A. For each current expression above, it is assumed that time is
expressed in microseconds.

To create a sketch, we first realise that the m aximum current for any of the th ree cases
will be 1.732 A, and af ter one time constant (10, 1, or 0.1 μs), the current will d rop to
36.79% of this value (637.2 m A); after approximately 5 time constants, the curren t will
be close to zero.

Sketch based on hand analysis Circuit used for PSpice
verification
As can be seen by comparing the two plots, which probably should
have the same x-axis scale
labels for easier
comparison, the PSpice
simulation results obtained
using a parametric sweep
do in fact agree with our
hand calculations.

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Engineering Circuit Analysis, 7
th
Edition Chapter Eight Solutions 10 March 2006

14.

(a)
12
6
6
103.3
101
103.3
−
−
×=
×
×
=τ

(b)

() A 1.121.55
A 1.5
103.3
10432
..
2
1
6126
103.3/105101
6
6
0
2
0
==
=
×
××
=
=
−−
××××−
−
−
epsi
I
IL
ω

(c)
From the PSpice
simulation, we see that
the inductor current is
1.121 A at t = 5 ps, in
agreement with the hand
calculation.

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Engineering Circuit Analysis, 7
th
Edition Chapter Eight Solutions 10 March 2006

15. Assume the source Thévenin resistance is zero, and assume the transient is measured to
5 τ. Then,

9
6
7L5L
5 100 10 secs
RR
(5)(125.7)10
R
10
−
−
−
τ= ∴ τ= = ×
∴>
so R must be greater than 6.285 kΩ.

(If 1τ assumed then
6.285
R 125.7
5
>= Ω)
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Engineering Circuit Analysis, 7
th
Edition Chapter Eight Solutions 10 March 2006

16. For t < 0, we have a current divider with i L(0
-
) = ix(0
-
) = 0.5 [ 10 (1/ (1 + 1.5)] mA
= 2 mA. For t > 0, the resistor through which i
x flows is shorted, so that i x(t > 0) = 0.
The remaining 1-kΩ resistor and 1
-mH inductor network exhibits a decaying current such
that i
L(t) = 2e
-t/τ
mA where τ = L/R = 1 μS.

(a)

(b)
t (μs)
i
x (mA)
2

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Engineering Circuit Analysis, 7
th
Edition Chapter Eight Solutions 10 March 2006

17. () ()
3
2
3
6
12
0 10 so 0 44.72 V
21
Cv v
−
−
−
×
⎡⎤ ===
⎣⎦
10
0
.01
v e e
−−

(a) τ = RC = 100 s

(b) vt
/0
( ) (0) 44.72 V
tRC t
==
= 36.62 Ve
−

Thus, v (20) =
44.72= .
0.01(20)
Since
6
36.62
100 10
v
i 366 nA
R
== =
×

(c)

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Engineering Circuit Analysis, 7
th
Edition Chapter Eight Solutions 10 March 2006

18. If i(0) = 10 A, v(0) = 10 V.
//
() (0) 10 V
tRC t
vt v e e
−−
==
2
−
=
−
=
−
=
−
=

(a) At t = 1 s, ve
1/ 2
(1) 10 = 6.065 V

(b) At t = 2 s, ve
1
(2) 10 = 3.679 V

(c) At t = 5 s, ve
2.5
(5) 10 = 821 mV

(d) At t = 10 s, ve
5
(10) 10 = 67.4 mV

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Engineering Circuit Analysis, 7
th
Edition Chapter Eight Solutions 10 March 2006

19. Referring to Fig. 8.62, we note that τ = RC = 4 s. Thus,

.
/0 .25
() (0) 5 V
tRC t
vt v e e
−−
==

(a) v(1 ms) =
(
)0.25 0.001
5 = 4.999 Ve
−

(b) v(2 m
s) =
(
)0.25 0.002
5 = 4.998 Ve
−

Therefore i(2 ms
) = 4.998 / 1000 = 4.998 mA
(c) v(4 ms) =
(
)0.25 0.004
5 = 4.995 Ve
−

21
2
WCv== 49.9 mJ

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Engineering Circuit Analysis, 7
th
Edition Chapter Eight Solutions 10 March 2006

20. (a)
//
( ) (0) 1.5 V
tRC tRC
vt v e e
−−
==

()
ln
1.5
tvt
RC
− ⎡⎤
=
⎢⎥
⎣⎦

Thus,

9
10
210
7.385
() 0.1
ln 10 ln
1.5 1.5
t
R
vt
C
−
−
−−×
== =
⎡⎤ ⎡⎤
⎢⎥ ⎢⎥
⎣⎦ ⎣⎦
Ω

(b)

We see from the PSpice simulation that our predicted voltage at 2 ns agrees with the information used to
calculate R = 7.385 Ω.

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Engineering Circuit Analysis, 7
th
Edition Chapter Eight Solutions 10 March 2006

21. The film acts as an intensity integrator. Assuming that we may model the intensity as a
simple decaying exponential,

φ(t) = φ o e
-t/ τ

where the time constant τ = RTHC represents the effect of the Thévenin equivalent
resistance of the equipment as it drains the energy stored in the two capacitors, then the
intensity of the image on the film Φ is actually proportional to the integrated exposure:

Φ = dte
t
∫
−
timeexposure
0
/
o
K
τ
φ

where K is some constant. Solving the integral, we find that

Φ =
[ ]1 - K -
time)/ osure(exp
oτ
τφ
−
e

The maximum value of this intensity function is –Kφ oτ.

With 150 m s yielding an image intensity of approxim ately 14% of the m aximum
observed and the knowledge that at 2 s no further increase is seen leads us to estim ate
that 1 – e
–150×10
–3
/ τ
= 0.14, assuming that we are observing single-exponential decay
behavior and that the response speed of the film is not affecting the m easurement. Thus,
we may extract an estimate of the circuit time constant as
τ = 994.5 ms.

This estim ate is consis tent with the addition al observation that a t t = 2 s, the im age
appears to be saturated.

With two 50-m F capacitors connected in parallel for a total capacitance of 100 mF, we
may estimate the Thévenin equivalent resistance from
τ = RC as Rth = τ / C
= 9.945 Ω.

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Engineering Circuit Analysis, 7
th
Edition Chapter Eight Solutions 10 March 2006

22.

3000 / 24 125
30
(a) (v0) 8(50 200) 192V
50
( )
192
c
c
vt e e
−−
=×=
==

(b)
125
0.1 18.421ms
t
et
−
=∴=
192 V
tt

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Engineering Circuit Analysis, 7
th
Edition Chapter Eight Solutions 10 March 2006

23.

(a)
64 4
10 /100 10 10
80 80 V, 0; 0.5 69.31
−− −
== >=∴=
tt t
c
ve e t e t s
μ

(b)
2 20,000 211
80 80 34.66
24
−
==∴ =
t
c
wCe C t s
μ
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Engineering Circuit Analysis, 7
th
Edition Chapter Eight Solutions 10 March 2006

24.

46
4
/2 10 2 10
25 25 25
3 2
0: ( ) 0, 10 5000 10 mA
3
20
( ) 6.667V
3
0: 0 ( ) 6.667
6.667
( ) 6.667 V ( ) 0.3333 mA
20 10
−
−×××
−− −
<==+∴=
∴==
>=∴=
−
∴= ∴= =
×
cs s s
c
t
sc
tt t
cctit i ii
vt
ti vt e
vt e it e e

v
C(t)
i
C(t)

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Engineering Circuit Analysis, 7
th
Edition Chapter Eight Solutions 10 March 2006

25.

()
()
()
()
()
()
Amsi
Vemsv
Amsi
Vemsv
Ai
Vv
03
1203
05.1
5.4205.1
1.00
200
63
63
102050/103
102050/105.1
=
==
=
==
=
=
−−
−−
×××−
×××−
+
+

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Engineering Circuit Analysis, 7
th
Edition Chapter Eight Solutions 10 March 2006

26.

(a)
12
(0 ) 60 30 mA, (0 ) 30 20 mA
23
Lx
ii
−−
=× = =× =

(b)
(0 ) 30 mA, (0 ) 30 mA
Lx
ii
++
== −

(c)
250 / 0.05 5000
1.5
( ) 30 30 mA, (0.3ms)
30 6.694mA
tt
LL
x
it e e i
ei
−−
−
==
== =−

Thus, i
x = –6.694 mA.

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Engineering Circuit Analysis, 7
th
Edition Chapter Eight Solutions 10 March 2006

27.

(a)
ii
500
0.4
(0)4A ()4 A(0 1ms)
(0.8ms) 4 2.681A
t
LL
L
tet
ie
−
−
=∴ = ≤≤
==

(b)
0.5
250( 0.001)
0.25
(1ms) 4 2.426A
( ) 2.426
(2ms) 2.426 1.8895 A
L
t
L
L
ie
it e
ie
−
−−
==

∴ =

−−
==∴

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Engineering Circuit Analysis, 7
th
Edition Chapter Eight Solutions 10 March 2006

28.

(a)
50,000 50,000
1
40 mA 10 40 , 27.73
−−
=∴=∴=
tt
L
ie e t s
μ

(b)
6
0.5
(1000 )50
(1000 )5 10
33
(10 ) 40 24.26mA
24.26 ( 10 )
10 24.26 2.426 0.8863
0.25(1000 R)10 ,1000 R 0.8863 4 10 R 2545
LL
Rt
Ris e i
ets
enμ
μ
−
−
−+
−+×
== ∴
=>

∴
−+
=∴=
=+ +=×× Ω l
∴=

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Engineering Circuit Analysis, 7
th
Edition Chapter Eight Solutions 10 March 2006

29.

(a)
12(0) 20mA, (0) 15mA==ii

50000 100000
( ) 40 45 V (0) 85V
−−
∴ =+ ∴=
tt
vt e e v

(b)
0.75 1.5
(15 ) 40 45 28.94Vvs e eμ
−−
=+=

(c)
50000 100000 . 50000
2
50000
40 45 Let
10
45 40 8.5 0
40 1600 1530
0.17718, 0
90
0.17718, 34.61
ttt
t
ee ex
xx
x
ets
μ
−− −
−
=
85
=+

∴ +−=
−± +
∴==<
==∴
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Engineering Circuit Analysis, 7
th
Edition Chapter Eight Solutions 10 March 2006

30.

22
2
12 1
12 1 2
50 5011 2
12 12
12
12
12
50 /1000
22
1
12R R 2R
0: , (0)
R+R R R
2R 2R R
0: ( )
RR R+R
2R R
(0 ) 10 R R 5 . Also, (1ms)
R+R
5 10 0.05R 0.6931 R 13.863
111
R 7.821
13.863 R 5
−−
+
−
<= ↓=
+
>= ∴=
+
∴ == ∴ =Ω
== ∴ = ∴ = Ω
∴ +=∴= Ω
RL
Rt Rt
LR
RR
Rtv i
tit e v e
vv
e

31.
(a)
75024
(0) 0.4A ( ) 0.4 A, 0
60
−
== ∴ = >
t
LL
ii t e t

(b)
750
5
24 20V, 0
6
3
(0 ) 50 0.4 7.5V
8
() 7.5 V, 0
+
−
=× = <
=× ×=
∴ =>
x
x
t
x
vt
v
vt e t

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Engineering Circuit Analysis, 7
th
Edition Chapter Eight Solutions 10 March 2006

32.
25 / 0.5 50
3
20 10 25
4
25 10 10 A, 0
−−
=×+ =
∴=Ω∴= = >
L
in L L
ttin
in L
Li
vii
v
viee
i

t

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Engineering Circuit Analysis, 7
th
Edition Chapter Eight Solutions 10 March 2006

33.

= 2.5 A
24 / 8 3
3
11
11
64 40
(0) 5A
440848
55A
() 2.5 A, 0; ( 0.1)
(0.03) 2.285 A, (0.1) 1.
8
−−
−
−
=×=
+
∴= =
∴= > −
==
L
tt
L
t
i
ie e
it e t i
ii
52 A

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Engineering Circuit Analysis, 7
th
Edition Chapter Eight Solutions 10 March 2006

34.

(a)
100
1.5
(0) 4A 4 A, 0 15ms
(15ms) 4 0.8925 A
−
−+
=∴= <<
∴ ==
t
LL
L
ii et
ie

(b)
20( 0.015)
0.3
15ms: 0.8925 A
(30ms) 0.8925 0.6612A
t
L
L
ti e
ie
+− −
+−
>=

∴ ==
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Engineering Circuit Analysis, 7
th
Edition Chapter Eight Solutions 10 March 2006

35.

(a)
11 2 2(0 ) (0 ) 10A, (0 ) (0 ) 20A (0 ) 30Aii ii i
+− + − +
== ==∴=

(b)
0.08 5
L / R ms 1.6667ms;
48 3
eq eqτ====

(c) ;
12(0 ) 10A, (0 ) 20Aii
−−
==
600
() 30 A
t
it e
−
=

(d)
600
48 1440 V
t
vi e
−
=− =−

(e)

600 600 600
10
0
600
2
0
600 600
0
10( 440) 10 24 10 24 14A
2.5( 1440) 20
620614A
−−−
−
−−
=− += += −
=− +
=+=+∫
∫
t
tt tt
t
t
tt t
ie dtee
dt
ee
ie

22
22
L
2 1200
R
0011
W (0) 0.1 10 0.4 20 5 80 85J
22
11
W ( ) 0.1 14 0.4 14 9.8 39.2 49J
22
900 48
W 48 900 48 ( 1) 36J
1200
49 36 85 checks
L
t
idt e dt
∞∞
−
=× × +× × =+ =
∞= × × + × × = + =
×
==× =−=
−
(f)
+=
∫∫
∴

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Engineering Circuit Analysis, 7
th
Edition Chapter Eight Solutions 10 March 2006

36.

(a)
1
c1
2 2 100 2
(0) 100 33.33V; (0 ) 16.667mA
223 223
v (9:59) 33.33V, (9:59) 16.667mA
c
vi
i
−
=× ×= = ×=
++

∴ == (b)
/ 400 300 / 400
1
( ) 33.33 , 10:00 (10:05) 33.33
15.745
15.745 V, (10:05) 3.936mA
4000
t
cc
e t v e
i
−−
+
=>∴=
== =

(c)
τ = 400 s, so 1.2τ = 480 s. v C(1.2τ) = 33.33 e
–1.2
= 10.04 V.
Using Ohm’s law, we find that i
1(1.2τ) = v C(1.2τ)/ 4000 = 2.51 mA.

(d) PSpice Verification:
vt

We see from the DC analysis of the circuit that
our initial value is correct; the Probe
output confirms our hand calculation
s, especially for part (c).
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Engineering Circuit Analysis, 7
th
Edition Chapter Eight Solutions 10 March 2006

37.

25
0: 1.25 34 100(1.25 0.8 ) 25 0.2A
20
x
xx x x x xi
ti ii ii i>=∴= −++∴=

(a) (0 ) (1.25 0.8 1)0.2 0.290A
si
−
=−+=

(b) (0 ) 0.2A
xi
−
=

(c)
5
( ) 25 0.2 5 V (0 ) 0.05A
100
tt
cx
vt e e i
−− +
=× = ∴ = =

(d)
34 20 33.2
0.8 (0 ) 0.04A (0 ) 0.04 0.2767A
120 120 120
xsii
++
=∴=−×==

(e)
0.41
(0.4) 5 0.03352A
100
xie
−
=× =

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Engineering Circuit Analysis, 7
th
Edition Chapter Eight Solutions 10 March 2006

38.
(a)
6
10 /(10 50 200) 20000
(0) 10V ( ) 10 10 V
t t
cc
vvte e
−+ −
=∴ = =

(b)
2
10
( 100 ) (0 ) 50mA
200
150
(100 ) 10 5.413mA
10 40 250
AA
Ais i
iseμ
μ
−
−
−==
⎛⎞
==
⎜⎟
+⎝⎠

(c) PSpice Verification.
=

From the DC simulation, we see that PSpice verifies our hand calculation of i
A = 50 mA.
The transient response plotted using Probe indicates that at 100 μs, the current is approx
imately
5.46 mA, which is within acceptable round-off error compared to the hand calculated value.
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Engineering Circuit Analysis, 7
th
Edition Chapter Eight Solutions 10 March 2006

39.
(a)
1
12
() 8( 1) 6mA( 0)
12 4
it t=− =− <
+

(b)
63
10 / 5 2 10 100
100
4 12 6
() , 0
()
2
48
48 V
12 mA, 0
−×× −
−
=Ω =

∴ == >
∴=>
c
tt
tkv
e e
e t
cvt t
it

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Engineering Circuit Analysis, 7
th
Edition Chapter Eight Solutions 10 March 2006

40.

(a)
66
10 / 8 10 / 0.8
1,250,000 125,000
(0) 20V, (0) 80V
20 , 80
80 20 V, 0
−−
−−
==
∴==
∴=−= − >
CLeft CRIGHT
tt
CL CR
tt
out CR CL
vv
(b)
veve
vvv e e t
1.25 0.125
6.25 0.625
(0 ) 60V; (1 ) 80 20 5.270V
(5 ) 80 20 10.551V
+− −
−−
==−=
=− =−
out out
out see
vse eμ
μvv

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Engineering Circuit Analysis, 7
th
Edition Chapter Eight Solutions 10 March 2006

41. (a)

6
10 / 4 250,000
0.25 40
0: 0 20V( 0)
5104
10.25
0: Apply 1V 0.1 0.25A
5
1
R4
0.25
( ) 20 20 V( 0)
−−
−−
<++= ∴=<
−
>=∴+−=
∴= =Ω
∴= = >
cccc
c
ci n
eq
tt
cv vvv
tv t
i
vt e
e t

(b) v
C(3 μs) = 9.447 V

(c) PSpice verification. Note that the switch parameters had to be changed in order to
perform this simulation.
tv

As can be seen from the simulation results, our hand calculations are accurate.
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Engineering Circuit Analysis, 7
th
Edition Chapter Eight Solutions 10 March 2006

42.
6
63
10 /( 1000) 500 /( 1000)
1000 / 2742.4
10 ( 10 )
1
10
0: (0) 60V
50
01ms:60
60
R500
1.2 0.18232 2 5.4848, 1742.4
R 1000 500
(1ms) 60 41.67V
1ms : 41.67 /(1742.4 R 1000)
25 41.67
oo
c
tR R
c
o
o
o
c
t
ctv
tve e
nR
ve
tv e
e
−
−+ −+
−
−−
−
<=
<< = ∴
∴==∴+==Ω
+
∴= =
>= +
∴=
l

00( )
1
1
3
11
1 .1000
0.5108 ,1742.4 R 1000
1742.4 R 1000
11
1957.6, R 1000 215.2 10 R 274.2
R 215.2
−
∴= +
+
==+=∴ =Ω

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Engineering Circuit Analysis, 7
th
Edition Chapter Eight Solutions 10 March 2006

43.

(a) With the switch closed, define a nodal voltage V
1 at the top of the 5-kΩ resistor.
Then,

0 = (V
1 – 100)/ 2 + (V1 – VC)/ 3 + V1/ 5 [1]
0 = V
C/ 10 + (VC – V1)/ 3 + (VC – 100) [2]

Solving, we find that V
C = vC(0
-
) = 99.76 V.

(b)
7
10 / 3939 2539
0: R 10 6.5 3.939 87.59 87.59 V( 0)
−−
>==Ω∴= = >
tt
eq c
tk v ee t
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Engineering Circuit Analysis, 7
th
Edition Chapter Eight Solutions 10 March 2006

44. t < 0:

39 6
111 11 1
111 1
/ 75 10 2 10 10 /150 6667
6667
1 412 4 20 0.5mA (0) 6 20 26
(0v) 13V
0:
Apply 1mA 1 0.6 2.5mA; 30 75V R 75
( ) 13 13 13
( ) 0.4333 mA ( 0)
310
−
−××× − −
−
=+ ∴= ∴ =+ =
=
>←∴+=∴=±==∴=Ω
∴= = =
∴= = >
×
c
c
in eq
ttt
c
toiii v iii
ti ii v ik
vt e e e
v
it e t

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Engineering Circuit Analysis, 7
th
Edition Chapter Eight Solutions 10 March 2006

45.

(a)
12(0 ) 100V. (0 ) 0, (0 ) 0
Rvvv
−− −
===

(b)
12(0 ) 100V. (0 ) 0, (0 ) 100V
Rvvv
−++
===

(c)
6420 5
10 2 10 8 10
20 5
s
τ
−−×
=××=×
+
2

(d)
12.5
( ) 100 V, 0
t
R
vt e t
−
=>

(e)
12.5
4()
() 5 mA
210
tRvt
it e
−
==
×

(f)
63
3 12.5 12.5 12.5
1
12.5 12.5 12.5
2
10 10
( ) 5 10 100 100 20 80V
20 50
1000
() 5 0 80 0 80 80V
5
−− − −
−−−
=−× += +=− +
=+ =−+ =−+∫
∫
t
tt tt
o
o
t
tt tt
o
o
e dt e e
vt e dt e e
vt

(g)
62 62
12
62
12
625 4 4 611
( ) 20 10 80 64mJ, ( ) 5 10 80 16mJ
22
1
(0) 20 10 100 100mJ, (0) 0
2
25
25 10 2 10 2 10 ( 1)10 20mJ
25
64 16 20 100 checks
−−
−
∞
−− −
∞= × × × = ∞ ×× × =
=× × × = =
=× ××=××− =
−
++ =
∫
cc
cc
t
R
o
ww
wed t
ww

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Engineering Circuit Analysis, 7
th
Edition Chapter Eight Solutions 10 March 2006

46.

(a) 0: 1mA (0) 10V, (0) 1mA (0) 10V, 0
sc L xti v i v t<=∴± =↓=−∴= <

(b) 0:t>
49
/10 20 10 5000
3 3 / 0.1 3 10000 10000
L
5000 10000
( ) 10 10 V
i() 10 10 10 A () V, 0
()
tt
c
tt t
L
xcL
vt e e
te e vtet
vvvt e e t
−
−×× −
−− − −
−−
= =
=− − =− ∴± = >

10 V,
0
tt

∴=− = − >
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Engineering Circuit Analysis, 7
th
Edition Chapter Eight Solutions 10 March 2006

47.

(a) 0: 20V 20V, 20mA ( ) 20mA, 0<= ∴= = ∴= <
scL xtv v i it t

(b) 0:
84
10000 / 2 10 10 5000
8 5000 5000
10000 5000 10000 5000
0 ( ) 0.02 A; (
( ) 2 10 20( 5000)
( ) ( ) ( ) 0.02 0.002
) 20 20 V
2 m
A
A 20 2 m
A
tt
c
tt
ttt
t
sL
c
t
xLc
it e vt e e
it e e
it it it e e e e
−
−− × −
−−−
−−−−
>=∴= = =
↓=××− =−
=+= − = −
tv

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Engineering Circuit Analysis, 7
th
Edition Chapter Eight Solutions 10 March 2006

48.

PSpice verification: Note that the switch properties were changed.
()
blow.not does fuse thems, 100 prior toA 1 than less todropped hascurrent thesince
A .868501.11.0
A )( :0
A 1.1
909.0
1
R
V
)0(
1.0363.2
363.2
=
∴
==
=>
===
×−
−
−
t
esi
etit
i
L
t
L
L

We see from the simulation result that the current through the fuse (R3) is 869 mA
, in
agreement with our hand calculation.
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Engineering Circuit Analysis, 7
th
Edition Chapter Eight Solutions 10 March 2006

49. ()6()6(2)3(4) Vvt ut ut ut=− −+−

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Engineering Circuit Analysis, 7
th
Edition Chapter Eight Solutions 10 March 2006

50. ( ) 2 ( ) 2 ( 2) 8 ( 3) 6 ( 4) Ait ut ut ut ut=+ −−−+−

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Engineering Circuit Analysis, 7
th
Edition Chapter Eight Solutions 10 March 2006

51. (a) f(–1) = 6 + 6 – 3 = 9

(b) f(0
–
) = 6 + 6 – 3 = 9

(c) f(0
+
) = 6 + 6 – 3 = 9

(d) f(1.5) = 0 + 6 – 3 = 3

(e) f(3) = 0 + 6 – 3 = 3

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Engineering Circuit Analysis, 7
th
Edition Chapter Eight Solutions 10 March 2006

52. (a) g(–1) = 0 – 6 + 3 = –3

(b) g(0
+
) = 9 – 6 + 3 = 6

(c) g(5) = 9 – 6 + 3 = 6

(d) g(11) = 9 – 6 + 3 = 6

(e) g(30) = 9 – 6 + 3 = 6

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Engineering Circuit Analysis, 7
th
Edition Chapter Eight Solutions 10 March 2006

53. (a)
1
1
11
300 ( 1)V, 120 ( 1)V; 3 ( )A
100
1.5: ( 1.5) 3 1A
300
120
0.5: ( 0.5) 1 0.6A;
300
120 300 120
0.5: 0.4A; 1.5: 0.6A
300 300 300
=− =−+=−
=− − = × =
−
=−=+=
==−=−==−=
AB cvu tv utiut
ti
ti
ti t
i

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Engineering Circuit Analysis, 7
th
Edition Chapter Eight Solutions 10 March 2006

54.

600 ( 1) , 600( 1) ( )V, 6( 1) ( 1)A
AB cvtutvv tutitut
↑=+=+ =−−
(a)
11
1
11.5: 0; 0.5: 600( 0.5) /300 1A
600(0.5) 600(1.5)
0.5: 4A
300 300
600(1.5) 600(2.5) 1
1.5: 6 0.5 3 5 1 9A
300 300 3
=− = =− = − =−
== + =
= = + + ×× =++=
ti ti
ti
ti

(b)

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Engineering Circuit Analysis, 7
th
Edition Chapter Eight Solutions 10 March 2006

55.

(a) 2(1) 3(1) 4(3) 3 4 1uu u−−
+ =−+=

(b)
[5uu−(2)] [2 (1)] [1 ( 1)]
431
12
u+−−
=
××=

(c)
(1) 1
4 (1) 4 1.4715
u
eu e
−− +
==
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Engineering Circuit Analysis, 7
th
Edition Chapter Eight Solutions 10 March 2006

56.

(a)
100 20
0: 0 10 6A
50 50
60
0: 0 0 2A
30
x
xti

(b) t < 0: The voltage source is shorting out the 30-Ω re
sistor, so i x = 0.
t > 0: i
x = 60/ 30 = 2 A.
ti
<=++×=
>=++=

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Engineering Circuit Analysis, 7
th
Edition Chapter Eight Solutions 10 March 2006

57. 0.5:t=−
200 1/50 1
50 25 16.667, 2 3 2.5A
66.67 1/50 1/ 25 1/50 2
xi==− =−=
++

0.5:t=
200
3A
66.67
xi==

1.5:t=
100 1
3 2
66.67
xi=− .5A
3
×=

2.5:t=
200 100
2A
50
xi
−
==

3.5:t=
100
2A
50
xi=− =−
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Engineering Circuit Analysis, 7
th
Edition Chapter Eight Solutions 10 March 2006

58. ( ) 4 16 ( ) 20 ( 4) 6 ( 6)Vvt ut ut ut=− + − − −
t (s)
v(t) (V)
4
-12
8
2
0 4 6

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Engineering Circuit Analysis, 7
th
Edition Chapter Eight Solutions 10 March 2006

59. (a) 7()0.2()8( 2)3
(1
) 9.8 volts
−+−
=
ut ut t
v
+

(b) Resistor of value 2Ω

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Engineering Circuit Analysis, 7
th
Edition Chapter Eight Solutions 10 March 2006

60.
()( ) 1 ( ) A and ( )
R
t
o L
R
V
it e ut v t it R
R −⎛⎞
=− =⎜⎟
⎝⎠

(a)
()
1000
() 1 () V = 1.21 () V
R
t
tL
Ro
vt V e ut e ut
−
−⎛⎞
=− −⎜⎟
⎝⎠

(b)
()
32
(2 10 ) = 1.2 1 V = 1.038 V
R
ve
−−
×−

(c)

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Engineering Circuit Analysis, 7
th
Edition Chapter Eight Solutions 10 March 2006

61.

(a)
200000
() (2 2 ) ()
m
t
L
it e ut
−
=−
mA

(b)
vt
33
200000 200000
() L 15 10 10 ( 2)
( 200000 ) ( ) 6 ( )V
−−
−−
′==×× −
−=
LL
tt
i
euteut

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Engineering Circuit Analysis, 7
th
Edition Chapter Eight Solutions 10 March 2006

62.

(a)
2.5
1
() 2 2(1 ) ()A ( 0.5) 2A
t
L
it e ut i
−
=+ − ∴− =

(b)
1.25
(0.5) 2 2(1 ) 3.427A
Lie
−
=+ − =

(c)
3.75
(1.5) 2 2(1 ) 3.953A
Lie
−
=+ − =

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Engineering Circuit Analysis, 7
th
Edition Chapter Eight Solutions 10 March 2006

63.

(a)
it

(b)
1000
1
( ) (100 80 ) ( )V
t
vt e ut
−
=−
20 / 0.02
1000
() (4 4 ) ()
() 4(1 ) ()A
t
L
t
L
e ut
it e ut
−
−
=−
∴ =−

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Engineering Circuit Analysis, 7
th
Edition Chapter Eight Solutions 10 March 2006

64. (a) 0 W

(b) The total inductance is 30 || 10 = 7.5 mH. The Thévenin equivalent resistance is
12 || 11 = 5.739 kΩ. Thus, the circuit time
constant is L/R = 1.307 μs. The final value of
the total current flowing into the parallel inductor combination is 50/12 mA = 4.167 mA.
This will be divided between the two inductors, so that i(∞) = (4.167)(30)/ (30 + 10) =
3.125 mA.

We may therefore write i(t) = 3.125[1 – e-
10
6
t/ 1.307
] A. Solving at t = 3 μs,
we find 2.810 A.

(c) PSpice verification

We see from the Probe output that our hand calculations are correct by verifying using
the cursor
tool at t = 3 μs.

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Engineering Circuit Analysis, 7
th
Edition Chapter Eight Solutions 10 March 2006

65. τ = L/RTH =
66
6
30 10 30 10
910 s
5||10 3.333
−−
−
××
==×

(a) ( ) ( ) ( )
fn
it i t i t=+

6
10
9 9
and
5t
nf
iAe i
−
==
Thus,
6
10
9
9
()
5 t
it Ae
−
=+
At t = 0, i(0
-
) = i(0
+
) = 4.5/5. Thus, A = –4.5/5 = 0.9
so

6
10
9
9
() 0.9
5 t
it e
−
=− A

(b) At t = 1.5 μs, i =
1.5
9
9
( ) 0.9 1.038 A
5 t
it e
−
=− =

(c)

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Engineering Circuit Analysis, 7
th
Edition Chapter Eight Solutions 10 March 2006

66. τ = L/Req =
33
45 10 45 10
0.0135 s
10 || 5 3.333
−−
××
==

(a) v ( )
Rf n
t v v=+
v
f(t) = 0 since inductor acts as a short circuit.
Thus, .
74.07
()
t
Rn
vt v Ae
−
==

At t = 0, i
L = 12/10 = 1.2 A = i L(0
-
) = iL(0
+
).

Writing KVL for this instant in time, 16 – 10(1.2 + v
R/5) = v R
74.0744
Therefore (0 ) V and hence ( ) V
33
t
RR
vv te
+−
==

(b) At t = 2 ms
,
( )
3
74.07 2 104
(2 ms) V = 1.15 V
3
R
−
−×
=ve

(c)

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Engineering Circuit Analysis, 7
th
Edition Chapter Eight Solutions 10 March 2006

67. τ =
3
510
50 s
100
eq
L
R
μ
−
×
==

11
()
1
f n
vt v v=+ where v
1
6 V since the inductor acts as a short circuit
f
=

Therefore
6
10
50
1
() 6
t
vt . Ae
−
=+

At t = 0
-
, iL = 0 = i L(0
+
). Thus, v 1(0
+
) = 0 since no current flows through the resistor.

Hence
6
10
50
1
() 6 1
t
vt V. e
−⎛⎞
=−⎜⎟
⎜⎟
⎝⎠

At t = 27 μs,

27
6 50
1
(27 10 ) 6 1 = 2.5 Vve
−
−⎛⎞
×=− ⎜⎟
⎝⎠

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Engineering Circuit Analysis, 7
th
Edition Chapter Eight Solutions 10 March 2006

68.

(a)
() 10 , 0=<
L
it At

(b)
()
L
it
5/0.5
10
8 2
() 8 2 A, 0
t
t
L
e
it e t
−
−
=+

∴ =+ >

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Engineering Circuit Analysis, 7
th
Edition Chapter Eight Solutions 10 March 2006

69.

(a)
() 2A, 0
L
it t=>

(b)
()
L
it
4/0.1
40
5
() 5 3 A, 0
t
t
L
e
it e t
−
−
=−

∴ =− >

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Engineering Circuit Analysis, 7
th
Edition Chapter Eight Solutions 10 March 2006

70.

(a) 0, 0

(b) 0, 200V

(c) 1A, 100V

(d)

3
4000
4000
11
50 10 1
ms 1(1 ) ( )A, (0.2ms) 0.5507A
200 4
( ) (100 100 ) ( )V, (0.2ms) 144.93V
t
LL
t
ieuti
vt e ut vτ
−
−
−
×
==∴=− =
=+ =

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Engineering Circuit Analysis, 7
th
Edition Chapter Eight Solutions 10 March 2006

71.

P Q, Q , R 125 , L 5H
L LP LQLP5PR125P25
−−
= + = Ω=
∴=∴===∴=∫
Pt Pt Pt
iieedtAe
dt
di
i
dt+=
di
(a)

25 25 25 25 25 25
2510 2
) 2 2 A A
L2 5
21 022
A , (0) A 0 0.08A
125 25 25
−− − −
−
==∴= + = × +Q(
(b)
(c)
2510 10 ( )
Q( ) 2 2 ( ) 0.16 0.08 A, 0
5
−+
==+∴=− >
tut
tu tie t

(d)

25
=+ = =∴=∴==
∫
t
tt tt tt t
o
o
t
tieedteeee
iei i
∴
25 25 25 25
2510 ( ) 2
) 2 ( ) 2 A A
52 5
2
(0) 0 A ( ) 0.08(1 )A, 0
25
−−−
−
==∴= +=+
=∴ =− ∴ = − > ∫
t
tt t t
o
tut
tu tieed tee
ii te t
Q(
25 25 25
25
25 25
22
25
25 2510 ( )cos50
( ) 2 ( )cos50 2cos50
5
2 (25cos50 50sin50 ) A
50 25
1
2 (25cos50 50sin50 ) 25 A
3125 3125
242
cos50 sin50
125 125 125
−−
−−
− −
==∴ = ×+
⎡⎤
Q
∴=+ +
⎢⎥
+⎣⎦
⎡⎤
=+ − ×+
⎢⎥
⎣⎦
=+−
∫
t
ttt
o
t
t
tt
o
t
t tut t
tu t tiet edtAe
e
ie t t e
e
ette
tte
25 25
25
A
22
(0) 0 0 A A 0
125 125
( ) 0.016cos50 0.032sin50 0.016 A, 0
−−
−
+
=∴= − + ∴ =
∴=+− >
tt
t
e
i
it t t e t
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Engineering Circuit Analysis, 7
th
Edition Chapter Eight Solutions 10 March 2006

72.

(a)
100 100
( ) 15A, 0
20 5
Lit t=−=− <

(b) (0 ) (0 ) 15A
LLii
+−
==−

(c)
100
() 5A
20
Li∞= =

(d)
40
() 5 20 A, 0
t
L
it e t
−
=− >

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Engineering Circuit Analysis, 7
th
Edition Chapter Eight Solutions 10 March 2006

73.

L
9000
9000
9000
18 1
(0 ) 0.1A (0 ) 0.1A
60 30 2
i () 0.1 0.1 0.2A
()0
.2 0.1 A, 0
( ) 0.1 ( ) (0.2 0.1 ) ( )A
or, ( ) 0.1 (0.1 0.1 ) ( )A
−+
−
−
−
=×=∴=
+
∞= + =
∴=− >
∴= −+−
=+ −
LL
t
L
t
L
t
L
ii
it e t
it u t e ut
it e ut

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Engineering Circuit Analysis, 7
th
Edition Chapter Eight Solutions 10 March 2006

74.
(a)
30 3
(0 ) 3A, (0 ) 4A
7.5 4
xLii
−−
=×= =

(b) (0 ) (0 ) 4A
xLii
++
==

(c)
10 / 0.5 20
0.8
() () 3A
() 3 1 3 A (0.04)
3 3.449A
−−
−
∞= ∞=
∴=+ =+ ∴
=+ =
xL
tt
xxii
it e e i
e

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Engineering Circuit Analysis, 7
th
Edition Chapter Eight Solutions 10 March 2006

75.

