Engineering Circuit Analysis 7th Ed. Solutions - Chapter 2

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Engineering Circuit Analysis, 7th Edition
1. (a)
12 μs
(b) 750 mJ
(c) 1.13 kΩ
(d) 3.5 Gbits
(e) 6.5 nm
(f) 13.56 MHz
Chapter Two Solutions
10 March 2006
(g) 39 pA
(h) 49 kΩ
(i) 11.73 pA
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Engineering Circuit Analysis, 7th Edition
2.
(a) 1 MW
(e) 33 μJ
(b) 12.35 mm (f) 5.33 nW
(c) 47. kW
(g) 1 ns
(d) 5.46 mA (h) 5.555 MW
Chapter Two Solutions
10 March 2006
(i) 32 mm
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Engineering Circuit Analysis, 7th Edition
3. (a)
Chapter Two Solutions
⎛ 745.7 W ⎞
⎟ = 298.3
⎝ 1 hp ⎠
( 400 Hp ) ⎜
10 March 2006
kW
⎛ 12 in ⎞ ⎛ 2.54 cm ⎞ ⎛ 1 m ⎞
(b) 12 ft = (12 ft) ⎜
⎟⎜
⎟⎜
⎟ = 3.658 m
⎝ 1 ft ⎠ ⎝ 1 in ⎠ ⎝ 100 cm ⎠
(c) 2.54 cm =
25.4 mm
⎛ 1055 J ⎞
(d) ( 67 Btu ) ⎜
⎟ = 70.69
⎝ 1 Btu ⎠
(e) 285.4´10-15 s =
kJ
285.4 fs
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Engineering Circuit Analysis, 7th Edition
4.
Chapter Two Solutions
10 March 2006
(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, 7th Edition
5.
Chapter Two Solutions
10 March 2006
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 = (130.5×103 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, 7th Edition
6.
Chapter Two Solutions
10 March 2006
The 400-mJ pulse lasts 20 ns.
(a) To compute the peak power, we assume the pulse shape is square:
Energy (mJ)
400
20
Then
t (ns)
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, 7th Edition
7.
Chapter Two Solutions
10 March 2006
The 1-mJ pulse lasts 75 fs.
(a) To compute the peak power, we assume the pulse shape is square:
Energy (mJ)
1
75
Then
t (fs)
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, 7th Edition
8.
Chapter Two Solutions
10 March 2006
The power drawn from the battery is (not quite drawn to scale):
P (W)
10
6
t (min)
5
7
17
24
(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, 7th Edition
9.
Chapter Two Solutions
10 March 2006
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 minutes is given by
∫
300 s
0
300
10 2
⎡ 10
⎤
⎢⎣ − 300 t + 10 ⎥⎦ dt = − 600 t + 10t
0
= 1.5 kJ
(a) The total energy transferred is 288 + 1.5 = 289.5 kJ
(b) The energy transferred in the last five minutes is 1.5 kJ
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Engineering Circuit Analysis, 7th Edition
Chapter Two Solutions
10 March 2006
charge q = 18t2 – 2t4 C.
10. Total
(a) q(2 s) = 40 C.
(b) To find the m aximum charge within 0 ≤ t ≤ 3 s, we need to take the
second derivitives:
f irst and
dq/dt = 36t – 8t3 = 0, leading to roots at 0, ± 2.121 s
d2q/dt2 = 36 – 24t2
substituting t = 2.121 s into the expression for d2q/dt2, we obtain a value of
–14.9, so that this root represents a maximum.
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).
50
70
(b)
(a)
60
50
0
30
tim e (t)
q (C)
40
20
-50
10
0
-100
-10
-20
0
0.5
1
1.5
tim e (s )
2
2.5
3
-150
0
0.5
1
1.5
i (A )
2
2.5
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3
Engineering Circuit Analysis, 7th Edition
11.
Chapter Two Solutions
10 March 2006
Referring to Fig. 2.6c,
⎧- 2 + 3e −5t A,
i1 (t ) = ⎨
3t
⎩ - 2 + 3e A,
t<0
t>0
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 + 3e3t = 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
i (t )dt
−0.8 1
0
−5 t
∫−0.8 ⎡⎣ −2 + 3e ⎤⎦ dt
+
0
3
⎛
⎞
= ⎜ −2t − e−5t ⎟
5
⎝
⎠ -0.8
+
0.1
∫ ⎡⎣ −2 + 3e ⎤⎦ dt
3t
0
( −2t + e )
3t
0.1
0
= 33.91 C
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Engineering Circuit Analysis, 7th Edition
12.
Chapter Two Solutions
10 March 2006
Referring to Fig. 2.28,
(a) The average current over one period (10 s) is
iavg = [-4(2) + 2(2) + 6(2) + 0(4)]/10
= 800 mA
(b) The total charge transferred over the interval 1 < t < 12 s is
qtotal =
12
∫ i(t )dt = -4(1) + 2(2) + 6(2) + 0(4) – 4(2) = 4 C
1
(c) See Fig. below
q (C)
8
-8
16
2
4
10
6
8
12
14
t(s)
16
-16
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Engineering Circuit Analysis, 7th Edition
13. (a)
VBA = –
(b) VED =
2 pJ
-1.602 × 10-19 C
Chapter Two Solutions
=
12.48 MV
0
-1.602 × 10-19 C
=
0
3 pJ
1.602 × 10-19 C
=
–18.73 MV
(c) VDC = –
10 March 2006
(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
–19
C from B to C, or +2 pJ to move +1.602x10–19 C from C to B.
+1.602x10
Thus, it requires –3 pJ + 2 pJ = –1 pJ to m ove +1.602x10 –19 C from D to C to
B.
Hence,
VDB =
−1 pJ
= –6.242 MV.
1.602 × 10-19 C
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Engineering Circuit Analysis, 7th Edition
Chapter Two Solutions
10 March 2006
14.
+
V1
–
+
–
V2
+
-
–
+
Voltmeter
Voltmeter
From the diagram, we see that V2 = –V1 = +2.86 V.
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Engineering Circuit Analysis, 7th Edition
Chapter Two Solutions
15. (a)
Pabs = (+3.2 V)(-2 mA) = –6.4 mW
(b)
Pabs = (+6 V)(-20 A) = –120 W
10 March 2006
(or +6.4 mW supplied)
(or +120 W supplied)
(d) Pabs = (+6 V)(2 ix) = (+6 V)[(2)(5 A)] = +60 W
(e) Pabs = (4 sin 1000t V)(-8 cos 1000t mA)| t = 2 ms
= +12.11 W
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Engineering Circuit Analysis, 7th Edition
16.
Chapter Two Solutions
10 March 2006
i = 3te-100t mA and v = [6 – 600t] e-100t mV
(a) The power absorbed at t = 5 ms is
[
]
Pabs = (6 − 600t ) e −100t ⋅ 3te −100t t = 5 ms
=
μW
0.01655 μW = 16.55 nW
(b) The energy delivered over the interval 0 < t < ∞ is
∞
∫ P dt
0
=
abs
∞
∫ 3t (6 − 600t ) e
− 200 t
0
dt
μJ
Making use of the relationship
∞
∫ x e
0
n
− ax
dx
=
n!
a n +1
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, 7th Edition
17.
Chapter Two Solutions
(a) Pabs = (40i)(3e-100t)| t = 8 ms =
[
[
]
2
360 e −100t t =8 ms
10 March 2006
= 72.68 W
]
2
di ⎞
⎛
(b) Pabs = ⎜ 0.2 ⎟ i = - 180 e −100t t =8 ms = - 36.34 W
dt ⎠
⎝
(
)
t
(c) Pabs = ⎛⎜ 30∫ idt + 20 ⎞⎟ 3e −100t
⎝ 0
⎠
t = 8 ms
t
= ⎛⎜ 90e −100t ∫ 3e −100t ′ dt ′ + 60e −100t ⎞⎟
= 27.63 W
0
⎝
⎠ t =8 ms
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Engineering Circuit Analysis, 7th Edition
18.
Chapter Two Solutions
10 March 2006
(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 voltage 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 corres ponds to zero voltage , and likewise, the
maximum voltage occu rs at zero curren t. The m aximum power point, therefore,
occurs somewhere between these two points. By trial and error,
Pmax is roughly (375 mV)(2.5 A) = 938 mW, or just under 1 W.
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Engineering Circuit Analysis, 7th Edition
19. (a)
Chapter Two Solutions
10 March 2006
P first 2 hours = ( 5 V )( 0.001 A ) = 5 mW
P next 30 minutes = ( ? V )( 0 A )
= 0 mW
P last 2 hours = ( 2 V )( −0.001 A ) = −2 mW
(b) Energy = (5 V)(0.001 A)(2 hr)(60 min/ hr)(60 s/ min) = 36 J
(c) 36 – (2)(0.001)(60)(60) = 21.6 J
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Engineering Circuit Analysis, 7th Edition
20.
Chapter Two Solutions
10 March 2006
Note that in the table below, only th e –4-A sou rce and the –3-A source are actually
“absorbing” power; the remaining sources are supplying power to the circuit.
Source
2 V source
8 V source
-4 A source
10 V source
-3 A source
Absorbed Power
(2 V)(-2 A)
(8 V)(-2 A)
(10 V)[-(-4 A)]
(10 V)(-5 A)
(10 V)[-(-3 A)]
The 5 powe r quantities sum to –4 – 16 +
conservation of energy.
Absorbed Power
-4W
- 16 W
40 W
- 50 W
30 W
40 – 50 + 30 = 0, as de
manded from
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Engineering Circuit Analysis, 7th Edition
Chapter Two Solutions
10 March 2006
21.
20 A
32 V
8V
–16 A
40 V
–12 A
40 V
P8V supplied
= (8)(8)
=
64 W
(source of energy)
P32V supplied
= (32)(8)
=
256 W
(source of energy)
P–16A supplied
= (40)(–16)
=
–640 W
P40V supplied
= (40)(20)
=
800 W
P–12A supplied
= (40)( –12) =
(source of energy)
–480 W
Check: ∑ supplied power = 64 + 256 – 640 + 800 – 480 = 0 (ok)
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Engineering Circuit Analysis, 7th Edition
22.
Chapter Two Solutions
10 March 2006
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 a ppropriate voltage (p assive sign conven tion), 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,
expect from the principle of conservation of energy.
as we
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Engineering Circuit Analysis, 7th Edition
23.
Chapter Two Solutions
10 March 2006
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, 5v1.
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, 7th Edition
24.
Chapter Two Solutions
10 March 2006
The voltage across the dependent source = v2 = –2ix = –2(–0.001) = 2 mV.
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Engineering Circuit Analysis, 7th Edition
25.
Chapter Two Solutions
10 March 2006
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, 7th Edition
26.
Chapter Two Solutions
10 March 2006
The supply voltage is 110 V, and the m aximum dissipated power is 500 W. The fuse s
are specified in te rms of current, so we need to determine the m aximum current th at
can flow through the fuse.
P=VI
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
delivered 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 m aximum current of 4.5 A. This
leads to a maximum power of (110)(4.5) = 495 W delivered to the application.
Although 495 W is less than the m aximum po wer allowed, this fuse will provid e
adequate protection for the app lication circuitry. If a faul t 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, 7th Edition
27. (a)
(b)
Chapter Two Solutions
imax = 5/ 900 =
5.556 mA
imin = 5/ 1100 =
4.545 mA
10 March 2006
p = v2 / R so
pmin = 25/ 1100
= 22.73 mW
pmax = 25/ 900
= 27.78 mW
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Engineering Circuit Analysis, 7th Edition
28.
Chapter Two Solutions
10 March 2006
p = i2 R, so
pmin = (0.002)2 (446.5) = 1.786 mW and (more relevant to our discussion)
pmax = (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 more appropriate.
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Engineering Circuit Analysis, 7th Edition
29. (a)
Chapter Two Solutions
10 March 2006
Pabs = i2R = [20e-12t] 2 (1200) μW
= [20e-1.2] 2 (1200) μW
=
(b)
43.54 mW
Pabs = v2/R = [40 cos 20t] 2 / 1200 W
2
=
[40 cos 2] / 1200 W
=
230.9 mW
(c)
=
Pabs = v i=
keep in mind we
are using radians
8t 1.5 W
253.0 mW
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Engineering Circuit Analysis, 7th Edition
30.
Chapter Two Solutions
10 March 2006
It’s probably best to begin this problem by sketching the voltage waveform:
v (V)
+10
40
60
t (ms)
20
-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) pabs max =
2
vmax
= (10)2 / 50 = 2 W
R
(e) pabs avg =
⎤
1 ⎡ (+10) 2
(−10) 2
⋅
20
+
⋅ 20⎥ = 2 W
⎢
R
40 ⎣ R
⎦
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Engineering Circuit Analysis, 7th 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 term inal of the
voltage
source.
The power supplied by the voltage source is Vs I.
The power absorbed by resistor R1 is I2R1.
The power absorbed by resistor R2 is I2R2.
Since we know that the total power supplied is equal to the total power absorbed,
we may write:
Vs I = I2R1 + I2R2
or
Vs = I R1 + I R2
Vs = I (R1 + R2)
By Ohm’s law,
I = VR2 / R2
so that
Vs =
VR2
R2
(R1 + R2 )
Solving for VR2 we find
VR2 = Vs
R2
(R1 + R2 )
Q.E.D.
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Engineering Circuit Analysis, 7th Edition
Chapter Two Solutions
10 March 2006
32. (a)
6
5
4
c urrent (m A )
3
2
1
0
-1
-2
-3
-4
-1.5
-1
-0.5
0
0.5
voltage (V )
1
1.5
2
2.5
(b) W e see from our answer to part (a ) that this device has a reasonably linear
characteristic (a not unreasonable degree of e xperimental error 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)
2
1.5
c urrent (m A )
1
0.5
0
-0.5
-1
-1.5
-1.5
-1
-0.5
0
0.5
voltage (V )
1
1.5
2
2.5
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Engineering Circuit Analysis, 7th Edition
33.
Chapter Two Solutions
10 March 2006
Top Left Circuit:
I = (5/10) mA = 0.5 mA, and P10k = V2/10 mW = 2.5 mW
Top Right Circuit:
I = -(5/10) mA = -0.5 mA, and P10k = V2/10 mW = 2.5 mW
Bottom Left Circuit: I = (-5/10) mA = -0.5 mA, and P10k = V2/10 mW = 2.5 mW
Bottom Right Circuit: I = -(-5/10) mA = 0.5 mA, and P10k = V2/10 mW = 2.5 mW
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34. The
Chapter Two Solutions
10 March 2006
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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35.
Chapter Two Solutions
10 March 2006
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, 7th Edition
36.
Chapter Two Solutions
10 March 2006
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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37.
tem
Chapter Two Solutions
10 March 2006
We need to create a 47 0-Ω resistor from 28 AWG wire , knowing that the am bient
perature is 108oF, or 42.22oC.
Referring to Table 2.3, 28 AWG wire is 65.3 m Ω/ft at 20 oC, and using the equation
provided we compute
R2/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 mΩ/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 shoul d 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, 7th Edition
38.
Chapter Two Solutions
10 March 2006
We are given that the conductivity σ of copper is 5.8×107 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 r2 = π (5.12×10-4 m)2 = 8.236×10-7 m2
Thus, R = (15.24 m)/[( 5.8×107)( 8.236×10-7)] = 319.0 mΩ
(b) We assume that the conductivity value specified also holds true at 50oC.
The cross-sectional area of the foil is
A = (33 μm)(500 μm)(1 m/106 μm)( 1 m/106 μm) = 1.65×10-8 m2
So that
R = (15 cm)(1 m/100 cm)/[( 5.8×107)( 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
I2R = (3)2 (156.7) mW = 1.410 W
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Engineering Circuit Analysis, 7th Edition
39.
Chapter Two Solutions
10 March 2006
Since R = ρ l / A, it follows that ρ = R A/ l .
From Table 2.4, we see that 28 AWG soft copper wire (cross-sectional area = 0.080 4
mm2) 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 mm2.
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, 7th Edition
40.
Chapter Two Solutions
10 March 2006
(a)
From the text,
(1) Zener diodes,
(2) Fuses, and
(3) Incandescent (as opposed to fluorescent) light bulbs
This last on e requires a few facts to be put together. We have st ated 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, converted 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, 7th Edition
41.
Chapter Two Solutions
10 March 2006
The quoted resistivity ρ of B33 copper is 1.7654 μΩ.cm.
A = πr2 = π(10–3)2 = 10–6π m2. l = 100 m.
(1.7654 ×10 Ωcm ) (1 m/100 cm )(100 m ) = 0.5619 Ω
Thus, R = ρ l / A =
−6
10−6 π
And P = I2R = (1.5)2 (0.5619) = 1.264 W.
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42.
Chapter Two Solutions
10 March 2006
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 l will change the resistance of our “device.”
A simple variable resistor concept, then:
Leads to
connect to
circuit
Copper wire
Rotating sh ort
wire de
termines length
of long wire
used in circuit.
But this is s omewhat impractical, as the l eads may turn out to have alm ost the sam e
resistance unless we have a very long wi
re, which can also be im practical. One
improvement would be to replace the coppe r 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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43.
Chapter Two Solutions
10 March 2006
(a) We need to plot the negative and positive voltage ranges separately, as the positive
voltage range is, after all, exponential!
x 10
-6
10
16
-4
14
12
10
-5
c urrent (A )
c urrent (A)
10
8
6
10
-6
4
10
2
-7
0
-2
-0.7
10
-0.6
-0.5
-0.4
-0.3
-0.2
voltage (V )
-0.1
0
-8
0.1
0
(b) To dete rmine the resist ance of the device at
corresponding current:
0.01
0.02
0.03
0.04
voltage (V)
V = 550 mV, we com
0.05
0.06
pute the
I = 10-9 [e39(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 [e39I – 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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0.07
Engineering Circuit Analysis, 7th Edition
44.
Chapter Two Solutions
10 March 2006
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 o f the values listed e liminates 2, 4 a nd 6 AW G w ire as the ir respec tive
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-AW G wire weight less? Again referring to T able 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 mm2. The volume of 12-AWG wire required is therefore 6345900 mm 3, and that
of 28-AWG wire required is only 3750 mm3.
The best (but not the only) choice for a portable application is clear: 28-AWG wire!
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45.
cm
Chapter Two Solutions
10 March 2006
Our target is a 100-Ω resistor. We see from the plot that at ND = 1015 cm-3, μn ~ 2x103
2
/V-s, yielding a resistivity of 3.121 Ω-cm.
At ND = 1018 cm-3, μn ~ 230 cm2/ V-s, yielding a resistivity of 0.02714 Ω-cm.
Thus, we s ee that the lower doping level cl
early provides m aterial 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 1015 cm-3, we need only define the resistor geom etry to complete the
design.
We choose a geometry as shown in the figure; our contact area is arbitrarily chosen as
100 μm by 250 μm, so that only the length L remains to be specified. Solving,
L=
R
ρ
A=
(100 Ω)(100 μ
m )(250 μ
m )
= 80.1 μ m
4
( 3.121 Ω cm ) 10 μ m/cm
(
)
Design summary (one possibility):
Contact
ND = 1015 cm-3
L = 80.1 μm
width
= 100 μm
(Note: this is som ewhat atypical; in the semiconductor 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.)
100 μm
Wafer surface
250 μm
contact
80.1 μm
contact
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Chapter Three Solutions
10 March 2006
1.
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2.
Chapter Three Solutions
10 March 2006
(a) six nodes; (b) nine branches.
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3.
Chapter Three Solutions
10 March 2006
(a) Four nodes; (b) five branches; (c) path, yes – loop, no.
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4.
Chapter Three Solutions
10 March 2006
(a) Five nodes;
(b) seven branches;
(c) path, yes – loop, no.
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5.
Chapter Three Solutions
10 March 2006
(a) The number of nodes remains the same – four (4).
(b) The number of nodes is increased by one – to five (5).
(c)
i)
ii)
iii)
iv)
v)
YES
NO
YES
NO
NO
– does not return to starting point
– does not return to starting point
– point B is crossed twice
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6.
(a) By KCL at the bottom node:
So
iZ = 9 A.
Chapter Three Solutions
10 March 2006
2 – 3 + iZ – 5 – 3 = 0
(b) If the left-most resistor has a value of 1 Ω, then 3 V appears across the parallel
network (the ‘+’ reference terminal being the bottom node) Thus, the value of the
other resistor is given by
R=
3
= 600 mΩ .
−(−5)
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7.
Chapter Three Solutions
10 March 2006
(a) 3 A;
(b) –3 A;
(c) 0.
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8.
Chapter Three Solutions
10 March 2006
By KCL, we may write:
5 + iy + iz = 3 + ix
(a) ix = 2 + iy + iz = 2 + 2 + 0 = 4 A
(b) iy = 3 + ix – 5 – iz
iy = –2 + 2 – 2 iy
Thus, we find that iy = 0.
(c) 5 + iy + iz = 3 + ix
→ 5 + ix + ix = 3 + ix so ix = 3 – 5 = -2A.
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9.
Chapter Three Solutions
10 March 2006
Focusing our attention on the bottom left node, we see that ix = 1 A.
Focusing our attention next on the top right node, we see that iy = 5 A.
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10.
Chapter Three Solutions
10 March 2006
We obtain the current each bulb draws by dividing its power rating by the operating
voltage (115 V):
I100W = 100/115 = 896.6 mA
I60W = 60/115 = 521.7 mA
I40W = 347.8 mA
Thus, the total current draw is 1.739 A.
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11.
Chapter Three Solutions
10 March 2006
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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12.
Chapter Three Solutions
10 March 2006
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 the sign could be read and customers will know the restaurant is open for
business.
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13.
Chapter Three Solutions
10 March 2006
(a) vy = 1(3vx + iz)
vx = 5 V and given that iz = –3 A, we find that
vy = 3(5) – 3 = 12 V
(b) vy = 1(3vx + iz) = –6 = 3vx + 0.5
Solving, we find that vx = (–6 – 0.5)/3 = –2.167 V.
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14.
Chapter Three Solutions
10 March 2006
(a) ix = v1/10 + v1/10 = 5
2v1 = 50
so
v1 = 25 V.
By Ohm’s law, we see that iy = v2/10
also, using Ohm’s law in combination with KCL, we may write
ix = v2/10 + v2/10 = iy + iy = 5 A
Thus,
iy = 2.5 A.
(b) From part (a), ix = 2 v1/ 10. Substituting the new value for v1, we find that
ix = 6/10 = 600 mA.
Since we have found that iy = 0.5 ix,
iy = 300 mA.
(c) no value – this is impossible.
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15.
Chapter Three Solutions
10 March 2006
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 IX of 12.5 A.
(1) By KVL, –40 + (–110) + R(5) – 20 = 0
Thus, R = 34 Ω.
(2) By KVL, -VG – (-110) + 40 = 0
So
VG = 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 means to find this current: The current flowing into the “+”
terminal of the –110-V source is 12.5 + 6 = 18.5 A.
Then,
Ix = 18.5 – 5 = 13.5 A
By Ohm’s law,
I x = G · VG
So G = 13.5/ 150
or
G = 90 mS
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16.
Chapter Three Solutions
10 March 2006
(a) -1 + 2 + 10i – 3.5 + 10i = 0
Solving,
i = 125 mA
(b) +10 + 1i - 2 + 2i + 2 – 6 + i = 0
Solving, we find that 4i = -4 or i = - 1 A.
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17.
Chapter Three Solutions
10 March 2006
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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18.
Chapter Three Solutions
10 March 2006
Begin by defining a clockwise current i.
-vS + v1 + v2 = 0
and hence i =
so
vS = v1 + v2 = i(R1 + R2)
vS
.
R1 + R2
Thus, v1 = R1i =
R1
R2
vS and v2 = R2i =
vS .
R1 + R2
R1 + R2
Q.E.D.
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19.
Chapter Three Solutions
10 March 2006
Given: (1) Vd = 0 and (2) no current flows into either terminal of Vd.
Calculate Vout by writing two KVL equations.
Begin by defining current i1 flowing right through the 100 Ω resistor, and i2 flowing
right through the 470 Ω resistor.
-5 + 100i1 + Vd = 0
[1]
-5 + 100i1 + 470i2 + Vout = 0 [2]
Making use of the fact that in this case Vd = 0, we find that i1 = 5/100 A.
Making use of the fact that no current flows into the input terminals of the op amp,
i1 = i2. Thus, Eq. [2] reduces to
-5 + 570(5/100) + Vout = 0 or
Vout = -23.5 V
(hence, the circuit is acting as a voltage amplifier.)
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20.
Chapter Three Solutions
10 March 2006
(a) By KVL, -2 + vx + 8 = 0
so that
vx = -6 V.
(b) By KCL at the top left node,
iin = 1 + IS + vx/4 – 6
or
iin = 23 A
(c) By KCL at the top right node,
IS + 4 vx = 4 - vx/4
So
IS = 29.5 A.
(d) The power provided by the dependent source is 8(4vx) = -192 W.
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21.
(a) Working from left to right,
v1 = 60 V
v2 = 60 V
i2 = 60/20 = 3 A
i4 = v1/4 = 60/4 = 15 A
v3 = 5i2 = 15 V
By KVL, -60 + v3 + v5 = 0
v5 = 60 – 15 = 45 V
v4 = v5 = 45
Chapter Three Solutions
v1 = 60 V
v2 = 60 V
v3 = 15 V
v4 = 45 V
v5 = 45 V
10 March 2006
i1 = 27 A
i2 = 3 A
i3 = 24 A
i4 = 15 A
i5 = 9 A
i5 = v5/5 = 45/5 = 9 A
i3 = i4 + i5 = 15 + 9 = 24 A
i1 = i2 + i3 = 3 + 24 = 27
(b) It is now a simple matter to compute the power absorbed by each element:
p1
p2
p3
p4
p5
= -v1i1
= v2i2
= v3i3
= v4i4
= v5i5
= -(60)(27)
= (60)(3)
= (15)(24)
= (45)(15)
= (45)(9)
= -1.62 kW
= 180 W
= 360 W
= 675 W
= 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, 7th Edition
22.
Chapter Three Solutions
10 March 2006
Refer to the labeled diagram below.
Beginning from the left, we find
p20V = -(20)(4) = -80 W
v1.5 = 4(1.5) = 6 V
therefore p1.5 = (v1.5)2/ 1.5 = 24 W.
v14 = 20 – v1.5 = 20 – 6 = 14 V therefore p14 = 142/ 14 = 14 W.
i2 = v2/2 = v1.5/1.5 – v14/14 = 6/1.5 – 14/14 = 3 A
Therefore v2 = 2(3) = 6 V and p2 = 62/2 = 18 W.
v4 = v14 – v2 = 14 – 6 = 8 V therefore
p4 = 82/4 = 16 W
i2.5 = v2.5/ 2.5 = v2/2 – v4/4 = 3 – 2 = 1 A
Therefore v2.5 = (2.5)(1) = 2.5 V and so
p2.5 = (2.5)2/2.5 = 2.5 W.
I2.5 = - IS, thefore IS = -1 A.
KVL allows us to write
-v4 + v2.5 + vIS = 0
so VIS = v4 – v2.5 = 8 – 2.5 = 5.5 V and
pIS = -VIS IS = 5.5 W.
A quick check assures us that these power quantities sum to zero.
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23.
Chapter Three Solutions
10 March 2006
Sketching the circuit as described,
(a) v14 = 0.
v13 = v43
v23 = -v12 – v34 = -12 + 8
v24 = v23 + v34 = -4 – 8
= 8V
= -4 V
= -12 V
(b) v14 = 6 V. v13 = v14 + v43 = 6 + 8
v23 = v13 – v12 = 14 – 12
v24 = v23 + v34 = 2 – 8
= 14 V
= 2V
= -6 V
(c) v14 = -6 V. v13 = v14 + v43 = -6 + 8
v23 = v13 – v12 = 2 – 12
v24 = v23 + v34 = -10 – 8
= 2V
= -10 V
= -18 V
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24.
Chapter Three Solutions
10 March 2006
(a) By KVL, -12 + 5000ID + VDS + 2000ID = 0
Therefore,
VDS = 12 – 7(1.5) = 1.5 V.
(b) By KVL, - VG + VGS + 2000ID = 0
Therefore,
VGS = VG – 2(2) = -1 V.
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Engineering Circuit Analysis, 7th Edition
25.
Chapter Three Solutions
10 March 2006
Applying KVL around this series circuit,
-120 + 30ix + 40ix + 20ix + vx + 20 + 10ix = 0
where vx is defined across the unknown element X, with the “+” reference on top.
Simplifying, we find that 100ix + vx = 100
To solve further we require specific information about the element X and its
properties.
(a) if X is a 100-Ω resistor,
vx = 100ix so we find that 100 ix + 100 ix = 100.
Thus
ix = 500 mA and
px = vx ix = 25 W.
(b) If X is a 40-V independent voltage source such that vx = 40 V, we find that
ix = (100 – 40) / 100 = 600 mA and
px = vx ix = 24 W
(c) If X is a dependent voltage source such that vx = 25ix,
ix = 100/125 = 800 mA and
px = vx ix = 16 W.
(d) If X is a dependent voltage source so that vx = 0.8v1,
where v1 = 40ix, we have
100 ix + 0.8(40ix) = 100
px = vx ix = 0.8(40)(0.7576)2 = 18.37 W.
or ix = 100/132 = 757.6 mA and
(e) If X is a 2-A independent current source, arrow up,
100(-2) + vx = 100
so that vx = 100 + 200 = 300 V and
px = vx ix = -600 W
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26.
Chapter Three Solutions
10 March 2006
(a) We first apply KVL:
-20 + 10i1 + 90 + 40i1 + 2v2 = 0
where v2 = 10i1. Substituting,
70 + 70 i1 = 0
or
i1= -1 A.
(b) Applying KVL,
-20 + 10i1 + 90 + 40i1 + 1.5v3 = 0
[1]
where
v3 = -90 – 10i1 + 20 = -70 – 10 i1
alternatively, we could write
v3 = 40i1 + 1.5v3 = -80i1
Using either expression in Eq. [1], we find
i1 = 1 A.
(c) Applying KVL,
-20 + 10i1 + 90 + 40i1 - 15 i1 = 0
Solving, i1 = - 2A.
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27.
Chapter Three Solutions
10 March 2006
Applying KVL, we find that
-20 + 10i1 + 90 + 40i1 + 1.8v3 = 0
[1]
Also, KVL allows us to write
v3 = 40i1 + 1.8v3
v3 = -50i1
So that we may write Eq. [1] as
50i1 – 1.8(50)i1 = -70
or i1 = -70/-40 = 1.75 A.
Since v3 = -50i1 = -87.5 V, no further information is required to determine its value.
The 90-V source is absorbing (90)(i1) = 157.5 W of power and the dependent source
is absorbing (1.8v3)(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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28.
Chapter Three Solutions
10 March 2006
(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 may calculate the necessary resistance
R = [13 – 10.5 – 0.055(4)]/ 4 = 570 mΩ
(b) The total power delivered to the battery consists of the power absorbed by the
0.035-Ω resistance (0.035i2), and the power absorbed by the 10.5-V ideal battery
(10.5i). Thus, we need to solve the quadratic equation
0.035i2 + 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 the 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 means we need
R = [13 – 10.5 – 0.055(14.29)]/ 14.29 = 119.9 mΩ
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29.
Chapter Three Solutions
10 March 2006
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 Vbattery = 13 – (0.02 + 0.5)i = 10.43 V.
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30.
Chapter Three Solutions
10 March 2006
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 – (2v3 + v2) + (4v1 – v2) = 0
[1]
where we have defined i to be flowing in the clockwise direction, and
v1 = 5i, v2 = 25i, and v3 = 20i.
Performing the necessary substition, Eq. [1] becomes
50i - (40i + 25i) + (20i – 25i) = 40
so that i = 40/-20 = -2 A
Computing the absorbed power is now a straightforward matter:
p40V
p5Ω
p25Ω
p20Ω
pdepsrc1
pdepsrc2
= (40)(-i)
= 5i2
= 25i2
= 20i2
= (2v3 + v2)(-i) = (40i + 25i)
= (4v1 - v2)(-i) = (20i - 25i)
= 80 W
= 20 W
= 100 W
= 80 W
= -260 W
= -20 W
and we can easily verify that these quantities indeed sum to zero as expected.
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Engineering Circuit Analysis, 7th Edition
31.
Chapter Three Solutions
10 March 2006
We begin by defining a clockwise current i.
(a) i = 12/(40 + R) mA, with R expressed in kΩ.
We want i2 · 25 = 2
2
or
⎛ 12 ⎞
⎜
⎟ ⋅ 25 = 2
⎝ 40 + R ⎠
Rearranging, we find a quadratic expression involving R:
R2 + 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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32.
Chapter Three Solutions
10 March 2006
By KVL,
–12 + (1 + 2.3 + Rwire 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 negligible compared to the other resistances in the circuit!
Thus,
i = 12/(1 + 2.3 + 48.6) = 231.2 mA
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33.
Chapter Three Solutions
10 March 2006
We can apply Ohm’s law to find an expression for vo:
vo = 1000(–gm vπ)
We do not have a value for vπ, but KVL will allow us to express that in terms of vo,
which we do know:
–10×10–3 cos 5t + (300 + 50×103) i = 0
where i is defined as flowing clockwise.
Thus,
vπ = 50×103 i
= 50×103 (10×10–3 cos 5t) / (300 + 50×103)
= 9.940×10–3 cos 5t V
and we by substitution we find that
vo
= 1000(–25×10–3)( 9.940×10–3 cos 5t )
= –248.5 cos 5t mV
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34.
Chapter Three Solutions
10 March 2006
By KVL, we find that
–3 + 100 ID + VD = 0
–3
Substituting ID = 3×10–6(eVD / 27×10 – 1), we find that
–3
–3 + 300×10–6(eVD / 27×10 – 1) + VD = 0
This is a transcendental equation. Using a scientific calculator or a numerical
software package such as MATLAB®, we find
VD = 246.4 mV
Let’s assume such software-based assistance is unavailable. In that case, we need to
“guess” a value for VD, 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
0
1
0.5
0.25
0.245
0.248
0.246
RESULT
–3
3.648×1012
3.308×104
0.4001
–0.1375
0.1732
–0.0377
oops
better
At this point, the error is
getting much smaller, and
our confidence is increasing
as to the value of VD.
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Engineering Circuit Analysis, 7th Edition
35.
Chapter Three Solutions
10 March 2006
Define a voltage vx, “+” reference on the right, across the dependent current source.
Note that in fact vx 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 – vx / 100 – vx / 25 + 0.8 ix = 0
[1]
This represents one equation in two unknowns. A second equation to introduce at this
point is
ix = vx /25 so that Eq. [1] becomes
5 – vx / 100 – vx / 25 + 0.8 (vx / 25) = 0
Solving for vx, we find vx = 277.8 V. It is a simple matter now to compute the power
absorbed by each element:
P5A
= –5 vx
= –1.389 kW
P100Ω
= (vx)2 / 100
=
771.7 W
P25Ω
2
=
3.087 kW
Pdep
= (vx) / 25
2
= –vx(0.8 ix) = –0.8 (vx) / 25
= –2.470 kW
A quick check assures us that the calculated values sum to zero, as they should.
(b) Again summing the currents into the right–hand node,
5 – vx / 100 – vx / 25 + 0.8 iy = 0
[2]
where iy = 5 – vx/100
Thus, Eq. [2] becomes
5 – vx / 100 – vx / 25 + 0.8(5) – 0.8 (iy) / 100 =
0
Solving, we find that vx x = 155.2 V and iy = 3.448 A
So that
P5A
= –5 vx
= –776.0 W
P100Ω
= (vx)2 / 100
=
240.9 W
P25Ω
2
= (vx) / 25
=
963.5 W
Pdep
= –vx(0.8 iy)
= –428.1 W
A quick check shows us that the calculated values sum to 0.3, which is reasonably
close to zero compared to the size of the terms (small roundoff errors accumulated).
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Engineering Circuit Analysis, 7th Edition
36.
Chapter Three Solutions
10 March 2006
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 + 3i1 + 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
i1
or
i1
=
=
3 i1 + v/2,000
–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
i1 = 571.4 μA
Since ix = i1, we find that ix = 571.4 μA.
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37.
Chapter Three Solutions
10 March 2006
Define a voltage vx with its “+” reference at the center node. Applying KCL and
summing the currents into the center node,
8 – vx /6 + 7 – vx /12 – vx /4 = 0
Solving, vx = 30 V.
It is now a straightforward matter to compute the power absorbed by each element:
P8A
= –8 vx
P6Ω
2
= (vx) / 6
=
P8A
= –7 vx
= –210 W
P12Ω
2
= (vx) / 12
=
75 W
P4Ω
2
=
225 W
= (vx) / 4
= –240 W
150 W
and a quick check verifies that the computed quantities sum to zero, as expected.
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38.
Chapter Three Solutions
10 March 2006
(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)(4×106 / 5×103) = 40 V
P4kΩ = v2/4000 = 400 mW
and
(b) Once again, we first 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 – 20×10–3 = v/1000
Solving,
v = 30 V
P20mA = v · 20×10–3 = 600 mW
and
(c) Once again, we first 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 – 2ix = v/1000
where
ix = 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,
Pdep = v · 2ix
= 555.8 mW
(d) We note that ix = 60/1000 = 60 mA. KCL stipulates that (viewing currents
into and out of the top node)
80 – 30 + is = ix = 60
Thus, is = 10 mA
and
P60V = 60(–10) mW = –600 mW
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39.
Chapter Three Solutions
10 March 2006
(a) To cancel out the effects of both the 80-mA and 30-mA sources, iS must be set to
iS = –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/1×103 + iS = 0
Simplifying slightly, this becomes
50 - vS + 103 iS = 0
[1]
We are seeking a value for vS such that vS · iS = 0. Clearly, setting vS = 0 will achieve
this. From Eq. [1], we also see that setting vS = 50 V will work as well.
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40.
Chapter Three Solutions
10 March 2006
Define a voltage v9 across the 9-Ω resistor, with the “+” reference at the top node.
(a) Summing the currents into the right-hand node and applying KCL,
5 + 7 = v9 / 3 + v9 / 9
Solving, we find that v9 = 27 V. Since ix = v9 / 9,
ix = 3 A.
(b) Again, we apply KCL, this time to the top left node:
2 – v8 / 8 + 2ix – 5 = 0
Since we know from part (a) that ix = 3 A, we may calculate v8 = 24 V.
(c) p5A = (v9 – v8) · 5 = 15 W.
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41.
Chapter Three Solutions
10 March 2006
Define a voltage vx across the 5-A source, with the “+” reference on top.
Applying KCL at the top node then yields
5 + 5v1 - vx/ (1 + 2) – vx/ 5 = 0
[1]
where v1 = 2[vx /(1 + 2)] = 2 vx / 3.
Thus, Eq. [1] becomes
5 + 5(2 vx / 3) – vx / 3 – vx / 5 = 0
or 75 + 50 vx – 5 vx – 3 vx = 0, which, upon solving, yields vx = -1.786 V.
The power absorbed by the 5-Ω resistor is then simply (vx)2/5 = 638.0 mW.
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42.
Chapter Three Solutions
10 March 2006
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.
Summing 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 computed as:
p2A = -v(2) = 36.67 W
p6A = -v(6) = 110 W
p3A = -v(3) = 55 W
The power absorbed by each resistor is simply v2/5 = 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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43.
Chapter Three Solutions
10 March 2006
Defining a voltage Vx across the 10-A source with the “+” reference at the top node,
KCL tells us that 10 = 5 + I1Ω, where I1Ω is defined flowing downward through the
1-Ω resistor.
Solving, we find that I1Ω = 5 A, so that Vx = (1)(5) = 5 V.
So, we need to solve
Vx = 5 = 5(0.5 + Rsegment)
with Rsegment = 500 mΩ.
From Table 2.3, we see that 28-AWG solid copper wire has a resistance of 65.3
mΩ/ft. Thus, the total number of miles needed of the wire is
500 mΩ
(65.3 mΩ/ft)(5280 ft/mi)
= 1.450 × 10-3 miles
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44.
Chapter Three Solutions
10 March 2006
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 + iS = 0
or
iS = -10.2 A.
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45.
Chapter Three Solutions
(a) Applying KCL, 1 – i – 3 + 3 = 0
10 March 2006
so i = 1 A.
(b) Looking at the left part of the circuit, 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 = 9A
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46.
Chapter Three Solutions
10 March 2006
(a) We may redraw the circuit as
Then, we see that v = (1)(1) = 1 V.
(b) We may combine all sources to the right of the 1-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.
Chapter Three Solutions
10 March 2006
(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 top. 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 may 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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Engineering Circuit Analysis, 7th Edition
48.
Chapter Three Solutions
10 March 2006
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 may 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,
P16Ω = v2/16 =
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 goes for the 7 A
source). Thus,
P2 A abs = -15(2) = -30 W
P7 A abs = -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,
P5V abs = -5(6.875) = -34.38 W
P10V abs = -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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49.
Chapter Three Solutions
10 March 2006
We can combine the voltage sources such that
i = vS/ 14
(a) vS = 10 + 10 – 6 – 6 = 20 –12 = 8
Therefore
i = 8/14 = 571.4 mA.
(b) vS = 3 + 2.5 – 3 – 2.5 = 0
Therefore
i = 0.
(c) vS = -3 + 1.5 – (-0.5) – 0 = -1 V
Therefore
i = -1/14 = -71.43 mA.
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50.
Chapter Three Solutions
10 March 2006
We first simplify as shown, making use of the fact that we are told ix = 2 A to find the
voltage across the middle and right-most 1-Ω resistors as labeled.
By KVL, then, we find that
v1 = 2 + 3 = 5 V.
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51.
Chapter Three Solutions
10 March 2006
We see that to determine the voltage v we will need vx due to the presence of the
dependent current soruce. So, let’s begin with the right-hand side, where we find that
vx = 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 vx = v/10×103
or
v = -515 V.
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52.
Chapter Three Solutions
10 March 2006
(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
P2V
P4V
P-3V
= -2(-5)
= -(-4)(4)
= -(-9)(-3)
= 10 W
= 16 W
= -27 W
For the current sources,
So that the absorbed power is
P-5A
P-4A
P3A
P12A
= +(-5)(6)
= -(-4)(4)
= -(3)(7)
= -(12)(-3)
= -30 W
= 16 W
= -21 W
= 36 W
A quick check assures us that these absorbed powers sum to zero as they should.
(b) We need to change the 4-V source such that the voltage across the –5-A source
drops to zero. Define Vx across the –5-A source such that the “+” reference terminal is
on the left. Then,
-2 + Vx – Vneeded = 0
or Vneeded = -2 V.
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53.
Chapter Three Solutions
10 March 2006
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 equivalent 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
Req = 1 kΩ.
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54.
Chapter Three Solutions
10 March 2006
(a) We see 1Ω || (1 Ω + 1 Ω) || (1 Ω + 1 Ω + 1 Ω)
= 1Ω || 2 Ω || 3 Ω
= 545.5 mΩ
(b) 1/Req = 1 + 1/2 + 1/3 + … + 1/N
Thus, Req = [1 + 1/2 + 1/3 + … + 1/N]-1
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55.
Chapter Three Solutions
10 March 2006
(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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56.
Chapter Three Solutions
10 March 2006
(a) no simplification is possible using only source and/or resistor combination
techniques.
(b) We first simplify the circuit to
5Ω
0V
5Ω
1A
7Ω
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:
1A
2.917 Ω
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57.
Req
Chapter Three Solutions
10 March 2006
= 1 kΩ + 2 kΩ || 2 kΩ + 3 kΩ || 3 kΩ + 4 kΩ || 4 kΩ
= 1 kΩ + 1 k Ω + 1.5 kΩ + 2 kΩ
= 5.5 kΩ.
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58.
Chapter Three Solutions
10 March 2006
(a) Working 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:
100 Ω
Req →
100 Ω
100 Ω
100 Ω
100 Ω
100 Ω
250 Ω
100 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,
Req = 100 Ω || 273.1 Ω = 73.20 Ω.
(b) First, we combine 24 Ω || (50 Ω + 40 Ω) || 60 Ω = 14.4 Ω, which leaves us with
5Ω
10 Ω
Req →
14.4 Ω
20 Ω
Thus, Req = 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:
15 Ω
10 Ω
2Ω
50 Ω
8Ω
20 Ω
30 Ω
We now see we have (10 Ω + 15 Ω) || 50 Ω = 16.67 Ω. Redrawing once again,
16.67 Ω
2Ω
8Ω
50 Ω
where the equivalent resistance is seen to be 2 Ω + 50 Ω || 16.67 Ω + 8 Ω = 22.5 Ω.
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59.
Chapter Three Solutions
10 March 2006
(a) Req = [(40 Ω + 20 Ω) || 30 Ω + 80 Ω] || 100 Ω + 10 Ω
=
60 Ω.
(b) Req = 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 = R2 + 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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60.
Chapter Three Solutions
10 March 2006
(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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61.
Chapter Three Solutions
10 March 2006
Req = [(5 Ω || 20 Ω) + 6 Ω] || 30 Ω + 2.5 Ω
= 10 Ω
The source therefore provides a total of 1000 W and a current of 100/10 = 10 A.
P2.5Ω = (10)2 · 2.5 = 250 W
V30Ω = 100 - 2.5(10) = 75 V
P30Ω = 752/ 30 = 187.5 W
I6Ω = 10 – V30Ω /30 = 10 – 75/30 = 7.5 A
P6Ω = (7.5)2 · 6 = 337.5 W
V5Ω = 75 – 6(7.5) = 30 V
P5Ω = 302/ 5 = 180 W
V20Ω = V5Ω = 30 V
P20Ω = 302/20 = 45 W
We check our results by verifying that the absorbed powers in fact add to 1000 W.
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62.
Chapter Three Solutions
10 March 2006
To begin with, the 10-Ω and 15-Ω resistors are in parallel ( = 6 Ω), and so are the
20-Ω and 5-Ω resistors (= 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:
9A
-
vx
+
14 Ω
6Ω
4Ω
6Ω
Next, we note that (14 Ω + 6 Ω) || (4 Ω + 6 Ω) = 6.667 Ω
so that
vx = 9(6.667) = 60 V
and
ix = -60/10 = -6 A.
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63.
Chapter Three Solutions
10 March 2006
(a) Working from right to left, and borrowing x || y notation from resistance
calculations to indicate the operation xy/(x + y),
Gin
= {[(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
= {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,
22.22 mS
13.64 mS
Gin →
100 mS
we see that Gin = 10 + 22.22 + 13.64 = 135.9 mS.
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64.
Chapter Three Solutions
10 March 2006
The bottom four resistors between the 2-Ω resistor and the 30-V source are shorted
out. The 10-Ω and 40-Ω resistors 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
P1Ω = I2 · 1 = 18.37 W
To compute P10Ω, consider that since the 10-Ω and 40-Ω resistors are in parallel, the
same voltage Vx (“+” reference on the left) appears across both resistors. The current I
= 4.286 A flows into this combination. Thus, Vx = (8)(4.286) = 34.29 V and
P10Ω = (Vx)2 / 10 = 117.6 W.
P13Ω = 0 since no current flows through that resistor.
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65.
Chapter Three Solutions
10 March 2006
With the meter being a short circuit and no current flowing through it, we can write
R1i1 = R2i2
R3i3 = RiR
→
R1i1 R2i2
[1]
=
R3i3 RiR
And since i1 = i3, and i2 = iR, Eq [1] becomes
R
R1 R2
, or R = R2 3 . Q.E.D.
=
R3 R
R1
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66.
Chapter Three Solutions
10 March 2006
The total resistance in the series string sums to 75 Ω.
The voltage dropped across the 2.2 Ω resistor is
2.2
= 293.3 mV
75
and the voltage dropped across the 47 Ω resistor is
V2.2Ω = 10
V47Ω = 10
47
= 6.267 mV
75
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67.
Chapter Three Solutions
10 March 2006
We first note that the 4.7 kΩ and 2.2 kΩ resistors can be combined into a single 1.5
kΩ resistor, which is then in series with the 10 kΩ resistor. Next we note that the 33
kΩ / 47 kΩ parallel combination can be replaced by a 19.39 kΩ resistance, which is
in series with the remaining 33 kΩ resistor.
By voltage division, then, and noting that V47kΩ is the same voltage as that across the
19.39 kΩ resistance,
19.39
V47kΩ = 2
= 607.0 mV
10 + 1.5 + 33 + 19.39
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68.
Chapter Three Solutions
10 March 2006
(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,
1
100
= 571.4 mA
I 33Ω = 12
1 +1 +1
10 10 100
(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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69.
Chapter Three Solutions
10 March 2006
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,
1
10 + 15
= 22.12 A
I15Ω = ( 30 − 8 )
1
1
+
10 + 15 8.479
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Engineering Circuit Analysis, 7th Edition
70.
Chapter Three Solutions
10 March 2006
One possible solution of many:
vS = 2(5.5) = 11 V
R1 = R2 = 1 kΩ.
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71.
Chapter Three Solutions
10 March 2006
One possible solution of many:
iS = 11 mA
R1 = R2 = 1 kΩ.
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72.
Chapter Three Solutions
10 March 2006
p15Ω = (v15)2 / 15×103 A
v15 = 15×103 (-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.
Chapter Three Solutions
10 March 2006
Replace the top 10 kΩ, 4 kΩ and 47 kΩ resistors with 10 kΩ + 4 kΩ || 47 kΩ =
13.69 kΩ.
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/ 15×103 = 868.3 μW.
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74.
Chapter Three Solutions
10 March 2006
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 resistors 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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75.
Chapter Three Solutions
10 March 2006
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 kΩ 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-kΩ resistor:
0.5v1
47
47 + 100 || 20
=
0.5(2.5)(47)
47 + 16.67
= 0.9228 V
So that p47kΩ = (0.9928)2 / 47×103 = 18.12 μW
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76.
Chapter Three Solutions
10 March 2006
The temptation to write an equation such as
v1 = 10
20
20 + 20
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 kΩ resistors.
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77.
Chapter Three Solutions
10 March 2006
R 2 || (R 3 + R 4 )
R 1 + [R 2 || (R 3 + R 4 )]
R 2 (R 3 + R 4 ) (R 2 + R 3 + R 4 )
= VS
R 1 + R 2 (R 3 + R 4 ) (R 2 + R 3 + R 4 )
R 2 (R 3 + R 4 )
= VS
R 1 (R 2 + R 3 + R 4 ) + R 2 (R 3 + R 4 )
(a) v2 = VS
R1
R 1 + [R 2 || (R 3 + R 4 )]
R1
= VS
R 1 + R 2 (R 3 + R 4 ) (R 2 + R 3 + R 4 )
R 1 (R 2 + R 3 + R 4 )
= VS
R 1 (R 2 + R 3 + R 4 ) + R 2 (R 3 + R 4 )
(b) v1 = VS
⎞
⎛ v ⎞⎛
R2
⎟⎟
(c) i4 = ⎜⎜ 1 ⎟⎟ ⎜⎜
+
+
R
R
R
R
3
4 ⎠
⎝ 1 ⎠⎝ 2
R 1 (R 2 + R 3 + R 4 ) R 2
= VS
R 1 [R 1 (R 2 + R 3 + R 4 ) + R 2 (R 3 + R 4 )(R 2 + R 3 + R 4 )]
R2
= VS
R 1 (R 2 + R 3 + R 4 ) + R 2 (R 3 + R 4 )
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78.
Chapter Three Solutions
10 March 2006
(a) With the current source open-circuited, we find that
v1 = − 40
500
= -8V
500 + 3000 || 6000
(b) With the voltage source short-circuited, we find that
i2 = ( 3 × 10−3 )
1/3000
1/ 500 + 1/3000 + 1/6000
i3 = ( 3 × 10−3 )
500
500 + 3000 || 6000
=
400 μ A
= 600 μ A
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79.
Chapter Three Solutions
10 March 2006
(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
IS
1
1 + (2 + R)
=
IS
5.25
The current through the 5-Ω resistor is
IS ⎛ 3 ⎞
⎜
⎟ = 2A
5.25 ⎝ 3 + 9 ⎠
so that
IS = 42 A.
(b) Given that IS is now 50 A, the current through the 5-Ω resistor becomes
IS ⎛ 3 ⎞
⎜
⎟ = 2.381 A
5.25 ⎝ 3 + 9 ⎠
Thus, vx = 5(2.381) = 11.90 V
(c)
vx
IS
⎡ 5IS ⎛ 3 ⎞⎤
⎢ 5.25 ⎜ 3 + 9 ⎟⎥
⎝
⎠⎦
= ⎣
= 0.2381
IS
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80.
Chapter Three Solutions
10 March 2006
First combine the 1 kΩ and 3 kΩ resistors to obtain 750 Ω.
By current division, the current through resistor Rx is
IRx = 10 × 10 − 3
2000
2000 + R x + 750
and we know that Rx · IRx = 9
so
9 =
20 R x
2750 + R x
9 Rx + 24750 = 20 Rx
or Rx = 2250 W. Thus,
PRx = 92/ Rx = 36 mW.
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81.
Chapter Three Solutions
10 March 2006
Define R = R3 || (R4 + R5)
Then vR
⎛ R ⎞
⎟⎟
= VS ⎜⎜
⎝ R + R2 ⎠
⎛ R 3 (R 4 + R 5 ) (R 3 + R 4 + R 5 ) ⎞
⎟⎟
= VS ⎜⎜
(
)
R
(R
R
)
R
R
R
R
+
+
+
+
5
3
4
5
2 ⎠
⎝ 3 4
⎛
⎞
R 3 (R 4 + R 5 )
⎟⎟
= VS ⎜⎜
R
(R
+
R
+
R
)
+
R
(R
+
R
)
4
5
3
4
5 ⎠
⎝ 2 3
Thus,
v5
⎛ R5 ⎞
⎟⎟
= vR ⎜⎜
R
R
+
5 ⎠
⎝ 4
⎛
⎞
R3 R5
⎟⎟
= VS ⎜⎜
⎝ R 2 (R 3 + R 4 + R 5 ) + R 3 (R 4 + R 5 ) ⎠
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Engineering Circuit Analysis, 7th Edition
82.
Chapter Three Solutions
10 March 2006
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 I1/I2 = R2/R1
Therefore
I1 = R2I2/R1 = 30(15)/20 = 22.5 mA
Thus, Ix = I1 . 15/ 45 = 7.5 mA
(d) I1 = IS R2/ (R1 + R2) = 60 (30)/ 50 = 36 mA
Thus, Ix = I1 15/ 45 = 12 mA.
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Engineering Circuit Analysis, 7th Edition
83.
Chapter Three Solutions
10 March 2006
vout = -gm vπ (100 kΩ || 100 kΩ) = -4.762×103 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, 7th Edition
84.
Chapter Three Solutions
10 March 2006
vout = -1000gm vπ
where vπ = 3 sin 10t
15 || 3
(15 || 3) + 0.3
= 2.679 sin 10t V
therefore
vout = -(2.679)(1000)(38×10-3) sin 10t = -101.8 sin 10t V.
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Engineering Circuit Analysis, 7th Edition
1. (
-0.2
a)
0.1
-0.5
-0.3
0.1
-0.3
-0.4
0
0.4
v1
=
v2
v3
Chapter Four Solutions
10 March 2006
0
4
6
Solving this matrix equation using a scientific calculator, v2 = -8.387 V
(b) Using a scientific calculator, the determinant is equal to 32.
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Engineering Circuit Analysis, 7th Edition
2. (
-1
a)
1
2
1
2
0
1
3
4
Chapter Four Solutions
vA
=
vB
vC
10 March 2006
27
-16
-6
Solving this matrix equation using a scientific calculator,
vA = 19.57
vB = 18.71
vC = -11.29
(b) Using a scientific calculator,
-1
1
2
11
2
0
3=
4
16
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Engineering Circuit Analysis, 7th Edition
Chapter Four Solutions
10 March 2006
3.
(a)
We begin by simplifying the equations prior to solution:
4 = 0.08v1 − 0.05v2 − 0.02v3
8 = −0.02v1 − 0.025v2 + 0.045v3
−2 = −0.05v1 + 0.115v2 − 0.025v3
Then, we can solve the matrix equation:
⎡ 0.08 -0.05 -0.02 ⎤ ⎡ v1 ⎤ ⎡ 4 ⎤
⎢-0.02 -0.025 0.045 ⎥ ⎢ v ⎥ = ⎢ 8 ⎥
⎢
⎥⎢ 2⎥ ⎢ ⎥
⎢⎣ -0.05 0.115 -0.025⎦⎥ ⎣⎢ v3 ⎥⎦ ⎢⎣ −2⎥⎦
to
(b)
obtain v1 = 264.3 V, v2 = 183.9 V and v3 = 397.4 V.
We may solve the matrix equation d irectly using MATLAB, but a be tter check is to
invoke the symbolic 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, 7th 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:
v v −v
1= 1 + 1 2
2
3
Node 2:
−3 =
v2 v2 − v1
+
1
3
Which may be simplified to:
and
5v1 – 2v2 = 6
-v1 + 4v2 = -9
Solving, we find that v1 = 333.3 mV.
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Engineering Circuit Analysis, 7th Edition
5.
node,
Chapter Four Solutions
10 March 2006
We begin by selecting the bottom node as the reference term inal, and defining
two nodal voltages V A and V B, as shown. (Note if we cho ose the upp er right
v1 becomes a nodal voltage and falls directly out of the solution.)
VA
VB
Ref.
We note that after completing nodal analysis, we can find v1 as v1 = VA – VB.
At node A: 4 =
VA VA − VB
+
10
5
At node B: −(−6) =
Simplifying,
Solving,
[1]
VB VB − VA
[2]
+
8
5
3VA – 2VB = 40 [1]
–8VA + 13VB = 240 [2]
VA = 43.48 V and VB = 45.22 V, so v1 = –1.740 V.
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Engineering Circuit Analysis, 7th Edition
6.
Chapter Four Solutions
10 March 2006
By inspection, no current flows through the 2 Ω resistor, so i1 = 0.
We next designate the bottom node as the reference term inal, and define V A and
VB as shown:
VA
VB
Ref.
VA VA − VB
[1]
+
3
1
V V V − VA
At node B: −2 = B + B + B
6
6
1
At node A: 2 =
obtain
2]
Note this y ields V A and V B, not v1, due to our choice of reference node. So, we
v1 by KVL: v1 = VA – VB.
Simplifying Eqs. [1] and [2],
–3V
4VA – 3VB = 6
A + 4VB = –6
[1]
[2]
Solving, VA = 0.8571 V and VB = -0.8571 V, so v1 = 1.714 V.
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Engineering Circuit Analysis, 7th Edition
7.
Chapter Four Solutions
10 March 2006
The bottom node has the largest number of branch connections, so we choose that as
our reference node. This also m akes vP 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 = v1/ 20 + (v1 – vP)/ 40
[1]
NODE P:
0 = (vP – v1)/ 40 + vP/ 100 + (vP – v2)/ 50
[2]
NODE 2:
-2.5 + 2 = (v2 – vP)/ 50 + (v2 – v3)/ 10
[3]
NODE 3:
5 – 2 = v3/ 200 + (v3 – v2)/ 10
[4]
Simplifying,
-50
Solving,
60v1 - 20vP
= 8000
v1 + 110 vP - 40v2
=0
- vP
+ 6v2 - 5v3
= -25
-200v2 + 210v3 = 6000
[1]
[2]
[3]
[4]
vP = 171.6 V
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Engineering Circuit Analysis, 7th Edition
8.
Chapter Four Solutions
10 March 2006
The logical choice for a reference node is the bottom node, as then vx will
automatically become a nodal voltage.
NODE 1:
4 = v1/ 100 + (v1 – v2)/ 20 + (v1 – vx)/ 50
[1]
NODE x:
10 – 4 – (-2) = (vx – v1)/ 50 + (vx – v2)/ 40
[2]
NODE 2:
-2 = v2 / 25 + (v2 – vx)/ 40 + (v2 – v1)/ 20
[3]
Simplifying,
Solving,
4 = 0.0800v1 – 0.0500v2 – 0.0200vx [1]
8 = -0.0200v1 – 0.02500v2 + 0.04500vx [2]
-2 = -0.0500v1 + 0.1150v2 – 0.02500vx [3]
vx = 397.4 V.
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Engineering Circuit Analysis, 7th Edition
9.
Chapter Four Solutions
10 March 2006
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) Writing the two nodal equations, then
NODE X:
–10 = (vX – 240)/ 3 + (vX – vY)/ 6 [1]
NODE Y:
0 = (vY – vX)/ 6 + vY/ 30 + (vY – 60)/ 12
[2]
Simplifying, -180 + 1440 = 9 vX – 3 vY [1]
10800 = - 360 vX + 612 vY [2]
Solving,
vX = 181.5 V and
Thus,
v1 = 240 – vX = 58.50 V and
(b)
vY = 124.4 V
v2 = vY – 60 =
64.40 V
The power absorbed by the 6-Ω resistor is
(vX – vY)2 / 6 = 543.4 W
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Engineering Circuit Analysis, 7th Edition
10.
Chapter Four Solutions
10 March 2006
Only one nodal equation is required: At the node where three resistors join,
0.02v1 = (vx – 5 i2) / 45 + (vx – 100) / 30 + (vx – 0.2 v3) / 50
[1]
This, however, is one equation in four unknow ns, the other three resulting from the
presence of the dependent sources. Thus, we require three additional equations:
i2 = (0.2 v3 - vx) / 50
[2]
v1 = 0.2 v3 - 100
[3]
v3 = 50i2
Simplifying,
[4]
v1 – 0.2v3
= -100
– v3 + 50 i2 = 0
–vx
+ 0.2v3 – 50 i2 = 0
0.07556vx – 0.02v1 – 0.004v3 – 0.111i2 = 33.33
[3]
[4]
[2]
[1]
Solving, we find that v1 = -103..8 V and i2 = -377.4 mA.
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Engineering Circuit Analysis, 7th Edition
11. If
Chapter Four Solutions
10 March 2006
v1 = 0, the dependent source is a short circuit and we may redraw the circuit as:
.
20 Ω
40 Ω
At NODE 1:
Since
+
v1 = 0
-
10 Ω
4 - 6 = v1/ 40 + (v1 – 96)/ 20 + (v1 – V2)/ 10
v1 = 0, this simplifies to
-2 = -96 / 20 - V2/ 10
so that V2 = -28 V.
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Engineering Circuit Analysis, 7th Edition
12.
Chapter Four Solutions
10 March 2006
We choose the bottom node as ground to make calculation of i5 easier. The left-most
node is named “1”, the top node is nam ed “2”, the central node is nam ed “3” and the
node between the 4-Ω and 6-Ω resistors is named “4.”
NODE 1:
NODE 2:
NODE 3:
NODE 4:
- 3 = v1/2 + (v1 – v2)/ 1
[1]
2 = (v2 – v1)/ 1 + (v2 – v3)/ 3 + (v2 – v4)/ 4
3 = v3/ 5 + (v3 – v4)/ 7 + (v3 – v2)/ 3
0 = v4/ 6 + (v4 – v3)/ 7 + (v4 – v2)/ 4
[2]
[3]
[4]
Rearranging and grouping terms,
-12
3v1 – 2v=
2
v1 + 19v2 – 4v3 – 3v4
–35v2 + 71v3 – 15v4=
-42v2 – 24v3 + 94v4=
-6 [1]
= 24 [2]
315 [3]
0
[4]
Solving, we find that v3 = 6.760 V and so
i5 = v3/ 5 = 1.352 A.
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Engineering Circuit Analysis, 7th Edition
13.
Chapter Four Solutions
10 March 2006
We can redraw this circuit and eliminate the 2.2-kΩ resistor as no current flows
through it:
5 mA
v2
470 Ω
9V
10 kΩ
+ v1 -
7 mA
vx
0.2 v1 ↓
At NODE 2: 7×10-3 – 5×10-3 = (v2 + 9)/ 470 + (v2 – vx)/ 10×10-3
At NODE x: 5×10
-3
– 0.2v1 = (vx – v2)/ 10×103
[1]
[2]
The additional equation required by the presence of the dependent source and the fact
that its controlling variable is not one of the nodal voltages:
v1 = v2 – vx
[3]
Eliminating the variable v1 and grouping terms, we obtain:
and
Solving, we find
10,470 v2 – 470 vx = –89,518
1999 v2 – 1999 vx = 50
vx = –8.086 V.
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Engineering Circuit Analysis, 7th 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, 7th Edition
15.
Chapter Four Solutions
10 March 2006
We choose the bottom node as the reference terminal. Then:
Node
1:
v v −v
−2 = 1 + 1 2
2
1
Node
2:
4=
v2 − v1 v2 − v3 v2 − v4
[2]
+
+
1
2
4
Node
3:
2=
v3 − v2 v3 v3 − v4
[3]
+ +
2
5
10
Node
4:
0=
v4 v4 − v3 v4 − v2
[4]
+
+
6
10
4
Node
5:
−1 =
Node
6:
1=
v6 v6 − v7 v6 − v8
[6]
+
+
5
2
10
Node
7:
2=
v7 − v5 v7 − v6 v7 − v8
[7]
+
+
1
2
4
Node
8:
0=
v8 v8 − v6 v8 − v7
[8]
+
+
6
10
4
[1]
v5 v5 − v7
+
2
1
[5]
Note that Eqs. [1-4] may be solved independently of Eqs. [5-8].
Sim
plifying,
3v1
−4v1
and
−2v2
+7v2
−5v2
−15v2
−2v3
+8v3
−6v3
8v6
−2v6
−6v6
−2v7
−5v7
+7v7
−15v7
3v5
−4v5
−v4
−v4
+31v4
= −4 [1]
= 16 [2]
to yield
= 20 [3]
= 0 [4]
v1 = 3.370 V
v2 = 7.055 V
v3 = 7.518 V
v4 = 4.869 V
−v8
−v8
+31v8
= −2
= 10
=8
=0
v5 = 1.685 V
v6 = 3.759 V
v7 = 3.527 V
v8 = 2.434 V
[5]
[6]
to yield
[7]
[8]
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Engineering Circuit Analysis, 7th Edition
16.
Chapter Four Solutions
10 March 2006
We choose the center node for our common terminal, since it connects to the largest
number of branches. W e na me 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 = (VA – VD)/ 10 + VA/ 20 + (VB – VC)/ 12.5
[1]
At node C:
VC = 150
[2]
At node D:
-10 = VD/ 25 + (VD – VA)/ 10
Our supernode-related equation is VB – VA = 100
[3]
[4]
Simplifiying and grouping terms,
0.15 VA + 0.08 VB - 0.08 VC – 0.1 VD
VC
-25 VA
+ 35 V=
D
- VA
+ VB =
= 5
= 150
-2500
100 [4]
[1]
[2]
[3]
Solving, we find that VD = -63.06 V. Since v4 = - VD,
v4 = 63.06 V.
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Engineering Circuit Analysis, 7th Edition
17.
Chapter Four Solutions
10 March 2006
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 supe rnode about nodes 1 and
2, encompassing the dependent voltage source.
At the supernode,
At node 2,
5 – 8 = (v1 – v2)/ 2 + v3/ 2.5
8 = v2 / 5 + (v2 – v1)/ 2
[1]
[2]
Our supernode equation is v1 - v3 = 0.8 vA
[3]
Since
vA = v2, we can rewrite [3] as v1 – v3 = 0.8v2
Simplifying and collecting terms,
0.5 v1 - 0.5 v2 + 0.4 v3
-0.5 v1 + 0.7 v2
v1 - 0.8 v2
- v3
(a) Solving for v2 = vA, we find that
= -3
= 8
= 0
[1]
[2]
[3]
vA = 25.91 V
(b) The power absorbed by the 2.5-Ω resistor is
= 82.66 mW.
(v3)2/ 2.5 = (-0.4546)2/ 2.5
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Engineering Circuit Analysis, 7th Edition
18.
Chapter Four Solutions
10 March 2006
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 = (v1 – v2)/ 20 + (v1 – v3)/ 50
[1]
NODE
2:
v2 = 0.4 v1
[2]
NODE
3:
0.01 v1 = (v3 – v2)/ 30 + (v3 – v1)/ 50
[3]
Simplifying and collecting terms, we obtain
0.07 v1 – 0.05 v2 – 0.02 v3
0.4 v1
–=
v2
-0.03 v1 – 0.03333 v2 + 0.05333 v3=
=5
0
0
[1]
[2]
[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
might have expected.
Solving, we find that v1 = 148.2 V, v2 = 59.26 V, and v3 = 120.4 V.
The power supplied by the dependent current source is therefore
(0.01 v1) • v3 = 177.4 W.
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Engineering Circuit Analysis, 7th Edition
19.
At node x:
At node y:
Chapter Four Solutions
vx/ 4 + (vx – vy)/ 2 + (vx – 6)/ 1
(vy – kvx)/ 3 + (vy – vx)/ 2
=0
=2
10 March 2006
[1]
[2]
Our additional constraint is that vy = 0, so we may simplify Eqs. [1] and [2]:
14
-2k
vx = 48
[1]
vx - 3 vx = 12 [2]
Since Eq. [1] yields vx = 48/14 = 3.429 V, we find that
k = (12 + 3 vx)/ (-2 vx)
= -3.250
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Engineering Circuit Analysis, 7th Edition
20.
Chapter Four Solutions
10 March 2006
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 nam e the other node of the 4- Ω
resistor node “1”, and the node joining the 2- Ω r esistor and t he 2- A c urrent s ource
node “2.” Finally, we create a supernode with nodes “1” and “2.”
At the supernode:
Our remaining equations:
and
–2 = v1/ 4 + (v2 – 4)/ 2
v1 – v2 = –3 – 0.5i1
i1 = (v2 – 4)/ 2
Equation [1] simplifies to
Combining Eqs. [2] and [3,
v1 + 2 v2
4 v1 – 3 v=
2
= 0
–8
[1]
[2]
[3]
[1]
[4]
Solving these last two equations, we find that v2 = 727.3 mV. Making use of Eq. [3],
we therefore find that
i1 = – 1.636 A.
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Engineering Circuit Analysis, 7th Edition
21.
Chapter Four Solutions
10 March 2006
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 = v1/ 40 + (v1 – v3)/ 10 + (v3 – v1)/ 10 + v2/ 50 + (v3 – 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:
v4 = 200
[2]
Supernode KVL equation:
v1 – v3 = 400 + 4v20
[3]
Where the controlling voltage
v20 = v3 – v4 = v3 – 200
[4]
Thus, Eq. [1] becomes -6 = v1/ 40 + v2/ 50 + (v3 – 200)/ 20 or, more simply,
4 = v1/ 40 + v2/ 50 + v3/ 20 [1’]
and Eq. [3] becomes
v1 – 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 additi onal equation, possibly in terms of v2. Referring to the
circuit,
v1 - v2 = 400
[5]
Rewriting as a matrix equation,
⎡1
⎢ 40
⎢1
⎢
⎢⎣1
1
50
0
-1
1
⎤
⎡ 4⎤
20⎥ ⎡v1 ⎤
⎢
⎥
- 5 ⎥ ⎢v2 ⎥ = ⎢⎢- 400⎥⎥
⎥
⎢⎣ 400⎥⎦
0 ⎥ ⎢⎣v3 ⎥⎦
⎦
Solving, we find that
v1 = 145.5 V, v2 = -254.5 V, and v3 = 109.1 V. Since v20 = v3 – 200, we find that
v20 = -90.9 V.
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Engineering Circuit Analysis, 7th Edition
22.
Chapter Four Solutions
10 March 2006
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.” Th e reference node has
already been selected, and designated using a ground symbol.
By inspection, v2 = 5 V.
Forming a supernode with nodes 1 & 3, we find
At the supernode:
-2 = v3/ 1 + (v1 – 5)/ 10
[1]
At node 4:
2 = v4/ 2 + (v4 – 5)/ 4
[2]
Our supernode KVL equation:
v1 – v3 = 6
[3]
Rearranging, simplifying and collecting terms,
and
v1 + 10 v3 = -20 + 5 = -15
[1]
v1 - v3 = 6
[2]
Eq. [3] may be directly solved to obtain
v4 = 4.333 V.
Solving Eqs. [1] and [2], we find that
v1 = 4.091 V
and
v3 = -1.909 V.
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Engineering Circuit Analysis, 7th Edition
23.
Chapter Four Solutions
10 March 2006
We begin by selecting the bottom node as the reference, naming the nodes as shown
below, and forming a supernode with nodes 5 & 6.
2 A
v3
v2
4Ω
v5
v4
2Ω
1V
4A
v6
3V
4V
v1
1Ω
3Ω
By inspection,
v4 = 4 V.
By KVL, v3 – v4 = 1 so v3 = -1 + v4 = -1 + 4
At the supernode,
2 = v6/ 1 + (v5 – 4)/ 2
At node 1,
4 = v1/ 3
At node 2,
-4 – 2 = (v2 – 3)/ 4
Solving, we find that
or
v3 = 3 V.
[1]
therefore,
v1 = 12 V.
v2 = -21 V
Our supernode KVL equation is
v5 - v6 = 3 [2]
Solving Eqs. [1] and [2], we find that
v5 = 4.667 V and v6 = 1.667 V.
The power supplied by the 2-A source therefore is (v6 – v2)(2) = 45.33 W.
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Engineering Circuit Analysis, 7th Edition
24.
Chapter Four Solutions
10 March 2006
We begin by selecting the bottom node as the reference, naming each node as shown
below, and forming two different supernodes as indicated.
v2
v3
v5
v4
v8
v6
v7
Voltages in
volts.
Currents in
amperes.
Resistances
in ohms.
v1
By inspection, v7 = 4 V
At node 2:
and v1 = (3)(4) = 12 V.
-4 – 2 = (v2 – v3)/ 4
At the 3-4 supernode:
0 = (v3 – v2)/ 4 + (v4 – v5)/ 6
or
or
At node 5:
0 = (v5 – v4)/ 6 + (v5 – 4)/ 7 + (v5 – v6)/ 2
or
At the 6-8 supernode: 2 = (v6 – v5)/ 2 + v8/ 1
3-4 supernode KVL equation:
6-8 supernode KVL equation:
v3 - v4 = -1
v6 – v8 = 3
v2 -v3 = -24
[1]
-6v2 + 6v3 + 4v4 – 4v5 = 0
[2]
-14v4 + 68v5 – 42v6 = 48 [3]
or
-v5 + v6 + 2v8 = 4
[4]
[5]
[6]
Rewriting Eqs. [1] to [6] in matrix form,
0 0 0 0 ⎤ ⎡v2 ⎤
⎡- 24⎤
⎡1 - 1
⎢v ⎥
⎢ 0⎥
⎥
⎢- 6 6
4 - 4 0 0⎥ ⎢ 3 ⎥
⎥
⎢
⎢
⎢ 48⎥
⎢0
0 - 14 68 - 42 0 ⎥ ⎢v4 ⎥
⎥
⎥ ⎢ ⎥ = ⎢
⎢
0
0
0
1
1
2
⎢ 4⎥
⎥ ⎢v5 ⎥
⎢
⎢ -1 ⎥
⎢0
1 -1 0
0 0 ⎥ ⎢v6 ⎥
⎥
⎢
⎥ ⎢ ⎥
⎢
0 0
1 - 1 ⎥⎦ ⎢⎣v8 ⎥⎦
⎢⎣ 3 ⎥⎦
⎢⎣0 0
Solving, we find that
v2 = -68.9 V, v3 = -44.9 V, v4 = -43.9 V, v5 = -7.9 V, v6 = 700 mV, v8 = -2.3 V.
The power generated by the 2-A source is therefore (v8 – v6)(2) = 133.2 W.
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Engineering Circuit Analysis, 7th Edition
25.
Chapter Four Solutions
10 March 2006
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, v5 = -2.2 V.
Also, we see that v4 – v5 = 9, so that v4 = 9 – 2.2 = 6.8 V.
Proceeding with nodal analysis,
At node 1:
-3×10-3 = v1/ 10×103 + (v1 – v2)/ 2.2×103
[1]
At node 2:
0 = (v2 – v1)/ 2.2×103 + (v2 – v3)/ 4.7×103
[2]
At node 3:
1×103 + 3×103 = (v3 – v2)/ 4.7×103 + v3/3.3×103 [3]
Solving,
v1 = -8.614 V, v2 = -3.909 V and v3 = 6.143 V.
Note that we could also have made use of the supernode approach here.
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Engineering Circuit Analysis, 7th Edition
26.
Chapter Four Solutions
10 March 2006
Mesh 1: –4 + 400i1 + 300i1 – 300i2 – 1 = 0 or 700i1 – 300i2 = 5
Mesh 2: 1 + 500i2 – 300i1 +2 – 2 = 0
or –300i1 + 500i2 = –3.2
Solving,
i1 = 5.923 mA and
i2 = -2.846 mA.
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Engineering Circuit Analysis, 7th Edition
Chapter Four Solutions
10 March 2006
27.
(a) Define a clockwise m esh current i1 in th e lef t-most m esh; a clockw ise m esh
curren
t i2 in the central m esh, and note that iy can be used as a m esh current for the
rem
aining mesh.
Mesh 1: -10 + 7i1 – 2i2 = 0
Mesh 2:
-2i1 + 5i2 = 0
Mesh
y:
-2i2 + 9iy = 0
Solve the resulting matrix equation:
⎡ 7 −2 0 ⎤ ⎡ i1 ⎤ ⎡10 ⎤
⎢ −2 5 0 ⎥ ⎢ i ⎥ = ⎢ 0 ⎥ to find that i = 1.613 A, and i = 143.4 mA.
1
y
⎢
⎥⎢ 2⎥ ⎢ ⎥
⎢⎣ 0 −2 9 ⎥⎦ ⎢⎣i y ⎥⎦ ⎢⎣ 0 ⎥⎦
(b) The power supplied by the 10 V source is (10)(i1) = 10(1.613) = 16.13 W.
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Engineering Circuit Analysis, 7th Edition
28.
Chapter Four Solutions
10 March 2006
Define three mesh currents as shown:
(a) The current through the 2 Ω resistor is i1.
Mesh 1: 5i1 – 3i2 = 0
Mesh 2: –212 +8i2 –3i1 = 0
Mesh 3: 8i3 – 5i2 + 122 = 0
Solving,
or 5i1 – 3i2
= 0
or -3i1 +8i2
= 212
or
–5i2 + 8i3 = –122
i1 = 20.52 A, i2 = 34.19 A and i3 = 6.121 A.
(b) The current through the 5 Ω resistor is i3, or 6.121 A.
*** Note: since the problem statement did not specify a directi on, only the current
magnitude is relevant, and its sign is arbitrary.
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Engineering Circuit Analysis, 7th Edition
29.
(a)
Chapter Four Solutions
10 March 2006
We begin by defining three clockwise m esh currents i1, i2 and i3 in the lef t-most,
central, and right-most meshes, respectively. Then,
Note that ix = i2 – i3.
Mesh 1: i1 = 5 A (by inspection)
Mesh 3: i3 = –2 A (by inspection)
Mesh
–125
Thus,
2: –25i1 + 75i2 – 20i3 = 0, or, making use of the above,
+ 75i2 + 40= 0 so that i2 = 1.133 A.
ix = i2 – i3 = 1.133 – (–2) = 3.133 A.
(b) The power absorbed by the 25 Ω resistor is
P25Ω = 25 (i1 – i2)2 = 25 (5 – 1.133)2 = 373.8 W.
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Engineering Circuit Analysis, 7th Edition
30.
Chapter Four Solutions
10 March 2006
Define three mesh currents as shown. Then,
Mesh 1:
Mesh 2:
Mesh 3:
Solving,
–2 + 80i1 – 40i2 – 30i3 = 0
=0
–40i1 + 70i2
+70i3 = 0
–30i1
⎡ 80 −40 −30⎤ ⎡ i1 ⎤ ⎡ 2 ⎤
⎢
0 ⎥⎥ ⎢⎢i2 ⎥⎥ = ⎢⎢ 0 ⎥⎥
⎢ −40 70
70 ⎦⎥ ⎢⎣ i3 ⎥⎦ ⎢⎣ 0 ⎥⎦
⎣⎢ −30 0
we find that i2 = 25.81 mA and i3 = 19.35 mA. Thus, i = i3 – i2 = –6.46 mA.
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Engineering Circuit Analysis, 7th Edition
31.
Chapter Four Solutions
10 March 2006
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 meshes we define a clockwise curren t. The
remaining mesh current is clearly 8 A. We may then write:
MESH 1:
12 i1 - 4 i2
MESH 2:
-4 i1 + 9 i2 - 3 i3 = 0
MESH 3:
= 100
-3 i2 + 18 i3 = -80
Solving this system of three (independent) equations in three unknowns, we find that
i2 =
ix = 2.791 A.
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Engineering Circuit Analysis, 7th Edition
32.
Chapter Four Solutions
10 March 2006
We define four clockwise mesh currents. The top mesh current is labeled i4. The
bottom left mesh current is labeled i1, the bottom right m esh current is labeled i3, and
the remaining mesh current is labeled i2. Define a voltage “v4A” across the 4-A current
source with the “+” reference terminal on the left.
By inspection,
i3 = 5 A and ia = i4.
MESH 1: -60 + 2i1 – 2i4 + 6i4 = 0
or 2i1
MESH 2: -6i4 + v4A + 4i2 – 4(5) = 0
or
MESH 4: 2i4 – 2i1 + 5i4 + 3i4 – 3(5) – v4A = 0 or -2i1
+ 4i4
= 60
[1]
4i2 - 6i4 + v4A = 20
[2]
+ 10i4 - v4A = 15
[3]
At this point, we are short an equation. Returning to the circuit diagram, we note that
i2 – i4 = 4
[4]
Collecting these equations and writing in matrix form, we have
⎡60⎤
⎡ 2 0 4 0⎤ ⎡i1 ⎤
⎢ ⎥
⎢ 0 4 - 6 1 ⎥ ⎢i ⎥
⎥ ⎢ 2 ⎥ = ⎢20⎥
⎢
⎢15 ⎥
⎢- 2 0 10 - 1 ⎥ ⎢i4 ⎥
⎢ ⎥
⎥ ⎢ ⎥
⎢
⎣ 4⎦
⎣ 0 1 - 1 0 ⎦ ⎣v 4 A ⎦
Solving,
i1 = 16.83 A, i2 = 10.58 A, i4 = 6.583 A and v4A = 17.17 V.
Thus, the power dissipated by the 2-Ω resistor is
(i1 – i4)2 • (2) = 210.0 W
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Engineering Circuit Analysis, 7th Edition
33.
Chapter Four Solutions
10 March 2006
We begin our analysis by defining three clockwise mesh currents. We will call the top
mesh current i3, the bottom left mesh current i1, and the bottom right mesh current i2.
By inspection, i1 = 5 A [1]
MESH 3:
or
and
i2 = -0.01 v1
[2]
50 i3 + 30 i3 – 30 i2 + 20 i3 – 20 i1 = 0
-20 i1 – 30 i2 + 100 i3 = 0 [3]
These three equations are insufficient, how ever, to solve for the unknow ns. It would
be nice to be able to express the dependent source cont rolling variable v1 in term s of
the mesh currents. Returning to the diagra m, it can be seen that KVL around m esh 1
will yield
- v1 + 20 i1 – 20 i3 + 0.4 v1 = 0
or v1 = (20(5)/ 0.6 - 20 i3/ 0.6 [4]
or
v1 = 20 i1/ 0.6 – 20 i3/ 0.6
Substituting Eq. [4] into Eq. [2] and then the modified Eq. [2] into Eq. [3], we find
-20(5) – 30(-0.01)(20)(5)/0.6 + 30(-0.01)(20) i3/ 0.6 + 100 i3 = 0
Solving, we find that i3 = (100 – 50)/ 90 = 555.6 mA
Thus, v1 = 148.1 V, i2 = -1.481 A, and the power generated by the dependent voltage
source is
0.4 v1 (i2 – i1) = -383.9 W.
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Engineering Circuit Analysis, 7th Edition
34.
Chapter Four Solutions
10 March 2006
We begin by defining four clockwise mesh currents i1, i2, i3 and i4, in the meshes of
our circuit, starting at the left-m ost m esh. W e also define a voltage vdep acros s th e
dependent current source, with the “+” on the top.
i1 = 2A
By inspection,
and
i4 = -5 A.
At Mesh 2:
10 i2 - 10(2) + 20 i2 + vdep = 0
[1]
At Mesh 3:
- vdep + 25 i3 + 5 i3 –5(-5) = 0
[2]
Collecting terms, we rewrite Eqs. [1] and [2] as
30 i2
+ vdep = 20
[1]
30 i3 – vdep = -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 pos sible to express th e
current of the dependent sour ce in term s of m esh currents. (W e might also choose to
obtain an expression for vdep in term s of m esh currents using KVL around m esh 2 or
3.)
Thus, 1.5ix = i3 - i2 where ix = i1 – i2 so -0.5 i2 - i3 = -3
[3]
In matrix form,
⎡ 30 0 1⎤ ⎡i2 ⎤
⎡ 20⎤
⎢ 0 30 - 1⎥ ⎢i ⎥ = ⎢- 25⎥
⎢
⎥ ⎢3 ⎥
⎢
⎥
⎢⎣- 0.5 - 1 0⎥⎦ ⎢⎣vdep ⎥⎦
⎢⎣ - 3 ⎥⎦
Solving, we find that i2 = -6.333 A so that
ix = i1 – i2 = 8.333 A.
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Engineering Circuit Analysis, 7th Edition
35.
Chapter Four Solutions
10 March 2006
We define a clockwise mesh current i1 in the bottom left mesh, a clockwise mesh
current i2 in the top left mesh, a clockwise mesh current i3 in the top right mesh, and a
clockwise mesh current i4 in the bottom right mesh.
MESH
1:
-0.1 va + 4700 i1 – 4700 i2 + 4700 i1 – 4700 i4 = 0
[1]
MESH
2:
9400 i2 – 4700 i1 – 9 = 0
[2]
MESH 3:
MESH
4:
9 + 9400 i3 – 4700 i4
=
0
9400 i4 – 4700 i1 – 4700 i3 + 0.1 ix = 0
[3]
[4]
The presen ce of the two de pendent sources has led to the in troduction of tw o
additional unknowns (ix and va) besides our four m esh currents. In a perfect world, it
would simplify the solution if we could e xpress these tw o quantities in term s of the
mesh currents.
Referring to the circuit diagram, we see that ix = i2 (easy enough) and that
va = 4700 i3 (also straightforward). Thus, subs tituting these express ions into our
four mesh equations and creating a matrix equation, we arrive at:
Solving,
⎡ 0⎤
⎡ 9400 - 4700 - 470 - 4700⎤ ⎡i1 ⎤
⎢i ⎥
⎢ 9⎥
⎥
⎢- 4700 9400
0
0⎥ ⎢ 2 ⎥
⎢
= ⎢ ⎥
⎢- 9 ⎥
⎢
0
0 9400 - 4700 ⎥ ⎢i3 ⎥
⎢ ⎥
⎥ ⎢ ⎥
⎢
0.1 - 4700 9400 ⎦ ⎣i4 ⎦
⎣ 0⎦
⎣- 4700
i1 = 239.3 μA, i2 = 1.077 mA, i3 = -1.197 mA and i4 = -478.8 μA.
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Engineering Circuit Analysis, 7th Edition
36.
Chapter Four Solutions
10 March 2006
We define a clockwise mesh current i3 in the upper right mesh, a clockwise mesh
current i1 in the lower left m esh, and a clockwise m esh current i2 in th e lower right
mesh.
MESH 1:
-6 + 6 i1 - 2 = 0
[1]
MESH 2:
2 + 15 i2 – 12 i3 – 1.5 = 0
[2]
MESH
3:
i3 = 0.1 vx
[3]
Eq. [1] may be solved directly to obtain
i1 = 1.333 A.
It would help in the solution of Eqs. [2] and [3] if we could express the dependent
source controlling variable vx in term s of m esh curren ts. Referring to the circuit
diagram, we see that vx = (1)( i1) = i1, so Eq. [3] reduces to
i3 = 0.1 vx = 0.1 i1 = 133.3 mA.
As a result, Eq. [1] reduces to
i2 = [-0.5 + 12(0.1333)]/ 15 = 73.31 mA.
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Engineering Circuit Analysis, 7th Edition
37.
10 March 2006
(a) Define a mesh current i2 in the second mesh. Then KVL allows us to write:
MESH 1: -9 + R i1 + 47000 i1 – 47000 i2 = 0
[1]
MESH 2: 67000 i2 – 47000 i1 – 5 = 0
[2]
Given
that i1 = 1.5 mA, we may solve Eq. [2] to find that
i2 =
and
Chapter Four Solutions
so
R =
5 + 47(1.5)
mA = 1.127 mA
67
9 - 47(1.5) + 47(1.127)
= -5687 Ω.
1.5 × 10-3
(b) This value of R is unique; no other value will satisfy both Eqs. [1] and [2].
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Engineering Circuit Analysis, 7th Edition
38.
Chapter Four Solutions
10 March 2006
Define three clockwise mesh currents i1, i2 and i3. The bottom 1-kΩ resistor can be
ignored, as no current flows through it.
MESH 1:
-4 + (2700 + 1000 + 5000) i1 – 1000 i2 = 0
[1]
MESH 2:
(1000 + 1000 + 4400 + 3000) i2 – 1000 i1 – 4400 i3 + 2.2 – 3 = 0 [2]
MESH 3:
(4400 + 4000 + 3000) i3 - 4400 i2 – 1.5 = 0
[3]
Combining terms,
–1000
Solving,
8700 i1 – 1000 i2
= 4
i1 + 9400 i2 – 4400 i3 = 0.8
– 4400 i2 + 11400 i3 = 1.5
[1]
[2]
[3]
i1 = 487.6 μA, i2 = 242.4 μA and i3 = 225.1 μA.
The power absorbed by each resistor may now be calculated:
2
P5k
=
5000 (i=
1)
2
=
2700 (i=
P2.7k
1)
1000 (i1 – i2)2 =
P1ktop =
2
= 1000
(i=
P1kmiddle
2)
P1kbottom
= 0
=
=
4400 (i2 – i3)2 =
P4.4k
2
3000 (i=
P3ktop =
3)
2
=
4000 (i=
P4k
3)
2
= 3000
(i=
P3kbottom
2)
1.189 mW
641.9 μW
60.12 μW
58.76 μW
0
1.317 μW
152.0 μW
202.7 μW
176.3 μW
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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Engineering Circuit Analysis, 7th Edition
39.
Chapter Four Solutions
10 March 2006
(a) We begin by naming four mesh currents as depicted below:
Proceeding with mesh analysis, then, keeping in mind that Ix = -i4,
MESH 1:
(4700 + 300) i1 - 4700 i2
= 0
[1]
MESH 2:
(4700 + 1700) i2 – 4700 i1 – 1700 i3 = 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 vπ [3a]
MESH
4:
where
3000 i4 – 3000 i3 + 1
vπ = 4700(i1 – i2) [3b]
= 0
[4]
Simplifying and combining Eqs. 3a and 3b,
5000 i1 – 4700 i2
i1 + 6400 i2 – 1700 i3
–141 i1 + 141 i2 – i3
– 3000 i3 + 3000 i4
–4700
Solving, we find that i4 = -333.3 mA, so
(b) At node “π” :
= 0
= 0
=
= –1
0
Ix = 333.3 μA.
0.03 vπ = vπ / 300 + vπ / 4700 + vπ /1700
Solving, we find that vπ = 0, therefore no current flows through the dependent source.
Hence,
Ix = 333.3 μA as found in part (a).
(c) V s/ I x has units of resistance. It can be t hought of as the resistance “seen” by the
voltage source Vs…. more on this in Chap. 5….
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Engineering Circuit Analysis, 7th Edition
40.
Chapter Four Solutions
10 March 2006
We begin by naming each mesh and the three undefined voltage sources as shown
below:
MESH 1:
–Vz + 9i1 – 2i2
MESH 2:
– 7i4 = 0
–2i1 + 7i2 – 5i3
MESH 3:
Vx
MESH 4:
Vy – 7i1
= 0
– 5i2 + 8i3 – 3i4 = 0
– 3i3 + 10i4 = 0
Rearranging and setting i1 – i2 = 0, i2 – i3 = 0, i1 – i4 = 0 and i4 – i3 = 0,
9i1 - 2i2
-7i4 = Vz
-2i1 + 7i2 - 5i3
= 0
-5i2 + 8i3 – 3i4 = - Vx
-7i1 -3i3 + 10i4
= - Vy
Since
i1 = i2 = i3 = i4, these equations produce:
Vz = 0
0 = 0
-Vx = 0
-Vy = 0
This is a unique solution. Therefore,
the request that nonzero values be
found cannot be satisfied.
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Engineering Circuit Analysis, 7th Edition
41.
Chapter Four Solutions
10 March 2006
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 m esh currents i1, i2, i3 and i4. By inspection,
we see that i1 = 2 mA.
MESH 2:
-10 + 5000i2 + 4 + 1000i3 = 0
[1]
MESH 3:
-1000i3 + 6 + 10,000 – 10,000i4 = 0
[2]
MESH 4:
i4 = -0.5i2
[3]
Reorganising, we find
5000 i2 + 1000 i3
= 6
9000 i3 – 10,000 i4 = -6
+
i4 = 0
0.5 i2
[1]
[2]
[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]. Either way, we find that
i1 = 2 mA, i2 = 1.5 mA, i3 = -1.5 mA and i4 = -0.75 mA.
The power generated by each source is:
P2mA = 5000(i1 – i2)(i1)
P4V
= 4 (-i2)
= 6 (-i3)
P6V
1000 i3 (i3 – i2)
PdepV =
10,000(i3 – i4)(0.5 i2)
PdepI=
= 5 mW
= -6 mW
= 9 mW
= 4.5 mW
= -5.625 mW
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Engineering Circuit Analysis, 7th Edition
42.
Chapter Four Solutions
10 March 2006
This circuit does not require the supermesh technique, as it does not contain any
current sources. Redrawing the circuit so its planar nature and m esh structure are
clear,
1Ω
•
i2
i1
20 V
2.5 iA
•
MESH 1:
2Ω
4Ω
3Ω
i3
iA →
-20 + i1 – i2 + 2.5 iA = 0
[1]
MESH
2:
2 i 2 + 3 i 2 + i 2 – 3 i 3 – i1 = 0
[2]
MESH
3:
-2.5 iA + 7 i3 – 3 i2
[3]
= 0
Combining terms and making use of the fact that iA = - i3,
i1 – i2 – 2.5 i3 = 20
- i1 + 6i2 – 3 i3 = 0
–3 i2 + 9.5 i3 = 0
Solving,
[1]
[2]
[3]
i1 = 30.97 A, i2 = 6.129 A, and i3 = 1.936 A. Since iA = - i3,
iA = –1.936 A.
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Engineering Circuit Analysis, 7th Edition
43.
Define four mesh currents
Chapter Four Solutions
i1
i2
i3
i4
10 March 2006
By inspection, i1 = -4.5 A.
We form a supermesh with meshes 3 and 4 as defined above.
MESH 2:
SUPERME
Superm
SH:
2.2 + 3 i2 + 4 i2 + 5 – 4 i3 = 0
[1]
3 i 4 + 9 i4 – 9 i1 + 4 i3 – 4 i2 + 6 i3 + i3 – 3 = 0
[2]
esh KCL equation:
i4 - i3 = 2
[3]
Simplifying and combining terms, we may rewrite these three equations as:
7 i2 – 4 i3
-4 i2 + 11 i3 + 12 i4
- i3 + i4
= -7.2
= -37.5
= 2
[1]
[2]
[3]
Solving, we find that i2 = -2.839 A, i3 = -3.168 A, and i4 = -1.168 A.
The power supplied by the 2.2-V source is then 2.2 (i1 – i2) = -3.654 W.
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Engineering Circuit Analysis, 7th Edition
44.
Chapter Four Solutions
10 March 2006
We begin by defining six mesh currents as depicted below:
i1
•
•
i5
i6
i2
i3
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) i1 – 2200 i5 + (2200 + 1000 + 4700) i2 – 1000 i3 = 0
[1]
3, 4 Supermesh:
(4700 + 1000 + 2200) i3 – 1000 i2 – 2200 i6 + (4700 + 10,000 + 1000) i4 = 0
[2]
MESH 5:
(2200 + 4700) i5 – 2200 i2 + 3.2 – 1.5 = 0
[3]
MESH 6:
1.5 + (4700 + 4700 + 2200) c – 2200 i3 = 0
[4]
1, 2 Supermesh KCL equation:
i1 – i2 = 3×10-3
[5]
3, 4 Supermesh KCL equation:
i4 – i3 = 2×10-3
[6]
We can simplify these equations prior to solution in several ways. Choosing to retain
six equations,
-2200 i5
= 0
[1]
15,700 i1 + 7900 i2 - 1000 i3
-2200 i6 = 0
[2]
- 1000 i2 + 7900 i3 + 15,700 i4
- 2200 i2
+ 6900 i5
= -1.7
[3]
[4]
- 2200 i3
+ 11,600 i6 = -1.5
-3
– i2
= 3×10 [5]
i1
+ i4
= 2×10-3 [6]
- i3
Solving, we find that i4 = 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, 7th Edition
45.
Chapter Four Solutions
10 March 2006
We define a mesh current ia in the left-hand mesh, a mesh current i1 in the top right
mesh, and a mesh current i2 in the bottom right mesh (all flowing clockwise).
The left-most mesh can be analysed separately to determine the controlling voltage va,
as KCL assures us that no current flows through either the 1-Ω or 6-Ω resistor.
Thus, -1.8 + 3 ia – 1.5 + 2 ia = 0, which may be solved to find ia = 0.66 A. Hence,
va = 3 ia = 1.98 V.
Forming one supermesh from the remaining two meshes, we may write:
-3 + 2.5 i1 + 3 i2 + 4 i2 = 0
and the supermesh KCL equation: i2 – i1 = 0.5 va = 0.5(1.98) = 0.99
Thus, we have two equations to solve:
2.5 i1 + 7 i2 = 3
-i1 + i2 = 0.99
Solving, we find that i1 = -413.7 mA and the voltage across the 2.5Ω res istor
(arbitrarily assuming the left terminal is the “+” reference) is 2.5 i1 = -1.034 V.
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Engineering Circuit Analysis, 7th Edition
46.
Chapter Four Solutions
10 March 2006
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, i1, i2,
and i3, beginning with the left-most mesh.
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 i1 + 32×10-3 i2 = 11.8
– i2 = 100
i1
Solving, we find that i1 = 254.2 A and i2 = 154.2 A.
The final mesh current is easily found: i3 = 13×103/ (14 + 11.6 + 15) = 320.2 A.
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Engineering Circuit Analysis, 7th Edition
47.
MESH 1:
MESH
2:
MESH
3:
Chapter Four Solutions
-7 + i1 – i2 = 0
i2 – i1 + 2i2 + 3i2 – 3i3 = 0
3i3 – 3i2 + xi3 +2i3 – 7 = 0
10 March 2006
[1]
[2]
[3]
Grouping terms, we find that
i1 – i2
-i1 + 6i2 – 3i3
-3i2 + (5 + x)i3
= 7
=
=7
[1]
[2]
0
[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 i2 = 2.273 A. Solving these “four”
equations, we find that
x = (7 + 3 i2 – 5 i3)/ i3 = 4.498 Ω.
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Engineering Circuit Analysis, 7th Edition
48.
Chapter Four Solutions
10 March 2006
We begin by redrawing the circuit as instructed, and define three mesh currents:
300 mΩ
2Ω
3Ω
i2
1Ω
i1
i3
2Ω
7A
7V
B
y inspection, i3 = 7 A.
MESH 1:
-7 + i1 – i2 = 0
or
i1 – i2 = 7 [1]
MESH 2:
(1 + 2 + 3) i2 – i1 –3(7) = 0 or
-i1 + 6i2 = 21 [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 meshes.
Solving, we find that
i1 = 12.6 A, i2 = 5.6 A and i3 = 7 A.
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Engineering Circuit Analysis, 7th Edition
49. (
Chapter Four Solutions
10 March 2006
a) We are asked for a voltage, and have one current source and one voltage source.
Nodal analysis is probably best then- th e nodes can be nam ed 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” a nd 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 = v1/ 2 + vx/ 9
[1]
and the KVL Eqn:
v1 – vx = 22
[2]
Rearranging,
11(18) = 9 v1 + 2 vx
22 = v1 – vx [2]
[1]
Solving, vx = 0
(b) We are asked for a voltage, and so may suspect that nodal analysis is preferrable;
with two current sources a nd only one voltage source (eas ily dea lt 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:
and the KVL Eqn:
6 + 9 = vx / 10 + vy/ 20
vy – vx = 12
[1]
[2]
Rearranging, 15(20) = 2 vx + vy [1]
and
12 = - vx + vy [2]
Solving, we find that vx = 96 V.
(c) W e are asked for a voltage, but would ha ve 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, whic h would lead to the need for another
equation if invoking mesh analysis. Trying nodal analysis,
0.1 vx = (v1 – 50) / 2 + vx / 4
[1]
referring to the circuit w e see tha t vx = v1 – 100. Rearranging so that we m
eliminate v1 in Eq. [1], we obtain v1 = vx + 100. Thus, Eq. [1] becomes
ay
0.1 vx = (vx + 100 – 50)/ 2 + vx / 4
and a little algebra yields vx = -38.46 V.
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Engineering Circuit Analysis, 7th Edition
50.
Chapter Four Solutions
10 March 2006
v1
(a) (b)
Ref.
and
(a) We begin by noting that it is a voltage that is required; no current values are
requested. This is a three- mesh circuit, or a four-node circuit, depending on your
perspective. Either app roach requires th ree equations…. Except that applying the
supernode technique reduces the number of needed equations by one.
At the 1, 3 supernode:
0 = (v1 – 80)/ 10 + (v1 – v3)/ 20 + (v3 – v1)/ 20 + v3/ 40 + v3/ 30
v3 - v1 = 30
We simplify these two equations and collect terms, yielding
0.1 v1 + 0.05833 v3 = 8
- v1 + v3 = 30
Solving, we find that v3 = 69.48 V
Both ends of the
resistor are
connected to the
supernode, so we
could actually just
ignore it…
(b) Mesh analysis would be straightforward, 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 o nly two equations, and the desired voltage will be a nodal
voltage.
At the b, c, d supernode:
and:
vd – vb = 30
0 = (vb – 80)/ 10 + vd/ 40 + vc/ 30
vc – vd = 9
Simplify and collect terms:
Solving, vd (= v3) = 67.58 V
0.1 vb + 0.03333 vc + 0.025 vd = 80
-vb
+
vd = 30
vd = 9
vc -
(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 superm esh, 1 KCL equation
and Ohm’s law. Nodal analysis requires 1 supernode, 1 KVL equation, 1 other nodal
equation, and one equation to express
i1 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.”
and:
Mesh 1:
2, 3 supermesh:
Rewriting,
-80 + 10 i1 + 20 i1 – 20 i2 + 30 i1 – 30 i3 = 0
20 i2 – 20 i1 – 30 + 40 i3 + 30 i3 – 30 i1 = 0
i2 - i3 = 5 i1
60
i1 – 20 i2 – 30 i3 = 80
-50 i1 + 20 i2 + 70 i3 = 30
5 i1 – i2 + i3 = 0
Solving, i3 = 4.727 A
so
v3 = 40 i3 = 189 V.
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Engineering Circuit Analysis, 7th Edition
51.
Chapter Four Solutions
10 March 2006
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, form ing a supernode about the
30-V source would lead to 5 – 1 – 1 = 3 sim ulataneous equations as well, plus several
subtraction and division opera tions to find the currents. Thus, m esh a nalysis has a
slight edge here.
Define three clockwis e m esh currents: ia in the lef t-most m esh, ib in the top r ight
mesh, and ic in the bottom right mesh. Then our mesh equations will be:
Mesh a:
Mesh b:
Mesh c:
-80 + (10 + 20 + 30) ia – 20 ib – 30 ic = 0
-30 + (12 + 20) ib – 12 ic – 20 ia = 0
(12 + 40 + 30) ic – 12 ib – 30 ia = 0
[1]
[2]
[3]
Simplifying and collecting terms,
60 ia – 20 ib – 30 ic = 80
-20 ia + 32 ib – 12 ic = 30
-30 ia – 12 ib + 82 ic = 0
[1]
[2]
[3]
Solving, we find that ia = 3.549 A, ib = 3.854 A, and ic = 1.863 A. Thus,
i1 = ia = 3.549 A
and i2 = ia – ic = 1.686 A.
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Engineering Circuit Analysis, 7th Edition
52.
10 March 2006
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 actua lly find the current i10. Mesh analy sis, on the o ther hand, w ill
require 2 mesh/supermesh equations, 1 KCL equation, and one subtraction step to find
i10. Thus, mesh analys is has a clear edge. Define three clockwise m esh currents: i1 in
the bottom left mesh, i2 in the top mesh, and i3 in the bottom right mesh.
MESH 1:
i1 = 5 mA by inspection
SUPERMESH:
MESH
Sim
Chapter Four Solutions
3:
plify:
i1 – i2 = 0.4 i10
i1 – i2 = 0.4(i3 – i2)
i1 – 0.6 i2 – 0.4 i3 = 0
[1]
[2]
-5000 i1 – 10000 i2 + 35000 i3 = 0 [3]
0.6
i2 + 0.4 i3 = 5×10-3 [2]
-10000 i2 + 35000 i3 = 25
[3]
Solving, we find i2 = 6.6 mA and i3 = 2.6 mA. Since i10 = i3 – i2, we find that
i10 = -4 mA.
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Engineering Circuit Analysis, 7th Edition
53.
Chapter Four Solutions
10 March 2006
For this circuit problem, nodal analysis will require 3 simultaneous nodal equations,
then subtraction/ division steps to obtain th e desired curren ts. Mesh analysis requ ires
1 mesh equation, 1 superm esh equation, 2 simple KCL equations and one subtraction
step to dete rmine the currents. If either te chnique has an edge in this situa tion, it’s
probably mesh analysis. Thus, define four clockwise mesh equations: ia in the bottom
left mesh, ib in the top lef t mesh, ic in the top right m esh, and id in the bottom right
mesh.
At the a, b, c supermesh:
-100 + 6 ia + 20 ib + 4 ic + 10 ic – 10 id = 0 [1]
Mesh d:
100 + 10 id – 10 ic + 24 id = 0
[2]
KCL:
and
- ia + ib = 2
ib + ic = 3 i3 = 3 ia
[3]
[4]
-
Collecting terms & simplifying,
Solving,
6 ia + 20 ib + 14 ic – 10 id = 100
-10 ic + 34 id = -100
- ia + ib
= 2
= 0
-3 ia – ib + ic
[1]
[2]
[3]
[4]
ia = 0.1206 A, ib = 2.121 A, ic = 2.482 A, and id = -2.211 A. Thus,
i3 = ia = 120.6 mA and i10 = ic – id = 4.693 A.
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Engineering Circuit Analysis, 7th Edition
54.
Chapter Four Solutions
10 March 2006
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 KVL equation. Mesh analysis will require the solution of three simultaneous
mesh equations (one m esh current can be found by inspecti
on), 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 requi re 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:
v6
v3
v4
v1
v5
v2
-2×10-3 = (v1 – 1.3)/ 1.8×103
NODE 1:
→ v1 = -2.84 V.
2, 4 Supernode:
2.3×10-3 = (v2 – v5)/ 1x103 + (v4 – 1.3)/ 7.3×103 + (v4 – v5)/ 1.3×103 + v4/ 1.5×103
KVL equation:
Node 5:
-v2 + v4 = 5.2
0 = (v5 – v2)/ 1x103 + (v5 – v4)/ 1.3x103 + (v5 – 2.6)/ 6.3x103
Simplifying and collecting terms,
14.235 v2 + 22.39 v4 – 25.185 v5 = 35.275 [1]
-v2 +
v4
= 5.2
[2]
-8.19 v2
– 6.3 v4 + 15.79 v5 = 3.38
[3]
Solving, we find the voltage at the central node is v4 = 3.460 V.
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Engineering Circuit Analysis, 7th Edition
55.
Chapter Four Solutions
10 March 2006
Mesh analysis yields current values directly, so use that approach. We therefore
define four clockwise mesh currents, starting with i1 in th e left-most mesh, then i2, i3
and i4 moving towards the right.
Mesh
1:
-0.8ix + (2 + 5) i1 – 5 i2 = 0 [1]
Mesh
2:
i2 = 1 A by inspection
[2]
Mesh 3:
(3 + 4) i3 – 3(1) – 4(i4) = 0 [3]
Mesh 4:
(4 + 3) i4 – 4 i3 – 5 = 0
[4]
Simplify and collect terms, noting that ix = i1 – i2 = i1 – 1
-0.8(
i1 – 1) + 7 i1 – 5(1) = 0 yields i1 = 677.4 mA
Thus, [3] and [4] become:
7 i 3 – 4 i4 = 3
-4 i3 + 7 i4 = 5
[3]
[4]
Solving, we find that i3 = 1.242 A and i4 = 1.424 A. A map of individual branch
currents can now be drawn:
677.4 mA
→
-322.6 mA
↓
↑
677.4 mA
-242.0 mA
↓
↑
182.0 mA
↑
-1.424 A
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Engineering Circuit Analysis, 7th Edition
56.
Chapter Four Solutions
10 March 2006
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 em ploy 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 i1, i2, i3 and i4 star ting w ith the lef tmost mesh and moving towards the right of the circuit.
At the 1,2 supermesh:
and
2000 i1 + 6000 i2 – 3 + 5000 i2 = 0
i1 – i2 = 2×10-3
[1]
[2]
by inspection,
i4 = -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 i1 and i2
are sufficient to enable us to find the voltage across the current source.
Simplifying, we obtain
1000
Solving,
i1 = 1.923 mA and
2000 i1 + 11000 i2 = 3
i1 - 1000 i2 = 2
[1]
[2]
i2 = -76.92 μA.
Thus, the voltage across the 2-mA source (“+” reference at the top of the source) is
v = -2000 i1 – 6000 (i1 – i2) = -15.85 V.
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Engineering Circuit Analysis, 7th Edition
57.
Chapter Four Solutions
10 March 2006
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 cl ockwise mesh currents
ia, ib, ic, id starting with the left-most mesh and proceeding to the right of the circuit.
At the a, b supermesh:
-5 + 2 ia + 2 ib + 3 ib – 3 ic = 0
[1]
At the c, d supermesh:
3 ic – 3 ib + 1 + 4 id = 0
[2]
and
ia - ib = 3
ic - id = 2
[3]
[4]
Simplifying and collecting terms, we obtain
2 ia + 5 ib – 3 ic
= 5
-3 ib + 3 ic + 4 id = -1
ia - ib
= 3
ic - id = 2
[1]
[2]
[3]
[4]
Solving, we find ia = 3.35 A, ib = 350 mA, ic = 1.15 A, and id = -850 mA. As i1 = ib,
i1 = 350 mA.
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Engineering Circuit Analysis, 7th Edition
58.
Chapter Four Solutions
10 March 2006
Define a voltage vx at the top node of the current source I2, and a clockwise mesh
current ib in the right-most mesh.
We want 6 W dissipated in the 6-Ω resistor, which leads to the requirement ib = 1 A.
Applying nodal analysis to the circuit,
I1 + I2 = (vx – v1)/ 6 = 1
so our requirement is I1 + I2 = 1. There is no constraint on the value of v1 other than
we are told to select a nonzero value.
Thus, we choose I1 = I2 = 500 mA and v1 = 3.1415 V.
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Engineering Circuit Analysis, 7th Edition
59.
Chapter Four Solutions
10 March 2006
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 mesh currents i1, i2, i3,
i4 starting with the left-most mesh and proceeding towards the right of the circuit.
Mesh 1:
(2000 + 1000 + 5000) i1 – 6000 i2 – 2 = 0 [1]
2, 3 Supermesh:
2 + (5000 + 5000 + 1000 + 6000) i2 – 6000 i1 + (3000 + 4000 + 5000) i3 – 5000 i4
= 0 [2]
[3]
and
i2 - i3 = 7×10-3
Mesh 4:
i4 = -1 mA by inspection
[4]
Simplifying and combining terms,
8000
-6000
i1 –
6000 i2
1000 i2 – 1000 i3
i1 + 17000 i2 + 12000 i3
= 2
= 7
= -7
[1]
[4]
[2]
Solving, we find that
i1 = 2.653 A, i2 = 3.204 A, i3 = -3.796 A, i4 = -1 mA
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Engineering Circuit Analysis, 7th Edition
60.
Chapter Four Solutions
10 March 2006
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.
First, define clockwise m esh currents ix, i1, i2 and i3 starting from the lef t-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. Th en, define four nodal voltages, V 1, V 2, V 3 and
V4 moving from left to right along the top of the circuit.
At the left-most mesh, ix = –5 i1
[1]
For the supermesh, we can write 4i1 – 2ix + 2 + 2i3 = 0
and the corresponding KCL equation: i3 – i1 = 1
[2]
[3]
Substituting Eq. [1] into Eq. [2] and simplifying,
14
i1 + 2i3 = –2
–i1 + i3 = 1
Solving,
Then,
i1 = –250 mA and i3 = 750 mA.
ix = -5 i1 = 1.35 A and i2 = i1 – 2 = –2.25 A
Nodal voltages are straightfoward to find, then:
V4 = 2i3 = 1.5 V
V3 = 2 + V4 = 3.5 V
V2 – V3 = 2 i1 or V2 = 2 i1 + V3 = 3 V
V1 – V2 = 2 ix or V1 = 2 ix + V2 = 5.5 V
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Engineering Circuit Analysis, 7th Edition
61.
Chapter Four Solutions
10 March 2006
Hand analysis:
Define three clockwise m esh currents: i1 in the bottom left mesh, i2 in the top m esh,
and i3 in the bottom right mesh.
MESH 1:
i1 = 5 mA by inspection
SUPERMESH:
MESH
Sim
3:
plify:
i1 – i2 = 0.4 i10
i1 – i2 = 0.4(i3 – i2)
i1 – 0.6 i2 – 0.4 i3 = 0
[1]
[2]
-5000 i1 – 10000 i2 + 35000 i3 = 0 [3]
0.6
i2 + 0.4 i3 = 5×10-3 [2]
-10000 i2 + 35000 i3 = 25
[3]
Solving, we find i2 = 6.6 mA and i3 = 2.6 mA. Since i10 = i3 – i2, we find that
i10 = -4 mA.
PSpice simulation results:
i10
→
Summary: The current entering the right-hand node of the 10-k Ω resistor R2 is
equal to 4.000 mA. Since this current is –i10, i10 = -4.000 mA as found by hand.
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Engineering Circuit Analysis, 7th Edition
62.
Chapter Four Solutions
10 March 2006
Hand analysis:
Define the nodes as:
v6
v3
v4
v5
v2
v1
-2×10-3 = (v1 – 1.3)/ 1.8×103
NODE 1:
→ v1 = -2.84 V.
2, 4 Supernode:
2.3×10-3 = (v2 – v5)/ 1x103 + (v4 – 1.3)/ 7.3×103 + (v4 – v5)/ 1.3×103 + v4/ 1.5×103
KVL equation:
Node 5:
-v2 + v4 = 5.2
0 = (v5 – v2)/ 1x103 + (v5 – v4)/ 1.3x103 + (v5 – 2.6)/ 6.3x103
Simplifying and collecting terms,
14.235 v2 + 22.39 v4 – 25.185 v5 = 35.275 [1]
-v2 +
v4
= 5.2
[2]
-8.19 v2
– 6.3 v4 + 15.79 v5 = 3.38
[3]
Solving, we find the voltage at the central node is v4 = 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, 7th Edition
63.
Hand analysis:
At the 1,2 supermesh:
and
Chapter Four Solutions
10 March 2006
2000 i1 + 6000 i2 – 3 + 5000 i2 = 0
i1 – i2 = 2×10-3
[1]
[2]
by inspection,
i4 = -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 i1 and i2
are sufficient to enable us to find the voltage across the current source.
Simplifying, we obtain
1000
Solving,
i1 = 1.923 mA and
2000 i1 + 11000 i2 = 3
i1 - 1000 i2 = 2
[1]
[2]
i2 = -76.92 μA.
Thus, the voltage across the 2-mA source (“+” reference at the top of the source) is
v = -2000 i1 – 6000 (i1 – 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 hand calculation.
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Engineering Circuit Analysis, 7th Edition
64.
Chapter Four Solutions
10 March 2006
Hand analysis:
Define a voltage vx at the top node of the current source I2, and a clockwise mesh
current ib in the right-most mesh.
We want 6 W dissipated in the 6-Ω resistor, which leads to the requirement ib = 1 A.
Applying nodal analysis to the circuit,
I1 + I2 = (vx – v1)/ 6 = 1
so our requirement is I1 + I2 = 1. There is no constraint on the value of v1 other than
we are told to select a nonzero value.
Thus, we choose I1 = I2 = 500 mA and v1 = 3.1415 V.
PSpice simulation results:
Summary: We see from the labeled schematic above that our choice for I1, I2 and
V1 lead to 1 A through the 6-Ω resistor, or 6 W dissipated in that resistor, as desired.
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Engineering Circuit Analysis, 7th Edition
65.
Chapter Four Solutions
10 March 2006
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 righ t. The rightmost 2-Ω resistor has therefore been shorted out. Applying nodal analysis then,
Node 1:
-5 i1 = (v1 – v2)/ 2
[1]
Node 2:
0 = (v2 – v1)/ 2 + v2/ 2 + (v2 – 2)/ 2 [2]
and,
i1 = (v2 – 2)/ 2
[3]
Simplifying and collecting terms,
v1 + v2 = 10
-v1 + 3 v2 = 2
Solving, we find that
[1]
[2]
v1 = 3.143 V and v2 = 1.714 V.
Defining clockwise m esh currents i a, i b, i c, i d starting with the left-most m esh and
proceeding right, we may easily determine that
ia = -5 i1 = 714.3 mA
ib = -142.9 mA
ic = i1 – 2 = -2.143 A
id = 3 + ic = 857.1 mA
PSpice simulation results:
Summary:
The simulation results agree with the hand calculations.
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Engineering Circuit Analysis, 7th Edition
66.
Chapter Four Solutions
10 March 2006
(a) One possible circuit configuration of many that would satisfy the requirements:
100 Ω
+ vx
5V
50 Ω
-
20 Ω
3A
↑
2 vx
10 Ω
At node 1:
-3 = (v1 – 5)/ 100 + (v1 – v2)/ 50
[1]
At node 2:
2 vx = (v2 – v1)/ 50 + v2/ 30
[2]
and,
vx = 5 – v1 [3]
Simplifying and collecting terms,
150
v1 – 100 v2 = -14750
[1]
2970
v1 + 80 v2 = 15000
[2]
Solving, we find that v1 = 1.036 V and v2 = 149.1 V.
The current through the 100-Ω resistor is simply (5 – v1)/100 = 39.64 mA
The current through the 50-Ω resistor is (v1 – v2)/ 50 = -2.961 A,
and the current through the 20- Ω and 10- Ω ser ies com bination is v2/ 30 = 4.97 A.
Finally, the dependent source generates a current of 2 vx = 7.928 A.
(b) PSpice simulation results
Summary:
The simulated results agree with the hand calculations.
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Engineering Circuit Analysis, 7th Edition
67.
Chapter Four Solutions
10 March 2006
One possible solution of many:
+ 5V -
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 || 1Ω || 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, 7th Edition
68.
Chapter Four Solutions
We first name each node, resistor and voltage source:
R7
R4
R6
R3
R1
2
1
3
10 March 2006
R9
5
4
R2
V1
R5
R8
R10
0
We next write an appropriate input deck for SPICE:
* 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
And obtain the following output:
We see from this simulation result that the voltage v5 = 2.847 V.
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Engineering Circuit Analysis, 7th Edition
69.
Chapter Four Solutions
10 March 2006
One possible solution of many:
v1
R1
R2
v2
R3
v3
All resistors are
1 Ω, except R1,
which represents
5 1-Ω resistors
in series.
R4
R5
Verify:
v1 = 9(4/9) = 4 V
v2 = 9(3/9) = 3 V
v3 = 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
VOLTAGE
( 1) 9.0000 (
( 5) 1.0000
2)
NODE
VOLTAGE
NODE
VOLTAGE
NODE
4.0000 ( 3) 3.0000 ( 4) 2.0000
2
1
R1
R2
3
R3
4
R4
5
0
R5
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Engineering Circuit Analysis, 7th Edition
70.
Chapter Four Solutions
10 March 2006
(a) If only two bulbs are not lit (and thinking of each bulb as a resistor), the bulbs
must be in parallel- otherwise, the burne d 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 k Ω. A sim plified, elec trically-equivalent m odel f or this cir cuit
would be a 115 VAC source connected in parallel to a resistor Req such that
1/Req = 1/13.22×103 + 1/13.22×103 + …. + 1/13.22×103 (400 – 2 = 398 terms)
or Req = 33.22 Ω. We expect the source to provide 398 W.
* 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
( 1) 0.0000
NODE VOLTAGE
NODE VOLTAGE
VOLTAGE SOURCE CURRENTS
NAME
CURRENT
V1
0.000E+00
TOTAL POWER DISSIPATION 0.00E+00 WATTS
NODE VOLTAGE
This calculated power is not the value
sought. It is an artifact of the use of
an ac s ource, which re quires that we
perform an ac analysis. T he supplied
power is th en separately computed as
(1.15×102)(3.462) = 398.1 W.
**** 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
(c) The in herent serie s res istance of th e wire connection s 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, 7th Edition
1.
Chapter Five Solutions
10 March 2006
Define percent error as 100 [ex – (1 + x)]/ ex
x
0.001
0.005
0.01
0.05
0.10
0.50
1.00
5.00
1+x
1.001
1.005
1.01
1.05
1.10
1.50
2.00
6.00
ex
1.001
1.005
1.010
1.051
1.105
1.649
2.718
148.4
% error
5×10-5
1×10-3
5×10-3
0.1
0.5
9
26
96
Of course, “reasonable” is a very subjective term. However, if we choose x < 0.1, we
ensure that the error is less than 1%.
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Engineering Circuit Analysis, 7th Edition
2.
Chapter Five Solutions
10 March 2006
(a) Short-circuit the 10 V source.
Note that 6 || 4 = 2.4 Ω. By voltage division, the voltage across the 6 Ω
resistor is then
4
2.4
= 1.778 V
3 + 2.4
So that i1′ =
1.778
= 0.2963 A .
6
(b) Short-circuit the 4 V source.
Note that 3 || 6 = 2 Ω. By voltage division, the voltage across the 6 Ω
resistor is then
−10
So that i1′′ =
2
= −3.333 V
6
−3.333
= −0.5556 A .
6
(c) i1 = i1′ + i2′′ = -259.3 mA
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Engineering Circuit Analysis, 7th Edition
3.
Chapter Five Solutions
10 March 2006
Open circuit the 4 A source. Then, since
(7 + 2) || (5 + 5) = 4.737 Ω, we can calculate v1′ = (1)(4.737) = 4.737 V.
To find the total current flowing through the 7 Ω resistor, we first determine the total
voltage v1 by continuing our superposition procedure. The contribution to v1 from the
4 A source is found by first open-circuiting the 1 A source, then noting that current
division yields:
5
20
4
=
= 1.053 A
5 + (5 + 7 + 2) 19
Then, v1′′ = (1.053)(9) = 9.477 V. Hence, v1 = v1′ + v1′′ = 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, 7th Edition
4.
Chapter Five Solutions
10 March 2006
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 Vx. We next define three clockwise mesh
currents in the bottom three meshes: i1, iy and i4. Finally, we define a clockwise mesh
current i3 in the top mesh, noting that it is equal to –4 A.
Our general mesh equations are then:
-Vx + 18i1 – 10iy = 0
–10i1 + 15iy – 3i4 = 0
–3iy + 16i4 – 5i3 = 0
** Set Vx = 10 V, i3 = 0. Our mesh equations then become:
18i1 – 10 i′y
= 10
–10i1 + 15 i′y – 3i4 = 0
– 3 i′y + 16i4 = 0
Solving, i′y = 0.6255 A.