(a)
30
(0 ) (0 ) 3A
10
xLii
−−
===

(b)
30 30 15
(0 ) 3 2.4A
30 7.5 40 10 15
+
=×+×=
++
xi
(c)
6/0.5
12 0.4830 30
() 3A () 30.6
7.5 40
3 0.6 (0.04) 3 0.6 2.629A
t
xx
t
x
ii t
e
−
−−
∴ = −
=− =
e
ei
∞= ×=
=− ∴

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Engineering Circuit Analysis, 7
th
Edition Chapter Eight Solutions 10 March 2006

76.

52 / 0.2 260
260
OC: 0, 4 ( )V
0.2
S : 0C .1 ( ) ,1
2 ( ) 0.6 2
40 60
12 ( ) 12 ( ) ( )
2.6 60 2.6 60 13
4() ()
413 52 (1 )() (1 )()
52 13
60 4.615 (1 ) ( )V
−−
+−
==
−
=+ =+
∴= ∴= = =
×
∴=×=Ω∴= − = −
∴= = −
xoc
xxx
xx
x
xa b
tt
th L
t
xLvvut
vvv
ut ut v v
vut ut ut
vi
ut ut

R ieu teu t
vi eut
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Engineering Circuit Analysis, 7
th
Edition Chapter Eight Solutions 10 March 2006

77.

(a)
OC:
11 1
1
40
100 30 20 0, 2A
80 ( )V
20 10
SC: 10A, 10 20A
20
R 4 () 20(1 ) ()A
−
−++ = =
∴=
×
=↓=+ =
∴=Ω∴ = −
oc
sc
t
th L
iii
vut
ii
it e ut

(b)
40 40
40
40
1
0.1 20 40 ( ) 80 ( )
100 ( ) 80 ( )
() 10 8 ()A
10
tt
L
t
t
ute ut
ut e ut
it e ut
−−
−
−
=×× =
−
ve

== −
∴

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Engineering Circuit Analysis, 7
th
Edition Chapter Eight Solutions 10 March 2006

78. (5)(2) 10 s
eq
RC τ== =
Thus, .
0.2
()
t
n
vt Ae
−
=

.
0.1 5tt
nf
vv v Ae Be
−−
=+= +
At t = 0, v(0) = 0 since no source exists prior to t = 0. Thus, A + B = 0 [1].
As t , . We
need another equation. →∞ ( ) 0v∞→

()
0
4.7 4.7 4.7
(0) 2 therefore or 0.1 5
55
t
dv dv
iC AB
dt dt
+
=
== = −−=
5
[2]

Solving our two equations, we find that A = –B = 0.192.

Thus, vt
()
0.1 5
( ) 0.192
tt
e e
−−
=−

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Engineering Circuit Analysis, 7
th
Edition Chapter Eight Solutions 10 March 2006

79. Begin by transforming the circuit such that it contains a 9.4 cos 4t u(t) V voltage source
in series with a 5 Ω resistor, in series with the 2 F capacitor.

Then we find that

9.4cos4 5
(5)(2)
tiv
dv
v
dt
=+
= +

or 0.1 0.94cos4
dv
vt
dt
+= , so that ( )
0.1 0.1 0.1
( ) 0.94cos4
tt
vt e t e dt Ae
−−
=+
t
∫

Performing the integration, we find that

0.110cos4 400sin 4
( ) 0.94
1 1600
ttt
vt Ae
−+⎡⎤
=+
⎢⎥
+⎣⎦
.

At t = 0, v
= 0, so that
()
0.94
10
1601
A=−

0.10.94
( ) 10 10cos4 400sin 4
1601
t
and vt e t t
−
⎡⎤=− + +
⎣⎦

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Engineering Circuit Analysis, 7
th
Edition Chapter Eight Solutions 10 March 2006

80. (a)
93
10 / 2 10 500000
1
6
(0 ) 3 2V (0 )
9
6
() 26(27) 6V
7
() 68 68 V, 0
( 2 ) (0 ) 2V, (2 ) 6 8 3.057V
−+
−× −
−−
=×= =
∞= − =−
∴ =− + =− + >
−= = =−+=−
cc
c
tt
c
cc c
vv

v
vt e e t
vsv vs e
μμ

(b) PSpice verification.

As can be seen from the plot above, the PSpice simulation results confirm our hand
calcu
lations of v
C(t < 0) = 2 V and v C(t = 2 μs) = -3.06 V
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Engineering Circuit Analysis, 7
th
Edition Chapter Eight Solutions 10 March 2006

81. ()
3
210 50 0.1RCτ
−
==× =

10t
n
vAe
−
=

()
10
( ) 4.5 since 4.5 V
t
CCnCf C
vt v v Ae v
−
=+= + ∞=

Since (0 ) (0 ) 0
CC
vv
−+
==

()
10 10
( ) 4.5 4.5 4.5 1
tt
C
vt e e
−−
=− + = −
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Engineering Circuit Analysis, 7
th
Edition Chapter Eight Solutions 10 March 2006

82.
83
10 /1.75 10 57140
10
(0 ) 10mA, ( ) 2.5mA, (0) 0
1
10 1
(0 ) 1.4286mA 10mA, 0
1.75 4
2.5 (1.4286 2.5) 2.5 1.0714 mA, 0
AAc
AA
tt
Aii v
ii t
ie e t
−
+
−× −
== ∞= =
=× ∴= <
=+ − =− >

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Engineering Circuit Analysis, 7
th
Edition Chapter Eight Solutions 10 March 2006

83.

83 5
10 /10 10
10
(0 ) 2.5mA, ( ) 10mA
4
10 7.5
(0) 7.5V (0 ) 17.5mA
11
10 7.5 10 7.5 mA, 0, 2.5mA 0
−
+
−−
== ∞=
=∴ =+=
=+ =+ > = <
AA
cA
tt
AAii
vi
ie etit

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Engineering Circuit Analysis, 7
th
Edition Chapter Eight Solutions 10 March 2006

84.

(a) i
in (-1.5) = 0

(b) i
in (1.5) = 0

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Engineering Circuit Analysis, 7
th
Edition Chapter Eight Solutions 10 March 2006

85.

(a)
37 3
500
500
12 ( ) 24 ( )V
0: (0 ) 8V (0 ) 8V
2
0: ( ) 24 16V
3
200
RC 10 3 10 2 10
30
( ) 16 24 V, 0
( ) 8 ( ) (16 24 ) ( )
s
cc
c
t
c
t
c
vu tut
tv v
tv
vt
e t
vt u t e ut
−+
−−
−
−
=− − +
<=−∴=−
>∞=×=
=×××=×
∴=− >
∴=−−+−

(b)
500
12 24 8
(0 ) 0.4mA, (0 ) 3.2mA
30 10
24
( ) 0.8mA
30
( ) 0.4 ( ) (0.8 2.4 ) ( )mA
in in
in
t
inii
i
it u
t e ut
−+
−−+
==− = =
∞= =
=− + +

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Engineering Circuit Analysis, 7
th
Edition Chapter Eight Solutions 10 March 2006

86.
6
/ 10 / 33.33
30,000
3
OC
−−
: 0 1
, 3 1 2V
100 100 100
SC: 3V 0.06A
100 100
R / 2/ 0.06 33.33
(1 ) 2(1 )
2(1 )V, 0
th
xx x
xoc
xx
xs c
th oc sc
tRC t
coc
tvv v
vv
vv
vi
vi
vv e e
et
− −
−
−+ =∴= =−=
=∴= + =
∴= = = Ω
∴= − = −
=− >

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Engineering Circuit Analysis, 7
th
Edition Chapter Eight Solutions 10 March 2006

87.

(0 ) 10V (0 ), (0 ) 0
(0 ) 0 ( ) 0 for all
cc in
in in
vvi
iit
−+
+
== =
=∴ = t
−

2.5
0.375 20( 0.5) 7.5( 0.5)
7.5(0.3)
0 0.5 : 10(1 )V
(0.4) 6.321V,(0.5) 7.135V
2010
5 5 50 8
0.5: A ( ) 10 8 V, 4 8
12 6 6 3 3
50 50
( ) 7.135 16.667 9.532 V
33
(0.8) 16.667 9.532
t
c
cc
c
tt
c
c
tsv e
vv
tv
vt e e
ve
−
−×− −−
−
<< = −
==
−
>=∴∞=++==Ω
⎛⎞
=+ − = −
⎜⎟
⎝⎠
15.
∴ =− =662V

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Engineering Circuit Analysis, 7
th
Edition Chapter Eight Solutions 10 March 2006

88.

(a) For t < 0, there are no active sou
rces, and so v C = 0.

For 0 < t < 1, only the 40-V source is active. R
th = 5k || 20 k = 4 kΩ and hence

τ = Rth C = 0.4 s. The “final” value (assuming no other source is ever added) is found
by voltage division to be v
C(∞) = 40(20)/(20 + 5) = 32 V. Thus, we may write
v
C(t) = 32 + [0 – 32] e
–t/ 0.4
V = 32(1 – e
-2.5t
) V.

For t > 1, we now have two sources operating, although the circuit time
constant remains
unchanged. We define a new time axis temporarily: t' = t – 1. Then v
C(t' = 0
+
) =
v
C(t = 1) = 29.37 V. This is the voltage across the capacitor when the second source
kicks on. The new final voltage is found to be v
C(∞) = 40(20)/ (20 + 5) +
100(5)/ (20 + 5) = 52 V.

Thus, v
C(t') = 52 + [29.37 – 52] e
-2.5t'
= 52 – 22.63 e
-2.5(t – 1)
V.
For t < 0,
v
C = 0.
We see from the simulation results that
our hand calculations and sketch are
indeed correct.

(b)

(c)

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Engineering Circuit Analysis, 7
th
Edition Chapter Eight Solutions 10 March 2006

89.

(a)
0: 8(10 20) 240V ( ) 80V, 0
R
tv t t<+= ==<

(b)

(c) 0: ( ) 80V
R
tvt<=

(d)

6
/10 10 100000
100000
0: ( ) 8 30 240V (0 ) 240V
1
( ) : ( ) 8(10 10) 80V
2
( ) 80 160 80 160 V
( ) 80 160 V, 0
cc
c
tt
c
t
R
t v
tv
vt e e
vt e t
−
+
−× −
−
<=×=∴=
=∞ ∞= × + =
tv
∴ =+ =+
∴ =+ >
6
/ 50 10 20000
20000
(0 ) 80V, ( ) 240V ( ) 240 160 240 160 V
20 80
(0 ) 80V, (0 ) 8 10 10 32 16 48V
30 20 50
() 80V () 8032 V, 0
tt
cc c
RR
t
RR
vte e
vv
vvtet
−
−− × −
−+
−
= ∞= ∴ =− =−
==×+×=+=
+
∞= ∴ = − >
vv
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Engineering Circuit Analysis, 7
th
Edition Chapter Eight Solutions 10 March 2006

90.

6
1
11
6
2
10 /( 100)
1000 /( 100) 1000 /( 100)
1
1
10 /( 100) 31 000
2
2
2
0: 0
01ms:9(1 )
1
89(1 ),
9
1000
2.197, R 355.1

R100
1m
s : 8 , 10 1 8 (R 100)
1000
2.079, R 480.9 100 380.9
R 100
c
tR
c
RR
tR
ctv
tve
ee
tve tt e
−+
−+ −+
′−+ −−
<=
<< = −
∴=− =
∴ ==Ω
′>= =−∴− +
+
==−=Ω
+
∴
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Engineering Circuit Analysis, 7
th
Edition Chapter Eight Solutions 10 March 2006

91.

2000
,
1000
,
,,
2000 1000
1000 1000 2
1000
1000
200 V
100(1 )V
0
200 100 100
100 200( ) 100 0,
100 10,000 80,000
0.25 0.75
400
0.5, 0.6931ms
−
−
−−
−−
−
−
=
=−
=−=
∴=−
∴+ −=
−± +
==
∴==
t
xL
t
xc
xxLxc
tt
tt
t
t
ve

−±
vv v
ee
ee
e
et
ve

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Engineering Circuit Analysis, 7
th
Edition Chapter Eight Solutions 10 March 2006

92.

()
W0.02
V 3.31000//900107
W0.08
V 6.3900107.
W0.00110001.00
2
3
2
3
322
==
=ΩΩ××=
==
=××==
=×==<
−
−
R
V
P
V
R
V
P
RIV
RItP
final
final
init
init

8
6
4
2
Power (W)
07
Time (ms)

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Engineering Circuit Analysis, 7
th
Edition Chapter Eight Solutions 10 March 2006

93. For t < 0, the voltage across all three capacitors is simply 9 (4.7)/ 5.7 = 7.421 V. The
circuit time constant is
τ = RC = 4700 (0.5455×10
-6
) = 2.564 ms.

When the circuit was first constructed, we assume no energy was stored in any of the
capacitors, and hence the voltage across each was zero. When the switch was closed, the
capacitors be
gan to charge according to ½ Cv
2
. The capacitors charge with the same
current flowing through each, so that by KCL we may write

dt
dv
C
dt
dv
C
dt
dv
C
3
3
2
2
1
1
==

With no initial energy stored, integration yields the relationship C 1v1 = C2v2 = C3v3
throughout the charging (i.e. un
til the switch is eventually opened). Thus, just prior to
the switch being thrown at what we now call t = 0, the total voltage across the capacitor
string is 7.421 V, and the individual voltages may be found by solving:

v 1 + v 2 + v 3 = 7.421
10
-6
v1 – 2×10
-6
v2 = 0
2×10
-6
v2 – 3×10
-6
v3 = 0

so that v 2 = 2.024 V.

With the initial voltage across the 2-uF capacitor now known, we may write

v(t) = 2.024 e
-t/ 2.564×10
-3
V

(a) v(t = 5.45 ms
) = 241.6 mV.

(b) The voltage across the entire capacitor string can be written as 7.421 e
-t/ 2.564×10
-3
V.
Thus, the voltage across the 4.7-kΩ resistor
at t = 1.7 ms = 3.824 V and the dissipated
power is therefore 3.111 mW.

(c) Energy stored at t = 0 is ½ Cv
2
= 0.5(0.5455×10
-6
)(7.421)
2
= 15.02 μJ.

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Engineering Circuit Analysis, 7
th
Edition Chapter Eight Solutions 10 March 2006

94. voltage follower
2() ()
ovt vt∴ =

6
2
10 / 0.5 200
10,000
( ) 1.25 ( )V ( )
( ) 1.25 ( )
1.25 ( )V
−×
−
==
=
=
o
x
t
vt ut vt
vt e
ut
eut

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Engineering Circuit Analysis, 7
th
Edition Chapter Eight Solutions 10 March 2006

95. This is a voltage follower
2() ()
ovt vt∴ = , where v 2(t) is defined at the non-inverting input.
The time constant of the RC input circuit is 0.008(1000+250) = 10 s.

Considering initial conditions: v
C(0
-
) = 0 ∴ v C(0
+
) = 0.
Applying KVL at t = 0
+
,

5 = v 250/250 + v 2/1000.

Since at t = 0
+
v250 = v2, we find that v 2(0
+
) = 1 V.
As t →
∞, v 2→ 0, so we may write,

v
o(t) = v 2(t) = 1.0e
-t/10
u(t) V.

PSpice verification:

In plotting both the hand-derived result and the PS
pice simulation result, we see that the
ideal op amp approximation holds very well for this particular circuit. Although the 741
contains internal capacitors, it does not introduce any shorter time constants than that of
the input circuit.
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Engineering Circuit Analysis, 7
th
Edition Chapter Eight Solutions 10 March 2006

96. For t < 0, the current source is an open circuit and so i 1 = 40/ 50 = 0.8 A.
The current through the 5-Ω resistor is [40 – 10(0.8)]/ 5 = 7.2 A, so the inductor current
is equal to – 7.2 A

PSpice Simulation

From the PSpice simulation, we see that our t < 0 calcu
lation is indeed correct, and find
that the inductor current at t = 50 ms is 7.82 A.

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Engineering Circuit Analysis, 7
th
Edition Chapter Eight Solutions 10 March 2006

97. (a)

And we may write v
o(t) = -0.2[1 + e
-20×10
3
t
]u(t) V.

(b) PSpice verification:
20,000
1 4
7 20,000 20,000
4
20,000
32 0,000
20,000 20,0004
0 (virtual gnd) ( )A
10
4
10 0.2
10
( ) 0.2(1 ) ( )
() 10 () 0.4 ()V
() () () ( 0.2 0.2 0.4 ) ()
t
t
tt t
co
o
t
c
t
R
tt
ocR
vi e ut
ved t
e
vt e ut
vt it e ut
vt vt vt e e ut
−
−−
−
−
−−
=∴ =
∴= =−
∴= −
∴= =
∴=−− =−+ −
∫

We can see from the simulation result that our ideal op amp approxim
ation is not
providing a great deal of accuracy in modeling the transient response of an op amp in this
particular circuit; the output was predicted to be negative for t > 0.

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Engineering Circuit Analysis, 7
th
Edition Chapter Eight Solutions 10 March 2006

98. One possible solution of many: implement a capacitor to retain charge; assuming the
light is left on long enough to fully charge the capacitor, th e stored ch arge will run the
lightbulb after the wall switch is turned off. Taking a 40-W light bulb connected to 115
V, we estimate th
e resistance of the light bu lb (which chan ges with its temperature) as
330.6 Ω. We define “on” for the light bulb som ewhat arbitrarily as 50% intensity, taking
intensity as proportional to the dissipated power. Thus, we need at least 20 W (246 mA
or 81.33 V) to the light bulb for 5 seconds after the light switch is turned off.

The circuit above contains a 1-MΩ resist
or in parallel with the capacitor to allow current
to flow through the light bulb when the light switch is on. In order to determ ine the
required capacitor size, we first recognise that it will see a Thevenin equivalent resistance
of 1 MΩ || 330.6 Ω = 330.5 Ω. We want v
C(t = 5s) = 81.33 = 115e
-5/τ
, so we need a
circuit time constant of t = 14.43 s and a capacitor value of
τ/ Rth = 43.67 mF.

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Engineering Circuit Analysis, 7
th
Edition Chapter Eight Solutions 10 March 2006

99. Assume at least 1 μA required otherwise alarm triggers.

Add capacitor C.

6
1
10
6
(1) 1 volt
1000
(0) .1.5 1.496 volts
1002.37
1
We have 1 1.496 or C 2.48 F
10 (1.496)
−
=
==
∴ ===
l
c
c
Cv
v
e
n
μ

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Engineering Circuit Analysis, 7
th
Edition Chapter Eight Solutions 10 March 2006

100. (a) Note that negativ e times are not perm itted in PSpice. The only way to m odel this
situation is to shift the time axis by a fixed amount, e.g.,
1tt
′=+.

(b) Negative times are not permitted in PSpice. The only way to model this situation is to
shift the time axis by a fixed amount, e.g., 2tt′=+.

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Engineering Circuit Analysis, 7
th
Edition Chapter Eight Solutions 10 March 2006

101. (a) / 0.1 sLRτ== . This is much less than either the period or pulsewidth.

(b)

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Engineering Circuit Analysis, 7
th
Edition Chapter Eight Solutions 10 March 2006

102. (a) /1 LR sτ= . This is much less than either the period or pulsewidth. =

(b)

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Engineering Circuit Analysis, 7
th
Edition Chapter Eight Solutions 10 March 2006

103.

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Engineering Circuit Analysis, 7
th
Edition Chapter Eight Solutions 10 March 2006

104.
6
3
500 10
34 ns
14.7 10
eq
L
R
τ
−
×
== =
×

The transient response will therefore have the form .
6
29.4 10t
Ae
−×

(a)

(b)

(c)

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Engineering Circuit Analysis, 7
th
Edition Chapter Nine Solutions 10 March 2006

1. Parallel RLC circuit:

(a)
() ()
31
6611 1
175 10 s
2 2 (4 ||10)(10 ) 2 (2.857)(10 )RC
α
−
−−
== = =×
(b) ()()
0
36
11
22.4 krad/s
210 10LC
ω
−−
== =
×

(c) The circuit is overdamped since
0
αω>.

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Engineering Circuit Analysis, 7
th
Edition Chapter Nine Solutions 10 March 2006

2. Parallel RLC circuit:

(a) For an underdamped response, we require
0
αω<, so that

12
11 1 12
or ;
22
L
RR
RC CLC
−
<>>
210
.

Thus, R > 707 kΩ .

(b) For critical damping, 1

707 k
2
L
R
C
== Ω

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Engineering Circuit Analysis, 7
th
Edition Chapter Nine Solutions 10 March 2006

3. Parallel RLC circuit:

(a)
() ()
81
6911 1
510 s
2 2 (4 || 10)(10 ) 2 (1)(10 )RC
α
−
−−
== = =×

()()
13
0
12 911
3.16 10 rad/s 31.6 Trad/s
10 10LC
ω
−−
== =× =

(b)
22 9 21 18
1,2 0
0.5 10 10 (0.25)(10 ) 0.5 31.62 Grad/sjjααω=− ± − =− × ± − =− ±s

(c) The circuit is underdamped since
0
αω<.

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Engineering Circuit Analysis, 7
th
Edition Chapter Nine Solutions 10 March 2006

4. Parallel RLC circuit:

(a) For an underdamped response, we require
0
αω<, so that

15
18
11 1 110
or ;
222
L
RR
RC CLC
−
−
<>>
×210
.

Thus, R > 11.18 Ω.

(b) For critical damping,
1
11.18
2
L
R
C
==
Ω

(c) For overdamped, R < 11.18 Ω .

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Engineering Circuit Analysis, 7
th
Edition Chapter Nine Solutions 10 March 2006

5. 11
12
22 22
1
22
L10, 6 , 8
6 , 8 adding,
14 2 7
1
6 7 49 48 , 6.928
LC
rad/s 6.928L 10, L 1.4434H,
11
C 14.434mF, 7 R 4.949
48L 2RC
o
oo
oo o ssss
sω
ααω ααω
αα
ωω ω
−−
−
=Ω =− =−
∴− = + − − =− − −
−=−∴=
∴− =− + − ∴ = =
∴==
== =∴= Ω

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Engineering Circuit Analysis, 7
th
Edition Chapter Nine Solutions 10 March 2006

6.
100 200
40 30 mA, C 1mF, (0) 0.25V
−−
=− = =−
tt
c
ie e v

(a)
100 200
100 200
100 2001
( ) 0.25 (40 30 ) 0.25
C
( ) 0.4( 1) 0.15( 1) 0.25
() 0.4 0.15 Vtt
tt
c
oo
tt
tt
vt=−idt e e dt
vt e e
vt e e
−−
−−
−−
= − −

(b)

(c)

∴ =− −+ −−
=− +
∫∫
∴
22 22
12
3
22
100 200
R
100 , 200
300 2 , 150 1
1500
150 ,R 3.333 Also,
2R10 150
200 150 22500 20000
1 100
20000 , L 0.5H
LC L
i ( ) 0.12 0.045 A
R
−
−−
=− =−α+ α −ω =− =−α− α −ω
∴−=−αα= −
∴ +==Ω
−=−− −ω∴ω=
∴ == =
== +
oo
oo
tt
s
v
tee
ss
∴
100 200
100 200
) ( ) ( ) (0.12 0.04) ( 0.045 0.03)
( ) 80 15 mA, 0
tt
Rc
tt
it i t i t e e
it e e t
−−
−−
=− − = − + − +
(
=− >∴
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Engineering Circuit Analysis, 7
th
Edition Chapter Nine Solutions 10 March 2006

7. Parallel RLC with ω o = 70.71 × 10
12
rad/s. L = 2 pH.

(a)
21 22
12 2 121
(70.71 10 )
1
So 100.0 aF
(70.71 10 ) (2 10 )
−
ω= = ×
==
××
o
LC
C
(b)
91
10 181
510
2
1
So 1 M
(10 )(100 10 )
−
−
α= = ×
==
×
s
RC
R

Ω

(c) α is the neper frequency: 5 Gs
-1

(d)
22 9 12 1
1
22 1
2
5 10 70.71 10

9 12
5 10 70.71 10
−
−
=−α+ α −ω =− × + ×
=−α− α −ω
o
oSj s
s
=− × − ×Sj
(e) 9
5
12
510
7.071 10
70.71 10
−α×
ζ= = = ×
ω×
o

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Engineering Circuit Analysis, 7
th
Edition Chapter Nine Solutions 10 March 2006

8. Given:
2 1
4,
2
=α=LRC
RC

Show that is a solution to
12() ( )
−α
=+
t
vt e At A

2
2
11
0[++=
dv dv
Cv
dt R dt L
1]

112
11 2
2
11 2 12
1211
121() ( )
()
() ()
()
(2 ) [3]
−α −α
−α
−α −α
−α
−α
=−α+
=−α−α
=−α−α−α −α
=−α −α + −α
=−α −α −α
tt
t
tt
t
tdv
eA eAtA
dt
AAtAe
dv
AAtA e Ae
dt
AAAAte
AAAte
[2]

Substituting Eqs. [2] and [3] into Eq. [1], and using the information initially provided,

2
11 2
12 1222
11 1
(2 ) ( ) ( )
22
11
() ()
24
0
−α −α −α
−α −α⎛⎞
−+++
⎜⎟
⎝⎠
−++ +
=tt
tt
Ae At Ae Ae
RC RC RC
At A e At A e
RC R C
1
t

Thus, is in fact a solutio
n to the differential equation.
12() ( )
−α
=+
t
vt e At A

Next, with
2
(0) 16==vA
and
121
0
()(16)
=
=−α=−α=
t
dv
AA A
dt
4

we find that
1
416=+ αA

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Engineering Circuit Analysis, 7
th
Edition Chapter Nine Solutions 10 March 2006

9. Parallel RLC with ω o = 800 rad/s, and α = 1000 s
-1
when R = 100 Ω.

2
1
so 5 F
2
1
so 312.5 mH
α= = μ
ω= =
o
C
RC
L
LC

Replace the resistor with 5 meters of 18 AWG copper wire. From Table 2.3, 18 AWG soft solid
copper wire has a resistance of 6.39 Ω/1000ft. Thus, the wire has a resistance of

100cm 1in 1ft 6.39
(5m)
1m 2.54cm 12in 1000ft
0.1048 or 104.8m
⎛⎞⎛ ⎞⎛⎞⎛ ⎞Ω
⎜⎟⎜ ⎟⎜⎟⎜
⎝⎠⎝ ⎠⎝⎠⎝
=Ω Ω
⎟ ⎠

(a) The resonant frequency is unchanged, so 800rad/sω=
o

(b)
311
954.0 10
2
−
α= = × s
RC

(c)

Define the percent change as
100
ζ−ζ
×
ζ
new old
old

100
95300%=
old
α−α
=×
α
new old

α
ζ=
ω
α
ζ=
ω
old
old
o
new
new
o
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Engineering Circuit Analysis, 7
th
Edition Chapter Nine Solutions 10 March 2006

10. 5H, R 8 , C 12.5mF, (0 ) 40V
+
==Ω= =Lv

(a)
2
28
1,2 1 2
12
12 1 2 2 2 11 1000 1
(0)i 8A: 5,
16,
2RC 2 8 12.5 LC
4 5 25 16 2, 8 ( ) A A
1000 40
40 A A (0 ) (0 ) 80( 8 5) 1040
12.5 8
/ 2A 8A 520 A 4A 3A 480, A 160,A 120
( ) 120
o
tt
o
L
sv t e e
vi
vs
vt
+
−−
++
=α= = =ω==
××
ω = =− ± − =− − ∴ = +
⎛⎞
′

∴=+ = − − = −−=−
⎜⎟
⎝⎠
=− − ∴− =− − ∴− =− = =−

=−
28
160 V, 0
tt
eet
−−
+>∴

(b)
28
34
34
34 3 4
28
443 (0 ) 40
(0 ) 8A Let ( ) A A ; (0 ) 5A
R8
(0 ) A A (0 ) (0 ) 8 5 13A;
40
(0 ) 2A 8A 8 A / 4 A 4A
5
3A 13 4, A 3, A 16 ( ) 16 3 A, 0
+
+− − +
++ +
+
−−
==+ ===
∴ = + =− − =− − =−
=− − = = ∴ =− −
∴−=−+ = =−∴=− + >
tt
cR
Rc
tt v
ii teei
iii
is
it
e e t

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and educators for course preparation. If you are a student using this Manual, you are using it without permission.

Engineering Circuit Analysis, 7
th
Edition Chapter Nine Solutions 10 March 2006

11. (a)
3
4
11101
10 1.581
22102
C
L
R
C
−
−
== == Ω
Therefore

R = 0.1R
C = 158.1 mΩ

(b)
41 3
011
3.162 10 s and 3.162 10 rad/s
2RC LC
αω
−
==× = =×

Thus,
22 1 4
1,2 0
158.5 s and 6.31 10 sααω
1
− −
=−s ± − = − − ×
Ae Ae
−− ×
=+
vv
−+ − +
== = =

So we may write it
4
158.5 6.31 10
12
()
tt

With ii (0 ) (0 ) 4 A and (0 ) (0 ) 10 V

A
1+ A2 = 4 [1]
Noting

0
(0 ) 10
t
di
vL

dt
+
=
==
[2] ()
34
12
10 158.5 6.31 10 10AA
−
−−×=

Solving Eqs. [1] and [2] yields A
1 = 4.169 A and A 2 = –0.169 A

So that
4
158.5 6.31 10
( ) 4.169 0.169 A
tt
it e e
−− ×
=−

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Engineering Circuit Analysis, 7
th
Edition Chapter Nine Solutions 10 March 2006

12. (a)
1
011
500 s and 100 rad/s
2RC LC
αω
−
== = =

Thus,
22 1
1,2 0
10.10 s and 989.9 sααω
1
− −
=−s ± − = − −
Ae Ae
−−
=+
vv
−+ − +
== = =

So we may write it [1]
10.1 989.9
12
()
tt
R

With ii (0 ) (0 ) 2 mA and (0 ) (0 ) 0

A 1+ A2 = 0 [2]

We need to find
0
R
t
di
dt
=
. Note that
()1
R
di t dv
dt R dt
= [3] and
CR
dv
iC ii
dt
= =− −.

Thus,
33
0(0 )
(0 ) (0 ) (0 ) 2 10 2 10
CR
t
dv v
iC [4] ii
dt R
+
+
++ + −−
=
==−−=−×−=−×

Therefore, we may write based on Eqs. [3] and [4]:

0
(50)( 0.04) 2
R
t
di
dt
=
=−=− [5]. Taking the derivative of Eq. [1] and combining with
Eq. [5] then yields:
ss [6].
11 2 2
2AA+=−

Solving Eqs. [2] and [6] yields A
1 = 2.04
− mA and A2 = 2.04 mA

So that
()
10.1 989.9
( ) 2.04 mA
tt
R
it e e
−−
=− −

(b) (c) We see that the simulation agrees.

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and educators for course preparation. If you are a student using this Manual, you are using it without permission.

Engineering Circuit Analysis, 7
th
Edition Chapter Nine Solutions 10 March 2006

13.
1
(0) 40A, (0) 40V, L H, R 0.1 , C 0.2F
80
====Ω=iv

(a)
2
1,2
10 40
12 12
12
12 2 2 1
10 4018 0
25, 400,
2 0.1 0.2 0.2
20, 25 625 400 10, 40
() A A 40 A A;
1(0)
(0 ) 10A 40A (0 ) (0) 2200
CR
A 4A 220 3A 180 A 60, A 20
( ) 20 60 V,
−−
++
−−
α= = ω = =
××
ω= =
− ± − = −

∴ =+∴=+
⎛⎞
′′=− − = − =−
⎜⎟
⎝⎠

∴− − =− ∴− =− ∴ = =−

(b) i(t) = – v/ R
– C
dt
dv
=
tt-tt
eeee
401040 10
-40)(0.2)(60)( 10)0.2(-20)(- 600 200
−−−
−−−
=
A
tt
ee
40 10
120 160
−−
−

∴ =− + >
o
o
tt
tt
s
vt e e
v
vvi
vt e e t 0
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Engineering Circuit Analysis, 7
th
Edition Chapter Nine Solutions 10 March 2006

14. (a)
81 5
011
6.667 10 s and 10 rad/s
2RC LC
αω
−
==× = =

Thus,
22 1 9
1,2 0
7.5 s and 1.333 10 sααω
1
− −
=−s . So we may write
[1] With ii ,
± − = − − ×
t×
vv
−+ − +
== = =
9
7.5 1.333 10
12
()
t
C
it Ae Ae
−−
=+ (0 ) (0 ) 0 A and (0 ) (0 ) 2 V
6
62
(0 ) (0 ) 0.133 10
15 10
CR
ii
++
−
=− =− =− ×
×
so that

A 1+ A2 = –0.133×10
6
[2]

We need to find
0
R
t
di
dt
=
. We know that
6
6
002
2 so 10
210
tt
di di
L
dt dt
−
==
==
×
= . Also,

1
and
R
C
didv dv
Ci
dt dt R dt
== so
( )
9
7.5 1.333 10
121
ttCRidi
Ae Ae
dt CR CR
−−×
== +
.

Using
96
12 1
0 1
0 so 7.5 1.33 10 10 ( )
CCR
t
di didi
2
AA AA
dt dt dt dt CR
=
= =− −× =−− +
di
++ [3]

Solving Eqs. [2] and [3] yields A
1 = 0.75
− mA and A 2 = –0.133 MA (very different!)

So that
()
9
3 7.5 6 1.333 10
( ) 0.75 10 0.133 10 A
tt
C
it e e
−− − ×
=− × + ×

(b)

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Engineering Circuit Analysis, 7
th
Edition Chapter Nine Solutions 10 March 2006

15. (a)
1
011
0.125 s and 0.112 rad/s
2RC LC
αω
−
== = =

Thus,
22 1
1,2 0
0.069 s and 0.181 sααω
1
− −
=−s . So we may write

vt [1]
± − = − −
Ae Ae
−−
=+
vv
−+ −+
==− ==
0.069 0.181
12
()
tt
With ii , (0 ) (0 ) 8 A and (0 ) (0 ) 0
CC
A1+ A2 = 0 [2]

We need to find
0
R
t
di
dt
=
. We know that

0.069 0.181
12
( ) 4 0.069 0.181
tt
Cdv
it C Ae Ae
dt
−−
⎡⎤==− −
⎣⎦
. So,

[
]
12
(0) 4 0.069 0.181 8
C
iA [3] A=− − =−

Solving Eqs. [2] and [3] yields A
1 = 17.89
− V and A2 = 17.89 V

So that

0.069 0.181
( ) 17.89 V
tt
vt e e
−−
⎡⎤=− −
⎣⎦

(b)
0.069 0.181
1.236 8.236
tt
ee
dt
−−
−
dv
= . We set this equ al to 0 and solve for t
m:

0.069
0.112
0.181
3.236
1.236
m
m
m
t
t
t
e
e
e
−
−
== , so that t m = 8.61 s.

Substituting into our expression for the voltage, the peak valu
e is

v(8.61) = –6.1 V

(c) The simulation agrees with the analytic re
sults.

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Engineering Circuit Analysis, 7
th
Edition Chapter Nine Solutions 10 March 2006

16.
66 3
26
3
1,2
2000 6000
12 1
3
12 12
100
(0) 2A, (0) 100V
50
10 3 10
4000, 12 10
2 50 2.5 100 2.5
16 12 10 200, 4000 2000
() A A , 0 A A 2
10 3
(0 ) 100 3000 2000A 6000A 1.5 A 3A 0.5 2A
100
Lc
o
tt
Ls
Liv
w
s
it e e t
i
+
−−
+
== =
×
α= = = = ×
×× ×
−× = =− ±

2
∴ =+ >∴+=
−×
′=×=−=−−∴−=−−∴=−
2000 6000
21
2000 6000
A 0.25, A 2.25 ( ) 2.25 0.25 A, 0
0: ( ) 2A ( ) 2 ( ) (2.25 0.25 ) ( )A
tt
L
tt
LL
it e e t
tit itut e e ut
−−
−−
∴=− = ∴ = − >
>=∴=−+ −

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Engineering Circuit Analysis, 7
th
Edition Chapter Nine Solutions 10 March 2006

17.