** Set Vx = 0 V, i3 = – 4 A. Our mesh equations then become:
18i1 – 10 i′′y
=0
–10i1 + 15 i′′y – 3i4 = 0
– 3 i′′y + 16i4 = -20
Solving, i′′y = –0.4222 A.
Thus, iy = i′y + i′′y = 203.3 mA.
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Engineering Circuit Analysis, 7th Edition
5.
Chapter Five Solutions
10 March 2006
We may solve this problem without writing circuit equations if we first realise that the
current i1 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 i1 = 4K1 + 10K2, where K1 and K2 are constants that depend
on the circuit topology and resistor values.
We may not change K1 or K2, as only the source voltages may be changed. If we
increase both sources by a factor of 10, then i1 increases by the same amount.
Thus, 4 V → 40 V and 10 V → 100 V.
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Engineering Circuit Analysis, 7th Edition
6.
Chapter Five Solutions
iA, vB “on”, vC = 0:
iA, vC “on”, vB = 0:
iA, vB, vC “on” :
ix = 20 A
ix = -5 A
ix = 12 A
so, we can write
ix’ + ix” + ix”’ = 12
ix’ + ix”
= 20
ix’
+ ix”’ = - 5
In matrix form,
10 March 2006
⎡12 ⎤
⎡1 1 1 ⎤ ⎡i′x ⎤
⎢1 1 0⎥ ⎢i′′⎥ = ⎢20⎥
⎢ ⎥
⎥ ⎢ x⎥
⎢
⎢⎣- 5 ⎥⎦
⎢⎣1 0 1⎥⎦ ⎢⎣i′x′′⎥⎦
(a) with iA on only, the response ix = ix’ = 3 A.
(b) with vB on only, the response ix = ix” = 17 A.
(c) with vC on only, the response ix = ix”’ = -8 A.
(d) iA and vC doubled, vB reversed: 2(3) + 2(-8) + (-1)(17) = -27 A.
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Engineering Circuit Analysis, 7th Edition
7.
Chapter Five Solutions
10 March 2006
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’ = 24
20
20
= 24
= 10 V
10 + 20 + 45 || 30
10 + 20 + 18
We find the contribution of the 2-A source by shorting both voltage sources and
applying current division:
10
⎡
⎤
vx” = 20 ⎢2
⎥ = 8.333 V
⎣ 10 + 20 + 18 ⎦
Finally, the contribution from the 45-V source is found by open-circuiting the 2-A
source and shorting the 24-V source. Defining v30 across the 30-Ω resistor with the
“+” reference on top:
0
= v30/ 20 + v30/ (10 + 20) + (v30 – 45)/ 45
solving, v30 = 11.25 V and hence vx”’ = -11.25(20)/(10 + 20) = -7.5 V
Adding the individual contributions, we find that vx = vx’ + vx” + vx”’ = 10.83 V.
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Engineering Circuit Analysis, 7th Edition
8.
Chapter Five Solutions
10 March 2006
The contribution of the 8-A source is found by shorting out the two voltage sources
and employing simple current division:
i3' = − 8
50
= -5A
50 + 30
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" =
100
= 6.25 A
(50 + 30) || 60 || 30
The contribution of the 60-V source is found in a similar way as i3"' = -60/30 = -2 A.
The total response is i3 = i3' + i3" + i3"' = -750 mA.
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9.
Chapter Five Solutions
10 March 2006
(a) By current division, the contribution of the 1-A source i2’ is
i2’ = 1 (200)/ 250 = 800 mA.
The contribution of the 100-V source is i2” = 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,
i2”’ = 0.5 (50)/ 250 = 100 mA
Thus, i2 = i2’ + i2” + i2”’ = 1.3 A
(b) P1A = (1) [(200)(1 – 1.3)] = 60 W
P200 = (1 – 1.3)2 (200) = 18 W
P100V = -(1.3)(100) = -130 W
P50 = (1.3 – 0.5)2 (50) = 32 W
P0.5A = (0.5) [(50)(1.3 – 0.5)] = 20 W
Check: 60 + 18 + 32 + 20 = +130.
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Engineering Circuit Analysis, 7th Edition
10.
Chapter Five Solutions
10 March 2006
We find the contribution of the 4-A source by shorting out the 100-V source and
analysing the resulting circuit:
'
'
'
4 = V1' / 20 + (V1' – V')/ 10
[1]
0.4 i1' = V1'/ 30 + (V' – V1')/ 10
[2]
where i1' = V1'/ 20
Simplifying & collecting terms, we obtain
30 V1' – 20 V' = 800
-7.2 V1' + 8 V' = 0
[1]
[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 ia flowing in the
left mesh and a clockwise mesh current ib flowing in the right mesh.
"
"
"
30 ia – 60 + 30 ia – 30 ib = 0 [1]
ib = -0.4 i1" = +0.4 ia
[2]
Solving, we find that ia = 1.25 A and so V" = 30(ia – ib) = 22.5 V.
Thus, V = V' + V" = 82.5 V.
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11.
Chapter Five Solutions
10 March 2006
(a) Linearity allows us to consider this by viewing each source as being scaled by
25/ 10. This means that the response (v3) will be scaled by the same factor:
25 iA'/ 10 + 25 iB'/ 10 = 25 v3'/ 10
B
∴ v3 = 25v3'/ 10 = 25(80)/ 10 = 200 V
(b)
iA' = 10 A, iB' = 25 A
iA" = 10 A, iB" = 25 A
iA = 20 A, iB = -10 A
→ v4' = 100 V
→ v4" = -50 V
→ v4 = ?
We can view this in a somewhat abstract form: the currents iA and iB multiply
the same circuit parameters regardless of their value; the result is v4.
⎡10 25⎤ ⎡a ⎤ ⎡100 ⎤
Writing in matrix form, ⎢
⎥⎢ ⎥ = ⎢
⎥ , we can solve to find
⎣25 10⎦ ⎣b ⎦ ⎣- 50⎦
a = -4.286 and b = 5.714, so that 20a – 10b leads to v4 = -142.9 V
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12.
Chapter Five Solutions
10 March 2006
With the current source open-circuited and the 7-V source shorted, we are left with
100k || (22k + 4.7k) = 21.07 kΩ.
Thus, V3V = 3 (21.07)/ (21.07 + 47) = 0.9286 V.
In a similar fashion, we find that the contribution of the 7-V source is:
V7V = 7 (31.97) / (31.97 + 26.7) = 3.814 V
Finally, the contribution of the current source to the voltage V across it is:
V5mA = (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, 7th Edition
13.
Chapter Five Solutions
10 March 2006
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,
3
⎛
⎞
i60μA = 60 × 10 − 6 ⎜
⎟ = 43.51 μA
⎝ 0.5 + 3 + 0.6368 ⎠
The contribution from the voltage source is found by first noting that 2.7M || 5M =
1.753 MΩ. The total current 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 i60μA + i1.5V = 43.74 μA and the
dissipated power is (43.74×10-9)2 (500×103) = 956.6 μW.
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14.
Chapter Five Solutions
10 March 2006
We first determine the contribution of the voltage source:
'
'
↑
I1'
Via mesh analysis, we write: 5 = 18000 I1' – 17000 Ix'
-6 Ix' = -17000 Ix' + 39000 Ix'
Solving, we find I1' = 472.1 mA and Ix' = 205.8 mA, so V' = 17×103 (I1' - Ix')
= 4.527 V. We proceed to find the contribution of the current source:
V"
Vx"
Via supernode: -20×10-3 = Vx"/ 22×103 + V"/ 0.9444×103
and
V" – Vx" = 6Ix" or V" – Vx" = 6 Vx"/ 22×103
[1]
[2]
Solving, we find that V" = -18.11 V. Thus, V = V' + V" = -13.58 V.
The maximum power is V2/ 17×103 = V2/ 17 mW = 250 mW, so
′ = 83.3 V.
V = (17)(250) = 65.19 = V′ - ( -18.11) . Solving, we find Vmax
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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15.
Chapter Five Solutions
10 March 2006
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 =
Thus,
5
2.1
- 2
= 195.1 mA
2 + 2.1
2 + 2.1
P2Ω = i2 . 2 = 76.15 mW
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16.
Chapter Five Solutions
10 March 2006
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 source acting alone (right).
V2
IB
vx'
V3
V1
IB
1, 3 supernode:
V1/ 100 + V1/ 17×103 + (V1 – 15)/ 33×103 + V3/ 103 = 20 IB [1]
and:
V1 – V3 = 0.7
Node 2:
[2]
-20 IB = (V2 – 15)/ 1000
[3]
We require one additional equation if we wish to have IB as an unknown:
20 IB + IB = V3/ 1000
[4]
Simplifying and collecting terms,
10.08912 V1 + V3 – 20×103 IB = 0.4545
V1 - V3
[1]
= 0.7
[2]
= 15
[3]
-V3 + 21×103 IB = 0
[4]
V2 + 20×103 IB
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 iB', noting that the 33-kΩ and 17-kΩ resistors are now in parallel. We
find that VTH = 16.85 cos 6t V by voltage division, and RTH = 100 || 17k || 33k =
99.12 Ω. We may now proceed:
20 iB' = vx' / 1000 + (vx' – 16.85 cos 6t)/ 99.12
20 iB' + iB'' = vx'/ 1000
B
B
[1]
[2]
Solving, we find that iB' = 798.6 cos 6t mA. Thus, adding our two results, we find the
complete current is
iB = iB' + IB = -31.04 + 798.6 cos 6t μA.
B
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Chapter Five Solutions
10 March 2006
17.
We first consider the effect of the 2-A source separately, using the left circuit:
3 ⎤
⎡
Vx' = 5 ⎢2
⎥ = 1.765 V
⎣ 3 + 14 ⎦
Next we consider the effect of the 6-A source on its own using the right circuit:
⎡ 9 ⎤
Vx" = 5 ⎢6
⎥ = 15.88 V
⎣ 9 + 8⎦
Thus, Vx = Vx' + Vx" = 17.65 V.
(b) PSpice verification (DC Sweep)
The DC
sweep
results
below
confirm
that Vx' =
1.765 V
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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
Vx:
Vx′ Vx′ − 2
− 4Vx′ =
+
100
50
which leads to Vx' = 9.926 mV.
The circuit on the right yields the contribution of the 6-A source to Vx:
− 4Vx′′ =
Vx′′ Vx′′
+
100 50
which leads to Vx" = 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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19.
Chapter Five Solutions
10 March 2006
Vx − 12 Vx Vx + 15
=0
+
+
1
3
2
so Vx = 2.455 V
Vx′ − 6 Vx′ Vx′ + 10
=0
+
+
1
3
2
so Vx′ = 0.5455 V
Vx′′ − 6 Vx′′ Vx′′ + 5
=0
+
+
1
3
2
so Vx′′ = 1.909 V
Adding, we find that Vx' + Vx" = 2.455 V = Vx as promised.
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20.
Chapter Five Solutions
10 March 2006
(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 therefore 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 transformations. 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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21.
Chapter Five Solutions
10 March 2006
(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 transformation, 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, 7th Edition
22.
Chapter Five Solutions
10 March 2006
(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.
1.6 cos 400t A
(b) 2k || 3k + 6k = 7.2 kΩ.
25 Ω
7.2k || 12k = 4.5 kΩ
3.5 kΩ
4.5 kΩ
(20)(4.5) = 90 V.
8 kΩ
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23.
Chapter Five Solutions
10 March 2006
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 5.8kΩ resistor is then simply given by voltage division:
i = 3 × 10-3
4.7
= 1.343 mA
4.7 + 5.8
The power dissipated by the 5.8-kΩ resistor is then i2 . 5.8×103 = 10.46 mW.
(Note that we did not “transform” either source, but rather drew on the relevant
discussion to understand why the 1-kΩ resistor could be omitted.)
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24.
Chapter Five Solutions
10 March 2006
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:
Ω
I →
Ω
I = 43/ 15.8 mA = 2.722 mA. Therefore, P5.8Ω = I2. 5.8×103 = 42.97 mW.
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25.
Chapter Five Solutions
10 March 2006
(100 kΩ)(6 mA) = 0.6 V
0.6 V
470 k || 300 k = 183.1 kΩ
(-3 – 0.6)/ 300×103 = -12 μA
(183.1 kΩ)(-12 μA) = -2.197 V
183.1
-2.197 V
← I
Solving, 9 + 1183.1×103 I – 2.197 = 0, so I = -5.750 μA. Thus,
P1MΩ = I2 . 106 = 33.06 μW.
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26.
Chapter Five Solutions
10 March 2006
(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 + 47×103 I1 – 4 I1 + 13.3×103 I1 + 200 = 0
Solving, we find that I1 = -247/ (60.3×103 – 4) = -4.096 mA.
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27.
Chapter Five Solutions
10 March 2006
(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)
V1 is the top-labelled voltage (12.77 mV).
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28.
Chapter Five Solutions
10 March 2006
(a) 12/ 9000 = 1.333 mA. 9k || 7k = 3.938 kΩ. → (1.333 mA)(3.938 kΩ) = 5.249
V.
3.938
5.249 V
5.249/ 473.938×103 = 11.08 μA
11.08 μA
473.9 kΩ
10 kΩ
473.9 k || 10 k = 9.793 kΩ. (11.08 mA)(9.793 kΩ) = 0.1085 V
0.1085 V
11.793 kΩ
17 kΩ
Ix = 0.1085/ 28.793×103 = 3.768 μA.
(b)
Ix
↓
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29.
Chapter Five Solutions
10 March 2006
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. Req = 6.75 MΩ
∴ Ix = 8.749/ (6.75 + 4.7) μA = 764.1 nA.
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30.
Chapter Five Solutions
10 March 2006
To begin, note that (1 mA)(9 Ω) = 9 mV, and 5 || 4 = 2.222 Ω.
15
9
2.222 Ω
The above circuit may not be further simplified using only source transformation
techniques.
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31.
Chapter Five Solutions
10 March 2006
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 may
therefore be neglected from the perspective of terminals a & b. Thus, we may draw:
2Ω
7.25 A
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32.
Chapter Five Solutions
10 March 2006
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 compute this current,
I = 1.8/ 1.840 + 0 + 0 μA = 978.3 nA.
Thus,
P1MΩ = (978.3×10-9)2 (106) = 957.1 nW.
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33.
Chapter Five Solutions
10 March 2006
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)
1.567
1.563
1.558
Current (mA)
1.6681
6.599
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 = Vsrc – 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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34.
Chapter Five Solutions
10 March 2006
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)
1.567
1.563
1.558
Current (mA)
1.6681
6.599
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 Rp
(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, 7th Edition
35.
Chapter Five Solutions
10 March 2006
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, 7th Edition
36.
Chapter Five Solutions
10 March 2006
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 resistor in the circuit to the right is:
(-1)
2 || 5
= -0.4167 V.
2 || 5 + 2
-1 V
±
Thus, the power dissipated by the 5-Ω resistor is (-0.4167)2 / 5 = 34.73 mW.
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37.
Chapter Five Solutions
10 March 2006
(a) We may omit the 10 Ω resistor from the circuit, as it does not affect the voltage or
current associated with RL 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 RL.
(b)
We see from simulating both circuits simultaneously that the
voltage across RL is the same (4 V).
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38.
Chapter Five Solutions
10 March 2006
(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 I2R = 987.6 mW.
(b) We note that PSpice will NOT allow the 7 Ω resistor to be left floating! For both
circuits 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 resistor; 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, 7th Edition
39.
Chapter Five Solutions
10 March 2006
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 –0.75v3 current source (arrow pointing upwards) in
parallel with 4 Ω. Selecting the bottom node as a reference terminal, and naming the
top left node Vx and the top right node Vy, we write the following equations:
–0.75v3 = Vx/4 + (Vx – Vy)/3
3 = Vy/2 + (Vy – Vx)/3
v3 = Vy – Vx
Solving, we find that v3 = –2 V.
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Engineering Circuit Analysis, 7th Edition
40.
Chapter Five Solutions
10 March 2006
(a) RTH = 25 || (10 + 15) = 25 || 25 = 12.5 Ω.
25
⎛
⎞
⎛ 15 + 10 ⎞
VTH = Vab = 50 ⎜
⎟ + 100 ⎜
⎟ = 75 V.
⎝ 10 + 15 + 25 ⎠
⎝ 15 + 10 + 25 ⎠
(b) If Rab = 50 Ω,
2
⎡ ⎛ 50 ⎞⎤ ⎛ 1 ⎞
P50Ω = ⎢75 ⎜
⎟⎥ ⎜ ⎟ = 72 W
⎣ ⎝ 50 + 12.5 ⎠⎦ ⎝ 50 ⎠
(c) If Rab = 12.5 Ω,
2
⎡ ⎛ 12.5 ⎞⎤ ⎛ 1 ⎞
P12.5Ω = ⎢75 ⎜
⎟⎥ ⎜
⎟ = 112.5 W
⎣ ⎝ 12.5 + 12.5 ⎠⎦ ⎝ 12.5 ⎠
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41.
Chapter Five Solutions
10 March 2006
(a) Shorting the 14 V source, we find that RTH = 10 || 20 + 10 = 16.67 Ω.
Next, we find VTH by determining VOC (recognising that the right-most 10 Ω resistor
carries no current, hence we have a simple voltage divider):
10 + 10
VTH = VOC = 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) I5Ω = 9.333/ (5 + 16.67) = 0.4307 A. Thus,
P5Ω = (0.4307)2 . 5 = 927.5 mW
(c)
We see from the PSpice simulation that keeping four significant digits in
calculating the Thévenin equivalent yields at least 3 digits’ agreement in the results.
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Engineering Circuit Analysis, 7th Edition
42.
Chapter Five Solutions
10 March 2006
(a) Replacing the 7 Ω resistor with a short circuit, we find
ISC = 15 (8)/ 10 = 12 A.
Removing the short circuit, and open-circuiting the 15 A source, we see that
RTH = 2 + 8 = 10 Ω.
Thus, VTH = ISCRTH = (12)(10) = 120 V.
Our Thévenin equivalent is therefore a 120 V source in series with 10 Ω.
(b) As found above, IN = 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 combination of current division and
Ohm’s law, we find
⎛ 10 ⎞
v1 = 7 ⎜ 12 ⎟ = 49.41 V
⎝ 17 ⎠
As expected, the results are equal.
(d) Employing the more convenient Thévenin equivalent model,
v1 = 120(1)/17 = 7.059 V.
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43.
Chapter Five Solutions
10 March 2006
(a) RTH = 10 mV/ 400 μA = 25 Ω
(b) RTH = 110 V/ 363.6×10–3 A = 302.5 Ω
(c) Increased current leads to increased filament 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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44.
Chapter Five Solutions
10 March 2006
(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
RTH = (5 + 10) || 10 + 10 = 16 Ω.
To continue, we return to the original circuit and replace the 1 Ω resistor 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,
+3=0
– 4 + 9i1 – 9i2
=0
– 9i1 + 34i2 – 10isc
–3
– 10i2 + 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 = solve(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, I1Ω = VTH/(RTH + 1) = 200 mA.
Thus, P1Ω = (0.2)2.1 = 40 mW.
Switching to the Norton equivalent, we find I1Ω by current division:
I1Ω = (0.2125)(16)/(16+1) = 200 mA. Once again, P1Ω = 40 mW (as
expected).
(c) As we can see from simulating the original circuit simultaneously with its
Thevenin and 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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45.
Chapter Five Solutions
10 March 2006
(a) Removing terminal c, we need write only one nodal equation:
Vb − 2 Vb − 5
, which may be solved to
+
12
15
yield Vb = 4 V. Therefore, Vab = VTH = 2 – 4
= -2 V.
RTH = 12 || 15 = 6.667 Ω. We may then
calculate IN as IN = VTH/ RTH
0.1 =
= -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 Vb = 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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46.
Chapter Five Solutions
10 March 2006
(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 parallel with 10 Ω in parallel with
(20 Ω + 40 Ω), so
RTH = 50 || 10 || (20 + 40) = 7.317 Ω
Using superposition to determine the voltage Vxx' across the 50-Ω resistor, we find
⎡
⎡
⎤
50 || (20 + 40) ⎤
40
Vxx' = VTH = ⎢88
⎥ + (1)(50 || 10) ⎢
⎥
⎣ 10 + [50 || (20 + 40)] ⎦
⎣ 40 + 20 + (50 || 10) ⎦
40
⎡ 27.27 ⎤
⎡
⎤
= ⎢88
+ (1)(8.333) ⎢
⎥
⎥ = 69.27 V
⎣ 37.27 ⎦
⎣ 40 + 20 + 8.333 ⎦
(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 parallel 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
27.27 ⎤ ⎛ 40 ⎞
⎡
Vyy' = VTH = (1)(RTH) + ⎢88
⎟
⎥⎜
⎣ 10 + 27.27 ⎦ ⎝ 20 + 40 ⎠
= 59.52 V
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47.
Chapter Five Solutions
10 March 2006
(a) Select terminal b as the reference terminal, and define a nodal voltage V1 at the
top of the 200-Ω resistor. Then,
0 =
V1 − 20 V1 − VTH
V
+
+ 1
40
100
200
[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 + 15/ 20) V1 – VTH = 0
[1]
[2]
Solving, we find that VTH = 38.89 V. To find RTH, we short the voltage source and
inject 1 A into the port:
V − Vin V1
V
V1
0 = 1
[1]
+
+ 1
100
40 200
+
Vin − V1
1.5 i1 + 1 =
[2]
100
Vin
i1 = V1/ 200
[3]
1A
-
Ref.
Combining Eqs. [2] and [3] yields
1.75 V1 – Vin = -100
[4]
Solving Eqs. [1] & [4] then results in Vin = 177.8 V, so that RTH = Vin/ 1 A = 177.8 Ω.
(b) Adding a 100-Ω load to the original circuit or our Thévenin equivalent, the
voltage across the load is
100
⎛
⎞
2
V100Ω = VTH ⎜
⎟ = 14.00 V , and so P100Ω = (V100Ω) / 100 = 1.96 W.
⎝ 100 + 177.8 ⎠
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48.
Chapter Five Solutions
10 March 2006
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 Vx across the 200-Ω resistor.
Then,
1 = V1/ 100 + (V1 – Vx)/ 50
-0.1 V1 = Vx/ 200 + (Vx – V1)/ 50
[1]
[2]
which may be simplified to
3 V1 – 2 Vx = 100
0
16 V1 + 5 Vx =
[1]
[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, IN = 0.
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49.
Chapter Five Solutions
10 March 2006
With no independent sources present, VTH = 0.
We decide to inject a 1-A current into the port:
vx
v•f
A
Ref.
Node ‘x’:
Supernode:
and:
0.01 vab = vx/ 200 + (vx – vf)/ 50
1 = vab/ 100 + (vf – vx)
vab – vf = 0.2 vab
[1]
[2]
[3]
Rearranging and collecting terms,
-2 vab + 5 vx – 4 vf = 0
vab – 2 vx + 2 vf = 100
- vf = 0
0.8 vab
[1]
[2]
[3]
Solving, we find that vab = 192.3 V, so RTH = vab/ (1 A) = 192.3 Ω.
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50.
Chapter Five Solutions
10 March 2006
We first find RTH by shorting out the voltage source and open-circuiting the current
source.
Looking into the terminals a & b, we see
RTH = 10 || [47 + (100 || 12)]
= 8.523 Ω.
Returning to the original circuit, we decide to perform nodal analysis to obtain VTH:
-12×103 = (V1 – 12)/ 100×103 + V1/ 12×103 + (V1 – VTH)/ 47×103
[1]
12×103 = VTH/ 10×103 + (VTH – V1)/ 47×103
[2]
Rearranging and collecting terms,
0.1146 V1 - 0.02128 VTH = -11.88 [1]
[2]
-0.02128 V1 + 0.02128 VTH = 12
Solving, we find that VTH = 83.48 V.
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51.
Chapter Five Solutions
10 March 2006
(a) RTH = 4 + 2 || 2 + 10 = 15 Ω.
(b) same as above: 15 Ω.
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52.
Chapter Five Solutions
10 March 2006
For Fig. 5.78a, IN = 12/ ~0 → ∞ A in parallel with ~ 0 Ω.
For Fig. 5.78b, VTH = (2)(~∞) → ∞ V in series with ~∞ Ω.
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53.
Chapter Five Solutions
10 March 2006
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
+
-
I = 0.5 Vx + 0.25 Vx = 0.5 + 0.25 = 0.75 A.
RTH = 1/ I = 1.333 Ω.
The Norton equivalent is 0 A in parallel with 1.333 Ω.
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54.
Chapter Five Solutions
10 March 2006
Performing nodal analysis to determine VTH,
100×10-3 = Vx/ 250 + Voc/ 7.5×103
and
[1]
Vx – Voc = 5 ix
where ix = Vx/ 250. Thus, we may write the second equation as
0.98 Vx – Voc = 0
[2]
Solving, we find that Voc = VTH = 23.72 V.
In order to determine RTH, we inject 1 A into the port:
Vab/ 7.5×103 + Vx/ 250
=1
[1]
and Vx – Vab = 5 ix = 5Vx/ 250
or
-Vab + (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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55.
Chapter Five Solutions
10 March 2006
We first note that VTH = Vx, so performing nodal analysis,
-5 Vx = Vx/ 19
Thus, VTH (and hence IN) = 0.
which has the solution Vx = 0 V.
(Assuming RTH ≠ 0)
To find RTH, 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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56.
Chapter Five Solutions
10 March 2006
Shorting out the voltage source, we redraw the circuit with a 1-A source in place of
the 2-kΩ resistor:
Ref.
1A
V
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,
–V = 0
[1]
0.01667 vgs + 0.00433 V = 1
[2]
vgs
Solving, we find that vgs = V = 47.62 V. Hence, RTH = V/ 1 A = 47.62 Ω.
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57.
Chapter Five Solutions
10 March 2006
We replace the source vs and the 300-Ω resistor with a 1-A source and seek its
voltage:
V2
V1
+
1A
V
Ref.
By nodal analysis,
1 = V1/ 2×106
so V1 = 2×106 V.
Since V = V1, we have Rin = V/ 1 A = 2 MΩ.
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58.
Chapter Five Solutions
10 March 2006
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 terminals:
+
V
-
1A
Ref.
We select the bottom node as our reference terminal, and define nodal voltages V1
and V2. Then,
1 = V1 / 2×106 + (V1 – V2)/ rπ
[1]
0.02 vπ = (V2 – V1)/ rπ + V2/ 1000 + V2/ 2000
[2]
where vπ = V1 – V2
Simplifying & collecting terms,
(2×106 + rπ) V1 – 2×106 V2 = 2×106 rπ
[1]
-(2000 + 40 rπ) V1 + (2000 + 43 rπ) V2 = 0 [2]
⎞
⎛
666.7 + 14.33 rπ
⎟⎟ .
Solving, we find that V1 = V = 2 × 106 ⎜⎜
6
⎝ 2 × 10 + 666.7 + 14.33 rπ ⎠
Thus,
RTH = 2×106 || (666.7 + 14.33 rπ) Ω.
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59.
Chapter Five Solutions
10 March 2006
(a) We first determine vout in terms of vin and the resistor values only; in this case,
VTH = vout. Performing nodal analysis, we write two equations:
0=
−vd ( −vd − vin ) ( −vd − vo )
+
+
Ri
R1
Rf
[1]
0=
and
( vo + vd ) + ( vo − Avd )
Rf
Ro
[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, VTH =
vin Ri ( Ro − AR f )
R1 Ro + Ri Ro + R1 R f + Ri R f + R1 Ri + AR1 Ri
approaches –Rf/R1.
, which in the limit of A → ∞,
To find RTH, 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:
0=
−vd vd ( −vd − vT )
− +
Ri
R1
Rf
[1]
and
1=
( vT + vd ) + ( vT − Avd )
Rf
Ro
[2]
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 VT/1 = VT, this is our Thévenin equivalent resistance (RTH).
+
vout
–
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60.
Chapter Five Solutions
10 March 2006
Such a scheme probably would lead to maximum or at least near-maximum power
transfer to our home. Since we pay the utility company based on the power we use,
however, this might not be such a hot idea…
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61.
Chapter Five Solutions
10 March 2006
We need to find the Thévenin equivalent resistance of the circuit connected to RL, 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:
V2
V1
+ VTH -
Ref.
V1 = 20 8/ 20 = 8 V;
V2 = -2 (6) = -12 V.
We define VTH = V1 – V2 = 8 + 12 = 20 V. Then,
2
VTH
400
PRL|max =
=
= 6.329 W
4 RL
4(15.8)
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Engineering Circuit Analysis, 7th Edition
62.
Chapter Five Solutions
10 March 2006
(a) RTH = 25 || (10 + 15) = 12.5 Ω
Using superposition, Vab = VTH = 50
25
15 + 10
= 75 V.
+ 100
15 + 10 + 25
50
(b) Connecting a 50-Ω resistor,
2
VTH
752
Pload =
=
= 90 W
R TH + R load
12.5 + 50
(c) Connecting a 12.5-Ω resistor,
Pload =
2
VTH
752
=
= 112.5 W
4 R TH
4 (12.5)
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Engineering Circuit Analysis, 7th Edition
63.
Chapter Five Solutions
10 March 2006
(a) By inspection, we see that i10 = 5 A, so
VTH = Vab = 2(0) + 3 i10 + 10 i10 = 13 i10 = 13(5) = 65 V.
To find RTH, we open-circuit the 5-A source, and connect a 1-A source between
terminals a & b:
+
1A
Vx
–
A simple KVL equation yields Vx = 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) Pmax =
2
VTH
4 R TH
=
652
= 70.42 W
4(15)
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Engineering Circuit Analysis, 7th Edition
64.
(a)
Chapter Five Solutions
10 March 2006
Replacing the resistor RL with a 1-A source, we seek the voltage that develops across
its terminals with the independent voltage source shorted:
−10i1 + 20ix + 40i1 = 0
[1] ⇒ 30i1 + 20ix = 0 [1]
and i1 − ix = 1
[2] ⇒
i1 − ix = 1
[2]
Solving, i1 = 400 mA
So V = 40i1 = 16 V and RTH =
(b)
V
= 16 Ω
1A
Removing the resistor RL from the original circuit, we seek the resulting open-circuit
voltage:
+
VTH
-
VTH − 10i1 VTH − 50
+
[1]
20
40
V − 50
where i1 = TH
40
1 ⎛ V − 50 ⎞ ⎛ VTH − 50 ⎞
V
so [1] becomes 0 = TH − ⎜ TH
⎟+⎜
⎟
20 2 ⎝ 40 ⎠ ⎝ 40 ⎠
V
V − 50
0 = TH + TH
20
80
0 = 4VTH + VTH − 50
0=
5VTH = 50
or VTH = 10 V
Thus, if
RL = RTH = 16 Ω,
VRL = VTH
RL
V
= TH = 5 V
RL + RTH
2
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Engineering Circuit Analysis, 7th Edition
Chapter Five Solutions
10 March 2006
65.
(a)
I N = 2.5 A
↓
2A
RN
20 Ω
20i 2 = 80
i = 2A
By current division,
RN
RN + 20
Solving, RN = RTH = 80 Ω
2 = 2.5
Thus, VTH = VOC = 2.5 × 80 = 200 V
2
VTH
2002
=
= 125 W
4 RTH 4 × 80
(b)
Pmax =
(c)
RL = RTH = 80 Ω
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Chapter Five Solutions
10 March 2006
66.
10 W to 250Ω corresp to 200 mA.
IN
By Voltage ÷,
So
RN
IR = IN
R
20 W to 80Ω corresp to 500 mA.
RN
R + RN
0.2 = I N
RN
250 + RN
[1]
0.5 = I N
RN
80 + RN
[2]
Solving, I N = 1.7 A and RN = 33.33 Ω
(a)
If vLiL is a maximum,
RL = RN = 33.33 Ω
33.33
= 850 mA
33.33 + 33.33
vL = 33.33iL = 28.33V
iL = 1.7 ×
(b)
If vL is a maximum
VL = I N ( RN RL )
So vL is a maximum when RN RL is a maximum, which occurs at RL = ∞.
Then iL = 0 and vL = 1.7 × RN = 56.66 V
(c)
If iL is a maximum
iL = iN
RN
; max when RL = 0 Ω
RN + RL
So iL = 1.7A
vL = 0 V
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Engineering Circuit Analysis, 7th Edition
67.
Chapter Five Solutions
10 March 2006
There is no conflict with our derivation concerning maximum power. While a dead
short across the battery terminals will indeed result in maximum current draw from
the battery, and power is indeed proportional to i2, the power delivered to the load is
i2RLOAD = i2(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, 7th Edition
68.
Chapter Five Solutions
10 March 2006
Remove RE: RTH = RE Rin
V − vπ V − vπ
+
[1]
300
70 × 103
vπ
v −V
v −V
[2]
+ π
+ π
0=
3
10 ×10
300
70 × 103
bottom node: 1 − 3 × 10−3 vπ =
at other node:
1A
V
Simplifying and collecting terms,
210×105 = 70×103 V + 300V + 63000 vπ – 70x103 vπ - 300 vπ
or 70.3 × 103V − 7300vπ = 210 ×105 [1]
0 = 2100vπ + 70 × 103 vπ − 70 × 103V + 300vπ − 300V
or − 69.7 × 103V + 72.4 × 103 vπ = 0
[2]
solving, V = 331.9V So RTH = RE 331.9Ω
Next, we determine vTH using mesh analysis:
and:
−vs + 70.3 × 103 i1 − 70 × 103 i2 = 0
[1]
80 × 103 i2 − 70 × 103 i1 + RE i3 = 0
[2]
−3
i3 − i2 = 3 × 10 vπ
−3
[3]
or i3 − i2 = 3 × 10 (10 × 10 )i2
or i3 − i2 = 30i2
or
−31i2 + i3 = 0
Solving :
⎡ 70.3 × 103
⎢
3
⎢ −70 ×10
⎢
0
⎣
3
−70 × 103
80 × 103
−31
[3]
0 ⎤ ⎡ i1 ⎤ ⎡vs ⎤
⎥
RE ⎥ ⎢⎢i2 ⎥⎥ = ⎢⎢ 0 ⎥⎥
1 ⎥⎦ ⎢⎣ i3 ⎥⎦ ⎢⎣ 0 ⎥⎦
We seek i3:
−21.7 × 103 vs
i3 =
7.24 ×106 + 21.79 × 103 RE
−21.7 × 103 RE
vs
So VOC = VTH = RE i3 =
7.24 ×106 + 21.79 ×103 RE
2
2
⎡ V
⎤ ⎡
⎤
8 vs 2
−21.7 ×103 RE
P8Ω = 8 ⎢ TH ⎥ = ⎢
⎥
2
6
3
⎣ RTH + 8 ⎦ ⎣ 7.24 ×10 + 21.79 × 10 RE ⎦ ⎡ 331.9 RE ⎤
⎢ 331.9 + R ⎥
⎣
E ⎦
11.35 ×106 (331.9 + RE )2
=
vs 2
6
3
2
(7.24 ×10 + 21.79 × 10 RE )
This is maximized by setting RE = ∞.
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Engineering Circuit Analysis, 7th Edition
69.
Chapter Five Solutions
10 March 2006
Thévenize the left-hand network, assigning the nodal voltage Vx at the free end of
right-most 1-kΩ resistor.
Vx oc
A single nodal equation: 40 × 10−3 =
7 × 103
So VTH = Vx oc = 280 V
RTH = 1 k + 7 k = 8 kΩ
Select R1 = RTH = 8 kΩ.
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Engineering Circuit Analysis, 7th Edition
Chapter Five Solutions
10 March 2006
100 kΩ
70.
1 MΩ
850 MΩ
D = RA + RB + RC = 1 + 850 + 0.1 = 851.1× 106
RA RB 106 ×105
=
= 117.5 Ω
D
D
R R
105 × 850 × 106
= 99.87 k Ω
R2 = B C =
D
851.1×106
RC RA 850 ×106 ×106
=
= 998.7 k Ω
R3 =
D
851.1×106
R1 =
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Engineering Circuit Analysis, 7th Edition
Chapter Five Solutions
10 March 2006
71.
R1
R2
R3
N = R1 R2 + R2 R3 + R3 R1
= 0.1× 0.4 + 0.4 × 0.9 + 0.9 × 0.1
= 0.49Ω
N
= 1.225 Ω
RA =
R2
RB =
N
= 544.4 mΩ
R3
RC =
N
= 4.9 Ω
R1
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Engineering Circuit Analysis, 7th Edition
Chapter Five Solutions
10 March 2006
72.
Δ1 : 1 + 6 + 3 = 10 Ω
6 ×1
6×3
3 ×1
= 0.6,
= 1.8 ,
= 0.3
10
10
10
Δ 2 : 5 + 1 + 4 = 10 Ω
5 ×1
1× 4
5× 4
= 0.5 ,
= 0.4 ,
=2
10
10
10
1.8 + 2 + 0.5 = 4.3 Ω
0.3 + 0.6 + 0.4 = 1.3 Ω
1.3 4.3 = 0.9982 Ω
0.9982 + 0.6 + 2 = 3.598 Ω
3.598 6 = 2.249 Ω
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Engineering Circuit Analysis, 7th Edition
Chapter Five Solutions
10 March 2006
73.
6 × 2 + 2 × 3 + 3 × 6 = 36 Ω 2
36
36
36
= 6Ω,
= 18 Ω ,
= 12 Ω
6
2
3
12 4 = 3 Ω, 6 12 Ω = 4 Ω
4 + 3 + 18 = 25 Ω
18
3 × = 2.16 Ω
25
18
4 × = 2.88 Ω
25
3
4 × = 0.48 Ω
25
9.48 × 2.16 + 9.48 × 2.88 + 2.88 × 2.16 = 54 Ω 2
54
54
= 18.75 Ω
= 5.696 Ω
2.88
9.48
54
= 25 Ω
2.16
75 18.75 = 15 Ω
96
100 25 = 20 Ω
(15 + 20) 5.696 = 4.899 Ω
∴ Rin = 5 + 4.899 = 9.899 Ω
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Engineering Circuit Analysis, 7th Edition
74.
Chapter Five Solutions
10 March 2006
We begin by converting the Δ-connected network consisting of the 4-, 6-, and 3-Ω
resistors to an equivalent Y-connected network:
RA
RB
RC
D = 6 + 4 + 3 = 13 Ω
R R
6× 4
= 1.846 Ω
R1 = A B =
D
13
R R
4×3
= 923.1mΩ
R2 = B C =
D
13
R R
3× 6
= 1.385 Ω
R3 = C A =
D
13
Then network becomes:
1.846 Ω
0.9231 Ω
1.385 Ω
Then we may write
Rin = 12 [13.846 + (19.385 6.9231)]
= 7.347Ω
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Engineering Circuit Analysis, 7th Edition
Chapter Five Solutions
75.
10 March 2006
1+1+ 2 = 4 Ω
1× 2 1
R1 =
= Ω
4
2
2 ×1 1
= Ω
R2 =
4
2
1× 1
= 0.25 Ω
R3 =
4
0.5
0.25
0.5
Next, we convert the Y-connected network on the left to a Δ-connected network:
1× 0.5 + 0.5 × 2 + 2 ×1 = 3.5 Ω 2
3.5
= 7Ω
0.5
3.5
= 1.75 Ω
RB =
2
3.5
= 3.5 Ω
RC =
1
RA =
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:
1.75
7
This circuit may be easily analysed to find:
0.25
1.458
12 × 1.458
= 5.454 V
1.75 + 1.458
RTH = 0.25 + 1.458 1.75
Voc =
= 1.045 Ω
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Engineering Circuit Analysis, 7th Edition
76.
Chapter Five Solutions
10 March 2006
We begin by converting the Y-network to a Δ-connected network:
N = 1.1 + 1.1 + 1.1 = 3 Ω 2
3
RA = = 3 Ω
1
3
RB = = 3 Ω
1
3
RC = = 3 Ω
1
3Ω
3
3
Next, we note that 1 3 = 0.75 Ω , and hence have a simple Δ-network. This is easily
converted to a Y-connected network:
0.75 + 3 + 3 = 6.75 Ω
0.75 × 3
= 0.3333 Ω
6.75
3× 3
= 1.333 Ω
R2 =
6.75
3 × 0.75
= 0.3333 Ω
R3 =
6.75
R1 =
Analysing this final
circuit,
1A
RN = 1.333 + 0.3333
= 1.667Ω
1/ 3
1/ 3 + 1 + 1/ 3
1
1
=
=
1+ 3 +1 5
= 0.2 A
= 200 mA
I N = I SC = 1×
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Engineering Circuit Analysis, 7th Edition
77.
Chapter Five Solutions
10 March 2006
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 terminals, the 10-Ω resistor in parallel
with the voltage source has no influence if we replace the “dependent” source with a
fixed 0.5-A source:
0.5 A
Then, we may write:
so that ia = 200 mA.
-1 + (10 + 10 + 10) ia – 10 (0.5) = 0
We next find that VTH = Vab = 10(-0.5) + 10(ia – 0.5) + 10(-0.5) = -13 V.
To determine RTH, we first recognise that with the 1-V source shorted, I1 = 0 and
hence the dependent current source is dead. Thus, we may write RTH from inspection:
RTH = 10 +
10 +
10 || 20 = 26.67 Ω.
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Engineering Circuit Analysis, 7th Edition
78.
Chapter Five Solutions
10 March 2006
(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 resistor as a leg.
Converting those networks into Δ-connected networks,
Σ = (17)(10) + (1)(4) + (4)(17) = 89×106 Ω2
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 resistor 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:
178 kΩ
27.49 kΩ
27.49 kΩ
3.912 kΩ
233.6 kΩ
We next attack the Y-connected network in the center:
Σ = (22.25)(22.25) + (22.25)(2.618) + (2.618)(22.25) = 611.6×106 Ω2
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,
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
312.6 Ω
21.33 kΩ
3.068 kΩ
(b)
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Engineering Circuit Analysis, 7th Edition
79.
Chapter Five Solutions
10 March 2006
(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, 7th Edition
80.
Chapter Five Solutions
10 March 2006
First, replace network to left of the 0.7-V source with its Thévenin equivalent:
15
= 2.609 V
100 + 15
RTH = 100k 15k = 13.04 k Ω
VTH = 20 ×
Redraw:
13.04 kΩ
2.609 V
Analysing the new circuit to find IB, we note that IC = 250 IB:
−2.609 + 13.04 × 103 I B + 0.7 + 5000( I B + 250 I B ) = 0
2.609 − 0.7
= 1.505 μA
13.04 × 103 + 251× 5000
I C = 250 I B = 3.764 × 10−4 A
IB =
= 376.4 μA
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Engineering Circuit Analysis, 7th Edition
81.
Chapter Five Solutions
10 March 2006
(a) Define a nodal voltage V1 at the top of the current source IS, and a nodal voltage
V2 at the top of the load resistor RL. Since the load resistor can safely dissipate 1 W,
and we know that
V22
PRL =
1000
then V2 max = 31.62 V . 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
[3]
or Vx = V1/ 6
Substituting Eq. [3] into Eqs. [1] and [2], we find that
7 V1 = 1200 IS
-5000 V1 + 6063.6 V2 = 0
[1]
[2]
Substituting V2 max = 31.62 V 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, 7th Edition
82.
Chapter Five Solutions
10 March 2006
Short out all but the source operating at 104 rad/s, and define three clockwise mesh
currents i1, i2, and i3 starting with the left-most mesh. Then
608 i1 – 300 i2
= 3.5 cos 104 t
-300 i1 + 316 i2 – 8 i3 = 0
-8 i2 + 322 i3 = 0
[1]
[2]
[3]
Solving, we find that i1(t) = 10.84 cos 104 t mA
i2(t) = 10.29 cos 104 t mA
i3(t) = 255.7 cos 104 t μA
Next, short out all but the 7 sin 200t V source, and and define three clockwise mesh
currents ia, ib, and ic starting with the left-most mesh. Then
608 ia – 300 ib
= -7 sin 200t [1]
-300 ia + 316 ib – 8 ic = 7 sin 200t [2]
[3]
-8 ib + 322 ic = 0
Solving, we find that ia(t) = -1.084 sin 200t mA
ib(t) = 21.14 sin 200t mA
ic(t) = 525.1 sin 200t μA
Next, short out all but the source operating at 103 rad/s, and define three clockwise
mesh currents iA, iB, and iC starting with the left-most mesh. Then
608 iA – 300 iB
=
0
-300 iA + 316 iB – 8 iC = 0
-8 iB + 322 iC = -8 cos 104 t
B
B
[1]
[2]
[3]
Solving, we find that iA(t) = -584.5 cos 103 t μA
iB(t) = -1.185 cos 103 t mA
iC(t) = -24.87 cos 103 t mA
B
We may now compute the power delivered to each of the three 8-Ω speakers:
p1 = 8[i1 + ia + iA]2 = 8[10.84×10-3 cos 104 t -1.084×10-3 sin 200t -584.5×10-6 cos 103 t]2
p2 = 8[i2 + ib + iB]2 = 8[10.29×10-3 cos 104 t +21.14×10-3 sin 200t –1.185×10-3 cos 103 t]2
p3 = 8[i3 + ic + iC]2 = 8[255.7×10-6 cos 104 t +525.1×10-6 sin 200t –24.87×10-3 cos 103 t]2
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Engineering Circuit Analysis, 7th Edition
83.
Chapter Five Solutions
10 March 2006
Replacing the DMM with a possible Norton equivalent (a 1-MΩ resistor in parallel
with a 1-A source):
1A
Vin
We begin by noting that 33 Ω || 1 MΩ ≈ 33 Ω. Then,
and
0 = (V1 – Vin)/ 33 + V1/ 275×103
[1]
1 - 0.7 V1 = Vin/ 106 + Vin/ 33×103 + (Vin – V1)/ 33
[2]
Simplifying and collecting terms,
(275×103 + 33) V1 - 275×103 Vin = 0
[1]
22.1 V1 + 1.001 Vin = 33
[2]
Solving, we find that Vin = 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, 7th Edition
84.
Chapter Five Solutions
10 March 2006
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 resistance, the small the power delivery to R:
⎡
⎤
120
PR = 10 i = 10 ⎢
⎥
⎢⎣ Rcylinder + 10 ⎥⎦
2
2
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, 7th Edition
85.
Chapter Five Solutions
10 March 2006
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, 7th Edition
86.
Chapter Five Solutions
10 March 2006
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.
3
New wiring scheme
Original wiring scheme
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,
2
213.3 ⎤
213.3 ⎤
⎡
⎡
⎢120 Rs + 213.3 ⎥
⎢60 Rs + 213.3 ⎥
⎣
⎦ +2 ⎣
⎦
320
320
2
= 30
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, 7th 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 Rlimiting and in series with the LED.
From KVL,
9
ILED =
R limiting + R LED
The maximum value of this current will occur at the minimum LED resistance, 47 Ω.
Thus, we solve
9
35×10-3 =
R limiting + 47
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, 7th Edition
1.
Chapter Six Solutions
This is an inverting amplifier, therefore, Vout = −
Rf
R1
10 March 2006
Vin
So:
100
× 3 = −30V
10
1M
b) Vout = −
× 2.5 = −2.5V
1M
4.7
c) Vout = −
× −1 = 1.42V
3.3
a) Vout = −
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2.
Chapter Six Solutions
10 March 2006
This is also an inverting amplifier. The loading resistance Rs only affects the output
current drawn from the op-amp. Therefore,
47
× 1.5 = −7.05V
10
b) Vout = 9V
c) Vout = −680mV
a) Vout = −
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Chapter Six Solutions
For this inverting amplifier, vout = −
3.
10 March 2006
10k
× vin = −10vin . Therefore,
1k
a) vout = −10vin = −20 sin 5t
2π / 5
4π /5
6π / 5
8π / 5
2π
25
20
15
10
5
0
-5
-10
-15
-20
-25
b) vout = −10vin = −10 − 5 sin 5t
0
-2
-4
-5 V
-6
-8
-10 V
-10
-12
-14
-15 V
-16
2π / 5
4π /5
6π / 5
8π / 5
2π
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Chapter Six Solutions
For this inverting amplifier, v out = −
4.
Rf
R1
10 March 2006
vin = −0.1vin , hence,
a) vout = −0.1vin = − cos 4t
1.5
π/2
π
3π / 2
2π
1
0.5
0
-0.5
-1
-1.5
b) vout = −0.1vin = −1.5 − 0.4 cos 4t
0
-0.2
-0.4
-0.6
-0.8
-1
-1.1 V
-1.2
-1.4
-1.5 V
-1.6
-1.8
-1.9 V
-2
π/2
π
3π / 2
2π
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5.
Chapter Six Solutions
10 March 2006
One possible solution is by using an inverting amplifier design, we have
⇔
V out = −
Rf
V in
R in
Rf
V out
9
=
V in
5
R in
=−
Using standard resistor values, we can have Rf=9.1kΩ and Rin=5.1kΩ
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6.
Chapter Six Solutions
10 March 2006
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:
R f 20
=
Rin
5
Giving possible resistor values Rf = 20 kΩ and Rin = 5.1 kΩ
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7.
Chapter Six Solutions
10 March 2006
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:
R f 1.5
=
Rin
5
Giving possible resistor values Rf = 1.5 kΩ and Rin = 5.1 kΩ
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Engineering Circuit Analysis, 7th Edition
8.
Chapter Six Solutions
10 March 2006
Similar to question 7, the following is proposed:
The resistances are given by:
Rf 3
=
Rin 9
Giving possible resistor values Rf = 3.0kΩ and Rin = 9.1 kΩ
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Engineering Circuit Analysis, 7th Edition
9.
Chapter Six Solutions
This circuit is a non-inverting amplifier, therefore, Vout = (1 +
10 March 2006
Rf
R1
)Vin
So:
47
) × 300m = 1.71 V
10
1M
b) Vout = (1 +
) × 1.5 = 3 V
1M
4.7
c) Vout = (1 +
) × −1 = −2.42 V
3.3
a) Vout = (1 +
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10.
Chapter Six Solutions
10 March 2006
This is again a non-inverting amplifier. Similar to question 9, we have:
a) Vout = 200m × (1 + 4.7) = 1.14 V
b) Vout = (1 + 1) × 9 = −18 V
c) Vout = 7.8 × 100m = 0.78 V
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Engineering Circuit Analysis, 7th Edition
Chapter Six Solutions
10 March 2006
1
vout = (1 + )vin = 2vin for this non inverting amplifier circuit, therefore:
1
a) vout = 2vin = 8 sin 10t
11.
10
2π / 5
4π /5
6π / 5
8π / 5
2π
8
6
4
2
0
-2
-4
-6
-8
-10
b) vout = 2vin = 2 + 0.5 sin 10t
2.6
2.5 V
2.4
2.2
2V
2
1.8
1.6
1.5 V
1.4
1.2
1
2π / 5
4π /5
6π / 5
8π / 5
2π
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12.
vout = (1 +
Rf
Rin
Chapter Six Solutions
10 March 2006
)vin = 1.5vin for this non inverting op-amp circuit. Hence,
a) vout = 1.5vin = 3 cos 2t
4
3
2
1
0
-1
-2
-3
π/2
2π
-4
b) vout = 1.5vin = 6 + 1.5 cos 2t
8
7.5 V
7
6V
6
5
5.5 V
4
3
2
1
0
π/2
2π
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13.
Chapter Six Solutions
10 March 2006
The first step is to perform a simple source transformation, so that a 0.15-V source in
series with a 150-Ω resistor is connected to the inverting pin of the ideal op amp.
Then, vout = −
2200
(0.15) = - 2.2 V
150
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14.
Chapter Six Solutions
10 March 2006
In order to deliver 150 mW to the 10-kΩ resistor, we need vout =
(0.15)(10 × 103 ) = 38.73 V. Writing a nodal equation at the inverting input,
we find
5 5 − vout
0 =
+
R 1000
Using vout = 38.73, we find that R = 148.2 Ω.
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Engineering Circuit Analysis, 7th Edition
15.
Chapter Six Solutions
10 March 2006
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 amp 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 Rf = 133 Ω, and a resistor R1 = 1 Ω.
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Engineering Circuit Analysis, 7th Edition
16.
Chapter Six Solutions
10 March 2006
We begin by labeling the nodal voltages v- and v+ at the inverting and non-inverting
input terminals, respectively. Since no current can flow 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
0 =
(v- - vS )
(v - v )
+ - out
100
22000
0 =
- vS
- vout
+
100
22000
or
Solving,
vout = -220 vS
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17.
Chapter Six Solutions
10 March 2006
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,
0 =
4 - 3 4 - Vo
+
1000 17000
Solving, we find that Vo = 21 V. Since no current can flow through the 300-kΩ
resistor, V1 = 21 as well.
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18.
Chapter Six Solutions
10 March 2006
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 mA)(1.875 kΩ) =
1.875 V, we may redraw the circuit as
Ω
Ω
V2
This is now a simple inverting amplifier with gain – Rf/ R1 = -75.33/ 1.975 = -38.14.
Thus,
V2 = -38.14(3.975) = -151.6 V.
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Engineering Circuit Analysis, 7th Edition
19.
Chapter Six Solutions
10 March 2006
This is a simple inverting amplifier, so we may write
vout =
- 2000
(2 + 2 sin 3t ) = - 4(1 + sin 3t ) V
1000
vout(t = 3 s) = -5.648 V.
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Engineering Circuit Analysis, 7th Edition
20.
Chapter Six Solutions
10 March 2006
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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21.
Chapter Six Solutions
10 March 2006
This is a simple non-inverting amplifier circuit, and so it has a gain of 1 + Rf/ 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: Rf = 236 kΩ and R1 = 1 kΩ.
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Engineering Circuit Analysis, 7th Edition
22.
Chapter Six Solutions
10 March 2006
Define a nodal voltage V- at the inverting input, and a nodal voltage V+ at the noninverting input. Then,
V+
At the non-inverting input: -3×10-6 =
[1]
1.5 × 106
Thus, V+ = -4.5 V, and we therefore also know that V- = -4.5 V.
At the inverting input:
0 =
VV -V
+ - out
R6
R7
[2]
Solving and making use of the fact that V- = -4.5 V,
vout = −
⎞
⎛
R7
(4.5) - 4.5 = - 4.5 ⎜⎜ R 7 + 1⎟⎟ V
R6
⎠
⎝ R6
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Engineering Circuit Analysis, 7th Edition
23.
Chapter Six Solutions
10 March 2006
(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) R1 = ∞, RA = 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, 7th Edition
24.
Chapter Six Solutions
10 March 2006
It is probably best to first perform a simple source transformation:
(1 mA)(2 kΩ) = 2 V.
13 kΩ
V-
1 kΩ
3V
2 kΩ
V+
vout
2V
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:
0 =
2 - 3 2 - vout
+
1000 13000
which yields vout = -11 V.
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25.
Chapter Six Solutions
10 March 2006
We begin by find the Thévenin equivalent to the left of the op amp:
Vth = -3.3(3) vπ = -9.9 vπ = − 9.9
1000 vS
= -9 vS
1100
Rth = 3.3 kΩ, so we can redraw the circuit as:
100 kΩ
3.3 kΩ
vout
-9 vS
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.
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26.
Chapter Six Solutions
10 March 2006
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
37.7 kΩ
vout
500 Ω
1.8 kΩ
370 Ω
23.899 mV
-6 V
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
23.899 × 10-3
23.899 × 10-3 - vout
+
0 =
500
37.7 × 103
Solving,
vout = 1.826 V
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Engineering Circuit Analysis, 7th Edition
27.
Chapter Six Solutions
10 March 2006
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
37.7 kΩ
vout
500 Ω
370 Ω
1.8 kΩ
55.1 mV
-6 V
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
55.1 × 10-3
55.1 × 10-3 - vout
+
0 =
500
37.7 × 103
Solving,
vout = 4.21 V
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28.
Chapter Six Solutions
10 March 2006
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,
100
= 14.29 V
21
100
= - (5)
= - 50 V
10
vout LEFTOPAMP = - (-3)
vout RIGH OPAMP
vx = vout LEFTOPAMP - vout RIGH OPAMP = 14.29 + 50 = 64.29 V.
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29.
Chapter Six Solutions
10 March 2006
A general summing amplifier with N input sources:
R1
va
Rf
R2
vb
v1
vout
v2
RN
vN
1. va = vb = 0
2. A single nodal equation at the inverting input leads to:
0 =
va − vout
v −v
v −v
v − vN
+ a 1 + a 2 + ... + a
Rf
R1
R2
RN
Simplifying and making use of the fact that va = 0, we may write this as
⎡ 1 N
⎤
v1 N
v2 N
vN N
v
−
R
=
R
+
R
+
...
+
Ri
∏
∏
∏
i
i
⎢ R ∏ i ⎥ out
R
R
R
i
=
i
=
i
=
i
=
1
1
1
1
N
2
1
f
⎣
⎦
or simply
v
v
v
v
− out = 1 + 2 + ... + N
Rf
R1
R2
RN
Thus,
N
v
vout = - R f ∑ i
i =1 R i
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30.
Chapter Six Solutions
A general difference amplifier:
10 March 2006
R4
R1
vout
R2
v1
v2
R3
Writing a nodal equation at the inverting input,
v -v
v -v
0 = a 1 + a out
R1
Rf
Writing a nodal equation at the non-inverting input,
v
v -v
0 = b + b 2
R3
R2
Simplifying and collecting terms, we may write
(Rf + R1) va – R1 vout = Rf v1
[1]
(R2 + R3) vb = R3 v2
[2]
From Eqn. [2], we have vb =
R3
v2
R2 + R3
Since va = vb, we can now rewrite Eqn. [1] as
− R 1 vout = R f v1 −
(R f + R1 )R 3 v
R2 + R3
2
and hence
vout = -
Rf
R ⎛ R + R1 ⎞
⎟ v2
v1 + 3 ⎜⎜ f
R1
R1 ⎝ R 2 + R 3 ⎟⎠
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31.
Chapter Six Solutions
10 March 2006
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 may compute the light-dependent
resistance (assuming a linear response in the range between 0 and 6 candela) as RCdS =
-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 than 1.5 V for intensities greater
than 2 candela.
Thus, vout(2 candela) = -RCdS vS/ R1 = -70 VS/ R1 = 1.5
(R1 in kΩ).
Pick R1 = 10 kΩ. Then vS = -0.2143 V.
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32.
Chapter Six Solutions
10 March 2006
We want Rf/ Rinstrument = 2K, and Rf/ Rvocal = 1K, where K is a constant not specified.
Assuming K = 1, one possible solution of many is:
Rf = 2 Ω
Rvocal = 1 Ω
vocals
microphone
Rinstruments = 2 Ω
vout
instruments
microphone
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33.
Chapter Six Solutions
10 March 2006
One possible solution of many:
99 kΩ
1 kΩ
vout
2V
vS
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34.
Chapter Six Solutions
10 March 2006
v +v +v
To get the average voltage value, we want vout = 1 2 3 . This voltage stays
3
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
v1 v2 v3 − vo
+
+
=
R1 R2 R3 R f1
If we assume R1=R2=R3=R, then
v 0 = − R f1
v1 + v 2 + v 3
R
Using the nodal equation at the inverting input of the second op-amp, we have:
− vout vo − R f1 v1 + v2 + v3
=
=
Rf2
R4
R4
R
Or,
vout =
R f 2 R f1 v1 + v2 + v3
R4
R
For simplicity, we can take Rf2 = Rf1 = R4= Rx, then, to give a voltage average,
v +v +v
v +v +v
vout = Rx 1 2 3 = 1 2 3
3
R
I.e. Rx/R = 3. Therefore, the circuit can be completed with R1 = R2= R3 = 30 kΩ and
Rf2 = Rf1 = R4 = 10 kΩ
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35.
Chapter Six Solutions
10 March 2006
The first stage is to subtract each voltage signal from the scale by the voltage
corresponding to the weight of the pallet (Vtare). 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 the
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, V1 – V4 (now gives
the weight of the items only), is then added by using a two stage summing amplifier:
The output is given by:
vout =
R f 2 R f1 v1 + v2 + v3 + v4
Rin
R
Therefore, to get the sum of the voltages v1 to v4, we only need to set all resistances to
be equal, so setting Rf2 = Rf1 = Rin = R =10 kΩ would give an output that is
proportional to the total weight of the items
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36.
Chapter Six Solutions
10 March 2006
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 hour), 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:
vout = (1 +
Rf
Rin
)vin = 1.609vin
This gives Rf/Rin =0.609 ≈ 61/100, i.e. Rf = 6.2 kΩ and Rin = 10 kΩ
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37.
Chapter Six Solutions
10 March 2006
vout of stage 1 is (1)(-20/ 2) = -10 V.
vout 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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38.
Chapter Six Solutions
10 March 2006
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: vout 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:
⎛ R3 ⎞
⎟⎟
V2 ⎜⎜
⎝ R2 + R3 ⎠
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:
0 =
⎤
⎤
1 ⎡ ⎛ R3 ⎞
1 ⎡ ⎛ R3 ⎞
⎟⎟ - V1 ⎥ +
⎟⎟ - Vout Stage 1 ⎥
⎢V2 ⎜⎜
⎢V2 ⎜⎜
R f ⎢⎣ ⎝ R 2 + R 3 ⎠
R 1 ⎢⎣ ⎝ R 2 + R 3 ⎠
⎥⎦
⎥⎦
which may be solved to obtain:
⎛R
⎞ R3
Vout = Vout Stage 1 = ⎜⎜ f + 1⎟⎟
V2
⎝ R1
⎠ R2 + R3
-
Rf
V1
R1
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39.
Chapter Six Solutions
10 March 2006
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:
0 =
0 - Vout stage 1
47
+
0-2
0-3
+
which leads to Vout steage 1 = -114.1 V
1
7
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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40.
Chapter Six Solutions
10 March 2006
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 = −
R
(− 5 − 10 − 15) = 10
100
Solving, we find that R = 33.33 Ω.
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41.
Chapter Six Solutions
10 March 2006
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
5
50
= 1.667 V
100 + 50
The second stage is wired as a voltage follower also, so
vout = 1.667 V.
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42.
Chapter Six Solutions
10 March 2006
a) Since the voltage supply is higher than the Zener voltage of the diode, the diode is
operating in the breakdown region. This means V2 = 4.7 V, and assuming ideal opamp, V1 = V2= 4.7 V. This gives a nodal equation at the inverting input:
4.7 V3 − 4.7
=
1k
1.1k
Solving this gives V3 = 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 V3 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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43.
Chapter Six Solutions
10 March 2006
The following circuit can be used:
The circuit is governed by the equations:
V3 = (1 +
Rf
)V1
Rin
And
V1 = V2 = Vdiode
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 Rf = 0Ω, and Rin can be
arbitary – Rin =100 kΩ would be sufficient.
The resistor value of R2 is determined by:
Vs − Vdiode
I ref
At a voltage of 5.1 V, the current is 76 mA, as described in the problem. This gives
R2≈ 51 Ω using standard resistor values.
R2 =
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44.
Chapter Six Solutions
10 March 2006
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:
Vout = −
Rf
Rin
Vin
Here, the input voltage is the diode voltage = 10 V, and the desired output voltage is 2.5 V. This gives Rf/Rin = 25 / 100 = 50 / 200, or Rf = 51 kΩ and Rin = 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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45.
Chapter Six Solutions
10 March 2006
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 noninverting 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 Vin = -20 V. Then, using the inverting amp
equation, we have Rf / Rin = 12/20 giving Rf = 120 kΩ and Rin = 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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46.
Chapter Six Solutions
10 March 2006
a) using inverting amplifier:
Rf/Rin = 5/3.3 giving Rf = 51 kΩ and Rin = 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 Rf = 33 kΩ and Rin = 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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47.
Chapter Six Solutions
10 March 2006
The following setup can be used:
I
Is
Is = 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 R1 = (12 – 10) V / 25
mA = 80 Ω = 82 Ω using standard values.
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48.
Chapter Six Solutions
10 March 2006
Using the following current source circuit, we have:
I
Is
Is = I = 12.5 mA = 20 V / R, assuming ideal op-amp. This gives R = 1.6 kΩ and R1 =
(27 – 20) / 12.5 mA = 560 Ω.
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49.
Chapter Six Solutions
10 March 2006
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 = (2720)/ 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 RL = (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 > RL > 66.67 Ω.
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Chapter Six Solutions
10 March 2006
50.
(a) va = vb = 1 nV ∴ vd = 0 and vout = 0. Thus,
P8Ω = 0 W.
(b) va = 0, vb = 1 nV ∴ vd = -1 nV
v2
8
= -19.28 μV. Thus, P8Ω = out = 46.46 pW.
vout = (2×105)(-1×10-9)
75 + 8
8
(c) va = 2 pV, vb = 1 fV ∴ vd = 1.999 pV
v2
8
vout = (2×105)(1.999×10-12)
= 38.53 nV. Thus, P8Ω = out = 185.6 aW.
75 + 8
8
(c) va = 50 μV, vb = -4 μV ∴ vd = 54 μV
8
vout = (2×105)(54×10-6)
= 1.041 V.
75 + 8
Thus, P8Ω =
2
vout
= 135.5 mW.
8
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Chapter Six Solutions
10 March 2006
51.
Writing a nodal equation at the “-vd” node,
0 =
or
-v -v
- vd - vd - VS
+
+ d out
Rf
R1
R in
[1]
(R1Rf + RinRf + RinR1) vd + RinR1vout = -RinRfVS
[1]
Writing a nodal equation at the “vout” node,
v - (-vd )
- vout - Avd
+ out
[2]
Rf
Ro
Eqn. [2] can be rewritten as:
- (R f + R o )
vd =
vout
[2]
R o - AR f
so that Eqn. [1] becomes:
R in (AR f - R o ) VS
vout = AR in R 1 + R f R 1 + R in R f + R in R 1 + R o R 1 + R o R in
0 =
where for this circuit, A = 106, Rin = 10 TΩ, Ro = 15 Ω, Rf = 1000 kΩ, R1 = 270 kΩ.
(a) –3.704 mV;
(b) 27.78 mV;
(c) –3.704 V.
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52.
vout = Avd = A
(
Chapter Six Solutions
10 March 2006
)
Ri
80 × 1015 sin 2t V
16 + R i
(a) A = 105, Ri = 100 MΩ, Ro value irrelevant. vout = 8 sin 2t nV
(b) A = 106, Ri = 1 TΩ, Ro value irrelevant. vout = 80 sin 2t nV
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Chapter Six Solutions
10 March 2006
53.
(a) Find vout/ vin if Ri = ∞, Ro = 0, and A is finite.
The nodal equation at the inverting input is
0 =
-v -v
- vd - vin
+ d out
100
1
[1]
At the output, with Ro = 0 we may write vout = Avd so vd = vout/ A. Thus, Eqn. [1]
becomes
v
v
v
0 = out + vin + out + out
A
100A 100
from which we find
vout
- 100A
=
[2]
vin
101 + A
(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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54.
Chapter Six Solutions
10 March 2006
(a) δ = 0 V ∴ vd = 0, and P8Ω = 0 W.
(b) δ = 1 nV, so vd = 5 – (5 + 10-9) = -10-9 V
Thus,
8
= -19.28 μV and P8Ω = (vout)2/ 8 = 46.46 pW.
vout = (2×105)vd
8 + 75
(c) δ = 2.5 μV, so vd = 5 – (5 + 2.5×10-6) = -2.5×10-6V
Thus,
8
= -48.19 mV and P8Ω = (vout)2/ 8 = 290.3 μW.
vout = (2×105)vd
8 + 75
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Chapter Six Solutions
10 March 2006
55.
AD549
Writing a single nodal equation at the output, we find that
0 =
vout - vin
v - Avd
+ out
Ro
Ri
[1]
Also, vin – vout = vd, so Eqn. [1] becomes
and
0 = (vout – vin) Ro + (vout – Avin + Avout) Ri
vout =
(R o + AR i )
R o + ( A + 1) R i
vin
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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56.
Chapter Six Solutions
10 March 2006
The Voltage follower with a finite op-amp model is shown below:
Nodal equation at the op-amp output gives:
Vout − Vin AVd − Vout
=
Ri
Ro
But in this circuit, Vd=Vin – Vout. Substitution gives:
Vout − Vin A(Vin − Vout ) − Vout AVin − ( A + 1)Vout
=
=
Ri
Ro
Ro
Further rearranging gives:
⇔
RoVout + Ri ( A + 1)Vout = RoVin + Ri AVin
Ro + Ri A
Vout =
Vin
Ro + Ri ( A + 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 Vout = Vin, which is exactly what we expected.
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57.
Chapter Six Solutions
10 March 2006
a) By definition, when the op-amp is at common mode, vout = ACMvin. Therefore, a model that can represent this is:
This model relies on that fact that ACM is much smaller than the differential gain A, and therefore when the inputs
are different, the contribution of ACM is negligible. When the inputs are the same, however, the differential term
Avd vanishes, and so vout = ACMv2, which is correct.
b) The voltage source in the circuit now becomes 105vd+10v2. Assuming Ro = 0, the circuit in figure 6.25 becomes:
The circuit is governed by the following equations:
105 vd + 10vb − va va − v1
=
R
R
vb = v2 / 2 (from voltage divider)
vd = vb − va
so
Rearranging gives:
105 (v2 / 2 − va ) + 10(v2 / 2) − va = va − v1
50000v2 − 105 va + 5v2 − va = va − v1
50005v2 + v1
va =
(105 + 2)
Then, the output is given by:
vout = 105 vd + 10vb = 105 × (v2 / 2 − va ) + 5v2
in this case, v1 = 5 + 2 sin t, v2 = 5. This gives va=0.50004v2+9.99980×10-6v1. Thus
vout = 105(-0.00004v2 - 9.99980×10-6v1)+5v2 = 1.00008v2 - 0.99998v1 = 0.0005 –
1.99996 sin t
c) If the common mode gain is 0, than the equation for va becomes
50000v2 + v1
(105 + 2)
Giving va=0.49999v2+9.99980×10-6v1. The output equation becomes
vout = 105 vd = 105 × (v2 / 2 − va ) =0.99998v2-0.99998v1 = 1.99996 sin t
va =
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58.
Chapter Six Solutions
10 March 2006
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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59.
Chapter Six Solutions
10 March 2006
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:
V3 − V1 V1
=
4.7k 1k
Solving gives V3 = 26.79 V
b) The simulation result is shown below
There are considerable discrepancies between calculated and simulated voltages. In particular, V1 = 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 V3 = 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. V2) 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 Vsupply = 5.28 V
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60.
Chapter Six Solutions
10 March 2006
The ideal op amp model predicts a gain vout/ vin = -1000/ 10 = -100, regardless of the
value of vin. In other words, it predicts an input-output characteristic such as:
vout (V)
1
-100
vin (V)
From the PSpice simulation result shown below, we see that the ideal op amp model
is reasonably accurate for |vin| × 100 < 15 V (the supply voltage, assuming both have
the same magnitude), but the onset of saturation is at ±14.5 V, or |vin| ~ 145 mV.
Increasing |vin| past this value does not lead to an increase in |vout|.
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61.
Chapter Six Solutions
10 March 2006
Positive voltage supply, negative voltage supply, inverting input, ground, output pin.
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62.
Chapter Six Solutions
10 March 2006
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 should be the open-loop gain
times the differential input voltage, minus any resistive losses.
From the simulation results below, we see that all three op amps saturate at a voltage
magnitude of approximately 14 V, corresponding to a differential input voltage of 50
to 100 μV, 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
ation
μA 741
LM 324
LF 411
-92 μV
41.3 μV
-31.77 μV
negative
-14.32 V
-14.71 V
-13.81 V
satur
ation
volta
ge
onset of
positi
ve
satur
ation
54.4 mV
337.2 mV
39.78 mV
positive
14.34 V
13.87 V
13.86 V
satur
ation
volta
ge
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Engineering Circuit Analysis, 7th Edition
63.
Chapter Six Solutions
10 March 2006
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. Performing 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 < Vin < 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 model.
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Engineering Circuit Analysis, 7th Edition
64.
Chapter Six Solutions
10 March 2006
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. Assuming the output
voltage from the op-amp before Ro 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 quite different to the 1 Ω figure given in the table.
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Engineering Circuit Analysis, 7th Edition
65.
Chapter Six Solutions
10 March 2006
Based on the detailed model of the LF 411 op amp , we can write the following nodal
equation at the inverting input:
0 =
- vd
v -v
Av - v
+ x 4 d + 6d d
R in
10
10 + R o
Substituting values for the LF 411 and simplifying, we make appropriate
approximations and then solve for vd in terms of vx, finding that
vd =
- 106
vx
199.9 × 106
= -
vx
199.9
With a gain of –1000/10 = -100 and supply voltage magnitudes of 15 V, we are
effectively limited to values of |vx| < 150 mV.
For vx = -10 mV, PSpice predicts vd = 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 μA, whereas the hand
calculations predict 99.5vx mA = -995 nA (which is reasonably close).
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Engineering Circuit Analysis, 7th Edition
66.
Chapter Six Solutions
10 March 2006
(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 < Vout < 15 V.
(b) The peak input voltage is 75 mV. Therefore, 15/ 75×10-3 = 200, and we should set
the resistance ratio Rf/ R1 < 199 to ensure the op amp does not saturate.
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Engineering Circuit Analysis, 7th Edition
67.
Chapter Six Solutions
10 March 2006
(a)
We see from the simulation result that negative saturation begins at Vin = –4.72 V,
and positive saturation begins at Vin = +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, 7th Edition
68.
Chapter Six Solutions
10 March 2006
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 resistances in ohms):
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Engineering Circuit Analysis, 7th Edition
69.
Chapter Six Solutions
10 March 2006
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: Vaverage = (1.45 + 3.95 + 7.82)/ 3 = 4.407 V.
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Engineering Circuit Analysis, 7th Edition
70.
Chapter Six Solutions
10 March 2006
Assuming ideal situations (ie slew rate = infinite)
a)
20
18 V
V out (V)
15
10
5
0
-5
-5
-3
-1
1
3
5
V active (V)
b)
20
18 V
V out (V)
15
10
5
0
-5
-5
-3
-1
1
3
5
V active (V)
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Engineering Circuit Analysis, 7th Edition
71.
Chapter Six Solutions
10 March 2006
a)
15
12 V
10
V out (V)
5
0
-2
-1
0
1
2
-5
-10
-12 V
-15
V active (V)
b) The simulation is performed using the following circuit:
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, 7th Edition
Chapter Six Solutions
10 March 2006
72. a)
15
10
V out (V)
5
0
-5
-3
-1
1
3
5
-5
-10
-15
V active (V)
b)
15
10
V out (V)
5
0
-5
-3
-1
1
3
5
-5
-10
-15
V 2 (V)
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73.
Chapter Six Solutions
10 March 2006
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, 7th Edition
74.
Chapter Six Solutions
10 March 2006
The voltage output of the circuit is given by
vout =
R4 1 + R2 / R1
R
(
) v + − 2 v−
R3 1 + R4 / R3
R1
a) When R1 = R3 and R2 = R4, the equation reduces to
R
vout = 4 (v+ − v− )
R3
When v+ = v-, vout = 0, thus ACM = 0. Hence CMRR = ∞
b) If R1, R2, R3 and R4 are all different, then when v+ = v- = v,
R 1 + R2 / R1
R
vout = ( 4 (
) − 2 )v
R3 1 + R4 / R3
R1
Simplifying the algebra gives
R R − R3 R4
vout = 1 4
v
R1 R3 + R1 R4
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, 7th Edition
75.
Chapter Six Solutions
10 March 2006
a) The voltage at node between R1 and R2 is
⎛
⎞
R2
⎟⎟
V 1 = V ref ⎜⎜
+
R
R
2 ⎠
⎝ 1
by treating it as a voltage divider. Similarly, the voltage at node between RGauge and
R3 is:
⎛
⎞
R3
⎟
V2 = Vref ⎜
⎜R +R
⎟
3
Gauge
⎝
⎠
Therefore, the output voltage is
⎛ R2
⎞
R3
⎟
−
Vout = V1 − V2 = Vref ⎜
⎜R +R R +R
⎟
2
3
Gauge ⎠
⎝ 1
b) If R1 = R2 = R3 = Rgauge then the two terms in the bracket cancels out, giving Vout =
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 > Vout > 0.001.
To simplify the situation, let R1 = R2 = R3 = R, then at maximum loading,
⎛ R
⎞
R
R
⎞
⎟ = 12⎛⎜ 1 −
−
Vout = Vref ⎜
⎟
⎜R+R R+R
⎟
⎝ 2 R + 5k + 50m ⎠
Gauge + ΔR ⎠
⎝
Using this we can set up two inequalities according to the two limits. The first one is:
12 R
⎛
⎞
⎜6 −
⎟ ≥ 0.001
R + 5k + 50m ⎠
⎝
Solving gives
12 R
5.999 ≥
R + 5000.05
4998.38 ≥ R
Similarly, the lower gain limit gives: 5.5 ≤
12 R
R + 5000.05
⇒ 4230 ≤ R
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
4.7k
⎛1
⎞
Vout = 12⎜ −
⎟ = 0.1855
⎝ 2 4.7k + 5k + 50m ⎠
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.
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Engineering Circuit Analysis, 7th Edition
1.
i=C
Chapter Seven Solutions
10 March 2006
dv
dt
(a)
i=0
(DC)
(b)
i=C
dv
= − 10 × 10−6 115 2 (120π ) sin120π t = −613sin120π t mA
dt
(c)
i=C
dv
= − (10 × 10−6 )( 4 ×10−3 ) e− t = −40e − t nA
dt
(
)(
)
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Engineering Circuit Analysis, 7th Edition
2.
i=C
v=
Chapter Seven Solutions
10 March 2006
dv
dt
6−0
dv
t + 6 = 6 − t , therefore i = C
= −4.7 × 10−6 μA
0−6
dt
i (μA)
t (s)
–4.7
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Engineering Circuit Analysis, 7th Edition
3.
(a)
i=C
Chapter Seven Solutions
10 March 2006
dv
dt
dv
= 30 ⎡⎣e − t − te− t ⎤⎦
dt
therefore i = 10−3
dv
= 30 (1 − t ) e− t mA
dt
(b)
dv
= 4 ⎡⎣ −5e −5t sin100t + 100e −5t cos100t ⎤⎦
dt
dv
therefore i = 10−3
= 4e −5t (100 cos100t − 5sin100t ) mA
dt
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Engineering Circuit Analysis, 7th Edition
4.
Chapter Seven Solutions
10 March 2006
1
W = CV 2
2
(a)
⎛1⎞
−6
⎜ ⎟ 2000 × 10 1600 = 1.6 J
⎝2⎠
(b)
2
⎛1⎞
−3
⎜ ⎟ 25 ×10 ( 35 ) = 15.3 J
⎝2⎠
(c)
2
⎛ 1 ⎞ −4
⎜ ⎟ 10 ( 63) = 198 mJ
⎝2⎠
(d)
⎛1⎞
−3
⎜ ⎟ 2.2 × 10 ( 2500 ) = 2.75 J
⎝2⎠
(e)
2
⎛1⎞
⎜ ⎟ ( 55 )( 2.5 ) = 171.9 J
⎝2⎠
(f)
2
⎛1⎞
−3
⎜ ⎟ 4.8 × 10 ( 50 ) = 6 J
⎝2⎠
(
)
(
(
(
(
)
)
)
)
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Engineering Circuit Analysis, 7th Edition
εA
Chapter Seven Solutions
8.854 × 10 −12 (78.54 × 10 −6 )
= 6.954 pF
100 × 10 − 6
5. (a)
C=
(b)
2 1×10−3
1
2E
2
=
= 16.96 kV
Energy, E = CV ∴V =
2
6.954 × 10−12
C
(c)
E=
d
=
(
C=
εA
d
10 March 2006
)
1
2 E 2(2.5 × 10 −6 )
= 500 pF
CV 2 ∴ C = 2 =
2
(100 2 )
V
∴ε =
Cd (500 × 10 −12 )(100 × 10 −6 )
= 636.62 pF .m −1
=
−6
A
(78.54 × 10 )
\Relative permittivity :
ε 636.62 × 10−12
=
= 71.9
ε 0 8.854 × 10−12
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Engineering Circuit Analysis, 7th Edition
6. (a)
For VA = –1V, W =
Chapter Seven Solutions
)
)
2K sε 0
× 10 −12
(Vbi − V A ) = 2(11.8) 8−.854
(0.57 + 1)
qN
1.6 × 10 19 1 × 10 24
(
= 45.281 × 10
Cj =
(
10 March 2006
(
−9
m
)(
11.8 8.854 × 10 −12 1 × 10 −12
45.281 × 10
)(
−9
) = 2.307 fF
(
)
)
2K sε 0
2(11.8) 8.854 × 10 −12
(Vbi − V A ) =
(0.57 + 5)
For VA = –5V, W =
qN
1.6 × 10 −19 1 × 10 24
(b)
(
)(
= 85.289 × 10 −9 m
Cj =
(c)
(
)(
11.8 8.854 × 10 −12 1 × 10 −12
85.289 × 10
For VA = –10V,
W=
(
−9
) = 1.225 fF
)
)
2K sε 0
× 10 −12
(Vbi − V A ) = 2(11.8) 8−.854
(0.57 + 10)
qN
1.6 × 10 19 1 × 10 24
(
)(
= 117.491 × 10 −9 m
Cj =
(
)(
11.8 8.854 × 10 −12 1 × 10 −12
117.491 × 10
−9
) = 889.239aF
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Engineering Circuit Analysis, 7th Edition
7.
Chapter Seven Solutions
10 March 2006
We require a capacitor that may be manually varied between 100 and 1000 pF by rotation
of a knob. Let’s choose an air di electric for simplicity of cons truction, and a series of 11
half-plates:
fixed
Top view
Side view with no
overlap between plates
Side view with a small
overlap between plates.
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 cm2)/ (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 fo r 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 pieshaped 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 180o, we need to set out minimum angle to be 18o.
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Engineering Circuit Analysis, 7th Edition
t
8. (a)
Energy stored = ∫ v ⋅ C
(b)
Vmax = 3 V
t0
Chapter Seven Solutions
10 March 2006
t
−
2×10−3
⎛ 3 −t ⎞
dv
= C∫
3e 5 ⋅ ⎜ − e 5 ⎟ dt = −1.080 μ J
0
dt
⎝ 5
⎠
1
CV 2 = 1.35mJ ∴ 37% E max = 499.5μJ
2
V at 37% Emax = 1.825 V
Max. energy at t=0, =
v (t ) = 1.825 = 3e
−
t
5 ∴ t = 2.486 s ⇒≈ 2 s
(c)
1.2 ⎞
⎛
−
dv
−6 ⎜ 3
= 300 × 10
− e 5 ⎟ = −141.593μA
i =C
⎜ 5
⎟
dt
⎝
⎠
(d)
P = vi = 2.011 − 120.658 × 10 −6 = −242.6μW
(
)
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Engineering Circuit Analysis, 7th Edition
Chapter Seven Solutions
10 March 2006
9. (a)
v=
2
(3.14159) 1 × 10 −3 2 = 33.421mV
1 π
1
. 1 × 10 −3 =
.
C 2
2
47 × 10 −6
(b)
v=
2
(3.14159 ) 1 × 10 −3 2 = 33.421mV
1 ⎛π
1
⎞
.⎜ 1 × 10 −3 + 0 ⎟ =
.
C ⎝2
2
⎠ 47 × 10 −6
(c)
v=
2
2⎞
π
1 ⎛π
1
⎛ 3π
−3 2 ⎞
1
10
.⎜ 1 × 10 −3 + 1 × 10 −3 ⎟ =
.
×
⎟ = 50.132mV
⎜
C ⎝2
4
⎠
⎠ 47 × 10 −6 ⎝ 4
(
)
(
(
)
(
)
)
(
(
)
)
(
)
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Engineering Circuit Analysis, 7th Edition
10.
Chapter Seven Solutions
⎞⎤
1 200ms
1 ⎡⎛ 7 × 10 −3
cos πt ⎟⎥
V= ∫
idt = ⎢⎜ −
⎟
π
C 0
C ⎣⎢⎜⎝
⎠⎦⎥
E=
200ms
=
0
10 March 2006
0.426
C
1
181.086 × 10 −9
181.086 × 10 −9
CV 2 = 3 × 10 −6 =
∴C =
= 30181μF
2
2C
2 3 × 10 −6
(
)
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Engineering Circuit Analysis, 7th Edition
Chapter Seven Solutions
10 March 2006
11.
(a)
c = 0.2 μ F, vc = 5 + 3cos 2 200tV; ∴ ic = 0.2 × 10−6 (3) (−2) 200 sin 200t cos 200t
∴ ic = −0.12sin 400tmA
(b)
1
1
wc = cvc2 = × 2 × 10−7 (5 + 3cos 2 200t ) 2 ∴ wc max = 10−7 × 64 = 6.4 μ J
2
2
(c)
vc =
(d)
vc = 500 − 400e −100t V
t
1
× 106 ∫ 8e −100t ×10−3 dt = 103 × 40(−0.01) (e−100t − 1) = 400(1 − e100t )V
0
0.2
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Engineering Circuit Analysis, 7th Edition
12.
Chapter Seven Solutions
10 March 2006
0.1
vc (0) = 250V, c = 2mF (a) vc (0.1) = 250 + 500∫ 5dt
0
0.2
∴ vc (0.1) = 500V; vc (0.2) = 500∫ 10dt = 1000V
0.1
∴ vc (0.6) = 1750V, vc (0.9) = 2000V
t
∴ 0.9 < t < 1: vc = 2000 + 500∫ 10dt = 2000 + 5000(t − 0.9)
0.9
∴ vc = 2100 = 2000 + 5000(t2 − 0.9) ∴ t2 = 0.92 ∴ 0.9 < t < 0.92s
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Engineering Circuit Analysis, 7th Edition
Chapter Seven Solutions
10 March 2006
13.
(a)
1
1
wc = Cv 2 = × 10−6 v 2 = 2 × 10−2 e −1000t ∴ v = ±200e −500t V
2
2
−6
i = Cv′ = 10 (±200) (−500)e−500t = m0.1e−500t
−v 200
∴R =
=
= 2k Ω
0.1
i
(b)
PR = i 2 R = 0.01× 2000e −1000t = 20e −1000t W
∞
∴ WR = ∫ 20e −1000t dt = −0.02e −1000t 0∞ = 0.02J
0
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Engineering Circuit Analysis, 7th Edition
14.
(a) Left circuit:
By
Voltage division, VC =
Right
circuit:
Chapter Seven Solutions
10 March 2006
1k
(5) = 0.877V
4.7k + 1k
2
V1 = 1(1 // 2 ) = V
3
1
1
Voltage Division, V2 = V ∴VC = − V
3
3
By
(b)
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Engineering Circuit Analysis, 7th Edition
15.
v=L
Chapter Seven Solutions
10 March 2006
di
dt
(a)
v = 0 since i = constant (DC)
(b)
v = −10−8 115 2 (120π ) sin120π t = −613sin120π t μ V
(c)
v = −10−8 115 2 ( 24 ×10−3 ) e−6t = − 240e −6t pV
(
)
(
)
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Engineering Circuit Analysis, 7th Edition
16.
v=L
Chapter Seven Solutions
10 March 2006
di
dt
⎡ ( 6 − 0 ) × 10−9 ⎤
i=⎢
t + 6 × 10−9 = 6 × 10−9 − 10−6 t , therefore
−3 ⎥
⎢⎣ ( 0 − 6 ) × 10 ⎥⎦
di
v = L = − (10−12 )(10−6 ) = −10−18 V = − 1 aV
dt
v (aV)
t (s)
–1
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Engineering Circuit Analysis, 7th Edition
17.
v=L
(a)
L
Chapter Seven Solutions
10 March 2006
di
dt
(
)
di
= 5 ×10−6 30 ×10−9 ⎣⎡e− t − te− t ⎦⎤ = 150 (1 − t ) e− t fV
dt
(b)
(
)(
)
di
= 5 ×10−6 4 × 10−3 ⎡⎣ −5e −5t sin100t + 100e −5t cos100t ⎤⎦
dt
therefore v = 100e −5t ( 20 cos100t − sin100t ) pV
L
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Engineering Circuit Analysis, 7th Edition
18.
W=
Chapter Seven Solutions
10 March 2006
1 2
LI . Maximum energy corresponds to maximum current flow, so
2
Wmax =
1
2
5 × 10−3 (1.5 ) = 5.625 mJ
2
(
)
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Engineering Circuit Analysis, 7th Edition
Chapter Seven Solutions
10 March 2006
19.
(a)
(b)
PL = vLiL ∴ PL max = (−100) (−5) = 500W at t = 40− ms
(c)
PL min = 100(−5) = −500W at t = 20+ and 40+ ms
(d)
WL =
1 2
1
Li L ∴ WL (40ms) = × 0.2(−5) 2 = 2.5J
2
2
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Engineering Circuit Analysis, 7th Edition
Chapter Seven Solutions
10 March 2006
20.
L = 50 × 10−3 , t < 0 : i = 0; t > 0 i = 80te −100t mA = 0.08te −100t A
∴ i′= 0.08e-100t − 8te −100t ∴ 0.08 = 8t , tm , = 0.01s, i max = 0.08 × 0.01e −1
∴ i max = 0.2943mA; v = 0.05i′ = e −100t (0.004 − 0.4t )
0.8
= 0.02 s
40
v = e −2 (0.004 − 0.008) = −0.5413mV this is minimum∴ v max = 0.004V at t=0
∴ v′ = e −100t (−0.4) − 100e −100t (0.004 − 0.4t ) ∴−0.4 = 0.4 − 40t , t =
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Engineering Circuit Analysis, 7th Edition
Chapter Seven Solutions
10 March 2006
21.
(a)
t > 0 : is = 0.4t 2 A ∴ vin = 10is + 5is′ = 4t 2 + 4tV
(b)
iin′ = 0.1vs +
1 t
40tdt + 5 = 4t + 4t 2 + 5A
5 ∫0
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Engineering Circuit Analysis, 7th Edition
Chapter Seven Solutions
22.
vL = 20 cos1000tV, L = 25mH, iL (0) = 0
(a)
iL = 40 ∫ 20 cos 1000tdt = 0.8sin 1000tA ∴ p = 8sin 2000t W
(b)
1
w = × 25 ×10−3 × 0.64sin 2 1000t = 8sin 2 1000t mJ
2
10 March 2006
t
0
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Engineering Circuit Analysis, 7th Edition
Chapter Seven Solutions
10 March 2006
23.
t
(a)
0 < t < 10 ms: iL = −2 + 5∫ 100dt = −2 + 500t ∴ iL (10ms) = 3A, iL (8ms) = 2A
(b)
iL (0) = 0 ∴ iL (10ms) = 500 × 0.01 = 5A ∴ iL (20ms) = 5 + 5∫
0
0.02
0.01
104 (0.02 − t )dt
4
∴ iL (20ms) = 5 + 5 × 104 (0.02t − 0.5t )0.02
0.01 = 5 + 5 × 10 (0.0002 − 0.00015) = 7.5A
1
∴ wL = × 0.2 × 7.52 = 5.625J
2
(c)
If the circuit has been connected for a long time, L appears like short circuit.
V8Ω =
8
(100V ) = 80V
2+8
I 2Ω =
20V
= 10 A
2Ω
∴i x =
80V
= 1A
80Ω
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Engineering Circuit Analysis, 7th Edition
24.
Chapter Seven Solutions
10 March 2006
After a very long time connected only to DC sources, the inductors act as short circuits.
The circuit may thus be redrawn as
⎛ 80 ⎞
⎜
⎟ ⎛ 100 ⎞
And we find that ix = ⎜ 9 ⎟ ⎜
⎟= 1A
⎜ 80 + 80 ⎟ ⎝ 2 + 8 ⎠
9 ⎠
⎝
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Engineering Circuit Analysis, 7th Edition
Chapter Seven Solutions
25.
L = 5H, VL = 10(e −t − e −2t )V, iL (0) = 0.08A
(a)
vL (1) = 10(e −1 − e−2 ) = 2.325+ V
(b)
iL = 0.08 + 0.2 ∫ 10(e − t − e −2t ) dt = 0.08 + 2(−e − t + 0.5e −2t )t0
10 March 2006
t
0
iL = 0.08 + 2(−e − t + 0.5e −2t + 1 − 0.5) = 1.08 + e −2t − 2e − t ∴ iL (1) = 0.4796A
(c)
iL (∞) = 1.08A
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Engineering Circuit Analysis, 7th Edition
Chapter Seven Solutions
10 March 2006
26.
(a)
(b)
40
12
+ 40 × 5 ×
12 + 20 + 40
12 + 20 + 40
200 100
=
+
= 100V
3
3
vx = 120 ×
vx =
15 12
120
15
×
× 40 + 40 × 5
12 + 15 60 15 + 60
15 12 + 60
120
1
6.667
× × 40 + 200
12 + 12 5
66.667
= 40 + 20 = 60V
=
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Engineering Circuit Analysis, 7th Edition
27.
Chapter Seven Solutions
(a)
1
wL = × 5 × 1.62 = 6.4J
2
(b)
1
wc = × 20 × 10−6 ×1002 = 0.1J
2
(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)
10 March 2006
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Engineering Circuit Analysis, 7th Edition
28.
(a)
Chapter Seven Solutions
10 March 2006
vs = 400t 2 V, t > 0; iL (0) = 0.5A; t = 0.4s
1
vc = 400 × 0.16 = 64V, wc × 10−5 × 642 = 20.48mJ
2
(b)
(c)
0.4
1
iL = 0.5 + 0.1∫ 400t 2 dt = 0.5 + 40 × × 0.43 = 1.3533A
0
3
1
∴ wL = × 10 × 1.35332 = 9.1581J
2
0.4
iR = 4t 2 , PR = 100 × 16t 4 ∴ wR = ∫ 1600t 4 dt = 320 × 0.45 = 3.277J
0
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Engineering Circuit Analysis, 7th Edition
29. (a)
(b)
P7Ω = 0W ; P10Ω =
Chapter Seven Solutions
10 March 2006
V 2 (2 )2
=
= 0.4W
10
R
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, 7th 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, RTH = 8.23 || 8 = 4.06 Ω. To find VTH, we remove the inductor:
+ VTH –
V1
V2
V3
REF
Writing the nodal equations required:
(V
(V
1 – 9)/3 + V1/6 + (V1 – V2)/4 = 0
V3/8 + (V3 – V2)/5 = 0
Solving,
(b)
2 – V1)/4 + V2/7 + (V2 – V3)/5 = 0
V3 = 1.592 V, therefore VTH = 9 – V3 = 7.408 V.
iL = 7.408/4.06 = 1.825 A (inductor acts like a short circuit to DC).
(c)
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Engineering Circuit Analysis, 7th Edition
Chapter Seven Solutions
10 March 2006
31.
⎛
⎞
⎛
⎞
⎜
⎟
⎜
⎟
1
1
⎜
⎟
⎜
⎟
C equiv ≡ 10 μ +
in series with 10μ in series with 10μ +
⎜ 1
⎜ 1
1 ⎟
1 ⎟
⎜ 10μ + 10μ ⎟
⎜ 10μ + 10μ ⎟
⎝
⎠
⎝
⎠
≡ 4.286μF
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Engineering Circuit Analysis, 7th Edition
32.
Chapter Seven Solutions
10 March 2006
Lequiv ≡ (77 p // (77 p + 77 p )) + 77 p + (77 p // (77 p + 77 p )) = 179.6& pH
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Engineering Circuit Analysis, 7th Edition
33.
(a)
have
(b)
Using
Chapter Seven Solutions
10 March 2006
Assuming all resistors have value R, all inductors have value L, and all capacitors
value C,
At dc, 20μF is open circuit; 500μH is short circuit.
10k
(9 ) = 3.6V
voltage division, V x =
10k + 15k
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Engineering Circuit Analysis, 7th Edition
34.
Chapter Seven Solutions
10 March 2006
(a) As all resistors have value R, all inductors value L, and all capacitors value C,
+
Vx
-
(b)
Vx = 0 V as L is short circuit at dc.
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Engineering Circuit Analysis, 7th Edition
35. C
equiv = { [(100 n + 40 n) || 12 n] + 75 n}
Chapter Seven Solutions
10 March 2006
|| {7 μ + (2 μ || 12 μ)}
C equiv ≡ 85.211nF
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Engineering Circuit Analysis, 7th Edition
36. L
equiv
Chapter Seven Solutions
10 March 2006
= {[ (17 p || 4 n) + 77 p] || 12 n} + {1 n || (72 p + 14 p)}
Lequiv ≡ 172.388 pH
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Engineering Circuit Analysis, 7th Edition
37.
Chapter Seven Solutions
10 March 2006
C T − C x = (7 + 47 + 1 + 16 + 100) = 171μF
E CT −C x =
1
(CT − C x )V 2 = 1 (171μ )(2.5)2 = 534.375μJ
2
2
E C x = E CT − E CT −C x = (534.8 − 534.375)μJ = 425nJ
1
425n (2 )
∴ E C x = 425n = C xV 2 ⇒ C x =
= 136nF
2
(2.5)2
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Engineering Circuit Analysis, 7th Edition
Chapter Seven Solutions
10 March 2006
38.
(a)
⎛
⎞ ⎛
⎞
⎜
⎟
⎜
⎟
1
1
⎜
⎟
⎜
⎟ = 2.75H
For all L = 1.5H, Lequiv = 1.5 +
+
⎜ 1 + 1 ⎟ ⎜ 1 + 1 + 1 ⎟
⎜
⎟ ⎜
⎟
⎝ 1.5 1.5 ⎠ ⎝ 1.5 1.5 1.5 ⎠
(b)
For a general network of this type, having N stages (and all L values equiv),
n
Lequiv = ∑
LN
N −1
N =1 NL
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Engineering Circuit Analysis, 7th Edition
Chapter Seven Solutions
10 March 2006
39.
(a)
⎞
⎛
⎞ ⎛
⎟
⎜ 1 ⎟ ⎜
1
⎜
⎟ = 3H
⎜
⎟
Lequiv = 1 +
+
⎜1 + 1⎟ ⎜1+1+1⎟
⎜
⎟ ⎜
⎟
⎝ 2 2⎠ ⎝3 3 3⎠
(b)
For a network of this type having 3 stages,
Lequiv = 1 +
1
1
+
2+ 2 3+3
(2 )
2
(3)
2
+
1
3
=1+
(2)2 + (3)3
2(2 ) 3(3)2
Extending for the general case of N stages,
1
1
+K+
1 1 1 1 1
1
1
+
+ +
+K
2 2 3 3 3
N
N
1
1
1
= 1+
+
+K+
= N
2(1 / 2) 3(1 / 3)
N(1/N)
Lequiv = 1 +
1
+
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Engineering Circuit Analysis, 7th Edition
40.
C equiv =
Chapter Seven Solutions
10 March 2006
(3 p )(0.25 p ) = 0.231 pF
3 p + 0.25 p
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Engineering Circuit Analysis, 7th Edition
41.
Lequiv =
Chapter Seven Solutions
10 March 2006
(2.3& n )(0.3& n ) = 0.2916& nH
2.6& n
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Engineering Circuit Analysis, 7th Edition
42.
Chapter Seven Solutions
(a)
Use 2 x 1μH in series with 4 x 1μH in parallel.
(b)
Use 2 x 1μH in parallel, in series with 4 x 1μH in parallel.
(c)
Use 5 x 1μH in parallel, in series with 4 x 1μH in parallel.
10 March 2006
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Engineering Circuit Analysis, 7th Edition
Chapter Seven Solutions
10 March 2006
43.
(a)
R = 10Ω :10 10 10 =
∴ R eq =
(b)
(c)
10 10
55
+ 10 + 10 10 =
,
3 3
3
55
30 = 11.379Ω
3
L = 10H ∴ Leq = 11.379H
1
= 5.4545
1/ 30 + 1/10 + 1/ 20
10
∴ Ceq = 5.4545 + = 8.788F
3
C = 10F :
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Engineering Circuit Analysis, 7th Edition
Chapter Seven Solutions
10 March 2006
44.
(a)
oc :L eq = 6 1 + 3 = 3.857H
sc : L eq = (3 2 + 1) 4 = 2.2
(b)
4 = 1.4194H
1
7
1
= , ceq =
= 1.3125F
1/ 4 + 1/ 2 3
3 / 7 + 1/ 2
1
5
5
= , Ceq = 4 + = 4.833F
sc :
1/ 5 + 1 6
6
oc :1 +
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Engineering Circuit Analysis, 7th Edition
Chapter Seven Solutions
10 March 2006
45.
(a)
(b)
(c)
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Engineering Circuit Analysis, 7th Edition
Chapter Seven Solutions
46.
is = 60e −200t mA, i1 (0) = 20mA
(a)
6 4 = 2.4H ∴ v = Leq is′ = 2.4 × 0.06(−200)e−200t
10 March 2006
or v = −28.8e −200t V
1 t
4.8 −200t
(e
−28.8e−200t dt + 0.02 =
− 1) + 0.02
∫
6 o
200
= 24e−200t − 4mA(t > 0)
(b)
i1 =
(c)
i2 = is − i1 = 60e−200t − 24e−200t + 4 = 36e−200t + 4mA(t > 0)
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Engineering Circuit Analysis, 7th Edition
Chapter Seven Solutions
47.
vs = 100e −80tV , v1 (0) = 20V
(a)
i = Ceq vs′ = 0.8 ×10−6 (−80)100e −80t = −6.4 × 10−3 e−80t A
(b)
v1 = 106 (−6.4 ×10−3 ) ∫ e −80t dt + 20 =
t
o
10 March 2006
6400 −80t
(e − 1) + 20
80
∴ v1 = 80e−80t − 60V
(c)
t
106
1600 −80t
(−6.4 ×10−3 ) ∫ e −80t dt + 80 =
(e − 1) + 80
v2
o
4
80
= 20e −80t + 60V
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Engineering Circuit Analysis, 7th Edition
Chapter Seven Solutions
10 March 2006
48.
(a)
vc − vs
v −v
+ 5 ×10−6 vc′ + c L = 0
20
10
t
vL − vc
1
+
vL dt + 2 = 0
10
8 × 10−3 ∫o
(b)
20i20 +
t
1
(i − iL )dt + 12 = vs
−6 ∫o 20
5 ×10
t
1
(i − i20 )dt − 12 + 10iL + 8 × 10−3 iL′ = 0
−6 ∫o L
5 × 10
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Engineering Circuit Analysis, 7th Edition
Chapter Seven Solutions
10 March 2006
49.
vc (t ): 30mA: 0.03 × 20 = 0.6V, vc = 0.6V
9V: vc = 9V, 20mA: vc = −0.02 × 20 = 0.4V
0.04 cos103 t : vc = 0
∴ vc (t ) = 9.2V
vL (t ): 30mA, 20mA,
9V: vL = 0; 0.04 cos103 t : vL = −0.06 × 0.04 (−1000) sin103 t = 2.4sin103 tV
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Engineering Circuit Analysis, 7th Edition
50.
Chapter Seven Solutions
10 March 2006
We begin by selecting the bottom node as the reference and assigning four nodal
voltages:
V4
V2
V1
V3
Ref.
1, 4 Supernode:
and:
Node
V
2:
20×10-3 e-20t =
t
V1 - V2
+ 0.02 × 103 ∫ V4 − 40e − 20t dt ′ [1]
0
50
1 – V4 = 0.2 Vx
0 =
(
or
)
0.8V1 + 0.2 V2 – V4 = 0
V2 - V1
V - 40e −20t
dV2
+ 2
+ 10- 6
50
100
dt
[2]
[3]
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Engineering Circuit Analysis, 7th Edition
Chapter Seven Solutions
10 March 2006
R i = ∞, R o = 0, A = ∞∴ vi = 0 ∴ i = Cvs′
51. (a)
also 0 + Ri + vo = 0 ∴ vo = − RCvs′
1
idt + vi
c∫
−1
1+ A
vo = −Avi ∴ vi = vo ∴ i =
vi
A
R
1
1
1
1+ A
v
∴ vs = ∫ idt − vo = − vo +
− o dt
∫
c
A
A
RC
A
1+ A
1+ A
∴ Avs′ = −vo′ −
vo or vo′ +
vo + Avs′ = 0
RC
RC
−vi + Ri − Avi = 0, vs =
(b)
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Engineering Circuit Analysis, 7th Edition
52.
Chapter Seven Solutions
10 March 2006
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, R2, of
1-MΩ and feedback capacitor, Cf, of 1 μF.
dv c f
ions
i=Cf
= 1.602 × 10 −19 ×
dt
sec
0=
0=
Va − V
dv c f
1 × 10
dt
+Cf
6
=
Va − V
1 × 10
6
+ 1.602 × 10 −19
ions
sec
dv c f
−V
−V
−19 ions
+Cf
=
1
.
602
10
+
×
R2
dt
sec
1 × 10 6
Integrating current with respect to t,
(
)
1 t
vdt ' = C f Vc f − Vc f (0)
R2 ∫0
1.602 × 10 −19 × ions
= C f Vc f
R2
− R1
−1
× 1.602 × 10 −19 × ions ⇒ Vout =
× 1.602 × 10 −19 × ions
Vc f = Va − Vout ⇒ Vout =
R2 C f
Cf
R1 = 1 MΩ, Cf = 1μF
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Engineering Circuit Analysis, 7th Edition
Chapter Seven Solutions
10 March 2006
53.
R = 0.5MΩ, C = 2 μ F, R i = ∞, R o = 0, vo = cos10t − 1V
(a)
vo ⎞
1⎞
1 t⎛
⎛
Eq. (16) is: ⎜ 1 + ⎟ vo = −
⎜ vs + ⎟ dt − vc (0)
∫
o
RC ⎝
A⎠
⎝ A⎠
vo ⎞ ⎛
1⎞
1⎞
1 ⎛
1⎞
1
⎛
⎛
∴ ⎜1 + ⎟ vo′ = −
⎜ vs + ⎟ ∴ ⎜ 1 + ⎟ (−10sin10t ) = −1⎜ vs + cos10t − ⎟
A⎠
RC ⎝
A⎠ ⎝ A⎠
A
⎝ A⎠
⎝
1⎞
1 1
⎛
∴ vs = ⎜1 + ⎟10sin10t + − cos10t Let A = 2000
A A
⎝ A⎠
∴ vs = 10.005sin10t + 0.0005 − 0.0005cos10t
(b)
Let A = ∞∴ vs = 10sin10tV
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Engineering Circuit Analysis, 7th Edition
54.
Chapter Seven Solutions
10 March 2006
Create a op-amp based differentiator using an ideal op amp with input capacitor C1 and
feedback resistor Rf followed by inverter stage with unity gain.
1mV
dvs
R
= 60 ×
/ min
R f C1
dt
rpm
R
RfC1=60 so choose Rf = 6 MΩ and C1 = 10 μF.
Vout = +
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Engineering Circuit Analysis, 7th Edition
55. (a)
0=
10 March 2006
Va − Vout
1
+
vdt
Rf
L∫
Va = V = 0,∴
(b)
Chapter Seven Solutions
− Rf t
Vout
1
v s dt '
=
⇒
=
V
v
dt
L
out
L ∫0
Rf
L∫
In practice, capacitors are usually used as capacitor values are more readily
available than inductor values.
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Engineering Circuit Analysis, 7th Edition
56.
Chapter Seven Solutions
10 March 2006
One possible solution:
vout = −
1
vin dt
R1C f ∫
we want vout = 1 V for vin = 1 mV over 1 s.
vin
In other words, 1 = −
1 1 −3
10−3
10
dt
=
−
R1C f ∫0
R1C f
Neglecting the sign (we can reverse terminals of output connection if needed),
we therefore need R1Cf = 10–3.
Arbitrarily selecting Cf = 1 μF, we find R1 = 1 kΩ.
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Engineering Circuit Analysis, 7th Edition
57.
Chapter Seven Solutions
10 March 2006
One possible solution of many:
dvin
dt
⎛ dv ⎞ 100 mV
maximum ⎜ in ⎟ . =
60s
⎝ dt ⎠
vout = − RC
vin
⎛ 100 mV ⎞
In other words, vout = 1 V = RC ⎜
⎟
⎝ 60s ⎠
or RC = 600
Arbitrarily selecting C = 1000 μF, we find that R = 600 kΩ.
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Engineering Circuit Analysis, 7th Edition
58.
Chapter Seven Solutions
10 March 2006
One possible solution of many:
dvin
dt
⎛ dv ⎞ 100 mV
At 1 litre/s, ⎜ in ⎟ =
.
s
⎝ dt ⎠
vout = − RC
vin
⎛ 100 mV ⎞
In other words, vout = 1 V = RC ⎜
⎟
⎝ 1s ⎠
or RC = 10
Arbitrarily selecting C = 10 μF, we find that R = 1 MΩ.
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Engineering Circuit Analysis, 7th Edition
59.
Chapter Seven Solutions
10 March 2006
One possible solution:
vin
The power into a 1 Ω load is I2, therefore energy = W = I2Δt.
vout =
1
I 2 dt
∫
R1C f
we want vout = 1 mV for vin = 1 mV (corresponding to 1 A 2 ).
(
)
Thus, 10−3 = RC 10−3 , so RC = 1
Arbitrarily selecting C = 1 μF, we find that we need R = 1 MΩ.
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Engineering Circuit Analysis, 7th Edition
60.
Chapter Seven Solutions
10 March 2006
One possible solution of many:
vout = − RC
dvin
dt
vin
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/s2. Therefore,
vout = 2.237 RC = 1
so RC = 0.447
Arbitrarily selecting C = 1 μF, we find that R = 447 kΩ.
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Engineering Circuit Analysis, 7th Edition
Chapter Seven Solutions
10 March 2006
61.
(a)
(b)
20v20 +
t
1
(v20 − vc )dt + 12 = is
−6 ∫o
5 × 10
t
1
(vc − v20 )dt − 12 + 10vc + 8 × 10−3 vc′ = 0
−6 ∫o
5 × 10
(c)
iL − is
i −i
+ 5 × 10−6 iL′ + L c = 0
20
10
t
ic − iL
1
+
i dt + 2 = 0
−3 ∫o c
10
8 ×10
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Engineering Circuit Analysis, 7th Edition
Chapter Seven Solutions
10 March 2006
62.
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Engineering Circuit Analysis, 7th Edition
Chapter Seven Solutions
10 March 2006
63.
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Engineering Circuit Analysis, 7th Edition
Chapter Seven Solutions
10 March 2006
64.
(a)
(b) “Let
is = 100e-80t A and i1 (0) = 20 A in the circuit of (new) Fig. 7.62.
(a) Determine v(t) for all t.
(b) Find i1(t) for t ≥ 0.
(c) Find v2(t) for t ≥ 0.”
(c) (a)
L eq = 1 4 = 0.8μ H∴ v(t ) = Leq is′ = 0.8 × 10−6 × 100(−80)r −80t V
∴ v(t) = −6.43-80t mV
t
(b)
i1 (t ) = 106 ∫ −6.4 × 10−3 e −80t dt + 20 ∴ i1 (t ) =
(c)
i2 (t ) = is − i1 (t ) ∴ i2 (t ) = 20e−80t + 60A
o
6400 −80t
(e − 1) = 80e−80t − 60A
80
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Engineering Circuit Analysis, 7th Edition
Chapter Seven Solutions
10 March 2006
65.
In creating the dual of the original circuit, we have lost both vs and vout. However, we
may write the dual of the original transfer function: iout/ is. Performing nodal analysis,
1 t
V1dt ′ + G in (V1 - V2 )
L1 ∫0
[1]
iout = Aid = GfV2 + Gin (V2 – V1)
[2]
iS =
Dividing, we find that
iout
=
iS
G in (V2 - V1 ) + G f V2
1 t
V1dt ′ + G in (V1 - V2 )
L1 ∫0
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Engineering Circuit Analysis, 7th Edition
66. I
PSpice
L = 4/10 = 400 mA. W =
Chapter Seven Solutions
10 March 2006
1 2
LI L = 160 mJ
2
verification:
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Engineering Circuit Analysis, 7th Edition
67. I
L = 4/(4/3) = 3 A. W =
Chapter Seven Solutions
10 March 2006
1 2
LI L = 31.5 J
2
PSpice verification:
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Engineering Circuit Analysis, 7th Edition
68.
Chapter Seven Solutions
10 March 2006
We choose the bottom node as the reference node, and label the nodal voltage at the top
of the dependent source VA.
Then, by KCL,
VA − 4 VA VA
V
+
+
= 0.8 A
100
20 25
25
Solving, we find that VA = 588 mV.
Therefore,
VC, 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
1
1
2
CV 2 = 10−3 ( 0.588 ) = 173 μJ
2
2
(
)
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Engineering Circuit Analysis, 7th Edition
69.
Chapter Seven Solutions
10 March 2006
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 =
PSpice
1
1
2
CV 2 = 5 × 10−6 ( 32 ) = 2.56 mJ
2
2
(
)
verification:
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Engineering Circuit Analysis, 7th Edition
70. C
Chapter Seven Solutions
10 March 2006
1 = 5 nF, Rf = 100 MΩ.
vout = − R f C1
(
)( )
dvs
= − 5 × 10−9 108 ( 30 cos100t ) = −15cos10t V
dt
Verifying with PSpice, choosing the LF411 and ±18 V supplies:
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Engineering Circuit Analysis, 7th Edition
71. PSpice
Chapter Seven Solutions
10 March 2006
verification
w = ½ Cv2 = 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, 7th Edition
72. PSpice
Chapter Seven Solutions
10 March 2006
verification
w = ½ Li2 = 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, 7th Edition
73.
Chapter Seven Solutions
10 March 2006
Va − V s 1
+ ∫ v L f dt
R1
L
− Vs 1
Va = Vb = 0,
0=
+ ∫ v L f dt
R1
L
0=
V L f = Va − Vout = 0 − Vout =
Vout = −
L f dVs
R1 dt
=−
(
L dVs
R1 dt
)
Lf d
A cos 2π 10 3 t ⇒ L f = 2 R1 ; Let _R = 1 Ω and L = 1 H.
R1 dt
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 combination of several
sinusoids.
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Engineering Circuit Analysis, 7th Edition
74. PSpice
Chapter Seven Solutions
10 March 2006
verification
w = ½ Cv2 = 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, 7th Edition
75. PSpice
Chapter Seven Solutions
10 March 2006
verification
w = ½ Li2 = 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, 7th Edition
1.
i (t ) = i (0)e
−R t
L
10 March 2006
9
= 2e −4.7×10 t mA
(
−4.7×109 100×10−12
(a)
i(100 ps) = 2e
(b)
i(212.8 ps) = 2e
(c)
vR = –iR
(
)
= 1.25 mA
−4.7×109 212.8×10−12
vR(75 ps) = −2 ( 4700 ) e
(d)
Chapter Eight Solutions
)
= 736 μ A
(
−4.7×109 75×10−12
)
= − 6.608 V
vL(75 ps) = vR(75 ps) = –6.608 V
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Engineering Circuit Analysis, 7th Edition
2.
Chapter Eight Solutions
10 March 2006
1 2
Li = 100 mJ at t = 0.
2
Thus, i( 0) = 0.1 = 316 mA
W=
R
− t
and i( t ) = i (0)e L = 316e− t / 2 mA
(a)
At t = 1 s, i (t ) = 316e −1/ 2 mA = 192 mA
(b)
At t = 5 s, i (t ) = 316e −5/ 2 mA = 25.96 mA
(c)
At t = 10 s, i (t ) = 316e −10 / 2 mA = 2.13 mA
(d)
At t = 2 s, i (t ) = 316e −1 mA = 116.3 mA . Thus, the energy remaining is
1
W (2) = Li (2) 2 = 13.53 mJ
2
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Engineering Circuit Analysis, 7th Edition
3.
Thus,
We know that i (t ) = i (0)e
L=
(
R
− t
L
−100 500 ×10−6
Chapter Eight Solutions
−3
= 2 × 10 e
−
100
t
L
10 March 2006
, and that i(500 μs) = 735.8 μA.
) = −100 ( 500 ×10 ) = 50 mH
⎛ i (500 × 10−6 ) ⎞
ln ⎜
⎟
−3
⎝ 2 × 10
⎠
−6
⎛ 735.8 ×10−6 ⎞
ln ⎜
⎟
−3
⎝ 2 ×10
⎠
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Engineering Circuit Analysis, 7th Edition
4.
Thus,
We know that i (t ) = i (0)e
R=−
R
− t
L
= 1.5e
Chapter Eight Solutions
−
R
t
3×10−3
10 March 2006
, and that i(2) = 551.8 mA.
3 × 10−3 ⎛ i (2) ⎞
3 × 10−3 ⎛ 0.5518 ⎞
ln ⎜
=
−
ln ⎜
⎟ = 1.50 mΩ
⎟
2
2
⎝ 1.5 ⎠
⎝ 1.5 ⎠
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Engineering Circuit Analysis, 7th Edition
5.
At
At
Thus,
R
− t
L
Chapter Eight Solutions
−
R
10 March 2006
t
We know that i (t ) = i (0)e = 1.5e
, and that W(0) = 1 J; W(10–3) = 100 mJ.
2
1
t = 0,
3 ×10−3 ⎡⎣i ( 0 ) ⎤⎦ = 1 therefore i (0) = 25.82 A.
2
2
1
t = 1 ms,
3 ×10−3 ⎡⎣i 10−3 ⎤⎦ = 0.1 therefore i (10−3 ) = 8.165 A.
2
(
)
(
R=−
3×10−3
) (
)
3 × 10−3 ⎛ i (t ) ⎞
3 × 10−3 ⎛ 8.165 ⎞
ln ⎜
ln ⎜
=
−
⎟
⎟ = 3.454 Ω
t
0.001 ⎝ 25.82 ⎠
⎝ i (0) ⎠
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Engineering Circuit Analysis, 7th Edition
Chapter Eight Solutions
10 March 2006
6.
(a)
Since the inductor current can’t change instantaneously, we simply need to find iL while
the switch is closed. The inductor is shorting out both of the resistors, so iL(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, 7th Edition
7.
Chapter Eight Solutions
10 March 2006
(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 acting 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 iL 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 remain the same,
so
iL = 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, 7th Edition
Chapter Eight Solutions
10 March 2006
8.
(a)
iL (0) = 4.5mA, R/L =
103
106
=
4 × 10−3
4
6
∴ iL = 4.5e −10 t / 4 mA ∴ iL (5μ s ) = 4.5e −1.25
= 1.289 mA.
(b) iSW( 5 μs) = 9 – 1.289 = 7.711 mA.
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Engineering Circuit Analysis, 7th Edition
9.
(a)
Chapter Eight Solutions
10 March 2006
100
= 2A ∴ iL (t ) = 2e−80t / 0.2
50
−400 t
= 2e A, t > 0
iL (0) =
(b)
iL (0.01) = 2e −4 = 36.63mA
(c)
2e −400t1 = 1, e 400t1 = 2, t1 = 1.7329ms
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Engineering Circuit Analysis, 7th Edition
Chapter Eight Solutions
10 March 2006
10. (a)
di
+ 5i = 0
[1]
dt
vR = −2i so Eq. [1] can be written as
L
⎛ −v ⎞
d⎜ R ⎟
2 ⎠
⎛ −v ⎞
− 5 ⎜ R ⎟ = 0 or
L ⎝
dt
⎝ 2 ⎠
dv
2.5 R + 2.5vR = 0
dt
(b) Characteristic equation is 2.5s + 2.5 = 0, or s + 1 = 0
Solving,
s = –1, vR (t ) = Ae −t
(c)
At t = 0–, i(0–) = 5 A = i(0+). Thus, vR (t ) = −10e − t , t > 0
2
vR (0− ) = (10 ) = 6.667 V
3
+
vR (0 ) = −10 V
vR (1) = −10e −1 = − 3.679 V
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Engineering Circuit Analysis, 7th Edition
Chapter Eight Solutions
10 March 2006
11.
(a)
I I
i
t
t
= e − t /τ , = ln o , o = 10 ∴ = ln10 = 2.303;
i i
Io
τ
τ
t
t
Io
I
= 100, = 4.605; o = 1000, = 6.908
i
i
τ
τ
(b)
i
d (i / Io )
d ()
= e− t /τ ,
= −et /τ ; t t / τ = 1,
=a−e −1
Io
d (t / τ )
d ()
⎛t
⎞
i
Now, y = m( x − 1) + b = −e −1 ( x − 1) + e−1 ⎜ = x, = y ⎟
Io
⎝τ
⎠
−1
−1
At y = 0, e ( x − 1) = e ∴ x = 2 ∴ t / τ = 2
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Engineering Circuit Analysis, 7th Edition
12.
Chapter Eight Solutions
10 March 2006
Reading from the graph current is at 0.37 at 2 ms
∴ τ = 2 ms
I 0 = 10 A
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Engineering Circuit Analysis, 7th Edition
Chapter Eight Solutions
10 March 2006
w = ½ Li2, 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 iL(t) = 1.732 e–0.1t A.
For R = 10 kΩ, τ = 1 μs so iL(t) = 1.732 e-t. For R = 100 kΩ, τ = 100 ns or 0.1 μs so
iL(t) = 1.732 e-10t A. For each current expression above, it is assumed that time is
expressed
in microseconds.
13.
To create a sketch, we firs t 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 tim e 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, 7th Edition
Chapter Eight Solutions
10 March 2006
14.
(a)
3.3 × 10 −6
τ=
= 3.3 × 10 −12
6
1 × 10
(b)
1
2
ω = .L.I 0 2
I0 =
2 × 43 × 10− 6
= 5.1 A
3.3 × 10− 6
i (5 ps ) = 5.1e −1×10 ×5×10
6
− 12
/ 3.3×10 − 6
= 1.12 A
(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, 7th Edition
Chapter Eight Solutions
10 March 2006
15. Assume the source Thévenin resistance is zero, and assume the transient is measured to
5τ. Then,
τ=
L
5L
∴ 5τ =
= 100 × 10−9 secs
R
R
(5)(125.7)10−6
∴R >
10−7
(If 1τ assumed then R >
so R must be greater than 6.285 kΩ.
6.285
= 125.7Ω )
5
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Engineering Circuit Analysis, 7th Edition
16. For
that
Chapter Eight Solutions
10 March 2006
t < 0, we have a current divider with iL(0-) = ix(0-) = 0.5 [ 10 (1/ (1 + 1.5)] mA
= 2 mA. For t > 0, the resistor through which ix flows is shorted, so that ix(t > 0) = 0.
The remaining 1-kΩ resistor and 1-mH inductor network exhibits a decaying current such
iL(t) = 2e-t/τ mA where τ = L/R = 1 μS.
(a)
(b)
ix (mA)
2
t (μs)
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Engineering Circuit Analysis, 7th Edition
17.
Chapter Eight Solutions
10 March 2006
2
1
2 ×10−3
C ⎡⎣ v ( 0 ) ⎤⎦ = 10−3 so v ( 0 ) =
= 44.72 V
2
10−6
(a)
τ = RC = 100 s
(b)
v(t ) = v(0)e −t / RC = 44.72e −0.01t V
Thus, v(20) = = 44.72e −0.01(20) = 36.62 V .
v
36.62
Since i = =
= 366 nA
R 100 × 106
(c)
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Engineering Circuit Analysis, 7th Edition
Chapter Eight Solutions
18. If
i(0) = 10 A, v(0) = 10 V. v(t ) = v(0)e −t / RC = 10e− t / 2 V
(a)
At t = 1 s, v(1) = 10e −1/ 2 = 6.065 V
(b)
At t = 2 s, v(2) = 10e −1 = 3.679 V
(c)
At t = 5 s, v(5) = 10e −2.5 = 821 mV
(d)
At t = 10 s, v(10) = 10e −5 = 67.4 mV
10 March 2006
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Engineering Circuit Analysis, 7th Edition
19.
Chapter Eight Solutions
10 March 2006
Referring to Fig. 8.62, we note that τ = RC = 4 s. Thus,
v(t ) = v(0)e −t / RC = 5e−0.25t V.
(a)
v(1 ms) = 5e −0.25( 0.001) = 4.999 V
(b)
v(2 ms) = 5e −0.25( 0.002 ) = 4.998 V
Therefore i(2 ms) = 4.998 / 1000 = 4.998 mA
(c)
v(4 ms) = 5e −0.25( 0.004 ) = 4.995 V
1
W = Cv 2 = 49.9 mJ
2
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Engineering Circuit Analysis, 7th Edition
20.