2
21
1, 2
50 450
12 12 1
12 2 2 1
12
(0) 2A, (0) 2V
51
1000 1000 45
= 250, 22500
212 2
250 250 22500 50, 450
A A A A 2; (0 ) 45( 2) 50A 450A
A 9A 1.8 8A 0.2 A 0.025, A 2.025(A)
( ) 2.025
−
−− +
−
== =
+
×
== =
××
=− ± − =− −
′∴= + ∴ + = = −=− −
∴+ =∴− = ∴ =− =
∴=
Lc
o
tt
LL
L
iv
ss
ie e i
it e
αω
50 450
0.025 A, 0
−
−>
tt
et
2

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and educators for course preparation. If you are a student using this Manual, you are using it without permission.

Engineering Circuit Analysis, 7
th
Edition Chapter Nine Solutions 10 March 2006

18.

(a)
2
43 6
1, 2 1 2
12
1212 22 1
43 61 1440 1440
20, 144
2RC 72 10
20 400 144 4, 36: A A
118
(0) 18 , (0) 1440 0
236
0 4A 36A A 9A 18 8A , A 2.25, A 20.25
( ) 20.25 2.25 V, 0
o
tt
t
sv e e
vAAv
vt e e t
αω
+
−−
−−
=== ==
=− ± − =− − = +
⎛⎞
′==+ = − =
⎜⎟
⎝⎠

∴=− − =− − =∴ =− =− =
(b) (c) Solving using a scientific calculator, we find that t
s = 1.181 s.
∴ =− >
43 6 4 3 6
43 61
( ) 0.5625 0.0625 0.05625 0.05625
36 1440
( ) 0.50625 0.00625 A, 0
tt t t
ttv
v e e e e
it e e t
−− − −
−−
′= − − +it=+
∴=− >
43 6
max max
at 0 18V 0.18 20.25 2.25
s stt
v e e
−−
=∴ = ∴ = −vt
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Engineering Circuit Analysis, 7
th
Edition Chapter Nine Solutions 10 March 2006

19. Referring to Fig. 9.43,
L = 1250 mH

so Since α > ω
o, this circuit is over damped.
1
1 4rad/s
1
5
2
−
ω= =
α= =
o
LC
s
RC

The capacitor stores 390 J at t = 0
−
:

2
11
2
2
So ( 0 ) 125 V (0 )
+
=
===
cc
c
ccWCv
W
vv
C

The inductor initially stores zero energy,

so
22
1,2
(0 ) (0 ) 0
53 8, 2
−+
==
=−α± α −ω =− ± =− −
LL
oii
S

Thus,
82
()
−−
=+
tt
vt Ae Be

Using the initial conditions, (0) 125 [1]== +vA
B
38 2
3
(0 )
(0 ) (0 ) (0 ) 0 (0 ) 0
2
(0 ) 125
So ( 0 ) 62.5 V
22
50 10 [ 8 2 ]
(0 ) 62.5 50 10 (8 2 ) [2]
+
+++ +
+
+
−− −
+−
++=++=
=− =− =−
==×− −
=− =− × +
LRc c
c
tt
c
c
v
iii i
v
i
dv
i C Ae Be
dt
iA B

Solving Eqs. [1] and [2], A = 150 V
B = −25 V

Thus,
82
( ) 166.7 41.67 , 0
−−
=− >
tt
vt e e t

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Engineering Circuit Analysis, 7
th
Edition Chapter Nine Solutions 10 March 2006

20. (a) We want a response
46− −
=+
tt
vAe Be

1
22 2
1
22
21
5
2
4525
6525
−
α= =
=−α+ α −ω =− =− + −ω
=−α− α −ω =− =− − −ω
oo
oo
s
RC
S
S
2

Solving either equation, we obtain ω o = 4.899 rad/s

Since
2
211
, 833.3 mHω= = =
ω
o
o L
LC C
++
==
Rc

(b) If .ii (0 ) 10 A and (0 ) 15 A, find A and B

with
34 6
3
46
(0 ) 10 A, (0 ) (0 ) (0 ) 20 V
(0) 20 [1]
50 10 ( 4 6 )
(0 ) 50 10 ( 4 6 ) 15 [2]
Solving, 210 V, 190 V
Thus, 210 190 , 0
++ ++
−− −
+−
−−
====
=+=
==×− −
=× −− =
==−
=− >
RR c
tt
c
c
ttiv vv
vAB
dv
i
C Ae Be
dt
iA B
AB
ve et

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and educators for course preparation. If you are a student using this Manual, you are using it without permission.

Engineering Circuit Analysis, 7
th
Edition Chapter Nine Solutions 10 March 2006

21. Initial conditions:
50
(0) (0) 0 (0) 2 A
25
−+ +
== ==
LL Rii i
(a) (0 ) (0 ) 2(25) 50 V
+−
===
ccvv

(b) (0 ) (0 ) (0 ) 0 2 2 A
+++
=− − = − =−
cLRiii

(c) t > 0: parallel (source-free) RLC circuit

11
4000
2
1
3464 rad/s
−
α= =
ω= =
o
s
RC
LC

Since α > ω0, this system is overdamped. Thus,

Solving, we find A = −25 and
B = 75
so that
2000 6000
( ) 25 75 , 0
−−
=− + >
tt
c
vt e e t

(d)

(e)
2000 6000
25 75 0 274.7
−−
−+ =⇒=
tt
ee μst
using a scientific calculator
2000 6000
6 2000 6000
()
(5 10 )( 2000 6000 )
(0 ) 0.01 0.03 2 [1]
and ( 0 ) 50 [2]
−−
−− −
+
+
=+
==× − −
=− − =−
=+=
tt
c
tt
c
c
c
vt Ae Be
dv
iC Ae Be
dt
iAB
vAB
22
2000, 6000
=−α ± α −ω
=− −
o
s1,2

(f)
max
25 75 50 V=− + =
c
v
So, solving |−+
2000 6000
25 75
−−
s stt
ee | = 0.5 in view of the graph in part (d),
we find t
s = 1.955 ms using a scientific calculator’s equatio n solver routine.
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and educators for course preparation. If you are a student using this Manual, you are using it without permission.

Engineering Circuit Analysis, 7
th
Edition Chapter Nine Solutions 10 March 2006

22. Due to the presence of the inductor, (0 ) 0
−
=
cv . Performing mesh analysis,

12
21 2
1292 2 0 [1]
22370 [2]
and
−+− =
−++=
−=
A
A
ii
iiii
ii i

4.444 H
→i2
→i1

Rearranging, we obtain 2i
1 – 2i2 = 0 and –4i 1 + 6 i 2 = 0. Solving, i 1 = 13.5 A and i 2 = 9 A.

(a)
12 2(0 ) 4.5 A and (0 ) 9 A
−−
=−= = =
ALiii ii

(b) t > 0:

(0 ) 7 (0 ) 3 (0 ) 2 (0 ) 0
so, (0 ) 0
++++
+
− +−+
=
cAAA
Aviii
i

(c) due to the presence of the inductor. (0 ) 0
−
=
cv
=
around left mesh:
4.444 H

(d)
1 A
73(1)20
6
6 V 6
1
−+− +=
= ∴==Ω
LC
LC TH
v
vR

(e)
1
22
1, 21
3.333
2
1
3 ra
d/s
1.881, 4.785
−
α= =
ω= =
=−α± α −ω =− −
o
o
s
RC
LC
S

1.881 4.785
()
(0 ) 0 [1]
−−
+
=+
==+
tt
A
A
i t Ae Be
iAB
Thus,

To find the second equation required to determine the coefficients, we write:

=− −
=− −
LcR
c
Aiii
dv
Ci
dt

3 1.881 4.785
1.881 4.785
25 10 1.881(6 ) 4.785(6 )
−− −
−−
⎡ ⎤−× − −

3
(0 ) 9 25 10 [ 1.881(6 ) 4.785(6 )]
+−
==− × − − −−
LiA BAB
or 9 = -0.7178A – 0.2822B
[2]

Solving Eqs. [1] and [2], A = −20.66 and
B = +20.66
So that it
4.785 1.881
( ) 20.66[ ]
−−
=−
tt
A
e e
⎣ ⎦
−
tt
tt
Ae Be
Ae Be
-
=
PROPRIETARY MATERIAL . © 2007 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers
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Engineering Circuit Analysis, 7
th
Edition Chapter Nine Solutions 10 March 2006

23. Diameter of a dime: approximately 8 mm. Area =
22
0.5027cmπ=r

14 2
(88)(8.854 10 F/cm)(0.5027cm )
0.1cm
39.17pF
−
εε ×
==
=
roA
d
Capacitance

4H=μL

1
79.89Mrad/sω= =
o
LC

For an over damped response, we require α > ω
o.

6
12 61
79.89 10
2
1
2(39.17 10 )(79.89 10 )
−
>×
<
××
RC
R
Thus,

or 159.8<ΩR

*Note: The final answer depends quite strongly on the choice of ε
r.
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Engineering Circuit Analysis, 7
th
Edition Chapter Nine Solutions 10 March 2006

24. (a) For critical damping,
3
6
1110
4.564
2 2 12 10
L
R
C
−
−
== =
×
Ω.

(b)
() ()
31
611
9.129 10 s
2 2 4.564 12 10RC
α
−
−
=×
×
e AtA
−×
=+
A A=+=∴=
==
Thus, vt [1]
() ( )
3
9.129 10
12 t
C

At t = 0,
vA . ()
12 2
(0) 0 12 12 V
C

Taking the derivative of Eq. [1],

()
()
3
9.129 10 3 3
11
9.129 10 9.129 10 12
C t
dv t
eA tA
dt
−×
⎡⎤=−×+−×
⎣⎦

and also ii , so ( )
CR L
i=− +
()
3
16
0(0)11 12
0 0 9.129 10 12
12 10 4.565
CC
t
dv v
A
dt C R
−
=
⎛⎞ ⎛⎞
=− + =− + − ×
⎜⎟⎜⎟
× ⎝⎠⎝⎠

Solving,
, so we ma
y write
3
1
109.6 10 VA=− ×

.
() ()
3
9.129 10 3
109.6 10 12
t
C
vt e t
−×
=−×+

(c) We see from plotting both the analytic result in Probe and the simulated voltage, the
two are in excellent ag
reement (the curves lie on top of one another).

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Engineering Circuit Analysis, 7
th
Edition Chapter Nine Solutions 10 March 2006

25. (a) For critical damping,
8
3
1110
1.581 m
2210
L
R
C
−
−
== = Ω.

(b)
()()
51
3311
3.162 10 s
2 2 1.581 10 10
RC
α
−
−−
== =×
×

Thus, [1]
() ( )
5
3.162 10
12 t
L
it e AtA
−×
=+

At t = 0, .
()
12 2
(0) 0 10 10 A
L
iAA A=+=∴=

Taking the derivative of Eq. [1],

()
()
5
3.162 10 5 5
11
3.162 10 3.162 10 10
L t
di t
eA tA
dt
−×
⎡⎤=−×+−×
⎣⎦
[2]

and also
0
(0) 0
L
C
t
di
Lv
dt
=
== [3], so

Solving Eqs. [2] and [3],

()
()
5
1
3.162 10 10 = 3.162 10 VA=× ×
6
, so we may write

.
() ()
5
3.162 10 6
3.162 10 10
t
L
it e t
−×
=× +

(c) We see from plotting both the analytic result in Probe and the simulated voltage, the
two are in reasonable agreem
ent (some numerical error is evident).

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and educators for course preparation. If you are a student using this Manual, you are using it without permission.

Engineering Circuit Analysis, 7
th
Edition Chapter Nine Solutions 10 March 2006

26. It is unlikely to observe a critically damped response in real-life circuits, as it would be
virtually impossible to obtain the exact values required for R, L and C. However, using
carefully chosen components, it is possible to obtain a response which is for all intents
and purposes very close to a critically damped response.

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Engineering Circuit Analysis, 7
th
Edition Chapter Nine Solutions 10 March 2006

27.
L

crit. damp. (a)
23
L 4RC 41210 8mH
−
==×××=

(b)

(c)
maL
t
x
250
max: (250 2) 0, 1 250 2, 0 No!
0, 2A 0.02 (250 2); SOLVE: 23.96ms
s
m mm
t
mL sit tt
i e t t
−
+= = + <
∴

s
==∴= + =
250
12
250
1
3 3 1.25
11 1000
250 (A A )
2RC 2 1 2
(0)2A,(0)2V (A 2)
Then 8 10 (0 ) 2 8 10 (A 500), (1.25 2) 0.9311A
t
oL
t
Lc L
L
ie t
iv iet
ieαω
−
−
−+ − −
===∴= +
××
==∴=+
′×=−=×−=+=

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Engineering Circuit Analysis, 7
th
Edition Chapter Nine Solutions 10 March 2006

28.
R

crit. damp. (a)
232 6100
L 4R C 10 4R 10 R 57.74
3
−−
==×=×∴= Ω

(b)
31
3464
12
2
6
5
121
3464 51
10 / 2.5 3464
30
( ) (A A ) (0) 100V
100
(0) 1.7321A 100 A
57.74
10 100
(0 ) 1.7321 0 A 3464A A 3.464 10
2.5 57.74
( ) (3.464 10 100)V, 0
o
t
cc
L
c
t
c
s
vt e t v
i
v
vt e t tωα
−
−
+
−
== × =
∴ =+=
== ∴=
⎛⎞
′=−==−∴=
⎜⎟
⎝⎠
×
∴ =×+>

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Engineering Circuit Analysis, 7
th
Edition Chapter Nine Solutions 10 March 2006

29. Diameter of a dime is approximately 8 mm. The area, therefore, is πr
2
= 0.5027 cm
2
.

The capacitance is
14
(88)(8.854 10 )(0.5027)
0.1
39.17 pF
−
εε ×
=
=
ro
A
d

w ith
1
4 H, 79.89 Mrad/s=μ ω= =
oL
LC

For critical damping, we require
1
2
=ω
o
RC

or
1
159.8
2
==
ω
o
R
C
Ω
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Engineering Circuit Analysis, 7
th
Edition Chapter Nine Solutions 10 March 2006

30.
8
L 5mH, C 10 F, crit. damp. (0) 400V, (0) 0.1A
−
== =− =vi

(a)
2328
L 4R C 5 10 4R 10 R 353.6
−−
==×= ∴= Ω

(b)

(c)
max(0) 0.1Aii
∴==
8
141,420
12
141,421 3
21
11
141,421
10
141,420 (A A )
2 353.6
A 0.1 (A 0.1), 5 10
(A 141,420 0.1) 400 A 65,860
( 65,860 0.1). 0
( 65860) 141,420 ( 65,860 0.1) 0
8.590 ( )
t
t
t
tt
m
mm
ie t
et
ie t i
eet
tsit
αα
α
μ
−
−−
−
−−
==∴= +
×
∴=∴= + ×
−×=−∴=−
′∴=−+=
∴ ++ − +=
∴=∴=
6
141,420 8.590 10
6
max
( 65,860 8.590 10 0.1) 0.13821A
i ( ) 0.13821A
m
e
it
−
−××
−
−××+=−
∴ ==

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Engineering Circuit Analysis, 7
th
Edition Chapter Nine Solutions 10 March 2006

31. Critically damped parallel RLC with
31
10 , 1M
−−
α==sR Ω.
We know
3
3
6
11 0
10 , so 500 F
22 10
−
===
×
C
RC
μ

Since α = ω o,
3
61
10
1
or 10
so 2 GH (!)
−
−
=
=
=
o
LC
LC
L

2
9
2
72
9
29 22 9
13
210
50 turns 1 m
(4 10 H/m) . (0.5cm) . .
cm 100 cm
So
210
(4 10 )(50) (0.5) 2 10
So 8.106 10 cm
−
−
μ
==×
⎡⎤⎛⎞ ⎛
π× π
⎜⎟ ⎜⎢⎥
⎝⎠ ⎝⎣⎦
=×
π× = ×
=×
NA
L
S
s
s
s
s
ω=
⎞
⎟
⎠
If

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Engineering Circuit Analysis, 7
th
Edition Chapter Nine Solutions 10 March 2006

32.

2
1214 1413
1, 26, 26 1 5
2RC 2 2 LC 2
() ( cos5 sin5)
od
t
c
vt e B t B t
αω ω
−
×
===== = =−=
×
∴= +

(a) (0 ) (0) 4A
LLii
+
==

(b) (0 ) (0) 0
ccvv
+
==

(c)
1
(0 ) (0 ) 0
L
Lciv
++
′==

(d)
(0 )1
(0 ) [ (0 ) (0 )] 4 4 4( 4 0) 16
2
c
cLR
v
vii
c
+
+++
⎡⎤
′=− − =−− =−+=−
⎢⎥
⎣⎦
V/s

(e)

(f)
11 2 2
2
()0 1(B) B 0, () B sin5, (0) B(5) 16
B 3.2, ( ) 3.2 sin5 V, 0
t
cc
t
c
ev t e tv
vt e t t
−+
−
′
∴ =∴= = = =−
∴=− =− >

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and educators for course preparation. If you are a student using this Manual, you are using it without permission.

Engineering Circuit Analysis, 7
th
Edition Chapter Nine Solutions 10 March 2006

33.

66 3
27
66
4000
12
6
110 110
4000, 2 10
2RC 100 2.5 LC 50
20 10 16 10 2000
(B cos2000 B sin 2000 )
(0) 2A, (0) 0 (0 ) 2A; (0 ) (0 ) (0 )
11 1 210
(0 ) (0) (0 ) 0 (0 )
L R RC 125
o
d
t
c
Lcc cLR
ccc c
ie t t
ivi iii
ivv i
αω
ω
+
−
+++
+++
== = ===×
×
=×−×=
∴= +
′′ ′==∴=− =−−
×
′′ − − =− =
∴
+
∴=
6
12
4000
210
B 2A, 16,000 2000B ( 2)( 4000) B 4
125
( ) ( 2cos2000 4sin 2000 )A, 0
t
c
it e t t t
−
×
=− = = + − − ∴ =
∴= − + >

2

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Engineering Circuit Analysis, 7
th
Edition Chapter Nine Solutions 10 March 2006

34.

(a)
22
21 100 1 100
8, , 36 64
2
RC 12.5 LC L
100
100 L 1H
L
odo
o
= = ω = = ω = =ω −
2
α=
(b) (c)
∴ω= = ∴ =
8
12
8
12
22
8
0: ( ) 4A; 0: ( ) (B cos6 B sin6 )
(0) 4A B 4A, (4cos6 B sin 6 ) (0) 0
(0 ) (0 ) 0 6B 8(4) 0, B 16/3
( ) 4 ( ) (4cos6 5.333sin 6 ) ( )A
−
−
++
−
<=>= +
=∴= = + =
′==∴−==
∴ =−+ +
t
LL
t
LL
Lc
t
L
c
t tite t t
ii ett v
itv
it u t e t tut
ti
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and educators for course preparation. If you are a student using this Manual, you are using it without permission.

Engineering Circuit Analysis, 7
th
Edition Chapter Nine Solutions 10 March 2006

35.

(a)
93 9
28
22 6 6
5000 4 4
12
5000 4 4
2
9
110 110
5000, 1.25 10
2RC 2 20 5 LC 1.6 5
125 10 25 10 10,000
() (Bcos10 B sin10 )
(0) 200V, (0) 10mA ( ) (200cos10 B sin10 )
110
(0 ) (0 ) (0)
5
−
−
−
++
α= = ω== =×
×× ×
ω= ω−α= × − × =
∴ =+
==∴= +
′== −
o
do
t
c
t
cL c
c
cc L
vt e t t
vi vte t
v
vi i
c

t
9
24
2
5000 4 4
2
(0)
20,000
10 200
10 0 10 B 200(5000)
5 20,000
B 100V ( ) (200cos10 100sin10 ) V, 0
−
−
⎡⎤
⎢⎥
⎣⎦
⎛⎞
=−==−
⎜⎟
⎝⎠
∴=∴= + >
t
c
vt e t t t

(b)
2
5000 4
500 6 6 5000 4
5000 9 6 5000 4
5 1
10 , C
R
[10 ( 200sin 100cos] 5000(200cos 100sin)]
[10 ( 2sin 0.5cos)] 2.5 10 sin10 /
1
(200cos 100sin) 5 10 2.5 10 sin10
20,000
−
−
−−
−−
−
′=− = +
′=−+− +
=−−=−×
⎡⎤
−
∴=+ −×××
⎢⎥
⎣⎦
=
sw L L c c
t
c
tt
t t
Liiivv
ve
ee tv
ie e t
e
000 4 4
5000 4 4
(0.01cos10 0.0075sin10 )A
10 (10cos10 7.5sin10 )mA, 0
−
−

s
∴=− − >
t
t
sw
tt
ie t tt

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and educators for course preparation. If you are a student using this Manual, you are using it without permission.

Engineering Circuit Analysis, 7
th
Edition Chapter Nine Solutions 10 March 2006

36.

(a)
66
2
22
20
12
1
20
2
22
6
1 10 1 1.01 10
20, 40,400
2RC 2000 25 LC 25
40,400 400 200
(A cos200 A sin 200 )
(0) 10V, (0) 9mA A 10V
(10cos200 A sin 200 ) V, 0
1
(0 ) 200A 20 10 200(A 1) (0 )
C
10
2
o
do
t
L
t
o
ve t t
vi
ve t t t
vi
−
−
++
×
α= = = ω = = =
×
ω=ω
−α= − =

∴=+
==∴=

∴=+ >
′=−×= −=
=

3
2
20
( 10 ) 40 A 1 0.2 0.8
5
( ) (10cos200 0.8sin 200 ) V, 0
t
vt e t t t
−
−
−=−∴=−=

∴ =+ > (b)
20
10.032 cos (200 4.574 )V
2
T3.42ms
200
−
=− °
==
t
ve t
π

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Engineering Circuit Analysis, 7
th
Edition Chapter Nine Solutions 10 March 2006

37.

63
12 6
44
100
12
100
12
66
1
110 1
100 , 1.01 10
2RC 2 5 LC
60
101 10 10 100; (0) 6mA
10
(0) 0 ( ) (A cos1000 A sin1000 ), 0
A 0, ( ) A sin1000
11
(0) (0) 10[ (0) (0)] 10
C 5000
(610
−
−
−
−
++ + +
−
α= = = ω = = ×
×
∴ω= × − = = =
=∴ = + >
∴==
′==−− =
−×
o
dL
t
cc
t
c
cc c
s
i
vvte t tt
vt e t
vi i v

3
22
100
1 4
4 100
100
1
) 6000 1000A A 6
1
( ) 6 sin1000 V, 0 ( )
10
( ) 10 ( 6) sin1000 A
( ) 0.6 sin1000 mA, 0
−
−−
−
=− = ∴ =−
∴ =− > ∴ =−
=− −
∴ =>
t
c
t
c
t
vt e t t it
vt e t
it e t t
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Engineering Circuit Analysis, 7
th
Edition Chapter Nine Solutions 10 March 2006

38. We replace the 25-Ω resistor to obtain an underdamped response:

LC
1
and
2RC
1

0==ωα ; we require α < ω0.

Thus,
3464
R 1010
1
6
<
×
−
or R > 34.64 mΩ.

For R = 34.64 Ω (1000× the minimum required value), the response is:

v(t) = e
-αt
(A cos ωdt + B sin ωdt) where α = 2887 s
-1
and ωd = 1914 rad/s.
i
L(0
+
) = iL(0
-
) = 0 and v C(0
+
) = v C(0
-
) = (2)(25) = 50 V = A.

i
L(t) =
dt
dv
dt
dv
CL
L L=
=
()
( )[ ]tBtAettBttAe
dd
t
dddd
tωωαωωωω
αα
sin cos - cos sin L ++−
−−

i L(0
+
) = 0 = [ A - B
3
1050
d
3αω
−
×
], so that B = 75.42 V.

Thus, v(t) =
e
-2887t
(50 cos 1914t + 75.42 sin 1914t) V.

From PSpice the settling time using R = 34.64 Ω is approximately 1.6 ms.

Sketch of v(t). PSpice schematic for t > 0 circuit.

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and educators for course preparation. If you are a student using this Manual, you are using it without permission.

Engineering Circuit Analysis, 7
th
Edition Chapter Nine Solutions 10 March 2006

39.

11
1
1
21 1
/
11 2
/2
1
1
(0) 0; (0) 10A
(Acos Bsin ) A 0,
Bsin
[Bsin Bcos ]0
1
tan , tan
1
T;
2
Bsin B
sin ; let
V
mm d
d
t
dd
t
d
t
dd d
dd
dm
d
mm dm
d
tt
md mm
m
dm
m
vi
ve t t
ve t
ve t t
tt
tt t
ve tv e
v
te
α
−α
−α
−
−α −α −απ ω
−απ ω
==
=ω+ω∴=
=ω
′=−αω+ω ω=
ωω
∴ω= =
αωα
π
=+ =+
ω
=ω=−
ω∴=−
2
1
/
22
0
2
2
1
21
100
121
100, 100; ,
2RC R
1 21 100
6 6 441/ R 6R 441
LC R R
21
R 1/ 6 441 10.3781 To keep
100
0.01, chose
2
(0 )
21 0
BB6 4R10
10.378 10.
d
m
m
d
d
m
d
m
v
v
en
n
v
v
v
απ ω
+
=
ω
∴ =α= α= =
π
ω= = ∴ω= − ∴ −
π
⎡⎤ π⎛⎞
R 10.3780
=+=Ω⎢⎥⎜⎟
⎝⎠⎢⎥⎣⎦
′Ω= ω
⎛⎞
=−
= +
⎜⎟
⎝⎠
l
l
<=
∴
2
2.02351
1
21
B 1.380363
3780
21 21
2.02351; 6 1.380363
10.378 10.378
304.268 sin 1.380363 0.434 ,
71.2926 Computed values show
0.7126 0.01
d
t
m
sm m
ve tvts
vv
tv v
−
⎛⎞
∴=
⎜⎟ ⎝⎠
⎛⎞
α= = ω = − =
⎜⎟ ⎝⎠
∴
1
2.145sec;
m
==
=
==<

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and educators for course preparation. If you are a student using this Manual, you are using it without permission.

Engineering Circuit Analysis, 7
th
Edition Chapter Nine Solutions 10 March 2006

40. (a) For t < 0 s, we see from the circuit that the capacitor and the resistor are shorted by
the presence of the inductor. Hence, i
L(0
-
) = 4 A and v C(0
-
) = 0 V.

When the 4-A source turns off at t = 0 s, we are left with a parallel RLC circuit such that
α = 1/2RC = 0.4 s
-1
and ω0 = 5.099 rad/s. Since α < ω0, the response will be
underdamped with
ωd = 5.083 rad/s. Assume the form i L(t) = e
-αt
(C cos ωdt + D sin ωdt)
for the response.

W ith i L(0
+
) = iL(0
-
) = 4 A, we find C = 4 A. To find D, we first note that
v
C(t) = v L(t) = L
dt
di
L

and so v
C(t) = (2/13) [e
-αt
(-Cωd sin ωdt + Dωd cos ωdt) - αe
-αt
(C cos ωdt + D sin ωdt)]

W ith v C(0
+
) = 0 = (2/13) (5.083D – 0.4C), we obtain D = 0.3148 A.

Thus, i L(t) = e
-0.4t
(4 cos 5.083t + 0.3148 sin 5.083t) A and i L(2.5) = 1.473 A.

(b)
α = 1/2RC = 4 s
-1
and ω0 = 5.099 rad/s. Since α < ω0, the new response will still be
underdamped, but with
ωd = 3.162 rad/s. We still may write

vC(t) = (2/13) [e
-αt
(-Cωd sin ωdt + Dωd cos ωdt) - αe
-αt
(C cos ωdt + D sin ωdt)]

and so with v C(0
+
) = 0 = (2/13) (3.162D – 4C), we obtain D = 5.06 A.

Thus, i L(t) = e
-4t
(4 cos 3.162t + 5.06 sin 3.162t ) A and i L(.25) = 2.358 A.

(c)
We see from the simulation result below that our hand calculations are correct; the slight disagreement is due to numerical inaccuracy. Changing the step ceiling from the 10-ms value employed to a smaller value
will improve the accuracy.

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Engineering Circuit Analysis, 7
th
Edition Chapter Nine Solutions 10 March 2006

41. (a,b) For t < 0 s, we see from the circuit below that the capacitor and the resistor are
shorted by the presence of the inductor. Hence, i
L(0
-
) = 4 A and v C(0
-
) = 0 V.

When the 4-A source turns off at t = 0 s, we are left with a parallel RLC circuit such that
α = 1/2RC = 1 s
-1
and ω0 = 5.099 rad/s. Since α < ω0, the response will be underdamped
with
ωd = 5 rad/s. Assume the form i L(t) = e
-αt
(C cos ωdt + D sin ωdt) for the response.

W ith i L(0
+
) = iL(0
-
) = 4 A, we find C = 4 A. To find D, we first note that
v
C(t) = v L(t) = L dt
di
L

and so v
C(t) = (2/13) [e
-αt
(-Cωd sin ωdt + Dωd cos ωdt) - αe
-αt
(C cos ωdt + D sin ωdt)]

W ith v C(0
+
) = 0 = (2/13) (5D – 4), we obtain D = 0.8 A.
Thus, i
L(t) = e
-t
(4 cos 5t + 0.8 sin 5t) A
(c) Using the cursor tool, the settling time is approximately 4.65 s.

We see that the simulation
result confirms our hand
analysis; there is only a
slight difference due to
numerical error between
the simulation result and
our exact expression.
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and educators for course preparation. If you are a student using this Manual, you are using it without permission.

Engineering Circuit Analysis, 7
th
Edition Chapter Nine Solutions 10 March 2006

42.
22
20
12 1
20
2
20
22
R80
(0) 50 80 2 210V, (0) 0, 20
2L 4
100
500: 500 20 10
2
( ) (A cos10 A sin10 ) A 210V
1
( ) (210cos10 A sin10 ); (0 ) (0 ) 0
C
0 10A 20(210), A 420 ( ) (210cos10
cL
od
t
c
t
cc c
t
c
vi
vt e t t
vt e t t v i
vt e t α
ωω
−
−+ +
−
=+×= = = = =
== = −=
∴= + ∴=
′∴= + = =
∴= − = ∴ = +
0.8
20
12
22
20
20
420sin10 )
(40ms) (210cos0.4 420sin 0.4) 160.40V
Also, ( B cos10 B sin10 ),
11 1
(0 ) (0 ) [0 (0 )] 210
L2 2
(0 ) 105 10B B 10.5
( ) 10.5 sin10 A, 0
( ) 80 840 sin
c
t
L
LL c
L
t
L
t
RL
t
ve
ie t t
iv v
i
it e t t
vt i e
−
−
++ +
+
−
−
∴= + =
=+
==−=×
′∴=−=∴=
∴=− >
∴==
0.8
20
20
0.8
10 V
(40ms) 840 sin 0.4 146.98V
() () () ()
(40ms) 160.40 146.98 13.420V
[check: ( 210cos 420sin 840sin)
( 210cos10 420sin10 )V, 0
(40ms) ( 210cos 420sin 840
R
LccRL
t
L
t
L
t
ve
vt vt vt vt v
ve
ettt
ve
−
−
−
−
∴=− =−
=− − − ∴
=− + =−
=− − +
=− + >
∴=−−+
20
0.8
sin)
( 210cos10 420sin10 )V, 0
V (40ms)
(420sin 0.4 210cos0.4) 13.420VChecks]
t
L
e
ttt
e
−
−
=
−+ >
∴=
−=−

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and educators for course preparation. If you are a student using this Manual, you are using it without permission.

Engineering Circuit Analysis, 7
th
Edition Chapter Nine Solutions 10 March 2006

43. Series:
2
4
12
1
22
4214
4, 20, 20 16 2
2L 1/ 2 LC 0.2
(A cos2 A sin 2 ); (0) 10A, (0) 20V
1
A 10; (0 ) (0 ) 4(20 20) 0
L
(0 ) 2A 4 10 A 20
() (10cos2 20sin2)A, 0
od
t
LL
LL
L
t
L
R
ie t ti v
iv
i
it e t t t
αω ω
−
++
+
−
== = === = −=
∴= + = =
′∴= = = − =
′∴=−×∴=
∴= + >

c

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and educators for course preparation. If you are a student using this Manual, you are using it without permission.

Engineering Circuit Analysis, 7
th
Edition Chapter Nine Solutions 10 March 2006

44. (a)
2
22 2
2
46 4
10000
12
10000
21
6
1
R11
crit. damp; LR
C
4L LC 4
1 200
L 4 10 0.01H, 10
40 .02
( ) (A A ); (0) 10V, (0) 0.15A
1
A10,() (A0);(0)
C
(0) 10 ( 0.15) 150,000
Now, ( 0 ) A
−
−
−+
+
===∴=

∴=×× = = = =

∴ = + =− =−
′

∴=− = − =−
=− − =
′=
o
o
t
cc L
t
cc
L
c
vt e t v i
vt e t v
i
v
αω
αω
5
1
10,000
10 150,000 A 50,000
( ) (50,000 10) V, 0
−
+= ∴=

=−>
t
c
vt e t t
∴

(b)
10,000
33
max
( ) [50,000 10,000(50,000 10)]
15
5 50,000 10 0.3ms
50,000
( ) (15 10) 5 0.2489V
(0) 10V 10V
t
c
mm
cm
cc
e t
tt
vt e e
vv
−
−−
′=− −= ∴
=−∴==
vt

∴

(c)
,max0.2489V
cv=
=−==
=− ∴ =

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and educators for course preparation. If you are a student using this Manual, you are using it without permission.

Engineering Circuit Analysis, 7
th
Edition Chapter Nine Solutions 10 March 2006

45. “Obtain an expression for v c(t) in the circuit of Fig. 9.8 (dual) that is valid for all t′′.

66
27
66
1, 2
2000 6000
12
12
3
R0.0210 103
4000, 1.2 10
2L 2 2.5 2.5 10
4000 16 10 12 10 2000, 6000
1
( ) A A ; (0) 100 2V
50
1
(0) 100A 2 A A , (0 )
C
3
( (0)) 10 100 3000 /
100
3000 200A
o
tt
cc
Lc
L
s
vt e e v
iv
iv s
αω
−−
+
××
== = = =×
××
∴=− ± × − × =− −
∴ =+ =×=
′=∴=+ =
−=−××=−
∴−=−
12 12
21
200 6000600A , 1.5 A 3A
0.5 2A , 0.25, A 2.25
( ) (2.25 0.25 ) ( ) 2 ( )V (checks)
tt
c
vt e e ut u t
−−
−−=−−
∴=− =− =
∴ =− +−

A Ω
μF
mF

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and educators for course preparation. If you are a student using this Manual, you are using it without permission.

Engineering Circuit Analysis, 7
th
Edition Chapter Nine Solutions 10 March 2006

46. (a) 22
12
12
12
2R2 1
1, 5, 2
2L 2 LC
(B cos2 B sin2 ), (0) 0, (0) 10V
B0, B sin2
1
(0) (0 ) (0 ) V (0 ) 0 10 2B
1
B5 5sin2A,0
odo
t
LL
t
L
LR c
t
L
ie t ti v
iet
ivv
iett
2
c
ωωα
−
−
+++
−
=== == = −=αω

∴=+ ==

∴==
==−=−=

∴=∴ =− >
(b)
11
22
2 max
max
5[ (2cos2 sin2 )] 0
2cos2 sin 2 ,tan 2 2
0.5536 , ( ) 2.571A
2 2 0.5536 , 2.124,
( ) 0.5345 2.571A
and 0.5345A
−
′=− − =

∴ ==

∴== −
=× + =
=∴=
=
t
L
L
LL
L
tt
ttt
tsit
tt
it i
i
π
ie

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and educators for course preparation. If you are a student using this Manual, you are using it without permission.