Thus,
Chapter Eight Solutions
10 March 2006
(a) v(t ) = v(0)e− t / RC = 1.5e− t / RC V
−t
⎡ v(t ) ⎤
= ln ⎢
⎥
RC
⎣ 1.5 ⎦
R=
−t
−2 ×10−9
=
= 7.385 Ω
⎡ v(t ) ⎤
⎡ 0.1 ⎤
−10
10 ln ⎢ ⎥
C ln ⎢
⎣ 1.5 ⎥⎦
⎣ 1.5 ⎦
(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, 7th Edition
21.
Chapter Eight Solutions
10 March 2006
The film acts as an intensity integrator. Assuming that we may model the intensity as a
simple decaying exponential,
φ(t) = φo 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 = K∫
exposure time
0
φo e − t / τ dt
where K is some constant. Solving the integral, we find that
[
]
Φ = - K φoτ e − (exp osure time)/ τ - 1
The maximum value of this intensity function is –Kφoτ.
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
–3
that 1 – e–150×10 / τ = 0.14, assuming that we are observing single-exponential decay
behavior and that the response speed of the fil m 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 pa rallel 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, 7th Edition
Chapter Eight Solutions
10 March 2006
22.
(a)
(b)
30
= 192V
50
vc (t ) = 192e −3000t / 24 = 192e −125t V
vc (0) = 8(50 200) ×
0.1 = e −125t ∴ t = 18.421ms
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Engineering Circuit Analysis, 7th Edition
Chapter Eight Solutions
10 March 2006
23.
6
4
4
(a)
vc = 80e −10 t /100 = 80e−10 t V, t > 0; 0.5 = e−10 t ∴ t = 69.31μ s
(b)
wc =
1
1
C 802 e −20,000t = C 802 ∴ t = 34.66 μ s
2
4
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Engineering Circuit Analysis, 7th Edition
Chapter Eight Solutions
10 March 2006
24.
t < 0 : ic (t ) = 0, 10 = 5000is + 104 is ∴ is =
∴ vc (t ) =
20
= 6.667V
3
4
t > 0 : is = 0 ∴ vc (t ) = 6.667e − t / 2×10 ×2×10
∴ vc (t ) = 6.667e−25t V ∴ ic (t ) =
2
mA
3
−6
−6.667 −25t
e = 0.3333e−25t mA
3
20 ×10
vC(t)
iC(t)
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Engineering Circuit Analysis, 7th Edition
Chapter Eight Solutions
10 March 2006
25.
( )
i (0 ) = 0.1A
v 0 + = 20V
+
v(1.5ms ) = 20e −1.5×10 / 50×20×10 = 4.5V
i (1.5ms ) = 0 A
−3
−6
v(3ms ) = 20e −3×10 / 50×20×10 = 1V
i (3ms ) = 0 A
−3
−6
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Engineering Circuit Analysis, 7th Edition
Chapter Eight Solutions
10 March 2006
26.
(a)
1
2
iL (0− ) = × 60 = 30 mA, ix (0− ) = × 30 = 20 mA
2
3
(b)
iL (0+ ) = 30 mA, ix (0+ ) = −30 mA
(c)
iL (t ) = 30e−250t / 0.05 = 30e−5000t mA, iL (0.3ms)
= 30e−1.5 = 6.694mA = −ix
Thus,
ix = –6.694 mA.
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Engineering Circuit Analysis, 7th Edition
Chapter Eight Solutions
10 March 2006
27.
(a)
iL (0) = 4A ∴ iL (t ) = 4e−500t A (0 ≤ t ≤ 1ms)
iL (0.8ms) = 4e−0.4 = 2.681A
(b)
iL (1ms) = 4e −0.5 = 2.426A
∴ iL (t ) = 2.426e −250( t −0.001)
∴ iL (2ms) = 2.426e −0.25 = 1.8895− A
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Engineering Circuit Analysis, 7th Edition
Chapter Eight Solutions
10 March 2006
28.
(a)
iL = 40e−50,000t mA ∴10 = 40e−50,000t , ∴ t1 = 27.73μ s
(b)
iL (10 μ s) = 40e −0.5 = 24.26mA ∴ iL
= 24.26e− (1000+ R )50t (t > 10 μ s)
−6
∴10 = 24.26e− (1000+ R )5×10 ∴ ln2.426 = 0.8863
= 0.25(1000 + R)10−3 ,1000 + R = 0.8863 × 4 ×103 ∴ R = 2545+ Ω
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Engineering Circuit Analysis, 7th Edition
Chapter Eight Solutions
10 March 2006
29.
(a)
i1 (0) = 20mA, i2 (0) = 15mA
∴ v(t ) = 40e −50000t + 45e −100000t V ∴ v(0) = 85V
(b)
v(15μ s ) = 40e −0.75 + 45e−1.5 = 28.94V
(c)
85
= 40e −50000t + 45e −100000t . Let e −50000t = x
10
∴ 45 x 2 + 40 x − 8.5 = 0
−40 ± 1600 + 1530
= 0.17718, < 0
90
∴ e −50000t = 0.17718, t = 34.61μ s
∴x =
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Engineering Circuit Analysis, 7th Edition
30.
t < 0 : vR =
Chapter Eight Solutions
2R1R 2
2R1
, ↓ iL (0) =
R1 +R 2
R1 + R 2
t > 0 : iL (t ) =
2R1
2R1R 2 −50 R2t
e−50 R2t ∴ vR =
e
R1 + R 2
R1 +R 2
∴ vR (0+ ) = 10 =
2R1R 2
∴ R1 R 2 = 5Ω. Also, vR (1ms)
R1 +R 2
= 5 = 10e−50 R2 /1000 ∴ 0.05R 2 = 0.6931∴ R 2 = 13.863Ω
∴
1
1 1
+
= ∴ R1 = 7.821Ω
13.863 R1 5
10 March 2006
31.
24
= 0.4A ∴ iL (t ) = 0.4e−750t A, t > 0
60
(a)
iL (0) =
(b)
5
vx = × 24 = 20V, t < 0
6
3
vx (0+ ) = 50 × 0.4 × = 7.5V
8
−750 t
∴ vx (t ) = 7.5e V, t > 0
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Engineering Circuit Analysis, 7th Edition
32.
Chapter Eight Solutions
10 March 2006
3iL
× 20 + 10iL = 25iL
4
v
vin ∴ in = 25Ω∴ iL = 10e−25t / 0.5 = 10e −50t A, t > 0
iL
v in =
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33.
iL (0) =
Chapter Eight Solutions
10 March 2006
64
40
×
= 5A
4 + 40 8 48
∴ iL = 5e −24t / 8 = 5e −3t A
∴ i1 (t ) = 2.5e −3t A, t > 0; i1 (−0.1)= 2.5 A
i 1 (0.03) = 2.285− A, i1 (0.1) = 1.852 A
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Engineering Circuit Analysis, 7th Edition
Chapter Eight Solutions
10 March 2006
34.
(a)
iL (0) = 4A ∴ iL = 4e−100t A, 0 < t < 15ms
∴ iL (15ms) = 4e −1.5 = 0.8925+ A
(b)
t >15ms: iL = 0.8925+ e−20(t −0.015) A
∴ iL (30ms) = 0.8925+ e −0.3 = 0.6612A
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Engineering Circuit Analysis, 7th Edition
Chapter Eight Solutions
10 March 2006
35.
(a)
i1 (0+ ) = i1 (0− ) = 10A, i2 (0+ ) = i2 (0− ) = 20A ∴ i (0+ ) = 30A
(b)
τ = L eq / R eq =
(c)
i1 (0− ) = 10A, i2 (0− ) = 20A ; i (t ) = 30e −600t A
(d)
v = −48i = −1440e−600t V
(e)
i1 = 10(−440) ∫ e −600t dt + 10 = 24e −600t t0 +10 = 24e −600t − 14A
0.08 5
= ms = 1.6667ms;
48 3
t
0
t
i2 = 2.5(−1440) ∫ e −600t dt + 20
0
= 6e −600t t0 +20 = 6e −600t + 14A
(f)
1
1
WL (0) = × 0.1×102 + × 0.4 × 202 = 5 + 80 = 85J
2
2
1
1
WL (∞) = × 0.1× 142 + × 0.4 × 142 = 9.8 + 39.2 = 49J
2
2
∞
∞
900 × 48
WR = ∫ i 2 48dt = ∫ 900 × 48e−1200t dt =
(−1) = 36J
0
0
−1200
∴ 49 + 36 = 85 checks
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Engineering Circuit Analysis, 7th Edition
Chapter Eight Solutions
10 March 2006
36.
(a)
(b)
2
2
100 2
× = 33.33V; i1 (0− ) =
× = 16.667mA
2+2 3
2+2 3
∴ vc (9 : 59) = 33.33V, i1 (9 : 59) = 16.667mA
vc (0) = 100 ×
vc (t ) = 33.33e − t / 400 , t > 10 : 00 ∴ vc (10 : 05) = 33.33e −300 / 400
= 15.745+ V, i1 (10 : 05) =
(c)
15.745
= 3.936mA
4000
τ = 400 s, so 1.2τ = 480 s. vC(1.2τ) = 33.33 e–1.2 = 10.04 V.
Using Ohm’s law, we find that i1(1.2τ) = vC(1.2τ)/ 4000 = 2.51 mA.
(d) PSpice
Verification:
We see from the DC analysis of the circuit that our initial value is correct; the Probe
output confirms our hand calculations, especially for part (c).
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Engineering Circuit Analysis, 7th Edition
Chapter Eight Solutions
10 March 2006
37.
t> 0:
25ix
= 1.25ix ∴ 34 = 100(1.25ix − 0.8ix + ix ) + 25ix ∴ ix = 0.2A
20
(a)
is (0− ) = (1.25 − 0.8 + 1)0.2 = 0.290 A
(b)
ix (0− ) = 0.2A
(c)
vc (t ) = 25 × 0.2e −t = 5e− t V ∴ ix (0+ ) =
(d)
0.8ix (0+ ) = 0.04A ∴ is (0+ ) =
(e)
ix (0.4) =
5
= 0.05A
100
34
20 33.2
− 0.04 ×
=
= 0.2767A
120
120 120
1
× 5e−0.4 = 0.03352A
100
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Engineering Circuit Analysis, 7th Edition
Chapter Eight Solutions
10 March 2006
38.
(a)
vc (0) = 10V ∴ vc (t ) = 10e
(b)
iA (−100μ s) = iA (0− ) =
−106 t /(10 + 50 200)
= 10e −20000t V
10
= 50mA
200
⎛ 1 ⎞ 50
iA (100 μ s ) = 10e−2 ⎜
= 5.413mA
⎟
⎝ 10 + 40 ⎠ 250
(c) PSpice
Verification.
From the DC simulation, we see that PSpice verifies our hand calculation of iA = 50 mA.
The transient response plotted using Probe indicates that at 100 μs, the current is approximately
5.46 mA, which is within acceptable round-off error compared to the hand calculated value.
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Engineering Circuit Analysis, 7th Edition
39.
Chapter Eight Solutions
10 March 2006
12
= −6mA (t < 0)
12 + 4
(a)
i1 (t ) = 8(−1)
(b)
4 12 6 = 2k Ω, vc (0) = 48V
6
3
∴ vc (t ) = 48e−10 t / 5×2×10 = 48e−100t V, t > 0
∴ i (t ) = 12e−100t mA, t > 0
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Engineering Circuit Analysis, 7th Edition
Chapter Eight Solutions
10 March 2006
40.
(a)
vCLeft (0) = 20V, vCRIGHT (0) = 80V
6
6
∴ vCL = 20e−10 t / 8 , vCR = 80e−10 t / 0.8
∴ vout = vCR − vCL = 80e−1,250,000t − 20e−125,000t V, t > 0
(b)
vout (0+ ) = 60V; vout (1μ s) = 80e−1.25 − 20e −0.125 = 5.270V
vout (5μ s) = 80e−6.25 − 20e −0.625 = −10.551V
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Engineering Circuit Analysis, 7th Edition
41. (a)
Chapter Eight Solutions
10 March 2006
vc − 0.25vc vc vc − 40
+ +
= 0 ∴ vc = 20V (t < 0)
5
10
4
1 − 0.25
+ 0.1 − iin = 0.25A
t > 0: Apply vc = 1V ∴
5
1
∴ R eq =
= 4Ω
0.25
t < 0:
6
∴ vc (t ) = 20e −10 t / 4 = 20e −250,000t V (t > 0)
(b)
(c)
vC(3 μs) = 9.447 V
PSpice verification. Note that the switch parameters had to be changed in order to
perform this simulation.
As can be seen from the simulation results, our hand calculations are accurate.
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Engineering Circuit Analysis, 7th Edition
42.
Chapter Eight Solutions
10 March 2006
t < 0 : vc (0) = 60V
50 −500 /( Ro +1000)
e
60
R
500
∴
= ln 1.2 = 0.18232 ∴ o + 2 = 5.4848, Ro = 1742.4Ω
R o + 1000
500
6
0 < t < 1ms: vc = 60e−10 t /( Ro +1000) ∴
∴ vc (1ms) = 60e−1000 / 2742.4 = 41.67V
6
−3
t > 1ms : vc = 41.67e −10 (t −10 ) /(1742.4 + R1 1000)
∴ 25 = 41.67e −1000( ) ∴ 0.5108 =
= 1957.6, R1 1000 = 215.2
.1000
,1742.4 + R1 1000
1742.4 + R1 1000
1
1
+ 10−3 =
∴ R1 = 274.2Ω
R1
215.2
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Engineering Circuit Analysis, 7th Edition
Chapter Eight Solutions
10 March 2006
43.
(a)
With the switch closed, define a nodal voltage V1 at the top of the 5-kΩ resistor.
Then,
0 = (V1 – 100)/ 2 + (V1 – VC)/ 3 + V1/ 5
0 = VC/ 10 + (VC – V1)/ 3 + (VC – 100)
[1]
[2]
Solving, we find that VC = vC(0-) = 99.76 V.
(b)
7
t > 0 : R eq = 10 6.5 = 3.939k Ω∴ vc = 87.59e −10 t / 3939 = 87.59e −2539t V (t > 0)
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Engineering Circuit Analysis, 7th Edition
44.
Chapter Eight Solutions
10 March 2006
t < 0:
12 = 4i1 + 20i1 ∴ i1 = 0.5mA ∴ vc (0) = 6i1 + 20i1 = 26i1
vc (0) = 13V
t > 0: Apply ← 1mA ∴1 + 0.6i1 = i1 ∴ i1 = 2.5mA; ± vin = 30i1 = 75V ∴ R eq = 75k Ω
3
−9
6
∴ vc (t ) = 13e −t / 75×10 ×2×10 = 13e−10 t /150 = 13e−6667 t
∴ i1 (t ) =
vo
= 0.4333e −6667 t mA (t > 0)
3 × 104
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Engineering Circuit Analysis, 7th Edition
Chapter Eight Solutions
10 March 2006
45.
(a)
v1 (0− ) = 100V. v2 (0− ) = 0, vR (0− ) = 0
(b)
v1 (0− ) = 100V. v2 (0+ ) = 0, vR (0+ ) = 100V
(c)
τ=
(d)
vR (t ) = 100e −12.5t V, t > 0
(e)
i (t ) =
(f)
v1 (t ) =
(g)
20 × 5
×10−6 × 2 104 = 8 × 10−2 s
20 + 5
vR (t )
= 5e −12.5t mA
4
2 ×10
106 t
103 −12.5t t
−3 −12.5t
−12.5t
−
×
e
dt
+
=
e
+ 80V
5
10
100
o +100 = −20e
∫
o
20
50
1000 t −12.5t
v2 (t ) =
dt + 0 = −80e −12.5t to +0 = −80e−12.5t + 80V
5e
∫
o
5
1
1
wc1 (∞) = × 20 ×10−6 × 802 = 64mJ, wc 2 (∞) × 5 × 10−6 × 802 = 16mJ
2
2
1
wc1 (0) = × 20 ×10−6 × 1002 = 100mJ, wc 2 (0) = 0
2
∞
25
wR = ∫ 25 × 10−6 e −25t × 2 × 104 dt =
× 2 × 104 (−1)10−6 = 20mJ
o
−25
64 + 16 + 20 = 100 checks
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Engineering Circuit Analysis, 7th Edition
Chapter Eight Solutions
10 March 2006
46.
(a)
t < 0: is = 1mA ∴± vc (0) = 10V, ↓ iL (0) = −1mA ∴ vx (0) = 10V, t < 0
(b)
t > 0 : vc (t ) = 10e −t /10 ×20×10 = 10e−5000t V
4
−9
i L (t ) = −10−3 e − 103t / 0.1 = −10−3 e −10000t A ∴ ± vL (t ) = e −10000t V, t > 0
∴ vx = vc − vL (t ) = 10e −5000t − e−10000t V, t > 0
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Engineering Circuit Analysis, 7th Edition
Chapter Eight Solutions
10 March 2006
47.
(a)
t < 0: vs = 20V ∴ vc = 20V, iL = 20mA ∴ ix (t ) = 20mA, t < 0
(b)
t > 0: vs = 0 ∴ iL (t ) = 0.02e −10000t A; vc (t ) = 20e− t / 2×10 10 = 20e−5000t V
−8
4
↓ ic (t ) = 2 × 10−8 × 20(−5000) e −5000t = −2e−5000t mA
ix (t ) = iL (t ) + ic (t ) = 0.02e −10000t − 0.002e −5000t A = 20e −10000t − 2e−5000t mA
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Engineering Circuit Analysis, 7th Edition
48.
Chapter Eight Solutions
10 March 2006
V
1
=
= 1.1 A
R 0.909
t > 0 : iL (t ) = e − 2.363t A
iL ( 0 − ) =
iL (0.1s ) = 1.1e − 2.363×0.1 = 0.8685 A
∴since the current has dropped to less than 1 A prior to t = 100 ms, the fuse does not blow.
PSpice verification: Note that the switch properties were changed.
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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49.
Engineering Circuit Analysis, 7th Edition
Chapter Eight Solutions
v(t ) = 6u (t ) − 6u (t − 2) + 3u (t − 4)
V
10 March 2006
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Engineering Circuit Analysis, 7th Edition
50.
Chapter Eight Solutions
10 March 2006
i (t ) = 2u (t ) + 2u (t − 2) − 8u (t − 3) + 6u (t − 4) A
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Engineering Circuit Analysis, 7th Edition
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
Chapter Eight Solutions
10 March 2006
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Engineering Circuit Analysis, 7th Edition
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
Chapter Eight Solutions
10 March 2006
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Engineering Circuit Analysis, 7th Edition
53.
Chapter Eight Solutions
10 March 2006
(a) v A = 300u (t − 1) V, vB = −120u (t + 1) V; ic = 3u (−t )A
100
= 1A
300
−120
t = 0.5 : i1 (−0.5) =
+ 1 = 0.6A;
300
120
300 120
t = 0.5 : i1 = −
= −0.4A; t = 1.5 : i1 =
−
= 0.6A
300
300 300
t = −1.5 : i1 (−1.5) = 3 ×
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Engineering Circuit Analysis, 7th Edition
Chapter Eight Solutions
10 March 2006
54.
↑ v A = 600tu (t + 1) v, vB = 600 (t + 1) u (t ) V, ic = 6(t − 1) u (t − 1)A
(a)
t = −1.5 : i1 = 0; t = −0.5 : i1 = 600(−0.5) / 300 = −1A
600(0.5) 600(1.5)
+
= 4A
300
300
600(1.5) 600(2.5) 1
t = 1.5 : i1 =
+
+ × 6 × 0.5 = 3 + 5 + 1 = 9A
300
300
3
t = 0.5 : i1 =
(b)
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Engineering Circuit Analysis, 7th Edition
Chapter Eight Solutions
10 March 2006
55.
(a)
2u (−1) − 3u (1) + 4u (3) = −3 + 4 = 1
(b)
[5 − u (2)] [2 + u (1)] [1 − u (−1)]
(c)
4e −u (1)u (1) = 4e −1 = 1.4715+
= 4 × 3 ×1 = 12
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Engineering Circuit Analysis, 7th Edition
Chapter Eight Solutions
10 March 2006
56.
100
20
+ 0 + 10 ×
= 6A
50
50
60
t > 0: ix = 0 + + 0 = 2A
30
(a)
t < 0: ix =
(b)
t < 0: The voltage source is shorting out the 30-Ω resistor, so ix = 0.
t > 0: ix = 60/ 30 = 2 A.
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Engineering Circuit Analysis, 7th Edition
57.
t = −0.5 :
50 25 = 16.667, ix =
t = 0.5 :
ix =
t = 1.5 :
ix = 3 −
t = 2.5 :
ix =
t = 3.5 :
ix = −
Chapter Eight Solutions
10 March 2006
200
1/ 50
1
−2
= 3 − = 2.5A
66.67
1/ 50 + 1/ 25 + 1/ 50
2
200
= 3A
66.67
100 1
× = 2.5A
66.67 3
200 − 100
= 2A
50
100
= −2A
50
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Engineering Circuit Analysis, 7th Edition
58.
Chapter Eight Solutions
10 March 2006
v(t ) = 4 − 16u (t ) + 20u (t − 4) − 6u (t − 6)V
v(t) (V)
8
4
2
0
4
6
t (s)
-12
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Engineering Circuit Analysis, 7th Edition
59. (a)
Chapter Eight Solutions
10 March 2006
7 u (t ) − 0.2 u (t ) + 8(t − 2) + 3
v (1) = 9.8 volts
(b)
Resistor of value 2Ω
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Engineering Circuit Analysis, 7th Edition
Chapter Eight Solutions
10 March 2006
60.
i (t ) =
R
− t ⎞
Vo ⎛
L
−
e
1
⎜
⎟ u (t ) A and vR (t ) = i ( t ) R
R⎝
⎠
(a)
R
− t ⎞
⎛
vR (t ) = Vo ⎜1 − e L ⎟ u (t )
⎝
⎠
(b)
vR (2 ×10−3 ) = 1.2 1 − e−2
(
(
)
V =
1.2 1 − e −1000t u (t )
V
) V = 1.038 V
(c)
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Engineering Circuit Analysis, 7th Edition
Chapter Eight Solutions
10 March 2006
61.
(a)
iL (t ) = (2 − 2e −200000t ) u (t )mmA
(b)
vL (t ) = LiL′ = 15 × 10−3 × 10−3 (−2)
(−200000e −200000t ) u (t ) = 6e −200000t u (t )V
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Engineering Circuit Analysis, 7th Edition
Chapter Eight Solutions
10 March 2006
62.
(a)
iL (t ) = 2 + 2 (1 − e −2.5t ) u (t ) A ∴ i1 (−0.5) = 2A
(b)
iL (0.5) = 2 + 2 (1 − e−1.25 ) = 3.427A
(c)
iL (1.5) = 2 + 2 (1 − e−3.75 ) = 3.953A
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Engineering Circuit Analysis, 7th Edition
Chapter Eight Solutions
10 March 2006
63.
(a)
iL (t ) = (4 − 4e−20t / 0.02 )u (t )
∴ iL (t ) = 4(1 − e −1000t )u (t )A
(b)
v1 (t ) = (100 − 80e−1000t )u (t )V
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Engineering Circuit Analysis, 7th Edition
64.
Chapter Eight Solutions
10 March 2006
(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.
6
We may therefore write i(t) = 3.125[1 – e-10 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, 7th Edition
65.
(a)
τ = L/RTH =
Chapter Eight Solutions
30 × 10−6 30 ×10−6
=
= 9 ×10−6
5 ||10
3.333
10 March 2006
s
i (t ) = i f (t ) + in (t )
in = Ae
−106 t
9
and i f =
9
5
6
Thus,
At
so
10
−
t
9
i (t ) = + Ae 9
5
t = 0, i(0 ) = i(0+) = 4.5/5. Thus, A = –4.5/5 = 0.9
6
10
−
t
9
i (t ) = − 0.9e 9 A
5
(b)
At t = 1.5 μs, i = i (t ) =
1.5
− t
9
− 0.9e 9 = 1.038 A
5
(c)
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Engineering Circuit Analysis, 7th Edition
66.
Chapter Eight Solutions
10 March 2006
45 ×10−3 45 ×10−3
=
= 0.0135 s
10 || 5
3.333
vR (t ) = v f + vn
τ = L/Req =
(a)
vf(t) = 0 since inductor acts as a short circuit.
Thus,
vR (t ) = vn = Ae−74.07 t .
t = 0, iL = 12/10 = 1.2 A = iL(0-) = iL(0+).
At
Writing KVL for this instant in time, 16 – 10(1.2 + vR/5) = vR
4
4
Therefore
vR (0+ ) = V and hence vR (t ) = e−74.07 t V
3
3
(b)
At t = 2 ms, vR (2 ms) =
4 −74.07( 2×10−3 )
V = 1.15 V
e
3
(c)
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Engineering Circuit Analysis, 7th Edition
67.
τ=
Chapter Eight Solutions
10 March 2006
L 5 ×10−3
=
= 50 sμ
Req
100
v1 (t ) = v1 f + v1n where v1 f = 6 V since the inductor acts as a short circuit
Therefore
At
v1 (t ) = 6 + Ae
−
106 t
50
.
t = 0-, iL = 0 = iL(0+). Thus, v1(0+) = 0 since no current flows through the resistor.
10 t ⎞
⎛
−
v1 (t ) = 6 ⎜ 1 − e 50 ⎟ V.
⎜
⎟
⎝
⎠
6
Hence
At
27
−
⎛
⎞
t = 27 μs, v1 (27 ×10−6 ) = 6 ⎜1 − e 50 ⎟ = 2.5 V
⎝
⎠
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Engineering Circuit Analysis, 7th Edition
Chapter Eight Solutions
10 March 2006
68.
(a)
iL (t ) = 10 A, t < 0
(b)
iL (t ) = 8 + 2e−5t / 0.5
∴ iL (t ) = 8 + 2e−10t A, t > 0
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Engineering Circuit Analysis, 7th Edition
Chapter Eight Solutions
10 March 2006
69.
(a)
iL (t ) = 2A, t > 0
(b)
iL (t ) = 5 − e−4t / 0.1
∴ iL (t ) = 5 − 3e−40t A, t > 0
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Engineering Circuit Analysis, 7th Edition
Chapter Eight Solutions
10 March 2006
70.
(a) 0,
0
(b)
0, 200V
(c) 1A,
(d)
100V
50 × 10−3 1
= ms ∴ iL = 1(1 − e −4000t ) u (t )A, iL (0.2ms) = 0.5507A
200
4
v1 (t ) = (100 + 100e −4000t ) u (t )V, v1 (0.2ms) = 144.93V
τ=
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Engineering Circuit Analysis, 7th Edition
71.
(a)
(b)
Chapter Eight Solutions
10 March 2006
di
+ Pi = Q, i = e − Pt ∫ Qe Pt dt + Ae − Pt , R = 125Ω, L = 5H
dt
di
∴ L LPi = LQ ∴ LP = 5P = R = 125 ∴ P = 25
dt
t
10
2
= 2 ∴ i = e −25t ∫ 2e 25t dt + Ae −25t = e−25t × e 25t to + Ae−25t
o
L
25
2
10
2
2
∴i =
+ Ae −25t , i (0) =
= ∴ A = 0∴i =
= 0.08A
125 25
25
25
Q(t ) =
t
10u (t )
2
= 2u (t ) ∴ i = e −25t ∫ 2e 25t dt + Ae −25t =
+ Ae −25t
o
5
25
2
i (0) = 0 ∴ A = − ∴ i (t ) = 0.08(1 − e −25t )A, t > 0
25
Q(t ) =
(c)
Q(t ) =
10 + 10u (t )
= 2 + 2 u (t ) ∴ i = 0.16 − 0.08e −25t A, t > 0
5
(d)
Q(t ) =
t
10u (t ) cos 50t
= 2u (t ) cos 50t ∴ i = e −25t ∫ 2 cos 50t × e 25t dt + Ae −25t
o
5
t
∴ i = 2e
−25t
⎡ e 25t
⎤
−25t
⎢ 502 + 252 (25cos 50t + 50sin 50t ) ⎥ + Ae
⎣
⎦o
⎡ e 25t
⎤
1
= 2e −25t ⎢
× 25⎥ + Ae −25t
(25cos 50t + 50sin 50t ) −
3125
⎣ 3125
⎦
2
4
2 −25t
e + Ae −25t
=
cos 50t +
sin 50t −
125
125
125
2
2
i (0) = 0 ∴ 0 =
−
+ A∴A = 0
125 125
∴ i (t ) = 0.016 cos 50t + 0.032sin 50t − 0.016e−25t A, t > 0
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Engineering Circuit Analysis, 7th Edition
Chapter Eight Solutions
10 March 2006
72.
100 100
−
= −15A, t < 0
20
5
(a)
iL (t ) =
(b)
iL (0+ ) = iL (0− ) = −15A
(c)
iL (∞) =
(d)
iL (t ) = 5 − 20e −40t A, t > 0
100
= 5A
20
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Engineering Circuit Analysis, 7th Edition
73.
Chapter Eight Solutions
10 March 2006
18
1
× = 0.1A ∴ iL (0+ ) = 0.1A
60 + 30 2
i L (∞) = 0.1 + 0.1 = 0.2A
iL (0− ) =
∴ iL (t ) = 0.2 − 0.1e −9000t A, t > 0
∴ iL (t ) = 0.1u (−t ) + (0.2 − 0.1e −9000t ) u (t )A
or, iL (t ) = 0.1 + (0.1 − 0.1e −9000t ) u (t )A
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Engineering Circuit Analysis, 7th Edition
74.
Chapter Eight Solutions
10 March 2006
30 3
× = 3A, iL (0− ) = 4A
7.5 4
(a)
ix (0− ) =
(b)
ix (0+ ) = iL (0+ ) = 4A
(c)
ix (∞) = iL (∞) = 3A
∴ ix (t ) = 3 + 1e−10t / 0.5 = 3 + e−20t A ∴ ix (0.04)
= 3 + e−0.8 = 3.449A
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Engineering Circuit Analysis, 7th Edition
Chapter Eight Solutions
10 March 2006
75.
(a)
(b)
(c)
ix (0− ) = iL (0− ) =
ix (0+ ) =
30
= 3A
10
30
30
15
× + 3×
= 2.4A
30 + 7.5 40
10 + 15
30 30
×
= 3A ∴ ix (t ) = 3 − 0.6e −6t / 0.5
7.5 40
= 3 − 0.6e −12t ∴ ix (0.04) = 3 − 0.6e −0.48 = 2.629A
ix ( ∞ ) =
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Engineering Circuit Analysis, 7th Edition
76.
Chapter Eight Solutions
10 March 2006
OC : vx = 0, voc = 4 u (t )V
vx − 0.2vx vx
+ ,12 u (t ) = 0.6vx + 2vx
40
60
v
12 u (t )
12 u (t ) u (t )
∴ vx =
∴ iab = x =
=
2.6
60 2.6 × 60 13
4 u (t )
u (t )
(1 − e−52t / 0.2 ) u (t ) =
(1 − e−260t ) u (t )
∴ Rth = 4 × 13 = 52Ω∴ iL =
52
13
+
−260 t
) u (t )V
∴ vx = 60iL = 4.615 (1 − e
SC : 0.1u (t ) =
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Engineering Circuit Analysis, 7th Edition
Chapter Eight Solutions
10 March 2006
77.
(a)
OC : − 100 + 30i1 + 20i1 = 0, i1 = 2A
∴ voc = 80 u (t )V
20 × 10
= 20A
20
∴ R th = 4Ω∴ iL (t ) = 20(1 − e −40t ) u (t )A
SC : i1 = 10A, ↓ isc = 10 +
(b)
vL = 0.1× 20 × 40e −40t u (t ) = 80e −40t u (t )
∴ i1 (t ) =
100u (t ) − 80e −40t u (t )
= 10 − 8e −40t u (t )A
10
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Engineering Circuit Analysis, 7th Edition
78.
Chapter Eight Solutions
10 March 2006
τ = Req C = (5)(2) = 10 s
Thus, vn (t ) = Ae−0.2t .
v = vn + v f = Ae −0.1t + Be −5t .
At t = 0, v(0) = 0 since no source exists prior to t = 0. Thus, A + B = 0 [1].
As t → ∞ , v(∞) → 0 . We need another equation.
i (0) = C
dv 4.7
=
( 2)
dt
5
therefore
dv
4.7
4.7
or − 0.1A − 5 B =
=
[2]
dt t =0+
5
5
Solving our two equations, we find that A = –B = 0.192.
Thus, v(t ) = 0.192 ( e−0.1t − e−5t )
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Engineering Circuit Analysis, 7th Edition
79.
Chapter Eight Solutions
10 March 2006
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
9.4 cos 4t = 5i + v
= (5)(2)
dv
+v
dt
dv
+ 0.1v = 0.94 cos 4t , so that v(t ) = e−0.1t ∫ ( 0.94 cos 4t ) e0.1t dt + Ae−0.1t
dt
Performing the integration, we find that
or
⎡10 cos 4t + 400sin 4t ⎤
v(t ) = 0.94 ⎢
+ Ae−0.1t .
⎥
1 + 1600
⎣
⎦
At t = 0, v = 0, so that A = −
and v(t ) =
0.94
(10 )
1601
0.94
⎡ −10e−0.1t + 10 cos 4t + 400sin 4t ⎤⎦
1601 ⎣
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Engineering Circuit Analysis, 7th Edition
80. (a)
Chapter Eight Solutions
10 March 2006
6
vc (0− ) = × 3 = 2V = vc (0+ )
9
6
vc (∞) = 2 − 6 (2 7) = −6V
7
9
3
∴ vc (t ) = −6 + 8e−10 t / 2×10 = −6 + 8e−500000t V, t > 0
vc (−2μ s) = vc (0− ) = 2V, vc (2μ s) = −6 + 8e−1 = −3.057V
(b) PSpice verification.
As can be seen from the plot above, the PSpice simulation results confirm our hand
lations of vC(t < 0) = 2 V and vC(t = 2 μs) = -3.06 V
calcu
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Engineering Circuit Analysis, 7th Edition
81.
Chapter Eight Solutions
10 March 2006
τ = RC = 2 ×10−3 ( 50 ) = 0.1
vn = Ae−10t
vC (t ) = vCn + vCf = Ae −10t + 4.5
Since
since vC ( ∞ ) = 4.5 V
vC (0− ) = vC (0+ ) = 0
vC (t ) = −4.5e−10t + 4.5 = 4.5 (1 − e−10t )
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Engineering Circuit Analysis, 7th Edition
Chapter Eight Solutions
10 March 2006
82. i (0− ) = 10 = 10mA, i (∞) = 2.5mA, v (0) = 0
A
A
c
1
10 1
iA (0+ ) =
× 1.4286mA ∴ iA = 10mA, t < 0
1.75 4
8
3
iA = 2.5 + (1.4286 − 2.5) e−10 t /1.75×10 = 2.5 − 1.0714e−57140t mA, t > 0
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Engineering Circuit Analysis, 7th Edition
Chapter Eight Solutions
10 March 2006
10
= 2.5mA, iA (∞) = 10mA
4
10 7.5
vc (0) = 7.5V ∴ iA (0+ ) = +
= 17.5mA
1
1
−
83. iA (0 ) =
8
3
5
iA = 10 + 7.5e −10 t /10 = 10 + 7.5e−10 t mA, t > 0, iA = 2.5mA t < 0
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Engineering Circuit Analysis, 7th Edition
Chapter Eight Solutions
10 March 2006
84.
(a)
iin (-1.5) = 0
(b)
iin (1.5) = 0
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Engineering Circuit Analysis, 7th Edition
Chapter Eight Solutions
10 March 2006
85.
(a)
vs = −12u (−t ) + 24 u (t )V
t < 0: vc (0− ) = −8V ∴ vc (0+ ) = −8V
2
t > 0 : vc (∞) = × 24 = 16V
3
200
RC =
× 103 × 3 × 10−7 = 2 × 10−3
30
∴ vc (t ) = 16 − 24e −500t V, t > 0
∴ vc (t ) = −8u (−t ) + (16 − 24e −500t ) u (t )
−12
24 + 8
= −0.4mA, iin (0+ ) =
= 3.2mA
30
10
24
= 0.8mA
iin (∞) =
30
iin (t ) = −0.4u (t ) + (0.8 + 2.4e −500t ) u (t )mA
(b) iin (0− ) =
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Engineering Circuit Analysis, 7th Edition
86.
Chapter Eight Solutions
10 March 2006
−vx vx 3 − vx
−
+
= 0 ∴ vx = 1, voc = 3 − 1 = 2V
100 100 100
v
v
SC : vx = 3V ∴ isc = x + x = 0.06A
100 100
∴ R th = voc / isc = 2 / 0.06 = 33.33Ω
OC :
6
∴ vc = voc (1 − e − t / RthC ) = 2(1 − e −10 t / 33.33 )
= 2(1 − e −30,000t ) V, t > 0
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Engineering Circuit Analysis, 7th Edition
87.
Chapter Eight Solutions
10 March 2006
vc (0− ) = 10V = vc (0+ ), iin (0− ) = 0
iin (0+ ) = 0 ∴ iin (t ) = 0 for all t
0 < t < 0.5s : vc = 10(1 − e −2.5t )V
vc (0.4) = 6.321V, vc (0.5) = 7.135V
5 50
8
20 − 10 5
= A ∴ vc (∞) = 10 + 8 + = V, 4 8 = Ω
12
6
6 3
3
50 ⎛
50 ⎞
+ ⎜ 7.135 − ⎟ e −0.375×20( t −0.5) = 16.667 − 9.532e −7.5(t −0.5) V
vc (t ) =
3 ⎝
3 ⎠
∴ vc (0.8) = 16.667 − 9.532e −7.5(0.3) = 15.662V
t > 0.5:
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Engineering Circuit Analysis, 7th Edition
Chapter Eight Solutions
10 March 2006
88.
(a) For
t < 0, there are no active sources, and so vC = 0.
For 0 < t < 1, only the 40-V source is active. Rth = 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 vC(∞) = 40(20)/(20 + 5) = 32 V. Thus, we may write
vC(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 vC(t' = 0+) =
vC(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 vC(∞) = 40(20)/ (20 + 5) +
100(5)/ (20 + 5) = 52 V.
Thus, vC(t') = 52 + [29.37 – 52] e-2.5t' = 52 – 22.63 e-2.5(t – 1) V.
(b)
For t < 0,
vC = 0.
(c)
We see from the simulation results that
our hand calculations and sketch are
indeed correct.
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Engineering Circuit Analysis, 7th Edition
Chapter Eight Solutions
10 March 2006
89.
(a)
t < 0 : 8(10 + 20) = 240V = vR (t ) = 80V, t < 0
(b)
t < 0 : vc (t ) = 8 × 30 = 240V ∴ vc (0+ ) = 240V
1
t = (∞) : vc (∞) = × 8(10 + 10) = 80V
2
−6
∴ vc (t ) = 80 + 160e − t /10×10 = 80 + 160e −100000t V
∴ vR (t ) = 80 + 160e −100000t V, t > 0
(c)
(d)
t < 0 : vR (t ) = 80V
−6
vc (0− ) = 80V, vc (∞) = 240V ∴ vc (t ) = 240 − 160e − t / 50×10 = 240 − 160e −20000t V
20
80
×10 + × 10 = 32 + 16 = 48V
30 + 20
50
−20000 t
V, t > 0
vR (∞) = 80V ∴ vR (t ) = 80 − 32e
vR (0− ) = 80V, vR (0+ ) = 8
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Engineering Circuit Analysis, 7th Edition
90.
Chapter Eight Solutions
10 March 2006
t < 0 : vc = 0
6
0 < t < 1ms : vc = 9 (1 − e −10 t /( R1 +100) )
∴ 8 = 9 (1 − e −1000 /( R1 +100) ),
∴
1
= e−1000 /( R1 +100)
9
1000
= 2.197, R1 = 355.1Ω
R1 + 100
6
t > 1ms : vc = 8e −10 t ′ /( R2 +100) , t ′ = t − 10−3 ∴1 − 8e −1000 (R 2 + 100)
∴
1000
= 2.079, R 2 = 480.9 − 100 = 380.9Ω
R 2 + 100
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Engineering Circuit Analysis, 7th Edition
91.
Chapter Eight Solutions
10 March 2006
vx , L = 200e −2000t V
vx ,c = 100(1 − e −1000t )V
vx = v x , L − vx , c = 0
∴ 200e −2000t = 100 − 100e−1000t
∴100e −1000t + 200(e −1000t ) 2 − 100 = 0,
−100 ± 10, 000 + 80, 000
= −0.25 ± 0.75
400
∴ e −1000t = 0.5, t = 0.6931ms
e −1000t =
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Engineering Circuit Analysis, 7th Edition
Chapter Eight Solutions
10 March 2006
92.
P(t < 0 ) = I 2 R = 0.0012 × 103 = 0.001 W
Vinit = I .R = 7 × 10 − 3 × 900 = 6.3 V
Pinit =
V2
= 0.08 W
R
V final = 7 × 10 − 3 × 900Ω // 1000Ω = 3.3 V
Pfinal =
V2
= 0.02 W
R
Power (W)
8
6
4
2
0
7
Time (ms)
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Engineering Circuit Analysis, 7th Edition
93. For
Chapter Eight Solutions
10 March 2006
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 began to charge according to ½ Cv2. The capacitors charge with the same
current flowing through each, so that by KCL we may write
C1
dv1
dv
dv
= C2 2 = C3 3
dt
dt
dt
With no initial energy stored, integration yields the relationship C1v1 = C2v2 = C3v3
throughout the charging (i.e. until 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:
v1
+ v2
+ v3 = 7.421
=0
10-6 v1 – 2×10-6 v2
-6
-6
2×10 v2 – 3×10 v3 = 0
so
that v2 = 2.024 V.
With the initial voltage across the 2-uF capacitor now known, we may write
-3
v(t) = 2.024 e-t/ 2.564×10 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 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 Cv2 = 0.5(0.5455×10-6)(7.421)2 = 15.02 μJ.
-3
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Engineering Circuit Analysis, 7th Edition
94. voltage
Chapter Eight Solutions
10 March 2006
follower ∴ vo (t ) = v2 (t )
v2 (t ) = 1.25 u (t )V = vo (t )
6
vx (t ) = 1.25e −10 / 0.5×200u (t )
= 1.25e −10,000t u (t )V
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Engineering Circuit Analysis, 7th Edition
95.
Chapter Eight Solutions
10 March 2006
This is a voltage follower ∴ vo (t ) = v2 (t ) , where v2(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: vC(0-) = 0 ∴ vC(0+) = 0.
Applying KVL at t = 0+,
5
v250/250 + v=
2/1000.
at t = 0+ v250 = v2, we find that v2(0+) = 1 V.
t → ∞, v2→ 0, so we may write,
Since
As
vo(t) = v2(t) = 1.0e-t/10 u(t) V.
PSpice
verification:
In plotting both the hand-derived result and the PSpice 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, 7th Edition
96. For
Chapter Eight Solutions
10 March 2006
t < 0, the current source is an open circuit and so i1 = 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 calculation is indeed correct, and find
that the inductor current at t = 50 ms is 7.82 A.
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Engineering Circuit Analysis, 7th Edition
97. (a)
v1 = 0 (virtual gnd) ∴ i =
Chapter Eight Solutions
10 March 2006
4 −20,000t
e
u (t )A
104
4 −20,000t
e
dt = −0.2e −20,000t to
o 10 4
∴ vc (t ) = 0.2(1 − e −20,000t ) u (t )
∴ vc = 107 ∫
t
∴ vR (t ) = 103 i (t ) = 0.4e −20,000t u (t )V
∴ vo (t ) = −vc (t ) − vR (t ) = (−0.2 + 0.2e −20,000t − 0.4e −20,000t ) u (t )
3
And we may write vo(t) = -0.2[1 + e-20×10 t]u(t) V.
(b) PSpice verification:
We can see from the simulation result that our ideal op amp approximation 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, 7th Edition
98.
Chapter Eight Solutions
10 March 2006
One possible solution of many: implement a capacitor to retain charge; assuming the
light is left on long enough to fully charge th e 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 res istance of the light bu lb (which chan ges with its tem perature) as
330.6 Ω. We define “on” for the light bulb som ewhat arbitrarily as 50% intensity, taking
intensity as proportional to the dissipated pow er. 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-M Ω resistor 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 M Ω || 330.6 Ω = 330.5 Ω. We want vC(t = 5s) = 81.33 = 115 e-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, 7th Edition
99.
Chapter Eight Solutions
10 March 2006
Assume at least 1 μA required otherwise alarm triggers.
Add capacitor C.
vc (1) = 1 volt
vc (0) =
1000
.1.5 = 1.496 volts
1002.37
∴ We have 1 = 1.496e
−
1
106 C
or C =
1
= 2.48μF
10 ln(1.496)
6
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Engineering Circuit Analysis, 7th Edition
100.
Chapter Eight Solutions
10 March 2006
(a) Note that negativ e tim es are not perm itted in PSpice. The only way to m
situation is to shift the time axis by a fixed amount, e.g., t ′ = t + 1 .
odel this
(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., t ′ = t + 2 .
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Engineering Circuit Analysis, 7th Edition
101. (a)
Chapter Eight Solutions
10 March 2006
τ = L / R = 0.1 s . This is much less than either the period or pulsewidth.
(b)
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Engineering Circuit Analysis, 7th Edition
102. (a)
Chapter Eight Solutions
10 March 2006
τ = L / R = 1 s . This is much less than either the period or pulsewidth.
(b)
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Chapter Eight Solutions
10 March 2006
103.
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Engineering Circuit Analysis, 7th Edition
104.
τ=
Chapter Eight Solutions
10 March 2006
L 500 ×10−6
=
= 34 ns
Req 14.7 ×103
6
The transient response will therefore have the form Ae −29.4×10 t .
(a)
(b)
(c)
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Engineering Circuit Analysis, 7th Edition
1.
Chapter Nine Solutions
10 March 2006
Parallel RLC circuit:
1
1
1
=
=
= 175 × 103 s −1
−6
2 RC ( 2 ) (4 ||10)(10 ) ( 2 ) (2.857)(10−6 )
(a)
α=
(b)
ω0 =
1
=
LC
1
( 2 ×10 )(10 )
−3
−6
=
22.4 krad/s
(c) The circuit is overdamped since α > ω0 .
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Engineering Circuit Analysis, 7th Edition
2.
Chapter Nine Solutions
10 March 2006
Parallel RLC circuit:
(a) For an underdamped response, we require α < ω0 , so that
1
<
2 RC
1
LC
or R >
1 L
1
2
; R>
.
2 C
2 10−12
Thus, R > 707 kΩ.
(b) For critical damping,
1 L
R=
= 707 kΩ
2 C
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Engineering Circuit Analysis, 7th Edition
3.
(a)
10 March 2006
Parallel RLC circuit:
α=
ω0 =
(b)
Chapter Nine Solutions
1
1
1
=
=
=
−6
2 RC ( 2 ) (4 ||10)(10 ) ( 2 ) (1)(10−9 )
5 × 108
1
=
LC
31.6 Trad/s
1
(10 )(10 )
−12
−9
= 3.16 × 1013 rad/s =
−1
s
s1,2 = −α ± α 2 − ω02 = −0.5 × 109 ± j 1021 − (0.25)(1018 ) = −0.5 ± j 31.62 Grad/s
(c) The circuit is underdamped since α < ω0 .
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Engineering Circuit Analysis, 7th Edition
4.
Chapter Nine Solutions
10 March 2006
Parallel RLC circuit:
(a) For an underdamped response, we require α < ω0 , so that
1
<
2 RC
1
LC
or R >
1 L
1
10−15
.
; R>
2 C
2 2 ×10−18
Thus, R > 11.18 Ω.
(b) For critical damping,
1 L
R=
= 11.18 Ω
2 C
(c) For overdamped, R < 11.18 Ω.
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Engineering Circuit Analysis, 7th Edition
5.
Chapter Nine Solutions
10 March 2006
ω o L = 10Ω, s1 = −6s −1 , s2 = −8s −1
∴−6 = α + α 2 − ω o2 , − 8 = −α − α 2 − ω o2 adding,
−14 = −2α ∴α = 7 s −1
∴−6 = −7 + 49 − ω o2 ∴ω o2 = 48
1
, ω o = 6.928
LC
rad/s∴ 6.928 L = 10, L = 1.4434H,
1
1
C=
= 14.434mF,
= 7 ∴ R = 4.949Ω
48L
2RC
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Engineering Circuit Analysis, 7th Edition
6.
(a)
Chapter Nine Solutions
10 March 2006
ic = 40e −100t − 30e −200t mA, C = 1mF, v(0) = −0.25V
t
1 t
0.25
(40e −100t − 30e−200t ) dt − 0.25
i
dt
−
=
c
∫
∫
o
o
C
∴ v(t ) = −0.4(e −100t − 1) + 0.15(e −200t − 1) − 0.25
v(t ) =
∴ v(t ) = −0.4e−100t + 0.15e−200t V
(b)
s1 = −100 = −α + α 2 − ωo2 , s2 = −200 = −α − α 2 − ωo2
∴−300 = −2α, α = 150s − 1
∴150 +
1
500
= 3.333Ω Also,
,R =
−3
2R10
150
−200 = −150 − 22500 − ωo2 ∴ωo2 = 20000
∴ 20000 =
∴ i R (t ) =
(c)
1 100
=
, L = 0.5H
LC
L
v
= 0.12e −100t + 0.045e −200t A
R
(i)t = −iR (t ) − ic (t ) = (0.12 − 0.04)e −100t + (−0.045 + 0.03)e−200t
∴ i (t ) = 80e−100t − 15e−200t mA, t > 0
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Engineering Circuit Analysis, 7th Edition
7.
(a)
(b)
Chapter Nine Solutions
10 March 2006
Parallel RLC with ωo = 70.71 × 1012 rad/s. L = 2 pH.
1
= (70.71× 1012 ) 2
LC
1
So 100.0
C=
=
12 2
(70.71×10 ) (2 ×10−12 )
ωo2 =
aF
1
= 5 × 109 s −1
2 RC
1
So 1R = 10
= MΩ
(10 ) (100 ×10−18 )
α=
(c)
α is the neper frequency: 5 Gs-1
(d)
S1 = −α + α 2 − ωo2 = −5 × 109 + j 70.71× 1012 s −1
S 2 = −α − α 2 − ωo2 = −5 × 109 − j 70.71× 1012 s −1
(e)
ζ=
α
5 ×109
=
= 7.071× 10−5
12
ωo 70.71×10
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Engineering Circuit Analysis, 7th Edition
8. Given:
L = 4 R 2C , α =
Chapter Nine Solutions
10 March 2006
1
2 RC
Show that v(t ) = e −αt ( A1t + A2 ) is a solution to
C
d 2 v 1 dv 1
+
+ v=0
dt 2 R dt L
[1]
dv
= e −αt ( A1 ) − αe −αt ( A1t + A2 )
dt
= ( A1 − α A1t − α A2 ) e −αt
[2]
2
d v
= ( A1 − α A1t − αA2 ) (−αe −αt ) − α A1e −αt
2
dt
= −α ( A1 − α A2 + A1 − α A1t ) e −αt
= −α (2 A1 − α A2 − α A1t )e −αt
[3]
Substituting Eqs. [2] and [3] into Eq. [1], and using the information initially provided,
2
1
1
⎛ 1 ⎞
−αt
(2 A1 ) e −αt + ⎜
( A1 ) e −αt
⎟ ( A1t + A2 ) e +
RC
2 RC
⎝ 2 RC ⎠
1
1
( A1t + A2 ) e −αt + 2 2 ( A1t + A2 ) e−αt
−
2 RC
4R C
=0
−
Thus, v(t ) = e −αt ( A1t + A2 ) is in fact a solution to the differential equation.
Next, with v(0) = A2 = 16
dv
= ( A1 − α A2 ) = ( A1 − 16α ) = 4
and
dt t =0
we find that A1 = 4 + 16α
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Engineering Circuit Analysis, 7th Edition
9.
Chapter Nine Solutions
10 March 2006
Parallel RLC with ωo = 800 rad/s, and α = 1000 s-1 when R = 100 Ω.
1
2 RC
1
ωo2 =
LC
α=
so C = 5μF
so
L = 312.5 mH
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
⎛ 100 cm ⎞ ⎛ 1in ⎞ ⎛ 1ft ⎞⎛ 6.39 Ω ⎞
(5 m) ⎜
⎟⎜
⎟⎜
⎟⎜
⎟
⎝ 1m ⎠ ⎝ 2.54 cm ⎠ ⎝ 12in ⎠⎝ 1000 ft ⎠
= 0.1048 Ω or 104.8 mΩ
(a)
The resonant frequency is unchanged, so ωo = 800 rad/s
(b)
α=
(c)
ζ old =
α old
ωo
ζ new =
α new
ωo
1
= 954.0 × 103 s −1
2 RC
Define the percent change as
=
ζ new − ζ old
× 100
ζ old
α new − α old
× 100
α old
= 95300%
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Engineering Circuit Analysis, 7th Edition
Chapter Nine Solutions
10.
L = 5H, R = 8Ω, C = 12.5mF, v(0+ ) = 40V
(a)
i (0+ ) = 8A: α =
10 March 2006
1
1000
1
=
= 5, ωo2 =
= 16,
2RC 2 × 8 × 12.5
LC
ωo = 4 s1,2 = −5 ± 25 − 16 = −2, − 8 ∴ v(t ) = A1e −2t + A 2 e −8t
1000 ⎛
40 ⎞
+
⎜ −iL (0 ) − ⎟ = 80 (−8 − 5) = −1040
12.5 ⎝
8 ⎠
v / s = −2A1 − 8A 2 ∴−520 = − A1 − 4A 2 ∴−3A 2 = −480, A 2 = 160, A1 = −120
∴ 40 = A1 + A 2 v′(0+ ) =
∴ v(t ) = −120e −2t + 160e −8t V, t > 0
(b)
v(0+ ) 40
=
= 5A
R
8
∴ i (0+ ) = A 3 + A 4 = −iR (0+ ) − ic (0+ ) = −8 − 5 = −13A;
ic (0+ ) = 8A Let i( t ) = A 3e −2t + A 4 e −8t ; iR (0+ ) =
40
= 8 A / s ∴ 4 = − A 3 − 4A 4
5
∴−3A 4 = −13 + 4, A 4 = 3, A 3 = −16 ∴ i (t ) = −16e −2t + 3e −8t A, t > 0
i (0+ ) = −2A 3 − 8A 4 =
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Engineering Circuit Analysis, 7th Edition
11. (a)
RC =
Therefore
Chapter Nine Solutions
10 March 2006
1 L 1 10−3 1
10 = 1.581 Ω
=
=
2 C 2 10−4 2
R
= 0.1RC = 158.1 mΩ
(b)
α=
1
= 3.162 ×104 s −1 and ω0 =
2RC
1
= 3.162 × 103 rad/s
LC
s1,2 = −α ± α 2 − ω02 = − 158.5 s −1 and − 6.31× 104 s −1
Thus,
4
So we may write i (t ) = A1e−158.5t + A2 e−6.31×10 t
With i(0− ) = i(0+ ) = 4 A and v(0− ) = v(0+ ) = 10 V
A1+ A2 = 4
Noting
v(0+ ) = L
(
di
= 10
dt t =0
[1]
)
10−3 −158.5 A1 − 6.31× 104 A2 = 10 [2]
Solving Eqs. [1] and [2] yields A1 = 4.169 A and A2 = –0.169 A
So that
4
i (t ) = 4.169e −158.5t − 0.169e −6.31×10 t A
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Engineering Circuit Analysis, 7th Edition
12. (a)
α=
1
= 500 s −1 and ω0 =
2RC
Chapter Nine Solutions
10 March 2006
1
= 100 rad/s
LC
s1,2 = −α ± α 2 − ω02 = − 10.10 s −1 and − 989.9 s −1
Thus,
So we may write iR (t ) = A1e−10.1t + A2 e −989.9t [1]
With i(0− ) = i (0+ ) = 2 mA and v(0− ) = v(0+ ) = 0
A1+ A2 = 0
We need to find
iC (0+ ) = C
Thus,
Therefore,
[2]
di ( t ) 1 dv
diR
dv
[3] and iC = C
. Note that R
= −i − iR .
=
dt t =0
dt
dt
R dt
dv
v(0+ )
= −i (0+ ) − iR (0+ ) = −2 × 10−3 −
= −2 × 10−3 [4]
dt t =0+
R
we may write based on Eqs. [3] and [4]:
diR
= (50)(−0.04) = −2 [5]. Taking the derivative of Eq. [1] and combining with
dt t =0
[6].
Eq. [5] then yields: s1 A1 + s 2 A2 = −2
Solving Eqs. [2] and [6] yields A1 = −2.04 mA and A2 = 2.04 mA
So that
(b)
(
iR (t ) = −2.04 e −10.1t − e−989.9t
(c)
) mA
We see that the simulation agrees.
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Engineering Circuit Analysis, 7th Edition
13.
i (0) = 40A, v(0) = 40V, L =
(a)
α=
Chapter Nine Solutions
10 March 2006
1
H, R = 0.1Ω, C = 0.2F
80
1
80
= 25, ωo2 =
= 400,
2 × 0.1× 0.2
0.2
ωo = 20, s1,2 = −25 ± 625 − 400 = 10, − 40
∴ v(t ) = A1 e −10t + A 2 e−40t ∴ 40 = A1 + A 2 ;
1⎛
v(0) ⎞
⎜ i (0) −
⎟ = −2200
C⎝
R ⎠
∴−A1 − 4A 2 = −220 ∴ − 3A 2 = −180 ∴ A 2 = 60, A1 = −20
v′(0+ ) = −10A1 − 40A 2 v′(0+ ) =
∴ v(t ) = −20e −10t + 60e −40t V, t > 0
(b)
dv
= 200e −10t − 600e −40t − 0.2(-20)(-10)e -10t − (0.2)(60)(-40)e −40t
dt
−10 t
= 160e
− 120e −40t A
i(t) = – v/ R – C
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Engineering Circuit Analysis, 7th Edition
α=
14. (a)
Chapter Nine Solutions
1
= 6.667 × 108 s −1 and ω0 =
2RC
10 March 2006
1
= 105 rad/s
LC
s1,2 = −α ± α 2 − ω02 = − 7.5 s −1 and − 1.333 × 109 s −1 . So we may write
Thus,
9
iC (t ) = A1e −7.5t + A2 e−1.333×10 t [1] With i (0− ) = i (0+ ) = 0 A and v(0− ) = v(0+ ) = 2 V ,
2
iC (0+ ) = −iR (0+ ) = −
= −0.133 × 106 so that
−6
15 × 10
A1+ A2 = –0.133×106 [2]
We need to find
C
diR
di
di
2
. We know that L
= 2 so
=
= 106 . Also,
dt t =0
dt t =0 2 ×10−6
dt t =0
(
)
9
i
di
di
1 dv
1
dv
= iC and R =
so R = C =
A1e−7.5t + A2 e−1.333×10 t .
dt
dt R dt
dt CR CR
di
di diR diC
1
( A1 + A2 ) [3]
+
+
= 0 so C
= −7.5 A1 − 1.33 × 109 A2 = −106 −
dt dt
dt
dt t =0
CR
Using
Solving Eqs. [2] and [3] yields A1 = −0.75 mA and A2 = –0.133 MA (very different!)
So that
(
9
iC (t ) = − 0.75 ×10−3 e −7.5t + 0.133 × 106 e−1.333×10 t
) A
(b)
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Engineering Circuit Analysis, 7th Edition
15. (a)
Thus,
α=
1
= 0.125 s −1 and ω0 =
2RC
Chapter Nine Solutions
10 March 2006
1
= 0.112 rad/s
LC
s1,2 = −α ± α 2 − ω02 = − 0.069 s −1 and − 0.181 s −1 . So we may write
v(t ) = A1e−0.069t + A2 e −0.181t [1]
With iC (0− ) = iC (0+ ) = −8 A and v(0− ) = v(0+ ) = 0 ,
A1+ A2 = 0
[2]
We need to find
diR
. We know that
dt t =0
dv
= 4 ⎡⎣ −0.069 A1e −0.069t − 0.181A2 e−0.181t ⎤⎦ . So,
dt
[3]
iC (0) = 4 [ −0.069 A1 − 0.181A2 ] = −8
iC (t ) = C
Solving Eqs. [2] and [3] yields A1 = −17.89 V and A2 = 17.89 V
So that
(b)
v(t ) = −17.89 ⎣⎡ e −0.069t − e−0.181t ⎦⎤ V
dv
= 1.236e −0.069t − 8.236e−0.181t . We set this equal to 0 and solve for tm:
dt
3.236 e −0.069tm
=
= e0.112tm , so that tm = 8.61 s.
1.236 e −0.181tm
Substituting into our expression for the voltage, the peak value is
v(8.61) = –6.1 V
(c) The simulation agrees with the analytic results.
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Engineering Circuit Analysis, 7th Edition
16.
Chapter Nine Solutions
10 March 2006
100
= 2A, vc (0) = 100V
50
106
3 × 106+3
α=
= 4000, wo2 =
= 12 × 106
2 × 50 × 2.5
100 × 2.5
3
16 − 12 × 10 = 200, s1,2 = −4000 ± 2000
iL (0) =
∴ iL (t ) = A1e −2000t + A 2 e −6000t , t > 0 ∴ A1 + A s = 2
−103 × 3
× 100 = −3000 = −2000A1 − 6000A 2 ∴−1.5 = − A1 − 3A 2 ∴ 0.5 = −2A 2
100
∴ A 2 = −0.25, A1 = 2.25 ∴ iL (t ) = 2.25e −2000t − 0.25e−6000t A, t > 0
iL′ (0+ ) =
t > 0: iL (t ) = 2A ∴ iL (t ) = 2u (−t ) + (2.25e−2000t − 0.25e−6000t ) u (t )A
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Engineering Circuit Analysis, 7th Edition
17.