Engineering Circuit Analysis, 7
th
Edition Chapter Nine Solutions 10 March 2006

47. (a)
6
2
22
1,2
10 40
12
12
12
111
11
R 250 1 10
25, 400
2L 10 LC 2500
25 15 10, 40
A A , (0) 0.5A, (0) 100V
11
0.5 A A , (0 ) (0 )
55
(100 25 100) 5 A / 10A 40A
5 10A 40 (0.5 A ) 10A 40
A20 30A
o
o
tt
LL c
LL
s
ie ei v
iv
s
αω
ααω
−−
++
== = == =
=− ± − =− ± =− −

∴=+ = =
′

∴=+ = =
− − =− =− −

∴=+ −=−
+∴−
12
1015, A 0.5, A 0
() 0.5 A, 0
t
L
it e t
−
=− = =
(b)
∴ =>
10 40
34 34
6
34 4 4 3
10
A A 100 A A ;
110
(0 ) ( 0.5) 1000
500
10A 40A 1000 3A 0, A 0, A 100
() 100 V 0
tt
c
cc
t
c
e
vi
c
vt e t
−−
+
−
=+∴=+
′′=−= −
ve
∴−− =−∴−= = =
=>∴
PROPRIETARY MATERIAL . © 2007 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers
and educators for course preparation. If you are a student using this Manual, you are using it without permission.

Engineering Circuit Analysis, 7
th
Edition Chapter Nine Solutions 10 March 2006

48. Considering the circuit as it exists for t < 0, we conclude that v C(0
-
) = 0 and i L(0
-
) = 9/4 =
2.25 A. For t > 0, we are left with a parallel RLC circuit having α = 1/2RC = 0.25 s
-1
and
ωo = 1/
LC = 0.3333 rad/s. Thus, we expect an underdam ped response with ωd =
0.2205 rad/s:
i
L(t) = e
-αt
(A cos ωdt + B sin ωdt)

iL(0
+
) = iL(0
-
) = 2.25 = A

so iL(t) = e
–0.25t
(2.25 cos 0.2205t + B sin 0.2205t)

In order to determine B, we must invoke the remaining boundary condition. Noting that
v
C(t) = v L(t) = L
dt
di
L

= (9)(-0.25)e
-0.25t
(2.25 cos 0.2205t + B sin 0.2205t)
+ (9) e
-0.25t
[-2.25(0.2205) sin 0.2205t + 0.2205B cos 0.2205t]
v
C(0
+
) = v C(0
-
) = 0 = (9)(-0.25)(2.25) + (9)(0.2205B)
so B = 2.551 and
i
L(t) = e
-0.25t
[2.25 cos 0.2205t + 2.551 sin 0.2205t] A

Thus, i
L(2) = 1.895 A
This answer is borne out by PSpice simulation:

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and educators for course preparation. If you are a student using this Manual, you are using it without permission.

Engineering Circuit Analysis, 7
th
Edition Chapter Nine Solutions 10 March 2006

49. We are presented with a series RLC circuit having
α = R/2L = 4700 s
-1
and ωo = 1/
LC = 447.2 rad/s; therefore we expect an
overdamped response with s
1 = -21.32 s
-1
and s2 = -9379 s
-1
.

From the circuit as it exists for t < 0, it is evident that i L(0
-
) = 0 and v C(0
-
) = 4.7 kV

Thus, v L(t) = A e
–21.32t
+ B e
-9379t
[1]

With i L(0
+
) = iL(0
-
) = 0 and i R(0
+
) = 0 we conclude that v R(0
+
) = 0; this leads to v L(0
+
) =
-v
C(0
-
) = -4.7 kV and hence A + B = -4700 [2]
Since v
L = L
dt
di
, we may integrate Eq. [1] to find an expression for the inductor current:

iL(t) =
⎥
⎦
⎤
⎢
⎣
⎡
−− tt
ee
937932.21
9379
B
-
21.32
A
-
L
1

At
t = 0
+
, iL = 0 so we have
0
9379
B
-
21.32
A
-
10500
1
3-
=
⎥
⎦
⎤
⎢
⎣
⎡
×
[3]

Simultaneous solution of Eqs. [2] and [3] yields A = 10.71 and B = -4711. Thus,

vL(t) = 10.71e
-21.32t
- 4711 e
-9379t
V, t > 0

and the peak inductor voltage magnitude is 4700 V.

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and educators for course preparation. If you are a student using this Manual, you are using it without permission.

Engineering Circuit Analysis, 7
th
Edition Chapter Nine Solutions 10 March 2006

50. With the 144 mJ originally stored via a 12-V battery, we know that the capacitor has a
value of 2 mF. The initial inductor current is zero, and the initial capacitor voltage is 12
V. We
begin by seeking a (painful) current response of the form

ibear = Ae
s
1
t
+ Be
s
2
t

Using our first initial condition,
ibear(0
+
) = iL(0
+
) = iL(0
-
) = 0 = A + B

di/dt = As1 e
s
1
t
+ Bs2 e
s
2
t

vL = Ldi/dt = ALs1 e
s
1
t
+ BLs2 e
s
2
t

vL(0
+
) = ALs1 + BLs2 = vC(0
+
) = vC(0
-
) = 12

What else is known? We know that the bear stops reacting at
t = 18 μ s, meaning that the
current flowing through its fur coat has dropped just below 100 mA by then (not a long
shock).

Thus, A exp[(18×10
-6
)s1] + B exp[(18×10
-6
)s2] = 100×10
-3

Iterating, we find that R
bear = 119.9775 Ω .

This corresponds to A = 100 mA, B = -100 mA, s
1 = -4.167 s
-1
and s2 = -24×10
6
s
-1

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and educators for course preparation. If you are a student using this Manual, you are using it without permission.

Engineering Circuit Analysis, 7
th
Edition Chapter Nine Solutions 10 March 2006

51. Considering the circuit at t < 0, we note that iL(0
-
) = 9/4 = 2.25 A and vC(0
-
) = 0.
For a critically damped circuit, we require α =
ωo, or
LC
1

RC2
1
= , which, with
L = 9 H and C = 1 F, leads to the requirement that R = 1.5 Ω (so α = 0.3333 s
-1
).

The inductor energy is given by
wL = ½ L [iL(t)]
2
, so we seek an expression for iL(t):

iL(t) = e
-αt
(At + B)

Noting that iL(0
+
) = iL(0
-
) = 2.25, we see that B = 2.25 and hence

iL(t) = e
-0.3333t
(At + 2.25)

Invoking the remaining initial condition requires consideration of the voltage across the
capacitor, which is equal in this case to the inductor voltage, given by:

vC(t) = vL(t) =
dt
di
L
L = 9(-0.3333) e
-0.3333t
(At + 2.25) + 9A e
-0.3333t

vC(0
+
) = vC(0
-
) = 0 = 9(-0.333)(2.25) + 9A so A = 0.7499 amperes and

iL(t) = e
-0.3333t
(0.7499t + 2.25) A

Thus, iL(100 ms) = 2.249 A and so wL(100 ms) = 22.76 J

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and educators for course preparation. If you are a student using this Manual, you are using it without permission.

Engineering Circuit Analysis, 7
th
Edition Chapter Nine Solutions 10 March 2006

52. Prior to t = 0, we find that
11
50
(10 ) and
15 5 v
vi i⎛⎞
=+ =
⎜⎟
⎝⎠

Thus,
10 500
1 so 100 V
15 15
vv
⎛⎞
−= =
⎜⎟ ⎝⎠
.

Therefore, (0 ) (0 ) 100 V, and (0 ) (0 ) 0.
CC LL
vv ii
+− +−
== ==

The circuit for
t > 0 may be reduced to a simple series circuit consisting of a 2 mH
inductor, 20 nF capacitor, and a 10 Ω resistor; the dependent source delivers exactly the
current to the 5 Ω that is required.

Thus,
()
31
310
2.5 10 s
222 10
R
L
α
−
−
== =×
×

and
()( )
5
0
3911
1.581 10 rad/s
210 2010LC
ω
−−
== =×
××

W ith
0
αω< we find the circuit is underdamped, with

22 5
0
1.581 10 rad/s
d
ωωα=−= ×

We may therefore write the response as
( )
12
() cos sin
t
Ld
it e B t B t
α d
ω ω
−
=+
At
t = 0, iB .
1
0 0
L
=∴=

Noting that
() ()
22
sin sin cos
ttL
dd d dd
eB t Be t t
dt dt
ααdi
ω αωω ω
−−
==−+ and

0
100
L
t
di
L
dt
=
=− we find that BB2 = -0.316 A.

Finally,
it
2500 5
( ) 316 sin1.581 10 mA
t
L
e t
−
=− ×
PROPRIETARY MATERIAL . © 2007 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers
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Engineering Circuit Analysis, 7
th
Edition Chapter Nine Solutions 10 March 2006

53. Prior to t = 0, we find that vC = 100 V, since 10 A flows through the 10 Ω resistor.

Therefore, (0 ) (0 ) 100 V, and (0 ) (0 ) 0.
CC LL
vv ii
+− +−
== ==

The circuit for
t > 0 may be reduced to a simple series circuit consisting of a 2 mH
inductor, 20 nF capacitor, and a 10 Ω resistor; the dependent source delivers exactly the
current to the 5 Ω that is required to maintain its current.

Thus,
()
31
310
2.5 10 s
222 10
R
L
α
−
−
== =×
×

and
()( )
5
0
3911
1.581 10 rad/s
210 2010LC
ω
−−
== =×
××

W ith
0
αω< we find the circuit is underdamped, with

22 5
0
1.581 10 rad/s
d
ωωα=−= ×

We may therefore write the response as
( )
12
() cos sin
t
Cd
vt e B tB t
α d
ω ω
−
=+
At
t = 0, vB .
1
100 100 V
C
=∴=

Noting that
C
L
dv
Ci
and
dt
=
()
()
2
22
100cos sin
100cos sin 100 sin cos
t
dd
t
dddddd
etBt
dt
et Btt B
α
α
ωω
αω ωωωωω
−
−
⎡⎤ +
⎣⎦
⎡⎤=− + − +
⎣⎦
d
t

which is equal to zero at
t = 0 (since iL = 0)

we find that
BB2 = 1.581 V .

Finally,
( ) ( )
2500 5 5
( ) 100cos 1.581 10 1.581sin 1.581 10 V
t
C
vt e t t
−
⎡⎤=× +×
⎣⎦

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and educators for course preparation. If you are a student using this Manual, you are using it without permission.

Engineering Circuit Analysis, 7
th
Edition Chapter Nine Solutions 10 March 2006

54. Prior to t = 0, i1 = 10/4 = 2.5 A, v = 7.5 V, and vg = –5 V.
Thus,
vC(0
+
) = vC(0
–
) = 7.5 + 5 = 12.5 V and iL = 0

After
t = 0 we are left with a series RLC circuit where
1
4
L
i
i=−. We may replace the
dependent current source with a 0.5 Ω resistor. Thus, we have a series RLC circuit with R
= 1.25 Ω , C = 1 F, and L = 3 H.

Thus,
11.25
0.208 s
26R
L
α
−
== =
and
0
11
577 mrad/s
3
LC
ω===

W ith
0
αω< we find the circuit is underdamped, so that

22
0
538 mrad/s
d
ωωα=−=

We may therefore write the response as
( )
12
() cos sin
t
Ld
it e B t B t
α d
ω ω
−
=+
At
t = 0, iB .
1
0 0 A
L
=∴=

Noting that
0
(0)
L
C
t
di
Lv
and
dt
=
=−

() []
22
()12.5
sin sin cos ( 0)
3
tt CL
dd d d vtdid
eB t Be t t t
dt dt L
αα
ωαωωω
−− −
⎡⎤== − += =
⎣⎦
=
,
we find that
BB2 = –7.738 V.

Finally,
it for t > 0 and 2.5 A, t < 0
0.208
( ) 1.935 sin 0.538 A
t
L
e t
−
=

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and educators for course preparation. If you are a student using this Manual, you are using it without permission.

Engineering Circuit Analysis, 7
th
Edition Chapter Nine Solutions 10 March 2006

55. Prior to t = 0, i1 = 10/4 = 2.5 A, v = 7.5 V, and vg = –5 V.
Thus,
vC(0
+
) = vC(0
–
) = 12.5 V and iL = 0

After
t = 0 we are left with a series RLC circuit where
1
4
L
i
i=−. We may replace the
dependent current source with a 0.5 Ω resistor. Thus, we have a series RLC circuit with R
= 1.25 Ω , C = 1 mF, and L = 3 H.

Thus,
11.25
0.208 s
26R
L
α
−
== =
and
()
0
3
11
18.26 rad/s
310LC
ω
−
== =
×

W ith
0
αω< we find the circuit is underdamped, so that

22
0
18.26 rad/s
d
ωωα=−=

We may therefore write the response as
( )
12
() cos sin
t
Cd
vt e B tB t
α d
ω ω
−
=+
At
t = 0, vB .
1
12.5 12.5 V
C
=∴=
Noting that

()
[] [
12
22
cos sin
12.5cos sin 12.5 sin cos
tC
dd
tt
dd ddddvd
eB tB t
dt dt
etBte tB
α
αα
ωω
]
d
t
α ωω ωωω
−
−−
⎡⎤=+
⎣⎦
=− + + − +
ω

and this expression is equal to 0 at
t = 0,
we find that
BB2 = 0.143 V.

Finally,
[
]
0.208
( ) 12.5cos18.26 0.143sin18.26 V
t
C
vt for t > 0 and 12.5 V, t < 0 e t t
−
=+

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and educators for course preparation. If you are a student using this Manual, you are using it without permission.

Engineering Circuit Analysis, 7
th
Edition Chapter Nine Solutions 10 March 2006

56. (a)

6
2
500
12 22
1
5000 500
1
R 100
Series, driven: 500,
2L 0.2
11010
250,000
LC 40
Crit. damp ( ) 3(1 2) 3,
(0) 3, (0) 300V
3(AA)33A,A
1
(0 ) A 300 [ (0) (0 )] 0
L
A 3000 ( ) 3
o
L
Lc
t
L
Lc R
t
L
if
6Aie t
iv v
ei
t e
α
ω
−
++
−−
== =
×
== =
∴
iv
=−=−
==

∴=− + + ∴ =− + =
=− = − =

∴=∴=−+
500
(3000 6), 0
( ) 3 ( ) [ 3 (3000 6)] ( )A
t
t
L
tt
it u t e t ut
−
+>

(b)
500
(3000 6) 3; by SOLVE, 3.357ms
ot
oo
et t
−
+= =
=−+−+ +
∴

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and educators for course preparation. If you are a student using this Manual, you are using it without permission.

Engineering Circuit Analysis, 7
th
Edition Chapter Nine Solutions 10 March 2006

57.

2
4
12
4
,1 2
4
11 2R2 1
(0) 0, (0) 0, 4, 4 5 20
2L 0.5 LC
20 16 2 ( ) (A cos2 A sin 2 )
10A ( ) 10 (A cos2 A sin 2 )
010A,A 10,()10 (Asin2 10cos2)
1
(0) (0) 4 0 0 (0)
L
cL o
t
dL
t
Lf L
t
L
LL L
vi
it e t t i
iitett
it e t t
iv i αω
ω
−
−
−
++ +
=======×=
∴= −=∴ = + +
=∴=+ +
∴= + =− = + −
==×=∴
2202A 40,A 20
,Lf
==+ =−

i L(t) = 10 - e
-4t
(20 sin 2t + 10 cos 2t) A, t > 0

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and educators for course preparation. If you are a student using this Manual, you are using it without permission.

Engineering Circuit Analysis, 7
th
Edition Chapter Nine Solutions 10 March 2006

58.
6
2
1,2
,
10 40
12
12
R 250 1 10
25, 400
2L 10 LC 2500
25 625 400 10, 40
(0) 0.5A, (0) 100V, 0.5A
( ) 0.5 A A A
0 : (0 ) 100 50 1 200 0.5 50V 50 5 (0 )
(0 ) 10 10 10A 40A , 0.5 0.5
o
LcL f
tt
L
L L
L
s
ivi
it e e
tv i
i
−−
++ +
+
α= = = ω = = =
=− ± − =− −
===−
∴=−+ +
′==−×−×=−∴−=
′∴=−∴−=−− =−
12
12 2 1 1 1 2
10AA
A A 1 10 10A 40( 1+A ) 50A 40, A 1,A 0
( ) 0.5 1 A, 0; ( ) 0.5A, 0
t
LL
it e t it t
−
++
∴+ =∴−=− − − =− + = =
∴=−+ > = >

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and educators for course preparation. If you are a student using this Manual, you are using it without permission.

Engineering Circuit Analysis, 7
th
Edition Chapter Nine Solutions 10 March 2006

59.

66 3
26
22
400
,, 1 2
4000
12 1
4000
2
110 110
4000, 20 10
2RC 100 2.5 LC 50
2000, (0) 2A, (0) 0
0, ( 0) (A cos2000 A sin 2000 )
work with : ( ) (B cos2000 B sin 2000 ) B 0
B sin 2000
o
do L c
t
cf cf c
t
cc
t
c
iv
iv ie t t
vvt e t t
ve t
+
−
−
−
α= = = ω = = = ×
×
∴ω = ω −α = = =
==∴= +
=+∴ =
∴=
6
5
54 000
22
6 4000
633 4000
4000
110
,(0) (0) (21)810
C 2.5
8 10 2000B , B 400, 400 sin 2000
( ) C 2.5 10 400 ( 4000sin 200 2000cos200 )
10 ( 4sin 2000 2cos2000 )
(2cos2000 4sin 2000
cc
t
c
tt
cc
t
tvi
ve t
it v e t
ett
ett
++
−
−−
−++ −
−
′==×=×
∴× = = =
′∴= =×× − +
=−+
=− )A, 0t>

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and educators for course preparation. If you are a student using this Manual, you are using it without permission.

Engineering Circuit Analysis, 7
th
Edition Chapter Nine Solutions 10 March 2006

60. (a)

66
26
3
3
,
1000
12
6
1
22
1810 81013
1000, 26 10
2RC 2 4 10 4
26 1 10 5000, (0) 8V
(0) 8mA, 0
(A cos1000 A sin5000 )
18
A 8; (0 ) (0 ) 8 10 (0.01 0.008) 0
C 4000
5000A 1000 8 0, A 1.6
o
dc
Lc f
t
c
cc
v
iv
ve t t
vi
−
++
×× ×
α= = = ω = = ×
××
∴ω= −× = =
==
∴=+
′∴===×−−=
∴

So v c(t) = e
-1000t
(8 cos 1000t + 1.6 sin 1000t) V, t > 0
(b)
−×= =

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Engineering Circuit Analysis, 7
th
Edition Chapter Nine Solutions 10 March 2006

61.
2
,
1
11R1 1
1, 1 crit. d amp
2L 1 LC
5
(0) 12 10V, (0) 2A, 12V
6
11
( ) 12 (A 2); (0 ) (0 ) (0 ) 1
C2
1A 2;A 1 ()12 ( 2)V, 0
−++ +
α= = = ω = = ∴
=× = = =
′∴=+ − = =× =
∴= + =−
o
cL cf
t
cc c L
viv
vt e t v i i

−
∴ = − + >
t
c
vt e t t

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and educators for course preparation. If you are a student using this Manual, you are using it without permission.

Engineering Circuit Analysis, 7
th
Edition Chapter Nine Solutions 10 March 2006

62. (a)

(b) vu
500 1500
,3 4
6
34
6
34
34 4 4 3
500 1500
10 ( )V, 10, 10 A A ,
(0) 0, (0) 0 A A 10V, (0 ) 2 10
[ (0) (0 )] 2 10 (0 0) 0 500A 1500A
A 3A 0, add: 2A 10, A 5 A 15
() 10 15 5 V, 0
()
tt
sc fc
cL c
LR
tt
c
R
tv v e e
vi v
ii
vt e e t
it
−−
+
+
−−
===++
′==∴+=− =×
−=×−==−−

∴− − = − =− = ∴ =−

∴ =− + >

500 1500
10 15 5 mA, 0
tt
ee t
−
=− + >∴
6
6
26
1,2
500 1500
12
66
12
12
110
10 ( ) V : 1000
2RC 2000 0.5
12103
0.75 10 500, 1500
LC 8
A A , (0) 10V, (0) 10mA
A 10, (0 ) 2 10 [ (0) (0 )] 2 10
10
0.01 0 500A 1500A 0,
1000
A
s
o
tt
co L
cL Rvut
s
ve ev i
Av ii
−−
++
=− α= = =
×
××
ω= = = × ∴ =− −
∴=+ = =
′∴+= =× − =×
⎛⎞
−=∴−− =
⎜⎟
⎝⎠
−
12 2 2 1
500 1500
500 15003A 0; add: 2 A 10, A 5, A 15
() 15 5 V 0
() 15 5 mA, 0
tt
c
tt
R
vt e e t
it e e t
−−
−−
−= − = =− =
∴ =− >
∴ =− >

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and educators for course preparation. If you are a student using this Manual, you are using it without permission.

Engineering Circuit Analysis, 7
th
Edition Chapter Nine Solutions 10 March 2006

63. (a)
6
6
26 6
1,2
500 1500
,12
66 4
12
4
11
110
( ) 10 ( )V: 1000
2RC 1000
1103 3
1000 10 10 500, 1500
LC 4 4
0A A,(0)10V,(0)0
10
10 A A , 10 (0 ) 10 0 2 10
500
2 10 500A 1500A 40 A
s
o
tt
cf c c L
cc
vt u t
s
vveev i
vi
−−
+
=−α= = =
×
ω= = ∴ =− ± − × =− −
=∴ = + = =
⎡⎤
′
∴=+ = = − =−×
⎢⎥
⎣⎦
∴−× =− − ∴ =
22 21
500 1500
6 500 1500
500 1500
3A 30 2A , A 15, A 5
515V,0 C
10 (2500 22,500 )
2.5 22.5 mA, 0
tt
cs c c
tt
s
tt
ve e tiiv
iee
eet
−−
−− −
−−
+∴= = =−
′∴=− + > ∴ = =
∴=−
=− >

(b)
,
500 1500
34 34
66 4
34
34 4 4 3
500 1500
6() 10 ()V 10V, (0) 0, (0) 0
10 A A A 10
10
(0 ) 10 (0 ) 10 0 2 10 500A 1500A
500
A 3A 40, add: 2 A 30, A 15, A 5,
10 5 15 V,
10 (
sc f c L
tt
c
cc
tt
cs cvt ut v v i
vAe
e
vi
veeii
−−
++
−−
−
=∴= ==

∴=+ + ∴+ =−
⎛⎞
′==+=×=−−
⎜⎟
⎝⎠

∴−− = − = =− =
=+ − ==
500 1500 500 1500
2500 22,500 ) 25 22.5 mA, 0
tttt
eeeet
− −−−
−+ =+ >

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and educators for course preparation. If you are a student using this Manual, you are using it without permission.

Engineering Circuit Analysis, 7
th
Edition Chapter Nine Solutions 10 March 2006

64. Considering the circuit at t < 0, we see that i L(0
-
) = 15 A and v C(0
-
) = 0.
The circuit is a series RLC with α = R/2L = 0.375 s
-1
and ω0 = 1.768 rad/s. We therefore
expect an underdam ped response with
ωd = 1.728 rad/s. The general form of the
response will be

vC(t) = e
-αt
(A cos ωdt + B sin ωdt) + 0 (v C(∞) = 0)

vC(0
+
) = v C(0
-
) = 0 = A and we may therefore write v C(t) = Be
-0.375t
sin (1.728t) V

iC(t) = -i L(t) = C
dt
dv
C
= (80×10
-3
)(-0.375B e
-0.375t
sin 1.728t

At t = 0
+
, iC = 15 + 7 – i L(0
+
) = 7 = (80×10
-3
)(1.728B) so that B = 50.64 V.

Thus, v
C(t) = 50.64 e
–0.375t
sin 1.807t V and v C(t = 200 ms) = 16.61 V.

The energy stored in the capacitor at
that instant is ½ Cv C
2 = 11.04 J

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Engineering Circuit Analysis, 7
th
Edition Chapter Nine Solutions 10 March 2006

65. (a) v S(0
-
) = v C(0
-
) = 2(15) = 30 V

(b) i L(0
+
) = iL(0
-
) = 15 A
Thus, i
C(0
+
) = 22 – 15 = 7 A and v S(0
+
) = 3(7) + v C(0
+
) = 51 V

(c) As t → ∞, the current through the inductor approaches 22 A, so v S(t→ ∞,) = 44 A.

(d) We are presented with a s eries RLC circuit having α = 5/2 = 2.5 s
-1
and ωo = 3.536
rad/s. The natural response will therefore be underdamped with
ωd = 2.501 rad/s.

iL(t) = 22 + e
-αt
(A cos ωdt + B sin ωdt)
i
L(0
+
) = iL(0
-
) = 15 = 22 + A so A = -7 amperes

Thus, i L(t) = 22 + e
-2.5t
(-7 cos 2.501t + B sin 2.501t)
v
S(t) = 2 i L(t) +
dt
di
i
dt
di
L
L
L
2 L+= = 44 + 2e
-2.5t
(-7cos 2.501t + Bsin 2.501t)
– 2.5e
-2.5t
(-7cos 2.501t + Bsin 2.501t) + e
-2.5t
[7(2.501) sin 2.501t + 2.501B cos 2.501t)]

vS(t) = 51 = 44 + 2(-7) – 2.5(-7) + 2.501B so B = 1.399 amperes and hence

vS(t) = 44 + 2e
-2.5t
(-7cos 2.501t + 1.399sin 2.501t )
-2.5e
-2.5t
(-7cos 2.501t + 1.399sin 2.501t ) + e
-2.5t
[17.51sin 2.501t + 3.499cos 2.501t)]

and v S(t) at t = 3.4 s = 44.002 V. This is borne out by PSpice simulation:

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Engineering Circuit Analysis, 7
th
Edition Chapter Nine Solutions 10 March 2006

66. For t < 0, we have 15 A dc flowing, so that i L = 15 A, v C = 30 V, v 3Ω = 0 and v S = 30 V.
This is a series RLC circuit with α = R/2L = 2.5 s
-1
and ω0 = 3.536 rad/s. We therefore
expect an underdamped response with
ωd = 2.501 rad/s.

0 < t < 1
vC(t) = e
-αt
(A cos ωdt + B sin ωdt)

vC(0
+
) = v C(0
-
) = 30 = A so we may write v C(t) = e
-2.5t
(30 cos 2.501t + B sin 2.501t)

C
=
dt
dv-2.5e
-2.5t
(30 cos 2.501t + B sin 2.501t)
+ e
-2.5t
[-30(2.501)sin 2.501t + 2.501B cos 2.501t ]

iC(0
+
) =

0
C
+
=t
dt
dv
C
= 80×10
-3
[-2.5(30) + 2.501B] = -i L(0
+
) = -i L(0
-
) = -15 so B = -44.98 V

Thus, v C(t) = e
-2.5t
(30 cos 2.501t – 44.98 sin 2.501t ) and
i
C(t) = e
-2.5t
(-15 cos 2.501t + 2.994 sin 2.501t ).
Hence, v
S(t) = 3 i C(t) + v C(t) = e
-2.5t
(-15 cos 2.501t – 36 sin 2.501t)

Prior to switching, v C(t = 1) = -4.181 V and i L(t = 1) = -i C(t = 1) = -1.134 A.

t > 2:
Define t ' = t – 1 for notational simplicity. Then, with the fact that v C(∞) = 6 V,
our response will now be v
C(t') = e
-αt'
(A' cos ωdt' + B' sin ωdt') + 6.
With v
C(0
+
) = A' + 6 = -4.181, we find that A' = -10.18 V.
i
C(0
+
) =

0
C
+
=′
′
t
td
dv
C
= (80×10
-3
)[(-2.5)(-10.18) + 2.501B')] = 3 – i L(0
+
) so B' = 10.48 V
Thus, v
C(t') = e
-2.5t
(-10.18 cos 2.501t' + 10.48 sin 2.501t') and
i
C(t') = e
-2.5t
(4.133 cos 2.501t' – 0.05919 sin 2.501t' ).
Hence, v
S(t') = 3 i C(t') + v C(t') = e
-2.5t
(2.219 cos 2.501t' + 10.36 sin 2.501t')

We see that our hand calculations are supported by
the PSpice simulation.
PROPRIETARY MATERIAL . © 2007 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers
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Engineering Circuit Analysis, 7
th
Edition Chapter Nine Solutions 10 March 2006

67. It’s probably easiest to begin by sketching the waveform v x:

(a) The source current ( = i
L(t) ) = 0 at t = 0
-
.

(b) iL(t) = 0 at t = 0
+

(c) We are faced with a series RLC circuit having α = R/2L = 2000 rad/s and ω0 = 2828
rad/s. Thus, an underdamped response is expected with
ωd = 1999 rad/s.

The general form of the expected response is i L(t) = e
-αt
(A cos ωdt + B sin ωdt)

i
L(0
+
) = iL(0
-
) = 0 = A so A = 0. This leaves i L(t) = B e
-2000t
sin 1999t

v
L(t) = L
dt
di
L
= B[(5×10
-3
)(-2000 e
-2000t
sin 1999t + 1999 e
-2000t
cos 1999t)]

v
L(0
+
) = v x(0
+
) – vC(0
+
) – 20 i L(0
+
) = B (5×10
-3
)(1999) so B = 7.504 A.

Thus, i
L(t) = 7.504 e
-2000t
sin 1999t and i L(1 ms) = 0.9239 A.

(d)
Define t ' = t – 1 ms for notational convenience. With no source present, we expect a
new response but with the same general form:

iL(t') = e
-2000t '
(A' cos 1999t' + B' sin 1999t')
v
L(t) = L
dt
di
L
, and this enables us to calculate that v L(t = 1 ms) = -13.54 V. Prior to the
pulse returning to zero volts, -75 + v
L + vC + 20 i L = 0 so v C(t' = 0) = 69.97 V.

i L(t' = 0) = A' = 0.9239 and –v x + vL + vC + 20 i L = 0 so that B' = -7.925.
Thus, i
L(t') = e-2000 t' (0.9239 cos 1999t' – 7.925 sin 1999t') and
hence i
L(t = 2 ms) = i L(t' = 1 ms) = -1.028 A.
1 2 3 4
t (s)
v
x (V)
75
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and educators for course preparation. If you are a student using this Manual, you are using it without permission.

Engineering Circuit Analysis, 7
th
Edition Chapter Nine Solutions 10 March 2006

68. The key will be to coordinate the decay dictated by α, and the oscillation period
determined by
ωd (and hence partially by α). One possible solution of many:

Arbitra rily set ωd = 2π rad/s.
We want a capacitor voltage v
C(t) = e
-αt
(A cos 2πt + B sin 2πt). If we go ahead and
decide to set v
C(0
-
) = 0, then we can force A = 0 and simplify some of our algebra.

Thus, v C(t) = B e
-αt
sin 2πt. This function has max/min at t = 0.25 s, 0.75 s, 1.25 s, etc.
Designing so that there is no strong damping for several seconds, we pick α = 0.5 s
-1
.
Choosing a series RLC circuit, this now establishes the following:

R/2L = 0.5 so R = L and

ωd =
2
2
0
2
1
- ⎟
⎠
⎞
⎜
⎝
⎛
ω = 39.73 rad/s =
LC
1

Arbitrarily selecting R = 1 Ω, we find that L = 1 H and C = 25.17 mF. We need the first
peak to be at least 5 V. Designing for B = 10 V, we ∴need iL(0
+
) = 2π(25.17×10
-3
)(10) =
1.58 A. Our final circuit, then is:

And the operation is verified by a simple PSpice simulation:

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Engineering Circuit Analysis, 7
th
Edition Chapter Nine Solutions 10 March 2006

69. The circuit described is a series RLC circuit, and the fact that osci llations are detected
tells us that it is an underdamped response that we are modeling. Thus,

iL(t) = e
-αt
(A cos ωdt + B sin ωdt) where we were given that ωd = 1.825×10
6
rad/s.

ω0 =
LC
1
= 1.914×10
6
rad/s, and so ωd
2 = ω0
2 – α
2
leads to α
2
= 332.8×10
9

Thus,
α = R/2L = 576863 s
-1
, and hence R = 1003 Ω.
Theoretically, this value must include the “radiation resistance” that accounts for the
power lost from the circuit and received by the radio; there is no way to separate this
effect from the resistance of the rag with the information provided.

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Engineering Circuit Analysis, 7
th
Edition Chapter Nine Solutions 10 March 2006

70. For t < 0, iL(0
-
) = 3 A and vC(0
-
) = 25(3) = 75 V. This is a series RLC circuit with α =
R/2L = 5000 s
-1
and ω0 = 4000 rad/s. We therefore expect an overdamped response with
s
1 = -2000 s
-1
and s2 = -8000 s
-1
. The final value of vC = -50 V.

For t > 0, vC(t) = A e
-2000t
+ B e
-8000t
- 50

vC(0
+
) = vC(0
-
) = 75 = A + B – 50
so A + B = 125 [1]

dt
dv
C
= -2000 Ae
-2000t
– 8000 Be
-8000t

iC(0
+
) =
+
=0
Ctdt
dv
C
= 3 – 5 – iL(0
-
) = -5 = -25×10
-6
(2000A + 8000B)

Thus, 2000A + 8000B = 5/25×10
-6
[2]

Solving Eqs. [1] and [2], we find that A = 133.3 V and B = -8.333 V. Thus,

vC(t) = 133.3 e
-2000t
– 8.333 e
-8000t
– 50

and vC(1 ms) = -31.96 V. This is confirmed by the PSpice simulation shown below.

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Engineering Circuit Analysis, 7
th
Edition Chapter Nine Solutions 10 March 2006

71. α = 0 (this is a series RLC with R = 0, or a parallel RLC with R = ∞)

ωo
2 = 0.05 therefore ωd = 0.223 rad/s. We anticipate a response of the form:
v(t) = A cos 0.2236t + B sin 0.2236t

v(0
+
) = v(0
-
) = 0 = A therefore v(t) = B sin 0.2236t

dv/dt = 0.2236B cos 0.2236t; iC(t) = Cdv/dt = 0.4472B cos 0.2236t

iC(0
+
) = 0.4472B = -iL(0
+
) = -iL(0
-
) = -1×10
-3
so B = -2.236×10
-3
and thus

v(t) = -2.236 sin 0.2236t mV

In designing the op amp stage, we first write the differential equation:
)0( 0 2 10
10
1
3-
0
=+=++′∫ LC
tii
dt
dv
tdv
and then take the derivative of both sides:
v
dt
vd
20
1
-
2
2
=
With
43
0
105)10236.2)(2236.0(
−−
=
×−=×−=
+
tdt
dv
, one possible solution is:

PSpice simulations are very sens itive to par ameter values; better results were obta ined
using LF411 instead of 741s (both we
re compared to the simple LC circuit simulation.)

Simulation using 741 op amps Simulation using LF411 op amps

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Engineering Circuit Analysis, 7
th
Edition Chapter Nine Solutions 10 March 2006

72. α = 0 (this is a series RLC with R = 0, or a parallel RLC with R = ∞)

ωo
2 = 50 therefore ωd = 7.071 rad/s. We anticipate a response of the form:
v(t) = A cos 7.071t + B sin 7.071t, knowing that iL(0
-
) = 2 A and v(0
-
) = 0.

v(0
+
) = v(0
-
) = 0 = A therefore v(t) = B sin 7.071t

dv/dt = 7.071B cos 7.071t; iC(t) = Cdv/dt = 0.007071B cos 7.071t

iC(0
+
) = 0.007071B = -iL(0
+
) = -iL(0
-
) = -2 so B = -282.8 and thus

v(t) = -282.8 sin 7.071t V

In designing the op amp stage, we first write the differential equation:
)0( 0 10 2
20
1
3-
0
=+=++′∫ LC
tii
dt
dv
tdv
and then take the derivative of both sides:
v
dt
vd05-
2
2
=
With
2178)8.282)(071.7(
0
−=−=
+
=tdt
dv
, one possible solution is:

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Engineering Circuit Analysis, 7
th
Edition Chapter Nine Solutions 10 March 2006

73.
(a)
v
dt
dv
dt
dvv
3.3
1
-
or
0 103.3
1000
3-
=
=×+

(b) One possible solution:

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Engineering Circuit Analysis, 7
th
Edition Chapter Nine Solutions 10 March 2006

74. We see either a series RLC with R = 0 or a parallel RLC with R = ∞; either way, α = 0.
ω0
2 = 0.3 so ωd = 0.5477 rad/s (com bining the two inductors in parallel for the
calculation). We expect a response of the form i(t) = A cos
ωdt + B sin ωdt.

i(0
+
) = i(0
-
) = A = 1×10
-3
di/dt = -Aωd sin ωdt + Bωd cos ωdt
v
L = 10di/dt = -10A ωd sin ωdt + 10Bωd cos ωdt
v
L(0
+
) = v C(0
+
) = v C(0
-
) = 0 = 10B(0.5477) so that B = 0
and hence i(t) = 10
-3
cos 0.5477t A

The differential equation for this circuit is

and
+
=0 tdt
di
= 0

v
dt
vd
tvdtvd
tt
3.0
or

2
1
10
10
1
2
2
0
3-
0
−=
+′++′
∫∫
dt
dv
0 2 =

1 Ω
↓
i
One possible solution is:
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Engineering Circuit Analysis, 7
th
Edition Chapter Nine Solutions 10 March 2006

75. (a) v R = vL
20(- i L) = 5
dt
di
L
or
L
L4- i
dt
di
=

(b) We expect a response of the form i
L(t) = A e
-t/ τ
where τ = L/R = 0.25.