Chapter Nine Solutions
10 March 2006
12
= 2A, vc (0) = 2V
5 +1
1000
1000 × 45
= 250, ω o2 =
= 22500
α=
2 × 1× 2
2
iL (0) =
s1,2 = −250 ± 2502 − 22500 = −50, − 450 s −1
∴ iL = A1e −50t + A 2 e −450t ∴ A1 + A 2 = 2; iL′ (0+ ) = 45(−2) = −50A1 − 450A 2
∴ A1 + 9A 2 = 1.8 ∴−8A 2 = 0.2 ∴ A 2 = −0.025, A1 = 2.025(A)
∴ iL (t ) = 2.025e −50t − 0.025e −450t A, t > 0
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Engineering Circuit Analysis, 7th Edition
Chapter Nine Solutions
10 March 2006
18.
(a)
1
1440
1440
=
= 20, ω o2 =
= 144
2RC
72
10
s1,2 = −20 ± 400 − 144 = −4, − 36: v = A1e −4t + A 2 e −36t
α=
⎛ 1 18 ⎞
v(0) = 18 = A1 + A2 , v′(0) = 1440 ⎜ − ⎟ = 0
⎝ 2 36 ⎠
∴ 0 = −4A1 − 36A 2 = − A1 − 9A 2 = ∴ 18 = −8A 2 , A 2 = −2.25, A1 = 20.25
+
∴ v(t ) = 20.25e −4 − 2.25e −36t V, t > 0
(b)
v
1
v′ = 0.5625e−4t − 0.0625e−36t − 0.05625e−4t + 0.05625e−36t
+
36 1440
∴ i (t ) = 0.50625e−4t − 0.00625e −36t A, t > 0
(c)
vmax at t = 0 ∴ vmax = 18V ∴ 0.18 = 20.25e−4ts − 2.25e −36ts
i (t ) =
Solving using a scientific calculator, we find that ts = 1.181 s.
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Engineering Circuit Analysis, 7th Edition
19.
so
Referring to Fig. 9.43,
ωo = 1
α=
LC
= 4 rad/s
Chapter Nine Solutions
10 March 2006
L = 1250 mH
Since α > ωo, this circuit is over damped.
1
= 5 s −1
2 RC
The capacitor stores 390 J at t = 0−:
1
Wc = C vc2
2
2Wc
= 125 V = vc (0+ )
So (vc 01 ) =
C
The inductor initially stores zero energy,
so
iL (0− ) = iL (0+ ) = 0
S1,2 = −α ± α 2 − ωo2 = −5 ± 3 = −8, − 2
Thus, v(t ) = Ae −8t + Be−2t
Using the initial conditions, v (0) = 125 = A + B
v(0+ )
+ ic (0+ ) = 0
iL (0+ ) + iR (0+ ) + ic (0+ ) = 0 +
2
125
v(0+ )
+
=−
= −62.5 V
So (ic 0 ) = −
2
2
dv
= 50 ×10−3[−8 Ae −8t − 2 Be −2t ]
ic = C
dt
+
[2]
ic (0 ) = −62.5 = −50 × 10−3 (8 A + 2 B)
Solving Eqs. [1] and [2],
[1]
A = 150 V
B = −25 V
Thus, v(t ) = 166.7e −8t − 41.67e −2t , t > 0
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Engineering Circuit Analysis, 7th Edition
20.
Chapter Nine Solutions
10 March 2006
We want a response v = Ae −4t + Be−6t
1
α=
= 5 s −1
2 RC
(a)
S1 = −α + α 2 − ωo2 = −4 = −5 + 25 − ωo2
S2 = −α − α 2 − ωo2 = −6 = −5 − 25 − ωo2
Solving either equation, we obtain ωo = 4.899 rad/s
Since
(b)
with
ωo2 =
1
1
, L = 2 = 833.3 mH
LC
ωoC
If iR (0+ ) = 10 A and ic (0+ ) = 15 A, find A and B.
iR (0+ ) = 10 A, vR (0+ ) = v(0+ ) = vc (0+ ) = 20 V
v(0) = A + B = 20
[1]
dv
ic = C
= 50 × 10−3 (−4 Ae −4t − 6 Be−6t )
dt
+
ic (0 ) = 50 × 10−3 (−4 A − 6 B) = 15
[2]
Solving, A = 210 V, B = −190 V
v=
Thus, 210
e −4t − 190e −6t , t > 0
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Engineering Circuit Analysis, 7th Edition
21. Initial
conditions:
iL (0− ) = iL (0+ ) = 0
(a)
vc (0+ ) = vc (0− ) = 2(25) = 50 V
(b)
ic (0+ ) = −iL (0+ ) − iR (0+ ) = 0 − 2 = −2 A
(c)
t > 0: parallel (source-free) RLC circuit
1
α=
= 4000 s −1
2 RC
1
ωo =
= 3464 rad/s
LC
Since
Chapter Nine Solutions
iR (0+ ) =
10 March 2006
50
=2 A
25
s1,2 = −α ± α 2 − ωo2
= −2000, − 6000
α > ω0, this system is overdamped. Thus,
vc (t ) = Ae −2000t + Be −6000t
dv
= (5 × 10−6 ) (−2000 Ae −2000t − 6000 Be −6000t )
dt
[1]
ic (0+ ) = −0.01A − 0.03B = −2
ic = C
and v( c 0+ ) = A + B = 50
[2]
Solving, we find A = −25 and B = 75
so that vc (t ) = −25e −2000t + 75e−6000t , t > 0
(d)
(e)
−25e −2000t + 75e −6000t = 0 ⇒ t = 274.7 μs
using a scientific calculator
(f)
vc max = −25 + 75 = 50 V
So,
we
solving | −25e −2000ts + 75e −6000ts | = 0.5 in view of the graph in part (d),
find ts = 1.955 ms using a scientific calculator’s equation solver routine.
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Engineering Circuit Analysis, 7th Edition
22.
Chapter Nine Solutions
10 March 2006
Due to the presence of the inductor, vc (0− ) = 0 . Performing mesh analysis,
→i1
→i2
4.444 H
Rearranging, we obtain 2i1 – 2i2 = 0 and
−9 + 2i1 − 2i2 = 0
[1]
2i2 − 2i1 + 3iA + 7i2 = 0
[2]
and i1 − i2 = iA
–4i1 + 6 i2 = 0. Solving, i1 = 13.5 A and i2 = 9 A.
(a)
iA (0− ) = i1 − i2 = 4.5 A and iL (0− ) = i2 = 9 A
(b)
t > 0:
around left mesh:
−vc (0+ ) + 7iA (0+ ) − 3iA (0+ ) + 2iA (0+ ) = 0
4.444 H
so, iA (0+ ) = 0
(c)
vc (0− ) = 0 due to the presence of the inductor.
(d)
−vLC + 7 − 3(1) + 2 = 0
1A
(e)
vLC = 6 V ∴ RTH =
1
= 3.333 s −1
2 RC
1
ωo =
= 3 rad/s
LC
6
= 6Ω
1
α=
S1,2 = −α ± α 2 − ωo2 = −1.881, − 4.785
Thus,
iA (t ) = Ae −1.881t + Be−4.785t
iA (0+ ) = 0 = A + B
[1]
To find the second equation required to determine the coefficients, we write:
iL = −ic − iR
= −C
dvc
− iA = −25 ×10−3 ⎡ −1.881(6 A)e−1.881t − 4.785(6 B)e−4.785t ⎤
⎣
⎦
dt
−1.881t
−4.785t
− Be
- Ae
iL (0+ ) = 9 = −25 × 10−3[−1.881(6 A) − 4.785(6 B)] − A − B
or
9 = -0.7178A – 0.2822B
[2]
So
Solving Eqs. [1] and [2], A = −20.66 and B = +20.66
that iA (t ) = 20.66[e −4.785t − e−1.881t ]
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Engineering Circuit Analysis, 7th Edition
23.
Chapter Nine Solutions
Diameter of a dime: approximately 8 mm.
Capacitance
L = 4μH
ωo =
10 March 2006
Area = π r 2 = 0.5027cm 2
ε r εo A (88) (8.854 ×10−14 F/cm) (0.5027cm 2 )
=
=
d
0.1cm
= 39.17pF
1
= 79.89 Mrad/s
LC
For an over damped response, we require α > ωo.
Thus,
1
> 79.89 × 106
2 RC
1
R<
−12
2(39.17 × 10 ) (79.89 × 106 )
or R < 159.8 Ω
*Note: The final answer depends quite strongly on the choice of εr.
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Engineering Circuit Analysis, 7th Edition
24.
(a) For critical damping, R =
α=
(b)
Chapter Nine Solutions
1 L 1
10−3
=
= 4.564 Ω .
2 C 2 12 × 10−6
1
1
=
= 9.129 ×103 s −1
−6
2 RC 2 ( 4.564 ) 12 × 10
(
)
vC ( t ) = e −9.129×10 t ( A1t + A2 )
3
Thus,
t = 0, vC (0) = A1 ( 0 ) + A2 = 12
At
10 March 2006
[1]
∴ A2 = 12 V .
Taking the derivative of Eq. [1],
dvC ( t )
dt
and
= e −9.129×10 t ⎡⎣ −9.129 × 103 A1t + A1 − 9.129 × 103 (12 ) ⎤⎦
3
also iC = −(iR + iL ) , so
Solving,
dvC
1 ⎛ v (0)
1
⎞
⎛ 12
⎞
=− ⎜ C
+ 0⎟ = −
+ 0 ⎟ A1 − 9.129 × 103 (12 )
−6 ⎜
dt t =0
C⎝ R
12 × 10 ⎝ 4.565
⎠
⎠
A1 = −109.6 × 103 V , so we may write
(
)
vC ( t ) = e −9.129×10 t −109.6 ×103 t + 12 .
3
(c) We see from plotting both the analytic result in Probe and the simulated voltage, the
two are in excellent agreement (the curves lie on top of one another).
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Engineering Circuit Analysis, 7th Edition
25.
(b)
(a) For critical damping, R =
α=
10 March 2006
1 L 1 10−8
=
= 1.581 mΩ .
2 C 2 10−3
1
1
=
= 3.162 ×105 s −1
−3
−3
2 RC 2 1.581× 10
10
(
)(
)
iL ( t ) = e−3.162×10 t ( A1t + A2 )
5
Thus,
At
Chapter Nine Solutions
[1]
t = 0, iL (0) = A1 ( 0 ) + A2 = 10
∴ A2 = 10 A .
Taking the derivative of Eq. [1],
diL ( t )
dt
and
= e −3.162×10 t ⎡⎣ −3.162 × 105 A1t + A1 − 3.162 × 105 (10 ) ⎤⎦ [2]
also L
5
diL
= vC (0) = 0 [3], so
dt t =0
Solving Eqs. [2] and [3],
A1 = 3.162 × 105 (10 ) = 3.162 × 106 V , so we may write
(
(
)
)
iL ( t ) = e −3.162×10 t 3.162 ×106 t + 10 .
5
(c) We see from plotting both the analytic result in Probe and the simulated voltage, the
two are in reasonable agreement (some numerical error is evident).
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Engineering Circuit Analysis, 7th Edition
26.
Chapter Nine Solutions
10 March 2006
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, 7th Edition
27.
crit. damp.
(b)
Chapter Nine Solutions
10 March 2006
L
(a)
L = 4R 2 C = 4 × 1× 2 × 10−3 = 8mH
1
1000
=
= 250 ∴ iL = e −250t (A1t + A 2 )
2RC 2 ×1× 2
iL (0) = 2A, vc (0) = 2V ∴ iL = e−250t (A1t + 2)
α = ωo
Then 8 × 10−3 iL′ (0+ ) = −2 = 8 × 10−3 (A1 − 500), = e −1.25 (1.25 + 2) = 0.9311A
(c)
iL max : (250tm + 2) = 0, 1 = 250tm + 2, tm < 0 No!
∴ tm = 0, iL max = 2A ∴ 0.02 = e −250ts (250ts + 2); SOLVE: ts = 23.96ms
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Engineering Circuit Analysis, 7th Edition
Chapter Nine Solutions
10 March 2006
28.
R
crit. damp.
(b)
(a)
L = 4R 2 C =
100
× 10−3 = 4R 2 ×10−6 ∴ R = 57.74Ω
3
1
× 2.5 = 3464 s −1
30
∴ vc (t ) = e −3464t (A1t + A 2 ) vc (0) = 100V
ω o = α = 103 /
100
= 1.7321A ∴100 = A 2
57.74
106 ⎛
100 ⎞
5
vc′ (0+ ) =
⎜ 1.7321 −
⎟ = 0 = A1 − 3464A 2 ∴ A1 = 3.464 × 10
2.5 ⎝
57.74 ⎠
t
3464
∴ vc (t ) = e − (3.464 × 105 t + 100) V, t > 0
iL (0) =
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Engineering Circuit Analysis, 7th Edition
29.
Chapter Nine Solutions
10 March 2006
Diameter of a dime is approximately 8 mm. The area, therefore, is πr2 = 0.5027 cm2.
ε r εo A (88) (8.854 × 10−14 ) (0.5027)
=
The capacitance is d
0.1
= 39.17 pF
w
ith L = 4μH, ωo =
1
= 79.89 Mrad/s
LC
For critical damping, we require
or R =
1
= 159.8 Ω
2ωoC
1
= ωo
2 RC
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Engineering Circuit Analysis, 7th Edition
Chapter Nine Solutions
30.
L = 5mH, C = 10−8 F, crit. damp. v(0) = −400V, i (0) = 0.1A
(a)
L = 4R 2 C = 5 × 10−3 = 4R 2 10−8 ∴ R = 353.6Ω
(b)
10 March 2006
108
α=
= 141, 420 ∴ i = e −141,420t (A1t + A 2 )
2 × 353.6
∴ A 2 = 0.1∴= e −141,421t (A1t + 0.1), 5 × 10−3
(A1 − 141, 420 × 0.1) = −400 ∴ A1 = −65,860
∴ i = e −141,421t (−65,860t + 0.1). i′ = 0
∴ e −α t (+65860) + 141, 420e −α t (−65,860tm + 0.1) = 0
∴ tm = 8.590 μ s ∴ i (tm ) = e −141,420×8.590×10
−6
(−65,860 × 8.590 ×10−6 + 0.1) = −0.13821A
∴ i max = i (tm ) = 0.13821A
(c)
∴imax = i (0) = 0.1A
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Engineering Circuit Analysis, 7th Edition
31.
Chapter Nine Solutions
10 March 2006
Critically damped parallel RLC with α = 10−3 s −1 , R = 1MΩ .
1
103
= 10−3 , so 500
C=
=
2 RC
2 ×106
1
Since α = ωo, ωo =
= 10−3
LC
1
or 10 = −6
LC
so L = 2 GH (!)
We know
μF
μN 2 A
= 2 × 109
L=
S
2
⎡⎛ 50 turns ⎞ ⎤
⎛ 1m ⎞
2
(4π× 10 H/m) ⎢⎜
⎟
⎟ . s ⎥ (0.5cm) .π . ⎜
⎝ 100 cm ⎠
⎣⎝ cm ⎠ ⎦
If So
s
= 2 ×109
−7
(4π2 ×10−9 ) (50) 2 (0.5) 2 s = 2 × 109
So 8.106
s=
×1013 cm
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Engineering Circuit Analysis, 7th Edition
Chapter Nine Solutions
10 March 2006
32.
1
4
1
4 × 13
=
= 1, ω o2 =
=
= 26, ω d = 26 − 1 = 5
2RC 2 × 2
LC
2
∴ vc (t ) = e − t ( B1 cos 5t + B2 sin 5t )
α=
(a)
iL (0+ ) = iL (0) = 4A
(b)
vc (0+ ) = vc (0) = 0
(c)
iL′ (0+ ) =
1
vc (0+ ) = 0
L
(d)
vc′ (0+ ) =
⎡
vc (0+ ) ⎤
1
+
+
[−iL (0 ) − iR (0 )] = 4 ⎢ −4 −
⎥ = 4 (−4 + 0) = −16 V/s
c
2 ⎦
⎣
(e)
∴ (e) 0 = 1(B1 )∴ B1 = 0, vc (t ) = B2 e − t sin 5t , vc′ (0 + ) = B2 (5) = −16
∴ B2 = −3.2, vc (t ) = −3.2e − t sin 5t V, t > 0
(f)
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Engineering Circuit Analysis, 7th Edition
33.
α=
Chapter Nine Solutions
10 March 2006
1
106
1 106+3
=
= 4000, ω o2 =
=
= 2 × 107
2RC 100 × 2.5
LC
50
ω d = 20 × 106 − 16 × 106 = 2000
∴ ic = e −4000t (B1 cos 2000t + B2 sin 2000t )
iL (0) = 2A, vc (0) = 0 ∴ ic (0+ ) = −2A; ic′ (0+ ) = −iL′ (0+ ) − iR′ (0+ )
1
1
1
2 ×106
vc (0) − vc′ (0+ ) = 0 −
ic (0+ ) =
L
R
RC
125
6
2 ×10
∴ B1 = −2A,
= 16, 000 = 2000B2 + (−2) (−4000) ∴ B2 = 4
125
∴ ic (t ) = e −4000t (−2 cos 2000t + 4sin 2000t )A, t > 0
∴ ic′ (0+ ) = −
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Engineering Circuit Analysis, 7th Edition
Chapter Nine Solutions
10 March 2006
34.
(a)
(b)
1
100
1 100 2
=
= 8, ωo2 =
=
, ωd = 36 = ωo2 − 64
2RC 12.5
LC
L
100
∴ωo2 = 100 =
∴ L = 1H
L
α=
t < 0: iL (t ) = 4A; t > 0: iL (t ) = e −8t (B1 cos 6t + B2 sin 6t )
iL (0) = 4A ∴ B1 = 4A, iL = e −8t (4 cos 6t + B2 sin 6t ) vc (0) = 0
iL′ (0+ ) = t vc (0+ ) = 0 ∴ 6B2 − 8(4) = 0, B2 = 16 / 3
∴ iL (t ) = 4u (−t ) + e −8t (4 cos 6t + 5.333sin 6t ) u (t ) A
(c)
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Engineering Circuit Analysis, 7th Edition
Chapter Nine Solutions
10 March 2006
35.
(a)
α
1
109−3
1
109
=
= 5000, ωo2 =
=
= 1.25 × 108
2RC 2 × 20 × 5
LC 1.6 × 5
ωd = ωo2 − α 2 = 125 × 106 − 25 × 106 = 10, 000
∴ vc (t ) = e −5000t (B1 cos104 t + B2 sin104 t )
vc (0) = 200V, iL (0) = 10mA ∴ vc (t ) = e −5000t (200 cos104 t + B2 sin104 t )
v (0) ⎤
1
109 ⎡
vc′ (0+ ) = ic (0+ ) =
iL (0) − c
⎢
5 ⎣
c
20, 000 ⎥⎦
=
109 ⎛ −2
200 ⎞
= 0 = 104 B2 − 200 (5000)
10 −
⎜
⎟
5 ⎝
20, 000 ⎠
∴ B2 = 100V ∴ vc (t ) = e −5000t (200 cos104 t + 100sin104 t ) V, t > 0
(b)
isw = 10−2 − iL , iL =
1
vc + Cvc′
R
vc′ = e −5000t [104 (−200sin + 100 cos] − 5000 (200 cos + 100sin)]
= e −500t [106 (−2sin − 0.5cos)] = −2.5 × 106 e−5000t sin104 t v / s
⎡ 1
⎤
(200 cos + 100sin) − 5 × 10−9 × 2.5 × 106 e−5000t sin104 t ⎥
∴ iL = e −5000t ⎢
⎣ 20, 000
⎦
4
4
−5000 t
=e
(0.01cos10 t − 0.0075sin10 t ) A
∴ isw = 10 − e −5000t (10 cos104 t − 7.5sin104 t ) mA, t > 0
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Engineering Circuit Analysis, 7th Edition
Chapter Nine Solutions
10 March 2006
36.
(a)
α=
1
106
1 1.01× 106
=
= 20, ωo2 =
=
= 40, 400
2RC 2000 × 25
LC
25
ωd = ωo2 − α 2 = 40, 400 − 400 = 200
∴ v = e −20t (A1 cos 200t + A 2 sin 200t )
v(0) = 10V, iL (0) = 9mA ∴ A1 = 10V
∴ v = e −20t (10 cos 200t + A 2 sin 200t ) V, t > 0
v′(0+ ) = 200A 2 − 20 ×10 = 200 (A 2 − 1) =
1
io (0+ )
C
106
(−10−3 ) = −40 ∴ A 2 = 1 − 0.2 = 0.8
25
∴ v(t ) = e −20t (10 cos 200t + 0.8sin 200t ) V, t > 0
=
(b)
v = 10.032e−20t cos (200t − 4.574°)V
T=
2π
= 3.42ms
200
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Engineering Circuit Analysis, 7th Edition
37.
Chapter Nine Solutions
10 March 2006
1
106−3
1
=
= 100 s −1 , ωo2 =
= 1.01× 106
2RC 2 × 5
LC
60
∴ωd = 101× 104 − 104 = 100; iL (0) =
= 6mA
10
vc (0) = 0 ∴ vc (t ) = e −100t (A1 cos1000t + A 2 sin1000t ), t > 0
α=
∴ A1 = 0, vc (t ) = A 2 e −100t sin1000t
1
1
ic (0+ ) = 106 [−i1 (0+ ) −
vc (0+ )] = 106
C
5000
(−6 ×10−3 ) = −6000 = 1000 A 2 ∴ A 2 = −6
vc′ (0+ ) =
∴ vc (t ) = −6e −100t sin1000tV, t > 0 ∴ i1 (t ) = −
1
104
vc (t ) = −10−4 (−6) e −100t sin1000tA
∴ i1 (t ) = 0.6e −100t sin1000t mA, t > 0
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Engineering Circuit Analysis, 7th Edition
38.
Chapter Nine Solutions
10 March 2006
We replace the 25-Ω resistor to obtain an underdamped response:
α =
1
2RC
and
1
10 × 10− 6 R
Thus,
1
; we require α < ω0.
LC
ω0 =
< 3464
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.
iL(0+) = iL(0-) = 0 and vC(0+) = vC(0-) = (2)(25) = 50 V = A.
dvL
dv
= L C
dt
dt
−αt
= L e (− Aω d t sin ω d t + Bω d t cos ω d t ) - αe −αt ( A cos ω d t + Bsinω d t )
iL(t) = L
[
]
50 × 10−3
iL(0 ) = 0 =
[B ω d - αA ], so that B = 75.42 V.
3
+
Thus,
v(t) = e-2887t (50 cos 1914t + 75.42 sin 1914t) V.
Sketch of v(t).
PSpice schematic for t > 0 circuit.
From PSpice the settling time using R = 34.64 Ω is approximately 1.6 ms.
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Engineering Circuit Analysis, 7th Edition
39.
Chapter Nine Solutions
10 March 2006
v(0) = 0; i (0) = 10A
v = eαt (A cos ωd t + Bsin ωd t ) ∴ A = 0,
v = Be −αt sin ωd t
v′ = e −αt [−α Bsin ωd t + ωd B cos ωd t ] = 0
∴ tan ωd t =
tm 2 = tm1 +
ωd
1
ω
, tm1 =
tan −1 d
α
ωd
α
1
π
Td = tm1 +
;
2
ωd
vm1 = Be −αtm1 sin ωd tm1 vm 2 = − Be −αtm1 −απ / ωd
v
v
1
sin ωd tm1 ∴ m 2 = −e −απ / ωd ; let m 2 =
Vm1
vm1 100
ωd
1
21
ln 100; α =
= ,
2RC R
π
1
21 ln100
6R 2 − 441
ω02 =
= 6 ∴ωd = 6 − 441/ R 2 ∴
LC
R πR
∴ eαπ / ωd = 100, α =
2
⎡
⎛ 21π ⎞ ⎤
∴ R = 1/ 6 ⎢ 441 + ⎜
⎟ ⎥ = 10.3781Ω To keep
⎝ 100 ⎠ ⎥⎦
⎢⎣
vm 2
< 0.01, chose R = 10.3780Ω v′(0+ ) = ωd
vm1
2
0
⎛ 21 ⎞
⎛
⎞
B = B 6−⎜
⎟ = 4R ⎜10 +
⎟ ∴ B = 1.380363
10.3780 ⎠
⎝ 10.378 ⎠
⎝
2
α=
21
⎛ 21 ⎞
= 2.02351; ωd = 6 − ⎜
⎟ = 1.380363
10.378
⎝ 10.378 ⎠
∴ v = 304.268e −2.02351t sin 1.380363t v tm1 = 0.434 s,
vm1 = 71.2926v Computed values show
ts = 2.145sec; vm 2 = 0.7126 < 0.01vm1
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Engineering Circuit Analysis, 7th Edition
40. (a)
Chapter Nine Solutions
10 March 2006
For t < 0 s, we see from the circuit that the capacitor and the resistor are shorted by
the presence of the inductor. Hence, iL(0-) = 4 A and vC(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 iL(t) = e-αt (C cos ωdt + D sin ωdt)
for the response.
and
ith iL(0+) = iL(0-) = 4 A, we find C = 4 A. To find D, we first note that
di
vC(t) = vL(t) = L L
dt
so vC(t) = (2/13) [e-αt (-Cωd sin ωdt + Dωd cos ωdt) - αe-αt (C cos ωdt + D sin ωdt)]
W
ith vC(0+) = 0 = (2/13) (5.083D – 0.4C), we obtain D = 0.3148 A.
W
Thus,
iL(t) = e-0.4t (4 cos 5.083t + 0.3148 sin 5.083t) A and iL(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 vC(0+) = 0 = (2/13) (3.162D – 4C), we obtain D = 5.06 A.
Thus,
(c)
iL(t) = e-4t (4 cos 3.162t + 5.06 sin 3.162t) A and iL(.25) = 2.358 A.
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, 7th 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, iL(0-) = 4 A and vC(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 iL(t) = e-αt (C cos ωdt + D sin ωdt) for the response.
and
ith iL(0+) = iL(0-) = 4 A, we find C = 4 A. To find D, we first note that
di
vC(t) = vL(t) = L L
dt
so vC(t) = (2/13) [e-αt (-Cωd sin ωdt + Dωd cos ωdt) - αe-αt (C cos ωdt + D sin ωdt)]
W
Thus,
ith vC(0+) = 0 = (2/13) (5D – 4), we obtain D = 0.8 A.
iL(t) = e-t (4 cos 5t + 0.8 sin 5t) A
W
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.
(c)
Using the cursor tool, the settling
time is approximately 4.65 s.
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Engineering Circuit Analysis, 7th Edition
42.
vc (0) = 50 + 80 × 2 = 210 V, iL (0) = 0, α =
Chapter Nine Solutions
10 March 2006
R 80
=
= 20
2L 4
100
= 500 : ω d = 500 − 202 = 10
2
∴ vc (t ) = e −20t (A1 cos10t + A 2 sin10t ) ∴ A1 = 210 V
ω o2 =
1
ic (0+ ) = 0
C
−20 t
∴ 0 = 10A 2 − 20 (210), A 2 = 420 ∴ vc (t ) = e (210 cos10t + 420sin10t )
∴ vc (t ) = e −20t (210 cos10t + A 2 sin10t ); vc′ (0+ ) =
∴ vc (40ms) = e −0.8 (210 cos 0.4 + 420sin 0.4) = 160.40 V
Also, i(L = e −20t B1 cos10t + B2 sin10t ),
iL (0+ ) =
1
1
1
vL (0+ ) = [0 − vc (0+ )] = × 210
L
2
2
∴ iL′ (0+ ) = −105 = 10B2 ∴ B2 = 10.5
∴ iL (t ) = −10.5e −20t sin10t A, t > 0
∴ vR (t ) = 80iL = 840e −20t sin 10tV
∴ vR (40ms) = −840e −0.8 sin 0.4 = −146.98 V
vL (t ) = −vc (t ) − vc (t ) − vR (t ) ∴ vL
(40ms) = −160.40 + 146.98 = −13.420 V
[check: vL = e −20t (−210 cos− 420sin + 840sin)
= e −20t (−210 cos10t + 420sin10t ) V, t > 0
∴ vL (40ms) = e −0.8 (−210 cos− 420sin + 840 sin) = e −20t
(−210 cos10t + 420sin10t )V, t > 0
∴ VL (40ms) = e −0.8
(420sin 0.4 − 210 cos 0.4) = −13.420 V Checks]
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Engineering Circuit Analysis, 7th Edition
43. Series:
Chapter Nine Solutions
10 March 2006
R
2
1
4
=
= 4, ω o2 =
=
= 20, ω d = 20 − 16 = 2
2L 1/ 2
LC 0.2
∴ iL = e −4t (A1 cos 2t + A 2 sin 2t ); iL (0) = 10A, vc (0) = 20V
α=
∴ A1 = 10; iL′ (0+ ) =
1
vL (0+ ) = 4 (20 − 20) = 0
L
∴ iL′ (0+ ) = 2A 2 − 4 × 10 ∴ A 2 = 20
∴ iL (t ) = e −4t (10 cos 2t + 20sin 2t )A, t > 0
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Engineering Circuit Analysis, 7th Edition
44. (a)
Chapter Nine Solutions
10 March 2006
R2
1
1
= ω o2 =
∴ L = R 2C
2
4L
LC
4
1
200
∴ L = × 4 × 104 −6 = 0.01H, α =
= 104 = ω o
4
0.02
∴ vc (t ) = e −10000t (A1t + A 2 ); vc (0) = −10V, iL (0) = −0.15A
crit. damp; α 2 =
∴ A 2 = −10, vc (t ) = e −10000t (A1t − 0); vc′ (0+ ) = −
1
C
iL (0) = −106 (−0.15) = 150, 000
Now, (vc′ 0+ ) = A1 + 105 = 150, 000 ∴ A1 = 50, 000
∴ vc (t ) = e −10,000t (50, 000t − 10) V, t > 0
(b)
vc′ (t ) = e −10,000t [50, 000 − 10, 000 (50, 000t − 10)] =∴
5 = 50, 000tm − 10 ∴ tm =
15
= 0.3ms
50, 000
∴ vc (tm ) = e −3 (15 − 10) = 5e −3 = 0.2489V
vc (0) = −10V ∴ vc max = 10V
(c)
vc ,max = 0.2489V
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Engineering Circuit Analysis, 7th Edition
45.
Chapter Nine Solutions
10 March 2006
“Obtain an expression for vc(t) in the circuit of Fig. 9.8 (dual) that is valid for all t′′.
μF
Ω
α=
mF
A
R 0.02 ×106
106 × 3
=
= 4000, ω o2 =
= 1.2 × 107
2L
2 × 2.5
2.5 × 10
∴ s1,2 = −4000 ± 16 ×106 − 12 × 106 = −2000, − 6000
1
× 100 = 2V
50
1
iL (0) = 100A ∴ 2 = A1 + A 2 , vc′ (0+ ) =
C
3
(−iL (0)) = −
× 103 × 100 = −3000v / s
100
∴−3000 = −200A1 − 600A 2 , − 1.5 = − A1 − 3A 2
∴ vc (t ) = A1e −2000t + A 2 e−6000t ; vc (0) =
∴ 0.5 = −2A 2 , = −0.25, A1 = 2.25
∴ vc (t ) = (2.25e −200t − 0.25e−6000t ) u (t ) + 2u (−t ) V (checks)
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Engineering Circuit Analysis, 7th Edition
46. (a)
Chapter Nine Solutions
10 March 2006
R 2
1
= = 1, ω o2 =
= 5, ω d = ω o2 − α 2 = 2
2L 2
LC
−t
∴ iL = e (B1 cos 2t + B2 sin 2t ), iL (0) = 0, vc (0) = 10V
α=
∴ B1 = 0, iL = B2 e− t sin 2t
1
i1 (0) = vL (0+ ) = vR (0+ ) − Vc (0+ ) = 0 − 10 = 2B2
1
∴ B2 = 5 ∴ iL = −5e− t sin 2tA, t > 0
(b)
iL′ = −5[e − t (2 cos 2t − sin 2t )] = 0
∴ 2 cos 2t = sin 2t , tan 2t = 2
∴ t1 = 0.5536 s, iL (t1 ) = −2.571A
2t2 = 2 × 0.5536 + π , t2 = 2.124,
iL (t2 ) = 0.5345 ∴ iL max = 2.571A
and i0.5345A
L max =
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Engineering Circuit Analysis, 7th Edition
47. (a)
α=
Chapter Nine Solutions
10 March 2006
R 250
1
106
=
= 25, ω o2 =
=
= 400
2L 10
LC 2500
s1,2 = −α ± α 2 − ω o2 = −25 ± 15 = −10, −40
∴ iL = A1 e −10t + A 2 e−40t , iL (0) = 0.5A, vc (0) = 100V
1
1
∴ 0.5 = A1 + A 2 , iL′ (0+ ) = vL (0+ ) =
5
5
(100 − 25 − 100) = −5 A / s = −10A1 − 40A 2
∴ 5 = 10 A1 + 40 (0.5 − A1 ) = 10A1 − 40
A1 + 20 ∴−30A1 = −15, A1 = 0.5, A 2 = 0
∴ iL (t ) = 0.5e−10t A, t > 0
(b)
vc = A 3e −10t + A 4 e−40t ∴100 = A 3 + A 4 ;
1
106
vc′ = ic′ (0+ )
(−0.5) = −1000
c
500
∴−10A 3 − 40A 4 = −1000 ∴−3A 4 = 0, A 4 = 0, A 3 = 100
∴ vc (t ) = 100e−10t V t > 0
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Engineering Circuit Analysis, 7th Edition
48.
Chapter Nine Solutions
10 March 2006
Considering the circuit as it exists for t < 0, we conclude that vC(0-) = 0 and iL(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:
iL(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
di
vC(t) = vL(t) = L L
dt
= (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]
vC(0+) = vC(0-) = 0 = (9)(-0.25)(2.25) + (9)(0.2205B)
so B = 2.551 and
iL(t) = e-0.25t [2.25 cos 0.2205t + 2.551 sin 0.2205t] A
Thus, iL(2) = 1.895 A
This answer is borne out by PSpice simulation:
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Engineering Circuit Analysis, 7th Edition
49.
Chapter Nine Solutions
10 March 2006
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 s1 = -21.32 s-1 and s2 = -9379 s-1.
From the circuit as it exists for t < 0, it is evident that iL(0-) = 0 and vC(0-) = 4.7 kV
Thus,
vL(t) = A e–21.32t + B e-9379t
[1]
With iL(0+) = iL(0-) = 0 and iR(0+) = 0 we conclude that vR(0+) = 0; this leads to vL(0+) =
-vC(0-) = -4.7 kV and hence A + B = -4700 [2]
di
Since
vL = L , we may integrate Eq. [1] to find an expression for the inductor current:
dt
B − 9379t ⎤
1 ⎡ A − 21.32t
iL(t) =
e
e
⎥
L ⎢⎣ 21.32
9379
⎦
1
B ⎤
⎡ A
t = 0+, iL = 0 so we have
= 0 [3]
At
-3 ⎢
500 × 10 ⎣ 21.32 9379 ⎥⎦
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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Engineering Circuit Analysis, 7th Edition
50.
Chapter Nine Solutions
10 March 2006
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 = Aes1t + Bes2t
Using our first initial condition, ibear(0+) = iL(0+) = iL(0-) = 0 = A + B
di/dt = As1 es1t + Bs2 es2t
vL = Ldi/dt = ALs1 es1t + BLs2 es2t
vL(0+) = ALs1 + BLs2 = vC(0+) = vC(0-) = 12
What else is known? We know that the bear stops reacting at t = 18 μs, 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 Rbear = 119.9775 Ω.
This corresponds to A = 100 mA, B = -100 mA, s1 = -4.167 s-1 and s2 = -24×106 s-1
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Engineering Circuit Analysis, 7th Edition
51.
Chapter Nine Solutions
10 March 2006
Considering the circuit at t < 0, we note that iL(0-) = 9/4 = 2.25 A and vC(0-) = 0.
1
1
=
, which, with
For a critically damped circuit, we require α = ωo, or
2RC
LC
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)
that iL(0+) = iL(0-) = 2.25, we see that B = 2.25 and hence
Noting
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) = L
diL
= 9(-0.3333) e-0.3333t (At + 2.25) + 9A e-0.3333t
dt
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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Engineering Circuit Analysis, 7th Edition
52. Prior
Thus,
Chapter Nine Solutions
10 March 2006
v
⎛ 50 ⎞
to t = 0, we find that v = (10 + i1 ) ⎜ ⎟ and i1 =
5
⎝ 15 ⎠
⎛ 10 ⎞ 500
so v = 100 V .
v ⎜1 − ⎟ =
⎝ 15 ⎠ 15
Therefore, vC (0+ ) = vC (0− ) = 100 V, and iL (0+ ) = iL (0− ) = 0.
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,
α=
and
ω0 =
R
10
=
= 2.5 × 103 s −1
2 L 2 2 ×10−3
(
1
=
LC
)
1
( 2 ×10 )( 20 ×10 )
−3
−9
= 1.581× 105 rad/s
ith α < ω0 we find the circuit is underdamped, with
W
ωd = ω02 − α 2 = 1.581× 105 rad/s
We may therefore write the response as
iL (t ) = e −α t ( B1 cos ωd t + B2 sin ωd t )
At
t = 0, iL = 0 ∴ B1 = 0 .
Noting
that
diL d −α t
= ( e B2 sin ωd t ) = B2 e −α t ( −α sin ωd t + ωd cos ωd t ) and
dt dt
L
diL
= −100 we find that B2 = -0.316 A.
dt t =0
Finally,
iL (t ) = −316e−2500t sin1.581× 105 t mA
B
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Engineering Circuit Analysis, 7th Edition
53. Prior
Chapter Nine Solutions
10 March 2006
to t = 0, we find that vC = 100 V, since 10 A flows through the 10 Ω resistor.
Therefore, vC (0+ ) = vC (0− ) = 100 V, and iL (0+ ) = iL (0− ) = 0.
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,
α=
and
ω0 =
R
10
=
= 2.5 × 103 s −1
−3
2 L 2 2 ×10
(
1
=
LC
)
1
( 2 ×10 )( 20 ×10 )
−3
−9
= 1.581× 105 rad/s
ith α < ω0 we find the circuit is underdamped, with
W
ωd = ω02 − α 2 = 1.581× 105 rad/s
We may therefore write the response as
vC (t ) = e −α t ( B1 cos ωd t + B2 sin ωd t )
At
t = 0, vC = 100 ∴ B1 = 100 V .
Noting
dvC
= iL and
dt
that C
d −α t
⎡e (100 cos ωd t + B2 sin ωd t ) ⎤⎦
dt ⎣
= e −α t ⎡⎣ −α (100 cos ωd t + B2 sin ωd t ) − 100ωd sin ωd t + B2ωd cos ωd t ⎤⎦
which is equal to zero at t = 0 (since iL = 0)
we find that B2 = 1.581 V .
B
Finally,
vC (t ) = e−2500t ⎡⎣100 cos (1.581×105 t ) + 1.581sin (1.581×105 t ) ⎤⎦ V
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Engineering Circuit Analysis, 7th Edition
54. Prior
Thus,
After
=
Chapter Nine Solutions
10 March 2006
to t = 0, i1 = 10/4 = 2.5 A, v = 7.5 V, and vg = –5 V.
vC(0+) = vC(0–) = 7.5 + 5 = 12.5 V and iL = 0
i
t = 0 we are left with a series RLC circuit where i1 = − L . We may replace the
4
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.
R 1.25
=
= 0.208 s −1
2L
6
1
1
ω0 =
=
= 577 mrad/s
3
LC
α=
Thus,
and
ith α < ω0 we find the circuit is underdamped, so that
W
ωd = ω02 − α 2 = 538 mrad/s
We may therefore write the response as
iL (t ) = e −α t ( B1 cos ωd t + B2 sin ωd t )
At
t = 0, iL = 0 ∴ B1 = 0 A .
Noting
,
diL
= −vC (0) and
dt t =0
that L
v (t ) −12.5
diL d −α t
= ⎣⎡ e ( B2 sin ωd t ) ⎦⎤ = B2 e−α t [ −α sin ωd t + ωd cos ωd t ] = C =
(t = 0)
dt dt
L
3
we find that B2 = –7.738 V.
B
Finally,
iL (t ) = 1.935e −0.208t sin 0.538t A for t > 0 and 2.5 A, t < 0
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Engineering Circuit Analysis, 7th Edition
55. Prior
Thus,
After
=
Chapter Nine Solutions
10 March 2006
to t = 0, i1 = 10/4 = 2.5 A, v = 7.5 V, and vg = –5 V.
vC(0+) = vC(0–) = 12.5 V and iL = 0
i
t = 0 we are left with a series RLC circuit where i1 = − L . We may replace the
4
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.
R 1.25
=
= 0.208 s −1
2L
6
1
1
ω0 =
=
= 18.26 rad/s
−3
LC
3
10
×
(
)
α=
Thus,
and
ith α < ω0 we find the circuit is underdamped, so that
W
ωd = ω02 − α 2 = 18.26 rad/s
We may therefore write the response as
vC (t ) = e −α t ( B1 cos ωd t + B2 sin ωd t )
At
t = 0, vC = 12.5 ∴ B1 = 12.5 V .
Noting
that
dvC d −α t
= ⎡⎣ e ( B1 cos ωd t + B2 sin ωd t ) ⎤⎦
dt
dt
−α t
= −α e [12.5cos ωd t + B2 sin ωd t ] + e−α t [ −12.5ωd sin ωd t + ωd B2 cos ωd t ]
and this expression is equal to 0 at t = 0,
we find that B2 = 0.143 V.
B
Finally,
vC (t ) = e −0.208t [12.5cos18.26t + 0.143sin18.26t ] V for t > 0 and 12.5 V, t < 0
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Engineering Circuit Analysis, 7th Edition
56. (a)
Series, driven: α =
Chapter Nine Solutions
10 March 2006
R 100
=
= 500,
2L 0.2
1 10 × 106
=
= 250, 000
LC
40
∴ Crit. damp iL ( f ) = 3(1 − 2) = −3,
ω o2 =
iL (0) = 3, vc (0) = 300V
∴ iL = −3 + e −500t (A1t + A 2 ) ∴ 3 = −3 + A 2 , A 2 = 6A
1
[vc (0) − vR (0+ )] = 0
L
−5000 t
∴ A1 = 3000 e
∴ iL (t ) = −3 + e −500t
iL (0+ ) = A1 − 300 =
(3000t + 6), t > 0
∴ iL (t ) = 3u (−t ) + [−3 + e −500t (3000t + 6)] u (t )A
(b)
e−500to (3000to + 6) = 3; by SOLVE, to = 3.357ms
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Engineering Circuit Analysis, 7th Edition
57.
vc (0) = 0, iL (0) = 0, α =
Chapter Nine Solutions
10 March 2006
R
2
1
=
= 4, ω o2 =
= 4 × 5 = 20
2L 0.5
LC
∴ω d = 20 − 16 = 2 ∴ iL (t ) = e −4t (A1 cos 2t + A 2 sin 2t ) + iL , f
iL , f = 10A ∴ iL (t ) = 10 + e −4t (A1 cos 2t + A 2 sin 2t )
∴ 0 = 10 + A1 , A1 = −10, iL (t ) = 10 + e −4t (A 2 sin 2t − 10 cos 2t )
iL (0+ ) =
1
vL (0+ ) = 4 × 0 = 0 ∴ iL (0+ ) = 0 = 2A 2 + 40, A 2 = −20
L
iL(t) = 10 - e-4t (20 sin 2t + 10 cos 2t) A, t > 0
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Engineering Circuit Analysis, 7th Edition
Chapter Nine Solutions
10 March 2006
58.
R 250
1
106
=
= 25, ωo2 =
=
= 400
2L 10
LC 2500
s1,2 = −25 ± 625 − 400 = −10, −40
α=
iL (0) = 0.5A, vc (0) = 100V, iL , f = −0.5A
∴ iL (t ) = −0.5 + A1e −10t + A 2 e −40t A
t = 0+ : vL (0+ ) = 100 − 50 × 1 − 200 × 0.5 = −50V ∴−50 = 5iL′ (0+ )
∴ iL′ (0+ ) = −10 ∴−10 = −10A1 − 40A 2 , 0.5 = −0.5 + A1 + A 2
∴ A1 + A 2 = 1∴−10 = −10A 2 − 40 (−1+A1 ) = −50A1 + 40, A1 = 1, A 2 = 0
∴ iL (t ) = −0.5 + 1e −10t A, t > 0; iL (t ) = 0.5A, t > 0
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Engineering Circuit Analysis, 7th Edition
59.
α=
Chapter Nine Solutions
10 March 2006
1
106
1 106+3
=
= 4000, ωo2 =
=
= 20 × 106
2RC 100 × 2.5
LC
50
∴ωd = ωo2 − α 2 = 2000, iL (0) = 2A, vc (0) = 0
ic , f = 0, (vc , f = 0) ∴ ic = e −400t (A1 cos 2000t + A 2 sin 2000t )
work with vc : vc (t ) = e −4000t (B1 cos 2000t + B2 sin 2000t ) ∴ B1 = 0
1
106
+
(2 × 1) = 8 × 105
∴ vc = B2 e
sin 2000t , vc′ (0 ) = ic (0 ) =
C
2.5
−4000 t
5
∴ 8 × 10 = 2000B2 , B2 = 400, vc = 400e
sin 2000t
−4000 t
+
∴ ic (t ) = Cvc′ = 2.5 × 10−6 × 400e −4000t (−4000sin 200t + 2000 cos 200t )
= 10−6+3+3 e −4000t (−4sin 2000t + 2 cos 2000t )
= e −4000t (2 cos 2000t − 4sin 2000t ) A, t > 0
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Engineering Circuit Analysis, 7th Edition
60. (a)
α=
Chapter Nine Solutions
10 March 2006
1
8 × 106
8 ×106 × 13
2
=
=
1000,
ω
=
= 26 × 106
o
3
2RC 2 × 4 ×10
4
∴ωd = 26 − 1 × 103 = 5000, vc (0) = 8V
iL (0) = 8mA, vc , f = 0
∴ vc = e−1000t (A1 cos1000t + A 2 sin 5000t )
1
8
− 0.008) = 0
ic (0+ ) = 8 × 106 (0.01 −
C
4000
∴ 5000A 2 − 1000 × 8 = 0, A 2 = 1.6
∴ A1 = 8; vc′ (0+ ) =
So vc(t) = e-1000t (8 cos 1000t + 1.6 sin 1000t) V, t > 0
(b)
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Engineering Circuit Analysis, 7th Edition
Chapter Nine Solutions
10 March 2006
61.
R 1
1
= = 1, ωo2 =
= 1∴ crit. damp
2L 1
LC
5
vc (0) = × 12 = 10V, iL (0) = 2A, vc , f = 12V
6
1
1
∴ vc (t ) = 12 + e − t (A1t − 2); vc′ (0+ ) = ic (0+ ) = × iL (0+ ) = 1
C
2
−t
∴1 = A1 + 2; A1 = −1∴ vc (t ) = 12 − e (t + 2) V, t > 0
α=
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Engineering Circuit Analysis, 7th Edition
62. (a)
Chapter Nine Solutions
10 March 2006
1
106
vs = 10u (−t ) V : α =
=
= 1000
2RC 2000 × 0.5
1
2 ×106 × 3
ωo2 =
=
= 0.75 × 106 ∴ s1,2 = −500, − 1500
LC
8
∴ vc = A1e −500t + A 2 e −1500t , vo (0) = 10V, iL (0) = 10mA
∴ A1 + A 2 = 10, vc′ (0+ ) = 2 ×106 [iL (0) − iR (0+ )] = 2 × 106
10 ⎞
⎛
⎜ 0.01 −
⎟ = 0 ∴−500A1 − 1500 A 2 = 0,
1000 ⎠
⎝
− A1 − 3A 2 = 0; add: − 2 A 2 = 10, A 2 = −5, A1 = 15
∴ vc (t ) = 15e −500t − 5e−1500t V t > 0
∴ iR (t ) = 15e −500t − 5e−1500t mA, t > 0
(b)
vs = 10u (t ) V, vc , f = 10, vc = 10 + A 3e−500t + A 4 e−1500t ,
vc (0) = 0, iL (0) = 0 ∴ A 3 + A 4 = −10V, vc′ (0+ ) = 2 × 106
[iL (0) − iR (0+ )] = 2 × 106 (0 − 0) = 0 = −500A 3 − 1500A 4
∴− A 3 − 3A 4 = 0, add: − 2A 4 = −10, A 4 = 5 ∴ A 3 = −15
∴ vc (t ) = 10 − 15e −500t + 5e −1500t V, t > 0
∴ iR (t ) = 10 − 15e −500t + 5e1500t mA, t > 0
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Engineering Circuit Analysis, 7th Edition
63. (a)
Chapter Nine Solutions
10 March 2006
1
106
=
= 1000
2RC 1000
1 106 × 3
3
ωo2 =
=
∴ s1,2 = −1000 ± 106 − × 106 = −500, − 1500
LC
4
4
−500 t
−1500 t
vc , f = 0 ∴ vc = A1e
+ A2e
, vc (0) = 10V, iL (0) = 0
vs (t ) = 10u (−t ) V: α =
10 ⎤
⎡
∴10 = A1 + A 2 , vc′ = 106 ic (0+ ) = 106 ⎢0 −
= −2 × 104
⎥
⎣ 500 ⎦
4
∴−2 × 10 = −500A1 − 1500A ∴ 40 = A1 + 3A 2 ∴ 30 = 2A 2 , A 2 = 15, A1 = −5
∴ vc = −5e −500t + 15e −1500t V, t > 0 ∴ is = ic = Cvc′
∴ is = 10−6 (2500e −500t − 22,500e −1500t )
= 2.5e −500t − 22.5e −1500t mA, t > 0
(b)
vs (t ) = 10u (t ) V ∴ vc , f = 10V, vc (0) = 0, iL (0) = 0
∴ vc = 10 + A3 e −500t + A 4 e−1500t ∴ A 3 + A 4 = −10
10 ⎞
⎛
4
vc′ (0+ ) = 106 ic (0+ ) = 106 ⎜ 0 +
⎟ = 2 × 10 = −500 A 3 − 1500 A 4
500
⎝
⎠
∴− A 3 − 3A 4 = 40, add: 2− A 4 = 30, A 4 = −15, A 3 = 5,
vc = 10 + 5e −500t − 15e−1500t V, is = ic =
10−6 (−2500e −500t + 22,500e −1500t ) = 25e −500t + 22.5e−1500t mA, t > 0
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Engineering Circuit Analysis, 7th Edition
64.
Chapter Nine Solutions
10 March 2006
Considering the circuit at t < 0, we see that iL(0-) = 15 A and vC(0-) = 0.
The circuit is a series RLC with α = R/2L = 0.375 s -1 and ω0 = 1.768 rad/s. W e therefore
of the
expect an underdam ped response with ωd = 1.728 rad/s. The general form
response will be
vC(t) = e-αt (A cos ωdt + B sin ωdt) + 0
(vC(∞) = 0)
vC(0+) = vC(0-) = 0 = A and we may therefore write vC(t) = Be-0.375t sin (1.728t) V
iC(t) = -iL(t) = C
dvC
= (80×10-3)(-0.375B e-0.375t sin 1.728t
dt
At
t = 0+, iC = 15 + 7 – iL(0+) = 7 = (80×10-3)(1.728B) so that B = 50.64 V.
Thus,
vC(t) = 50.64 e–0.375t sin 1.807t V and vC(t = 200 ms) = 16.61 V.
The energy stored in the capacitor at that instant is ½ CvC2 = 11.04 J
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Engineering Circuit Analysis, 7th Edition
Chapter Nine Solutions
65. (a)
vS(0-) = vC(0-) = 2(15) = 30 V
(b)
Thus,
iL(0+) = iL(0-) = 15 A
iC(0+) = 22 – 15 = 7 A and vS(0+) = 3(7) + vC(0+) = 51 V
10 March 2006
(c) As t → ∞, the current through the inductor approaches 22 A, so vS(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)
iL(0+) = iL(0-) = 15 = 22 + A so A = -7 amperes
Thus, iL(t) = 22 + e-2.5t (-7 cos 2.501t + B sin 2.501t)
di
di
vS(t) = 2 iL(t) + L L = 2iL + L = 44 + 2e-2.5t (-7cos 2.501t + Bsin 2.501t)
dt
dt
-2.5t
– 2.5e (-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 vS(t) at t = 3.4 s = 44.002 V. This is borne out by PSpice simulation:
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Engineering Circuit Analysis, 7th Edition
66. For
Chapter Nine Solutions
10 March 2006
t < 0, we have 15 A dc flowing, so that iL = 15 A, vC = 30 V, v3Ω = 0 and vS = 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+) = vC(0-) = 30 = A so we may write vC(t) = e-2.5t (30 cos 2.501t + B sin 2.501t)
dvC
= -2.5e-2.5t(30 cos 2.501t + B sin 2.501t)
dt
+ e-2.5t [-30(2.501)sin 2.501t + 2.501B cos 2.501t]
iC(0+) = C dvC
dt
= 80×10-3[-2.5(30) + 2.501B] = -iL(0+) = -iL(0-) = -15 so B = -44.98 V
t = 0+
Thus, vC(t) = e-2.5t (30 cos 2.501t – 44.98 sin 2.501t) and
iC(t) = e-2.5t (-15 cos 2.501t + 2.994 sin 2.501t).
Hence, vS(t) = 3 iC(t) + vC(t) = e-2.5t (-15 cos 2.501t – 36 sin 2.501t)
Prior to switching, vC(t = 1) = -4.181 V and iL(t = 1) = -iC(t = 1) = -1.134 A.
t > 2: Define t' = t – 1 for notational simplicity. Then, with the fact that vC(∞) = 6 V,
our response will now be vC(t') = e-αt' (A' cos ωdt' + B' sin ωdt') + 6.
With vC(0+) = A' + 6 = -4.181, we find that A' = -10.18 V.
iC(0+) = C dv C
dt ′ t ′ = 0 +
= (80×10-3)[(-2.5)(-10.18) + 2.501B')] = 3 – iL(0+) so B' = 10.48 V
vC(t') = e-2.5t (-10.18 cos 2.501t' + 10.48 sin 2.501t') and
iC(t') = e-2.5t (4.133 cos 2.501t' – 0.05919 sin 2.501t').
Hence, vS(t') = 3 iC(t') + vC(t') = e-2.5t (2.219 cos 2.501t' + 10.36 sin 2.501t')
Thus,
We see that our hand
calculations are supported by
the PSpice simulation.
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Engineering Circuit Analysis, 7th Edition
67.
Chapter Nine Solutions
10 March 2006
It’s probably easiest to begin by sketching the waveform vx:
vx (V)
75
1
2
(a) The source current ( = iL(t) ) =
3
4
0
at t = 0-.
t (s)
(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 iL(t) = e-αt (A cos ωdt + B sin ωdt)
iL(0+) = iL(0-) = 0 = A so A = 0. This leaves iL(t) = B e-2000t sin 1999t
vL(t) = L
diL
= B[(5×10-3)(-2000 e-2000t sin 1999t + 1999 e-2000t cos 1999t)]
dt
vL(0+) = vx(0+) – vC(0+) – 20 iL(0+) = B (5×10-3)(1999) so B = 7.504 A.
Thus, iL(t) = 7.504 e-2000t sin 1999t and iL(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')
diL
, and this enables us to calculate that vL(t = 1 ms) = -13.54 V. Prior to the
dt
pulse returning to zero volts, -75 + vL + vC + 20 iL = 0 so vC(t' = 0) = 69.97 V.
vL(t) = L
iL(t' = 0) = A' = 0.9239 and –vx + vL + vC + 20 iL = 0 so that B' = -7.925.
Thus, iL(t') = e-2000 t' (0.9239 cos 1999t' – 7.925 sin 1999t') and
hence iL(t = 2 ms) = iL(t' = 1 ms) = -1.028 A.
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Engineering Circuit Analysis, 7th Edition
68.
Chapter Nine Solutions
10 March 2006
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 vC(t) = e-αt (A cos 2πt + B sin 2πt). If we go ahead and
decide to set vC(0-) = 0, then we can force A = 0 and simplify some of our algebra.
Thus, vC(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
⎛1⎞
⎝ 2⎠
2
ωd = ω 02 - ⎜ ⎟ = 39.73 rad/s =
1
LC
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, 7th Edition
69.
Chapter Nine Solutions
10 March 2006
The circuit described is a series RLC circuit, and the fact that oscillations 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 = 1.825×106 rad/s.
ω0 =
1
= 1.914×106 rad/s, and so ωd2 = ω02 – α2 leads to α2 = 332.8×109
LC
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, 7th 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
s1 = -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]
dvC
= -2000 Ae-2000t – 8000 Be-8000t
dt
iC(0+) = C
dvC
= 3 – 5 – iL(0-) = -5 = -25×10-6 (2000A + 8000B)
dt t = 0 +
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, 7th Edition
71.
Chapter Nine Solutions
10 March 2006
α = 0 (this is a series RLC with R = 0, or a parallel RLC with R = ∞)
ωo2 = 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:
dv
1 t
v dt ′ + 10-3 + 2
= 0
(iC + iL = 0)
∫
0
dt
10
and then take the derivative of both sides:
d 2v
1
= - v
2
dt
20
dv
With
= (0.2236)(−2.236 × 10− 3 ) = −5 × 10 − 4 , one possible solution is:
dt t = 0 +
PSpice sim ulations ar e very sens itive to par ameter v alues; bette r re sults were obta ined
using LF411 instead of 741s (both were compared to the simple LC circuit simulation.)
Simulation using 741 op amps
Simulation using LF411 op amps
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Engineering Circuit Analysis, 7th Edition
72.
Chapter Nine Solutions
10 March 2006
α = 0 (this is a series RLC with R = 0, or a parallel RLC with R = ∞)
ωo2 = 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:
dv
1 t
v dt ′ + 2 + 10-3
= 0
(iC + iL = 0)
∫
0
dt
20
and then take the derivative of both sides:
d 2v
= - 50v
dt 2
dv
With
= (7.071)(−282.8) = −2178 , one possible solution is:
dt t = 0 +
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Engineering Circuit Analysis, 7th Edition
Chapter Nine Solutions
10 March 2006
73.
(a)
v
dv
+ 3.3 × 10- 3
1000
dt
or
= 0
dv
1
= v
dt
3.3
(b) One possible solution:
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Engineering Circuit Analysis, 7th Edition
74.
Chapter Nine Solutions
10 March 2006
We see either a series RLC with R = 0 or a parallel RLC with R = ∞; either way, α = 0.
ωd = 0.5477 rad/s (com bining the tw o inductors in parallel for the
calculation). We expect a response of the form i(t) = A cos ωdt + B sin ωdt.
ω02 = 0.3 so
i(0+) = i(0-) = A = 1×10-3
di/dt = -Aωd sin ωdt + Bωd cos ωdt
vL = 10di/dt = -10Aωd sin ωdt + 10Bωd cos ωdt
vL(0+) = vC(0+) = vC(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
1
1
dv
vdt ′ + 10-3 + ∫ vdt ′ + 2
∫
10 0
20
dt
t
di
=0
dt t = 0 +
t
or
d 2v
= − 0.3v
dt 2
= 0
↓
i
1Ω
One possible solution is:
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Engineering Circuit Analysis, 7th Edition
75. (a)
vR = vL
20(-
iL) = 5
diL
dt
Chapter Nine Solutions
10 March 2006
diL
= - 4iL
dt
or
(b) We expect a response of the form iL(t) = A e-t/ τ where τ = L/R = 0.25.
We know that iL(0-) = 2 amperes, so A = 2 and iL(t) = 2 e-4t
diL
= -4(2) = -8 A/s.
dt t = 0 +
One possible solution, then, is
8V
1 μF
1 MΩ
i
↓
1Ω
4 kΩ
1 kΩ
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Engineering Circuit Analysis, 7th Edition
Chapter Ten Solutions
10 March 2006
1.
(a)
2π103
= 290.9t rad/s
21.6
∴ f (t ) = 8.5sin (290.9t + Φ ) ∴ 0 = 8.5sin (290.9 × 2.1× 10−3 + Φ )
T = 4 (7.5 − 2.1)10−3 = 21.6 × 10−3 , ω =
∴Φ = −0.6109rad + 2π = 5.672rad or 325.0 °
∴ f (t ) = 8.5sin (290.9t + 325.0°)
(b)
8.5sin (290.9t + 325.0°) = 8.5
cos(290.9t + 235°) = 8.5 cos (290.9t − 125°)
(c)
8.5 cos (−125°) cos ωt + 8.5sin125°
sin ωt = −4.875+ cos 290.9t + 6.963sin 290.9t
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Engineering Circuit Analysis, 7th Edition
Chapter Ten Solutions
10 March 2006
2.
(a)
−10 cos ωt + 4sin ωt + ACos ( wt + Φ ), A > 0, − 180° < Φ ≤ 180°
A = 116 = 10.770, A cos Φ = −10, A sin Φ = −4 ∴ tan Φ = 0.4, 3d quad
∴Φ = 21.80° = 201.8°, too large ∴Φ = 201.8° − 360° = −158.20°
(b)
200 cos (5t + 130°) = Fcos 5t + G sin 5t ∴ F = 200cos130° = −128.6
G = −200sin130° = −153.2
(c)
(d)
i (t ) = 5cos10t − 3sin10t = 0, 0 ≤ t ≤ 1 s
sin10t 5
∴
= , 10t = 1.0304,
cos10t 3
t = 0.10304 s; also, 10t = 1.0304 + π, t = 0.4172 s; 10t = 1.0304 + 2π, t = 0.7314 s
0 < t < 10ms, 10 cos100πt ≥ 12sin100πt ; let 10cos100πt =12sin100πt
10
∴ tan100πt = , 100πt = 0.6947 ∴ t = 2.211 ms ∴ 0 < t < 2.211 ms
12
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Engineering Circuit Analysis, 7th Edition
Chapter Ten Solutions
10 March 2006
3.
⎛
⎛ −B ⎞ ⎞
that A cos x + B sin x = A2 + B 2 cos ⎜ x + tan −1 ⎜
⎟ ⎟ . For f(t), the angle is in the
⎝ A ⎠⎠
⎝
second quadrant; most calculators will return –30.96o, which is off by 180o.