We know that i
L(0
-
) = 2 amperes, so A = 2 and i L(t) = 2 e
-4t

+
=0
L
tdt
di
= -4(2) = -8 A/s.
One possible solution, then, is
1 MΩ
4 kΩ
1 kΩ
1 Ω
1 μF
8 V
i
↓

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Engineering Circuit Analysis, 7
th
Edition Chapter Ten Solutions 10 March 2006

1.
3
33
3
rad rad
210
(a) T=−4(7.5 2.1)10 21.6 10 , 290.9 ra
d/s
21.6
( ) 8.5sin (290.9 ) 0 8.5sin (290.9 2.1 10 )
0.6109 2 5.672 or 325.0
( )
−−
− π
=×ω= =
(b) (c)
8.5sin (290.9 325.0 )
∴ =+ Φ∴=××+ Φ
∴Φ=− + π= °
=+°
t
ft t
ft t∴
8.5sin (290.9 325.0 ) 8.5
cos(290.9 235 ) 8.5cos(290.9 125 )
t
tt
+°=
+°= −°
8.5cos( 125 )cos 8.5sin125
sin
4.875 cos290.9 6.963sin 290.9
+
−° ω+ °
ω=− +
t
tt t
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Engineering Circuit Analysis, 7
th
Edition Chapter Ten Solutions 10 March 2006

2.

(a)
10cos 4sin ACos( ), A 0, 180 180
A 116 10.770, Acos 10, Asin 4 tan 0.4, 3 quad
21.80 201.8 ,too large 201.8 360 158.20
d
tt wt−ω +ω+ +Φ>−°<Φ≤°
== Φ
=− Φ=−∴Φ=
(b) (c) (d)
∴Φ= °= ° ∴Φ= °− °=− °
200cos(5 130 ) Fcos5 Gsin5 F 200cos130 128.6
G 200
sin130 153.2
ttt+°= + ∴= °=−
=− °=−
( ) 5cos10 3sin10 0, 0 1 s
sin10 5
, 10 1.0304,
cos10 3
0.10304 s; also, 10 1.0304 , 0.4172 s; 10 1.0304 2 , 0.7314 s
it=− t t t
t
t
t
ttt
tt
=≤≤
∴ ==
== +π== +π=
0<<10ms, 10cos100 12sin100 ; let 10cos100 =12sin100
10
tan100 = , 100 0.6947 2.211 ms 0 2.211 ms
12
ttttt
tt t t
π≥ π π π
∴ ππ=∴=∴<<
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Engineering Circuit Analysis, 7
th
Edition Chapter Ten Solutions 10 March 2006

3.

(a) Note that
22 1
cos sin cos tan
B
AxBx AB x
A
−⎛ −⎛⎞
+=+ +
⎜⎟⎜
⎝⎠⎝⎠
⎞
⎟
. For f(t), the angle is in th
e
second quadrant; most calculators will return –30.96
o
, which is off by 180
o
.

(b)
( ) leads ( ) by 149.04 15.255 133.8ft gt °− °= °
( ) 50cos 30sin 58.31cos( 149.04 )
( ) 55cos 15sin 57.01cos( 15.255 )
ampl. of ( ) 58.31, ampl. of ( ) 57.01
=− ω − ω = ω + °
=ω−ω= ω+°
∴= =
ft t t t
gt t t t
ft gt
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Engineering Circuit Analysis, 7
th
Edition Chapter Ten Solutions 10 March 2006

4. it=ω( ) Acos( ), and
L( /
) R V cos
L[Asin()]RAcos()Vcos
LAsin cos LA cos sin RA cos cos RAsin sin
V cos
LAcos RAsin
and sin RAcos V
L
, tan
m
m
m
m
t
di dt i t
ttt
tttt
t
LA
Thus
−θ
+= ω
∴−ω ω−θ+ ω−θ= ω
−ω ω θ+ ω ω θ+ ω θ+ ω θ
=ω
∴ω θ= θ
ωθ+ θ=
ω
θ=

( ) ()
222 222
22 2
222 222
222
222
LR
and LA RA V
RL RL
LR
so that A V
RLRL
V
, R L A V and therefore we may write A
RL
m
m
m
m
R
Thus
ω
ω+ =
+ω +ω
⎛⎞ω
+=⎜⎟
+ω +ω⎝⎠
+ω = =
+ω*
*
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Engineering Circuit Analysis, 7
th
Edition Chapter Ten Solutions 10 March 2006

5. f = 13.56 MHz so ω = 2πf = 85.20 Mrad/s.
Delivering 300 W (peak) to a 5-Ω loa
d implies that
300
5
V
2
m
= so V m = 38.73 V.

Finally, (85.2×10
6
)(21.15× 10
–3
) + φ = nπ, n = 1, 3, 5, …

Since (85.2×10
6
)(21.15× 10
–3
) = 1801980, which is 573588π, we find that

φ = 573589π - (85.2×10
6
)(21.15× 10
–3
) = 573589π - 573588π = π
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Engineering Circuit Analysis, 7
th
Edition Chapter Ten Solutions 10 March 2006

6. (a) -33 sin(8t – 9
o
) → -33∠(-9-90)
o
= 33∠81
o
12 cos (8 t – 1
o
) → 12∠-1
o

(b) 15 cos (1000t + 66
o
) → 15 ∠ 66
o
-2 cos (1000 t + 450
o
) → -2 ∠ 450
o
= -2 ∠90
o
= 2 ∠ 270
o

(c) sin (t – 13
o
) → 1∠-103
o
cos ( t – 90
o
) → 1 ∠ -90
o

(d) sin t → 1 ∠ -90
o
cos ( t – 90
o
) → 1 ∠ -90
o

These two waveforms are in phase. Neither leads the oth
er.
33∠81
o
12∠-1
o
-33 sin(8t – 9
o
) leads 12 cos (8t – 1
o
) by
81 – (-1) = 82
o
.
1∠-103
o
1 ∠ -90
o
cos (t – 90
o
) leads sin (t – 13
o
)
by 66 – -90 = 156
o
.
15∠66
o
2∠270
o
15 cos (1000t + 66
o
) leads -2 cos (1000t + 450
o
)
by 66 – -90 = 156
o
.

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Engineering Circuit Analysis, 7
th
Edition Chapter Ten Solutions 10 March 2006

7. (a) 6 cos (2 π60t – 9
o
) → 6∠-9
o

-6 cos ( 2
π60t + 9
o
) → 6∠189
o

6∠-9
o
6∠189
o
-6 cos (2π60t + 9
o
) lags 6 cos (2π60t – 9
o
)
by 360 – 9 – 189 = 162
o
.

(b) cos ( t - 100
o
) → 1 ∠ -100
o
-cos ( t - 100
o
) → -1 ∠ -100
o
= 1 ∠80
o

1∠80
o
1∠-100
o
-cos (t - 100
o
) lags cos (t - 100
o
) by 180
o
.

(c) -sin t → -1∠-90
o
= 1∠90
o
sin t → 1∠ -90
o

1∠90
o
1 ∠ -90
o
-sin t lags sin t by 180
o
.

(d) 7000 cos (t –
π) → 7000 ∠ - π = 7000 ∠ -180
o
9 cos ( t – 3.14
o
) → 9 ∠ -3.14
o

7000 cos (t –
π) lags 9 cos (t – 3.14
o
)
by 180 – 3.14 = 176.9
o
.

9 ∠ -3.14
o
7000 ∠ -180
o
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Engineering Circuit Analysis, 7
th
Edition Chapter Ten Solutions 10 March 2006

8. v(t) = V 1 cos ωt - V2 sin ωt [1]

We assume this can be written as a single cosine such that

v(t) = V m cos (ωt + φ) = Vm cos ωt cos φ - Vm sin ωt sin φ [2]

Equating terms on the right hand sides of Eqs. [1] and [2],

V 1 cos ωt – V2 sin ωt = (Vm cos φ) cos ωt – (Vm sin φ) sin ωt

yields
V1 = Vm cos φ and V2 = Vm sin φ

Dividing, we find that φ
φ
φ
tan
cos V
sin V

V
V
m
m
1
2
==
and φ = tan
-1
(V2/ V1)

Next, we see from the above sketch that we may write V
m = V1/ cos φ or

2
2
2
11
1
m
V VV
V
V
+
=
=
2
2
2
1
V V+
φ
V2
V1
2
2
2
1
V V+

Thus, we can write v (t) = V m cos (ωt + φ) =
2
2
2
1
V V+ cos [ ωt + tan
-1
(V2/ V1)].
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Engineering Circuit Analysis, 7
th
Edition Chapter Ten Solutions 10 March 2006

9. (a) In the range 0 ≤ t ≤ 0.5, v(t) = t/0.5 V.
Thus, v (0.4) = 0.4/0.5 = 0.8 V.

(b) Remembering to set the calculator to radians, 0.7709 V.

(c) 0.8141 V.

(d) 0.8046 V.
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Engineering Circuit Analysis, 7
th
Edition Chapter Ten Solutions 10 March 2006

10. (a) V rms =
2
1
T
0
2
2
m
cos
T
V
⎥
⎦
⎤
⎢
⎣
⎡∫
dttω
=
2
1
T
0
2
2
m

T
2
cos
T
V
⎥
⎦
⎤
⎢
⎣
⎡∫
dt
t
π

=
2
1
T
0
2
m

T
4
cos 1
2T
V
⎥
⎦
⎤
⎢
⎣
⎡
⎟
⎠
⎞
⎜
⎝
⎛
+∫
dt
t
π

=
2
1
T
0
2
m
T
0
2
m

T
4
cos
2T
V

2T
V
⎥
⎦
⎤
⎢
⎣
⎡
+∫∫
dt
t
dt
π

=
2
1
4
0
2
m
2
m
cos
8
V
T
2T
V
⎥
⎦
⎤
⎢
⎣
⎡
+
π
π
u
= 2
V
m

*

(b) V
m =
V 155.6 2 110=, V 162.6 2 115=, V 169.7 2 201=

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and educators for course preparation. If you are a student using this Manual, you are using it without permission.

Engineering Circuit Analysis, 7
th
Edition Chapter Ten Solutions 10 March 2006

11. We begin by defining a clockwise current i. Then, KVL yields

–2×10
–3
cos5t + 10i + v C = 0.
Since
C
C
dv
ii C
dt
==
, we may rewrite our KVL equation as
3
30 2 10 cos5
C
C
dv
v
dt
−
+=× t [1]
We anticipate a response of the form v
C(t) = Acos(5t + θ). Since
5sin(5 )
C
dv
At
dt
θ=− + ,
we now may write Eq. [1] as –150Asin(5t + θ) + Acos(5t + θ) = 2×10
–3
cos5t. Using a
common trigonometric identity, we may combine the two terms on the left hand side into
a single cosine function:

()
2
21 150
150 cos 5 ta n 2 10 cos5A
3
AAt
A θ
−−⎛⎞
+++ =×
⎜⎟
⎝⎠
t

Equating terms, we find that A = 13.33 μV and θ = –tan
–1
150 = –89.62
o
. Thus,

v
C(t) = 13.33 cos (5t – 89.62
o
) μV.

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and educators for course preparation. If you are a student using this Manual, you are using it without permission.

Engineering Circuit Analysis, 7
th
Edition Chapter Ten Solutions 10 March 2006

12. KVL yields
–6cos400t + 100i + v
L = 0.
Since
2
L
di di
vL
dt dt
== , we may rewrite our KVL equation as
2 100 6cos400
di
i
dt
+= t [1]
We anticipate a response of the form i(t) = Acos(400t + θ). Since

400 sin(400 )
di
At
dt
θ=−+ ,
we now may write Eq. [1] as

–800Asin(400t + θ) + 100Acos(400t + θ) = 6 cos400t.

Using a common trigonometric identity, we may co
mbine the two terms on the left hand
side into a single cosine function:

()()
22
1 800
800 100 cos 400 t an 6cos400
100A
A At
A θ
−⎛⎞
++ +=
⎜⎟
⎝⎠
t

Equating terms, we find that A = 7.442 mA
and θ = –tan
–1
8 = –82.88
o
. Thus,

i (t) = 7.442 cos (400t – 82.88
o
) mV, so
2
L
di di
vL
dt dt
== = 5.954cos (400t + 7.12
o
)

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and educators for course preparation. If you are a student using this Manual, you are using it without permission.

Engineering Circuit Analysis, 7
th
Edition Chapter Ten Solutions 10 March 2006

13. 20cos500t V → 20∠0
o
V. 20 mH → j10 Ω.

Performing a quick source transformation, we replace the voltage source/20-Ω re
sistor
series combination with a 1∠0
o
A current source in parallel with a 20-Ω resistor.

20 || 60k = 19.99 Ω. By curren
t division, then,

IL =
19.99
19.99 5 10j++
= 0.7427∠-21.81
o
A. Thus, i L(t) = 742.7 cos (500t – 21.81
o
) mA.

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Engineering Circuit Analysis, 7
th
Edition Chapter Ten Solutions 10 March 2006

14.

At : R 80 20 16
80
0.4(15 85) cos500
85
4.8cos500 V
th
oc
oc
xx
vt
vt
−==
=−
Ω
∴=
(a)
1
224.8 10
cos 500 tan
1516 10
0.2544cos(500 32.01 )A
−⎛⎞
=−
⎜⎟
⎝⎠+
=−°
Lit

t
(b)
L 0.02 0.02544( 500)
sin (500 32.01 ) 2.544sin (500 32.01 )V
2.544cos(500 57.99 )V,

31.80cos(500 57.99 )mA
LL
Lx
tt
vti
t
′==× −
−°=− −°
vi
∴=+°
=+°
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Engineering Circuit Analysis, 7
th
Edition Chapter Ten Solutions 10 March 2006

15.

(a)
55
22
5100 800
cos 10 0.10600cos(10 57.99 )A
500500 800
57.99
0 when 0 10 , 25.83
180 2
R
it t
pit s
⎛⎞
=− = − °
⎜⎟
⎝⎠+
°π
==∴−π= μ

t=
(b)
355
5
5
55
5
L 8 10 0.10600( 10 ) sin (10 57.99 )
84.80sin (10 57.99 )
8.989sin (10 57.99 )
cos(10 57.99 ) 4.494 sin (2 16 115.989 )
0 when 2 10 115.989 0 , 180 ,
10.121 or 25.83
L
L
LL
L
vi t
vt
pvi t
tt
pt
ts
−
′±= =× × − − °

∴=− − °

∴==− − °
−°=− ×− °

∴=×−°=°°
=μ∴

(c)
55
5
10.600cos10 cos(10 57.99 )
0 when 10 , 15.708 and also 25.83
2
== − °
π
∴===μ=
ssL
s μ
i t tpv

p tt s st
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Engineering Circuit Analysis, 7
th
Edition Chapter Ten Solutions 10 March 2006

16.
55
5
5
5
51 5
22
3cos10 V, 0.1cos10 A
in series with 30 0.1cos10 A 30
Add, getting 0.2cos10 A 30
change to 6 cos 10 Vin series with 30 ; 30 20 50
61 0
cos 10 tan 0.11767cos(10 11.310
5050 10
−
==
Ω→ Ω
Ω
ΩΩ+Ω=Ω
⎛⎞
∴=−=−
⎜⎟
⎝⎠+
ss
s
Lvti t
vt
t
t
it t
5r ad
rad
)A
At 10 , 10 1 0.1167cos(1 11.310 ) 81.76mA
0.11767 10cos(1 11.30 90 ) 0.8462V
=μ =∴= − °=

°
∴=× −°+°=−
L
L
tsti
v

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Engineering Circuit Analysis, 7
th
Edition Chapter Ten Solutions 10 March 2006

17. cos500t V → 1∠0
o
V. 0.3 mH → j0.15 Ω.

Performing a quick source transformation, we replace the voltage source-resist
or series
combination with at 0.01∠0
o
A current source in parallel with a 100-Ω resistor. Current
division then leads to
()
LL
100
0.01 0.2 =
100 0.15j
+
+II

1 + 20
IL = (100 + j0.15) IL

Solving, we find that
IL = 0.0125∠ -0.1074
o
A,

so that i
L(t) = 12.5cos(500t – 0.1074
o
) mA.

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Engineering Circuit Analysis, 7
th
Edition Chapter Ten Solutions 10 March 2006

18.
12
1
22
V 120cos120 V
120 120
2A, 1A, 2 1 3A, 60 120 40
60 12
3 40 120V, L 12 37.70
120 37.70
cos 120 tan
4040 37.70
2.183cos(120 43.30 )A
ss
L
vt
it
t
−
== π
==+= =Ω
×= ω=π= Ω
⎛⎞
∴= π−
⎜⎟
⎝⎠+
=π −°

(a)
22
21
0.1 2.183 cos (120 43.30 )
2
0.2383cos (120 43.30 )J
ω= × × π−
°
°
L
t
∴

(b)
,
1
0.2383 0.11916J
2
ω=× =
Lav

=π −t
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Engineering Circuit Analysis, 7
th
Edition Chapter Ten Solutions 10 March 2006

19.
12
120cos400 V, 180cos200 V
Performing two quick source transformations,
120 180
2 A, 1.5 A, and noting that 60 120 40 ,
60 120
results in two current sources (with different frequencies) in parallel,
a
ss
vt vt==
== =Ω
nd a
lso in parallel with a 40 resistor and the 100 mH inductor.
Next we employ superposition. Open-circuiting the 200 rad/s source first,
we perform a source transformation to obtain a voltage source
Ω

1
222
having
magnitude 2 40 80 V. Applying Eqn. 10.4,
80 400(0.1)
cos(400 tan )
4040 400 (0.1)
, we open-circuit the 400 rad/s current source, and perform a
source transformation to obtain a voltage
L
it
Next
−
×=
′=−
+
1
222
source with magnitude
1.5 40 60 V. Its contribution to the inductor current is
60 200(0.1)
cos(200 tan )A
4040 200 (0.1)
so that 1.414cos(400 45 ) 1.342cos(200 26.57 )A
L
L
it
it t
−
×=
′′=−
+
=− °+−°
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Engineering Circuit Analysis, 7
th
Edition Chapter Ten Solutions 10 March 2006

20.
11
1
11
1
11
L
R,R0,A,ideal, RC
R
Vcos
,
RR
(Vcos)C
R
L
Vcos RC
R
For RL circuit, V cos L
R
L
Vcos
R
By c
io
m out
upper lower
c upper lower out m out
m out out out out
R
mr
mRR
tv
ii
i
ii i v t v
tv v v v
dv
tv
dt
tv vω
ω
ω
ω
ω
=∞ = =∞ =
=− =
′∴= + = − =−
′′∴=+=+
⎛⎞
=+
⎜⎟
⎝⎠
′∴=+
omparison,
R out
vv=

*

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Engineering Circuit Analysis, 7
th
Edition Chapter Ten Solutions 10 March 2006

21.
1
(a) V cos R (ignore I.C)
C
1
Vsi
n R
=+
′

∴−=+
∫m
m ti idt
ti i
Cω
ωω

(b)
()
1
ssume A cos( )
A
Vsin R Asin( ) cos( )
C
AA
V sin R Acos sin R Asin cos cos cos sin sin
CC
Equating terms on the left and right side,
A1
[1] R Asin cos tan so t an 1 CR ,
CC R
−
=+Φ
∴− =− +Φ + +Φ
∴− =− Φ Φ + Φ− Φ
Φ= Φ∴ Φ= Φ=
m
m
it
tt t
ttttt
A
ω

ωωωω ω
ωωω ωω ω ω ω
ωω
ω

22 2 22 2
222
22 2
22 2 22 2
1
22 2
and
CR A 1
[2] V R A
C1CR 1CR
CVAR C1 A
V1 CRA
CC1CR 1CR
CV 1
cos tan
CR1CR
−
−=− −
++
⎡⎤ +
∴= = + ∴=⎢⎥
++⎣⎦
⎛⎞
∴= +⎜⎟
+ ⎝⎠
m
m
m
m
it
ω
ωω
ωω
ωω
ωω
ωω
ω
ω
ωω
-

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Engineering Circuit Analysis, 7
th
Edition Chapter Ten Solutions 10 March 2006

22. (a) 7 ∠ -90
o
= -j 7

(b) 3 + j + 7 ∠ -17
o
= 3 + j + 6.694 – j 2.047 = 9.694 – j 1.047

(c) 14e
j15
o
= 14 ∠ 15
o
= 14 cos 15
o
+ j 14 sin 15
o
= 13.52 + j 3. 623

(d) 1 ∠ 0
o
= 1

(e) –2 (1 + j 9) = -2 – j 18 = 18.11 ∠ - 96.34
o

(f) 3 = 3 ∠ 0
o

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Engineering Circuit Analysis, 7
th
Edition Chapter Ten Solutions 10 March 2006

23. (a) 3 + 15 ∠ -23
o
= 3 + 13.81 – j 5.861 = 16.81 – j 5.861

(b) (j 12)(17
∠ 180
o
) = (12 ∠ 90
o
)(17 ∠ 180
o
) = 204 ∠ 270
o
= –j 204

(c) 5 – 16(9 – j 5)/ (33 ∠ -9
o
) = 5 – (164 ∠ -29.05
o
)/ (33 ∠ -9
o
)

= 5 – 4.992 ∠ -20.05
o
= 5 – 4.689 – j 1.712 = 0.3109 + j 1.712

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Engineering Circuit Analysis, 7
th
Edition Chapter Ten Solutions 10 March 2006

24. (a) 5 ∠ 9
o
– 9 ∠ -17
o
= 4.938 + j 0.7822 – 8.607 + j 2.631 = -3.668 + j 3.414

= 5.011 ∠ 137.1
o

(b) (8 – j 15)(4 + j 16) – j = 272 + j 68 – j = 272 + j 67 = 280.1 ∠ 13.84
o

(c) (14 – j 9)/ (2 – j 8) +
5 ∠ -30
o
= (16.64 ∠ -32.74
o
)/ (8.246 ∠ - 75.96
o
) + 4.330 – j 2.5

= 1.471 + j 1.382 + 4.330 – j 2.5 = 5.801 – j 1.118 = 5.908 ∠ -10.91
o

(d) 17 ∠ -33
o
+ 6 ∠-21
o
+ j 3 = 14.26 – j 9.259 + 5.601 – j 2.150 + j 3

= 19.86 – j 8.409 = 21.57 ∠ -22.95
o
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Engineering Circuit Analysis, 7
th
Edition Chapter Ten Solutions 10 March 2006

25. (a) e
j14
o
+ 9 ∠ 3
o
– (8 – j 6)/ j
2
= 1 ∠ 14
o
+ 9 ∠ 3
o
– (8 – j 6)/ (-1)

= 0.9703 + j 0.2419 + 8.988 + j 0.4710 + 8 – j 6 = 17.96 – j 5.287 = 18.72 ∠ -16.40
o

(b) (5 ∠ 30
o
)/ (2 ∠ -15
o
) + 2 e
j5
o
/ (2 – j 2)

= 2.5 ∠ 45
o
+ (2 ∠ 5
o
)/ (2.828 ∠ -45
o
) = 1.768 + j 1.768 + 0.7072 ∠ 50
o

= 1.768 + j 1.768 + 0.4546 + j 0.5418

= 2.224 + j 2.310 = 3.207 ∠ 46.09
o
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Engineering Circuit Analysis, 7
th
Edition Chapter Ten Solutions 10 March 2006

26.

(a)
5 110 1.7101 4.698∠− °=− − j

(b)
160
6 5.638 2.052
°
=− +
j
ej

(c)
(3 6)(2 50 ) 5.336 12.310+∠°=−+jj

(d)
100 40 107.70 158.20−− = ∠−j °
∠°+∠− °

(e)
2 50 3 120 1.0873 101.37= ∠− °
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Engineering Circuit Analysis, 7
th
Edition Chapter Ten Solutions 10 March 2006

27.

(a)
40 50 18 25 39.39 76.20∠− °− ∠ °= ∠− °

(b)
22 5
3 4.050 69.78
12
j
jj
−−
++ = ∠− °
+

(c)
3
(2.1 25 ) 9.261 75 2.397 8.945
+
∠°= ∠°= + j
(d)
0.3 rad
0.7 0.7 0.3 0.6687 0.2069=∠ = +
j
ej
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Engineering Circuit Analysis, 7
th
Edition Chapter Ten Solutions 10 March 2006

28.
(40 30 ) (40 30 )
(40 30) (40 30)
(40 30 ) (40 30 )
(40 30 ) (40 30 )
(40 53
20 A 100 20
50 , 10 A
(20 10) , 40 0.08(20 10)
(32 64) V (32 64 50)
34.93
+° +°
+° +°
+° +°
+° +°
−
=∴=
=− =−
∴= − = × −
∴= + ∴= + −
∴= ∫
tj t
cc
jt jt
cR
jt jt
LL
jt jt
Ls
jt
s
ie v e dt
vje ije
ijevj je
vje vjje
ve
.63 )
V
°

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Engineering Circuit Analysis, 7
th
Edition Chapter Ten Solutions 10 March 2006

29.
(10 25 )
(10 25 ) (10 25 )
(10 25 )
(1025) (1025)
(10 25 ) (10 25 )
(10 125.62 )
20 A
0.2 [20 ] 40
80
(80 40) , 0.08(80 40) 10
( 32 64) ( 12 64)
65.12 A
jt
L
jt t
L
jt
R
jt jt
sc
jt jt
cs
jt
s
ie
d
veje
dt
ve
vjei jje
ijeije
ie
+°
+° =°
+°
+° +°
+° +°
+°
=
==
=
=+ = +
∴=−+ ∴=−+
∴=

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Engineering Circuit Analysis, 7
th
Edition Chapter Ten Solutions 10 March 2006

30. 80cos(500 20 )V 5cos(500 12 )A−° → +°tt

(a) 40cos(500 10 ) 2.5cos(500 42 )A
so utvti t=+°∴= + °
+°
=−°

(b) =40sin (500 10 ) 40cos(500 80 )
2.5c
os(500 48 )A

(c)

(d)

∴=− °
s
out t
it
vt
(500 10 )
(500 42 )
40 40cos(500 10 )
40sin (500 10 ) 2.5 A
+°
+°
==+°
++° ∴=
jt
s
jt
out
ve t
jt ie
500 21.80 500
(500 53.80 )
(50 20) 53.85
3.366 A
+°+
+°
=+ =
∴=
jtj
s
jt
out
jt
e evj
ie
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Engineering Circuit Analysis, 7
th
Edition Chapter Ten Solutions 10 March 2006

31.

(a)
12sin (400 110 )A 12 20 A+°→∠°t

(b) 7sin800 3cos800 7 3
3 7 7.616 113.20 A−− →−
=− +
= ∠ °ttj
j

(c)

(d)

(e)

4cos(200 30 ) 5cos(200 20 )
4 30
5 20 3.910 108.40 A−°− +°
→∠− °−∠ °= ∠− °tt
3rad
600, 5ms : 70 30 V
70cos(600 5 10 30 ) 64.95 V
t
−
ω= = ∠ °
→××
+°=−
rad
600, 5ms : 60 40V 72.11 146.3
72.11cos(3 146.31 ) 53.75V
tjω= = + = ∠ °
→+° =
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Engineering Circuit Analysis, 7
th
Edition Chapter Ten Solutions 10 March 2006

32. 4000, 1mstω= =

(a) I5=

(b)

(c)

(d)

(e)

rad
80A
5cos(4 80 ) 4.294A
∠− °
∴=−°=−
x
x
i
I=−
rad
4 1.5 4.272 159.44 A
4.272cos(4 159.44 )
+ = ∠ °
3.750 A
−
∴=+
x
x j
i
°=
( ) 50sin (250 40 )
50cos(250 130 ) V 50 130 V
=−°
=−°→=∠−
x
xvt
°
t
t20cos108 30sin108
20 30 36.06 56.31 V
=−
→+ = ∠ °
xvt t
j
33cos(80 50 ) 41cos(80 75 )
33 50 41 75 72.27 63.87 V
=−°+−°
→∠−°+∠−°= ∠− °
xvt t
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Engineering Circuit Analysis, 7
th
Edition Chapter Ten Solutions 10 March 2006

33. V
12
3rad
rad rad
10 90mV, 500;V 8 90mV,
1200, M by 5, 0.5ms
( 5) [10cos(500 0.5 10 90 )
8cos(1.2 0.5 90 )]
50sin 0.25 40sin 0.6 34.96mV
out
t
v
−
=∠° ω= =∠°
ω= − =
=− × × + °
+×+°
=+=

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Engineering Circuit Analysis, 7
th
Edition Chapter Ten Solutions 10 March 2006

34. Begin with the inductor:
(2.5 ∠40
o
) (j500) (20× 10
-3
) = 25∠130
o
V across the inductor and the 25-Ω resistor.
The current through the 25-Ω resis
tor is then (25∠130
o
) / 25 = 1∠130
o
A.

The current through the unknown element is therefore 2.5∠40
o
+ 1∠130
o
=
2.693 ∠61.80
o
A; this is the same current through the 10-Ω resistor as well.
Armed with this information, KVL provides that

Vs = 10(26.93∠61.8
o
) + (25∠ -30
o
) + (25∠130
o
) = 35.47 ∠58.93
o

and so v
s(t) = 35.47 cos (500t + 58.93
o
) V.
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Engineering Circuit Analysis, 7
th
Edition Chapter Ten Solutions 10 March 2006

35. ω = 5000 rad/s.

(a) The inductor voltage = 48∠ 30
o
= jωL IL = j(5000)(1.2×10
-3
) IL

So IL = 8∠-60
o
and the total current flowing through the capacitor is
10 ∠ 0
o
- IL = 9.165∠49.11
o
A and the voltage V1 across the capacitor is

V1 = (1/jωC)(9.165∠49.11
o
) = -j2 (9.165∠49.11
o
) = 18.33∠-40.89
o
V.

Thus, v 1(t) = 18.33 cos (5000t – 40.89
o
) V.

(b)
V2 = V1 + 5(9.165∠ 49.11
o
) + 60∠120
o
= 75.88∠ 79.48
o
V

2( ) 75.88cos(5000 79.48 )V
∴ =+°vt t
(c)
V3 = V2 – 48∠30
o
= 75.88 ∠ 79.48
o
– 48∠30
o
= 57.70∠ 118.7
o
V

3( ) 57.70cos(5000 118.70 )V
∴ =+°vt t

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Engineering Circuit Analysis, 7
th
Edition Chapter Ten Solutions 10 March 2006

36. VR = 1∠0
o
V, Vseries = (1 + j ω –j/ω)(1∠0
o
)

V R = 1 and Vseries = () 1/ - 1
2
ωω+

We desire the frequency w at which Vseries = 2VR or Vseries = 2
Thus, we need to solve the equation ( ) 4 1/ - 1
2
=+ωω
or 0 1 - 3 -
2
=ωω

Solving, we find that ω = 2.189 rad/s.

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Engineering Circuit Analysis, 7
th
Edition Chapter Ten Solutions 10 March 2006

37. With an operating frequency of ω = 400 rad/s, the impedance of the 10-mH inductor is
j
ωL = j4 Ω, and the impedance of the 1-mF capacitor is –j/ ωC = -j2.5 Ω.

V 2 40 ( 2.5) 5 50 A
I 3 2 40 1.9513 41.211 A
V 4 1.9513 90 4.211 7.805 48.79 V
V V V 7.805 48.79 5 50
V 9.892 78.76 V, 9.892cos(400 78.76 ) V
+
+
∴=∠ °− =∠− °
∴=−∠ °= ∠− °
∴=× ∠ °− °= ∠ °
∴=−= ∠ °−∠−°
∴=∠° = +°
c
L
L
xLc
xx
j
vt

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Engineering Circuit Analysis, 7
th
Edition Chapter Ten Solutions 10 March 2006

38.

2
12
12
out 1 2If I 2 20 A, I 3 30 A V 80 10 V
II440AV 9030V
Now let I 2.5 60 A and I 2.5 60 A
Let V AI BI 80 10 A(2 20 ) B(3 30 )
90 30
and 90 30 (A B)(4 40 ) A B 12.415 20.2
440
+
=∠ ° =∠− ° → = ∠ °
==∠°→ =−
=∠−° =∠°
=+∴∠°=∠°+∠−°
−
−=+ ∠°∴+= = −
∠°
si s out
s s out
ss
ss
j
j
jj
1
80 10 3 30
AB A40 10 B(1.5 50)
220 220
12.415 20.21 B 40 10 B(1.5 50 )
12.415 20.21 40 10 B(1 1.5 50 )
B(1.1496 88.21 )
30.06 153.82
B 26.148 117.97
1.1496 88.21
A 12.415 2
+
+
+
∠° ∠− °
∴ =+ ∴=∠−°− ∠−°
∠° ∠
∴ −−=∠−°−∠−°
∴ −−∠−°=−∠−°
=∠+°
∠− °
∴==∠°
∠+ °
∴=−
j
j
j0.21 10.800 23.81
49.842 60.32
V (49.842 60.32 )(2.5 60 )
(26.15 117.97 )(2.5 60 )
165.90 140.63 V
−+
=∠−°
=∠−°∠−°
+∠°∠°
=∠−°
out
j
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Engineering Circuit Analysis, 7
th
Edition Chapter Ten Solutions 10 March 2006

39. We begin by noting that the series connection of capacitors can be replaced by a single
equivalent capacitance of value
1
545.5 F
11
1
23
C μ==
++
. Noting ω = 2πf,

(a)
ω = 2π rad/s, therefore ZC = –j/ωC =
()
6
10
291.8
2 545.5j
j
π
−
=−Ω .
(b)
ω = 200π rad/s, therefore ZC = –j/ωC =
()
6
10
2.918
200 545.5j
j
π
−
=−Ω .
(c)
ω = 2000π rad/s, therefore ZC = –j/ωC =
()
6
10
291.8 m
2000 545.5j
j
π
−
=−Ω .
(d)
ω = 2×10
9
π rad/s, therefore ZC = –j/ωC =
()
6
9
10
291.8 n
2 10 545.5j
j
π
−
=−Ω
×
.
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Engineering Circuit Analysis, 7
th
Edition Chapter Ten Solutions 10 March 2006

40. We begin by noting that the parallel connection of inductors can be replaced by a single
equivalent inductance of value
15
nH
15 6
L==
+ . In terms of impedance, then, we have

9
95
510
6
5
51
6
j
j
ω
ω
−
0
−
⎛⎞
×
⎜⎟
⎝⎠
=
+×
Z

Noting
ω = 2πf,

(a)
ω = 2π rad/s, therefore Z = j5.236×10
–9
Ω (the real part is essentially zero).

(b)
ω = 2×10
3
π rad/s, therefore Z = 5.483×10
–12
+ j5.236×10
–6
Ω.

(c)
ω = 2×10
6
π rad/s, therefore Z = 5.483×10
–6
+ j5.236×10
–6
Ω.

(d)
ω = 2×10
9
π rad/s, therefore Z = 2.615 + j2.497 Ω .

(e)
ω = 2×10
12
π rad/s, therefore Z = 5 + j4.775×10
–3
Ω .
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Engineering Circuit Analysis, 7
th
Edition Chapter Ten Solutions 10 March 2006

41.
(a) 800: 2 F 625, 0.6H 480
300( 625) 600( 480)
Z
300 625 600 480
478.0 175.65
in
jj
jj
jj
j
ω= μ →− →
−

∴=+
−+
=+ Ω

(b)
300( 312.5)
1600: Z
300 312.5
600( 960)
587.6 119.79j
600 960
in
j
j
j
j
−
=
−
+= Ω
+
ω=
+
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Engineering Circuit Analysis, 7
th
Edition Chapter Ten Solutions 10 March 2006

42.

(a)
At 100 rad/s, 2 mF - 5 ; 0.1 H 10 .
50 50 10 10 2 1
(10 10) ( 5)
10 5 2 1 2 1
26 Z 2026226
in
j j
jjj
jj
jjj
jjj
ω= → Ω → Ω
−−−
+−= =
++−
=− Ω∴ = +− = − Ω

(b) SC, : 20 10 6.667, (6.667 5) 10
50 66.67
150 200 30 40 4 3
6.667 5 20 15 4 3 4 3
(1.2 1.6)(4 3)
in in
ab j j

j jjj

j

9.6 2.8
jjj
ZZ jj
=−
+++−
===×
++ +−
=∴ + − j=+ Ω
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Engineering Circuit Analysis, 7
th
Edition Chapter Ten Solutions 10 March 2006

43.

800: 2 F 625, 0.6H 480
300( 625) 600( 480)
Z
300 625 600 480
478.0 175.65
in
jj
jj
jj
j
ω= μ →− →
−
∴=+
−+
=+ Ω
120 625
I
478.0 175.65 300 625
or I 0.2124 45.82 A
−
∴= ×
+−
=∠−°
j
jj

Thus, i (t) = 212.4 cos (800t – 45.82
o
) mA.

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Engineering Circuit Analysis, 7
th
Edition Chapter Ten Solutions 10 March 2006

44.
(a)
3 2mH : V (3 20 ) (3 4) 15 33.13 VjΩ+ = ∠− ° + = ∠ °

(b)
3 125 F: V (3 20 ) (3 4) 15 73.3 VjΩ+ μ = ∠− ° − = ∠− °

(c)
3 2mH 125 F: V (3 20 ) 3 9 20 VΩ μ =∠−°=∠−°

(d)
same: 4 000 V (3 20 ) (3 8 2)
V (
3 20 ) (3 6) 20.12 43.43 V
ω= ∴ = ∠− ° + −∴=∠− ° + = ∠ °
jj
j

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Engineering Circuit Analysis, 7
th
Edition Chapter Ten Solutions 10 March 2006

45.

(a)
C=μ

6
20 F, 100
11
11 0.005 0.01 0.002
1000 20 10
200 1000
1
196.12 11.310
0.005 0.001
in
in
jj
j
j
j−
ω=
==
−+
++××
(b) (c)

∴==∠ −° Ω
+
Z
Z
()
()
2
2
2
-6 2
-3 -6
1
100 rad/s
0.005 0.001 100
1
125
0.005 100 0.001
64 10 0.005 100 0.001
6.245 10 39 10 100 0.001
72.45 F
in
in
j jC
C
or C
so C
or C
ω= ∴ =
−+
==
+−
×= +
−
×=×= −
=μ
Z
Z
5
22
25 5 5
55 525
55
11
C=μ20 F 100
0.00050.1
/ 2 10 0.01
0.1 0.1
0.005 2 10 0.0001, 2 10 7.5 10
0.01
2 10 866.0 10 0 2 10 866.0 10 0.1 0
866.0 10 7.5 10 8 10
use sign:
in
jj
−
−− −
−− −−
−−−
∴ = = ∠=
−ω+×ω ∠
⎛⎞⎛⎞
∴ +× ω− = × − = ×
⎜⎟⎜⎟
ωω⎝⎠⎝⎠
×− ×=∴×ω ×ω−=
ω
×± ×+×
−ω=
Z
∓∓∴
6
5
556
5
444.3 and 0
410
866.0 10 7.5 10 8 10
use + sign: 11.254 and <0
410
=11.254 and 444.3rad/s
−
−−−
−
=<
×
−×±×+×
ω= =
×
∴ω
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Engineering Circuit Analysis, 7
th
Edition Chapter Ten Solutions 10 March 2006

46.

(a)
2
1111
25 0.0016
11 0.04 900
30
X 45.23 0.002 , 2261rad/s
x
jx
== ∴ +=
+
∴=Ω=ωω=
(b)
111 3
Y25 of tan
30
64.34 0.02 , 3217rad/s
−0
−⎛⎞
∠=−°=∠ − =
⎜⎟
⎝⎠
==ωω=
in j
x x
x

∴

2
2
22
22
30( 0.02 ) 30 0.092 0.012 18(c)
Z=×
30 0.02 30 0.02 900 0.0004
0.01225
(900 0.0004 )
0.012 0.01 22,500,
ω− ω ω+ω
=
+ω−ω + ω

∴ ω= + ω

3354rad/s
(d)
∴ ω= ω+ ω
in
jj j
jj
=
22
26
66
1810(900 0.0004 ),0.004 18 9000 0,
4500 2.25
10 0
4500 20.25 10 9 10 4500 3354
22
ω= + ω ω − ω+ =
ω− ω+ × =
±×−× ±
ω= = =572.9, 3927rad/s
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Engineering Circuit Analysis, 7
th
Edition Chapter Ten Solutions 10 March 2006

47. With an operating frequency of ω = 400 rad/s, the impedance of the 10-mH inductor is
j/ j
ωL = j4 Ω, and the impedance of the 1-mF capacitor is –ωC = -j2.5 Ω.
V 2 40 ( 2.5) 5 50 A
I 3 2 40 1.9513 41.211 A
∴=∠ °− =∠− °
∴=−∠ °= ∠− °
c
L j

2
1
2
1
2
2
22
1
240(R 2.5)
I
R4
240(R 2.5)
R4
1.9513 41.21
1.0250 81.21 ( 2.5)
R (1.0250 81.21 ) 2.562 8.789
0.15662R 1.0130R 2.532 0.3915
R 2.532 0.
L
j
j
j
j
Rj
jj
∠° −
=
+
∠° −
∴+=
∠− °
=∠°−
=∠°+∠−°
=++−
∴= +
22
21
15662R , 4 1.0130R 0.395
R 4.335 , R 3.211
−
+
=−
∴= Ω = Ω

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Engineering Circuit Analysis, 7
th
Edition Chapter Ten Solutions 10 March 2006

48. ω = 1200 rad/s.

(a)
2
2
1200
(200 80) (80 200 )[200 ( 80)]
200 (80 ) 40,000 6400 160
X 0 40,000 80 6400 0
1
46,400 80 , 580 1.437 F
1200
in
in
jjxjxjx
jx xx
xx x
xx C
C
ω=
−× + − + −
==
+− +
−+
=∴− + − =
∴==Ω=∴=μ
Z

(b)
22
2
222
2
80X 200X
100
200 (80 X)
6400X 40,000X
10,000
40,000 6400 160X X
0.64X 4X X 160X 46,400
3.64X 160 46,400 0,
160 25,600 675,600 160 837.4
X
7.28 7.28
1
X 93.05 ( 0)

C 8.956 F∴= μ
1200C
in in
j
j
X
−
−
==
+−
+
∴=
+− +
∴+=−+
∴+−=
−± + −±
==
∴= > =
ZZ
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Engineering Circuit Analysis, 7
th
Edition Chapter Ten Solutions 10 March 2006

49. At ω = 4 rad/s, the 1/8-F capacitor has an impedance of –j/ ωC = -j2 Ω, and the 4-H
inductor has an impedance of j
ωL = j16 Ω.

(a) Term inals ab open circuited: Z
in = 8 + j16 || (2 – j2) = 10.56 – j1.92 Ω

(b) Term inals ab short-
circuited: Z in = 8 + j16 || 2 = 9.969 + j0.2462 Ω

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Engineering Circuit Analysis, 7
th
Edition Chapter Ten Solutions 10 March 2006

50. f = 1 MHz, ω = 2πf = 6.283 Mrad/s
2 μF
→ -j0.07958 Ω = Z 1
3.2 μH → j20.11 Ω = Z 2
1 μF → -j0.1592 Ω = Z 3
1 μH → j6.283 Ω = Z 4
20 μH → j125.7 Ω = Z 5
200 pF → -j795.8 Ω = Z 6

The three impedances at the upper right, Z 3, 700 kΩ, and Z 3 reduce to –j0.01592 Ω

Then we form Z 2 in series with Z eq: Z2 + Zeq = j20.09 Ω .

Next we see 10
6
|| (Z 2 + Zeq) = j20.09 Ω.

Finally, Z in = Z 1 + Z4 + j20.09 = j 26.29 Ω.

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Engineering Circuit Analysis, 7
th
Edition Chapter Ten Solutions 10 March 2006

51. As in any true design problem, there is more than one possible solution. Model answers
f ollow:

(a) Using at least 1 inductor,
ω = 1 rad/s. Z = 1 + j4 Ω.
Construct this using a single 1 Ω resistor in series with a 4 H inductor.

(b) Force jL = j/C, so that C = 1/L. Then we construct the network using

a single 5 Ω resistor, a 2 H inductor, and a 0.5 F capacitor, all in series (any values for
these last two will suf
fice, provided they satisfy the C = 1/L requirement).

(c) Z = 7∠80
o
Ω. R = Re{Z} = 7cos80
o
= 1.216 Ω, and X = Im{Z} = 7sin80
o
= 6.894 Ω .
We can obtain this impedance at 100 rad/s by placing a resistor of value 1.216 Ω in
series with an inductor having a value of L = 6.894/
ω = 68.94 mH.

(d) A single resistor having value R = 5 Ω is the simp
lest solution.

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Engineering Circuit Analysis, 7
th
Edition Chapter Ten Solutions 10 March 2006

52. As in any true design problem, there is more than one possible solution. Model answers
f ollow:

(a) 1 + j4 kΩ at
ω = 230 rad/s may be constructed using a 1 kΩ resistor in series with an
inductor L and a capacitor C such that j230L – j /(230C) = 4000. Selecting arbitrarily C =
1 F yields a required inductance value of L = 17.39 H.

Thus, one design is a 1 kΩ resi
stor in series with 17.39 H in series with 1 F.

(b) To obtain a purely real impedance, the reactance of the inductor m
ust cancel the
reactance of the capacitor, In a series string, this is obtained by meeting the criterion
ωL
= 1/
ωC, or L = 1/ω
2
C = 1/100C.

Select a 5 MΩ resistor in series with 1 F in series with 100 mH.

(c) If Z = 80∠
–22
o
Ω is constructed using a series combination of a single resistor R and
single capacitor C, R = Re{Z} = 80cos(–22
o
) = 74.17 Ω. X = –1/ ωC = Im{Z} =
80sin(–22
o
) = –29.97 Ω. Thus, C = 667.3 μF.

(d) The simplest solution, independent of frequency, is a single 300 Ω resistor
.

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Engineering Circuit Analysis, 7
th
Edition Chapter Ten Solutions 10 March 2006

53. Note that we may replace the three capacitors in parallel with a single capacitor having
value .
333
10 2 10 4 10 7 mF
−−−
+× +× =

(a)
ω = 4π rad/s. Y = j4πC = j87.96 mS

(b)
ω = 400π rad/s. Y = j400πC = j8.796 S

(c)
ω = 4π×10
3
rad/s. Y = j4π×10
3
C = j879.6 S

(d)
ω = 4π×10
11
rad/s. Y = j4π×10
11
C = j8.796×10
9
S

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Engineering Circuit Analysis, 7
th
Edition Chapter Ten Solutions 10 March 2006

54. (a) Susceptance is 0

(b) B =
ωC = 100 S

(c) Z = 1 + j100
Ω, so Y =
2
1 1 100
=G + B
1 100 1+100
j
j
j
−
=
+
, where B = –9.999 mS.

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Engineering Circuit Analysis, 7
th
Edition Chapter Ten Solutions 10 March 2006

55.
2H 2,1F 1Let 1 0 A
2V 0.5 1 1
2(11)(1)11
10 1 1 1
0.5 0.5
1111
1
Now 0.5 S 2 , 0.5 S 2 H
2
Lci nL
in
in
in
jj
j j
jjjj
j
j
jj
j
j
∈
→→− =∠°
∴=∴=+ =+
∴=++ −=+
∠° −
∴== =−
+−
→Ω− = →
I
VIIV
V
V
V

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Engineering Circuit Analysis, 7
th
Edition Chapter Ten Solutions 10 March 2006

56.
(a) 500, Z 5 10 1 5 9
159 9
Y Y 500C
5 9 106 106
9
C 169.8 F
53,000
inRLC
inRLC c
jj j
j
j
ω= = + − = +
−

∴ ==∴==
+

(b)
,
106
R2
5
==
in ab 1.2Ω

(c)

∴==μ
o
C
o
,
C
1000 rad/s
5 2 5 5 3 5.831 30.96
and = 58.89 .
Thus,
11
0.1808 35.58 S
147.1 105.2 mS
S
in ab
S
jj j
j
j
ω= ∴
=+−= −= ∠− Ω
−Ω
=+= ∠
=+
Z
Z
Y
ZZ
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Engineering Circuit Analysis, 7
th
Edition Chapter Ten Solutions 10 March 2006

57.
62
462
62
6
462
42 6 42
42
0.1
(a) R=Ω550 : Z 500
100 0.001
50
,000 0.6 100 0.001
Z
100 0.001 100 0.001
5 10 0.0006 (60 50 )
Z
10 10
5 10 0.006
R 550 5.5 10
10 10
5.5 10 5 10 10
0.5 10
in in
in
in
in
j
j
jj
jj
j
−
−
−−
−
ω
=+
+ω
+ω − ω

∴=×
+ω−ω
×+ ω+ ω−ω

∴=
+ω
×+ ω

∴== ∴×
+ω
+× ω=×× ω
×ω=
62 10
0.5 10 , 10 ,×ω=
5
10 rad/sω=∴

64 2
462
251 0
51 01 0
5510
(b) X=Ω50 0.5 10 0.5 10 10
10 10
0, 2 10 10 0
210 410 410
10
2
−
−ω
= = × + × ω−ω
+ω
=ω−× ω+ =
×±× −×

10 rad/sω= = ∴ω=
in
∴

(c)
3
642
82
642
3
82
642 6 62
66 2 100 0.001 50,000 0.6
G1.810:Y
50,
000 0.6 50,000 0.6
510 610 (50 6)
25 10 0.36
510 610
1.8 10
25 10 0.36
5 10 6 10 4.5 10 648 10
0.5 10 48 10
in in
jj
jj
j
−
−
−
−−
− +ω −ω
=× = ×
+ω−ω
×+× ω+ ω−ω
=
×+ ω
×+× ω

∴×=
×+ ω
∴×+× ω=×+× ω

102.06×=×ω∴ ω= kra
d/s
∴

4
82
46 2
62 4
6 10
B1.510
25 10 0.36
10 37.5 10 54 10
54 10 10 37.5 10 0,
100 81
10 52.23 and 133.95krad/s
108 10
−
−
−
− −ω
=× =
×+ ω
∴ω= × + × ω
∴×ω−ω+ ×=
−
ω= ± =
×
in
(d)
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Engineering Circuit Analysis, 7
th
Edition Chapter Ten Solutions 10 March 2006

58.

(a)
1
11 3
1
I0.130
V 20 23.13 V 20V
Y(34)10
−
∠°
== =∠− °∴=
+j

(b)
21 2VVV 20V=∴ =

(c)

(d)

3
222
312
3
33 3
3
Y V (5 2)10 20 23.13 0.10770 1.3286 A
I I I 0.1 30 0.10770 1.3286 0.2 13.740 A
I0.2 13.740
V 44.72 77.18 V V
Y(24)10
j
j
−
−
==+ ×∠−°= ∠−°I
44.72V
∴=+= ∠°+ ∠− °= ∠ °
∠°
∴== = ∠ °∴
−
=
13
V=+V V 20 23.13 44.72 77.18 45.60 51.62
V
+∠− °+ ∠ °= ∠ °
45.60
V
∴ =
in
in
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Engineering Circuit Analysis, 7
th
Edition Chapter Ten Solutions 10 March 2006

59.

(a) 50 F

1
1
11
20 Y 0.1 0.05
1 1000 1
YR 8 4
1000 C 0.1 0.05
R
C
1
R 8 and C 250 F
4
in
in
jj
j j
j
j
μ
μ→− Ω∴ = +
=∴−==−
+
−

(b)
ω
∴=Ω = =
1
1
11 1
100 F
1
1
2000: 50 F 10 Y 0.1 0.1
500
R
C
500
R 5 5 R 5 , C
C
injj
j
jj
=→ −Ω ∴=+=
−ωμ
μ
−=−∴=Ω =
∴
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Engineering Circuit Analysis, 7
th
Edition Chapter Ten Solutions 10 March 2006

60.

2
2
2
22
10 10
(a)
ω G
in BBin

0 1 2 5 10 20
∞

0
0.0099
0.0385
0.2
0.5
0.8
1

0
0.0099
0.1923
0.4
0.5
0.4
0

Z1
10
Y
10 10
in
in
j
jj
jj
jj
10
Y
100
10
G, B
100
100
in
in in
j
ω
ωω
ωω
ωω
ωω
ω
ω ω
ωω
+
=+ =
−
∴ ×
+−
+
∴=
+
==
++
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Engineering Circuit Analysis, 7
th
Edition Chapter Ten Solutions 10 March 2006

61. As in any true design problem , there is m ore than one possible solution. Model answers
f ollow:

(a) Y = 1 – j4 S at
ω = 1 rad/s.
Construct this using a 1 S conductance in parallel with an inductance L such that 1/
ωL
= 4, or L = 250 mH.

(b) Y = 200 mS
(purely real at ω = 1 rad/s). This can be constructed using a 200 mS
conductance (R = 5 Ω), in pa rallel with an ind uctor L and capacitor C such th at
ωC –
1/
ωL = 0. Arbitrarily selecting L = 1 H, we find that C = 1 F.

One solution therefore is a 5 Ω resist
or in parallel with a 1 F capacitor in parallel with a 1
H inductor.

(c) Y = 7∠80
o
μS = G + jB at ω = 100 rad/s. G = Re{Y} = 7cos80
o
= 1.216 S (an 822.7
m Ω resistor). B = Im {Y} = 7sin8 0
o
= 6.894 S. W e m ay realize th is suscep tance by
placing a capacito r C in parallel with the resis tor such th at j
ωC = j6.894, or C = 68.94
mF.

One solution therefore is an 822.7 mΩ resistor in parallel with a 68.94 mF.

(d) The simplest solution is a single conductance G = 200 mS (a 5 Ω resistor).

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and educators for course preparation. If you are a student using this Manual, you are using it without permission.

Engineering Circuit Analysis, 7
th
Edition Chapter Ten Solutions 10 March 2006

62. As in any true design problem , there is m ore than one possible solution. Model answers
f ollow:

(a) Y = 1 – j4 pS at
ω = 30 rad/s.

Construct this using a 1 pS conductance (a 1 TΩ resistor) in parallel with an inductor L
such that –j4×
10
–12
= –j/ωL, or L = 8.333 GH.

(b) We may realise a pu rely real admittance of 5 μS by plac
ing a 5 μS conductance (a
200 kΩ resistor) in parallel with a capacitor C and inductance L such that
ωC – 1/ωL =
0. Arbitrarily selecting a value of L = 2 H, we find a value of C = 1.594 μF.

One possible solution, then, is a 200 kΩ resistor in parallel with a 2 H inductor and a
1.594 μF capacito
r.

(c) Y = 4∠–10
o
nS = G + jB at ω = 50 rad/s. G = Re{Y} = 4×10
–9
cos(–10
o
) = 3.939 nS
(an 253.9 MΩ resistor). B = Im{Y} = 4×10
–9
sin(–10
o
) = –6.946×10
–10
S. We may realize
this susceptance by placing an inductor L in parallel with the resistor such that –j/
ωL =
– j6.946×10
–10
, or L = 28.78 μH.

One possible solution, then, is a 253.9 MΩ resistor in parallel with a 28.78 μH inductor.
(d) The simplest possible solution is a 60 nS resistor (a 16.67 MΩ resi
stor).
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Engineering Circuit Analysis, 7
th
Edition Chapter Ten Solutions 10 March 2006

63.

11212
1121 2
12
21 21 2
21212 1 2
2VV V
5 , 75 5V 3V 3V 5V 5V
353
(5 2)V 2V 75
VVVV
10
356
10V 10V 6V 6V 5V 300 4V (5 4)V 300
52 75
4 300 1500 600 300 1200
V
52 2 17 30 8
454
vv
jj j j j
jj
jj j
v
jj
jjjj
j
jj
j j
jj j
jj
−−
−=
+ + −=+−−+
−
∴− + =−
−−
++=
−
−++−+
−−
∴= =
−
−
j j=∴+− =
−−
−
=
−+
600
34.36 23.63 V
25 30
j
j
= ∠°
−
(1)
(2)

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and educators for course preparation. If you are a student using this Manual, you are using it without permission.

Engineering Circuit Analysis, 7
th
Edition Chapter Ten Solutions 10 March 2006

64.

3I 5(I I ) 0 2I 5I 0
3(I 5) 5(I I ) 6(I 10) 0
5I (9 5)I 60 15
05
60 15 9 5 75 300
I
25 15 18
59 5
13.198 154.23 A
BBD BD
DD BD
BD
Bjj j
jj
jj j
j
jj j
jj j
jj
−−=∴−+=
+− −+ +=
∴ +− =−−
−− − −+
==
− −
−
=∠°
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and educators for course preparation. If you are a student using this Manual, you are using it without permission.

Engineering Circuit Analysis, 7
th
Edition Chapter Ten Solutions 10 March 2006

65.
12
12
20cos1000 V, 20sin1000 V
V200V,V 20V
0.01H 10 , 0.1mF 10
20 20
0, 0.04 2 2 0,
10 25 10
V 25(2 2) 70.71 45 V
( ) 70.71cos(1000 45 )V
ss
ss
xxx
x
x
x
vtvt
j
jj
vvvj
vj
jj
j
vt t
==
∴=∠° =−
→Ω →−Ω
−+
∴++ = +−=
−
=−=∠−°
∴= −°

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and educators for course preparation. If you are a student using this Manual, you are using it without permission.

Engineering Circuit Analysis, 7
th
Edition Chapter Ten Solutions 10 March 2006

66.
(a)
A
32 2
1
1
3ssume V 1V V 1 0.5V, I 1 0.5mA
V 1 0.5 (2 0.5)( 0.5) 0.75 1.5V
I 0.75 1.5mA, I 0.75 1.5 2 0.5 2.75 2mA
V 0.75 1.5 1.5 (2.75 2)( 0.5)
100
0.25 2.875V V
0.25 2.875
in
in
jj
jjj j
jj jj
jj jj
j
jj
= ∴ =− =−

∴=− + − − = −

∴=− ∴=−+−=−

∴

34.65 94.
+
=−−+ −−
=− − ∴ =
−−
= ∠ 97V°

(b)
33
22 1 2
22
11
223 233
2
0.5 Assume 1 V 1A,
1X, 1X, 2X
1 X (2 X)( X) 1 X 3X, I 1 X 3X, 3 X 4
1 X 3X 4X X 3X 1 5X (X 6X) X 6X 0
X6,X 6, 2.449k
in
in
c
jjx
jj j

2
j jj j j j
jjj j
j
X
− =∴=
=− =− → = −
−→
∴=− + − − =− − =− − =− −
∴=− − − + − =− + − ∴ − =
∴== Ω
VI
VI I
VI
V
Z=−
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Engineering Circuit Analysis, 7
th
Edition Chapter Ten Solutions 10 March 2006

67. Define three clockwise mesh currents i 1, i2, i3 with i 1 in the left mesh, i 2 in the top right
m esh, and i
3 in the bottom right mesh.

Mesh 1: -10∠0
o
+ (1 + 1 – j 0.25) I1 – I2 – (-j0.25)I3 = 0

Mesh 2: –
I1 + (1 + 1 + j4) I2 – I3 = 0

Mesh 3: (-j0.25 + 1 + 1)
I3 – I2 – (-j0.25I1) = 0

20.25 110
1240
0.25 1 0
2 0.25 1 0.25
124 1
0.25 1 2 0.25
10(1 1 0.5)
0.25(2 0.5) ( 2 0.25 0.25) (2 0.25)(4 1 0.5 8 1)
20 5
1.217 75.96 A, ( ) 1.2127cos(100 75.96 )A
815
−−
−+
−
=
−−
−+−
−−
+−
∴=
−+−+ + +− +−+−
−
=∴=∠−°= −°
+
x
x
xx
j
j
j
jj
j
jj
j
jj jj j jj
j
it t
j
I
I
I

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and educators for course preparation. If you are a student using this Manual, you are using it without permission.

Engineering Circuit Analysis, 7
th
Edition Chapter Ten Solutions 10 March 2006

68.

11 1 2
12
212 2
12
12
12V 10 0.25V 0.25V V V 0
(2 0.25)V V 0.25V 10
VVVV 4V0
V(2 4)VV 0
0.25V 0.25V V V V
0.25V V (2 0.25)V 0
20.25 110
1240
0.25 1 0
V
0.25 1 0.25
124 1
0.25 1 2
x
x
x
x
xx x
x
xjj
jj
j
j
jj
jj
j
j
j
jj
j
jj
−− + +− =
∴−−+=
−+−+ =
−++ − =
− + ++−
∴ −+− =
−−
−+
−
=
−−
−+ −
−− 0.25
10(1 1 0.5)
0.25(2 0.5) ( 2 0.25 0.25) (2 0.25)(4 1 0.5 8 1)
20 5
1.2127 75.96 V
815
1.2127cos(100 75.96 )V
x
j
jj jj j jj
j
j
vt
+−
=
−+−+ + +− +−+−
−
==∠−°
+
∴=−°

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and educators for course preparation. If you are a student using this Manual, you are using it without permission.

Engineering Circuit Analysis, 7
th
Edition Chapter Ten Solutions 10 March 2006

69.
(a)
1
1
1
11
11
1
1
1R, R0,AV/V0
VAV
IC (VV)
R
V(
1 A CR ) CR V
V
VAV (1A CR) CRV
VV
As A , C R
V
oo i
i
si
f
iff s
o
oi f fs
oo
(b)

A
CRA
V1
A CR
f
f
sf
j
jω
ω
s
j
jj
jj
jω
ωω
ωω
ω
=∞ = =− >>
+
==−
∴ ++ =
=− ∴− + + =
∴=−
++
→∞ →−
1
1
11
11
11
1
R1
RC
11CR
C
R
(V AV )
I( 1CR)( VV)C,VA V
R
V(1 A)(1 C R ) V CR CR V,
V [(1 A) (1 C R ) C R ] C R V
V
[(1 A)(1 C R ) C R ] C R V
A
CR AV
V(1A)(1
f
ff
ff
f
f
i
ff s i o i
f
if f s ff i
if fff s
o
ff f fs
fo
s
j
j
jj
jjj
jjj
jjj
j
jω
ω
ωω
ωωω
ωωω
ωωω
ω
==
+
+
+
=+=− = −
∴ ++ = −
++ + =
∴−++ + =
−
∴=
++
1
1CRV
As A ,
CR) CR V 1 CR
fo
fff s
j
jjω
ωω ω−
→∞ →
++
ff
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Engineering Circuit Analysis, 7
th
Edition Chapter Ten Solutions 10 March 2006

70. Define the nodal voltage v 1(t) at the junction between the two dependent sources.
The voltage source may be replaced by a 3∠-3
o
V source, the 600-μF capacitor by a
– j/ 0.6 Ω impedance, the 500-μF capacito
r by a –j2 Ω impedance, and the inductor by a
j2 Ω
impedance.

5 V2 + 3V2 =
2-
) - (

6.0/100
3-3 -
21
o
1
jj
VVV
+
−
∠
[1]

-5 V2 =
()
21 2

22jj
−
+
−
VV V [2]

Solving, we find that V2 = 9.81 ∠ – 13.36
o
mV.
Converting back to the time domain,

v2(t) = 9.81 cos (10
3
t – 13.36
o
) mV
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Engineering Circuit Analysis, 7
th
Edition Chapter Ten Solutions 10 March 2006

71. Define three clockwise mesh currents: i 1(t) in the left-most mesh, i 2(t) in the bottom right
m esh, and i
3(t) in the top right mesh. The 15-μF capacitor is replaced with a –j/ 0.15 Ω
impedance, the inductor is replaced by a j20 Ω
impedance, the 74 μF capacitor is
replaced by a –j1.351 Ω
impedance, the current source is replaced by a 2∠0
o
mA source,
and the voltage source is replaced with a 5∠0
o
V source.

Around the 1, 2 supermesh: (1 + j20) I1 + (13 – j1.351) I2 – 5 I3 = 0
and
–
I1 + I2 = 2×10
–3

Mesh 3: 5 ∠0
o
– 5 I2 + (5 – j 6.667) I3 = 0

Solving, we find that I1 = 148.0∠179.6
o
mA. Converting to the time domain,

i1(t) = 148.0cos (10
4
t + 179.6
o
) μA

Thus, P 1Ω = [i 1(1 ms)]
2

•
1
= (16.15×10
–3
)(1) W = 16.15 mW.

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Engineering Circuit Analysis, 7
th
Edition Chapter Ten Solutions 10 March 2006

72. We define an additional clockwise mesh current i 4(t) flowing in the upper right-hand
mesh. The inductor is replaced by a j0.004 Ω
impedance, the 750 μF capacitor is
replaced by a –j/ 0.0015 Ω im
pedance, and the 1000 μF capacitor is replaced by a –j/ 2
Ω im
pedance. We replace the left voltage source with a a 6 ∠ -13
o
V source, and the
right voltage source with a 6 ∠ 0
o
V source.

(1 – j/ 0.0015) I1 + j/0.0015I2 – I3 = 6 ∠ –13
o
[1]

(0.005 + j / 0.0015) I1 + (j0.004 – j/0.0015) I2 – j0.004 I4 = 0 [2]

–I1 + (1 – j500) I3 + j 500 I4 = –6 ∠ 0
o
[3]

–j0.004 I2 + j500I3 + (j0.004 – j500) I4 = 0 [4]

Solving, we find that

I1 = 2.002∠ –6.613
o
mA, I2 = 2.038 ∠ –6.500
o
mA, and I3 = 5.998 ∠ 179.8
o
A.

Converting to the time domain,

i 1(t) = 1.44 cos (2t – 6.613
o
) mA
i
2(t) = 2.038 cos (2t – 6.500
o
) mA
i
3(t) = 5.998 cos (2t + 179.8
o
) A

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Engineering Circuit Analysis, 7
th
Edition Chapter Ten Solutions 10 March 2006

73. We replace the voltage source with a 2115∠0
o
V source, the capacitor with a
– j/ 2
πC1 Ω impedance, and the inductor with a j0.03142 Ω impedance.

Defi ne Z such that Z
-1
= 2πC1 - j/0.03142 + 1/20

By voltage division, we can write that 6.014 ∠85.76
o
=
2115
20 +Z
Z

Thus, Z = 0.7411 ∠ 87.88
o
Ω. This allows us to solve for C1:

2πC1 – 1/0.03142 = -1.348 so that C1 = 4.85 F.

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Engineering Circuit Analysis, 7
th
Edition Chapter Ten Solutions 10 March 2006

74. Defining a clockwise mesh current i 1(t), we replace the voltage source with a
2115∠0
o
V source, the inductor with a j2 πL Ω impedance, and the capacitor with a
– j1.592 Ω
impedance.

Ohm’s law then yields
()
o
1
08.132
592.1220
2115
∠=
−+
=
Lj
π
I

Thus, 20 = ()
22
592.12 20−+Lπ and we find that L = 253.4 mH.
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Engineering Circuit Analysis, 7
th
Edition Chapter Ten Solutions 10 March 2006

75. (a) By nodal analysis:

0 = (Vπ – 1)/ Rs + Vπ / RB + Vπ / rπ + jωCπ Vπ + (Vπ – Vout) jωCμ [1]

-g mVπ = (Vout – Vπ) jωCμ + Vout / RC + Vout / RL [2]

Simplify and collect terms:

()
S
out
BS
R
1
C - CC
r
1
R
1
R
1
=
⎥
⎥
⎦
⎤
⎢
⎢
⎣
⎡
++
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
++ VV
μπμπ
πωωjj [1]

(-gm + jωCμ) Vπ - (jωCμ + 1/RC + 1/RL) Vout = 0 [2]

Defi ne
π
r
1
R
1
R
1

R
1
BS
S
++=
′ and ′
LR = RC || RL
Then ()
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
′
+
′
+
+++
′′
=Δ
SL
22
LSR
C
R
CC
C - CC 2C
R R
1-

μπμ
μπμμ
ωω
m
gj

And Vout =
()
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
′
+
′
+
+++
′′
−
SL
22
LS
SSm
R
C
R
CC
C - CC 2C
R R
1-
RCRg
μπμ
μπμμ
μ
ωω
ω
mgj
j

Therefore, ang(
Vout) =
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛−
−
2
1
tan
SmRg
Cj
μω
-
()
⎟
⎟
⎟
⎟
⎟
⎟
⎠
⎞
⎜
⎜
⎜
⎜
⎜
⎜
⎝
⎛
++
′′
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
′
+
′
+
+−
−
πμμ
μπμ
μω
ω CC 2C
R R
1-
R
C
R
CC
C
tan
22
LS
SL1
m
g

(b)

(c) The output is ~180
o
out of phase with the input for f < 10
5
Hz; only for f = 0 is it
exactly 180
o
out of phase with the input.
PROPRIETARY MATERIAL . © 2007 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers
and educators for course preparation. If you are a student using this Manual, you are using it without permission.

Engineering Circuit Analysis, 7
th
Edition Chapter Ten Solutions 10 March 2006

76.
,
V100V
OC: 0.02V 0
20 10
10
10 (0.05 0.1 0.02) V ,V
0.07 0.1
V 67.11 46.98
V 100 V 32.89 46.98 57.35 55.01 V
100
SC:V 100 I 0.02 100 7A
20
57.35 55.01
Z 4.698 6.711
7
xx
x
xx
x
ab oc x
xS C
th
j
j
jj
j
j
j
j
−
−+ − =
−
=++ =
+
∴= +
∴=−= − =∠−°
=∴↓= ×+ =
∠− °
∴= = − Ω

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Engineering Circuit Analysis, 7
th
Edition Chapter Ten Solutions 10 March 2006

77.

Let 1 0. Then 2 2 0.5
1
(1 ) 2
1
12
1
12
1
At 1, 1 1 2 1
1
0.5 0.5
11
in L in L
in
in
in
in
in
j jj
jj
j
j
j
j
j
jj j
j
j
ω ωω
ωω
ω
ω
ω
ω
ω
ω=∠ = = ∴ =
∴=+ +
=+ +
∴= =++
==−+=+
∴= =−
+IVIV
V
V
Z
Z
Y

so
Yin =
()
12
2
−+ωω
ω
j

R = 1/0.5 = 2 Ω and L = 1/0.5 = 2 H.

PROPRIETARY MATERIAL . © 2007 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers
and educators for course preparation. If you are a student using this Manual, you are using it without permission.

Engineering Circuit Analysis, 7
th
Edition Chapter Ten Solutions 10 March 2006

78.