(a) Note
f (t ) = −50 cos ωt − 30sin ωt = 58.31cos (ωt + 149.04°)
g (t ) = 55cos ωt − 15sin ωt = 57.01cos (ωt + 15.255°)
∴ ampl. of f (t ) = 58.31, ampl. of g (t ) = 57.01
(b)
f (t ) leads g (t ) by 149.04° − 15.255° = 133.8°
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Engineering Circuit Analysis, 7th Edition
4.
Chapter Ten Solutions
10 March 2006
i (t ) = A cos (ωt − θ), and
L(di / dt ) + Ri = Vm cos ωt
∴
[−ωA sin (ω
L t − θ)] + RA cos (ωt − θ) = Vm cos ωt
−ωLA sin ωt cos θ + ωLA cos ωt sin θ + RA cos ωt cos θ + RA sin ωt sin θ
= Vm cos ωt
∴ ωLA cos θ = RA sin θ
and ωLA sin θ + RA cos θ = Vm
ωL
R
ωL
Thus, tan θ =
*
and ωLA
+ RA
R 2 + ω2 L2
R
R 2 + ω2 L2
= Vm
⎛
⎞
R2
ω2 L2
so that ⎜
+
⎟ A = Vm
2
2 2
R 2 + ω2 L2 ⎠
⎝ R +ω L
Thus,
( R + ω L ) A = ( V ) and therefore we may write A = R V+ ω L *
2
2 2
m
m
2
2 2
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Engineering Circuit Analysis, 7th Edition
5.
Chapter Ten Solutions
f = 13.56 MHz so ω = 2πf = 85.20 Mrad/s.
Delivering 300 W (peak) to a 5-Ω load implies that
Finally,
Since
10 March 2006
Vm2
= 300 so Vm = 38.73 V.
5
(85.2×106)(21.15×10–3) + φ = nπ, n = 1, 3, 5, …
(85.2×106)(21.15×10–3) = 1801980, which is 573588π, we find that
φ = 573589π - (85.2×106)(21.15×10–3) = 573589π - 573588π = π
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Engineering Circuit Analysis, 7th Edition
6.
(a)
Chapter Ten Solutions
10 March 2006
-33 sin(8t – 9o) → -33∠(-9-90)o = 33∠81o
12 cos (8t – 1o) → 12∠-1o
33∠81o
-33 sin(8t – 9o) leads 12 cos (8t – 1o) by
81 – (-1) = 82o.
12∠-1o
(b)
15 cos (1000t + 66o)
-2 cos (1000t + 450o)
→
15 ∠ 66o
→ -2
∠ 450o = -2 ∠90o = 2 ∠ 270o
15 cos (1000t + 66o) leads -2 cos (1000t + 450o)
by 66 – -90 = 156o.
15∠66o
2∠270o
(c)
sin (t – 13o)
cos (t – 90o)
→
→1
1∠-103o
∠ -90o
cos (t – 90o) leads sin (t – 13o)
by 66 – -90 = 156o.
1∠-103o
(d)
1 ∠ -90o
sin t
cos (t – 90o)
→1
→1
∠ -90o
∠ -90o
These two waveforms are in phase. Neither leads the other.
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Engineering Circuit Analysis, 7th Edition
7.
(a)
6 cos (2π60t – 9o)
-6 cos (2π60t + 9o)
Chapter Ten Solutions
10 March 2006
→ 6∠-9o
→ 6∠189o
-6 cos (2π60t + 9o) lags 6 cos (2π60t – 9o)
by 360 – 9 – 189 = 162o.
6∠-9o
6∠189o
cos (t - 100o)
-cos (t - 100o)
(b)
→
1 ∠ -100o
→ -1
∠ -100o = 1 ∠80o
-cos (t - 100o) lags cos (t - 100o) by 180o.
1∠80o
1∠-100o
(c)
-sin t
sin t
∠-90o = 1∠90o
1∠ -90o
→ -1
→
-sin t lags sin t by 180o.
1∠90o
1 ∠ -90o
(d)
7000 cos (t – π)
9 cos (t – 3.14o)
→ 7000
∠ -π = 7000 ∠ -180o
→9
∠ -3.14o
7000 cos (t – π) lags 9 cos (t – 3.14o)
by 180 – 3.14 = 176.9o.
7000 ∠ -180o
9 ∠ -3.14o
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Engineering Circuit Analysis, 7th Edition
8.
Chapter Ten Solutions
v(t) = V1 cos ωt - V2 sin ωt
10 March 2006
[1]
We assume this can be written as a single cosine such that
v(t) = Vm 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],
V1 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
V2
V sin φ
= m
= tan φ and φ = tan-1(V2/ V1)
V1
Vm cos φ
V12 + V22
V2
φ
V1
Next, we see from the above sketch that we may write Vm = V1/ cos φ or
Vm =
V1
V1
2
1
2
2
V +V
Thus, we can write v(t) = Vm cos (ωt + φ) =
=
V12 + V22
V12 + V22 cos [ ωt + tan-1(V2/ V1)].
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Engineering Circuit Analysis, 7th Edition
9.
Chapter Ten Solutions
10 March 2006
(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, 7th Edition
Chapter Ten Solutions
10 March 2006
1
10. (a)
Vrms
⎡ V2 T
⎤ 2
= ⎢ m ∫ cos 2 ωt dt ⎥
⎣T 0
⎦
1
⎡ V2 T
2πt ⎤ 2
= ⎢ m ∫ cos 2
dt ⎥
0
T
T
⎣
⎦
1
⎡ V2 T ⎛
4πt ⎞ ⎤ 2
= ⎢ m ∫ ⎜1 + cos
⎟ dt ⎥
T ⎠ ⎦
⎣ 2T 0 ⎝
1
⎡ V2 T
V2 T
4πt ⎤ 2
= ⎢ m ∫ dt + m ∫ cos
dt ⎥
0
0
2T
2T
T
⎣
⎦
1
2
⎡ V2
V2
4π ⎤
= ⎢ m T + m cos u 0 ⎥
8π
⎣ 2T
⎦
V
= m
2
*
(b)
Vm = 110 2 = 155.6 V , 115 2 = 162.6 V , 120 2 = 169.7 V
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Engineering Circuit Analysis, 7th Edition
11.
Chapter Ten Solutions
10 March 2006
We begin by defining a clockwise current i. Then, KVL yields
–2×10–3cos5t + 10i + vC = 0.
Since
i = iC = C
dvC
, we may rewrite our KVL equation as
dt
dv
30 C + vC = 2 × 10−3 cos 5t
dt
[1]
dvC
= −5 A sin(5t + θ ) ,
dt
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:
We anticipate a response of the form vC(t) = Acos(5t + θ). Since
(150 A) + A2 cos ⎛⎜ 5t + θta+
2
⎝
n −1
150 A ⎞
−3
⎟ = 2 × 10 cos 5t
A ⎠
Equating terms, we find that A = 13.33 μV and θ = –tan–1 150 = –89.62o. Thus,
vC(t) = 13.33 cos (5t – 89.62o) μV.
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Engineering Circuit Analysis, 7th Edition
12. KVL
Chapter Ten Solutions
10 March 2006
yields
–6cos400t + 100i + vL = 0.
di
di
Since
vL = L = 2 , we may rewrite our KVL equation as
dt
dt
di
2 + 100i = 6 cos 400t
dt
We anticipate a response of the form i(t) = Acos(400t + θ). Since
di
= −400 A sin(400t + θ ) ,
dt
we now may write Eq. [1] as
[1]
–800Asin(400t + θ) + 100Acos(400t + θ) = 6 cos400t.
Using a common trigonometric identity, we may combine the two terms on the left hand
side into a single cosine function:
(800 A) + (100 A) cos ⎛⎜ 400t + θt + an −1
2
2
⎝
800 A ⎞
⎟ = 6 cos 400t
100 A ⎠
Equating terms, we find that A = 7.442 mA and θ = –tan–1 8 = –82.88o. Thus,
i(t) = 7.442 cos (400t – 82.88o) mV, so vL = L
di
di
= 2 = 5.954cos (400t + 7.12o)
dt
dt
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Engineering Circuit Analysis, 7th Edition
13. 20cos500
Chapter Ten Solutions
10 March 2006
t V → 20∠0o V. 20 mH → j10 Ω.
Performing a quick source transformation, we replace the voltage source/20-Ω resistor
series combination with a 1∠0o A current source in parallel with a 20-Ω resistor.
20 || 60k = 19.99 Ω. By current division, then,
IL =
19.99
= 0.7427∠-21.81o A. Thus, iL(t) = 742.7 cos (500t – 21.81o) mA.
19.99 + 5 + j 10
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Engineering Circuit Analysis, 7th Edition
Chapter Ten Solutions
10 March 2006
14.
At x − x : R th = 80 20 = 16Ω
80
cos 500t
85
∴ voc = 4.8cos 500t V
voc = −0.4 (15 85)
4.8
10 ⎞
⎛
cos ⎜ 500t − tan −1 ⎟
15 ⎠
⎝
162 + 102
= 0.2544 cos (500t − 32.01°) A
(a)
iL =
(b)
vL = LiL′ = 0.02 × 0.02544 (−500)
sin (500t − 32.01°) = − 2.544sin (500t − 32.01°) V
∴ vL = 2.544 cos (500t + 57.99°) V, ix
= 31.80 cos (500t + 57.99°) mA
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Engineering Circuit Analysis, 7th Edition
Chapter Ten Solutions
10 March 2006
15.
100
800 ⎞
⎛
5
cos ⎜105 t −
⎟ = 0.10600 cos (10 t − 57.99°) A
2
2
500 ⎠
⎝
500 + 800
π
57.99°
pR = 0 when i = 0 ∴105 t −
π = , t = 25.83μs
180
2
(a)
i=
(b)
± vL = Li′ = 8 × 10−3 × 0.10600 (−105 ) sin (105 t − 57.99°)
∴ vL = −84.80sin (105 t − 57.99°)
∴ pL = vL i = −8.989sin (105 t − 57.99°)
cos (105 t − 57.99°) = −4.494 sin (2 × 165 t − 115.989°)
∴ pL = 0 when 2 × 105 t − 115.989° = 0°, 180°,
∴ t = 10.121 or 25.83μs
(c)
ps = vs iL = 10.600 cos105 t cos (105 t − 57.99°)
∴ ps = 0 when 105 t =
π
, t = 15.708μs and also t = 25.83μs
2
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Engineering Circuit Analysis, 7th Edition
16.
Chapter Ten Solutions
10 March 2006
vs = 3cos105 t V, is = 0.1cos105 t A
vs in series with 30Ω → 0.1cos105 t A 30Ω
Add, getting 0.2 cos105 t A 30 Ω
change to 6 cos 105 t V in series with 30Ω; 30Ω + 20Ω = 50Ω
∴ iL =
6
10 ⎞
⎛
cos ⎜105 t − tan −1 ⎟ = 0.11767 cos (105 t − 11.310°) A
50 ⎠
⎝
502 + 102
At 10
t=
μs, 105 t = 1∴ iL = 0.1167 cos (1rad − 11.310°) = 81.76mA
∴ vL = 0.11767 × 10 cos (1rad − 11.30° + 90°) = −0.8462V
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Engineering Circuit Analysis, 7th Edition
17. cos500
Chapter Ten Solutions
10 March 2006
t V → 1∠0o V. 0.3 mH → j0.15 Ω.
Performing a quick source transformation, we replace the voltage source-resistor series
combination with at 0.01∠0o A current source in parallel with a 100-Ω resistor. Current
division then leads to
100
= IL
( 0.01 + 0.2I L )
100 + j 0.15
1 + 20IL = (100 + j0.15) IL
Solving, we find that IL = 0.0125∠-0.1074o A,
so that iL(t) = 12.5cos(500t – 0.1074o) mA.
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Engineering Circuit Analysis, 7th Edition
18.
Chapter Ten Solutions
10 March 2006
vs1 = Vs 2 = 120 cos120 π t V
120
120
= 2A,
= 1A, 2 + 1 = 3A, 60 120 = 40Ω
60
12
3 × 40 = 120 V, ωL = 12π = 37.70Ω
37.70 ⎞
⎛
cos ⎜120 πt − tan −1
⎟
40 ⎠
⎝
402 + 37.702
= 2.183cos (120 π t − 43.30°) A
∴ iL =
(a)
(b)
120
1
∴ ωL = × 0.1× 2.1832 cos 2 (120π t − 43.30°)
2
= 0.2383cos 2 (120π t − 43.30°) J
ωL , av =
1
× 0.2383 = 0.11916 J
2
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Engineering Circuit Analysis, 7th Edition
Chapter Ten Solutions
10 March 2006
19. vs1 = 120 cos 400t V, vs 2 = 180 cos 200t V
Performing two quick source transformations,
120
180
= 2 A,
= 1.5 A, and noting that 60 120 = 40 Ω,
60
120
results in two current sources (with different frequencies) in parallel,
and also in parallel with a 40 Ω 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 having
magnitude 2 × 40 = 80 V. Applying Eqn. 10.4,
80
400(0.1)
cos (400t − tan −1
)
iL′ =
40
402 + 4002 (0.1) 2
Next , we open-circuit the 400 rad/s current source, and perform a
source transformation to obtain a voltage source with magnitude
1.5 × 40 = 60 V. Its contribution to the inductor current is
60
200(0.1)
cos (200t − tan −1
)A
iL′′ =
40
402 + 2002 (0.1) 2
so that iL = 1.414 cos (400t − 45°) + 1.342 cos (200t − 26.57°) A
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Engineering Circuit Analysis, 7th Edition
20.
R i = ∞, R o = 0, A = ∞, ideal,
iupper = −
v
Vm cos ω t
, ilower = out
R
R1
∴ ic1 = iupper + ilower =
Chapter Ten Solutions
=
C1 R
1
10 March 2006
L
R
i
′
(vout − Vm cos ω t ) = −C1vout
R1
L
′
vout
R
d ⎛v ⎞
For RL circuit, Vm cos ω t = vr + L ⎜ R ⎟
dt ⎝ R ⎠
L
∴ Vm cos ω t = vR + v′R
R
By comparison, vR = vout
′ = vout +
∴ Vm cos ω t = vout + R1C1vout
*
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Engineering Circuit Analysis, 7th Edition
Chapter Ten Solutions
21.
(a)
1
idt (ignore I.C)
C∫
1
∴−ω Vm sin ω t = Ri′ + i
C
(b)
Assume iA=
10 March 2006
Vm cos ω t = Ri +
cos (ω t + Φ )
∴−ω Vm sin ω t = −Rω A sin (ω t + Φ ) +
A
cos (ω t + Φ )
C
∴−ω Vm sin ω t = − Rω A cos Φ sin ω t-Rω A sin Φ cos ω t +
A
A
cos ω t cos Φ − sin ω t sin Φ
C
C
Equating terms on the left and right side,
A
1
[1] Rω A sin Φ = cos Φ∴ tan Φ =
so tΦ = an −1 (1 ω CR ) , and
ω CR
C
ω CR
[2] −ω Vm = −Rω A
∴ω Vm =
∴i =
1+ ω C R
2
2
2
−
A
1
C 1 + ω 2C2 R 2
ω CVm
A ⎡ R ω 2C2 + 1 ⎤ A
=
1 + ω 2 C2 R 2 ∴ A =
⎢
⎥
C ⎣ 1 + ω 2 C2 R 2 ⎦ C
1 + ω 2C2 R 2
2
ω CVm
⎛
1 ⎞
cos ⎜ ω t + tan −1
⎟
ω CR ⎠
1 + ω 2 C2 R 2
⎝
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Engineering Circuit Analysis, 7th Edition
22. (a)
Chapter Ten Solutions
10 March 2006
7 ∠ -90o = -j 7
(b) 3 + j + 7 ∠ -17o = 3 + j + 6.694 – j 2.047 = 9.694 – j 1.047
o
(c)
14ej15 = 14 ∠ 15o = 14 cos 15o + j 14 sin 15o = 13.52 + j 3. 623
(d)
1 ∠ 0o = 1
(e) –2 (1 + j 9) = -2 – j 18 = 18.11 ∠ - 96.34o
(f) 3 = 3 ∠ 0o
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Engineering Circuit Analysis, 7th Edition
23.
(b)
Chapter Ten Solutions
10 March 2006
(a) 3 + 15 ∠ -23o = 3 + 13.81 – j 5.861 = 16.81 – j 5.861
(j 12)(17 ∠ 180o) = (12 ∠ 90o)(17 ∠ 180o) = 204 ∠ 270o = –j 204
(c) 5 – 16(9 – j 5)/ (33 ∠ -9o) = 5 – (164 ∠ -29.05o)/ (33 ∠ -9o)
= 5 – 4.992 ∠ -20.05o = 5 – 4.689 – j 1.712 = 0.3109 + j 1.712
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Engineering Circuit Analysis, 7th Edition
24.
Chapter Ten Solutions
10 March 2006
(a) 5 ∠ 9o – 9 ∠ -17o = 4.938 + j 0.7822 – 8.607 + j 2.631 = -3.668 + j 3.414
= 5.011∠ 137.1o
(b) (8 – j 15)(4 + j 16) – j = 272 + j 68 – j = 272 + j 67 = 280.1 ∠ 13.84o
(c) (14 – j 9)/ (2 – j 8) + 5 ∠ -30o = (16.64 ∠-32.74o)/ (8.246 ∠ - 75.96o) + 4.330 – j 2.5
= 1.471 + j 1.382 + 4.330 – j 2.5 = 5.801 – j 1.118 = 5.908 ∠ -10.91o
(d)
17 ∠ -33o + 6 ∠-21o + j 3 = 14.26 – j 9.259 + 5.601 – j 2.150 + j 3
= 19.86 – j 8.409 = 21.57 ∠ -22.95o
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Engineering Circuit Analysis, 7th Edition
25.
Chapter Ten Solutions
10 March 2006
o
(a) ej14 + 9 ∠ 3o – (8 – j 6)/ j2 = 1 ∠ 14o + 9 ∠ 3o – (8 – j 6)/ (-1)
= 0.9703 + j 0.2419 + 8.988 + j 0.4710 + 8 – j 6 = 17.96 – j 5.287 = 18.72 ∠ -16.40o
(b)
=
o
(5 ∠ 30o)/ (2 ∠ -15o) + 2 e j5 / (2 – j 2)
2.5 ∠ 45o + (2 ∠ 5o)/ (2.828 ∠ -45o) = 1.768 + j 1.768 + 0.7072 ∠ 50o
= 1.768 + j 1.768 + 0.4546 + j 0.5418
= 2.224 + j 2.310 = 3.207 ∠ 46.09o
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Engineering Circuit Analysis, 7th Edition
Chapter Ten Solutions
10 March 2006
26.
(a)
5∠ − 110° = −1.7101 − j 4.698
(b)
6e j160° = −5.638 + j 2.052
(c)
(3 + j 6) (2∠50°) = −5.336 + j12.310
(d)
−100 − j 40 = 107.70∠ − 158.20°
(e)
2∠50° + 3∠ − 120° = 1.0873∠ − 101.37°
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Engineering Circuit Analysis, 7th Edition
Chapter Ten Solutions
10 March 2006
27.
(a)
40∠ − 50° − 18∠25° = 39.39∠ − 76.20°
(b)
3+
(c)
(2.1∠25°)3 = 9.261∠75° = 2.397 + j8.945+
(d)
0.7e j 0.3 = 0.7∠0.3rad = 0.6687 + j 0.2069
2 2 − j5
+
= 4.050− ∠ − 69.78°
j 1+ j2
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Engineering Circuit Analysis, 7th Edition
Chapter Ten Solutions
10 March 2006
28.
ic = 20e(40t +30°) A ∴ vc = 100∫ 20e j (40t +30°) dt
vc = − j 50e j (40t +30°) , iR = − j10e j (40t + 30°) A
∴ iL = (20 − j10) e j (40t +30°) , vL = j 40 × 0.08(20 − j10) e j (40t + 30°)
∴ vL = (32 + j 64) e j (40t +30°) V ∴ vs = (32 + j 64 − j 50) e j (40t + 30°)
∴ vs = 34.93e j (40t −53.63° ) V
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Engineering Circuit Analysis, 7th Edition
Chapter Ten Solutions
10 March 2006
29.
iL = 20e j (10t + 25°) A
vL = 0.2
d
[20e j (10t + 25°) ] = j 40e(10t = 25° )
dt
vR = 80e j (10t + 25°)
vs = (80 + j 40) e j (10t + 25°) , ic = 0.08(80 + j 40) j10e j (10t + 25°)
∴ ic = (−32 + j 64) e j (10t + 25°) ∴ is = (−12 + j 64) e j (10t + 25°)
∴ is = 65.12e j (10t +125.62°) A
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Engineering Circuit Analysis, 7th Edition
Chapter Ten Solutions
30.
80 cos(500t − 20°) V → 5 cos (500t + 12°) A
(a)
vs = 40 cos (500t + 10°) ∴ iout = 2.5cos (500t + 42°) A
(b)
vs = 40sin (500t + 10°) = 40 cos (500t − 80°)
10 March 2006
∴ iout = 2.5cos (500t − 48°) A
(c)
vs = 40e j (500t +10°) = 40 cos (500t + 10°)
+ j 40sin (500t + 10°) ∴ iout = 2.5e j (500t + 42°) A
(d)
vs = (50 + j 20) e j 500t = 53.85+ e j 21.80°+ j 500t
∴ iout = 3.366e j (500t +53.80° ) A
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Engineering Circuit Analysis, 7th Edition
Chapter Ten Solutions
10 March 2006
31.
(a)
12 sin (400t + 110°) A → 12∠20°A
(b)
−7 sin 800t − 3cos 800t → j 7 − 3
= −3 + j 7 = 7.616∠113.20° A
(c)
4 cos (200t − 30°) − 5cos (200t + 20°)
→ 4∠ − 30° − 5∠20° = 3.910∠ − 108.40° A
(d)
ω = 600, t = 5ms : 70∠30° V
→ 70 cos (600 × 5 × 10−3rad + 30°) = −64.95 V
(e)
ω = 600, t = 5ms : 60 + j 40 V = 72.11∠146.3°
→ 72.11cos (3rad + 146.31°) = 53.75 V
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Engineering Circuit Analysis, 7th Edition
32.
ω = 4000, t = 1ms
(a)
I x = 5∠ − 80° A
Chapter Ten Solutions
10 March 2006
∴ ix = 5cos (4rad − 80°) = −4.294 A
(b)
I x = −4 + j1.5 = 4.272∠159.44° A
∴ ix = 4.272 cos (4rad + 159.44°) = 3.750− A
(c)
vx (t ) = 50sin (250t − 40°)
= 50 cos (250t − 130°) → Vx = 50∠ − 130° V
(d)
vx = 20 cos108t − 30sin108t
→ 20 + j 30 = 36.06∠56.31° V
(e)
vx = 33cos (80t − 50°) + 41cos (80t − 75°)
→ 33∠ − 50° + 41∠ − 75° = 72.27∠ − 63.87° V
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Engineering Circuit Analysis, 7th Edition
Chapter Ten Solutions
10 March 2006
33. V1 = 10∠90° mV, ω = 500; V2 = 8∠90° mV,
ω = 1200, M by − 5, t = 0.5ms
vout = (−5) [10 cos (500 × 0.5 ×10−3rad + 90°)
+ 8cos (1.2 × 0.5 + 90°)]
= 50sin 0.25rad + 40sin 0.6rad = 34.96mV
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Engineering Circuit Analysis, 7th Edition
34.
(2.5
Chapter Ten Solutions
10 March 2006
Begin with the inductor:
∠40o) (j500) (20×10-3) = 25∠130o V across the inductor and the 25-Ω resistor.
The current through the 25-Ω resistor is then (25∠130o) / 25 = 1∠130o A.
The current through the unknown element is therefore 2.5∠40o + 1∠130o =
2.693
∠61.80o A; this is the same current through the 10-Ω resistor as well.
Armed with this information, KVL provides that
Vs = 10(26.93∠61.8o) + (25∠ -30o) + (25∠130o) = 35.47 ∠58.93o
and
so vs(t) = 35.47 cos (500t + 58.93o) V.
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Engineering Circuit Analysis, 7th Edition
Chapter Ten Solutions
10 March 2006
35.
ω = 5000 rad/s.
(a)
The inductor voltage = 48∠ 30o = jωL IL = j(5000)(1.2×10-3) IL
So IL = 8∠-60o and the total current flowing through the capacitor is
∠ 0o - IL = 9.165∠49.11o A and the voltage V1 across the capacitor is
10
V1 = (1/jωC)(9.165∠49.11o) = -j2 (9.165∠49.11o) = 18.33∠-40.89o V.
Thus,
v1(t) = 18.33 cos (5000t – 40.89o) V.
(b)
V2 = V1 + 5(9.165∠49.11o) + 60∠120o = 75.88∠79.48o V
∴ v2 (t ) = 75.88cos (5000t + 79.48°) V
(c)
V3 = V2 – 48∠30o = 75.88 ∠79.48o – 48∠30o = 57.70∠118.7o V
∴ v3 (t ) = 57.70 cos (5000t + 118.70°) V
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Engineering Circuit Analysis, 7th Edition
36.
Chapter Ten Solutions
10 March 2006
VR = 1∠0o V, Vseries = (1 + jω –j/ω)(1∠0o)
VR = 1 and Vseries = 1 + (ω - 1/ω )
2
We desire the frequency w at which Vseries = 2VR or Vseries = 2
2
Thus, we need to solve the equation 1 + (ω - 1/ω ) = 4
or ω 2 - 3 ω - 1 = 0
Solving, we find that ω = 2.189 rad/s.
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Engineering Circuit Analysis, 7th Edition
37.
Chapter Ten Solutions
10 March 2006
With an operating frequency of ω = 400 rad/s, the impedance of the 10-mH inductor is
jωL = j4 Ω, and the impedance of the 1-mF capacitor is –j/ωC = -j2.5 Ω.
∴ Vc = 2∠40° (− j 2.5) = 5∠ − 50° A
∴ I L = 3 − 2∠40° = 1.9513∠ − 41.211° A
∴ VL = 4 × 1.9513∠90° − 4.211° = 7.805+ ∠48.79° V
∴ Vx = VL − Vc = 7.805+ ∠48.79° − 5∠ − 50°
∴ Vx = 9.892∠78.76° V, vx = 9.892 cos (400t + 78.76°) V
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Engineering Circuit Analysis, 7th Edition
Chapter Ten Solutions
10 March 2006
38.
If I si = 2∠20° A, I s 2 = 3∠ − 30° A → Vout = 80∠10° V
I s1 = I s 2 = 4∠40° A → Vout = 90 − j 30 V
Now let I s1 = 2.5∠ − 60° A and I s 2 = 2.5∠60° A
Let Vout = AI s1 + BI s 2 ∴ 80∠10° = A(2∠20°) + B(3∠ − 30°)
and 90 − j 30 = (A + B) (4∠40°) ∴ A + B =
90 − j 30
= 12.415+ − j 20.21
4∠40°
80∠10°
3∠ − 30°
=A+B
∴ A = 40∠ − 10° − B(1.5∠ − 50°)
2∠20°
2∠20
∴12.415+ − j 20.21 − B = 40∠ − 10° − B(1.5∠ − 50°)
∴
∴12.415+ − j 20.21 − 40∠ − 10° = B (1 − 1.5∠ − 50°)
= B (1.1496∠ + 88.21°)
30.06∠ − 153.82°
= 26.148∠117.97°
1.1496∠ + 88.21°
∴ A = 12.415+ − j 20.21 − 10.800 + j 23.81
∴B =
= 49.842∠ − 60.32°
Vout = (49.842∠ − 60.32°) (2.5∠ − 60°)
+ (26.15∠117.97°) (2.5∠60°)
= 165.90∠ − 140.63°V
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Engineering Circuit Analysis, 7th Edition
39.
Chapter Ten Solutions
10 March 2006
We begin by noting that the series connection of capacitors can be replaced by a single
1
equivalent capacitance of value C =
= 545.5 F
μ . Noting ω = 2πf,
1+ 1 + 1
2
3
− j106
= − j 291.8 Ω .
2π ( 545.5 )
(a)
ω = 2π rad/s, therefore ZC = –j/ωC =
(b)
ω = 200π rad/s, therefore ZC = –j/ωC =
(c)
− j106
ω = 2000π rad/s, therefore ZC = –j/ωC =
= − j 291.8 mΩ .
2000π ( 545.5 )
(d)
ω = 2×109π rad/s, therefore ZC = –j/ωC =
− j106
= − j 2.918 Ω .
200π ( 545.5 )
− j106
= − j 291.8 nΩ .
2 × 109 π ( 545.5 )
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Engineering Circuit Analysis, 7th Edition
40.
Chapter Ten Solutions
10 March 2006
We begin by noting that the parallel connection of inductors can be replaced by a single
1
5
equivalent inductance of value L =
= nH . In terms of impedance, then, we have
1+ 5 6
Noting
ω = 2πf,
5
⎛
⎞
5 ⎜ jω × 10−9 ⎟
6
⎠
Z= ⎝
5
−9
5 + jω ×10
6
(a)
ω = 2π rad/s, therefore Z = j5.236×10–9 Ω (the real part is essentially zero).
(b)
ω = 2×103π rad/s, therefore Z = 5.483×10–12 + j5.236×10–6 Ω.
(c)
ω = 2×106π rad/s, therefore Z = 5.483×10–6 + j5.236×10–6 Ω.
(d)
ω = 2×109π rad/s, therefore Z = 2.615 + j2.497 Ω.
(e)
ω = 2×1012π rad/s, therefore Z = 5 + j4.775×10–3 Ω .
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Engineering Circuit Analysis, 7th Edition
41.
(a)
Chapter Ten Solutions
10 March 2006
ω = 800 : 2μF → − j 625, 0.6H → j 480
300(− j 625) 600( j 480)
+
300 − j 625 600 + j 480
= 478.0 + j175.65Ω
∴ Zin =
(b)
ω = 1600 : Zin =
+
300(− j 312.5)
300 − j 312.5
600( j 960)
= 587.6 + j119.79Ω
600 + j 960
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Engineering Circuit Analysis, 7th Edition
Chapter Ten Solutions
10 March 2006
42.
(a)
At ω = 100 rad/s, 2 mF → - j 5 Ω; 0.1 H → j10 Ω.
50 − j 50 10 − j10 2 − j1
(10 + j10) (− j 5) =
=
10 + j 5
2 + j1 2 − j1
= 2 − j 6 Ω∴ Zin = 20 + 2 − j 6 = 22 − j 6 Ω
(b)
SCa, b : 20 10 = 6.667, (6.667 − j 5) j10
50 + j 66.67 150 + j 200 30 + j 40 4 − j 3
=
=
×
6.667 + j 5
20 + j15
4 + j3 4 − j3
= Z in ∴ Z in (1.2 + j1.6) (4 − j 3) = 9.6 + j 2.8 Ω
=
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Engineering Circuit Analysis, 7th Edition
Chapter Ten Solutions
10 March 2006
43.
ω = 800 : 2μF → − j 625, 0.6H → j 480
300(− j 625) 600( j 480)
+
300 − j 625 600 + j 480
= 478.0 + j175.65Ω
∴ Zin =
− j 625
120
×
478.0 + j175.65 300 − j 625
or I = 0.2124∠ − 45.82° A
∴I =
Thus, i(t) = 212.4 cos (800t – 45.82o) mA.
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Engineering Circuit Analysis, 7th Edition
Chapter Ten Solutions
44.
(a)
3 Ω + 2mH : V = (3∠ − 20°) (3 + j 4) = 15∠33.13° V
(b)
3 Ω + 125μF : V = (3∠ − 20°) (3 − j 4) = 15∠ − 73.3° V
(c)
3 Ω 2mH 125μF : V = (3∠ − 20°) 3 = 9∠ − 20° V
(d)
same: ω
4 = 000 ∴ V = (3∠ − 20°) (3 + j8 − j 2)
∴ V = (3∠ − 20°) (3 + j 6) = 20.12∠43.43° V
10 March 2006
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Engineering Circuit Analysis, 7th Edition
Chapter Ten Solutions
10 March 2006
45.
(a)
C = 20μF, ω = 100
1
Zin =
=
1
0.005 − j 0.01 + j 0.002
1
1
+
+ j1000 × 20 × 10−6
200 j1000
1
∴ Zin =
= 196.12∠ − 11.310°Ω
0.005 + j 0.001
(b)
ω = 100 rad/s ∴ Z in =
Z in = 125 =
(c)
1
0.005 − j 0.001 + j100C
1
0.0052 + (100C − 0.001)
2
or
64 × 10-6 = 0.0052 + (100C − 0.001)
so
or
6.245 ×10-3 = 39 × 10-6 = 100C − 0.001
C = 72.45 μF
C = 20μF ∴ Zin =
2
1
1
= 100∠ =
−5
0.01∠
0.0005 − j 0.1/ ω + j 2 ×10 ω
2
2
0.1 ⎞
0.1 ⎞
⎛
⎛
−5
−5
∴ 0.0052 + ⎜ 2 ×10−5 ω −
⎟ = 0.0001, ⎜ 2 × 10 −
⎟ = 7.5 × 10
ω ⎠
ω ⎠
⎝
⎝
0.01
∴ 2 ×10−5 −
∓ 866.0 × 10−5 = 0 ∴ 2 × 10−5 ω2 ∓ 866.0 × 10−5 ω − 0.1 = 0
ω
use − sign: ω =
866.0 × 10−5 ± 7.5 × 10−5 + 8 × 10−6
= 444.3 and < 0
4 × 10−5
−866.0 × 10−5 ± 7.5 × 10−5 + 8 × 10−6
= 11.254 and <0
4 × 10−5
∴ω =11.254 and 444.3rad/s
use + sign: ω =
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Engineering Circuit Analysis, 7th Edition
Chapter Ten Solutions
10 March 2006
46.
(a)
1
1
1
1
= 25 =
∴
+ 2 = 0.0016
1
1
0.04 900 x
+
jx 30
∴ X = 45.23 Ω = 0.002ω, ω = 2261rad/s
(b)
1⎞
−3 0
⎛ 1
∠Yin = −25° = ∠ f ⎜ o − j ⎟ = tan −1
x⎠
x
⎝ 30
∴ x = 64.34 = 0.02ω, ω = 3217rad/s
(c)
Zin =
30( j 0.02ω) 30 − j 0.092ω 0.012ω2 + j18ω
×
=
30 + j 0.02ω 30 − j 0.02ω 900 + 0.0004ω2
∴ 0.012ω2 = 25 (900 + 0.0004ω2 )
∴ 0.012ω2 = 0.01ω2 + 22,500, ω = 3354rad/s
(d)
18ω = 10 (900 + 0.0004ω2 ), 0.004ω2 − 18ω + 9000 = 0,
ω2 − 4500ω + 2.25 ×106 = 0
ω=
4500 ± 20.25 ×106 − 9 ×106 4500 ± 3354
=
= 572.9, 3927rad/s
2
2
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Engineering Circuit Analysis, 7th Edition
47.
Chapter Ten Solutions
10 March 2006
With an operating frequency of ω = 400 rad/s, the impedance of the 10-mH inductor is
jωL = j4 Ω, and the impedance of the 1-mF capacitor is –j/ωC = -j2.5 Ω.
∴ Vc = 2∠40° (− j 2.5) = 5∠ − 50° A
∴ I L = 3 − 2∠40° = 1.9513∠ − 41.211° A
IL =
2∠40° (R 2 − j 2.5)
R1 + j 4
2∠40° (R 2 − j 2.5)
1.9513∠ − 41.21°
= 1.0250∠81.21° ( R2 − j 2.5)
∴ R1 + j 4 =
= R 2 (1.0250∠81.21°) + 2.562∠ − 8.789°
= 0.15662R 2 + j1.0130 R 2 + 2.532 − j 0.3915
∴ R1 = 2.532 + 0.15662R 2 , 4 = 1.0130R 2 − 0.395−
∴ R 2 = 4.335+ Ω, R1 = 3.211 Ω
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Engineering Circuit Analysis, 7th Edition
Chapter Ten Solutions
48.
ω = 1200 rad/s.
(a)
ω = 1200
− j × (200 + j80) (80 x − j 200 x)[200 + j ( x − 80)]
=
Zin =
200 + j (80 − x)
40, 000 + 6400 − 160 x + x 2
10 March 2006
X in = 0 ∴−40, 000 x + 80 x 2 − 6400 x = 0
∴ 46, 400 = 80 x, x = 580 Ω =
(b)
Zin =
∴
1
∴ C = 1.437 μ
F
1200C
80X − j 200X
Zin = 100
200 + j (80 − X)
6400X 2 + 40, 000X 2
= 10, 000
40, 000 + 6400 − 160X + X 2
∴ 0.64X 2 + 4X 2 = X 2 − 160X + 46, 400
∴ 3.64X 2 + 160 X − 46, 400 = 0,
−160 ± 25, 600 + 675, 600 −160 ± 837.4
=
7.28
7.28
1
∴ C = 8.956μF
∴ X = 93.05− (> 0) =
1200C
X=
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Engineering Circuit Analysis, 7th Edition
49. At
Chapter Ten Solutions
10 March 2006
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 jωL = j16 Ω.
(a) Term
inals ab open circuited: Zin = 8 + j16 || (2 – j2) = 10.56 – j1.92 Ω
(b) Term
inals ab short-circuited: Zin = 8 + j16 || 2 = 9.969 + j0.2462 Ω
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Engineering Circuit Analysis, 7th Edition
50.
2
3.2
1
1
20
f = 1 MHz, ω = 2πf = 6.283 Mrad/s
μF
→ -j0.07958 Ω
μH
→ j20.11 Ω
μF
→ -j0.1592 Ω
μH
→ j6.283 Ω
μH
→ j125.7 Ω
200 pF
→ -j795.8 Ω
Chapter Ten Solutions
10 March 2006
= Z1
= Z2
= Z3
= Z4
= Z5
= Z6
The three impedances at the upper right, Z3, 700 kΩ, and Z3 reduce to –j0.01592 Ω
Then we form Z2 in series with Zeq: Z2 + Zeq = j20.09 Ω.
Next we see 106 || (Z2 + Zeq) = j20.09 Ω.
Finally,
Zin = Z1 + Z4 + j20.09 = j26.29 Ω.
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Engineering Circuit Analysis, 7th Edition
51.
f
Chapter Ten Solutions
10 March 2006
As in any true design problem, there is more than one possible solution. Model answers
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 suffice, provided they satisfy the C = 1/L requirement).
(c)
Z = 7∠80o Ω. R = Re{Z} = 7cos80o= 1.216 Ω, and X = Im{Z} = 7sin80o = 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 simplest solution.
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Engineering Circuit Analysis, 7th Edition
52.
f
Chapter Ten Solutions
10 March 2006
As in any true design problem, there is more than one possible solution. Model answers
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 resistor in series with 17.39 H in series with 1 F.
(b) To obtain a purely real impedance, the reactance of the inductor must cancel the
reactance of the capacitor, In a series string, this is obtained by meeting the criterion ωL
1/ωC, or L = 1/ω2C = 1/100C.
=
Select a 5 MΩ resistor in series with 1 F in series with 100 mH.
If Z = 80∠–22o Ω is constructed using a series combination of a single resistor R and
single capacitor C, R = Re{Z} = 80cos(–22o) = 74.17 Ω. X = –1/ωC = Im{Z} =
o
80sin(–22
) = –29.97 Ω. Thus, C = 667.3 μF.
(c)
(d) The simplest solution, independent of frequency, is a single 300 Ω resistor.
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Engineering Circuit Analysis, 7th Edition
Chapter Ten Solutions
10 March 2006
53.
Note that we may replace the three capacitors in parallel with a single capacitor having
value
10 −3 + 2 × 10−3 + 4 × 10−3 = 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×103 rad/s. Y = j4π×103C = j879.6 S
(d)
ω = 4π×1011 rad/s. Y = j4π×1011C = j8.796×109 S
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Engineering Circuit Analysis, 7th Edition
54.
Chapter Ten Solutions
10 March 2006
(a) Susceptance is 0
(b) B = ωC = 100 S
(c)
Z = 1 + j100 Ω, so Y =
1
1 − j100
=G + jB , where B = –9.999 mS.
=
1 + j100
1+1002
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Engineering Circuit Analysis, 7th Edition
Chapter Ten Solutions
10 March 2006
55.
2 H → j 2, 1F → − j1 Let I1∈ = ∠0° A
∴ VL = j 2V ∴ I c = I in + 0.5 VL = 1 + j1
∴ Vin = j 2 + (1 + j1) (− j1) = 1 + j1
∴ Vin =
1∠0°
1 1 − j1
=
= 0.5 − j 0.5
1 + j1 1 − j1
Vin
Now 0.5 S → 2 Ω, − j 0.5 S =
1
→2H
j2
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Engineering Circuit Analysis, 7th Edition
56.
(a)
Chapter Ten Solutions
10 March 2006
ω = 500, ZinRLC = 5 + j10 − j1 = 5 + j 9
1
5 − j9
9
=
∴ Yc =
= 500C
5 + j9
106
106
9
∴C =
= 169.8 μ
F
53, 000
∴ YinRLC =
106
= 21.2 Ω
5
(b)
R in , ab =
(c)
ω = 1000 rad/s ∴
Z S = 5 + j 2 − j 5 = 5 − j 3 = 5.831∠ − 30.96o Ω
and Z C = − j 58.89 Ω.
Thus,
Yin ,ab =
1
1
+
= 0.1808∠35.58o S
Z S ZC
= 147.1 + j105.2 mS
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Engineering Circuit Analysis, 7th Edition
57.
(a)
Chapter Ten Solutions
10 March 2006
j 0.1ω
100 + j 0.001ω
50, 000 + j 0.6ω 100 − j 0.001ω
∴ Zin =
×
100 + j 0.001ω 100 − j 0.001ω
R in = 550 Ω : Zin = 500 +
5 × 106 + 0.0006ω2 + j (60ω − 50ω)
104 + 10−6 ω2
5 × 106 + 0.006ω2
∴ R in =
= 550 ∴ 5.5 ×106
4
−6 2
10 + 10 ω
−4 2
+ 5.5 × 10 ω = 5 ×106 × 10−4 ω2
∴ Zin =
∴ 0.5 × 10−4 ω2 = 0.5 ×106 , ω2 = 1010 , ω = 105 rad/s
(b)
10ω
= 0.5 × 106 + 0.5 × 10−4 ω2 − 10ω
−6 2
10 + 10 ω
2
= 0, ω − 2 × 105 ω + 1010 = 0
X in = 50 Ω =
∴ω =
(c)
4
2 × 105 ± 4 × 1010 − 4 × 1010
= 105 ∴ω = 105 rad/s
2
G in = 1.8 × 10−3 : Yin =
100 + j 0.001ω 50, 000 − j 0.6ω
×
50, 000 + j 0.6ω 50, 000 − j 0.6ω
5 × 106 + 6 × 10−4 ω2 + j (50ω − 6ω)
25 × 108 + 0.36ω2
5 × 106 + 6 × 10−4 ω2
∴1.8 × 103 =
25 × 108 + 0.36ω2
∴ 5 × 106 + 6 ×10−4 ω2 = 4.5 × 106 + 648 × 10−6 ω2
=
∴ 0.5 ×106 = 48 × 10−6 ω2 ∴ω = 102.06 krad/s
(d)
−10ω
25 ×108 + 0.36ω2
∴10ω = 37.5 × 104 + 54 × 10−6 ω2
Bin = 1.5 × 10−4 =
∴ 54 × 10−6 ω2 − 10ω + 37.5 × 104 = 0,
ω = 10 ±
100 − 81
= 52.23 and 133.95 krad/s
108 × 10−6
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Engineering Circuit Analysis, 7th Edition
Chapter Ten Solutions
10 March 2006
58.
I1
0.1∠30°
=
= 20∠ − 23.13°∴ V1 = 20 V
Y1 (3 + j 4)10−3
(a)
V1 =
(b)
V2 = V1 ∴ V2 = 20V
(c)
I 2 = Y2 V2 = (5 + j 2)10−3 × 20∠ − 23.13° = 0.10770∠ − 1.3286° A
∴ I3 = I1 + I 2 = 0.1∠30° + 0.10770∠ − 1.3286° = 0.2∠13.740° A
∴ V3 =
(d)
I3 0.2∠13.740°
=
= 44.72∠77.18° V ∴ V3 = 44.72V
Y3 (2 − j 4)10−3
Vin = V1 + V3 + 20∠ − 23.13° + 44.72∠77.18° = 45.60∠51.62°
∴ Vin = 45.60V
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Engineering Circuit Analysis, 7th Edition
Chapter Ten Solutions
10 March 2006
59.
(a)
50 μ F → − j 20 Ω∴ Yin = 0.1 + j 0.05
1
1000
1
∴ R1 − j
=
= 8 − j4
1000
C
0.1 + j 0.05
R1 − j
C
1
∴ R 1 = 8 Ω and C1 =
= 250 μ F
4ω
Yin =
(b)
ω = 2000 : 50μ F → − j10 Ω ∴ Yin = 0.1 + j 0.1 =
∴ R1 − j
1
R1 − j
500
C1
500
= 5 − j 5 ∴ R1 = 5 Ω, C1 = 100μ F
C1
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Engineering Circuit Analysis, 7th Edition
Chapter Ten Solutions
10 March 2006
60.
(a)
10 10 + jω
=
jω
jω
jω
10 − jω
∴ Yin
×
10 + jω 10 − jω
Zin = 1 +
ω 2 + j10ω
ω 2 + 100
10ω
ω2
G in = 2
, Bin = 2
ω + 100
ω + 100
∴ Yin =
ω
Gin
Bin
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
B
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Engineering Circuit Analysis, 7th Edition
Chapter Ten Solutions
10 March 2006
61.
f
As in any true design problem , there is m ore than one possible solution. Model answers
ollow:
(a)
Y = 1 – j4 S at ω = 1 rad/s.
Construct this using a 1 S conductance in
= 4, or L = 250 mH.
(b)
1/
Y = 200 m S (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 –
ωL = 0. Arbitrarily selecting L = 1 H, we find that C = 1 F.
H
One solution therefore is a 5 Ω resistor in parallel with a 1 F capacitor in parallel with a 1
inductor.
(c)
mF.
parallel with an induc tance L such that 1/ ωL
Y = 7∠80o μS = G + jB at ω = 100 rad/s. G = Re{ Y} = 7cos80 o = 1.216 S (an 822.7
mΩ resistor). B = Im {Y} = 7sin8 0o = 6.894 S. W e m ay rea lize th is suscep tance by
placing a capacito r C in parallel with the resis tor such th at jωC = j6.894, or C = 68.94
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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Engineering Circuit Analysis, 7th Edition
62.
f
(a)
Chapter Ten Solutions
10 March 2006
As in any true design problem , there is m ore than one possible solution. Model answers
ollow:
Y = 1 – j4 pS at ω = 30 rad/s.
Construct this using a 1 pS conductance (a 1 T Ω resistor) in parall el with an inductor L
such that –j4×10–12 = –j/ωL, or L = 8.333 GH.
200
(b) We may realise a pu rely real admittance of 5 μS by pla cing a 5 μS conductance (a
kΩ resistor) in parallel with a capa citor C and inducta nce 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 20 0 k Ω resistor in parallel w ith a 2 H inductor and a
1.594
μF capacitor.
(c)
Y = 4 ∠–10o nS = G + jB at ω = 50 rad/s. G = Re{ Y} = 4×10 –9cos(–10o) = 3.939 nS
(an 253.9 MΩ resistor). B = Im{Y} = 4×10 –9sin(–10o) = –6.946×10 –10 S. We may r ealize
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 resistor).
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Engineering Circuit Analysis, 7th Edition
63.
Chapter Ten Solutions
10 March 2006
v V − V2 v1 − V2
, − j 75 = 5V1 + j 3V1 − j 3V2 − j 5V1 + j 5V2
− j5 = 1 + 1
+
3
− j5
j3
∴ (5 − j 2) V1 + j 2V2 = − j 75 (1)
v2 − V1 V2 − V1 V2
+
+
= 10
6
− j5
j3
− j10V2 + j10V1 + j 6V2 − j 6V1 + 5V2 = 300 ∴ j 4V1 + (5 − j 4) V2 = 300
(2)
5 − j 2 − j 75
j4
300
1500 − j 600 − 300 1200 − j 600
= 34.36∠23.63° V
=
∴ V2 =
=
j2
5 − j2
25 − j 30
17 − j30 + 8
j4
5 − j4
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Engineering Circuit Analysis, 7th Edition
Chapter Ten Solutions
10 March 2006
64.
j 3I B − j 5(I B − I D ) = 0 ∴−2I B + j 5I D = 0
3(I D + j 5) − j 5(I D − I B ) + 6 (I D + 10) = 0
∴ j 5I B + (9 − j 5) I D = −60 − j15
0
j5
−60 − j15 9 − j 5 −75 + j 300
=
IB =
− j2
j5
15 − j18
j5 9 − j5
= 13.198∠154.23° A
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Engineering Circuit Analysis, 7th Edition
65.
Chapter Ten Solutions
10 March 2006
vs1 = 20 cos1000t V, vs 2 = 20sin1000t V
∴ Vs1 = 20∠0° V, Vs 2 = − j 20V
0.01H → j10 Ω, 0.1mF → − j10 Ω
v − 20 vx vx + j 20
∴ x
+ +
= 0, 0.04vx + j 2 − 2 = 0,
− j10
j10
25
Vx = 25(2 − j 2) = 70.71∠ − 45° V
∴ vx (t ) = 70.71cos(1000t − 45°) V
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Engineering Circuit Analysis, 7th Edition
66.
(a)
Chapter Ten Solutions
10 March 2006
Assume V3 = 1V ∴ V2 = 1 − j 0.5V, I 2 = 1 − j 0.5 mA
∴ V1 = 1 − j 0.5 + (2 − j 0.5) (− j 0.5) = 0.75 − j1.5V
∴ I1 = 0.75 − j1.5 mA, ∴ Iin = 0.75 − j1.5 + 2 − j 0.5 = 2.75 − j 2 mA
∴ Vin = 0.75 − j1.5 − j1.5 + (2.75 − j 2) (− j 0.5)
= −0.25 − j 2.875 V ∴ V3 =
(b)
100
= 34.65+ ∠94.97°V
− j 0.25 − j 2.875
− j 0.5 → − jx Assume 1V3 = V ∴ I 3 = 1A,
V2 = 1 − jX, I 2 = 1 − jX, → I12 = 2 − jX
∴ V1 = 1 − jX + (2 − jX) (− jX) = 1 − X 2 − j 3X, I1 = 1 − X 2 − j 3X, I in = 3 − X 2 − j 4 X
∴ Vin = 1 − X 2 − j 3X − 4X 2 + jX 3 − j3X = 1 − 5X 2 + j (X 3 − 6X) ∴ X 3 − 6X = 0
∴ X 2 = 6, X = 6, Z c = − j 2.449 kΩ
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Engineering Circuit Analysis, 7th Edition
67.
m
Chapter Ten Solutions
10 March 2006
Define three clockwise mesh currents i1, i2, i3 with i1 in the left mesh, i2 in the top right
esh, and i3 in the bottom right mesh.
Mesh 1: -10∠0o + (1 + 1 – j0.25)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
−1 10
2 − j 0.25
−1
2 + j4 0
−1
j 0.25
0
Ix =
2 − j 0.25
−1
j 0.25
−1
2 + j4
−1
j 0.25
−1
2 − j 0.25
10 (1 + 1 − j 0.5)
j 0.25(2 − j 0.5) + (−2 + j 0.25 + j 0.25) + (2 − j 0.25) (4 + 1 − j 0.5 + j8 − 1)
20 − j 5
=
∴ I x = 1.217∠ − 75.96° A, ix (t ) = 1.2127 cos (100t − 75.96°) A
8 + j15
∴Ix =
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Engineering Circuit Analysis, 7th Edition
68.
Chapter Ten Solutions
10 March 2006
V1 − 10 − j 0.25V1 + j 0.25Vx + V1 − V2 = 0
∴ (2 − j 0.25) V1 − V2 + j 0.25 Vx = 10
V2 − V1 + V2 − Vx + j 4V2 = 0
− V1 + (2 + j 4) V2 − Vx = 0
− j 0.25Vx + j 0.25V1 + Vx + Vx − V2
∴ j 0.25V1 − V2 + (2 − j 0.25) Vx = 0
−1 10
2 − j 0.25
−1
2 + j4 0
−1
j 0.25
0
Vx =
j 0.25
− j 0.25
−1
−1
−1
2 + j4
j 0.25
−1
2 − j 0.25
10 (1 + 1 − j 0.5)
j 0.25(2 − j 0.5) + (−2 + j 0.25 + j 0.25) + (2 − j 0.25) (4 + 1 − j 0.5 + j8 − 1)
20 − j 5
=
= 1.2127∠ − 75.96° V
8 + j15
∴ vx = 1.2127 cos(100t − 75.96°) V
=
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Engineering Circuit Analysis, 7th Edition
69.
(a)
Chapter Ten Solutions
10 March 2006
R1 = ∞, R o = 0, A = −Vo / Vi >> 0
I=
V1 + AVi
= jω C1 (Vs − Vi )
Rf
∴ Vi (1 + A + jω C1R f ) = jω C1R f Vs
Vo
(1 + A + jω C1R f ) = jω C1R f Vs
A
jω C1R f A
V
V
∴ o =−
As A → ∞, o → − jω C1R f
Vs
1 + A + jω C1R f
Vs
Vo = − AVi ∴−
(b)
R f Cf =
I=
1
jω C f +
1
Rf
=
Rf
1 + jω C f R f
(V1 + AVi )
(1 + jω C f R f ) = (Vs − Vi ) jω C1 , Vo = − AVi
Rf
∴ Vi (1 + A) (1 + jω C f R f ) = Vs jω C1R f − jω C1R f Vi ,
Vi [(1 + A) (1 + jω C f R f ) + jω C1R f ] = jω C1R f Vs
Vo
[(1 + A) (1 + jω C f R f ) + jω C1R f ] = jω C1R f Vs
A
− jω C1R f
− jω C1R f A
V
V
∴ o =
As A → ∞, o →
Vs
1 + jω C f R f
Vs (1 + A) (1 + jω C f R f ) + jω C1R f
∴−
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Engineering Circuit Analysis, 7th Edition
70.
Chapter Ten Solutions
10 March 2006
Define the nodal voltage v1(t) at the junction between the two dependent sources.
The voltage source may be replaced by a 3∠-3o V source, the 600-μF capacitor by a
–j/ 0.6 Ω impedance, the 500-μF capacitor by a –j2 Ω impedance, and the inductor by a
j2 Ω impedance.
V1 - 3∠ - 3o
(V - V )
5V2 + 3V2 =
+ 1 2 [1]
100 − j / 0.6
- j2
-5V2 =
( V2 − V1 ) +
− j2
V2
j2
[2]
Solving, we find that V2 = 9.81 ∠ – 13.36o mV.
Converting back to the time domain,
v2(t) = 9.81 cos (103t – 13.36o) mV
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Engineering Circuit Analysis, 7th Edition
71.
m
and
Mesh
Chapter Ten Solutions
10 March 2006
Define three clockwise mesh currents: i1(t) in the left-most mesh, i2(t) in the bottom right
esh, and i3(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∠0o mA source,
and the voltage source is replaced with a 5∠0o V source.
Around the 1, 2 supermesh: (1 + j20) I1 + (13 – j1.351) I2 – 5 I3 = 0
–I1 + I2 = 2×10–3
3:
5
∠0o – 5 I2 + (5 – j6.667) I3 = 0
Solving, we find that I1 = 148.0∠179.6o mA. Converting to the time domain,
i1(t) = 148.0cos (104t + 179.6o) μA
Thus, P1Ω = [i1(1 ms)]2 • 1
=
(16.15×10–3)(1) W = 16.15 mW.
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Engineering Circuit Analysis, 7th Edition
72.
Chapter Ten Solutions
10 March 2006
We define an additional clockwise mesh current i4(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 impedance, and the 1000 μF capacitor is replaced by a –j/ 2
Ω impedance. We replace the left voltage source with a a 6 ∠ -13o V source, and the
right voltage source with a 6 ∠ 0o V source.
= 6 ∠ –13o [1]
(1 – j/ 0.0015) I1 + j/0.0015I2 – I3
(0.005 + j/ 0.0015) I1 + (j0.004 – j/0.0015) I2
–I1
+ (1 – j500) I3
– j0.004 I4 = 0
+
[2]
j500 I4 = –6 ∠ 0o [3]
–j0.004 I2 + j500I3 + (j0.004 – j500) I4 = 0
[4]
Solving, we find that
I1 = 2.002∠ –6.613o mA, I2 = 2.038 ∠ –6.500o mA, and I3 = 5.998 ∠ 179.8o A.
Converting to the time domain,
i1(t) = 1.44 cos (2t – 6.613o) mA
i2(t) = 2.038 cos (2t – 6.500o) mA
i3(t) = 5.998 cos (2t + 179.8o) A
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Engineering Circuit Analysis, 7th Edition
73.
Defi
Chapter Ten Solutions
We replace the voltage source with a 115 2 ∠0o V source, the capacitor with a
–j/ 2πC1 Ω impedance, and the inductor with a j0.03142 Ω impedance.
ne Z such that Z-1 = 2πC1 - j/0.03142 + 1/20
By voltage division, we can write that 6.014 ∠85.76o = 115 2
Thus,
10 March 2006
Z
Z + 20
Z = 0.7411 ∠ 87.88o 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, 7th Edition
74.
Chapter Ten Solutions
10 March 2006
Defining a clockwise mesh current i1(t), we replace the voltage source with a
115 2 ∠0o V source, the inductor with a j2πL Ω impedance, and the capacitor with a
–j1.592 Ω impedance.
Ohm’s law then yields I1 =
Thus, 20 =
115 2
= 8.132∠0o
20 + j (2πL − 1.592 )
202 + (2πL − 1.592)
2
and we find that L = 253.4 mH.
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Engineering Circuit Analysis, 7th Edition
75.
Chapter Ten Solutions
10 March 2006
(a) By nodal analysis:
0 = (Vπ – 1)/ Rs + Vπ / RB + Vπ / rπ + jωCπ Vπ + (Vπ – Vout) jωCμ [1]
-gmVπ = (Vout – Vπ) jωCμ + Vout / RC + Vout / RL
[2]
Simplify and collect terms:
⎡⎛ 1
⎤
1
1⎞
1
[1]
+
+ ⎟⎟ + jω (Cπ + C μ )⎥ Vπ - jωC μ Vout =
⎢⎜⎜
RS
⎣⎢⎝ R S R B rπ ⎠
⎦⎥
(-gm + jωCμ) Vπ - (jωCμ + 1/RC + 1/RL) Vout = 0
Defi
ne
1
′
RS
1
1
1
+
+
R S R B rπ
′
and R L = RC || RL
⎛
C + Cπ C μ ⎞⎟
-1
+ ω 2 (2C2μ + C μ Cπ ) - jω ⎜ g mC μ + μ
+
′ ′
′
′
⎜
RS R L
R
R S ⎟⎠
L
⎝
Then
Δ =
And
Vout =
Therefore,
=
[2]
g m R S − jω C μ R S
⎛
C μ + Cπ C μ ⎞⎟
-1
+ ω 2 2C2μ + C μ Cπ - jω ⎜ g m C μ +
+
′ ′
′
′
⎜
RS R L
RL
R S ⎟⎠
⎝
(
)
⎛
⎛
⎞⎞
⎜ − ω ⎜ g C + C μ + Cπ + C μ ⎟ ⎟
⎜
′
′ ⎟
⎜ m μ
− jωCμ ⎞
RL
R S ⎟⎠ ⎟
−1 ⎛
⎝
−1 ⎜
⎟
tan
ang(Vout) = tan ⎜⎜
2 ⎟
⎜ -1
⎟
2
2
⎝ g m RS ⎠
⎜ ′ ′ + ω (2C μ + C μ Cπ ) ⎟
⎜ RS R L
⎟
⎝
⎠
(b)
(c) The output is ~180o out of phase with the input for f < 105 Hz; only for f = 0 is it
exactly 180o out of phase with the input.
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Engineering Circuit Analysis, 7th Edition
76.
OC : −
Chapter Ten Solutions
10 March 2006
Vx 100 − Vx
+
− 0.02Vx = 0
− j10
20
j10 = (0.05 + j 0.1 + 0.02) Vx , Vx =
j10
0.07 + j 0.1
∴ Vx = 67.11 + j 46.98
∴ Vab ,oc = 100 − Vx = 32.89 − j 46.98 = 57.35∠ − 55.01° V
SC :Vx = 100 ∴↓ I SC = 0.02 ×100 +
∴ Zth =
100
= 7A
20
57.35∠ − 55.01°
= 4.698 − j 6.711Ω
7
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Engineering Circuit Analysis, 7th Edition
77.
Chapter Ten Solutions
10 March 2006
Let I in = 1∠0. Then VL = j 2ω I in = j 2ω ∴ 0.5VL = jω
∴ Vin = (1 + jω )
1
+ j 2ω
jω
1
+ j 2ω
jω
ω
V
1
∴ Zin = in = 1 +
+ j 2ω so Yin =
1
jω
ω + j (2ω 2 − 1)
At ω = 1, Zin = 1 − j1 + j 2 = 1 + j
= 1+
∴ Yin =
1
= 0.5 − j 0.5
1 + j1
R = 1/0.5 = 2 Ω
and
L = 1/0.5 = 2 H.
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Engineering Circuit Analysis, 7th Edition
Chapter Ten Solutions
10 March 2006
78.
(a)
(b) I
−15
(1 − j1)1 2 + j1 3 − j1
×
=
∴ V1 =
× 0.6 − j 0.2
2 − j1 2 + j1
5
j 2 + 0.6 − j 0.2
∴ V1 = 5∠90°∴ v1 (t ) = 5cos (1000t + 90°) V
Vs :
s:
j2 1− j2
= 0.8 + j 0.4 ∴ V1
1+ j2 1− j2
−10 + j 20
0.8 + j 0.4
= j 25
=
= 11.785+ ∠135° V
1 − j1 + 0.8 + j 0.4 1.8 − j 0.6
j2 1=
so
v1(t) = 11.79 cos (1000t + 135o) V.
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Engineering Circuit Analysis, 7th Edition
Chapter Ten Solutions
10 March 2006
79.