(a)

(b) I
s:

1
21 2
21 0.8 0.4 V
1212
0.8 0.4 10 20
25 11.785 135 V
1 1 0.8 0.4 1.8 0.6
jj
jj
jj
jj
j
jj j
+
−
==+∴
+−
+−+
=== ∠
−+ + −
°

so v 1(t) = 11.79 cos (1000t + 135
o
) V.
1
11
(1 1)1 2 1 3 1 15
V 0.6 0.2
2121 5 20.60.2
V 5 90 ( ) 5cos(1000 90 )V
s
jjj
j
jj j j
vt t
−+− −
×=∴= ×−
−+ +−
V:
∴=∠ °∴ = + °
PROPRIETARY MATERIAL . © 2007 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers
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Engineering Circuit Analysis, 7
th
Edition Chapter Ten Solutions 10 March 2006

79.
,OC:V 0 V 1 0 V
SC: I V 2I 1 0 1[0.25( 2I ) I ] 2I
1(0.5 2)I (0.5 1)I
I1
I 0.4 0.8 0.4 0.8
0.5 1 1 0
111 1
R 2.5 , 0.8, L 1.25H
0.4 L L 0.8
I 0.4 0.8 0.8944 63.43 A
La boc
NL N NN N
NN
N
NN
NN
NN
N
jjjj
jj j
jY j
j
j
jj
j
ω
=∴ =∠°
↓∴ = ∴∠°=− + +
∴= − + = +
∴= =− ∴= =−
+∠ °
∴= =Ω = =− = =
=− = ∠− °

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Engineering Circuit Analysis, 7
th
Edition Chapter Ten Solutions 10 March 2006

80. To solve this problem , we e mploy superposition in order to separate sources having
different frequencies. First considering the sources operating at w = 200 rad/s, we open-
circuit the 100 rad/s current source. This leads to
L
′V = (j)(2∠0) = j2 V. Therefore, ()
L
vt′
= 2cos(200t + 90
o
) V. For the 100 rad/s source, we find
()
()10, 0.5cos(100 90)V
2
2cos(200 90 ) 0.5cos(100 90 )V
LL
L
j
vt
vt t t
′′ ′′=∠ = +°
∴= +°+ +°
V

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Engineering Circuit Analysis, 7
th
Edition Chapter Ten Solutions 10 March 2006

81.
100
Use superposition. Left: V 100
100 300
300
50 0 V Right: V 100 150V
300 100
V 50 150 158.11 108.43 V
30,000
Z 100 300 150
200
ab
ab
th
th
j
jj
j
jj
jj
j
jj j
j
=
−
−
=− ∠ ° = =
−+
∴=− + = ∠ °
=−= =Ω
−

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Engineering Circuit Analysis, 7
th
Edition Chapter Ten Solutions 10 March 2006

82. This problem is easily solved if we first perform two source transf ormations to yield a
circuit containing only voltage sources and impedances:

Then
I =
oo
5 17 0.240 -90 2.920 -45
73 10 13 4
o
jj
∠+ ∠ − ∠
++ −

= (4.264∠ 50.42
o
)/ (83.49 ∠ 6.189
o
) = 51.07 ∠ 44.23 mA

Converting back to the time domain, we find that

i(t) = 51.07 cos (10
3
t + 43.23
o
) mA

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Engineering Circuit Analysis, 7
th
Edition Chapter Ten Solutions 10 March 2006

83.

(a) There are a number of possible approaches: Thévenizing everyth
ing to the left of the
capacitor is one of them.

VTH = 6(j2)/ (5 + j2) = 2.228 ∠ 68.2
o
V

ZTH = 5 || j2 = j10/ (5 + j2) = 1.857 ∠ 68.2
o
Ω

Then, by simple voltage division, we find that
V C = (2.228 ∠ 68.2
o
)
7 3/ - 2.68857.1
3/
o
jj
j
+∠ −

= 88.21 ∠-107.1
o
mV

Converting back to the time domain, v C(t) = 88.21 cos (t – 107.1
o
) mV.

(b) PSpice verification.
Running an ac sweep at the
frequency f = 1/2
π = 0.1592 Hz,
we obtain a phasor magnitude of
88.23 mV, and a phasor angle of
–107.1
o
, in agreement with our
calculated result (the slight
disagreement is a combination
of round-off error in the hand
calculations and the rounding
due to expressing 1 rad/s in Hz.
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Engineering Circuit Analysis, 7
th
Edition Chapter Ten Solutions 10 March 2006

84. (a) Performing nodal analysis on the circuit,

Node 1: 1 = V1/ 5 + V1/ (-j10) + (V1 – V2)/ (-j5) + (V1 – V2)/ j10 [1]

Node 2: j 0.5 = V2/ 10 + (V2 – V1)/ (-j5) + (V2 – V1)/ j10 [2]

Simplifying and collecting terms,

(0.2 + j 0.2) V1 – j0.1 V2 = 1 [1]

- j V1 + (1 + j) V2 = j5 [2]

Solving, we find that V2 = VTH = 5.423 ∠ 40.60
o
V

ZTH = 10 || [(j10 || -j5) + (5 || -j10)] = 10 || (-j10 + 4 – j2) = 5.882 – j3.529 Ω.
FREQ VM($N 0002,0) VP($N 0002,0)

1.592E+01 4.474E+00 1.165E+02

FREQ VM($N_0005,0) VP($N_0005,0)

1.592E+01 4.473E+00 1.165E+02
(b)

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Engineering Circuit Analysis, 7
th
Edition Chapter Ten Solutions 10 March 2006

85. Consider the circuit below:

Using voltage division, we may write:

Vout = Vin
CjR
Cjωω
/1
/1
+
, or
RCjω+
=
1
1

in
out
V
V

The magnitude of this ratio (consider, for example, an input with unity magnitude and zero phase) is
()
2
in
out
1
1

RCω+
=
V
V

As
ω → 0, this magnitude → 1, its maximum value.

As ω → ∞, this magnitude → 0; the capacitor is acting as a short circuit to the ac signal.

Thus, low frequency signals are transferred from the input to the output relatively
unaffected by this circuit, but high frequency sig n
als are attenuated, or “filtered out.”
This is readily apparent if we plot the ma
gnitude as a function of frequency (assuming R
= 1 Ω and C = 1 F for convenience):

Cjω
1

Vin
Vout
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Engineering Circuit Analysis, 7
th
Edition Chapter Ten Solutions 10 March 2006

86. Consider the circuit below:

Vin
Vout1/jωC
R

Using voltage division, we may write:

Vout = Vin
CjR
Rω/1+
, or
RCj
RCjωω
+
=
1

in
out
V
V

The magnitude of this ratio (consider, for example, an input with unity magnitude and zero phase) is
()
2
in
out
1

RC
RCω
ω+
=
V
V

As
ω → ∞, this magnitude → 1, its maximum value.

As ω → 0, this magnitude → 0; the capacitor is acting as an open circuit to the ac signal.

Thus, high frequency signals are transferred from the input to the output relatively
unaffected by this circuit, but low frequency signals are atten
uated, or “filtered out.”
This is readily apparent if we plot the magnitude as a function of frequency (assumi
ng R
= 1
Ω and C = 1 F for convenience):

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Engineering Circuit Analysis, 7
th
Edition Chapter Ten Solutions 10 March 2006

87. (a) Removing the capacitor temporarily, we easily find the Thévenin equivalent:

Vth = (405/505) VS and Rth = 100 || (330 + 75) = 80.2 Ω

(b)
Cj
Cjωω
/12.80
1/

505
405

Sout
+
=
VV
so
ω
12
S
out
10532.21
1

505
405

−
×+
⎟
⎠
⎞
⎜
⎝
⎛
=
jV
V

and hence
224
S
out
10411.61
0.802
ω
−
×+
=V
V

(c)

S
505
405
V
80.2 Ω
31.57 fF
+
V
out
-
Both the MATLAB plot of the frequency response and the PSpice simulation show essentially the same behavior; at a frequency of approximately 20 MHz, there is a sharp roll-off in the transfer function ma
gnitude.
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Engineering Circuit Analysis, 7
th
Edition Chapter Ten Solutions 10 March 2006

88. From the derivation, we see that

μ
μωω
)CR||(R1
)CR||(R )R || (Rg-

LC
LCLCm
in
outj
j
+
+
=
V
V

so that
2
1
2
2
LC
LC2
2
2
LC
LC2
2
LC
LC2
m
in
out
C
RR
RR
1
C
RR
RR

RR
RR
g

⎥
⎥
⎥
⎥
⎥
⎦
⎤
⎢
⎢
⎢
⎢
⎢
⎣
⎡
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
+
+
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
+
+
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
+
=
μ
μ
ω
ω
V
V

This function has a maximum value of g
m (RC || RL) at ω = 0. Thus, the capacitors reduce
the gain at high frequencies; this is the frequency regime
at which they begin to act as
short circuits. Therefore, the maximum gain is obtained at f
requencies at which the
capacitors may be treated as open circuits. If we do
this, we may analyze the circuit
of Fig. 10.25
b without the capacitors, which leads to

( )
BBS
B
LC
LC
m
BS
B
LC
LC
m
frequency lowS
out
Rr)Rr(R
Rr

RR
RR
g-
R|| rR
R|| r

RR
RR
g-
ππ
π
π
π
++
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
+
=
+
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
+
=
V
V

The resistor network comprised of rπ, RS, and RB acts as a voltage divider, leading to a
reduction in the gain of the amplifier. In the situation where r
π || RB >> RS, then it has
minimal effect and the gain will equal its “maximum” value of –g
m (RC || RL).

(b) If we set R
S = 100 Ω, R L = 8 Ω, R C | max = 10 kΩ and r πgm = 300, then we find that

B
B
m
S
touR || r100
R || r
(7.994) g-
π
π+
=V
V

We seek to maximize this term within the stated constraints. This requires a large value
of g
m, but also a large value of rπ || RB. This parallel combination will be less than the
smaller of the two terms, so even if we allow R
B → ∞, we are left with
ππ
πr100
2398-

r100
rg
(7.994) -
m
S
tou
+
=
+
≈V
V

Considering this simpler expression, it is clear that if we select r
π to be small, (i.e.
r
π << 100), then gm will be large and the gain will have a maximum value of
approxim ately –23.98.

(c) Referring to our original expression in which the gain V
out/ Vin was computed, we
see that the critical frequency
ωC = [(RC || RL) Cμ]
-1
. Our selection of maximum RC,
R
B → ∞, and rπ << 100 has not affected this frequency.
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Engineering Circuit Analysis, 7
th
Edition Chapter Ten Solutions 10 March 2006

89. Considering the ω = 2×10
4
rad/ s source first, we make the following replacements:

100 cos (2×10
4
t + 3
o
) V → 100 ∠3
o
V
33 μF → -
j1.515 Ω 112 μH → j2.24 Ω 92 μF → - j0.5435 Ω

Then

( V 1´ – 100 ∠ 3
o
)/ 47×10
3
+ V1´/ (-j1.515) + (V 1´ – V 2´)/ (56×10
3
+ j4.48) = 0 [1]

( V 2´ – V 1´)/ (56×10
3
+ j4.48) + V 2´/ (-j0.5435) = 0 [2]

Solving, we find that

V1´ = 3.223 ∠ -87
o
mV and V 2´ = 31.28 ∠ -177
o
nV

Thus, v1´(t) = 3.223 cos (2×10
4
t – 87
o
) mV and v2´(t) = 31.28 cos(2×10
4
t – 177
o
) nV

Considering the effects of the ω = 2×10
5
rad/ s source next,

100 cos (2×10
5
t - 3
o
) V → 100 ∠-3
o
V
33 μF → -
j0.1515 Ω 112 μH → j22.4 Ω 92 μF → - j0.05435 Ω

Then
V
1"/ -j0.1515 + (V 1" – V 2")/ (56×10
3
+ j44.8) = 0 [3]

( V 2" – V 1")/ (56×10
3
+ j44.8) + (V 2" – 100 ∠ 3
o
)/ 47×10
3
+ V2"/ (-j0.05435) = 0 [4]

Solving, we find that

V1" = 312.8 ∠ 177
o
pV and V 2" = 115.7 ∠ -93
o
μV
Thus,

v1"(t) = 312.8 cos (2×10
5
t + 177
o
) pV and v2"(t) = 115.7 cos(2×10
5
t – 93
o
) μV

Adding, we find

v1(t) = 3.223×10
-3
cos (2×10
4
t – 87
o
) + 312.8×10
-12
cos (2×10
5
t + 177
o
) V and

v2(t) = 31.28×10
-9
cos(2×10
4
t – 177
o
) + 115.7×10
-12
cos(2×10
5
t – 93
o
) V
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Engineering Circuit Analysis, 7
th
Edition Chapter Ten Solutions 10 March 2006

90. For the source operating at ω = 4 rad/s,
7 cos 4
t → 7∠ 0
o
V, 1 H → j4 Ω, 500 mF → - j0.5 Ω, 3 H → j12 Ω, and 2 F → - j/ 8 Ω.

Then by mesh analysis, (define 4 clockwise mesh currents I 1, I2, I3, I4 in the top left, top
right, bottom left and bottom right meshes, respectiv
ely):

(9.5 + j4) I1 – j4 I2 – 7 I 3 - 4 I 4 = 0 [1]
-
j4 I1 + (3 + j3.5) I 2 – 3 I 4 = -7 [2]
-7 I
1 + (12 – j/ 8) I 3 + j/ 8 I4 = 0 [3]
-3 I
2 + j/ 8 I3 + (4 + j11.875) I 4 = 0 [4]

Solving, we find that I 3 = 365.3 ∠ -166.1
o
mA and I 4 = 330.97 ∠ 72.66
o
mA.

For the source operating at ω = 2 rad/s,
5.5 cos 2
t → 5.5∠ 0
o
V, 1 H → j2 Ω, 500 mF → - j Ω, 3 H → j6 Ω, and 2 F → - j/ 4 Ω.

Then by mesh analysis, (define 4 clockwise mesh currents I A, IB, IC, ID in the top left, top
right, bottom left and bottom right meshes, respectiv
ely):

(9.5 + j2) IA – j2 IB – 7 I C – 4 I D = 0 [1]
-
j2 IA + (3 + j) IB – 3 I D = -7 [2]
-7 I
A + (12 – j/ 4) I C + j/ 4 ID = 0 [3]
-3 I
2 + j/ 4 IC + (4 + j5.75) I D = 0 [4]

Solving, we find that I C = 783.8 ∠ -4.427
o
mA and I D = 134 ∠ -25.93
o
mA.

V 1´ = -j0.25 (I 3 – I4) = 0.1517∠131.7
o
V and V 1" = -j0.25(I C – ID) = 0.1652∠-90.17
o
V

V 2
´ = (1 + j6) I4 = 2.013∠ 155.2
o
V and V 2" = (1 + j6) ID = 0.8151∠ 54.61
o
V

Converting back to the time domain,

v 1(t) = 0.1517 cos (4t + 131.7
o
) + 0.1652 cos (2t - 90.17
o
) V
v 2(t) = 2.013 cos (4t + 155.2
o
) + 0.8151 cos (2t + 54.61
o
) V
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and educators for course preparation. If you are a student using this Manual, you are using it without permission.

Engineering Circuit Analysis, 7
th
Edition Chapter Ten Solutions 10 March 2006

91.
(a)
100 100(2 1)
I 57.26 76.76 (2.29 )
2 2.5 3
2.5
21
L
j
in
j
j
j−
===∠ −°
− +
+
−

1
I (57.26 76.76 ) 25.61 140.19 (1.02 )
21
R
j
in
j−
= ∠−° =∠− °
−

2
I (57.26 76.76 ) 51.21 50.19 (2.05 )
21
c in
j=∠−° =∠−°
−

V 2.5 57.26 90 76.76 143.15 13.24 (2.86 )
L
in=× ∠°− °= ∠ °

V 2 25.61 140.19 51.22 140.19 (1.02 )
R
in=× ∠− °= ∠− °

V 51.21 140.19 (1.02 )
c in=∠− °

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Engineering Circuit Analysis, 7
th
Edition Chapter Ten Solutions 10 March 2006

92.

(a)

(b)

(c)
1
2
3
120
330A
40 30
120
5030
120
30
40
j
j
∠−°
∠°
==∠°
−
==∠−°
+
I
I
II
==I
2.058 30.96 A
2.4 53.13 A
123
6.265 22.14 A
s
=++
=∠−°
II

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Engineering Circuit Analysis, 7
th
Edition Chapter Ten Solutions 10 March 2006

93.

12
12 1 2
12
22 2I5A,I7A
I I 10 0 , I lags V,I leads V
I lags I . Use 2.5A /
[Analytically: 5 7 10
5cos 5sin 7cos 7sin
sin 1.4sin
5 1 1.4 sin 7 1 1sin 10
By SOLVE, 40.54 27.66 ]
in
jj
αβ
α αβ
αβ
ββ
β
==
+=∠°
∠+∠=
=+ ++
αβ
∴ =−
∴ −+−=
=− ° = °

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Engineering Circuit Analysis, 7
th
Edition Chapter Ten Solutions 10 March 2006

94. V 1 = 100∠0
o
V, |V 2| = 140 V, |V 1 + V2| = 120 V.
Let 50 V = 1 inch. From the sketch, for ∠V
2 positive,
V
2 = 140∠122.5
o
. We may also have V 2 = 140∠-122.5
o
V

[Analytically: |100 + 140∠
α| = 120
so | 100 + 140 cos
α + j140 sin α | = 120
Using the “Solve” routine of a scientific calculator,
α = ±122.88
o
.]

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Engineering Circuit Analysis, 7
th
Edition Chapter Eleven Solutions 10 March 2006
1.
6
22
50( 80)10
80 , 42.40 32.01
500 25 50 80
84.80 32.01 V, 1.696 32.01 A
1.0600 57.99 A
( / 2ms) 84.80cos(45 32.01 )2cos45 116.85W
50 1.696 cos (45 32.01 ) 136.55W
84.80cos(45 32
c
R
c
s
R
c
j
j
jj
p
p
p
π
−
==−Ω =∠−°Ω
×−
∴=∠−° =∠−°
=∠°
=° −°° =
=× °− °=
=° −
Z
VI
I
.01 ) 1.060cos(45 57.99 ) 19.69W°= °+ °=−

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Engineering Circuit Analysis, 7
th
Edition Chapter Eleven Solutions 10 March 2006
2.

22 4
42 4 2 11
(a)
2
2 1 4(4 ) 16 , 4(4 4 1)
22
8 8 2 (3) (1) 8 3 8 3 2 8 1 8 1 2 576 J
L
LL L
it vLi t tw Li t t
wtt w w
′=−∴== = = =× −+4H :
(b)
∴= − +∴ − =×−×+−×+×−=
233
1
1122 2
: (
2 1) 2 5 2 5 5 1 2
0.2 3 3 3
10 10 61 61
(2) 8 10 5 2 V P (2) 7 142.33W
333 3
t
t
c
cc
vtdttt tt
v
⎛⎞ ⎛⎞⎛⎞
=−+=−+=−−−+
⎜⎟ ⎜⎟⎜⎟
⎝⎠ ⎝⎠⎝⎠
=×−−++= ∴ =×=∫
0.2F
∴
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Engineering Circuit Analysis, 7
th
Edition Chapter Eleven Solutions 10 March 2006
3.
2
1, 2R1
(0) 2V, (0) 4A, 2, 3, 2 1 1, 3
2L LC
covi s αω=− = = = = = =− ± =− −

(a)
3
3
33
3 1
AB AB4;(0) (0)(442)14
1
A38 14B5,A 1, 5 A
3( 5 ) 23( 5 ) 2 35 52
3 5 P (0 ) (3 5)( 1 5) 8W
tt
L
tt
t
tt ttt t t
co
o
tt
cc
3
e i v
ie e
veedtee ee
ve e
−− + +
−−
−− −− − −
−− +
=+∴+= = =−××+=−ie

∴− − =− ∴ = =− =− +

(b)

0.2 0.6 0.2 0.6
P (0.2) (3 5 )( 5 ) 0.5542W
c
eeee
−− −
=− −+=−

(c)

0.4 1.2 1.2 0.4
P (0.4) (3 5 )(5 ) 0.4220W
c
eeee
−−−−
=− −=

∴+= − + −= − −= −− +−
=−∴ =−−+=−
∫
∴
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Engineering Circuit Analysis, 7
th
Edition Chapter Eleven Solutions 10 March 2006
4. We assume the circuit has already reached sinusoidal steady state by t = 0.
2.5 k
Ω → 2.5 kΩ, 1 H → j1000 Ω, 4 μF → -j250 Ω, 10 kΩ → 10 kΩ
Zeq = j1000 || -j250 || 10000 = 11.10 – j333.0 Ω

Veq =
o(20 30)(11.10 333.0)
2.631 50.54 V
2500 11.10 333.0j
j
∠−
=∠−
+−

I10k =
mA 50.54- 2631.0
10000
oeq
∠=
V I1 H = mA 140.5- 631.2
1000
oeq
∠=
j
V
I 4 μF = mA 46.93 52.10
250
oeq
∠=
−j
V
V 2.5k =
V 55.3774.19
0.33310.112500
)2500)(3020(
o
∠=
−+
∠
j

Thus, P
2.5k =
[]
mW 97.97
2500
55.37cos74.19
2
o
=

P 1 H = () [ ][ ] mW 3.395- )5.140cos(102.631 54.50cos631.2
o-3o
=−×−

P
4 μF =
( )[ ][ ] mW 13.58 )cos(39.461010.52 54.50cos631.2
o-3o
=×−

P 2.5k =
()[]
W 279.6
10000
54.50cos631.2
2
o
μ=
−

FREQ IM(V_PRINT1) IP(V_PRINT1)
1.592E+02 7.896E-03 3.755E+01

FREQ VM(R2_5k,$N_0002)VP(R2_5k,$N_0002)
1.592E+02 1.974E+01 3.755E+01

FREQ IM(V_PRINT2) IP(V_PRINT2)
1.592E+02 2.628E-03 -1.405E+02

FREQ VM(L,0) VP(L,0)
1.592E+02 2.629E+00 -5.054E+01

FREQ IM(V_PRINT11) IP(V_PRINT11)
1.592E+02 1.052E-02 3.946E+01

FREQ IM(V_PRINT12) IP(V_PRINT12)
1.592E+02 2.629E-04 -5.054E+01

PROPRIETARY MATERIAL . © 2007 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers
and educators for course preparation. If you are a student using this Manual, you are using it without permission.

Engineering Circuit Analysis, 7
th
Edition Chapter Eleven Solutions 10 March 2006
5.

rad rad
,
8
8
40 53.13
50A,C 4, 8(3 4)
11 4
3.417 33.15 17.087 33.15 ,
17.087cos(25 33.15 )V
P (0.1) 17.087cos(2.5 33.147 ) 5cos2.5 23.51 W
17.087
cos(25 33.15 )
8
(0.1) 2.136c
si n
s
s
s abs
ijj
j
vt
it
i
∠−°
→∠° →− Ω = − =
−
=∠−°∴= ∠−°
=−° ∴
=− − ° × =−
=−° ∴
=
Z
V
rad
2
8,
3
rad
3
2
3,
os(2.5 33.15 ) 0.7338 A
P 0.7338 8 4.307 W ;
17.087 33.15
3.417 19.98 A
34
(0.1) 3.417cos(2.5 19.98 ) 3.272A
P 3.272 3 32.12W
4(3.417 19.983 ) 13.67 70.02 ,
(0.1) 13.67
abs
abc
c
c
j
i
j
v
−°=−
∴ =×=
∠− °
== ∠ °
−
∴ =+° =−∴
=×=
=− ∠ ° = ∠− °
=
I
V
rad
,
0cos(2.5 70.02 ) 3.946V
P 3.946( 3.272) 12.911 W ( 0)
c abc
−°=
∴ =−=− Σ=

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and educators for course preparation. If you are a student using this Manual, you are using it without permission.

Engineering Circuit Analysis, 7
th
Edition Chapter Eleven Solutions 10 March 2006
6. For t > 0, i(t) =
2
88
Rt
tL
ee
−
−
= .

(a) p(0
+
) = (8)
2
(1) = 64 W

(b) at t = 1 s, i = 8e
–2
= 1.083 A; p(1) = i
2
R = 1.723 W.

(c) at t = 2 s, i = 8e
–4
= 146.5 mA; p(2) = i
2
R = 21.47 mW

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and educators for course preparation. If you are a student using this Manual, you are using it without permission.

Engineering Circuit Analysis, 7
th
Edition Chapter Eleven Solutions 10 March 2006
7.
3
30 10
( ) (3)(6000)
t
vt e
−
−
×
=

(a) p(0
+
) = v
2
(0
+
)/R = (18×10
3
)
2
/ 6000 = 54 kW

(b) p(0.03) = v
2
(0.03)/R = (18×10
3
e
–1
)
2
/ 6000 = 7.308 kW

(c) p(0.09) = v
2
(0.09)/R = (18×10
3
e
–3
)
2
/ 6000 = 134 W

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and educators for course preparation. If you are a student using this Manual, you are using it without permission.

Engineering Circuit Analysis, 7
th
Edition Chapter Eleven Solutions 10 March 2006
8. (a) p = (30×10
3
)
2
(1.2×10
–3
) = 1.080 MW

(b) W = (1.080×10
6
)(150×10
–6
) = 162 J

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and educators for course preparation. If you are a student using this Manual, you are using it without permission.

Engineering Circuit Analysis, 7
th
Edition Chapter Eleven Solutions 10 March 2006
9. W =
21
2
CV. The initial voltage, v(0
+
), is therefore

3
3
2 2(100 10 )
(0 ) 2 V
100 10
W
v
C
−
+
−
×
== =
×
and so
0.12
() 2 2 V
tt
RC
vt e e
−−
== .

The instantaneous power dissipated at t = 120 mS
is therefore

p(120 ms) =
22
(120 ms) 2
226 mW
1.2
ve
R
−
==

The energy dissipated over the first second is given by

2
2
11 2
00
() 2 2
1 100 mJ
2
t
RC
RC
vt e RC
dt dt e
RR R
−
−
⎛⎞⎡⎤
==− −≈
⎜⎟
⎢⎥⎣⎦⎝⎠
∫∫

ΔT = Q/m
c, where Q = 100 mJ, c = 0.9 kJ/kg •K, and m = 10
–3
kg.

Thus, the final temperature

= 271.15 + 23 +
()
6
3
100 10 kJ
kJ
10 kG 0.9
kg K
−
−
×
=
⎛⎞
⎜⎟
⋅⎝⎠
271.15 + 23 + 0.1111

= 294.3 K, representing a temperature increase of 0.1111 K.

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and educators for course preparation. If you are a student using this Manual, you are using it without permission.

Engineering Circuit Analysis, 7
th
Edition Chapter Eleven Solutions 10 March 2006
10. (a) p = (276)(130) = 358.8 mW

(b) v(t) = 2.76cos1000t V (given)
; we need to know the I-V relationship for this
(nonlinear) device.

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and educators for course preparation. If you are a student using this Manual, you are using it without permission.

Engineering Circuit Analysis, 7
th
Edition Chapter Eleven Solutions 10 March 2006
11.

,
2
4,
10
2
10,
5(10 5)
4 4 2.5 5 6.5 5
10
100
12.194 37.57 A
6.5 5
1
P 100 12.194cos37.57 483.3W
2
1
P (12.194) 4 297.4W,
2
P0
100 5
6.097 52.43 so
6.5 5 10
1
P (6.097) 10 185.87
2
in
s
sabs
abs
cabs
abs
jj
jj
j
j
j
−
=+ =+ + = + Ω
∴= = ∠− °
+
∴=−×× °=−
==
=
==∠°
+
=×=
Z
I
I
W
P 0 ( 0)
L
=Σ=

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and educators for course preparation. If you are a student using this Manual, you are using it without permission.

Engineering Circuit Analysis, 7
th
Edition Chapter Eleven Solutions 10 March 2006
12.
10,
10,
2
550
2
820
40 30
(10 10) 52.44 69.18 V
550 8 20
1
P 10 52.44cos69.18 93.19 W
2
1
P 10 52.44cos(90 69.18 ) 245.1 W
2
1 52.44
P cos(50 ) 176.8 W
25
1 52.44
P cos( 20 ) 161
28
gen
jgen
abs
abs
j
∠
∠−
∠°
=+ = ∠ °
∠°+∠− °
=×× °=
=×× °− °=
⎛⎞
=° =
⎜⎟
⎝⎠
⎛⎞
=− °=
⎜⎟
⎝⎠
V
.5 W ( )
gen abs
Σ=Σ

PROPRIETARY MATERIAL . © 2007 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers
and educators for course preparation. If you are a student using this Manual, you are using it without permission.

Engineering Circuit Analysis, 7
th
Edition Chapter Eleven Solutions 10 March 2006
13.
1
33 134
0.1 0.3
25 529
Ignore 30 on , 5 ,
68 10
R
sR R
jj
j
j
j
=+ =++ =+ Ω
−
+
°= =
+Z
VI I 3

(a)
2
3
1529
P 310.875 W=
210
Ω
⎛⎞
=×⎜⎟
⎜⎟
⎝⎠

(b)
,
(2 5)(4 3)
V=∠5 0 13.46351.94
V
68
1
P 13.463 5cos 51.94 20.75 W
2
s
sgen
jj
j
++
° = ∠ °
+

=× × °=
∴

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and educators for course preparation. If you are a student using this Manual, you are using it without permission.

Engineering Circuit Analysis, 7
th
Edition Chapter Eleven Solutions 10 March 2006
14.

10 5
10 10 10
10
10
2
10
50
50
50
PP0,
50 50
0
10 10 5
( 0.1 0.1 0.2) 5 10 0
79.06 16.57 V
1 79.06
P 312.5 W;
210
79.06 161.57 50
12.75 78.69 A
10
1
P 50 12.748cos78.69 62.50 W
2
79.06 161.57
jj
V
j
j
jj
jjj
j
−
Ω
==
−−
++ =
−
∴ −++ ++=
∴ =∠°
==
∠°−
== ∠ °
∴ =× × °=
∠°
=
VVV
V
V
I
I
50
50
15.811 7.57 :
5
1
P 50 15.811cos(90 71.57 ) 375.0 W
2
j
j
j
−
=∠−°
−
=× × °+ °=−

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and educators for course preparation. If you are a student using this Manual, you are using it without permission.

Engineering Circuit Analysis, 7
th
Edition Chapter Eleven Solutions 10 March 2006
15.

20
2 [1]
23
0 [2]
23
which simplify to
5 14 = 60 [1] and
2 + (3 2) = 0 [2]
Solving,
9.233 83.88 V and 5.122 140.2 V
1
P9.
2
xxc
c
ccx
xc
xc
xc
gen
and
j
jj
−−
+=
−
=+
−
−
−
=∠−° =∠−°
=×
VVV
V
VVV
VV
VV
VV
()233 2 5.122 cos( 83.88 140.2 ) 26.22 W×× − °+ °=

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and educators for course preparation. If you are a student using this Manual, you are using it without permission.

Engineering Circuit Analysis, 7
th
Edition Chapter Eleven Solutions 10 March 2006
16.

(a) X 0 R 0
in L th

j=∴Z
LL
= +

(b) R ,X independent R X
L th th th
j
∗
∴= = −ZZ

(c)
2
22
V1
R fixed P R R X
LLL
2(R R ) (X X )
th
LL th
th L th L
j ∴= ×
+++
∴ = − Z

(d)
22
22
2
22
22 2 2
22 2 2
2P RL
2
fixed, Let X X
(R R )
RR 2R(RR)
0
(R R )
R 2RR R 2RR 2R 0
RR
L
LL th
th L
th
th L L th L
L
th L
th th L L th L L
Lth
af
a
adf
dR
a
a
a
+=∴= =

(e)
22
X0R RX
L L th th th
=∴ = + = Z

X
+
R(XX)
ththL
+
++− +
==
⎡⎤++
⎣⎦
+++−==
∴= +
V
= + +
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Engineering Circuit Analysis, 7
th
Edition Chapter Eleven Solutions 10 March 2006
17.
10
120 107.3 116.6 V
10 5
10(10 15)
814
10 5
th
th
j
j
jj
j
j
−
==∠−
+
−+
== −
+
V
Z
°
Ω
TH L
== ΩZZ

(a)
()
*
814 j+

(b)
()
()
()
() ()
()()
()
*
*
*
,max
107.3 116.6
.
16
107.3 116.6 16.12 60.26
16
107.3 16.121 107.3
P cos 116.6 60.26 116.6 179.8 W
216 10
TH
L
TH TH
TH
LTH
TH TH
L
∠− °
==I
+
∠− °
∠− °
==
+
⎡⎤ ⎡⎤
= − °− °+ ° =⎢⎥ ⎢⎥
⎣⎦⎣⎦
V
ZZ
Z
VV
ZZ

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and educators for course preparation. If you are a student using this Manual, you are using it without permission.

Engineering Circuit Analysis, 7
th
Edition Chapter Eleven Solutions 10 March 2006
18.
22
2
22
R R 8 14 16.125
1 107.33
P 16.125 119.38W
2(8 16.125) 14
Lth L
L
=∴=+= Ω
=× =
++
Z

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and educators for course preparation. If you are a student using this Manual, you are using it without permission.

Engineering Circuit Analysis, 7
th
Edition Chapter Eleven Solutions 10 March 2006
19.
9.6 4.8I 1.92 I 4.8I
9.6
I5
1.92
V (0.6 5)8 24V
1
P241.6596 W (
2
)
x xx
x
o
jj
−=− − −+
∴==

∴=×=
gen∴=× × × =

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and educators for course preparation. If you are a student using this Manual, you are using it without permission.

Engineering Circuit Analysis, 7
th
Edition Chapter Eleven Solutions 10 March 2006
20.

(a)
max
480 80 60
80 60
80 60 80 60
28.8 38.4 28.8 38.4
th
L
jj
j
jj
jj
−
==Z
+−
=+ Ω∴ =− Ω Z

(b)
22
,max 2
5(28.8 38.4) 144 192V,
144 192
2 28.8
1 144 192
and P 28.8 250 W
24 28.8
th
L
L
jj
j
=+ =+
+

∴=
×
+
=×=
×I
V

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and educators for course preparation. If you are a student using this Manual, you are using it without permission.

Engineering Circuit Analysis, 7
th
Edition Chapter Eleven Solutions 10 March 2006
21. Zeq = (6 – j8) || (12 + j9) = 8.321 ∠ -19.44
o
W

Veq = (5 ∠-30
o
) (8.321 ∠ -19.44
o
) = 41.61 ∠ -49.44
o
V

P
total = ½ (41.61)(5) cos (-19.44
o
) = 98.09 W

I6-j8 = Veq / (6 – j8) = 4.161 ∠ 3.69
o
A

I4+j2 = I8+j7 = Veq/ 12+j9 = 2.774 ∠ -86.31
o
A

P
6-j8 = ½ (41.61)(4.161) cos (-49.44
o
– 3.69
o
) = 51.94 W

P
4+j2 = ½ (2.774)2 (4) = 15.39 W
P
8+j7 = ½ (2.774)2 (8) = 30.78 W

Check: Σ = 98.11 W (okay)

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and educators for course preparation. If you are a student using this Manual, you are using it without permission.

Engineering Circuit Analysis, 7
th
Edition Chapter Eleven Solutions 10 March 2006
22.

22
,max 2
10(20)10
100 20 40, 4 8
20 10 20 10
R R 8.944
12040
P 8.944 38.63W
2(4 8.944) 64
th th
Lth L
L
jj
j j
jj
==+==
++
∴= ∴=
+Ω
+
∴= ×
++VZ
Z
Ω
=

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and educators for course preparation. If you are a student using this Manual, you are using it without permission.

Engineering Circuit Analysis, 7
th
Edition Chapter Eleven Solutions 10 March 2006
23. We may write a single mesh equation: 170 ∠0
o
= (30 + j10) I1 – (10 – j 50)(-λ I1)
Solving,
I1 =
λλ50101030
0170
o
jj−++
∠

(a)
λ = 0, so I1 = A 18.43-5.376
1030
0170
o
o
∠=
+
∠
j
and, with the same current flowing
through both resistors in this case,
P
20 = ½ (5.376)
2
(20) = 289.0 W
P
10 = ½ (5.376)
2
(10) = 144.5 W

(b) λ = 1, so I1 =
A 543.005
4040
0170
o
o
∠=
−
∠
j

P
20 = ½ (3.005)
2
(20) = 90.30 W
The current through the 10-Ω resistor
is I1 + λI1 = 2 I1 = 6.01 ∠ 45
o
so

P 10 = ½ (6.01)
2
(10) = 180.6 W
(a)
FREQ IM(V_PRINT3) IP(V_PRINT3)
6.000E+01 5.375E+00 -1.846E+01

FREQ IM(V_PRINT4) IP(V_PRINT4)
6.000E+01 5.375E+00 -1.846E+01

(b)
FREQ IM(V_PRINT3) IP(V_PRINT3)
6.000E+01 6.011E+00 4.499E+01

FREQ IM(V_PRINT4) IP(V_PRINT4)
6.000E+01 3.006E+00 4.499E+01

(c)

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and educators for course preparation. If you are a student using this Manual, you are using it without permission.

Engineering Circuit Analysis, 7
th
Edition Chapter Eleven Solutions 10 March 2006
24. (a) Waveform (a): Iavg = A 667.1
3
)1(0)1)(5()1)(10(
=
+−+

Waveform (b): Iavg = A 5
2
)1(0)1)(20(
2
1
=
+

Waveform (c):
I
avg =
()
3
3
10
0
3
3
3
10
0
33

102
cos
2
104
108-
104
2
8sin
101
1
−
−
⎟
⎠
⎞
⎜
⎝
⎛
×
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛×
×=
××
−
−
−−∫
t
dt
tπ
π
π

=
() A
16
10
16
ππ
=−−

(b) Waveform (a):
22
avg
A 41.67
3
(0)(1)(25)(1)(100)(1)
I =
++
=
Waveform (b):
i(t) = -20×10
3
t + 20

i
2
(t) = 4×10
8
t
2
– 8×10
5
t + 400

() dttt∫
+××
×
=
-3
10
0
528
3-
2
avg
400 108 - 104
102
1
I
=
() () ()
2
3-
3
2
3
5
3
3
8
3-
A 66.67
102
0.1333
10400 10
2
108
- 10
3
104

102
1
=
×
=
⎥
⎦
⎤
⎢
⎣
⎡
+
××
×
−−−

Waveform (c):

()
()
2
3
3
3
10
0
3
3
3
10
0
3
2
3
2
avg
A 32
102
sin

2
10
1064

102
10sin
-
2
1064
104
2
64sin
101
1
I
3
3
=
⎥
⎦
⎤
⎢
⎣
⎡
×
−×=
⎥
⎦
⎤
⎢
⎣
⎡
×
×
×=
××
=
−
−−
−
−
∫
π
π
π
ππ
tt
dt
t

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and educators for course preparation. If you are a student using this Manual, you are using it without permission.

Engineering Circuit Analysis, 7
th
Edition Chapter Eleven Solutions 10 March 2006
25. At ω = 120π, 1 H → j377 Ω , and 4 μ F → -j663.1 Ω
Define
Zeff = j377 || -j663.1 || 10 000 = 870.5 ∠ 85.01
o
Ω

V2.5k =
( )
V 27.61- 520.4
85.01 870.5 2500
2500 92400
o
o
o
∠=
∠+
−∠

V 10k =
( )()
V 40.75 181.2
85.01 870.5 2500
85.01 870.5 92400
o
o
oo
∠=
∠+
∠−∠

Thus, P 2.5k = ½ (520.4)2 / 2 500 = 54.16 W
P
10k = ½ (181.2)2 / 10 000 = 1.642 W
P
1H = 0
P
4μF = 0 (A total absorbed power of 55.80 W.)

To check, the average power delivered by the source:

Isource = A 27.61- 0.2081
85.01870.52500
92400
o
o
o
∠=
∠+
−∠

and P source = ½ (2400)(0.2081) cos (-9
o
+ 27.61
o
) = 55.78 W (checks out).

FREQ IM(V_PRINT1) IP(V_PRINT1)
6.000E+01 2.081E-01 -2.760E+01

FREQ VM(R2_5k,$N_0002) VP(R2_5k,$N_0002)
6.000E+01 5.204E+02 -2.760E+01
FREQ IM(V_PRINT2) IP(V_PRINT2)
6.000E+01 4.805E-01 -3.260E+01

FREQ VM(L,0) VP(L,0)
6.000E+01 1.812E+02 5.740E+01

FREQ IM(V_PRINT11) IP(V_PRINT11)
6.000E+01 2.732E-01 1.474E+02
FREQ IM(V_PRINT12) IP(V_PRINT12)
6.000E+01 1.812E-02 5.740E+01

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and educators for course preparation. If you are a student using this Manual, you are using it without permission.

Engineering Circuit Analysis, 7
th
Edition Chapter Eleven Solutions 10 March 2006
26. (a) ()
0
1 144 144
1 cos2000 8.485
22
T
tdt
T
+==∫

(b)
()
0
1 144 144
1 cos2000 8.485
22
T
tdt
T
−==∫

(c)
()
0
1 144 144
1 cos1000 8.485
22
T
tdt
T
+==∫

(d)
()
o
01 144 144
1 cos 1000 176 8.485
22
T
tdt
T
⎡⎤+−==
⎣⎦∫

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and educators for course preparation. If you are a student using this Manual, you are using it without permission.

Engineering Circuit Analysis, 7
th
Edition Chapter Eleven Solutions 10 March 2006
27. (a) ()
0
14 2
1 cos20 1.414
2 2
T
tdt
T
+==∫

(b)
()
0
14 2
1 cos20 1.414
2 2
T
tdt
T
−==∫

(c)
()
0
14 2
1 cos10 1.414
22
T
tdt
T
+==∫

(d)
()
o
014 2
1 cos 10 64 1.414
22
T
tdt
T
⎡⎤+− ==
⎣⎦∫

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Engineering Circuit Analysis, 7
th
Edition Chapter Eleven Solutions 10 March 2006
28. T = 3 s; integrate from 1 to 4 s; need only really integrate from 1 to 3 s as function is zero
between t = 3 and t = 4 s.

33
2
11
1 100 100(2)
(10) 8.165 V
333
rms
Vd tt====∫

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Engineering Circuit Analysis, 7
th
Edition Chapter Eleven Solutions 10 March 2006
29. T = 3 s; integrate from 2 to 5 s; need only really integrate from 2 to 3 s as function is zero
between t = 3 and t = 4 s.

33
2
22
1 49 49(1)
(7) 4.041 A
333
rms
Id tt====∫

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Engineering Circuit Analysis, 7
th
Edition Chapter Eleven Solutions 10 March 2006
30. (a) 1 V

(b)
2
22 2
12
1
1 1.225 V
2
eff eff
rms
VVV
⎛⎞
=+=+ =
⎜⎟
⎝⎠

(c)
2
22 2
12
1
1 1.225 V
2
eff eff
rms
VVV
⎛⎞
=+=+ =
⎜⎟
⎝⎠

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Engineering Circuit Analysis, 7
th
Edition Chapter Eleven Solutions 10 March 2006
31.

(a)
10 9cos100 6sin100
11
V 100 81 36 158.5 12.590V
22
eff
vt t=+ +

(b) 2221
F (10 20 10 ) 150 12.247
4
eff
=++==

(c) F
avg =
10
4
40
4
)1)(10()1)(20()1)(10(
==
++

∴ =+×+×= =
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Engineering Circuit Analysis, 7
th
Edition Chapter Eleven Solutions 10 March 2006
32.

(a) g(t) = 2 + 3cos100t + 4cos(100t – 120
o
)
3 ∠0 + 4∠-120
o
= 3.606 ∠-73.90
o
so Geff =
2
606.3
4
2
+ = 3.240

(b)
22 2
( ) 2 3cos100 4cos(101 120 )
11
H 2 3 4 16.5 4.062
22
eff
ht t t=+ + − °
∴ =++ = =

(c)
0.1
62
0
631

() 100 , 0 0.1 F 10
0.3
10 1
10
10 33.33
33
eff
fttt tdt
−
=<<∴=
=×××= ∫
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Engineering Circuit Analysis, 7
th
Edition Chapter Eleven Solutions 10 March 2006
33.
2
( ) (2 3cos100 )ftt=−

(a) ( )ft
2
4 12cos100 9cos 100
( ) 4 12cos100 4.5 4.5cos200 F 4 4.5 8.5
av
t t
ft t t
=− +

(b) 22 211
F 8.5 12 4.5 12.43
22
eff
=+×+×=

∴ =− + + ∴ =+ =
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Engineering Circuit Analysis, 7
th
Edition Chapter Eleven Solutions 10 March 2006
34. (a) i eff = () A 6.455 0)5(10
3
1
2
1
22
=
⎥
⎦
⎤
⎢
⎣
⎡
+−+

(b) i
eff =
[] A 2.236 5 0 2020
2
1
2
1
1
0
==
⎥
⎦
⎤
⎢
⎣
⎡
+⎟
⎠
⎞
⎜
⎝
⎛
+−∫
dtt

(c) i
eff =
A 2.257
2
cos
2
8-
4
2
sin8
1
1
1
0
2
1
1
0
=
⎥
⎦
⎤
⎢
⎣
⎡
⎟
⎠
⎞
⎜
⎝
⎛
⎟
⎠
⎞
⎜
⎝
⎛
=⎥
⎦
⎤
⎢
⎣
⎡
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
⎟
⎠
⎞
⎜
⎝
⎛∫
t
dttπ
π
π

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Engineering Circuit Analysis, 7
th
Edition Chapter Eleven Solutions 10 March 2006
35.

(a)
o
2
A==B 10V, C D 0 10 0 10 45 18.48 22.50
11
P 18.48
42.68 W
24
==∴∠°+∠−°= ∠−
(b) (c) (d)
(e)
2
10
// 10 P 55.18 80.18 W
4
av
dc+∴= +=

∴=×× =
A==
22
C 10V, B D 0, 10cos10 10cos40 ,
110 110
P2 5 W
24 24
s
vtt== = +
=+= o
2
10cos10 10sin (10 45 ) 10 10 45 7.654 67.50
1 7.654
P7 .322 W
24
s
vt t=− +°→−∠−°=∠
∴==
o
22
10cos10 10sin (10 45 ) 10cos40 ;
10 0 10 45 18.48 22.50
1111
P 18.48 10 55.18 W
2424
vt t t=+ +°+
∠°
+ ∠− °= ∠−
∴=× ×+× ×=
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Engineering Circuit Analysis, 7
th
Edition Chapter Eleven Solutions 10 March 2006
36. Zeq = R || j0.3ω =
ω
ω
RjR
Rj
3.0
3.0
+
. By voltage division, then, we write:

V100mH =
ωω
ωω
ω
ω
ω
ωRj
Rj
jR
Rj
j
j
4.003.0
1.00.03-
0120
3.0
3.0
1.0
1.0
0120
2
2
+−
+
∠=
+
+
∠
V 300mH =
Rω.jω.
Rj
jR
Rj
j
jR
Rj
40030
36
0120
3.0
3.0
1.0
3.0
3.0
0120
2
+−
∠=
+
+
+
∠
ω
ω
ω
ω
ω ω

(a) We’re interested in the value of R that would lead to equal voltage magnitudes, or

( ) R0.10.03- (120) 36
2
ωωωjRj +=

Thus, 36Rω =
224
14496.12Rωω+ or R = 0.1061 ω

(b) Substituting into the expression for V100mH, we find that V100mH = 73.47 V,
independent of frequency.
To verify with PSpice, simulate the circuit at 60 Hz, or ω = 120π rad/s, so R = 40 Ω.
We also include a miniscule (1 pΩ) resistor to avoid inductor loop warnings. We
see
from the simulation results that the two voltage magnitudes are indeed the same.

FREQ VM($N_0002,$N_0003)VP($N_0002,$N_0003)
6.000E+01 7.349E+01 -3.525E+01

FREQ VM($N_0001,$N_0002)VP($N_0001,$N_0002)
6.000E+01 7.347E+01 3.527E+01

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Engineering Circuit Analysis, 7
th
Edition Chapter Eleven Solutions 10 March 2006
37.

(a)
,1V30
av V=

,2
1
V (10 30 50) 30V=
3
av=++

3
2
(b)

(c) PSpice verification for Sawtooth waveform
of Fig. 11.40a:

,1
V
0
222
,2
111
(20 ) 400 27 1200 34.64V
333
11
V (10 30 50 ) 3500 34.16V
33
eff
eff
tdt==×× ×==
=++=×=∫
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Engineering Circuit Analysis, 7
th
Edition Chapter Eleven Solutions 10 March 2006
38. Zeff = R ||
6
66
103
10

3
10jR
jRj−
−
=
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛−ωω

ISRC =
( )
()
666
6
6
66
1010310
10-R3120

103
1010
0120
RjjRj
j
jR
R
jjωω
ωω
ωω−−−
=
−
−−
∠

I3μF = ISRC
ω3
10
6
jR
R
−
(a)
For the two current magnitudes to be equal, we must have
1
3
10
6
=
−
ω
jR
R
. This is
only true when R = ∞; otherwise, current is shunted through the resistor and the two
capacitor currents will be unequal.
(b)
In this case, the capacitor current is

A )90 cos( 90 or A, 90
3
1010
1
0120
o
66
μωωμω
ωω
+=
−−
∠ tj
jj

(c) PSpice verification: set f = 60 Hz, simulate a single 0.75-μF capacitor, and include a
100-MΩ resistor in parallel with the capacitor to prevent a floating node. This should
resit in a rms current amplitude of 33.93 mA, which it does.
FREQ IM(V_PRINT3) IP(V_PRINT3)
6.000E+01 3.393E-02 9.000E+01

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Engineering Circuit Analysis, 7
th
Edition Chapter Eleven Solutions 10 March 2006
39.
0.5( 3)
2
2
1,
0
5
(3) 3 5
2, 3
3
22
23
( ) 10 [ ( ) ( 2)] 16 [ ( 3) ( 5)] V
Find eff. value separately
12 0
V 100 8 7.303
53
1 256
V 256 ( ) 6.654
55
V 7.303 6.654 9.879
1
V 100 256
5
t
eff
tt
eff
eff
t
eff
vt tut ut e ut ut
tdt
edt ee
tdt ee
−−
−− −
−
=−−+ −−−
==×=
==− =
∴= + =
=+
∫
∫
25
03
33 5
2
1 100
8 256 ( )
53
1 800
256(1 ) 9.879VOK
53
dt
ee e
e
−−
−
⎡⎤
⎢⎥⎣⎦
⎡⎤
=×+ −
⎢⎥
⎣⎦
⎡⎤
=+−=
⎢⎥
⎣⎦
∫∫

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Engineering Circuit Analysis, 7
th
Edition Chapter Eleven Solutions 10 March 2006
40. The peak instantaneous power is 250 mW. The combination of elements yields

Z = 1000 + j 1000 Ω = 1414 ∠45
o
Ω.
Arbitrarily designate
V = Vm ∠0 , so that I =
1414
45V

0V
o
mm
−∠
=
∠
Z
A.

W e may write p(t) = ½ V m Im cos φ + ½ V m Im cos (2ωt + φ) where φ = the angle of the
curren t (-45
o
). This function has a maximum value of ½ VmIm cos φ + ½ V mIm.

Thus, 0.250 = ½ VmIm (1 + cos φ) = ½ (1414) I m
2 (1.707)
and I
m = 14.39 mA.

In terms of rms current, the largest rms current permitted is 14.39 /
2 = 10.18 mA rms.

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Engineering Circuit Analysis, 7
th
Edition Chapter Eleven Solutions 10 March 2006
41. 435Arms=∠ °I

(a)
,
20 80 35 Vrms, P 80 10cos35 655.3 W
sgen
=+∠° =× °=VI

(b)
2
P R 16 20 320 W
R
==×=I

(c) P 655.3 320 335.3 W
Load
=−=

(d)
,
AP 80 10 800 VA
sgen
=×=

(e)
AP P 320VA
RR
==

(f) 10
L
=

(g)

0 4 35 7.104 18.84 A rms
AP
80 7.104 568.3 V A
L
∠°−∠°= ∠− °
∴ =× =
I
P 335.3
PF cos 0.599
A
P 568.3
since I lagsV, PF is lagging
L
LL
L
LL
θ====
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Engineering Circuit Analysis, 7
th
Edition Chapter Eleven Solutions 10 March 2006
42.

(a)
120
I
s== 9.214 26.25 A rms
192
4
12 16
P
F cos26.25
s
j
j
∠−°
+
+
0.8969lag

(b) P 120 9.214 0.8969 991.7W
s=× × =

(c)

(d) PSpi ce verification

FREQ VM($N_0003,0) VP($N_0003,0)
6.000E+01 1.200E+02 0.000E+00

FREQ IM(V_PRINT1) IP(V_PRINT1)
6.000E+01 9.215E+00 -2.625E+01
; (a) and (b) are correct

Next, add a 90.09-μF capacitor in parallel with the source:

FREQ IM(V_PRINT1) IP(V_PRINT1)
6.000E+01 8.264E+00 -9.774E-05 ;(c) is correct (-9.8×10
-5
degrees
is essentially zero, for unity PF).

∴ ==
22
22
48 1
Z=+4 4 (192 144)
34 2
5
11.68 5.76
Z 11.68 5.76 , Y
11.68 5.76
5.76
120 C , C
11.68 5.76
L
LL
j
j
j
j
j
j
j
=+ +
+
−
90.09 F
πμ
∴=+Ω=
+
∴ ==
+

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Engineering Circuit Analysis, 7
th
Edition Chapter Eleven Solutions 10 March 2006
43.

1
2
5 2 , 20 10 , 10 30 8.660 5
10 60 5 8.660
200 20 10
0 33.66 13.660 7265 22.09
15.11 3.908 A rms
25 8 20 10 480.9 26.00
20 10 33.66 13.660
25 8 200
20 10 0 200
480.9 26.00
AB c
D
jj j
j
j
j
jj
jj
j
j=+ Ω = − Ω = ∠°Ω= + Ω
=∠−°=− Ω
−+
− ∠°
===
−−+ ∠− °
−+ −
−
−+
==
∠− °ZZ Z
Z
I
I
∠ °
2
2
1
2
12
2
2
2
2
21
1
(20 10)
9.300 0.5681 A rms
480.9 20.00
AP 15.108
AP 5.881
AP 2 9.3 10
AP 9.3 10
AP 200 200 15.108
AA
BB
CC
D
S
j−
=∠− °
∠°
==
=− =
==
==×
==×
IZ
IIZ
IZ
IZ
I

2
29 1229 VA
10 5 773.5VA
86.49VA
864.9VA
3022VA
=
× =
×=
=
=

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Engineering Circuit Analysis, 7
th
Edition Chapter Eleven Solutions 10 March 2006
44.
12
30 15 , 40 40= ∠ °Ω = ∠ °ΩZZ

(a)
tot
=∠Z 30 15 40 40 68.37 29.31
PF cos29.3 0.8719 lag
°+∠°= ∠ °Ω

(b)
V = IZtot = so
o
683.8 29.31∠Ω

S = VI* = () . ()
oo
683.8 29.31 10 0 6838 29.31 VA∠∠=∠

Thus, the apparent power = S = 6.838 kVA.

(c) The impedance has a positive angle; it theref
ore has a net inductive character.

∴=° =
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Engineering Circuit Analysis, 7
th
Edition Chapter Eleven Solutions 10 March 2006
45. θ 1 = cos
-1
(0.92) = 23.07
o
, θ 2 = cos
-1
(0.8) = 36.87
o
, θ 3 = 0

S 1 = VA 42.59 100
92.0
23.07 100
o
j+=
∠

S 2 =
VA 5.871 250
8.0
78.63 250
o
j+=
∠

S 3 =
VA 500
1
0 500
o
=
∠

Stotal = S1 + S2 + S3 = 500 + j230.1 VA = 550.4 ∠ 24.71
o
VA

(a) Ieff =
rmsA 4.786
115
550.4

V
S
eff
total
==

(b) PF of composite load = cos (24.71
o
) = 0.9084 lagging
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and educators for course preparation. If you are a student using this Manual, you are using it without permission.

Engineering Circuit Analysis, 7
th
Edition Chapter Eleven Solutions 10 March 2006
46.

2
1
AP 10,000 VA, PF 0.8lag, 40 A rms
Let 40 0 A rms; P 10,000 0.8 8000 W
8000
Let R X R 5
40
cos 0.8lag cos 0.8 36.87
X 5tan36.87 3.75 , 5 3.75, 5.2 3.75
40(5.2 3.75) 256.4
LLL
LL
LL LL
LL
LL tot
s
j
jj
j
θθ
−
===
=∠° = × =
=+ ∴= =Ω
=∴= =°
∴= °= Ω =+ =+ Ω
∴= + =
I
I
Z
ZZ
V
1
35.80 V;
5.2 3.75
0.12651 0.09124S, 0.12651 (120 C 0.09124),
PF 0.9 lag, 25.84 tan 25.84 0.4843
0.09124 120 C
0.12651
C 79.48 F
tot
new
new new
j
jj
π
θ
π
μ
∠° =
+
=− =+ −
==° ∴°=
−
=∴
=
Y
Y

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and educators for course preparation. If you are a student using this Manual, you are using it without permission.

Engineering Circuit Analysis, 7
th
Edition Chapter Eleven Solutions 10 March 2006
47. Zeff = j100 + j300 || 200 = 237 ∠54.25
o
. PF = cos 54.25
o
= 0.5843 lagging.

(a) Raise PF to 0.92
lagging with series capacitance

Znew = j100 + jXC + j300 || 200 = 138.5 + j(192.3 + XC) Ω

o1-C1
23.07 0.92 cos
138.5
X 192.3
tan ==⎟
⎠
⎞
⎜
⎝
⎛ +
−

Solving, we find that X
C = -133.3 Ω = -1/ ωC, so that C = 7.501 μF

(b) Raise PF to 0.92
lagging with parallel capacitance

Znew = j100 || jXC + j300 || 200 =
)X100(
X 100
C
C+
−
j
+138.5 + j92.31 Ω
= 138.5 +
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
+
+
C
C
X100
X100
31.92j
Ω

o1-C
C
1
23.07 0.92 cos
138.5
X100
100X
92.31
tan ==
⎟
⎟
⎟
⎟
⎠
⎞
⎜
⎜
⎜
⎜
⎝
⎛
+
+
−

Solving, we find that X
C = -25 Ω = -1/ωC, so that C = 40 μF

General circuit for simulations. Results agree with hand calculations

FREQ IM(V_PRINT1) IP(V_PRINT1) θ PF
With no compensation: 1.592E+02 4.853E-01 -5.825E+01 54.25
o
0. 5843 lag
With series compensation: 1.592E+02 7.641E-01 -2.707E+01 23.07
o
0. 9200 lag
With parallel compensation: 1.592E+02 7.641E-01 -2.707E+01 23.07
o
0. 9200 lag

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and educators for course preparation. If you are a student using this Manual, you are using it without permission.

Engineering Circuit Analysis, 7
th
Edition Chapter Eleven Solutions 10 March 2006
48.
20
2
20
20(1 2)
10 10.769 3.846 11.435 19.65
32
100
8.745 19.65
11.435 19.654
100 8.745 19.65 823.5 294.1VA
10 20
8.745 19.65 5.423 49.40
30 20
20 5.432 588.2 0 VA
in
s
sss
j
jj
j
j
j
j
j
+
∗+
=− + = − = ∠− °Ω
+
∴= = ∠ °
∠− °
∴=− =− × ∠− °=− +
+
=∠°× =∠°
+
∴=× = +
Z
I
SVI
I
S
10
2
10
2
20
2
10
20 5.423 49.40
4.851 14.04
10 20
10 4.851 235.3 0 VA
20 4.851 470.6 VA,
10 8.745 764.7 VA, 0
j
j
j
j
jj
jj
−
×∠
== ∠− °
+
=× = +
=× =
=− × = − Σ=
I
S
S
S

PROPRIETARY MATERIAL . © 2007 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers
and educators for course preparation. If you are a student using this Manual, you are using it without permission.

Engineering Circuit Analysis, 7
th
Edition Chapter Eleven Solutions 10 March 2006
49.

1
1.
2
6,
4,
100 100
0
64 10 5
1 100
0.1 0.2 20
64 64
53.35 42.66 V
100 53.35 42.66
9.806 64.44 A
64
1
100 9.806 64.44 211.5 442.3VA
2
1
6 9.806 288.5 0VA
2
xxx
x
x
gen
abs
j abs
j
jj
jj
jj
j
j
j
−
−
−−
++ =
+−
⎛⎞
∴++=+
⎜⎟
++⎝⎠
∴= ∠ °
−∠°
∴= = ∠− °
+
∴=××∠°= +
=×× = +
=VVV
V
V
I
S
S
S
2
2
2
5
2,
10,1
( 4)9.806 0 192.3VA
2
100 53.35 42.66
14.99 121.6 ,
5
1
5 14.99 561.5 0VA
2
1
( 100)14.99 121.57 638.4 392.3VA
2
1 53.35
( 10) 0 142.3VA 142.3 90 VA 0
210
abs
gen
jabs
jj
j
j
jj
jj
−
−
=+
−∠°
== ∠ °
=×× = +
=∠ −° =−
⎛⎞
=−=−=∠−°
⎜⎟
⎝⎠
I
S
S
S
Σ=

PROPRIETARY MATERIAL . © 2007 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers
and educators for course preparation. If you are a student using this Manual, you are using it without permission.

Engineering Circuit Analysis, 7
th
Edition Chapter Eleven Solutions 10 March 2006
50.

(a) 500

(b)

(c)

1
VA, PF 0.75 lead
500 cos 0.75 375 330.7 VA
j
−
=∴
=∠− = −
S
500
1
W, PF 0.75 lead
500
500 sin (cos 0.75) 500 441.0 VA
.075
j
j
−
=∴
=− = −S 1
500 VAR, PF 0.75(lead) cos 0.75 41.41
P500/ tan 41.41 566.9W,
566.9 500 VA
j
θ
−
−= ∴=−= − °
∴ °=
=−
S
PROPRIETARY MATERIAL . © 2007 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers
and educators for course preparation. If you are a student using this Manual, you are using it without permission.

Engineering Circuit Analysis, 7
th
Edition Chapter Eleven Solutions 10 March 2006
51. 1600 500VA(gen)
s
j=+S

(a)
1600 500
4 1.25 4 1.25
400
400
3.333A rms 4 1.25 3.333
120
4 4.583A rms
400(4 4.583) 1600 1833 VA
ss
cL s c
L
L
j
jj
jj
j
j
jj
∗ +
==+∴=−
== ∴=−=−−
−
∴=− ∴
=+ =+
II

jII II
I
S

(b)
11833.3
PF cos tan 0.6575 lag
1600
L
−+⎛⎞
==
⎜⎟
⎝⎠

(c) 1600 500 1676 17.35 VA PF cos17.35 0.9545 lag
ss
j=+=∠°∴= °=S

PROPRIETARY MATERIAL . © 2007 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers
and educators for course preparation. If you are a student using this Manual, you are using it without permission.

Engineering Circuit Analysis, 7
th
Edition Chapter Eleven Solutions 10 March 2006
52.
11
(cos 0.8 36.87 , cos 0.9 25.84 )
−−
=° = °

(a) 1200 36.87 1600 25.84 900
960 720 1440 697.4 900
3300 1417.4 3592 23.25 VA
3591.5
15.62 A rms
230
tot
s
jj
j
=∠S °+∠°+
=+ + + +
=+ =∠°
∴= =I

(b) PF cos23.245 0.9188
s=°=

(c) 3300 1417 VAj=+
S

PROPRIETARY MATERIAL . © 2007 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers
and educators for course preparation. If you are a student using this Manual, you are using it without permission.

Engineering Circuit Analysis, 7
th
Edition Chapter Eleven Solutions 10 March 2006
53.

(a)
,
P 20 25 0.8 30 0.75 70 kW
stot
=+×+× =

(b)
1
2
1o
22
33
1
33
20,000
80 0 A rms
250
25,000/ 250 100 A rms
cos 0.8 36.87 100 36.87 A rms
30,000 40,000
AP 40,000 VA, 160A rms
0.75 250
cos 0.75 41.41 160 41.41 A rms
80 0 100 36.87 160
s
−
−
==∠°
==
∠=− =− ∴= ∠−
== ==
∠=− =− °∴ = ∠− °
∴=∠°+ ∠− °+ ∠−
I

I
II
I
II
I 41.41 325.4 30.64 A rms
AP 250 325.4 81,360 VA
s
°= ∠− °
∴=× =

(c)
3
70,000
PF 0.8604lag
81,360
==

PROPRIETARY MATERIAL . © 2007 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers
and educators for course preparation. If you are a student using this Manual, you are using it without permission.

Engineering Circuit Analysis, 7
th
Edition Chapter Eleven Solutions 10 March 2006
54. 200 kW average power and 280 kVAR reactive result in a power factor of
PF = cos (tan
-1
(280/200) = 0.5813 lagging, which is pretty low.

(a) 0.65 peak = 0.65(200) = 130 kVAR
Excess = 280 – 130 = 150 kVAR, for a cost of (12)(0.22)(150) = $396 / year.

(b) Target = S = P +
j0.65 P
θ = tan
-1
(0.65P/P) = 33.02
o
, so target PF = cos θ = 0.8385

(c) A single 100-kVAR increment costs $200 to install. The excess kVAR would then be
280 – 100 – 130 = 50 kVAR, for an annual penalty of $332. This would result in a
first-year savings of $64.

A single 200-kVAR increment costs $395 to install, and would remove the entire excess
kVAR. The savings wou
ld be $1 (wow) in the first year, but $396 each year thereafter.

The single 200-kVAR increment is the most economical choice.

PROPRIETARY MATERIAL . © 2007 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers
and educators for course preparation. If you are a student using this Manual, you are using it without permission.

Engineering Circuit Analysis, 7
th
Edition Chapter Eleven Solutions 10 March 2006
55. Perhaps the easiest approach is to consider the load and the compensation capacitor
separately. The load draws a complex power S
load = P + jQ. The capacitor draws a
purely reactive complex power S
C = -jQC.

θ
load = tan
-1
(Q/P), or Q = P tan θ load

Q
C = SC = Vrms
rms
(/C)jω−
V
=
2
C
rms
ωV =
2
CV
rms
ω

S
total = S load + SC = P + j(Q – QC)

θ
new = ang(S total) =
1 CQ-Q
tan
P
−⎛
⎜
⎝⎠
⎞
⎟
, so that Q – Q
C = P tan θ new

Substituting, we find that Q
C = P tan θ load – P tan θ new
or
= P (tan θ
2
CV
rms
ω load – tan θ new)

Thus, noting that θ
old = θload,

( )
old new
2
rms
P tan - tan
C =
Vθθ
ω

PROPRIETARY MATERIAL . © 2007 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers
and educators for course preparation. If you are a student using this Manual, you are using it without permission.

Engineering Circuit Analysis, 7
th
Edition Chapter Eleven Solutions 10 March 2006
56. V = 339 ∠-66
o
V, ω = 100π rad/ s, connected to Z = 1000 Ω.

(a)
Veff =
rms V 239.7
2
339=
(b)
pmax = 339
2
/ 1000 = 114.9 W

(c)
pmin = 0 W

(d)
Apparent power = Veff Ieff =
VA 57.46
1000
V

1000
2
339

2
339
2
eff
==
⎟
⎟
⎟
⎠
⎞
⎜
⎜
⎜
⎝
⎛
⎟
⎠
⎞
⎜
⎝
⎛

(e)
Since the load is purely resistive, it draws zero reactive power.

(f)
S = 57.46 VA

PROPRIETARY MATERIAL . © 2007 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers
and educators for course preparation. If you are a student using this Manual, you are using it without permission.

Engineering Circuit Analysis, 7
th
Edition Chapter Eleven Solutions 10 March 2006
57. V = 339 ∠-66
o
V, ω = 100π rad/s to a purely inductive load of 150 mH ( j47.12 Ω)

(a)
I =
A 156- 7.194
12.47
66-339

o
o
∠=
∠
=
jZ
V

so
Ieff =
rmsA 5.087
2
194.7=
(b)
p(t) = ½ VmIm cos φ + ½ VmIm cos(2ωt + φ)
where
φ = angle of current – angle of voltage

pmax = ½ VmIm cos φ + ½ VmIm = (1 + cos(-90
o
)) (339)(7.194)/ 2 = 1219 W

(c)
pmin = ½ VmIm cos φ - ½ VmIm = -1219 W

(d) apparent power = V
eff Ieff =
() VA 1219 087.5
2
339=
(e)
reactive power = Q = Veff Ieff sin (θ – φ) = 1219 VA

(f)
complex power = j1219 VA

PROPRIETARY MATERIAL . © 2007 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers
and educators for course preparation. If you are a student using this Manual, you are using it without permission.

Engineering Circuit Analysis, 7
th
Edition Chapter Eleven Solutions 10 March 2006
58. 1 H → j Ω, 4 μF → –j250 Ω

Zeff = j || –j250 || 10
4
Ω = 1.004 ∠89.99
o
Ω

V10k =
o
o
o
(5 0) (1.004 89.99 )
2.008 89.97 mV
2500 (1.004 89.99 )
∠∠
=∠
+∠

(a)
pmax = (0.002)
2
/ 10×10
3
= 400 pW

(b) 0 W (purely resistive elements draw no reactive power)

(c) apparent power = V
effIeff = ½ VmIm = ½ (0.002)
2
/ 10000 = 200 pVA

(d)
Ssource =
()
o150
5 0 0.005 -0.02292 VA
2 2500 0.02292∠⎛⎞
∠= ∠
⎜⎟
∠⎝⎠

PROPRIETARY MATERIAL . © 2007 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers
and educators for course preparation. If you are a student using this Manual, you are using it without permission.

Engineering Circuit Analysis, 7
th
Edition Chapter Eleven Solutions 10 March 2006
59. (a) At ω = 400 rad/s, 1 μF → -j2500 Ω, 100 mH → j40 Ω
Defi ne
Zeff = -j2500 || (250 + j40) = 256 ∠ 3.287
o
Ω

IS =
rmsA 3.049- 43.48
287.325620
012000
o
o
∠=
∠+
∠

Ssource = (12000)(43.48) ∠ 3.049
o
= 521.8 ∠3.049
o
kVA

S20Ω = (43.48)
2
(20) ∠0 = 37.81 ∠0 kVA

Veff =
rms V 0.2381 11130
287.325620
)287.3256)(012000(
o
∠=
∠+
∠∠
o
o

I1μF =
rmsA 90.24 4.452
0250-
oeff
∠=
j
V

so
S1μF = (11130)(4.452) ∠-90
o
= 49.55 ∠-90
o
kVA

V100mH =
rms V 15.18 1758
40250
)40)(2381.011130(
o
∠=
+
∠
j
j
o

I 100mH =
rmsA 852.8- 43.96
40
o100mH
∠=
j
V

so
S100μΗ = (1758)(4.43.96) ∠90
o
= 77.28 ∠90
o
kVA

V250Ω =
rms V 852.8 10990
40250
)250)(2381.011130(
o
−∠=
+
∠
j
o

so
S250Ω = (10990)
2
/ 250 = 483.1 ∠0
o
kVA

(b) 37.81
∠0 + 49.55 ∠-90
o
+77.28 ∠90
o
+ 483.1 ∠0
o
= 521.6 ∠3.014
o
kVA,
which is within rounding error of the comp
lex power delivered by the source.

(c) The apparent power of the source is 521.8 kVA. The apparent powers of the passive
elements sum to 37.81 + 49.55 + 77.28 + 483.1 = 647.7 kVA, so NO! Phase angle is
im
portant!

(d)
P = Veff Ieff cos (ang VS – ang IS) = (12000)(43.48) cos (3.049
o
) = 521 kW

(e)
Q = Veff Ieff sin (ang VS – ang IS) = (12000)(43.48) sin (3.049
o
) = 27.75 kVAR

PROPRIETARY MATERIAL . © 2007 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers
and educators for course preparation. If you are a student using this Manual, you are using it without permission.

Engineering Circuit Analysis, 7
th
Edition Chapter Eleven Solutions 10 March 2006
60. (a) Peak current = 282 = 39.6 A

(b)
θload = cos
-1
(0.812) = +35.71
o
(since lagging PF). Assume ang (V) = 0
o
.

p(t) =
() ()2300 2 39.6 cos(120 ) cos (120 35 71 )
o
πt πt - .

at
t = 2.5 ms, then, p(t) = 71.89 kW

(c) P = V
eff Ieff cos θ = (2300)(28) cos (35.71
o
) = 52.29 kW

(d)
S = Veff Ieff ∠θ = 64.4 ∠ 35.71
o
kVA

(e) apparent power = |
S| = 64.4 kVA

(f) |
Zload| = |V/ I| = 2300/28 = 82.14 Ω. Thus, Zload = 82.14 ∠ 35.71
o
Ω

(g) Q = V
eff Ieff sin θ = 37.59 kVAR

PROPRIETARY MATERIAL . © 2007 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers
and educators for course preparation. If you are a student using this Manual, you are using it without permission.

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