OC :VL = 0 ∴ Vab ,oc = 1∠0° V
SC : ↓ I N ∴ VL = j 2I N ∴1∠0° = − j1[0.25( j 2I N ) + I N ] + j 2I N
∴1 = (0.5 − j + j 2) I N = (0.5 + j1) I N
I
1
= 0.4 − j 0.8 ∴ YN = N = 0.4 − j 0.8
0.5 + j1
1∠0°
1
1
1
1
∴RN =
= 2.5 Ω,
=
= − j 0.8, L N =
= 1.25H
0.4
0.8
jω L N
jL N
∴ IN =
I N = 0.4 − j 0.8 = 0.8944∠ − 63.43° A
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Engineering Circuit Analysis, 7th Edition
80.
=
Chapter Ten Solutions
10 March 2006
To solve this problem , we e mploy super position in order to separate sources having
different frequencies. First cons idering the sources operating at w = 200 rad/s, we opencircuit the 100 rad/s current source. This leads to VL′ = (j)(2∠0) = j2 V. Therefore, vL′ ( t )
2cos(200t + 90o) V. For the 100 rad/s source, we find
j
VL′′ = (1∠0 ) , vL′′ = 0.5cos (100t + 90°) V
2
∴ vL ( t ) = 2 cos (200t + 90°) + 0.5cos (100t + 90°) V
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Engineering Circuit Analysis, 7th Edition
Chapter Ten Solutions
10 March 2006
81.
j100
j100 − j 300
− j 300
= −50∠0° V Right: Vab = j100
= j150 V
− j 300 + j100
∴ Vth = −50 + j150 = 158.11∠108.43° V
Use superposition. Left: Vab = 100
Zth = j100 − j 300 =
30, 000
= j150 Ω
− j 200
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Engineering Circuit Analysis, 7th Edition
82.
Then
Chapter Ten Solutions
10 March 2006
This problem is easily solved if we first perform two source transformations to yield a
circuit containing only voltage sources and impedances:
5∠17o + 0.240∠-90o − 2.920∠-45o
73 + 10 + j13 − j 4
= (4.264∠ 50.42o)/ (83.49 ∠ 6.189o) = 51.07 ∠ 44.23 mA
I =
Converting back to the time domain, we find that
i(t) = 51.07 cos (103t + 43.23o) mA
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Engineering Circuit Analysis, 7th Edition
Chapter Ten Solutions
10 March 2006
83.
(a) There are a number of possible approaches: Thévenizing everything to the left of the
capacitor is one of them.
VTH = 6(j2)/ (5 + j2) = 2.228 ∠ 68.2o V
ZTH = 5 || j2 = j10/ (5 + j2) = 1.857 ∠ 68.2o Ω
Then, by simple voltage division, we find that
− j /3
VC = (2.228 ∠ 68.2o)
1.857∠68.2o - j / 3 + j 7
= 88.21 ∠-107.1o mV
Converting back to the time domain, vC(t) = 88.21 cos (t – 107.1o) mV.
(b)
PSpice verification.
Running an ac sweep at the
frequency f = 1/2π = 0.1592 Hz,
we obtain a phasor magnitude of
88.23 mV, and a phasor angle of
–107.1o, 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, 7th Edition
84.
Chapter Ten Solutions
10 March 2006
(a) Performing nodal analysis on the circuit,
Node 1:
Node
2:
1 = V1/ 5 + V1/ (-j10) + (V1 – V2)/ (-j5) + (V1 – V2)/ j10
j0.5 = V2/ 10 + (V2 – V1)/ (-j5) + (V2 – V1)/ j10
[1]
[2]
Simplifying and collecting terms,
(0.2 + j0.2) V1 – j0.1 V2 = 1
[1]
-j V1 + (1 + j) V2 = j5
[2]
Solving, we find that V2 = VTH = 5.423 ∠ 40.60o V
ZTH = 10 || [(j10 || -j5) + (5 || -j10)] = 10 || (-j10 + 4 – j2) = 5.882 – j3.529 Ω.
(b)
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
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Engineering Circuit Analysis, 7th Edition
85.
Chapter Ten Solutions
10 March 2006
Consider the circuit below:
Vout
Vin
1
jω C
Using voltage division, we may write:
Vout = Vin
V out
1 / jωC
, or
R + 1 / jωC
V in
=
1
1 + j ω RC
The magnitude of this ratio (consider, for example, an input with unity magnitude and
zero phase) is
1
Vout
=
2
Vin
1 + (ωRC )
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 signals are attenuated, or “filtered out.”
This is readily apparent if we plot the magnitude as a function of frequency (assuming R
1 Ω and C = 1 F for convenience):
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Engineering Circuit Analysis, 7th Edition
86.
Chapter Ten Solutions
10 March 2006
Consider the circuit below:
Vout
1/jωC
Vin
R
Using voltage division, we may write:
Vout = Vin
R
jωRC
V
, or out =
R + 1 / jωC
Vin
1 + jωRC
The magnitude of this ratio (consider, for example, an input with unity magnitude and
zero phase) is
ωRC
Vout
=
2
Vin
1 + (ωRC )
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 attenuated, or “filtered out.”
This is readily apparent if we plot the magnitude as a function of frequency (assuming R
1 Ω and C = 1 F for convenience):
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Engineering Circuit Analysis, 7th Edition
87.
Chapter Ten Solutions
10 March 2006
(a) Removing the capacitor temporarily, we easily find the Thévenin equivalent:
Vth = (405/505) VS and Rth = 100 || (330 + 75) = 80.2 Ω
80.2 Ω
405
VS
505
31.57 fF
so
(b)
Vout =
405
1/jωC
VS
505 80.2 + 1 / jωC
and
hence
Vout
=
VS
+
Vout
-
Vout
1
⎛ 405 ⎞
= ⎜
⎟
−12
VS
⎝ 505 ⎠ 1 + j 2.532 × 10 ω
0.802
1 + 6.411 × 10− 24 ω 2
(c)
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 magnitude.
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Engineering Circuit Analysis, 7th Edition
88.
Chapter Ten Solutions
10 March 2006
From the derivation, we see that
- g m (R C || R L ) + jω (R C || R L )C μ
Vout
=
Vin
1 + jω (R C || R L )Cμ
so
that
1
2
⎡ 2 ⎛ R R ⎞2
⎤ 2
2
2 ⎛ R CR L ⎞
C L
⎟ Cμ ⎥
⎟ + ω ⎜⎜
⎢ g m ⎜⎜
R C + R L ⎟⎠
R C + R L ⎟⎠
Vout
⎢
⎥
⎝
⎝
= ⎢
2
⎥
Vin
2
2 ⎛ R CR L ⎞
⎢
⎥
⎟⎟ C μ
1 + ω ⎜⎜
⎢
⎥
⎝ RC + RL ⎠
⎣
⎦
This function has a maximum value of gm (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 frequencies at which the
capacitors may be treated as open circuits. If we do this, we may analyze the circuit
of Fig. 10.25b without the capacitors, which leads to
⎛ R R ⎞ (rπ || R B )
⎛ R R ⎞
Vout
rπ R B
= - g m ⎜⎜ C L ⎟⎟
= - g m ⎜⎜ C L ⎟⎟
VS low frequency
⎝ R C + R L ⎠ R S + rπ || R B
⎝ R C + R L ⎠ R S (rπ + R B ) + rπ R B
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 rπ || RB >> RS, then it has
minimal effect and the gain will equal its “maximum” value of –gm (RC || RL).
(b) If we set RS = 100 Ω, RL = 8 Ω, RC | max = 10 kΩ and rπgm = 300, then we find that
Vou t
rπ || R B
= - g m (7.994)
VS
100 + rπ || R B
of
We seek to maximize this term within the stated constraints. This requires a large value
gm, 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 RB → ∞, we are left with
Vou t
- 2398
g r
≈ - (7.994) m π =
VS
100 + rπ
100 + rπ
Considering this simpler expression, it is clear that if we select rπ to be small, (i.e.
rπ << 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 Vout/ Vin was computed, we
see that the critical frequency ωC = [(RC || RL) Cμ]-1. Our selection of maximum RC,
RB → ∞, and rπ << 100 has not affected this frequency.
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Engineering Circuit Analysis, 7th Edition
89. Considering
33
Chapter Ten Solutions
10 March 2006
the ω = 2×104 rad/ s source first, we make the following replacements:
100 cos (2×104t + 3o) V → 100 ∠3o V
μF → -j1.515 Ω
112 μH → j2.24 Ω
92 μF → -j0.5435 Ω
Then
(V1´ – 100 ∠ 3o)/ 47×103 + V1´/ (-j1.515) + (V1´ – V2´)/ (56×103 + j4.48) = 0
(V2´ – V1´)/ (56×103 + j4.48) + V2´/ (-j0.5435) = 0
[1]
[2]
Solving, we find that
V1´ = 3.223 ∠ -87o mV and V2´ = 31.28 ∠ -177o nV
Thus,
v1´(t) = 3.223 cos (2×104t – 87o) mV and v2´(t) = 31.28 cos(2×104t – 177o) nV
Considering the effects of the ω = 2×105 rad/ s source next,
33
100 cos (2×105t - 3o) V → 100 ∠-3o V
μF → -j0.1515 Ω
112 μH → j22.4 Ω
Then
92 μF → -j0.05435 Ω
V1"/ -j0.1515 + (V1" – V2")/ (56×103 + j44.8) = 0
[3]
(V2" – V1")/ (56×103 + j44.8) + (V2" – 100 ∠ 3o)/ 47×103 + V2"/ (-j0.05435) = 0 [4]
Solving, we find that
Thus,
V1" = 312.8 ∠ 177o pV and V2" = 115.7 ∠ -93o μV
v1"(t) = 312.8 cos (2×105t + 177o) pV and v2"(t) = 115.7 cos(2×105t – 93o) μV
Adding, we find
v1(t) = 3.223×10-3 cos (2×104t – 87o) + 312.8×10-12 cos (2×105t + 177o) V and
v2(t) = 31.28×10-9 cos(2×104t – 177o) + 115.7×10-12 cos(2×105t – 93o) V
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Engineering Circuit Analysis, 7th Edition
90.
Chapter Ten Solutions
10 March 2006
For the source operating at ω = 4 rad/s,
7 cos 4t → 7∠ 0o 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 I1, I2, I3, I4 in the top left, top
right, bottom left and bottom right meshes, respectively):
(9.5 + j4) I1 – j4 I2 – 7 I3
- 4 I4
-j4 I1 + (3 + j3.5) I2
– 3 I4
-7 I1 +
(12 – j/ 8) I3 + j/ 8 I4
-3 I2 + j/ 8 I3 + (4 + j11.875) I4
= 0
= -7
= 0
= 0
[1]
[2]
[3]
[4]
Solving, we find that I3 = 365.3 ∠ -166.1o mA and I4 = 330.97 ∠ 72.66o mA.
For the source operating at ω = 2 rad/s,
5.5 cos 2t → 5.5∠ 0o 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 IA, IB, IC, ID in the top left, top
right, bottom left and bottom right meshes, respectively):
(9.5 + j2) IA – j2 IB – 7 IC
– 4 ID
-j2 IA + (3 + j) IB
– 3 ID
-7 IA +
(12 – j/ 4) IC + j/ 4 ID
-3 I2 + j/ 4 IC + (4 + j5.75) ID
= 0
= -7
= 0
= 0
[1]
[2]
[3]
[4]
Solving, we find that IC = 783.8 ∠ -4.427o mA and ID = 134 ∠ -25.93o mA.
V1´ = -j0.25 (I3 – I4) = 0.1517∠131.7o V and V1" = -j0.25(IC – ID) = 0.1652∠-90.17o V
V2´ = (1 + j6) I4 = 2.013∠155.2o V and V2" = (1 + j6) ID = 0.8151∠54.61o V
Converting back to the time domain,
v1(t) = 0.1517 cos (4t + 131.7o) + 0.1652 cos (2t - 90.17o) V
v2(t) = 2.013 cos (4t + 155.2o) + 0.8151 cos (2t + 54.61o) V
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Engineering Circuit Analysis, 7th Edition
91.
(a)
IL =
100
j 2.5 +
−2
2 − j1
=
Chapter Ten Solutions
10 March 2006
100 (2 − j1)
= 57.26∠ − 76.76° (2.29in)
2.5 + j 3
I R = (57.26∠ − 76.76°)
− j1
= 25.61∠ − 140.19° (1.02in)
2 − j1
Ic = (57.26∠ − 76.76°)
2
= 51.21∠ − 50.19° (2.05in)
2 − j1
VL = 2.5 × 57.26∠90° − 76.76° = 143.15∠13.24° (2.86in)
VR = 2 × 25.61∠ − 140.19° = 51.22∠ − 140.19° (1.02in)
Vc = 51.21∠ − 140.19° (1.02in)
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Engineering Circuit Analysis, 7th Edition
Chapter Ten Solutions
10 March 2006
92.
(a)
120
= 3∠ − 30° A
40∠30°
120
I2 =
= 2.058∠30.96° A
50 − j 30
120
I3 =
= 2.4∠ − 53.13° A
30 + j 40
I1 =
(b)
(c)
I s = I1 + I 2 + I 3
= 6.265∠ − 22.14° A
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Engineering Circuit Analysis, 7th Edition
93.
Chapter Ten Solutions
10 March 2006
I1 = 5A, I 2 = 7A
I1 + I 2 = 10∠0°, I1 lags V, I 2 leads V
I1 lags I 2 . Use 2.5A / in
[Analytically: 5∠α + 7∠β = 10
= 5cos α + j 5sin α + 7 cos β + j 7 sin β
∴ sin α = −1.4sin β
∴ 5 1 − 1.42 sin 2 β + 7 1 − 1sin 2 β = 10
By SOLVE, α = −40.54° β = 27.66°]
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Engineering Circuit Analysis, 7th Edition
94.
Chapter Ten Solutions
10 March 2006
V1 = 100∠0o V, |V2| = 140 V, |V1 + V2| = 120 V.
Let 50 V = 1 inch. From the sketch, for ∠V2 positive,
V2 = 140∠122.5o. We may also have V2 = 140∠-122.5o V
[Analytically: |100 + 140∠α| = 120
so | 100 + 140 cos α + j140 sin α | = 120
Using the “Solve” routine of a scientific calculator,
α = ±122.88o.]
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Engineering Circuit Analysis, 7th Edition
Chapter Eleven Solutions
10 March 2006
1.
50 (− j80)
106
= − j80 Ω,
= 42.40∠ − 32.01°Ω
j 500 × 25
50 − j80
∴ V = 84.80∠ − 32.01° V, I R = 1.696∠ − 32.01° A
Zc =
I c = 1.0600∠57.99° A
ps (π / 2ms) = 84.80 cos (45° − 32.01°) 2 cos 45° = 116.85 W
pR = 50 × 1.6962 cos 2 (45° − 32.01°) = 136.55 W
pc = 84.80 cos (45° − 32.01°) = 1.060 cos (45° + 57.99°) = −19.69 W
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Engineering Circuit Analysis, 7th Edition
Chapter Eleven Solutions
10 March 2006
2.
(a)
1 2 1
Li = × 4 (4t 4 − 4t 2 + 1)
2
2
4
2
4
2
∴ wL = 8t − 8t + 2 ∴ wL (3) − wL (1) = 8 × 3 − 8 × 3 + 2 − 8 × 1 + 8 × 1 − 2 = 576 J
4H : i = 2t 2 − 1∴ v = Li′ = 4 (4t ) = 16t , wL =
1 t 2
⎛2
⎞
⎛2
⎞ ⎛2 ⎞
(2t − 1) dt + 2 = 5 ⎜ t 3 − t ⎟ + 2 = 5 ⎜ t 3 − t ⎟ − 5 ⎜ − 1⎟ + 2
0.2 F : vc =
∫
0.2 1
⎝3
⎠1
⎝3
⎠ ⎝3 ⎠
t
(b)
∴ vc (2) =
10
10
61
61
× 8 − 10 − + 5 + 2 = V ∴ Pc (2) = × 7 = 142.33 W
3
3
3
3
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Engineering Circuit Analysis, 7th Edition
Chapter Eleven Solutions
10 March 2006
R
1
= 2, ω o2 =
= 3, s1,2 = −2 ± 1 = −1, − 3
2L
LC
3.
vc (0) = −2V, i (0) = 4A, α =
(a)
1
i = Ae− t + Be−3t ∴ A + B = 4; i (0+ ) = vL (0+ ) = (−4 × 4 × +2) = −14
1
∴−A − 38 = −14 ∴ B = 5, A = −1, i = −e − t + 5e −3t A
t
∴+vc = 3∫ (−e− t + 5e−3t ) dt − 2 = 3(e− t − 5e −3t ) to − 2 = e −t − 3 − 5e −3t + 5 − 2
o
∴ vc = 3e − 5e−3t ∴ Pc (0+ ) = (3 − 5) (−1 + 5) = −8 W
−t
(b)
Pc (0.2) = (3e −0.2 − 5e−0.6 ) (−e0.2 + 5e −0.6 ) = −0.5542 W
(c)
Pc (0.4) = (3e −0.4 − 5e−1.2 ) (5e−1.2 − e−0.4 ) = 0.4220 W
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Engineering Circuit Analysis, 7th Edition
4.
2.5
Chapter Eleven Solutions
10 March 2006
We assume the circuit has already reached sinusoidal steady state by t = 0.
kΩ → 2.5 kΩ, 1 H → j1000 Ω, 4 μF → -j250 Ω, 10 kΩ → 10 kΩ
Zeq = j1000 || -j250 || 10000 = 11.10 – j333.0 Ω
(20∠30)(11.10 − j 333.0)
= 2.631∠ − 50.54o V
2500 + 11.10 − j 333.0
Veq
Veq
I10k =
= 0.2631 ∠ - 50.54o mA
I1 H =
= 2.631 ∠ - 140.5o mA
j1000
10000
Veq
(20∠30)(2500)
I4 μF =
= 10.52 ∠39.46o mA V2.5k =
= 19.74∠37.55o V
− j 250
2500 + 11.10 − j 333.0
Veq =
Thus,
[19.74 cos 37.55 ] =
P2.5k =
o 2
[
2500
97.97 mW
][
]
P1 H = 2.631cos(− 50.54 ) 2.631 × 10-3 cos(−140.5o ) = - 3.395 mW
o
[
][
[2.631cos(− 50.54 )] = 279.6 μW
P2.5k =
]
P4 μF = 2.631cos(− 50.54o ) 10.52 × 10-3 cos(39.46o ) = 13.58 mW
o
2
10000
FREQ
IM(V_PRINT1) IP(V_PRINT1)
1.592E+02 7.896E-03
3.755E+01
FREQ
VM(L,0)
1.592E+02 2.629E+00
FREQ
VM(R2_5k,$N_0002)VP(R2_5k,$N_0002)
1.592E+02 1.974E+01
3.755E+01
FREQ
IM(V_PRINT11) IP(V_PRINT11)
1.592E+02 1.052E-02
3.946E+01
FREQ
IM(V_PRINT2) IP(V_PRINT2)
1.592E+02 2.628E-03
-1.405E+02
FREQ
IM(V_PRINT12) IP(V_PRINT12)
1.592E+02 2.629E-04
-5.054E+01
VP(L,0)
-5.054E+01
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Engineering Circuit Analysis, 7th Edition
Chapter Eleven Solutions
5.
is → 5∠0° A, C → − j 4 Ω, Zin = 8 (3 − j 4) =
10 March 2006
40∠ − 53.13°
11 − j 4
= 3.417∠ − 33.15°∴ Vs = 17.087∠ − 33.15°,
vs = 17.087 cos (25t − 33.15°) V ∴
Ps ,abs (0.1) = −17.087 cos (2.5rad − 33.147°) × 5cos 2.5rad = −23.51 W
17.087
cos (25t − 33.15°) ∴
8
i8 (0.1) = 2.136 cos (2.5rad − 33.15°) = −0.7338 A
i8 =
∴ P8,abs = 0.73382 × 8 = 4.307 W ;
I3 =
17.087∠ − 33.15°
= 3.417∠19.98° A
3 − j4
∴ i3 (0.1) = 3.417 cos (2.5rad + 19.98°) = −3.272 A ∴
P3,abc = 3.2722 × 3 = 32.12 W
Vc = − j 4 (3.417∠19.983°) = 13.67∠ − 70.02°,
vc (0.1) = 13.670 cos (2.5rad − 70.02°) = 3.946 V
∴ Pc , abc = 3.946 (−3.272) = −12.911 W
(Σ = 0)
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Engineering Circuit Analysis, 7th Edition
−R t
L
Chapter Eleven Solutions
10 March 2006
= 8e −2t .
6. For
t > 0, i(t) = 8e
(a)
p(0+) = (8)2(1) = 64 W
(b)
at t = 1 s, i = 8e–2 = 1.083 A; p(1) = i2R = 1.723 W.
(c)
at t = 2 s, i = 8e–4 = 146.5 mA; p(2) = i2R = 21.47 mW
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Engineering Circuit Analysis, 7th Edition
7.
−t
v(t ) = (3)(6000)e 30×10
Chapter Eleven Solutions
10 March 2006
−3
(a)
p(0+) = v2(0+)/R = (18×103)2 / 6000 = 54 kW
(b)
p(0.03)
= v2(0.03)/R = (18×103e–1)2 / 6000 = 7.308 kW
(c) p(0.09) = v2(0.09)/R = (18×103e–3)2 / 6000 = 134 W
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Engineering Circuit Analysis, 7th Edition
8.
Chapter Eleven Solutions
10 March 2006
(a) p = (30×103)2 (1.2×10–3) = 1.080 MW
(b) W = (1.080×106)(150×10–6) = 162 J
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Engineering Circuit Analysis, 7th Edition
9. W =
Chapter Eleven Solutions
10 March 2006
1
CV 2 . The initial voltage, v(0+), is therefore
2
v(0+ ) =
2W
2(100 ×10−3 )
−t
−t
=
= 2 V and so v(t ) = 2e RC = 2e 0.12 V.
−3
C
100 × 10
The instantaneous power dissipated at t = 120 mS is therefore
p(120
=
ms)
v 2 (120 ms) 2e−2
=
= 226 mW
R
1.2
The energy dissipated over the first second is given by
−2 t
RC
1 2e
v 2 (t )
RC ⎛ 2 ⎞ −2
=
dt
∫0 R
∫0 R dt = − 2 ⎜⎝ R ⎟⎠ ⎡⎣⎢e RC − 1⎤⎦⎥ ≈ 100 mJ
1
ΔT = Q/mc, where Q = 100 mJ, c = 0.9 kJ/kg•K, and m = 10–3 kg.
Thus, the final temperature
= 271.15 + 23 +
100 ×10−6 kJ
= 271.15 + 23 + 0.1111
⎛
kJ ⎞
−3
(10 kG ) ⎜ 0.9 kg ⋅ K ⎟
⎝
⎠
= 294.3 K, representing a temperature increase of 0.1111 K.
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Engineering Circuit Analysis, 7th Edition
10. (a)
Chapter Eleven Solutions
10 March 2006
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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Engineering Circuit Analysis, 7th Edition
11.
Chapter Eleven Solutions
10 March 2006
j 5(10 − j 5)
= 4 + 2.5 + j 5 = 6.5 + j 5 Ω
10
100
∴ Is =
= 12.194∠ − 37.57° A
6.5 + j5
1
∴ Ps , abs = − × 100 × 12.194 cos 37.57° = −483.3 W
2
1
P4, abs = (12.194) 2 4 = 297.4 W,
2
Pcabs = 0
Zin = 4 +
100
j5
= 6.097∠52.43° so
6.5 + j 5 10
1
P10,abs = (6.097) 2 × 10 = 185.87 W
2
PL = 0
(Σ = 0)
I10 =
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Engineering Circuit Analysis, 7th Edition
12.
V = (10 + j10)
Chapter Eleven Solutions
10 March 2006
40∠30°
= 52.44∠69.18° V
5∠50° + 8∠ − 20°
1
P10, gen = × 10 × 52.44 cos 69.18° = 93.19 W
2
1
Pj10, gen = × 10 × 52.44 cos (90° − 69.18°) = 245.1 W
2
2
1 ⎛ 52.44 ⎞
P5∠50 abs = ⎜
⎟ cos (50°) = 176.8 W
2⎝ 5 ⎠
2
1 ⎛ 52.44 ⎞
P8∠− 20 abs = ⎜
⎟ cos (−20°) = 161.5 W
2⎝ 8 ⎠
(Σ gen = Σ abs )
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Engineering Circuit Analysis, 7th Edition
13.
ZR = 3 +
Chapter Eleven Solutions
10 March 2006
1
= 3 + 1 + j3 = 4 + j3 Ω
0.1 − j 0.3
Ignore 30° on Vs , I R = 5
2 + j5
5 29
, IR =
6 + j8
10
2
(a)
1 ⎛ 5 29 ⎞
P3 Ω = ⎜⎜
⎟ × 3 = 10.875 W
2 ⎝ 10 ⎟⎠
(b)
Vs = 5∠0°
∴ Ps , gen =
(2 + j 5) (4 + j 3)
= 13.463∠51.94° V
6 + j8
1
× 13.463 × 5cos 51.94° = 20.75 W
2
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Engineering Circuit Analysis, 7th Edition
14.
Chapter Eleven Solutions
10 March 2006
Pj10 = P− j 5 = 0,
V10 − 50 V10 V10 − j 50
+
+
=0
10
− j5
j10
∴ V10 (− j 0.1 + 0.1 + j 0.2) + j 5 + 10 = 0
∴ V10 = 79.06∠16.57° V
1 79.062
= 312.5 W;
2 10
79.06∠161.57° − 50
I 50 =
= 12.75∠78.69° A
j10
1
∴ P50V = × 50 × 12.748cos 78.69° = 62.50 W
2
79.06∠161.57° − j 50
= 15.811∠ − 7.57° :
I j 50 =
− j5
1
Pj 50 = × 50 × 15.811cos (90° + 71.57°) = −375.0 W
2
P10 Ω =
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Engineering Circuit Analysis, 7th Edition
15.
Vx − 20 Vx − Vc
+
= 2Vc
2
3
and
V
V − Vx
0= c + c
− j2
3
which simplify to
5Vx − 14Vc = 60
Chapter Eleven Solutions
10 March 2006
[1]
[2]
[1] and
j 2Vx + (3 − j 2)Vc = 0 [2]
Solving,
Vx = 9.233∠ − 83.88° V and Vc = 5.122∠ − 140.2° V
1
Pgen = × 9.233 × ( 2 × 5.122 ) cos (−83.88° + 140.2°) = 26.22 W
2
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Engineering Circuit Analysis, 7th Edition
Chapter Eleven Solutions
10 March 2006
16.
(a)
X in = 0 ∴ Z L = R th + j 0
(b)
R L , X L independent∴ Z L = Z∗th = R th − jX th
(c)
Vth
1
× R L ∴ Z L = R L − jX th
R L fixed∴ PL =
2 (R th + R L ) 2 + (X th + X L ) 2
(d)
X L fixed, Let X L + Xth = a ∴ f =
2
2PL
Vth
2
=
RL
(R th + R L ) 2 + a 2
R + R 2L + a 2 − 2R L (R th + R L )
df
= th
=0
2
dRL
⎡⎣(R th + R L ) 2 + a 2 ⎤⎦
R th2 + 2R th R L + R 2L + a 2 − 2R th R L = 2R 2L = 0
∴ R L = R th2 + a 2 =
(e)
R th2 + (Xth + X L ) 2
X L = 0 ∴ R L = R th2 + X th2 = Zth
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Engineering Circuit Analysis, 7th Edition
Chapter Eleven Solutions
10 March 2006
17.
− j10
= 107.3∠ − 116.6° V
10 + j 5
− j10 (10 + j15)
Zth =
= 8 − j14 Ω
10 + j 5
Vth = 120
(a)
ZTH = ( Z L ) = 8 + j14 Ω
(b)
IL =
*
VTH
ZTH + ( ZTH )
*
=
107.3∠ − 116.6°
.
16
( ZTH ) = (107.3∠ − 116.6° )(16.12∠ − 60.26° )
VL = VTH
*
16
ZTH + ( ZTH )
1 ⎡ (107.3)(16.12 ) ⎤ ⎡107.3 ⎤
PL ,max = ⎢
⎥⎢
⎥ cos ( −116.6° − 60.26° + 116.6° ) = 179.8 W
*
2 ⎣
16
⎦ ⎣ 10 ⎦
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Engineering Circuit Analysis, 7th Edition
18.
Chapter Eleven Solutions
10 March 2006
R L = Zth ∴ R L = 82 + 142 = 16.125 Ω
PL =
1
107.332
× 16.125 = 119.38 W
2 (8 + 16.125) 2 + 142
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Engineering Circuit Analysis, 7th Edition
Chapter Eleven Solutions
10 March 2006
19.
− j 9.6 = −4.8 I x − j1.92 I x − +4.8I x
9.6
=5
1.92
∴ V = (0.6 × 5)8 = 24 V
∴ Ix =
∴ Po =
1
× 24 × 1.6 × 5 = 96
2
)
( genW
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Engineering Circuit Analysis, 7th Edition
Chapter Eleven Solutions
10 March 2006
20.
(a)
(b)
j 480 80 − j 60
80 + j 60 80 − j 60
= 28.8 + j 38.4 Ω ∴ Z L max = 28.8 − j 38.4 Ω
Z th = 80 j 60 =
Vth = 5(28.8 + j38.4) = 144 + j192 V,
144 + j192
2 × 28.8
1 1442 + 1922
× 28.8 = 250 W
and PL ,max =
2 4 × 28.82
∴ IL =
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Engineering Circuit Analysis, 7th Edition
21.
Chapter Eleven Solutions
10 March 2006
Zeq = (6 – j8) || (12 + j9) = 8.321 ∠ -19.44o W
Veq = (5 ∠-30o) (8.321 ∠ -19.44o) = 41.61 ∠ -49.44o V
Ptotal = ½ (41.61)(5) cos (-19.44o) = 98.09 W
I6-j8 = Veq / (6 – j8) = 4.161 ∠ 3.69o A
I4+j2 = I8+j7 = Veq/ 12+j9 = 2.774 ∠ -86.31o A
P6-j8 = ½ (41.61)(4.161) cos (-49.44o – 3.69o) = 51.94 W
P4+j2 = ½ (2.774)2 (4) = 15.39 W
P8+j7 = ½ (2.774)2 (8) = 30.78 W
Check:
Σ = 98.11 W (okay)
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Engineering Circuit Analysis, 7th Edition
22.
Vth = 100
Chapter Eleven Solutions
10 March 2006
j10 (20)
j10
= 20 + j 40, Zth =
= 4 + j8 Ω
20 + j10
20 + j10
∴ R L = Zth ∴ R L = 8.944 Ω
∴ PL ,max =
1
202 + 402
× 8.944 = 38.63 W
2 (4 + 8.944) 2 + 64
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Engineering Circuit Analysis, 7th Edition
Chapter Eleven Solutions
10 March 2006
23.
We may write a single mesh equation: 170 ∠0o = (30 + j10) I1 – (10 – j50)(-λI1)
Solving,
170∠0 o
I1 =
30 + j10 + 10λ − j 50λ
170∠0 o
(a) λ = 0, so I1 =
= 5.376∠ - 18.43 o A and, with the same current flowing
30 + j10
through both resistors in this case,
P20 = ½ (5.376)2 (20) = 289.0 W
P10 = ½ (5.376)2 (10) = 144.5 W
170∠0 o
= 3.005∠45 o A
40 − j 40
P20 = ½ (3.005)2 (20) = 90.30 W
The current through the 10-Ω resistor is I1 + λI1 = 2 I1 = 6.01 ∠ 45o so
(b) λ = 1, so I1 =
(c)
P10 = ½ (6.01)2 (10) = 180.6 W
(a)
FREQ
IM(V_PRINT3)
6.000E+01 5.375E+00
IP(V_PRINT3)
-1.846E+01
FREQ
IM(V_PRINT4)
6.000E+01 5.375E+00
IP(V_PRINT4)
-1.846E+01
(b)
FREQ
IM(V_PRINT3)
6.000E+01 6.011E+00
IP(V_PRINT3)
4.499E+01
FREQ
IM(V_PRINT4)
6.000E+01 3.006E+00
IP(V_PRINT4)
4.499E+01
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Engineering Circuit Analysis, 7th Edition
24.
Chapter Eleven Solutions
10 March 2006
(10)(1) + (−5)(1) + 0(1)
= 1.667 A
3
1
(20)(1) + 0(1)
2
= 5A
Waveform (b): Iavg =
2
(a) Waveform (a): Iavg =
Waveform (c):
= −
16
π
(0 − 1) =
10 −3
−3
⎞ ⎛ πt ⎞
2πt
3 ⎛ 4 × 10
⎟⎟ cos⎜
8sin
dt = - 8 × 10 ⎜⎜
−3
−3 ⎟
4 × 10
⎝ 2π ⎠ ⎝ 2 × 10 ⎠ 0
10 − 3
1
Iavg =
1 × 10 −3 ∫ 0
(
16
π
)
A
(100)(1) + (25)(1) + (0)(1)
= 41.67 A 2
3
Waveform (b): i(t) = -20×103 t + 20
i2(t) = 4×108 t2 – 8×105 t + 400
10 -3
1
2
(4 × 108 t 2 - 8 × 10 5 t + 400) dt
I avg
=
-3 ∫ 0
2 × 10
5
⎡ 4 × 10 8
⎤
2
1
0.1333
−3 3 8 × 10
(
)
(
=
= 66.67 A 2
10
10 −3 ) + 400(10 −3 )⎥ =
-3 ⎢
-3
2
×
2 × 10 ⎣ 3
2
10
⎦
Waveform (c):
(b)
2
Waveform (a): I avg
=
10 − 3
1
2
=
I avg
1 × 10 −3 ∫ 0
10 −3
2πt
sin π × 10 3 t ⎤
3 ⎡t
dt
=
×
64sin 2
64
10
⎢
⎥
4 × 10 −3
2π × 10 3 ⎦ 0
⎣2
(
)
⎡10 −3
sin π ⎤
= 64 × 10 3 ⎢
−
= 32 A 2
3⎥
2π × 10 ⎦
⎣ 2
(
)
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Engineering Circuit Analysis, 7th Edition
25. At
Define
Chapter Eleven Solutions
10 March 2006
ω = 120π, 1 H → j377 Ω, and 4 μF → -j663.1 Ω
Zeff = j377 || -j663.1 || 10 000 = 870.5 ∠ 85.01o Ω
V2.5k
(400 2∠ − 9 ) 2500 = 520.4 ∠ - 27.61 V
=
o
o
o
2500 + 870.5 ∠85.01
400 2∠ − 9 o 870.5 ∠85.01o
V10k =
= 181.2 ∠57.40 o V
o
2500 + 870.5 ∠85.01
(
Thus,
)(
)
P2.5k = ½ (520.4)2 / 2 500
P10k = ½ (181.2)2 / 10 000
P1H
P4μF
=
=
=
=
54.16 W
1.642 W
0
0
(A total absorbed power of 55.80 W.)
To check, the average power delivered by the source:
Isource =
400 2∠ − 9 o
= 0.2081 ∠ - 27.61o A
2500 + 870.5∠85.01o
and Psource = ½ ( 400 2 )(0.2081) cos (-9o + 27.61o) = 55.78 W (checks out).
FREQ
IM(V_PRINT1)
6.000E+01 2.081E-01
IP(V_PRINT1)
-2.760E+01
FREQ
VM(L,0)
6.000E+01 1.812E+02
VP(L,0)
5.740E+01
FREQ
VM(R2_5k,$N_0002) VP(R2_5k,$N_0002)
6.000E+01 5.204E+02
-2.760E+01
FREQ
IM(V_PRINT11)
6.000E+01 2.732E-01
IP(V_PRINT11)
1.474E+02
FREQ
IM(V_PRINT2)
6.000E+01 4.805E-01
FREQ
IM(V_PRINT12)
6.000E+01 1.812E-02
IP(V_PRINT12)
5.740E+01
IP(V_PRINT2)
-3.260E+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, 7th Edition
Chapter Eleven Solutions
26. (a)
1 144
144
= 8.485
(1 + cos 2000t ) dt =
∫
2
T 0 2
(b)
1 144
144
= 8.485
(1 − cos 2000t ) dt =
∫
2
T 0 2
(c)
1 144
144
= 8.485
(1 + cos1000t ) dt =
∫
2
T 0 2
(d)
1 144 ⎡
144
= 8.485
1 + cos 1000t − 176o ⎤⎦ dt =
∫
⎣
2
T 0 2
10 March 2006
T
T
T
T
(
)
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Engineering Circuit Analysis, 7th Edition
Chapter Eleven Solutions
27. (a)
1 4
2
(1 + cos 20t ) dt = = 1.414
∫
T 02
2
(b)
1 4
2
(1 − cos 20t ) dt = = 1.414
∫
T 02
2
(c)
1 4
2
(1 + cos10t ) dt = = 1.414
∫
2
T 02
(d)
1 4⎡
2
= 1.414
1 + cos 10t − 64o ⎤⎦ dt =
∫
⎣
2
T 02
10 March 2006
T
T
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, 7th 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.
3
3
1
100
100(2)
(10) 2 dt =
Vrms =
t =
= 8.165 V
∫
31
3 1
3
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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, 7th 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.
3
3
1
49
49(1)
(7) 2 dt =
I rms =
t =
= 4.041 A
∫
32
3 2
3
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, 7th Edition
30.
Chapter Eleven Solutions
10 March 2006
(a) 1 V
2
(b)
2
1eff
Vrms = V
2
2eff
⎛ 1 ⎞
= 1 +⎜
⎟ = 1.225 V
⎝ 2⎠
2
2eff
⎛ 1 ⎞
= 1 +⎜
⎟ = 1.225 V
⎝ 2⎠
+V
2
2
(c)
2
1eff
Vrms = V
+V
2
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, 7th Edition
Chapter Eleven Solutions
10 March 2006
31.
(a)
v = 10 + 9 cos100t + 6sin100t
1
1
∴ Veff = 100 + × 81 + × 36 = 158.5 = 12.590 V
2
2
(b)
Feff =
1 2
(10 + 202 + 102 ) = 150 = 12.247
4
(c) F
avg =
(10)(1) + (20)(1) + (10)(1) 40
=
= 10
4
4
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Engineering Circuit Analysis, 7th Edition
Chapter Eleven Solutions
10 March 2006
32.
(a)
g(t) = 2 + 3cos100t + 4cos(100t – 120o)
o
o
∠0 + 4∠-120 = 3.606 ∠-73.90 so Geff =
3
3.606 2
4+
= 3.240
2
h (t ) = 2 + 3cos100t + 4 cos (101t − 120°)
(b)
1
1
∴ H eff = 22 + 32 + 42 = 16.5 = 4.062
2
2
(c)
f (t ) = 100t , 0 < t < 0.1∴ Feff =
=
1 0.1 6 2
10 t dt
0.3 ∫0
10
1
×106 × × 10−3 = 33.33
3
3
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Engineering Circuit Analysis, 7th Edition
Chapter Eleven Solutions
33.
f (t ) = (2 − 3cos100t ) 2
(a)
f (t ) = 4 − 12 cos100t + 9 cos 2 100t
∴ f (t ) = 4 − 12 cos100t + 4.5 + 4.5cos 200t ∴ Fav = 4 + 4.5 = 8.5
(b)
10 March 2006
1
1
Feff = 8.52 + × 122 + × 4.52 = 12.43
2
2
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Engineering Circuit Analysis, 7th Edition
Chapter Eleven Solutions
10 March 2006
1
⎡1
⎤ 2
ieff = ⎢ 102 + (−5) 2 + 0⎥ = 6.455 A
⎣3
⎦
(
34. (a)
)
1
⎡1 1
⎤ 2
(b) ieff = ⎢ ⎛⎜ ∫ [− 20t + 20] dt ⎞⎟ + 0⎥
=
⎠
⎣2 ⎝ 0
⎦
1
(c)
2
⎡1 ⎛ 1
⎛ 2π ⎞ ⎞⎤
ieff = ⎢ ⎜⎜ ∫ 8sin ⎜
t ⎟ dt ⎟⎟⎥
=
0
⎝ 4 ⎠ ⎠⎦
⎣1 ⎝
5 = 2.236 A
1
⎡ ⎛2⎞
⎛ πt ⎞ ⎤
⎢- 8 ⎜ π ⎟ cos ⎜ 2 ⎟⎥ = 2.257 A
⎝ ⎠⎦ 0
⎣ ⎝ ⎠
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Engineering Circuit Analysis, 7th Edition
Chapter Eleven Solutions
10 March 2006
35.
(a)
A = B = 10V, C = D = 0 ∴10∠0° + 10∠ − 45° = 18.48∠ − 22.50o
∴P =
(b)
A = C = 10V, B = D = 0, vs = 10 cos10t + 10 cos 40t ,
P=
(c)
1 1
× × 18.482 = 42.68 W
2 4
1 102 1 102
+
= 25
2 4
2 4
vs = 10 cos10t − 10sin (10t + 45°) → 10 − 10∠ − 45° = 7.654∠67.50o
1 7.6542
∴P =
= 7.322
2
4
(d)
W
W
v = 10 cos10t + 10sin (10t + 45°) + 10 cos 40t ;
10∠0° + 10∠ − 45° = 18.48∠ − 22.50o
1
1 1
1
∴ P = × 18.482 × + × 102 × = 55.18 W
2
4 2
4
(e)
102
= 80.18 W
// + 10dc ∴ Pav = 55.18 +
4
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Engineering Circuit Analysis, 7th Edition
36.
Zeq = R || j0.3ω =
Chapter Eleven Solutions
10 March 2006
j 0.3Rω
. By voltage division, then, we write:
R + j 0.3Rω
j 0.1ω
- 0.03ω 2 + j 0.1ωR
= 120∠0
V100mH = 120∠0
j 0.3Rω
− 0.03ω 2 + j 0.4 Rω
j 0.1ω +
R + j 0.3ω
j 0.3Rω
j 36 Rω
R + j 0.3ω
= 120∠0
V300mH = 120∠0
j 0.3Rω
− 0.03ω2 + j 0.4 Rω
j 0.1ω +
R + j 0.3ω
(a) We’re interested in the value of R that would lead to equal voltage magnitudes, or
j 36 Rω
Thus,
36Rω =
=
(
(120) - 0.03ω 2 + j 0.1ωR
)
12.96ω 4 + 144ω 2 R 2 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, 7th Edition
Chapter Eleven Solutions
10 March 2006
37.
(a)
Vav ,1 = 30V
1
Vav ,2 = (10 + 30 + 50) = 30V
3
(b)
Veff ,1 =
1 3
1
1
(20t ) 2 dt =
× 400 × × 27 = 1200 = 34.64V
∫
0
3
3
3
Veff ,2 =
1 2
1
(10 + 302 + 502 ) =
× 3500 = 34.16 V
3
3
(c) PSpice verification for Sawtooth waveform of Fig. 11.40a:
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Engineering Circuit Analysis, 7th Edition
38.
Chapter Eleven Solutions
10 March 2006
⎛ − j106 ⎞
− jR106
⎟⎟ =
Zeff = R || ⎜⎜
6
⎝ 3ω ⎠ 3ωR − j10
(
)
120ω 3ωR - j106
120∠0
ISRC =
=
106
R106
− j106 3ωR − j106 − jωR106
−j
−j
ω
3ωR − j106
R
I3μF = ISRC
106
R− j
3ω
(
)
(a) For the two current magnitudes to be equal, we must have
R
= 1 . This is
106
R− j
3ω
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
120∠0
1
106
−j
−j
3ω
ω
10
6
= j 90ω μA, or
90ω cos(ωt + 90o ) μA
(c) PSpice verification: set f = 60 Hz, simulate a single 0.75- μF capacitor, and include a
100-MΩ resistor in parallel w ith 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)
6.000E+01 3.393E-02
IP(V_PRINT3)
9.000E+01
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Engineering Circuit Analysis, 7th Edition
Chapter Eleven Solutions
10 March 2006
39.
v(t ) = 10t [u (t ) − u (t − 2)] + 16e −0.5(t −3) [u (t − 3) − u (t − 5)] V
Find eff. value separately
V1,eff =
1 2
20
× 8 = 7.303
100t 2 dt =
∫
0
5
3
V2,eff =
1 5
256 3 − t 5
e (−e )3 = 6.654
256e − ( t −3) dt =
∫
3
5
5
∴ Veff = 7.3032 + 6.6542 = 9.879
Veff =
5
1⎡ 2
2
+
t
dt
100
256e3e− t dt ⎤
∫
∫
⎢
⎥⎦
0
3
5⎣
=
1 ⎡100
⎤
× 8 + 256e3 (e −3 − e−5 ) ⎥
⎢
5⎣ 3
⎦
=
1 ⎡ 800
⎤
+ 256 (1 − e −2 ) ⎥ = 9.879 V OK
⎢
5⎣ 3
⎦
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Engineering Circuit Analysis, 7th Edition
40.
Chapter Eleven Solutions
The peak instantaneous power is 250 mW. The combination of elements yields
Z = 1000 + j1000 Ω = 1414 ∠45o Ω.
Arbitrarily
W
curren
10 March 2006
designate V = Vm ∠0 , so that I =
Vm ∠0 Vm ∠ − 45o
A.
=
Z
1414
e may write p(t) = ½ Vm Im cos φ + ½ Vm Im cos (2ωt + φ) where φ = the angle of the
t (-45o). This function has a maximum value of ½ VmIm cos φ + ½ VmIm.
Thus, 0.250 = ½ VmIm (1 + cos φ) = ½ (1414) Im2 (1.707)
and Im = 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, 7th Edition
Chapter Eleven Solutions
41.
I = 4∠35° A rms
(a)
V = 20I + 80∠35° Vrms, Ps , gen = 80 × 10 cos 35° = 655.3 W
(b)
PR = I R = 16 × 20 = 320 W
(c)
PLoad = 655.3 − 320 = 335.3 W
(d)
APs , gen = 80 × 10 = 800 VA
(e)
APR = PR = 320 VA
(f)
I L = 10∠0° − 4∠35° = 7.104∠ − 18.84° A rms
10 March 2006
2
∴ APL = 80 × 7.104 = 568.3 V A
(g)
PFL = cos θ L =
since I L lags V,
PL
335.3
=
= 0.599
APL 568.3
PFL is lagging
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Engineering Circuit Analysis, 7th Edition
Chapter Eleven Solutions
10 March 2006
42.
(a)
(b)
(c)
120
= 9.214∠ − 26.25° A rms
j192
4+
12 + j16
∴ PFs = cos 26.25 = 0.8969 lag
Is =
Ps = 120 × 9.214 × 0.8969 = 991.7W
j 48
1
(192 + j144)
= 4+
3 + j4
25
11.68 − j 5.76
∴ Z L = 11.68 + j 5.76 Ω, YL =
11.682 + 5.762
j 5.76
, C = 90.09μ F
∴ j120π C =
11.682 + 5.762
ZL = 4 +
(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).
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Engineering Circuit Analysis, 7th Edition
43.
Chapter Eleven Solutions
Z A = 5 + j 2 Ω, Z B = 20 − j10 Ω, Z c = 10∠30° Ω
8.660
=
10 March 2006
+ j5 Ω
Z D = 10∠ − 60° = 5 − j8.660 Ω
200
−20 + j10
0 33.66 − j13.660
7265∠22.09°
=
= 15.11∠3.908° A rms
I1 =
25 − j8
−20 + j10
480.9∠ − 26.00°
−20 + j10 33.66 − j13.660
25 − j8 200
−20 + j10 0
200 (20 − j10)
=
= 9.300∠ − 0.5681° A rms
I2 =
480.9∠ − 26.00° 480.9∠20.00°
2
APA = I1 Z A = 15.1082 29 = 1229 VA
APB = I1 − I 2
2
Z B = 5.8812 × 10 5 = 773.5 VA
APC = I 2 2 ZC = 9.32 × 10 = 86.49 VA
APD = I 2
2
Z1 = 9.32 × 10 = 864.9 VA
APS = 200 I1 = 200 × 15.108 = 3022 VA
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Engineering Circuit Analysis, 7th Edition
Chapter Eleven Solutions
44.
Z1 = 30∠15°Ω, Z 2 = 40∠40°Ω
(a)
Ztot = 30∠15° + 40∠40° = 68.37∠29.31°Ω
10 March 2006
∴ PF = cos 29.3° = 0.8719 lag
(b)
V = IZtot = 683.8∠29.31o Ω so
(
)
S = VI* = 683.8∠29.31o (10∠0 ) = 6838∠29.31o VA .
Thus, the apparent power = S = 6.838 kVA.
(c)
The impedance has a positive angle; it therefore has a net inductive character.
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Engineering Circuit Analysis, 7th Edition
45.
Chapter Eleven Solutions
10 March 2006
θ1 = cos-1(0.92) = 23.07o, θ 2 = cos-1 (0.8) = 36.87o, θ 3 = 0
100 ∠23.07o
S1 =
= 100 + j 42.59 VA
0.92
250 ∠36.87 o
S2 =
= 250 + j187.5 VA
0.8
500 ∠0o
S3 =
= 500 VA
1
Stotal = S1 + S2 + S3 = 500 + j230.1 VA = 550.4 ∠24.71o VA
(a)
Ieff =
Stotal
550.4
=
= 4.786 A rms
Veff
115
(b) PF of composite load = cos (24.71o) = 0.9084 lagging
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Engineering Circuit Analysis, 7th Edition
Chapter Eleven Solutions
10 March 2006
46.
APL = 10, 000 VA, PFL = 0.8lag, 40
IL =
Let I40
L =
A rms
∠0° A rms; PL = 10, 000 × 0.8 = 8000 W
8000
=5 Ω
402
cosθ L = 0.8lag∴θ L = cos −1 0.8 = 36.87°
Let R
ZL =
L
+ jX L ∴ R L =
∴ X L = 5 tan 36.87° = 3.75 Ω, Z L = 5 + j 3.75, Ztot = 5.2 + j 3.75 Ω
1
5.2 + j 3.75
= 0.12651 − j 0.09124S, Ynew = 0.12651 + j (120π C − 0.09124),
∴ Vs = 40 (5.2 + j 3.75) = 256.4∠35.80° V; Ytot =
PFnew = 0.9 lag,θ new = 25.84°∴ tan 25.84° = 0.4843
0.09124 − 120π C
∴
0.12651
C = 79.48μ F
=
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Engineering Circuit Analysis, 7th Edition
47.
Chapter Eleven Solutions
10 March 2006
Zeff = j100 + j300 || 200 = 237 ∠54.25o. PF = cos 54.25o = 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
⎛ 192.3 + X C ⎞
-1
o
tan −1 ⎜
⎟ = cos 0.92 = 23.07
⎝ 138.5 ⎠
Solving, we find that XC = -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 =
− 100 X C
+138.5 + j92.31 Ω
j (100 + X C )
⎛
100X C ⎞
⎟ Ω
= 138.5 + j ⎜⎜ 92.31 +
100 + X C ⎟⎠
⎝
100X C ⎞
⎛
⎜ 92.31 +
⎟
100 + X C ⎟
−1 ⎜
tan
= cos-1 0.92 = 23.07o
⎜
⎟
138.5
⎜
⎟
⎝
⎠
Solving, we find that XC = -25 Ω = -1/ωC, so that C = 40 μF
General circuit for simulations. Results agree with hand calculations
With no compensation:
With series compensation:
With parallel compensation:
FREQ
1.592E+02
1.592E+02
1.592E+02
IM(V_PRINT1) IP(V_PRINT1)
4.853E-01
-5.825E+01
7.641E-01
-2.707E+01
7.641E-01
-2.707E+01
θ PF
54.25o 0. 5843 lag
23.07o 0. 9200 lag
23.07o 0. 9200 lag
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Engineering Circuit Analysis, 7th Edition
48.
Chapter Eleven Solutions
10 March 2006
20 (1 + j 2)
= 10.769 − j 3.846 = 11.435+ ∠ − 19.65° Ω
3 + j2
100
∴ Is =
= 8.745∠19.65°
11.435∠ − 19.654°
∴ S s = − Vs I ∗s = −100 × 8.745∠ − 19.65° = −823.5 + j 294.1VA
Zin = − j10 +
I 20 = 8.745∠19.65° ×
10 + j 20
= 5.423∠49.40°
30 + j 20
∴ S 20 = 20 × 5.4322 = 588.2 + j 0 VA
I10 =
20 × 5.423∠49.40
= 4.851∠ − 14.04°
10 + j 20
S10 = 10 × 4.8512 = 235.3 + j 0 VA
S j 20 = j 20 × 4.8512 = j 470.6 VA,
S − j10 = − j10 × 8.7452 =
− j 764.7 VA,
Σ=0
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Engineering Circuit Analysis, 7th Edition
Chapter Eleven Solutions
10 March 2006
49.
Vx − 100
V
V − j100
+ x + x
=0
6 + j4
− j10
5
⎛ 1
⎞ 100
∴ Vx ⎜
+ j 0.1 + 0.2 ⎟ =
+ j 20
⎝ 6 + j4
⎠ 6 + j4
∴ Vx = 53.35− ∠42.66° V
100 − 53.35− ∠42.66°
∴ I1 =
= 9.806∠ − 64.44° A
6 + j4
1
∴ S1. gen = ×100 × 9.806∠64.44° = 211.5 + j 442.3VA
2
1
S 6, abs = × 6 × 9.8062 = 288.5 + j 0 VA
2
1
S j 4,abs = ( j 4) 9.8062 = 0 + j192.3VA
2
j100 − 53.35− ∠42.66°
= 14.99∠121.6°,
I2 =
5
1
S5 abs = × 5 × 14.992 = 561.5 + j 0 VA
2
1
S 2, gen = ( j100)14.99∠ − 121.57° = 638.4 − j 392.3VA
2
1 ⎛ 53.35 ⎞
S − j10, abs = ⎜
⎟ (− j10) = 0 − j142.3VA = 142.3∠ − 90° VA
2 ⎝ 10 ⎠
Σ=0
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Engineering Circuit Analysis, 7th Edition
Chapter Eleven Solutions
10 March 2006
50.
(a)
500 VA, PF = 0.75 lead∴
S = 500∠ − cos −1 0.75 = 375 − j 330.7 VA
(b)
(c)
500 W, PF = 0.75 lead∴
500
S = 500 −
sin (cos −1 0.75) = 500 − j 441.0 VA
j.075
−500 VAR, PF = 0.75(lead) ∴θ = − cos −1 0.75 = −41.41°
∴ P 500 / tan 41.41° = 566.9W,
S = 566.9 − j 500 VA
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Engineering Circuit Analysis, 7th Edition
51.
(a)
Chapter Eleven Solutions
10 March 2006
S s = 1600 + j 500 VA (gen)
1600 + j 500
= 4 + j1.25 ∴ I s = 4 − j1.25
400
400
Ic =
= j 3.333A rms∴ I L = I s − I c = 4 − j1.25 − j 3.333
− j120
∴ I L = 4 − j 4.583A rms∴
I ∗s =
S L = 400 (4 + j 4.583) = 1600 + j1833 VA
(b)
1833.3 ⎞
⎛
+
PFL = cos ⎜ tan −1
⎟ = 0.6575 lag
1600 ⎠
⎝
(c)
S s = 1600 + j 500 = 1676∠17.35° VA ∴ PFs = cos17.35° = 0.9545 lag
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Engineering Circuit Analysis, 7th Edition
Chapter Eleven Solutions
52.
(cos −1 0.8 = 36.87°, cos −1 0.9 = 25.84°)
(a)
S tot = 1200∠36.87° + 1600∠25.84° + 900
10 March 2006
= 960 + j 720 + 1440 + j 697.4 + 900
= 3300 + j1417.4 = 3592∠23.25° VA
∴ Is =
3591.5
= 15.62 A rms
230
(b)
PFs = cos 23.245° = 0.9188
(c)
S = 3300 + j1417 VA
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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, 7th Edition
Chapter Eleven Solutions
10 March 2006
53.
(a)
Ps ,tot = 20 + 25 × 0.8 + 30 × 0.75 = 70 kW
(b)
20, 000
= 80∠0° A rms
250
I 2 = 25, 000 / 250 = 100 A rms
I1 =
∠I 2 = − cos −1 0.8 = −36.87 ∴ I 2 = 100∠ − 36.87 o A rms
30, 000
40, 000
= 40, 000 VA, I 3 =
= 160 A rms
0.75
250
∠ I 3 = − cos −1 0.75 = −41.41° ∴ I 3 = 160∠ − 41.41° A rms
AP3 =
∴ I s = 80∠0° + 100∠ − 36.87° + 160∠ − 41.41° = 325.4∠ − 30.64° A rms
∴ APs = 250 × 325.4 = 81,360 VA
(c)
PF3 =
70, 000
= 0.8604 lag
81,360
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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, 7th Edition
54.
Chapter Eleven Solutions
10 March 2006
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) = 33.02o, 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 would 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, 7th Edition
55.
Chapter Eleven Solutions
10 March 2006
Perhaps the easiest approach is to consider the load and the compensation capacitor
separately. The load draws a complex power Sload = P + jQ. The capacitor draws a
purely reactive complex power SC = -jQC.
θload = tan-1(Q/P), or Q = P tan θload
QC = SC = Vrms
Vrms
2
2
= ω CVrms
= ω CVrms
(− j / ω C)
Stotal = Sload + SC = P + j(Q – QC)
⎛ Q-QC ⎞
θnew = ang(Stotal) = tan −1 ⎜
⎟ , so that Q – QC = P tan θnew
⎝ P ⎠
Substituting, we find that QC = P tan θload – P tan θnew
or
2
ω CVrms
= P (tan θload – tan θnew)
Thus, noting that θold = θload,
C =
P ( tan θ old - tan θ new )
2
ω Vrms
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, 7th Edition
56.
Chapter Eleven Solutions
10 March 2006
V = 339 ∠-66o V, ω = 100π rad/ s, connected to Z = 1000 Ω.
339
= 239.7 V rms
2
(b) pmax = 3392 / 1000 = 114.9 W
(a) Veff =
(c) pmin = 0 W
⎛ 339
⎞
2
⎟
Veff
⎛ 339 ⎞ ⎜
2
(d) Apparent power = Veff Ieff = ⎜
=
= 57.46 VA
⎟⎜
⎟
⎝ 2 ⎠ ⎜ 1000 ⎟ 1000
⎝
⎠
(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, 7th Edition
57.
(a)
(b)
(c)
Chapter Eleven Solutions
10 March 2006
V = 339 ∠-66o V, ω = 100π rad/s to a purely inductive load of 150 mH (j47.12 Ω)
339∠ - 66o
V
=
= 7.194 ∠ - 156o A
Z
j 47.12
7.194
= 5.087 A rms
so Ieff =
2
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(-90o)) (339)(7.194)/ 2 = 1219 W
I =
pmin = ½ VmIm cos φ - ½ VmIm = -1219 W
339
(5.087 ) = 1219 VA
2
(e) reactive power = Q = Veff Ieff sin (θ – φ) = 1219 VA
(d) apparent power = Veff Ieff =
(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, 7th Edition
58. 1
Chapter Eleven Solutions
10 March 2006
H → j Ω, 4 μF → –j250 Ω
Zeff = j || –j250 || 104 Ω = 1.004 ∠89.99o Ω
(5∠0) (1.004∠ 89.99o )
= 2.008 ∠89.97 o mV
2500 + (1.004∠89.99o )
pmax = (0.002)2 / 10×103 = 400 pW
V10k =
(a)
(b) 0 W (purely resistive elements draw no reactive power)
(c) apparent power = VeffIeff = ½ VmIm = ½ (0.002)2 / 10000 = 200 pVA
(d)
Ssource =
1
5∠0
⎞
( 5∠0 ) ⎜⎛
⎟
2
⎝ 2500∠0.02292 ⎠
= 0.005 ∠-0.02292o 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, 7th 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 ∠ 3.287o Ω
12000∠0
= 43.48 ∠ - 3.049o A rms
20 + 256∠3.287 o
Ssource = (12000)(43.48) ∠ 3.049o = 521.8 ∠3.049o kVA
IS =
S20Ω = (43.48)2 (20) ∠0 = 37.81 ∠0 kVA
(12000∠0)(256∠3.287 o )
Veff =
= 11130 ∠0.2381o V rms
o
20 + 256∠3.287
I1μF =
Veff
= 4.452 ∠90.24o A rms
- j 2500
so S1μF = (11130)(4.452) ∠-90o = 49.55 ∠-90o kVA
V100mH =
(11130∠0.2381o )( j 40)
= 1758 ∠81.15o V rms
250 + j 40
V100mH
= 43.96 ∠ - 8.852o A rms
j 40
so S100μΗ = (1758)(4.43.96) ∠90o = 77.28 ∠90o kVA
I100mH =
(11130∠0.2381o )(250)
= 10990 ∠ − 8.852o V rms
250 + j 40
so S250Ω = (10990)2 / 250 = 483.1 ∠0o kVA
V250Ω =
(b) 37.81 ∠0 + 49.55 ∠-90o +77.28 ∠90o + 483.1 ∠0o = 521.6 ∠3.014o kVA,
which is within rounding error of the complex 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
important!
(d) P = Veff Ieff cos (ang VS – ang IS) = (12000)(43.48) cos (3.049o) = 521 kW
(e) Q = Veff Ieff sin (ang VS – ang IS) = (12000)(43.48) sin (3.049o) = 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, 7th Edition
60.
(b)
10 March 2006
(a) Peak current = 28 2 = 39.6 A
θload = cos-1(0.812) = +35.71o (since lagging PF). Assume ang (V) = 0o.
p(t) =
at
Chapter Eleven Solutions
( 2300 2 ) (39.6) cos (120πt ) cos (120πt - 35.71 )
o
t = 2.5 ms, then, p(t) = 71.89 kW
(c) P = Veff Ieff cos θ = (2300)(28) cos (35.71o) = 52.29 kW
(d)
S = Veff Ieff ∠θ = 64.4 ∠ 35.71o kVA
(e) apparent power = |S| = 64.4 kVA
(f) |Zload| = |V/ I| = 2300/28 = 82.14 Ω. Thus, Zload = 82.14 ∠ 35.71o Ω
(g) Q = Veff 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.